Important Questions For CBSE Class 12 Maths Chapter 4 Determinates

CBSE Class 12 Maths Chapter 4 Determinates Important Questions

Question 1. Three points P(2x, x + 3), Q(0, x) and R(x + 3, x + 6) are eollinear. then x is equal to

  1. 0
  2. 2
  3. 3
  4. 1

Solution: 4. 1

As per the given condition \(\left|\begin{array}{ccc}
2 x & x+3 & 1 \\
0 & x & 1 \\
x+3 & x+6 & 1
\end{array}\right|=0\)

[ar(ΔPQR) = 0]

⇒ 2x (x – x – 6) — 0(x + 3 – x — 6) + (x + 3) {x + 3 – x} =0

or -12x + 3x + 9 = 0 => -9x = -9 => x = 1

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Question 2. If Cij denotes the cofactor of element pij of the matrix P = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & 2 & 4
\end{array}\right]\)
, then the value of C31, C23 is:

  1. 5
  2. 24
  3. -4
  4. -5

Solution: 1. 5

P = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & 2 & 4
\end{array}\right]\) (given)

⇒ C31, C23 = (3-4), {(-1)(2+3)} = 5

Question 3. The system of linear equations 5x + ky = 5; 3x + 3y = 5 will be consistent if:

  1. k ≠ -3
  2. k = -5
  3. k = 5
  4. k ≠ 5

Solution: 4. k ≠ 5

The system of linear equations is given as:

5x + ky = 5 and 3x + 3y = 5

It can be written in matrix form as \(\left[\begin{array}{ll}
5 & k \\
3 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
5 \\
5
\end{array}\right]\)

or A X = B

So; the given system of linear equations is consistent if |A| ≠ 0

⇒ \(\left|\begin{array}{ll}
5 & \mathrm{k} \\
3 & 3
\end{array}\right|\) 0

⇒15 – 3 k ≠ 0

⇒ k ≠ 5

Important Questions For CBSE Class 12 Maths Chapter 4

Question 4. If, for the matrix \(A=\left[\begin{array}{cc}
\alpha & -2 \\
-2 & \alpha
\end{array}\right]\)
,\(\left|\mathrm{A}^2\right|=125\); then the value of α is:

  1. 3
  2. -3
  3. 1
  4. 1

Solution: 1. 3

Given \(\left|A^3\right|=125\)

⇒ \((|A|)^3=125\)

(because \(\left|A^n\right|=|A|^n\))

⇒ \(\left(\alpha^2-4\right)^3=125\)

⇒ \(\left(\alpha^2-4\right)=5\)

⇒ \(\alpha^2=9\) or \(\alpha= \pm 3\)

Question 5. Let matrix X = [xij] is given by X = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right]\)
, Then, the matrix Y = [mij]. where mij = Minor of Xij is:

  1. \(\left[\begin{array}{ccc}
    7 & -5 & -3 \\
    19 & 1 & -11 \\
    -11 & 1 & 7
    \end{array}\right]\)
  2. \(\left[\begin{array}{ccc}
    7 & -19 & -11 \\
    5 & -1 & -1 \\
    3 & 11 & 7
    \end{array}\right]\)
  3. \(\left[\begin{array}{ccc}
    7 & 19 & -11 \\
    -3 & 11 & 7 \\
    -5 & -1 & -1
    \end{array}\right]\)
  4. \(\left[\begin{array}{ccc}
    7 & 19 & -11 \\
    -1 & -1 & 1 \\
    -3 & -11 & 7
    \end{array}\right]\)

Solution: 

X = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right]\)

Now, Y = [mij]

⇒ Y = \(\left[\begin{array}{lll}
m_{11} & m_{12} & m_{13} \\
m_{21} & m_{22} & m_{23} \\
m_{31} & m_{32} & m_{33}
\end{array}\right] \Rightarrow Y=\left[\begin{array}{ccc}
7 & 19 & -11 \\
-1 & -1 & 1 \\
-3 & -11 & 7
\end{array}\right]\)

Question 6. If x =-4 is a root of \(\left|\begin{array}{lll}
x & 2 & 3 \\
1 & x & 1 \\
3 & 2 & x
\end{array}\right|\) = 0
, then the sum of the other two roots is:

  1. 4
  2. -3
  3. 2
  4. 5

Solution: 1. 4

⇒ \(\left|\begin{array}{lll}
x & 2 & 3 \\
1 & x & 1 \\
3 & 2 & x
\end{array}\right|\) = 0 (given)

⇒ \(x\left(x^2-2\right)-2(x-3)+3(2-3 x)=0\)

⇒ \(x^3-2 x-2 x+6+6-9 x=0\)

⇒ \(x^3-13 x+12=0\)

⇒ \((x+4)\left(x^2-4 x+3\right)=0\)

(because x=-4 is a root)

⇒ \((x+4)(x-1)(x-3)=0\)

Hence; the sum of other two roots = 1 + 3 = 4

Question 7. The inverse of the matrix \(X=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]\)

  1. \(24\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right]\)
  2. \(\frac{1}{24}\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\)
  3. \(\frac{1}{24}\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]\)
  4. \(\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right]\)

Solution:

Given, \(X=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]\)

|X| = \(2(12)-0+0=24 \neq 0\)

⇒ \(X^{-1}\) exists

Now, adj X = \(\left[\begin{array}{ccc}12 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 6\end{array}\right]\)

⇒ \(X^{-1}=\frac{1}{|X|} \cdot \mathrm{adj} X\)

⇒ \(X^{-1}=\frac{1}{24}\left[\begin{array}{ccc}
12 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 6
\end{array}\right]=\left[\begin{array}{ccc}
1 / 2 & 0 & 0 \\
0 & 1 / 3 & 0 \\
0 & 0 & 1 / 4
\end{array}\right]\)

Question 8. If A is square matrix of order 3 such that A(adj A) = \(\left[\begin{array}{ccc}-2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2\end{array}\right]\), then find |A|.

Solution:

Given, A (adj A) = \(\left[\begin{array}{ccc}-2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2\end{array}\right]\)

We know that A(adj A)= |A| I

Now, A(adj A)= \(-2\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = -2I = |A| I\)

⇒ |A|=-2

Question 9. If A is a non-singular square matrix of order 3 such that A² = 3A, then the value of |A| is?

  1. -3
  2. 3
  3. 9
  4. 27

Solution:

Given, A² = 3A, |A| ≠ 0, order of A is 3

∴ |A²| = |3A|

⇒ |A|²= 3³|A|

(|A²| = |A|² and |KA| = Kn|A|)

or |A| = 27

Question 10. If A is a square matrix satisfying A’A = I, write the value of |A|.
Solution:

A’A = I (given)

⇒ \(\left|A^{\prime} A\right|=|I|\)

⇒ \(\left|A^{\prime}\right||A|=|I| \Rightarrow|A|^2=1\) (because \(\left|A^{\prime}\right|=|A|\))

⇒ \(|A|=1 \text { or }|A|=-1\)

Question 11. Given, \(A=\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right]\), compute A-1 and show that 2A-1 = 9I – A.

Solution:

Given, \(A=\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right]\)

|A| = \(14-(12)=2 \neq 0\)

Hence, A is invertible.

adj A = \(\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]\)

Hence, \(A^{-1}=\frac{1}{|A|} \cdot \mathrm{adj} \cdot(\mathrm{A})\)

⇒ \(\mathrm{A}^{-1}=\frac{1}{2}\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right] \text { or } 2 \mathrm{~A}^{-1}=\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]\)….(1)

Now, R.H.S. = \(9 \mathrm{I}-\mathrm{A}\)

⇒ \(9\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]=\left[\begin{array}{cc}
9 & 0 \\
0 & 9
\end{array}\right]-\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]=\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]=2 A^{-1}\)= L.H.S. [from (1)]

Hence proved.

Question 12. If \(A=\left[\begin{array}{ccc}
3 & 4 & 2 \\
0 & 2 & -3 \\
1 & -2 & 6
\end{array}\right]\)
; find A-1.

Hence solve the following system of equations: 3x + 4y + 2z = 8; 0x + 2y + 3z = 3 and x + 2y + 6z = -2 or,

OR

If \(A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]\) find \((A B)^{-1} \text {. }\)

Solution:

Given, \(A=\left[\begin{array}{ccc}
3 & 4 & 2 \\
0 & 2 & -3 \\
1 & -2 & 6
\end{array}\right]\)

⇒ |A| = 3(12 -6)-4(0 + 3) + 2(0-2)= 18- 12-4=2*0

Hence, A-1 exists.

Now, co-factors are given as:

C11 = 6, C12 = —3, C13 = -2,

C21=-28, C22 = 16, C23= 10,

C31= -16, C32 = 9, C33 = 6

Hence adj A = \(\left[\begin{array}{ccc}6 & -3 & -2 \\ -28 & 16 & 10 \\ -16 & 9 & 6\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc}6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6\end{array}\right]\)

⇒ \(A^{-1}=\frac{1}{|A|}(\mathrm{adj} A)=\frac{1}{2}\left[\begin{array}{ccc}
6 & -28 & -16 \\
-3 & 16 & 9 \\
-2 & 10 & 6
\end{array}\right]\)….(1)

The system of linear equations is given as:

3x + 4y + 2z = 8

0x + 2y —3 = 3

x – 2y + 6z = -2

This system is written in matrix form as

⇒ \(\left[\begin{array}{ccc}
3 & 4 & 2 \\
0 & 2 & -3 \\
1 & -2 & 6
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
8 \\
3 \\
-2
\end{array}\right]\)

⇒ AX = B

⇒ X = \(A^{-1} \cdot B\)

⇒ X = \(\frac{1}{2}\left[\begin{array}{ccc}
6 & -28 & -16 \\
-3 & 16 & 9 \\
-2 & 10 & 6
\end{array}\right] \cdot\left[\begin{array}{l}
8 \\
3 \\
-2
\end{array}\right]=\frac{1}{2}\left[\begin{array}{l}
48-84+32 \\
-24+48-18 \\
-16+30-12
\end{array}\right]\)

[from (1) and (2)]

⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{2}\left[\begin{array}{r}
-4 \\
6 \\
2
\end{array}\right]\)

x=-2, y=3, z=1

Or

Given, \(A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]\)…(1)

and \(B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]\)

|B| = 1 (3 – 0) – 2(- 1 – 0) – 2(2 – 0) = 3 + 2 – 4 = 1 ≠ 0

Now, the co-factors of matrix B are given as

⇒ \(C_{11}=3, C_{12}=1, C_{13}=2\)

⇒ \(C_{21}=2, C_{22}=1, C_{23}=2,\)

⇒ \(C_{31}=6, C_{32}=2, C_{33}=5\)

∴ (adj)(B) = \(\left[\begin{array}{lll}
3 & 1 & 2 \\
2 & 1 & 2 \\
6 & 2 & 5
\end{array}\right]^1=\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]\)

⇒ \(B^{-1}=\frac{1}{|B|}(\mathrm{adj} B)=\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]\)….(2)

∴ (AB)-1 = B-1 A-1

= \(\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right] \cdot\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\)

=\(\left[\begin{array}{ccc}
9-30+30 & -3+12-12 & 3-10+12 \\
3-15+10 & -1+6-4 & 1-5+4 \\
6-30+25 & -2+12-10 & 2-10+10
\end{array}\right]\)

⇒ \((A B)^{-1}=\left[\begin{array}{ccc}
9 & -3 & 5 \\
-2 & 1 & 0 \\
1 & 0 & 2
\end{array}\right]\)

(from (1) and (2))

⇒ \((\mathrm{AB})^{-1}=\left[\begin{array}{ccc}
9 & -3 & 5 \\
-2 & 1 & 0 \\
1 & 0 & 2
\end{array}\right]\)

Question 13. If \(A=\left[\begin{array}{ccc}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{array}\right]\), find A-1 and use it to solve the following system of equations.

5x – y + 4z = 5

2x + 3y + 5z = 2

5x – 2y + 6z = -1

Solution:

Given, \(A=\left[\begin{array}{ccc}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{array}\right]\)

∴ |A| = \(\left|\begin{array}{ccc}
5 & -1 & 4 \\
2 & 3 & 5 \\
5 & -2 & 6
\end{array}\right|\) = 5(18+10) + (12 – 25) + 4 (-4 – 15)= 140 – 13 – 76 = 51 ≠ 0.

Hence: A-1 exists

Now co-factors of elements of A are:

⇒ \(A_{11}=28, A_{12}=13, A_{13}=-19\)

⇒ \(A_{21}=-2, A_{22}=10, A_{23}=5\)

⇒ \(A_{31}=-17, A_{32}=-17, A_{33}=17\)

∴ adj A = \(\left[\begin{array}{ccc}
28 & 13 & -19 \\
-2 & 10 & 5 \\
-17 & -17 & 17
\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc}
28 & -2 & -17 \\
13 & 10 & -17 \\
-19 & 5 & 17
\end{array}\right]\)

⇒ \(A^{-1}=\frac{1}{|A|}(\mathrm{adj} A)=\frac{1}{51}\left[\begin{array}{ccc}
28 & -2 & -17 \\
13 & 10 & -17 \\
-19 & 5 & 17
\end{array}\right]\)

Given system of equations are

5x – y + 4z = 5

2x + 3y + 5z = 2

5x – 2y + 6z = -1

This system is written in matrix form as

⇒ \({\left[\begin{array}{ccc}
5 & -1 & 4 \\
2 & 3 & 5 \\
5 & -2 & 6
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
5 \\
2 \\
-1
\end{array}\right] }\)

⇒ AX = B ⇒ X=\(A^{-1}\) B

⇒ \({\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{51}\left[\begin{array}{ccc}
28 & -2 & -17 \\
13 & 10 & -17 \\
-19 & 5 & 17
\end{array}\right]\left[\begin{array}{r}
5 \\
2 \\
-1
\end{array}\right]}\)

or \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{51}\left[\begin{array}{l}
140-4+17 \\
65+20+17 \\
-95+10-17
\end{array}\right]\)

⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{51}\left[\begin{array}{r}
153 \\
102 \\
-102
\end{array}\right]=\left[\begin{array}{r}
3 \\
2 \\
-2
\end{array}\right]\)

⇒ x= 3, y = 2, z = -2

Question 14. Show that, for matrix A = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\)
, A³-6 A²+5 A+11 I=O, Hence, find \(A^{-1}\).

Or,

Using the matrix method, solve the following system of equations:

3x – 2y + 3z = 8

2x + y – z = 1

4x – 3y + 2z = 4

Solution:

Given A = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\)

To prove: A³ – 6A² + 5A + 11 I = O

⇒ \(A^2=A \cdot A=\left[\begin{array}{rrr}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right] \cdot\left[\begin{array}{rrr}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]=\left[\begin{array}{ccc}
1+1+2 & 1+2-1 & 1-3+3 \\
1+2-6 & 1+4+3 & 1-6-9 \\
2-1+6 & 2-2-3 & 2+3+9
\end{array}\right]\)

⇒ \(A^2=\left[\begin{array}{rrr}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]\)

and \(A^3= A^2 \cdot A=\left[\begin{array}{rrr}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right] \cdot\left[\begin{array}{rrr}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\)

–\(\left[\begin{array}{rrr}
4+2+2 & 4+4-1 & 4-6+3 \\
-3+8-28 & -3+16+14 & -3-24-42 \\
7-3+28 & 7-6-14 & 7+9+42
\end{array}\right]\)

⇒ \(A³=\left[\begin{array}{ccc}
8 & 7 & 1 \\
-23 & 27 & -69 \\
32 & -13 & 58
\end{array}\right]\)

L.H.S. = \(A^3-6 A^2+5 A+11I\)

= \(\left[\begin{array}{ccc}
8 & 7 & 1 \\
-23 & 27 & -69 \\
32 & -13 & 58
\end{array}\right]-6\left[\begin{array}{ccc}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]+5\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]+11\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)

= \(\left[\begin{array}{ccc}
8-24+5+11 & 7-12+5+0 & 1-6+5+0 \\
-23+18+5+0 & 27-48+10+11 & -69+84-15+0 \\
32-42+10+0 & -13+18-5+0 & 58-84+15+11
\end{array}\right]\)

= \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=\mathrm{O}\) Zero matrix

Now, A³-6A²+5A+11 I = O

A³A-1 – 6A²A -1+ 5AA-1 + 11 I A-1 = OA-1 (Post multiplying both sides by \(A^{-1}\)

A²(AA-1)-6A(AA-1)+5(AA-1)+11 I A-1 = OA-1

A²-6A +5I + 11A-1 = O

because AA-1=I and OA-1 =O

⇒ \(A^{-1}=\frac{-1}{11} \cdot\left(A^2-6 A+5I\right)=\frac{-1}{11}\left\{\left[\begin{array}{ccc}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]-6\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]+5\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\right\}\)

⇒ \(A^{-1}=\frac{-1}{11}\left[\begin{array}{ccc}
4-6+5 & 2-6+0 & 1-6+0 \\
-3-6+0 & 8-12+5 & -14+18+0 \\
7-12+0 & -3+6+0 & 14-18+5
\end{array}\right]\)

⇒ \(A^{-1}=\left[\begin{array}{ccc}
-3 / 11 & 4 / 11 & 5 / 11 \\
9 / 11 & -1 / 11 & -4 / 11 \\
5 / 11 & -3 / 11 & -1 / 11
\end{array}\right]\)

The given system of equations are:

3x -2y + 3 z = 8

2x + y – z = 1 and

4x – 3y + 2z = 4

By using the matrix method; the given system of equations can be written as; AX = B

where \(A=\left[\begin{array}{ccc}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]\) and \(B=\left[\begin{array}{l}8 \\ 1 \\ 4\end{array}\right]\)

Now; \(|A|=\left|\begin{array}{ccc}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{array}\right|\)=-3+16-30=-17 ≠0

Hence; \(\mathrm{A}^{-1}\) exists.

Now; \(A=\left[\begin{array}{ccc}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{array}\right]\)

Co-factors are given as: \(C_{11}=-1, \quad C_{12}=-8, C_{13}=-10, C_{21}=-5, C_{22}=-6, C_{23}=1, C_{31}=-1, C_{32}=9, C_{33}=7\)

Hence, (adj)(A)= \(\left[C_{i j}\right]^{\mathrm{T}}\)

adj(A)= \(\left[\begin{array}{ccc}
-1 & -5 & -1 \\
-8 & -6 & 9 \\
-10 & 1 & 7
\end{array}\right]\)

Now; AX = \(B \Rightarrow X=A^{-1} \cdot B\) = \(\frac{\mathrm{adj} A}{|A|} \cdot B\)

⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=-\frac{1}{17}\left[\begin{array}{ccc}
-1 & -5 & -1 \\
-8 & -6 & 9 \\
-10 & 1 & 7
\end{array}\right] \cdot\left[\begin{array}{l}
8 \\
1 \\
4
\end{array}\right]=-\frac{1}{17}\left[\begin{array}{c}
-8-5-4 \\
-64-6+36 \\
-80+1+28
\end{array}\right]=-\frac{1}{17}\left[\begin{array}{l}
-17 \\
-34 \\
-51
\end{array}\right]\)

⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)

x=1, y=2, z=3

Question 15. If \(A=\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\)
, find A-1. Hence, using A-1, solve the system of equations:

2x – 3y + 5z = 11,

3x + 2y – 4z = -5,

x + y -2z = -3.

Solution:

Given, \(A=\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\)

A is invertible if |A| ≠ 0

Now. |A| = 2(-4 + 4) + 3 (-6 + 4) + 5(3 – 2)= 0 – 6 + 5 =-1≠0

Co-factors are given as :

⇒ \(C_{11}=0, C_{12}=2, C_{13}=1\)

⇒ \(C_{21}=-1, C_{22}=-9, C_{23}=-5\)

⇒ \(C_{31}=2, C_{32}=23, C_{33}=13\)

⇒ (adj) \((\mathrm{A})=\left[\mathrm{C}_{1 \mathrm{ij}}\right]^{\mathrm{T}}\)

or (adj) \((\mathrm{A})=\left[\begin{array}{ccc}
0 & 2 & 1 \\
-1 & -9 & -5 \\
2 & 23 & 13
\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc}
0 & -1 & 2 \\
2 & -9 & 23 \\
1 & -5 & 13
\end{array}\right]\)

⇒ \(\mathrm{A}^{-1}=\frac{\mathrm{adj}(\mathrm{A})}{|\mathrm{A}|}=\left[\begin{array}{ccc}
0 & 1 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{array}\right]\)

Now, given equations are 2x-3y + 5z = 11; 3x + 2y -4z = -5 and x + y – 2z = -3

⇒ \({\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{l}
11 \\
-5 \\
-3
\end{array}\right]}\)

⇒ \(\mathrm{AX}=\mathrm{B} \text { or } \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B} \\
\Rightarrow \mathrm{X}=\left[\begin{array}{ccc}
0 & 1 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{array}\right]\left[\begin{array}{l}
11 \\
-5 \\
-3
\end{array}\right]\)

or \(\mathrm{X}=\left[\begin{array}{c}
0 \times 11+(1) \times(-5)+(-2) \times(-3) \\
(-2) \times 11+9 \times(-5)+(-23) \times(-3) \\
(-1) \times 11+5 \times(-5)+(-13) \times(-3)
\end{array}\right] \\
\Rightarrow \quad\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)

⇒ \(\mathrm{x}=1, \mathrm{y}=2, \mathrm{z}=3\)

Important Questions For CBSE Class 12 Maths Chapter 3 Matrices

CBSE Class 12 Maths Chapter 3 Matrices Important Questions

Question 1. The number of all possible matrices of order 2 x 3 with each country 1 or 2 is :

  1. 16
  2. 6
  3. 64
  4. 24

Solution: 3. 6

Required number of possible matrices

= (Number of entries)order

= (2)2×3 = (2)6 = 64

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Question 2. If a matrix A is both symmetric and skewed, symmetric. then A is necessarily a

  1. Diagonal matrix
  2. Zero square matrix
  3. Square matrix
  4. Identity matrix

Solution: 2. Zero square matrix

Given; AT = A (symmetric matrix)

and -AT= A (skew-symmetric matrix)

⇒ 2A = O or A = O

Hence, A is necessarily a zero-square matrix

Important Questions CBSE Class 12 Maths Chapter 3

Question 3. If \(\left[\begin{array}{cc}
3 c+6 & a-d \\
a+d & 2-3 b
\end{array}\right]=\left[\begin{array}{cc}
12 & 2 \\
-8 & -4
\end{array}\right]\)
, then the value of ab-cd is

  1. 4
  2. 16
  3. -4
  4. -16

Solution: 1. 4

Given: \(\left[\begin{array}{cc}
3 c+6 & a-d \\
a+d & 2-3 b
\end{array}\right]=\left[\begin{array}{cc}
12 & 2 \\
-8 & -4
\end{array}\right]\)

On comparing both sides, we get

a – d = 2 and a + d = -8

⇒  2a = -6 Or a=-3

⇒ d = -5

Also; 3c + 6 = 12, 2-3b = -4

⇒ c = 2, b = 2

Hence, ab- cd = (-3)2 – 2(-5) = -6 + 10 = 4

Question 4. For two matrices \(P=\left[\begin{array}{cc}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right] \text { and } Q^{\top}=\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\),
P-Q is

  1. \(\left[\begin{array}{cc}2 & 3 \\ -3 & 0 \\ 0 & -3\end{array}\right]\)
  2. \(\left[\begin{array}{cc}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]\)
  3. \(\left[\begin{array}{cc}4 & 3 \\ 0 & -3 \\ -1 & -2\end{array}\right]\)
  4. \(\left[\begin{array}{cc}2 & 3 \\ 0 & -3 \\ 0 & -3\end{array}\right]\)

Solution: 2. \(\left[\begin{array}{cc}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]\)

⇒ \(Q^{\top}=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right] \Rightarrow Q=\left[\begin{array}{cc}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]\)

Hence, \(P-Q=\left[\begin{array}{cc}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]=\left[\begin{array}{cc}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]\)

Question 5. A matrix A = [aij]3×3 is defined by \(a_{i j}=\left\{\begin{array}{ccc}
2 i+3 j & ; & i<j \\
5 & ; & i=j \\
3 i-2 j & ; & i>j
\end{array}\right.\)
, The number of elements in A, which are more than 5, is

  1. 3
  2. 4
  3. 5
  4. 6

Solution: 2. 4

Given; A = [aij]3×3

where \(a_{i j}=\left\{\begin{array}{ccc}
2 i+3 j & ; & i<j \\
5 & ; & i=j \\
3 i-2 j & ; & i>j
\end{array}\right.\)

⇒ A= \(\left[\begin{array}{lll}
5 & 8 & 11 \\
4 & 5 & 13 \\
7 & 5 & 5
\end{array}\right]\)

Hence; required number =4

Question 6. For the matrix X = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\), (X²-X) is:

  1. 21
  2. 31
  3. 1
  4. 51

Solution: 1. 21

⇒ \(X^2=\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right] \cdot\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]=\left[\begin{array}{lll}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{array}\right]\)

⇒ \(X^2-X=\left[\begin{array}{lll}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{array}\right]-\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]=\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}\right]=21\)

Question 7. Find the order of matrix A such that \(\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right]\), A = \(\left[\begin{array}{cc}-1 & -8 \\ 1 & -2 \\ 9 & 22\end{array}\right]\)

or,

If B = \(\left[\begin{array}{cc}
1 & -5 \\
0 & -3
\end{array}\right]\)
and A+2 B = \(\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]\)
, find matrix A.

Solution:

Let \(\mathrm{B}=\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right]\) and \(\mathrm{C}=\left[\begin{array}{cc}-1 & -8 \\ 1 & -2 \\ 9 & 22\end{array}\right]\)

We have \(\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right]_{3 \times 2} \quad A=\left[\begin{array}{cc}-1 & -8 \\ 1 & -2 \\ 9 & 22\end{array}\right]_{3 \times 2}\)

Let the order of A be m x n

BA will be possible if several columns in matrix B should be equal to several rows in matrix A ⇒ m = 2.

and the order of BA is 3 x n

Since, BA = C.

⇒ Order of BA will be the same as that of matrix C

⇒ 3 x n = 3 x 2

⇒ n = 2

Then, the order of matrix A is 2 x 2

A+2 B \(\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]\) (given)

⇒ A = \(\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]-2 B=\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]-2\left[\begin{array}{cc}
1 & -5 \\
0 & -3
\end{array}\right]=\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]-\left[\begin{array}{cc}
2 & -10 \\
0 & -6
\end{array}\right]\)

⇒ A = \(\left[\begin{array}{cc}
-2 & 14 \\
-7 & 11
\end{array}\right]\)

Question 8. If A = \(\left[\begin{array}{lll}
1 & 0 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{l}
2 \\
5 \\
6
\end{array}\right]\)
, find AB.

Solution:

AB= \(\left[\begin{array}{lll}
1 & 0 & 4
\end{array}\right]_{1 \times 3}\left[\begin{array}{l}
2 \\
5 \\
6
\end{array}\right]_{3 \times 1}\)

= \([1 \times 2+0 \times 5+4 \times 6]_{1 \times 1}=[2+0+24]=[26]\)

Question 9. Given, a skew-symmetric matrix \(A = \mathrm{A}=\left[\begin{array}{ccc}
0 & \mathrm{a} & 1 \\
-1 & \mathrm{~b} & 1 \\
-1 & \mathrm{c} & 0
\end{array}\right]\)
, then value of (a+b+c)² is

Solution:

Given \(A = \mathrm{A}=\left[\begin{array}{ccc}
0 & \mathrm{a} & 1 \\
-1 & \mathrm{~b} & 1 \\
-1 & \mathrm{c} & 0
\end{array}\right]\)

A is a skew-symmetric matrix, AT = -A

⇒ \(A^{\top}=\left[\begin{array}{ccc}
0 & -1 & -1 \\
a & b & c \\
1 & 1 & 0
\end{array}\right] \text { and }-A=\left[\begin{array}{ccc}
0 & -a & -1 \\
1 & -b & -1 \\
1 & -c & 0
\end{array}\right]\)

So, a = 1, b = 0 and c = -1

Now (a + b + c)² = (1 +0-1)² = 0

Question 10. If the matrices A = \(\left[\begin{array}{ccc}
0 & \mathrm{a} & -3 \\
2 & 0 & -\mathrm{t} \\
\mathrm{b} & 1 & 0
\end{array}\right]\)
 is skew symmetric, find the values of ‘a’ and ‘b’.

Solution:

A = \(\left[\begin{array}{ccc}
0 & \mathrm{a} & -3 \\
2 & 0 & -\mathrm{t} \\
\mathrm{b} & 1 & 0
\end{array}\right]\) (given)

A is skew-symmetric ⇒ AT = -A

⇒ \(\left[\begin{array}{ccc}
0 & 2 & b \\
a & 0 & 1 \\
-3 & -1 & 0
\end{array}\right]=\left[\begin{array}{ccc}
0 & -a & 3 \\
-2 & 0 & 1 \\
-b & -1 & 0
\end{array}\right]\)

⇒ a=-2, b=3

Question 11. If the matrix A = \(\left[\begin{array}{cc}
0 & 6-5 x \\
x^2 & x+3
\end{array}\right]\)
 is symmetric, find the values of x?

Solution:

A = \(\left[\begin{array}{cc}
0 & 6-5 x \\
x^2 & x+3
\end{array}\right]\) (given)

For a symmetric matrix, A = A1

⇒ \(\left[\begin{array}{cc}
0 & 6-5 x \\
x^2 & x+3
\end{array}\right]=\left[\begin{array}{cc}
0 & x^2 \\
6-5 x & x+3
\end{array}\right]\)….(1)

∴ 6-5x = x² [from (1)]

⇒ x² + 5x – 6 = 0

⇒ (x + 6) (x -1) = 0

⇒ x = – 6, 1

Question 12. If A = \(\left[\begin{array}{cc}
-3 & 2 \\
1 & -1
\end{array}\right] \text { and } I=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
, find scalar k such that A² + I = KA

Solution:

Given, A = \(\left[\begin{array}{cc}
-3 & 2 \\
1 & -1
\end{array}\right] \text { and } I=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

⇒ \(\left[\begin{array}{cc}
-3 & 2 \\
1 & -1
\end{array}\right]\left[\begin{array}{cc}
-3 & 2 \\
1 & -1
\end{array}\right]+\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]=K\left[\begin{array}{cc}
-3 & 2 \\
1 & -1
\end{array}\right]\)

(because \(A^2+I=K A\))

⇒ \(\left[\begin{array}{cc}
11 & -8 \\
-4 & 3
\end{array}\right]+\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]\) = \(\left[\begin{array}{cc}
-3 \mathrm{~K} & 2 \mathrm{~K} \\
\mathrm{~K} & -\mathrm{K}
\end{array}\right]\)

∴ \(\left[\begin{array}{cc}
12 & -8 \\
-4 & 4
\end{array}\right]\) = \(\left[\begin{array}{cc}
-3 \mathrm{~K} & 2 \mathrm{~K} \\
\mathrm{~K} & -\mathrm{K}
\end{array}\right]\)

⇒ K = -4 (on comparing both sides)

Question 13. If A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\)
, show that (A-2I)(A-3I) = O.

Solution:

A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\) (given)

⇒ A-2I = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]-2\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
2 & 2 \\
-1 & -1
\end{array}\right]\)

and A-3I = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]-3\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & 2 \\
-1 & -2
\end{array}\right]\)

⇒ (A-2I)(A-3I) = \(\left[\begin{array}{cc}
2 & 2 \\
-1 & -1
\end{array}\right]\left[\begin{array}{cc}
1 & 2 \\
-1 & -2
\end{array}\right]=\left[\begin{array}{cc}
2-2 & 4-4 \\
-1+1 & -2+2
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=O\)

Important Questions For CBSE Class 12 Maths Chapter 12 Linear Programming

CBSE Class 12 Maths Chapter 12 Linear Programming Important Questions

Question 1. A Linear programming problem is as follows: Minimize Z = 2x + y

Subject to the constraints:

  • x ≥ 3, x ≤ 9, y ≥ 0,
  • x-y ≥ 0, x + y ≤ 14

The feasible region has:

  1. 5 corner points including (0, 0) and (9, 5)
  2. 5 corner points including (7, 7) and (3, 3)
  3. 5 corner points including (14, 0) and (9, 0)
  4. 5 corner points including (3, 6) and (9, 5)

Solution: 2. 5 corner points including (7, 7) and (3, 3)

Given; Z = 2x + y subject to the constraints

x ≥ 3, x ≤ 9, y ≥ 0, x – y ≥ 0, x + y ≤ 14

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Important Questions For CBSE Class 12 Maths Chapter 12 Linear Programming Graph For 5 Corner Points

Hence, the feasible region has 5 corner points including (7, 7) and (3, 3)

Important Questions For CBSE Class 12 Maths Chapter 12

Question 2. A Linear Programming Problem is as follows: Maximise/Minimise objective function Z = 2x – y + 5

Subject to the constraints:

  • 3x + 4y ≤ 60,
  • x + 3y ≤ 30,
  • x ≥ 0, y ≥ 0

If the corner points of the feasible region are A(0, 10), B(12, 6), C(20, 0) and 0(0,0); then which of the following is true?

  1. The maximum value of Z is 40
  2. The minimum value of Z is -5
  3. The difference between the maximum and minimum values of Z is 35
  4. At two corner points, the value of Z is equal

Solution: 2. The minimum value of Z is -5.

Important Questions For CBSE Class 12 Maths Chapter 12 Linear Programming Corner Points Of The Feasible Region

Question 3. The corner points of the feasible region determined by a set of constraints (linear inequalities) are P(0, 5), Q(3, 5), R(5, 0), and S(4, 1) and the objective function is Z = ax + 2by; where a, b > 0. The condition on ‘a’ and ‘b’ such that the maximum of Z occurs at Q and S is:

  1. a-5b=0
  2. a-3b = 0
  3. a – 2b = 0
  4. a – 8b = 0

Solution: 4. a – 8b = 0

Given points arc P(0, 5), Q(3, 5), R(5, 0) and S(4, 1) and Z = ax + 2by

Since the maximum of Z occurs at Q and S;

⇒ 3a + 10b = 4a + 2b

⇒ 8b = a ⇒ a – 8b = 0

Question 4. For an L.P.P.. the objective function is Z = 4x + 3y and the feasible region is determined by a set of constraints (linear inequations) as shown in the graph.

Important Questions For CBSE Class 12 Maths Chapter 12 Linear Programming Graph For Set Of Constraints Linear Equations

Which one of the following statements is true?

  1. The maximum value of Z is at R
  2. The maximum value of Z is at Q
  3. The value of Z at R is less than the value of P
  4. The value of Z at Q is less than the value of R

Solution: 2. Maximum value of Z is at Q

Z = 4x + 3y

Important Questions For CBSE Class 12 Maths Chapter 12 Linear Programming Maximum Value At Q

A maximum value of Z is at Q.

Question 5. Solve the following problem graphically: Maximize Z = 3x + 9y

Subject to the constraints:

  • x + 3y ≤ 60,
  • x + y ≥ 10,
  • x ≤ y,
  • x ≥ 0, y ≥ 0

Solution:

First of all, let us graph the feasible region of the system of linear inequalities given above. The feasible region ABCD is shown.

Note that the region is bounded. The coordinates of the comer points A, B, C, and D are (0, 10), (5, 5), (15, 15), and (0, 20) respectively.

Important Questions For CBSE Class 12 Maths Chapter 12 Linear Programming Graph For Feasible Regio Of The Ssytem Of Linear Inequalities

Important Questions For CBSE Class 12 Maths Chapter 12 Linear Programming Line Segments Joining

The maximum value of Z is 180. which occurs at every point of the line segment joining the points C and D.

Important Questions For CBSE Class 12 Maths Chapter 13 Probability

CBSE Class 12 Maths Chapter 13 Probability Important Questions

Question 1. If A and B are two independent events and P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{2}\); find \(\mathrm{P}(\overline{\mathrm{A}} \mid \overline{\mathrm{B}})\).
Solution:

Given, A and B are two independent events and P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{2}\)

⇒ P(A∩B) = P(A).P(B) = \(\frac{1}{3}\) x \(\frac{1}{2}\) = \(\frac{1}{6}\)

⇒ P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{\frac{6}{1}}{\frac{1}{2}}=\frac{1}{3}\)

⇒ \(\mathrm{P}(\overline{\mathrm{A}} / \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A} / \mathrm{B}})=1-\mathrm{P}(\mathrm{A} / \mathrm{B}) \Rightarrow \mathrm{P}(\overline{\mathrm{A}} / \overline{\mathrm{B}})=1-\frac{1}{3}=\frac{2}{3}\)

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Important Questions For CBSE Class 12 Maths Chapter 13

Question 2. A coin is tossed once. If the head comes up, a die is thrown; blit if the tail comes up, the coin is tossed again. Find the probability of obtaining a head and number 6.
Solution:

S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, H), (T, T)}

Let event A denote “obtaining head” and B denote “obtaining number 6”

⇒ P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{6}\)

Hence, the probability of obtaining head and number 6

= P[{(H, 6)}]= P(A∩B) = P(A).P(B) = \(\frac{1}{2}\) x \(\frac{1}{6}\) = \(\frac{1}{12}\)

Question 3. Two cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of spade cards.
Solution:

Let random variable X denote the number of spade cards; then the possible values of X are 0, 1 or 2.

P(X = 0) = P(no spade and no spade) = \(\frac{39}{52} \times \frac{39}{52}=\frac{3}{4} \times \frac{3}{4}=\frac{9}{16}\)

P(X = 1) = P(spade and no spade or no spade and spade)

= \(\left(\frac{13}{52} \times \frac{39}{52}\right)+\left(\frac{39}{52} \times \frac{13}{52}\right)=\left(\frac{1}{4} \times \frac{3}{4}\right)+\left(\frac{3}{4} \times \frac{1}{4}\right)=\frac{3}{8}\)

P(X = 2) = P(spade and spade) = \(\left(\frac{13}{52} \times \frac{13}{52}\right)=\left(\frac{1}{4} \times \frac{1}{4}\right)=\frac{1}{16}\)

Hence, the probability distribution of X is:

Important Questions For CBSE Class 12 Maths Chapter 13 Probability Probability Distribution Of Number Of Spade Cards

Question 4. A pair of dice is thrown and the sum of the numbers appearing on the dice is observed to be 7. Find the probability that the number 5 has appeared on at least one die.

Or,

The probability that A hits the target is \(\frac{1}{3}\) and the probability that B hits it is \(\frac{2}{5}\). If both try to hit the target independently, find the probability that the target is hit.

Solution:

When a pair of dice is thrown, the sample space S contains 36 outcomes.

Let E: Event that number 5 has appeared on at least one die.

F: Event that the sum of the numbers on the dice is 7

⇒ E = {(5,1), (5,2), (5,3), (5, 4), (5, 5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)}, F = {(1, 6), (2,5), (3,4), (4,3), (5,2), (6,1)}

∴ E ∩ F = {(5, 2), (2,5)}

Now; P(E∩F) = \(\frac{2}{36}\),P(F) = \(\frac{6}{36}\)

Hence, required probability = \(P(E / F)=\frac{P(E \cap F)}{P(F)}=\frac{(2 / 36)}{(6 / 36)}=\frac{1}{3}\)

Given; P(E1) = \(\frac{1}{3}\), P(E2) =\(\frac{2}{5}\)

⇒ \(P\left(\overline{E_1}\right)=\frac{2}{3} \cdot P\left(\overline{E_2}\right)=\frac{3}{5}\)

⇒ Required Probability = P(target is hit)

= \(1-\mathrm{P}\left(\overline{\mathrm{E}}_1 \overline{\mathrm{E}}_2\right)=1-\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right)=1-\left(\frac{2}{3} \times \frac{3}{5}\right)=1-\frac{2}{5}=\frac{3}{5}\)

Or,

Let E1: Event that A hits the target

Let E2: Event that B hits the target

Given; \(P\left(E_1\right)=\frac{1}{3}, P\left(E_2\right)=\frac{2}{5} \Rightarrow P\left(\overline{E_1}\right)=\frac{2}{3} \cdot P\left(\overline{E_2}\right)=\frac{3}{5}\)

⇒ Required Probability = P (target is hit)

= \(1-\mathrm{P}\left(\overline{\mathrm{E}}_1 \overline{\mathrm{E}}_2\right)=1-\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right)=1-\left(\frac{2}{3} \times \frac{3}{5}\right)=1-\frac{2}{5}=\frac{3}{5}\)

Question 5. A random variable X has the probability distribution:

Important Questions For CBSE Class 12 Maths Chapter 13 Probability A Random Variable Of Probability Distribution

Find the value of K and P(X ≤ 2).

Solution:

We know that ∑ P(X) = 1 (for probability distribution)

⇒ 0 + K + 4K + 3K + 2K = 1

⇒ 10 K = 1

⇒ K = \(\frac{1}{10}\)…..(1)

∴ P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= 0 + k + 4K = 5K = \(\frac{5}{10}\) = \(\frac{1}{2}\) (from (1))

Question 6. A purse contains 3 silver and 6 copper coins and a second purse contains 4 silver and 3 copper coins. If a coin is drawn at random from one of the two purses, find the probability that it is a silver coin.
Solution:

Let E1: Event that first purse is selected. E2: Event that second purse is selected and A: Event that silver coin is drawn

⇒ \(P\left(E_1\right)=\frac{1}{2}, P\left(E_2\right)=\frac{1}{2}, P\left(A / E_1\right)=\frac{3}{9}, P\left(A / E_2\right)=\frac{4}{7}\)

∴ \(P(A)-P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)\) (by total probability theorem)

= \(\left(\frac{1}{2} \times \frac{3}{9}\right)+\left(\frac{1}{2} \times \frac{4}{7}\right)=\frac{1}{6}+\frac{2}{7}=\frac{19}{42}\)

Question 7. A coin is tossed 5 times. What is the probability of getting

  1. 3 heads,
  2. At most 3 heads?

Or,

Find the probability distribution of X, the number of heads in a simultaneous loss of two coins.

Solution:

1. P 3 Heads= \({ }^5 \mathrm{C}_3\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^2=10 \times \frac{1}{8} \times \frac{1}{4}=\frac{5}{16}\)

2. P (at most 3 Heads) =\(1-\mathrm{P}(4)-\mathrm{P}(5)\)

= \(1-{ }^3 C_4\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)-{ }^3 C_5\left(\frac{1}{2}\right)^5=1-\frac{5}{32}-\frac{1}{32}=\frac{26}{32}=\frac{13}{16}\)

Or,

Let X denotes the number of heads in a simultaneous toss of two coins, then the possible values of X are 0, 1 or 2.

P(X=0)=P(TT)= \(\left(\frac{1}{2} \cdot \frac{1}{2}\right)=\frac{1}{4}\),

P(X=1)=P(HT)+P(T H)=\(\left(\frac{1}{2} \cdot \frac{1}{2}\right)+\left(\frac{1}{2} \cdot \frac{1}{2}\right)=\frac{2}{4}\),

P(X=2)=P(HH)=\(\left(\frac{1}{2} \cdot \frac{1}{2}\right)=\frac{1}{4}\)

Hence; the probability distribution of X is given as :

Important Questions For CBSE Class 12 Maths Chapter 13 Probability Probability Distribution Of Number Of Heads In A Toss Of Two Coins

Question 8. If P(not A) = 0.7, P(B) = 0.7 and P(B/A) = 0.5, then find P( A/B)
Solution:

Given; \(\mathrm{P}(\overline{\mathrm{A}})\) = 0.7 ⇒ 1 – P(A) = 0.7 ⇒ P(A) = 0.3 and P(B) = 0.7

Also, P(B/A) = 0.5; P(A/B) = ?

We know that: P(B/A) = \(\frac{P(B \cap A)}{P(A)} \Rightarrow 0.5=\frac{P(B \cap A)}{0.3}\)

⇒ P(B ∩ A) = 0.5 x 0.3 = 0.15

Now, P(A ∩ B) = P(B ∩ A) = 0.15

⇒ P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{0.15}{0.70}=\frac{15}{70}=\frac{3}{14}\)

Question 9. A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Solution:

Let E: Event of obtaining the sum 8 on the dice.

F: Event that red die resulted in a number less than 4, and let first and second die represent the black and red die respectively.

⇒ E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

⇒ n(E) = 5,

F = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)}(3,1),(3,2),(3,3),(4,1),(4,2),(4,3),(5,1),(5,2),(5,3),(6,1),(6,2),(6,3)}

⇒ n(F) = 18

and E ∩ F = {(5, 3), (6, 2)}

⇒ n(E ∩F) = 2

Here, \(\mathrm{P}(\mathrm{F})=\frac{18}{36}=\frac{1}{2}$ and $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{2}{36}=\frac{1}{18}\)

∴ Required probability = \(P(E / F)=\frac{P(E \cap F)}{P(F)}=\frac{1 / 18}{1 / 2}=\frac{1}{9}\)

Question 10. A shopkeeper sells three types of flower seeds A1, A2, A3. They are sold in the form of a mixture, where the proportions of these seeds are 4:4:2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Based on the above information, answer the following questions

  1. Calculate the probability that a randomly chosen seed will germinate.
  2. Calculate the probability that the seed is of type A,, given that a randomly chosen seed germinates.

Solution:

Given, A1: A2: A3 = 4 : 4 : 2

⇒ \(P\left(A_1\right)=\frac{4}{10}, P\left(A_2\right)=\frac{4}{10}\) and \(P\left(A_3\right)=\frac{2}{10}\)

Let E be the event that a seed germinates.

∴ \(P\left(\frac{E}{A_1}\right)=\frac{45}{100} \cdot P\left(\frac{E}{A_2}\right)=\frac{60}{100}\) and \(P\left(\frac{E}{A_7}\right)=\frac{35}{100}\)

1. \(\mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{A}_1\right) \cdot \mathrm{P}\left(\mathrm{E} / \mathrm{A}_1\right)+\mathrm{P}\left(\mathrm{A}_2\right) \cdot \mathrm{P}\left(\mathrm{E} / \mathrm{A}_2\right)+\mathrm{P}\left(\mathrm{A}_3\right) \cdot \mathrm{P}\left(\mathrm{E} / \mathrm{A}_3\right)\)….(1)

= \(\left(\frac{4}{10} \cdot \frac{45}{100}\right)+\left(\frac{4}{10} \cdot \frac{60}{100}\right)+\left(\frac{2}{10} \cdot \frac{35}{100}\right)=\frac{490}{1000}=\frac{49}{100}=0.49\)

2. \(P\left(A_2 / E\right)=\frac{P\left(A_2\right) \cdot P\left(E / A_2\right)}{P\left(A_1\right) \cdot P\left(E / A_1\right)+P\left(A_2\right) \cdot P\left(E / A_2\right)+P\left(A_3\right) \cdot P\left(E / A_3\right)}\)

= \(\left(\frac{\frac{4}{10} \times \frac{60}{100}}{\frac{49}{100}}\right)\) (from (1))

= \(\frac{24}{49}=0.48\)

Question 11. A student either knows or guesses or copies the answer to a multiple-choice question with four choices. The probability that he makes a guess is 1/3 and the probability that he copies the answer is 1/6. The probability that his answer is correct given that he copied, is 1/8. Let E1, E2 and E3 are the events that the student guesses, copies or knows the answer respectively and A is the event that the student answers correctly.

1. What is the probability that he answers correctly, given that he knows the answer?

  1. 1
  2. 1/2
  3. 2/3
  4. 1/4

2. What is the probability that he answers correctly, given that he knows the answer?

  1. 1
  2. 0
  3. 1/4
  4. 1/8

3. What is the probability that he answers correctly, given that he had made a guess?

  1. 1/4
  2. 0
  3. 1
  4. 1/8

4. What is the probability that he knew the answer to the question, given that he answered it correctly?

  1. 24/29
  2. 4/29
  3. 1/29
  4. 3/29

5. \(\sum_{k=1}^3 P\left(E_1 \mid A\right) \text { is : }\)

  1. 0
  2. 1/3
  3. 1
  4. 11/8

Solution:

Given that:

E1 = Event that the student guesses the answer.

E2 = Event that the student copies the answer.

E3 = Event that the student knows the answer,

and A = event that the student answers correctly

Now, \(P\left(E_1\right)=\frac{1}{3}\) and \(P\left(E_2\right)=\frac{1}{6}\)

∴ \(P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=1 \Rightarrow \frac{1}{3}+\frac{1}{6}+P\left(E_3\right)=1\)

or \(P\left(E_3\right)=1-\frac{3}{6}=\frac{1}{2}\)….(1)

Also, \(\mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=\frac{1}{8}\) (given), \(\mathrm{P}\left(\mathrm{A} / \mathrm{E}_3\right)=1\)….(2)

(because his answer is correct, given that he knows it]

and \(P\left(A / E_1\right)=\frac{1}{4}\)…..(3)

(his answer is correct, given that he guesses and the multiple choice question has four choices)

1. (2)Required probability = P(E3) = 1/2 (from eq(1))

2. (1)Required probability = P( A/E3) = 1 (from eq(2))

3. (1) Required probability = P( A/E1) = 1/4 (from eq(3))

4. (1) Required probability = P(E3/A)

= \(\frac{P\left(E_3\right) \cdot P\left(A / E_3\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)+P\left(E_3\right) \cdot P\left(A / E_3\right)}\) (By Baye’s Theorem)

= \(\frac{\left(\frac{1}{2} \times 1\right)}{\left(\frac{1}{3} \times \frac{1}{4}\right)+\left(\frac{1}{6} \times \frac{1}{8}\right)+\left(\frac{1}{2} \times 1\right)}=\frac{\frac{1}{2}}{\left(\frac{1}{12}+\frac{1}{48}+\frac{1}{2}\right)}\)

= \(\frac{\frac{1}{2}}{\left(\frac{4+1+24}{48}\right)}=\frac{1}{2} \times \frac{48}{29}=\frac{24}{29}\)

5. (3) \(\sum_{k=1}^3 \mathrm{P}\left(\mathrm{E}_2 \mid \mathrm{A}\right)\)

= \(\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)+\mathrm{P}\left(\mathrm{E}_2 / \mathrm{A}\right)+\mathrm{P}\left(\mathrm{E}_2 / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)+\mathrm{P}\left(\mathrm{E}_3\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_3\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)+\mathrm{P}\left(\mathrm{E}_3\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_3\right)}\)

[By applying Baye’s theorem and then taking L.C.M.] =1

Question 12. Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4. 5 or 6. she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail1, what is the probability that she threw 3, 4.. 5 or 6 with the die?
Solution:

When a die is thrown then the sample space contains 6 outcomes i.e. S= {1,2, 4, 5, 6}

Let E1: Event that she gets 1 or 2 on the die,

E2: Event that she gets 3, 4, 5 or 6 on the die.

A: Event that she acts exactly one tail

Here, \(\mathrm{P}\left(\mathrm{E}_1\right)=\frac{2}{6}=\frac{1}{3}, P\left(\mathrm{E}_2\right)=\frac{4}{6}=\frac{2}{3} \text { and } \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=\frac{3}{8}, \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=\frac{1}{2}\)

When she gets 1 or 2, then she throws a coin three times, and the outcomes are {HHH, TTT, HHT, THH, HTH, TTH, HTT, THT}

⇒ Required probability = P(E2/A)

= \(\frac{P\left(E_2\right) \cdot P\left(A / E_2\right)}{P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)} \text { (Using Baye’s Theorem) } \)

= \(\frac{\frac{2}{3} \times \frac{1}{2}}{\left(\frac{1}{3} \times \frac{3}{8}\right)+\left(\frac{2}{3} \times \frac{1}{2}\right)}=\frac{\frac{1}{3}}{\left(\frac{1}{8}+\frac{1}{3}\right)}=\frac{8}{11}\)

Question 13. Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of the random variable X. and hence find the mean of the distribution. [CBSE 2018|
Solution:

S = {1, 2, 3, 4, 5, 6}

X denotes the larger of the two numbers (given)

If X = 2, then favourable cases are {(1, 2),(2,1)}

If X = 3, then favourable cases are {(1, 3),(2, 3), (3, 1),(3, 2)}

If X = 4, then favourable cases are {(1, 4),(2, 4),(3, 4),(4, 1),(4. 2),(4, 3)}

If X = 5, then favourable cases are {(1, 5),(2, 5),(3, 5),(4, 5),(5,1),(5, 2),(5, 3),(5, 4)}

If X = 6, then favourable cases are {(1,6),(2,6),(3,6),(4,6),(5,6),(6, 1),(6,2),(6,3),(6,4),(6,5)}

Important Questions For CBSE Class 12 Maths Chapter 13 Probability Two Numbers At Random Positive Integers

⇒ Mean = \(\Sigma X . P(X)=\frac{4+12+24+40+60}{30}=\frac{140}{30}=\frac{14}{3}\)

Question 14. A bag contains 5 red and 4 black balls, a second bag contains 3 red and 6 black balls. One of the two bags is selected at random and two balls are drawn at random (without replacement), both of which are found to be red. Find the probability that the balls are drawn from the second bag.
Solution:

Let E1: Event that first bag is selected,

E2: Event that the second bag is selected,

E: Event that both drawn balls are red

∴ Required probability = \(\mathrm{P}\left(\mathrm{E}_2 / \mathrm{E}\right)\)

= \(\frac{P\left(E_2\right) \cdot P\left(E_{/} / E_2\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)}\)(Using Bayes’ Theorem)

= \(\frac{\frac{1}{2} \times \frac{1}{12}}{\frac{1}{2} \times \frac{5}{18}+\frac{1}{2} \times \frac{1}{12}}=\left(\frac{\frac{1}{12}}{\frac{5}{18}+\frac{1}{12}}\right)=\frac{3}{13}\)

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals

CBSE Class 12 Maths Chapter 8 Application Of Integrals Important Questions

Question 1. Using integration, find the area bounded by the curve y ²= 4x, y-axis, and y = 3.

Or,

Using integration, find the region’s area bounded by the line 2y = – x + 8, x-axis. x = 2 and x = 4.

Solution:

Given curve is y² = 4x ….(1)

and given line is y = 3 …..(2)

From equations (1) and (2):

Point of intersection is B(9/4, 3)

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Integrals Area Bounded By The Curve

⇒ Required Area = \(\int_0^3\) x dy

= \(\frac{1}{4} \int_0^3 y^2, d y=\frac{1}{4}\left[\frac{y^3}{3}\right]_0^3=\frac{1}{4}\left[\frac{3^3}{3}-\frac{0}{3}\right]=\left(\frac{1}{4}, \frac{27}{3}\right)=\frac{9}{4} \text { sq. units }\)

Or,

Given lines are 2y + x = 8, x = 2 and x = 4

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Integrals Area Region Bounded By The Curve

⇒ Required area = \(\int_2^4 y \mathrm{dx}=\int_2^4\left(\frac{8-\mathrm{x}}{2}\right) \mathrm{dx}\)

= \(\int_2^4\left(4-\frac{1}{2} \mathrm{x}\right) \mathrm{dx}=\left[4 \mathrm{x}-\frac{1}{2}\left(\frac{\mathrm{x}^2}{2}\right)\right]_2^4\)

= \(\left[4 \times 4-\frac{1}{4}(4)^2\right]-\left[4 \times 2-\frac{1}{4}(2)^2\right]\)

=12-7=5

∴Required Area = 5 sq. units

Question 2. Using integration, find the area bounded by the circle x² + y² = 9.
Solution:

The whole area enclosed by the given circle will be 4 times the area of the region AOBA bounded by the curve, x-axis, and the ordinates x = 0 and x = 3 [as the circle is symmetrical about both the x-axis and y-axis]

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Are Bounded By The Circle

⇒ Required area = \(4 \int_0^3 y d x=4 \int_0^3 \sqrt{3^2-x^2} d x \quad\left[x^2+y^2=3^2 \text { gives } y= \pm \sqrt{3^2-x^2}\right]\)

As the region AOBA lies in the first quadrant, v is taken as positive. Integrating, we get the whole area enclosed by the given circle

⇒ Required area = \(4\left[\frac{x}{2} \sqrt{3^2-x^2}+\frac{3^2}{2} \sin ^{-1}\left(\frac{x}{3}\right)\right]_0^3\)

= \(4\left[\left(\frac{3}{2} \times 0+\frac{3^2}{2} \sin ^{-1}(1)\right)-0\right]=4\left(\frac{3^2}{2}\right)\left(\frac{\pi}{2}\right)=9 \pi \text { sq. units }\)

Question 3. Find the area of the region bounded by curve 4x²=y and the line y = 8x + 12. using integration.
Solution:

Given curve is 4x²= y….(1)

and given line is y = 8x + 12…..(2)

From equation (1) and (2), we get:

4x² – 8x – 12 = 0

⇒ x² – 2x – 3 = 0

⇒ (x-3) (x + 1) = 0 =3 x = 3, -1

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Are Region Bounded By The Curve And Line

From equation (1); when x = 3, y = 36 and when x = – 1 ⇒ y = 4.

So, point of intersection of the curve and line tire (3, 36) and (-1,4).

⇒ Required Area = \(\int_{-1}^3\left\{(8 x+12)-4 x^2\right\} d x=\left[\frac{8 x^2}{2}+12 x-\frac{4 x^3}{3}\right]_{-1}^3\)

= \((36+36-36)-\left(4-12+\frac{4}{3}\right)=36+\frac{20}{3}=\frac{128}{3} \text { sq. units }\)

Important Questions For CBSE Class 12 Maths Chapter 8

Question 4. Using integration, find the area of the region bounded by the curves x² + y² = 4, x =√3y, and the x-axis lying in the first quadrant.
Solution:

Given curve x² + y² = 4 is a circle with center (0, 0) and radius 2.

And line is x = √3y

Now, for the point of intersection of the line and circle, we have:

⇒ 4y² = 4 ⇒ y = ± 1

For y = 1; x = √3

So. point C is (3, 1)

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Area Bounded By The Curve And x axis Lying In First Quardant

Required area = area of OACO + Area of ABCA

= \(\int_0^{\sqrt{3}} y_{\text ({line })} d x+\int_{\sqrt{3}}^2 y_{\mid \text {circle } \mid} d x=\frac{1}{\sqrt{3}} \int_0^{\sqrt{3}} x d x+\int_{\sqrt{3}}^2 \sqrt{4-x^2} d x\)

= \(\frac{1}{2 \sqrt{3}}\left[x^2\right]_0^{\sqrt{3}}+\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_3^{-2}\)

= \(\frac{1}{2 \sqrt{3}}(3)+\left\{2 \sin ^{-1}(1)-\frac{\sqrt{3}}{2}-2 \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\right\}=\frac{\sqrt{3}}{2}+\left(2 \times \frac{\pi}{2}\right)-\frac{\sqrt{3}}{2}-2 \times \frac{\pi}{3}=\frac{\pi}{3} \text { sq.units }\)

Question 5. Using the method of integration, find the area of the triangle ABC. coordinates of whose vertices are A(2. 0), B(4, 5) and C(6, 3).
Solution:

Vertices of ΔABC are A(2,0), B(4,5) and C (6,3)

Equation of line AB: y = 5/2(x – 2)

Equation of line BC: y = 9-x

Equation of line AC: y= 3/4(x-2)

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Area Of The Triangle

⇒ Required Area of ΔABC = \(\int_2^4(\text { line } \mathrm{AB}) \mathrm{dx}+\int_4^6(\text { line } \mathrm{BC}) \mathrm{dx}-\int_2^6(\text { line } \mathrm{AC}) \mathrm{dx}\)

= \(\frac{5}{2} \int_2^4(x-2) d x+\int_4^6(9-x) d x-\frac{3}{4} \int_2^6(x-2) d x\)

= \(\frac{5}{2}\left[\frac{x^2}{2}-2 x\right]_2^4+\left[9 x-\frac{x^2}{2}\right]_4^6-\frac{3}{4}\left[\frac{x^2}{2}-2 x\right]_2^6\)

= \(\frac{5}{2}[(8-8)-(2-4)]+[(54-18)-(36-8)]-\frac{3}{4}[(18-12)-(2-4)]\)

= \(\left(\frac{5}{2} \times 2\right)+8-\left(\frac{3}{4} \times 8\right)=5+8-6=7 \text { sq. units }\)

Question 6. Using the method of integration, find the area of a triangle whose vertices arc (1, 0), (2, 2), and (3, 1).
Solution:

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Area Of ThE Triangle With Vertices

Equation of line AB is y – 0 = \(\frac{(2-0)}{(2-1)}(x-1) \Rightarrow y=2(x-1)\)

Equation of line BC is y – 2 = \(\frac{(1-2)}{(3-2)}(x-2) \Rightarrow y=(-x+1)\)

Equation of line AC is y – 0 = \(\frac{(1-0)}{(3-1)}(x-1) \Rightarrow y=1/2(x-1)\)

⇒ Required Area = \(\int_1^2(\text { line } A B) d x+\int_2^3(\text { line } B C) d x-\int_1^3(\text { line } A C) d x\)

= \(\int_1^2 2(\mathrm{x}-1) \mathrm{dx}+\int_2^3(-\mathrm{x}+4) \mathrm{dx}-\int_1^3 \frac{1}{2}(\mathrm{x}-1) \mathrm{dx}\)

= \(2\left(\frac{\mathrm{x}^2}{2}-\mathrm{x}\right)_1^2+\left(\frac{-\mathrm{x}^2}{2}+4 \mathrm{x}\right)_2^3-\frac{1}{2}\left(\frac{\mathrm{x}^2}{2}-\mathrm{x}\right)_1^3\)

= \(2\left[(2-2)-\left(\frac{1}{2}-1\right)\right]+\left[\left(\frac{-9}{2}+12\right)-(-2+8)\right]-\frac{1}{2}\left[\left(\frac{9}{2}-3\right)-\left(\frac{1}{2}-1\right)\right] \)

= \(\left(2 \times \frac{1}{2}\right)+\left(\frac{-9}{2}+6\right)-\frac{1}{2}(4-2)=1+\frac{3}{2}-1=\frac{3}{2} \text { sq. units }\)

Question 7. Using integration, find the region’s area in the first quadrant enclosed by the x-axis, the line y = x, and the circle x² + y² = 32.
Solution:

Given line is y = x ….(1)

and given circle is x² + y² = 32….(2)

From equations (1) and (2); we have

2x² = 32 ⇒ x = ± 4

∴ y = ± 4

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Area Of The Region In First Quadrant

Now; (4, 4) lies in 1st quadrant

⇒ Required Area  = \(\int_0^4 x d x+\int_1^{4 \sqrt{2}} \sqrt{32-x^2} d x\)

= \(\left(\frac{x^2}{2}\right)_0^4+\left[\frac{x}{2} \sqrt{32-x^2}+\frac{32}{2} \sin ^{-1}\left(\frac{x}{4 \sqrt{2}}\right)\right]_{-4}^{4 \sqrt{2}}\)

= \((8-0)+\left[\left(0+16 \times \frac{\pi}{2}\right)-\left(8+16 \times \frac{\pi}{4}\right)\right]=4 \pi \text { sq. units }\)

NCERT Class 6 Science Chapter 4 Getting To Know Plants Question And Answers

NCERT Class 6 Science Chapter 4 Getting To Know Plants Long Question And Answers

Question 1. Correct the following statements and rewrite them in your notebook.

  1. Stem absorbs water and minerals from the soil
  2. Leaves hold the plant upright
  3. Roots conduct water to the leaves
  4. The number of petals and sepals In a flower is always equal
  5. If the sepals of a flower are joined together, Its petals are also joined together
  6. If the petals of a flower are joined together, then the pistil is joined to the petal

Answer:

  1. Correct statements are as follows
  2. Roots absorb water and minerals from the soil.
  3. Roots hold the plant upright.
  4. The stem conducts water to the leaves.
  5. The number of petals and sepals in a flower can be equal or different.
  6. If the sepals of a flower are joined together, its petals are not necessarily joined together.
  7. If the petals of a flower are joined together, then the pistil may or may not be necessarily joined together.

Read and Learn More Class 6 Science Question And Answers

Question 2. Can you find a plant in your house or In your neighbourhood, which has a long, but weak stem? Write its name. In which category will you place it?
Answer: Yes, we can find money plants in our house as well as in our neighbourhood, which have a long, but weak stem. It is classified as a climber. It readily climbs up by taking support from neighbouring structures such as a tree or a rod, etc.

NCERT Class 6 Science Chapter 4 Getting To Know Plants Questions And Answers

Question 3. What is the function of a stem?
Answer: The functions of a stem in a plant are as follows

  1. It conducts water from the roots to the leaves and other parts of the plants.
  2. It conducts food from leaves to other parts of the plants.
  3. It bears leaves, flowers and fruits.
  4. It provides support to the plant.

Question 4. Which of the following leaves have reticulate venation? Wheat, Tulsi, Maize, Grass, Coriander (dhania), China rose.
Answer: Leaves of tulsi, coriander and China rose have reticulate venation.

Question 5. If a plant has fibrous roots, what type of venation do its leaves have?
Answer: If a plant has a fibrous root, its leaves have parallel venation.

Question 6. If a plant has leaves with reticulate venation, what kind of roots will it have? If a plant has leaves with reticulate venation, then the plant is likely to have tap roots. Is it possible for you to find out whether a plant has tap roots or fibrous roots by looking at the impression of its leaf on a sheet of paper?
Answer: Yes, it is possible to find whether a plant has a taproot or fibrous root by looking at the impression of its leaf on paper. If the leaf has parallel venation, the roots ofthe plant will be fibrous root the leaf has reticulate venation, the root will be tap root.

Question 7. What are the parts of a flower?
Answer: The parts of a flower are as follows

  1. Petals
  2. sepals
  3. Stamens
  4. Pistil

Question 8. From the following plants, which of them have flowers?
Grass, maize, wheat, chilli, tomato, tulsi, peepal, sheesham, banyan, mango, jamun, guava, pomegranate, papaya, banana, lemon, sugarcane, potato, groundnut.
Answer: All ofthe given plants produce flowers. But in some plants such as peepal, sugarcane, grass, etc. the flowers are very small The plants which produce prominent flowers are

Class 6 Science Chapter 4 Getting To Know Plants The Following Plants And Flowers

Question 9. Name the part of the plant which produces food. Name this process.
Answer: The part plant which produces food is the leaf. This process is known as photosynthesis.

Question 10. In which part of a flower, you will find the ovary?
Answer: Ovary is found in the lowermost part of the pistiL

Question 11. Name two plants in which one has joined sepals and the other has separate sepals.
Answer: Flowers with joined sepals are found in Hibiscus, Datura, cotton, tomato (anyone). Flowers with separate sepals are found in mustard, lotus, lily, jasmine, and rose (anyone).

Question 12. The table lists four different groups of plants and their features

Class 6 Science Chapter 4 Getting To Know Plants The Different Groups Of Plants And their features

The Picture Shows A Garden Plant.

Class 6 Science Chapter 4 Getting To Know Plants A Garden Plant

1. Which group does the garden plant belong to?

  1. Group 1
  2. Group 2
  3. Group 3
  4. Group 4

Answer: 2. Group 2

2. Which Group Of Plants Most likely needs support to grow?

  1. Group 1
  2. Group 2
  3. Group 3
  4. Group 4

Answer: 1. Group 1

Question 13. Three similar potted plants were taken to conduct an activity to determine the conditions essential for plant growth. Plant A was kept in sunlight, but not watered. Plant B was kept in sunlight and watered. Plant C was watered and kept in a dark room.

  1. Which plant will grow best and which plant(s) will not show proper growth?
  2. The above experiment lists two main factors necessary for plant growth
  3. Leaves were taken from each of the plants and boiled to remove the green colour. They then had an iodine solution placed on the leaves. Describe and explain what would be observed for leaves A and B.

Answer: Plant TV will grow best and plants ‘A‘ and ‘C’ will not show proper growth.

The two main factors necessary for plant growth are sunlight and water.

The result will not be the same as the leaf from plant ‘A’ will show no change In colour due to the absence of starch, but the leaf from plant ‘B’ will turn bluish-black because of the presence of starch.

Question 14. Ajit wants to test If plants need sunlight to make food. He keeps a potted plant in sunlight, After five days, he tests for the presence of starch In the leaves.

Class 6 Science Chapter 4 Getting To Know Plants If Plants Need Sunlight To Make Food

How can Ajit improve his test?

  1. He should choose a plant with larger green leaves.
  2. He should test another plant without leaves under the Sun.
  3. He should test a similar plant kept in the dark for five days,
  4. He should cover the plant with a transparent glass box to keep it warm.
  5. Which of the following statements Is true? Write ‘Yes’ or ‘No’ to mark your responses.

Class 6 Science Chapter 4 Getting To Know Plants The Following Statements Is True or not

Answer:

  1. He should test a similar plant kept in the dark for five days.
  2. No, Yes, Yes

Question 15. Identify the Incorrect statements and correct them.

  1. Anther Is a part of the pistil.
  2. The visible part of a bud is the petals.
  3. Lateral roots are present In a tap root.
  4. Leaves perform the function of transpiration only

Answer:

  1. It is an incorrect statement as another is a part of the stamen.
  2. It is an incorrect statement as the visible parts ofthe bud are sepals.
  3. It is a correct statement.
  4. It is an incorrect statement as leaves also perform photosynthesis along with transpiration.

Question 16. Read the functions of parts of a plant given below

  1. Fixes plant to the soil
  2. Prepares starch
  3. Takes part in reproduction
  4. Supports branches and bears flowers.
  5. In the diagram given in the below figure, write the names of the parts whose function you have just read in the appropriate space.

Class 6 Science Chapter 4 Getting To Know Plants plant

Answer: The functions of parts of a plant are

Class 6 Science Chapter 4 Getting To Know Plants Funtions And Parts Of A Plant

Question 17. Observe the figure and attempt the questions that follow it.

Class 6 Science Chapter 4 Getting To Know Plants leaf

  1. Label the parts 1, 2, 3 and 4 in the figure.
  2. What type of venation does the leaf have?
  3. What type of venation is seen in grass leaves

Answer: 2. In the given figure

  1. Petiole
  2. Midrib
  3. Lamina
  4. Vein

The leaf has reticulate venation, and the veins in the leaf occur in an irregular way forming a net-like pattern. Grass leaves have parallel types of venation. In which the veins run parallel to each other on both sides of the midrib.

Question 18. Draw the veins of leaves given in the figure below and write the type of venation.

Class 6 Science Chapter 4 Getting To Know Plants Types Of Venation

Answer:

The veins of leaves and their type of venation is

Class 6 Science Chapter 4 Chapter 4 Getting To Know Plants Types Of Venations

Question 19. Sudha fills two beakers with equal amounts of water she places a plant in beaker 1 with the roots of the plant dipped in water. She tightly covers the mouth of both beakers with plastic sheets. She leaves the beakers in sunlight and notes the amount of water in each beaker after 3 days

Class 6 Science Chapter 4 Getting To Know Plants Amount of water in the beaker at the start of the activity

What Is Sudha trying to find out?

  1. Do plants need water to live?
  2. Do plant roots absorb water?
  3. Does water help plants to stand straight?
  4. Does water evaporate faster in sunlight?

Suppose the beakers are not covered with plastic sheets. Will the amount of water in the beakers remain the same as shown in the table? Explain your answer.

Answer:

  1. Do plant roots absorb water?
  2. The amount of water in both beakers will not remain the same and will reduce because of the evaporation.

Question 20. The pictures show four different plants.

Class 6 Science Chapter 4 Getting To Know Plants The Four different plants

Which plants have the same type of roots?

  1. Only plant 1 and plant 2
  2. Only plant 2 and plant 3
  3. Plant 1, plant 2 and plant 3.
  4. Plant 2, plant 3 and plant 4
  5. Why is the root of plant1 thick and round?

Answer:

  1. Plant 1, plant 2 and plant 3
  2. The root of plant 1 is thick and round because these roots store food for the plant.

Question 21. The picture shows different parts of a flower.

Class 6 Science Chapter 4 Getting To Know Plants Types Of A Flower

  1. How many petals can be seen in the picture?
  2. What Is Label X?

Answer:

  1. 3 petals can be seen in the picture.
  2. Label X is the filament.

Question 22. Observe the picture of an activity given as a figure, carried out with leaves of plants and a polythene bag.

Class 6 Science Chapter 4 Getting To Know Plants Plants And Polythene Bag

  1. Now answer the following
  2. Which process is demonstrated in the activity?
  3. When will this activity show better results on a bright sunny day or a cloudy day?
  4. What will you observe in the polythene bag after a few hours of setting up the activity?
  5. Mention any one precaution you must take, while performing this activity.

Answer: The process demonstrated in the activity is transpiration.

  1. The activity will show the best results on a bright sunny day because transpiration is maximum in sunlight.
  2. After a few hours of setting up the activity, one observes small droplets of water inside the polythene bag.
  3. A major precaution one must take while performing the activity is that the polythene bag should be cleaned and its mouth should be sealed properly. Also, the twig should be fresh with 10-12 leaves.

NCERT Class 6 Science Chapter 4 Getting To Know Plants Short Question And Answers

Question 1. What kind of stem do the money plant, beanstalk, gourd plants and grapevines have?
Answer: All these above-mentioned plants, i.e. money plants, beanstalk, gourd plants and grapevines have soft, green and weak stems and are climbers.

Question 2. Do all the leaves have petioles?
Answer: No, all the leaves do not have petioles. In some plants, leaves are attached directly to the plant stem.

Question 3. Can you name the process that makes water drops appear on the polythene cover?
Answer: The process that makes water droplets appear inside the polythene cover is known as transpiration. This occurs due to the loss of water from the leaves.

Question 4. Look at the figure below. Who do you think is watering their plant correctly? Paheli or Boojho? Why

Class 6 Science Chapter 4 Chapter 4 Getting To Know Plants Watering Their Plants

Answer: I think that Paheli is watering the plants correctly. This is because she is sprinkling water on the roots, which helps in the upward conduction of water. How are the types of roots and leaf venation in a plant related to each other? Fill in the table below to justify your answer.

Class 6 Science Chapter 4 Getting To Know Plants The types of roots and leaf venation.

Answer: The table below shows the relation between the roots and leaf venation types in a plant.

Getting To Know Plants The types of roots and leaf venation.

Question 5. Write the name of the material that goes up in the stem and that which comes down.
Answer: The stem is like a street with two-way traffic. Through it, water and minerals go up while it conducts food from the leaves to other parts of the plant.

Question 6. Write down two examples of trees, shrubs, herbs and creepers growing near your area.
Answer: Two examples of trees – Oak and banyan (barged). Two examples of shrubs- the China rose and jasmine. Two examples of herbs are tomato and wheat. Two examples of creepers are pumpkin and watermelon.

Question 7. Can the stem of a plant be compared with a street with two-way traffic? Give reason.
Answer: Yes, the stem of a plant can be compared with a street with two ways traffic because

It carries water and minerals from the roots to the leaves and other parts ofplantin an upward direction.

It takes the food prepared by the leaves to other parts of the plant.

Question 8. Is it right to call the leaf as food factory of the plant? Justify your answer.
Answer: Yes, the leafs called as food factory of the plant. It is because the main function of a leaf is to synthesise food by the process of photosynthesis.

Question 9. Roots are necessary to keep the plants healthy and alive. Explain.
Answer: Roots absorb minerals and water from the soil. Both of these are needed for the manufacture of food from plant leaves. So, roots are necessary to keep the plant healthy and alive.

Question 10. Taproot is different from the fibrous root. Explain how.
Answer: The differences between tap and fibrous roots are

Class 6 Science Chapter 4 Getting To Know Plants The differents between tap root and fibrous root

Question 11. The type of leaf venation and root in plants are related interestingly. Explain the statement.
Answer: The relation between the type of leaf venation and the type of roots is as follows

  1. The plant having leaves with reticulate venation has tap roots, for Example sunflower plant.
  2. The plant having leaves with parallel venation has fibrous roots, Example wheat.

Question 12. Will a leaf taken from a potted plant kept in a dark room for a few days turn blue-black when tested for starch? Give a reason for your answer.
Answer: No, a leaf from a potted plant kept in the dark will not turn blue-black when tested for the presence of starch. This is because all the stored starch would have been used up by the plant. No fresh starch would be synthesised due to the non-availability of sunlight.

Question 13. Boojho wanted to test the presence of starch in leaves. He performed the following steps

  1. He took a leaf and boiled it in water.
  2. He placed the leaf in a Petri dish and poured some iodine over it
  3. He did not get the expected result. Which step did he miss? Explain.

Answer: Boojho did not get the expected results in his experiment because he missed an important step in the procedure. He did not boil the leaf in spirit to remove chlorophyll. It is necessary to remove chlorophyll because it interferes with the test for starch. It is also essential to remove chlorophyll from leaves so that the leaves get decolourised.

Question 14. Suggest the type of root system In grass. Also, explain that root system.
Answer: In grass, a fibrous root system is present. In such types of roots, there is no main root, instead many roots arise from one region. These grow horizontally in the soil and make a bushy/clustered appearance.

NCERT Class 6 Science Chapter 4 Getting To Know Plants Assertion-Reason Questions

Question 1. The following questions consist of two statements. Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A
  3. A is true, but R is false
  4. A is false, but R is true

1. Assertion (A) The Lamina of the leaf is green and helps in photosynthesis.
Reason (R) It is green due to the presence of green-coloured pigment.

2. Assertion (A) Roots are aerial parts of the plants.
Reason (R) Their main function is to absorb water and minerals from the soil.

3. Assertion (A) In tap root, the main root is present.
Reason (R) The smaller roots arising from the main root are called lateral roots.

4. Assertion (A) Transpiration helps in the movement of water from roots to leaves.
Reason (R) Transpiration is the loss of water from the stomata on leaves. As water evaporates from leaves, it creates a suction force that pulls water from the roots through the plant.

Answer:

Both A and R are true and R is the correct explanation of A.

A is false, but R is true. A can be corrected as Shoot is the aerial part ofthe plant whereas roots are present deep in the soil (underground)

Both A and R are true, but R is not the correct explanation of A. It can be corrected as In tap root one main root is present which grows vertically downward into the soil.

Both A and R are true and R is the correct explanation of A.

Question 2. Assertion (A) Plants with weak stems that cannot stand upright but spread on the ground are called creepers.
Reason (R) Plants that take support and climb up are called climbers.
Answer: Plants that take support and climb up are called climbers.

Question 3. Assertion (A) The Female part of the flower is called a pistil.
Reason (R) Transfer of pollen grains from the anther to the stigma is called pollination.
Answer: 2. Transfer of pollen grains from the anther to the stigma is called pollination.

 

NCERT Class 6 Science Chapter 4 Getting To Know Plants Very Short Questions and Answers

Question 1. Plants are classified into three main categories. Explain briefly.
Answer: Based on the size and nature of the stem, die plants are classified into three main categories, i.e. herbs, shrubs and trees.

Question 2. What is the usual name of medium-sized plants with hard and woody stems whose many branches arise just above the ground?
Answer: Shrubs

Question 3. All plants are of the same size. Do you agree?
Answer: No, all plants are of different sizes. According to their size, they are classified as herbs, shrubs and trees.

Question 4. Herb is different from a shrub. Explain.
Answer: The herb is a small plant with a tender and green stem. Whereas, the shrub is bigger than an herb and has a strong and thicker stem.

Question 5. Give the term for each of the following.

  1. These are small plants with green, soft, tender stems.
  2. These are bigger than herbs with thick and hard stems and branching at the base.

Answer: Herbs Shrubs

Question 6. It Is given that a plant Is very tall and has a hard, thick stem with branching on the upper part of the plant. Based on the given characteristics, categorise the plant.
Answer: Based on the given characteristics, it can be concluded that the given plant is a tree.

Question 7. Money plant Is an example of a creeper. Do you agree? Explain.
Answer: No, a money plant is not an example of a creeper. It is an example of a climber as it grows or climbs up by taking the support of neighbouring structures.

Question 8. Differentiate between vein and midrib.
Answer: A large number of linear structures that spread to all parts of the leaf are called veins. Whereas, the main vein (thick vein) in the middle of the leaf is called midrib.

Question 9. The leaves of grass are different from those of mango. Explain.
Answer: The leaves of grass show parallel venation whereas mango leaves have reticulate venation.

Question 10. leaves when treated with iodine give a blue-black colour. Give the reason.
Answer: The leaf gives a blue-black colour due to the presence of starch in it.

Question 11. In some plants, roots do not have any main root but all the roots are similar. Give some examples of such fibrous roots.
Answer: Examples of plants with fibrous roots are maize, wheat, rice, etc.

Question 12. Name the part of a plant which helps in holding the plant to the soil.
Answer: The roots anchor the plant to the soil.

Question 13. Roots absorb water and minerals from the soil. Why are these needed?
Answer: Roots absorb water and minerals from the soil. These are needed for the manufacture of food from plant leaves.

Question 14. Roots are helpful in the conservation of soil. Explain.
Answer: Roots help in holding the soil together. In this way, roots prevent the soil from being blown away by wind or washed away by water and thus, help in the conservation of soil.

Question 15. Differentiate between sepals and petals.
Answer: The small, green-coloured leaf-like structures seen in flowers are called sepals. Whereas, petals are the big, brightly coloured leaf-like structures seen in flowers.

Question 16. Plants also have reproductive organs like animals. Name the reproductive part of plants.
Answer: Stamens and pistil are the male and female reproductive parts of plants, respectively

NCERT Class 6 Science Chapter 4 Getting To Know Plants Fill In The Blanks 

1. The type of venation found in pea is Reticulate

2. Water comes out of the leaf by a process called transpiration

3. The small green leaves at the base of flowers are known as Sepals

4. The swollen basal part of the pistil is the Ovary, which bears the

5. Stamen has two parts called Anther, filaments.

6. The young unopened flower is termed a bud

7. Sepals, Petals stamens and pistils are the parts of a flower. Stamen Is made up of anther and filament and It represents the male part of the flower. The female part of the flower is called the pistil The basal, swollen part of the pistil is called the ovary which contains the ovules.

8. Loss of water through leaves is called Transpiration

9. The unwanted plants that grow in a field are called weeds

10. Mango leaves have venation reticulate

NCERT Class 6 Science Chapter 4 Getting To Know Plants True Or False

1. Stem bears branches, flowers and fruits. True

2. The arrangement of veins in the leaf lamina is called venation. True

3. Leaves are generally multicoloured. False, leaves are generally green in colour

4. Lateral roots are present in a tap root. True

5. Anther is a part of the pistil. False, another is a part of the stamen

6. Pollen grains are present in the anther of the pistil. False

7. Wheat has parallel venation and fibrous roots. True

NCERT Class 6 Science Chapter 4 Getting To Know Plants Match The Columns

Question 1. Match the Column 1 with Column 2.

Class 6 Science Chapter 4 Getting To Know Plants Match the coloumn 1 and column 2

Answer: A-2, B-1, C-4, D-3

Question 2. Match the Column 1 with Column 2.

Class 6 Science Chapter 4 Getting To Know Plants Match the coloumn 1 and column 2.2

Answer: A-3, B-2, C-4,D-5

Question 3. Match the Column 1 with Column 2.

Class 6 Science Chapter 4 Getting To Know Plants Match the column 1 and column 2.3

Answer: A-2, B-1, C-4, D-3

NCERT Class 6 Science Chapter 5 Body Movement Question And Answers

NCERT Class 6 Science Chapter 5 Class 6 Science Body Movement Long Questions And Answers

Question 1. Plants do not move from one place to another. Do they show any other kind of movement? What types of movements are shown by plants?
Answer: Plants do not move from one place to another, but they show other kinds of movements. For example, the movement of sunflowers toward the Sun, the movement of roots toward the water, and the wrinkling of leaves of touch-me-not plants when touched, etc.

Question 2. Fill in the table given below to explain how animals move from place to place.

The Living Organisms How Animals Move From Palce To Place

Answer: The table given below explains the movement of animals from place to place

The Living Organisms How Animals Move From Palce To Place.

Question 3. Why are there so many differences In the way that animals move from place to place?
Answer: There are so many differences in the way that animals move from place to place because of the differences in their body structure and also the different types of environments in which they live.

Read and Learn More Class 6 Science Question And Answers

Question 4. Fill In the table given below showing different types of movement in our body.

Class 6 Science Chapter 5 Body Movements Movements In Our Body

Answer:

Class 6 Science Chapter 5 Body Movements Movements In Our Body.

Question 5. Why is it that we can move a few parts of our body easily in various directions and some only in one direction? Why are we unable to move some parts at
Answer: We can move a few parts ofour body easily in various directions and some only in one direction. Some parts are fixed, i.e. they do not move at all. This is because the movement depends upon the type of joint that joins two or more bones together.

For example, the body parts having ball and socket joints move in all directions, while the body parts having hinge joints move only in one direction. The body parts having fixed joints are unable to move at all.

NCERT Class 6 Science Chapter 5 Body Movements Question And Answers

Question 6. How do we bend our elbows?
Answer: The elbow is not one long bone from the upper arm to our wrist. It consists of different bones joined together at the elbow. The joint present at the elbow is the hinge joint that helps in bending

Question 7. How many bones does the middle finger have?
Answer: The middle finger has three bones.

Question 8. Are muscles and bones always required for mo Pages
Answer: Yes, muscles and bones are always required for movement.

Question 9. Do all animals have bones?
Answer: No, all animals do not have bones. For example, earthworms.

Question 10. How does an earthworm fix parts of its body to the ground?
Answer: Earthworms possess a large number of tiny projections called bristles. These bristles are connected with muscles. They help to get a grip on the soil and thereby, help the earthworm in fixing parts of its body to the ground.

Question 11. Is the movement of a snail slow or fast as compared to an earthworm?
Answer: The movement snail is slow as compared to an earthworm.

Question 12. Why do different animals have the body parts that they do have and how do these body parts help animals to move the way they do?
Answer: Use of the differences in the way of their movement. For example, a snail has a thick muscular foot for movement Cockroach has three pairs of legs that help in walking.

Question 13. What are the differences in body parts between different animals used for movement? 
Answer: Animals of different kinds of body structures move in different ways using different body parts such as legs, wings, fins, tails, etc. These help them to walk, jump, fly, swim, run, etc.

Question 14. Why are there two legs for humans and four for cows and buffaloes?
Answer: Technically, humans do have four legs however when we evolved, the muscles in our legs grew stronger and longer, allowing us to stand upright. However, cows did not evolve with time and they still need four legs to balance themselves and around.

Question 15. Why is the bending of our legs different from that of our arms?
Answer: The bending ofour legs is different from that ofour arms because of the differences in their position and role in our body. Although they both are hinge joints, they bend in opposite directions.

Question 16. Distinguish between the following

  1. Movable joint and fixed joint
  2. Ball and socket joint and hinge joint

Answer: Differences between movable joints and fixed joints are as follows

Class 6 Science Chapter 5 Body Movements Movable Joint And Fixed Joint

Differences between Ball and socket joint and hinge joint are as follows-

Class 6 Science Chapter 5 Body Movements Ball And Socket Joint Hinge Joint

Question 17. Draw a diagram to show the Joint In the hand and answer the following questions.

  1. Can you bend your finger at every joint?
  2. How many bones are there in a human skull?
  3. Is your wrist flexible?
  4. What would happen if your hand had only one bone?

Answer:

  • We can bend our fingers at every joint.
  • The human skull has 22 bones
  • Yes, our wrist is flexible.
  • If our hand had only one bone, then we would not be able to bend our fingers and other parts of our hand

Question 18. How is the skeleton of a bird well-suited for flying?
Answer: The skeleton of a bird is well-suited for flying because

  • Bones are hollow and light in weight.
  • Bones of hindlimbs are used for walking and perching.
  • The bones of the forelimbs are modified as wings.
  • Shoulder bones are strong and helpful in flying.
  • Breast bones hold flight muscles and are used to move the wings up and down for flying.

Question 19. In the figure given below, there are two snakes of the same size, slithering on sand. Can you identify which of them would move faster and why?

Class 6 Science Chapter 5 Body Movements Two snakes of the same size slithering on sand

Answer:

A snake forms loops in its body while slithering. Each loop of the snake gives it a forward push by pressing against the ground.

The snake with a larger number of loops will move much faster than the snake with a lesser number of loops. Thus, snake A will move faster than snake B as we can see that loops in snake A are more than in snake B.

Question 20. Joints are the locations in the human body where two bones are connected. The picture shows three types of joints. The arrows show the movement of the bones in each Joint

Class 6 Science Chapter 5 Body Movements The Movement Of The Bones In each Joint

  1. In which joint can the pair of bones move in all directions?
  2. Which two joints shown in the picture are of the same type?
  3. Mention the name of that type of Joint.
  4. Some bone Joints are fixed and the bones at these joints cannot move. Which of these are fixed Joints?
  • Joints in the toe
  • Joints in the neck
  • Joints in the wrist
  • Joints in the skull.

Answer: The pair of bones can move in all directions in joint 3 as it is a ball and socket joint.

Joint 1 and joint 2, shown in the picture are the two joints that are of the same type. These are hinge joints.

Joints in the skull

21.

Class 6 Science Chapter 5 Body Movements Postion 1 And Postion 2

  1. In which position are the biceps most contracted?
  2. Which of these statements is true?
  3. The largest muscles of the human body are located in the arms
  4. The bending of the anus is controlled by muscle
  5. The muscles located in the arm have a feed shape
  6. Bending of arms involves the contraction and relaxation of a pair of muscles

Answer: The bending of arms involves the contraction and relaxation of a pair of muscles.

NCERT Class 6 Science Chapter 5 Body Movement Short Question And Answers

Question 1. Given below is a list of different types of movements in animals: Running, jumping, walking, slithering, crawling, flying swimming creeping. Write the types of movements seen in each animal.

  1. Duck 
  2. Horse 
  3. Kangaroo 
  4. Snail
  5. Snake
  6. Fish
  7. Human being
  8. Cockroach 

Answer: The types of movements seen in each animal are-

  1. Duck — Flying, walking, and swimming
  2. Horse — Running, jumping, and walking
  3. Kangaroo — Jumping
  4. Snail — Creeping
  5. Snake — Slithering
  6. Fish — Swimming
  7. Human being — Walking and running
  8. Cockroach — Walking and flying

Question 2. Which type of movement would have been possible if

  1. Our elbow had a fixed Joint.
  2. We were to have a ball and socket joint between our neck and head

Answer: If our elbow had a fixed joint, we would not be able to bend or fold our arms. A fixed joint does not allow any movement.

If we were to have a ball and socket joint between our neck and head, we would be able to rotate our head 360°. A ball and socket joint allows movement in all directions.

Question 3. Bones are hard structures and cannot be bent, but we can still bend our elbows, knees, etc. How is this possible?
Answer: The elbow and knee are not made up of a single bone rather they are made up of two or more bones that are joined to each other at a joint, i.e. hinge joint. This joint along with the muscles helps us to bend the elbow and knee.

Question 4. Study the classification chart given below

Class 6 Science Chapter 5 Body Movements Joints

What are A and B? Why?

Answer: A is the movable joint, which is further classified into

  • Ball and socket joint-e.g. hip joint and shoulder joint.
  • Pivotal joint – example between neck joint and head joint.
  • Hinge joint – Example elbow joint and knee joint.
  • B is the immovable or fixed joint. It is located within the skull. The upper jaw is fixed to the skull bones with the help of fixed joints.

Question 5. Write the type of joint which is used for each of the following movements

  1. A cricket bowler bowls the ball.
  2. A girl moves her head in the right and left direction.
  3. A person lifts weights to build up his biceps.

Answer:

  1. Ball and socket joint
  2. Pivotal joint
  3. Hinge joint

Question 6. Mention the number of bones present In the human skeletal system. Give its functions.
Answer: The human skeletal system is made up of 206 bones.

The functions of the skeletal system are

  1. To provide shape and support to the body.
  2. To protect internal organs.
  3. To make movements possible

Question 7. There are 12 pairs of curved bones called the Xln chest region. One end of bone X is joined to backbone and the other end is joined to bone Tin the front to form a box-like structure Z. This strong, box-like structure protects the delicate organs P, Q, and Rour body.

  1. What are (a) X, (b), and (c) Z?
  2. Name the organ
    • P
    • Q
    • R

Answer:

  1. Ribs (X)
  2. Breast bone (Y)
  3. Ribcage (Z)
  4. Heart (P)
  5. Lungs
  6. Liver (R)

Question 8. State the consequence if the backbone had only one long bone instead of many vertebrae.
Answer: Our backbone is comprised of 33 small bones called vertebrae. These are joined together with soft tissue in the form of cartilage discs. This arrangement makes these bony structures have some limited movements of the body.

If there is only one long bone instead of many in the backbone, the organism will not be able to avail the limited movements that are necessary for the organism for its survival.

Question 9. Write two ways by which we may know the shape of the human skeleton.
Answer: The two ways are given below

  • We can determine the shape of the human skeleton by feeling the bones ofour body.
  • We can also take X-ray images skeleton by using X-ray imaging

Question 10. Write the differences between bone and cartilage.
Answer: The differences between bone and cartilage are as follows

Class 6 Science Chapter 5 Body Movements The difference between Bone And Cartilage

Question 11. The skeleton is made up of bones, but it has some cartilage too. Do you agree?
Answer: Yes, cartilages are the additional parts of the skeleton that are not as hard as the bones and which can bent. So, we can say that though most of the skeleton is made up of bones, it has some cartilage too.

Question 12. Provide one-word answers to the statements given below.

  1. Joint which allows movement In all directions
  2. The hard structure that forms the skeleton
  3. Part of the body with a fixed joint.
  4. Help in the movement of the body through contraction and relaxation
  5. Bones that Join with the chest bone at one end and to the backbone at the other end
  6. Framework of bones which gives shape to our body
  7. Bones which enclose the organs of our body that lie below the abdomen our
  8. A joint where our neck joins the head
  9. Part of the skeleton that forms the earlobe

Answer:

  1. Ball and socket joint
  2. Bones
  3. Upper jaw with the skull
  4. Muscles
  5. Rib
  6. Skeleton
  7. Pelvic bones
  8. Pivotal joint
  9. Cartilage

Question 13. We need two muscles together to move a bone. Why?
Answer: A muscle can only pull, it cannot push. Thus, two muscles are required to work together to move a bone. When one muscle contracts, the bone is pulled. When another muscle of the pair pulls, it brings the bone back to its original position.

Question 14. Bhoojho fell off a tree and hurt his ankle. On examination, the doctor confirmed that the ankle was fractured. How was it detected?
Answer: The doctor must have observed swelling around his ankle and must have taken an X-ray of his ankle. X-ray images confirm any type of injury or fractures in the bones.

Question 15. The gait differs in different animals. Explain.
Answer: Different animals use different organs to move from one place to another. So, the manner of movement, i.e. the gait also differs in different animals.

Question 16. Earthworms are known as farmer’s friends why?
Answer: An earthworm is called a farmer’s friend as it eats its way through the soil thereby, loosening the soil. Also, the earthworm’s body excretes the undigested materials that increase the fertility of the soil.

Question 17. Can you imagine how an earthworm fixes parts of its body to the ground?
Answer: Under the earthworm’s body, it has a large number of tiny bristles projecting out. The bristles are connected with muscles and help to get a good grip on the ground.

Question 18. The skeleton of a cockroach is called an exoskeleton. Do you agree? Explain.
Answer: The body of a cockroach is covered with a hard outer protective covering called an exoskeleton. The skeleton of a cockroach is called the exoskeleton because it is present outside the body.

Question 19. The body of the fish is similar to that of the boat. Explain, how fish move in the water.
Answer: The streamlined shape of fish helps it to move in the water. The skeleton and muscles present on the front side of fish move on one side, while the tail muscles move the body on the other side. This makes a jerk and pushes the body forward. In this way, fish moves in water.

Question 20. Name the organisms that show these movements.

  • A thick muscular foot of the organism helps in dragging the shell.
  • The organism curves its body into various loops.
  • Their loops push the body forward.

Answer:

  1. Snail
  2. Snake

Question 22. The table shows how different animals move

Class 6 Science Chapter 5 Body Movements Animals How Does The Animal Move

Which animal uses all of its body parts to move?

  1. Frog
  2. Mouse
  3. Dragon
  4. Earthworm

Answer: 4. Earthworm

What helps a frog hop?

  1. Strong leg muscles
  2. Short body length
  3. Two pairs of legs
  4. Absence of a tail

Answer: 1. Strong leg muscles

Question 2. Rishi was playing in a garden. It was the rainy season. He observed an organism moving on the soil Based on his observation of the manner of movement, which organism is likely to be seen?

  1. Ant
  2. Cockroach
  3. Earthworm
  4. Snake

Answer: 3. Earthworm.

NCERT Class 6 Science Chapter 5 Body Movement Assertion-Reason Questions

The following questions consist of two statements. Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below.

  • Both A and R are true and R is the correct explanation of A
  • Both A and R are true, but R is not the correct explanation of A
  • A is true, but R is false
  • A is false, but R is true

Assertion (A) We can bend our body parts only at the joints.
Reason (R) Joints are those points where two or more bones are joined together.

Assertion (A) Pivot joint allows movement in all directions.
Reason (R) In a pivot joint, the rounded surface of one bone fits into a ring formed by the other bone.

Assertion (A) The bodies of the birds are well suited for flying.
Reason (R) The Presence of hollow bones makes the bones of birds very light

Answers

  1. Both A and R are true and R is the correct explanation of A.
  2. A is false, but R is true. A can be corrected as the Pivot joint allows only rotational movement.
  3. Both A and R are true and R is the correct explanation of A.

2. Direction The following questions consist of two statements. Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below.

  • Both A and R are true and R Is the correct explanation of A
  • Both A and R are true, but R is not the correct explanation of A
  • A is true, but R is false
  • A is false, but R is true

1. Assertion (A) An earthworm does not have bones. Reason (R) The body of an earthworm is made up of many rings, joined end to end.
Answer: 1. Both A and R are true and R Is the correct explanation of A

2. Assertion (A) Cockroaches can walk, climb and fly upto a short distance.
Reason (R) They have three pairs of jointed legs and two pairs of wings attached to the breast region which help them to walk and fly, respectively.
Answer: 1. Both A and R are true and R Is the correct explanation of A.

NCERT Class 6 Science Chapter 5 Body Movement Very Short Answer Type Questions

Question 1. Name the parts of our body responsible for the movements.
Answer: Bone and muscles ofour body are responsible for the movements.

Question 2. If there are no joints, then will it be possible to move? Explain.
Answer: No, movement in the body of an animal is possible in the presence of joints only.

Question 3. Can we bend our bodies at every part?
Answer: No, we can bend our body at every part, but only at joints.

Question 4. The arm rotates at a specific part of the body. Name the part at which the arms rotate.
Answer: Shoulders are that part of the body (or skeleton) to which our arms are attached and at which the arms rotate.

Question 5. Name the type of joints of your hand that help you
Answer: The hinge joint ofour hand helps us to grasp a badminton racquet.

Question 6. What do you mean by skeleton?
Answer: The bones and cartilage in our body form a framework that supports the whole body and gives a shape to it. This framework is called the skeleton.

Question 7. Name three components of the skeletal system.
Answer: A skeleton is made up of many bones, joints, and cartilage.

Question 8. Name the type of joints present mainly in the skull.
Answer: The skull bones are joined together by fixed joints.

Question 9. The hard and strong skull protects a delicate organ of our body. Name that organ.
Answer: The hard and strong skull protects a delicate organ of our body called the brain.

Question 10. Name the organs that are protected by the ribcage
Answer: The heart and lungs are protected by the ribcage.

Question 11. What would have happened if our backbone was
Answer: If our backbone was made up of one single bone, then we would not have been able to bend our body (from the waist).

Question 12. If an earthworm does not have bones or legs, then how does it move?
Answer: An earthworm does not have bones or legs. It has muscles which help it to lengthen and shorten the body to move.

Question 13. Comparing an earthworm and a snail, explain which of the two moves faster.
Answer: Earthworm moves faster than snails.

Question 14. Cockroaches can fly as well as walk. What helps the cockroach to walk?
Answer: Three pairs of legs help cockroaches to walk.

Question 15. Write any two adaptations of birds for flying in the air.
Answer: The following adaptations are seen in the body of birds.

  • Bones are hollow making the body lightweight for flying.
  • Forelimbs are modified into wings.

Question 16. Name three animals having streamlined bodies.
Answer: Fish, birds, and snakes have streamlined bodies.

Question 17. Some animals like fishes, birds, and snakes move due to their streamlined bodies. What do you mean by streamlined?
Answer: If the body tapers at both ends, then such a body shape is said to be streamlined.

Question 18. What is a ball and socket joint?
Answer: A joint in which the rounded end of one bone fits into the hollow space of another bone is called a ball and socket joint It allows movements in all directions. Joints between the upper arm and shoulder, thigh and hip are examples of ball and socket joints.

Question 19. Which of the skull bones are movable?
Answer: In the skull, only the lower jaw is movable. All the other bones of the skull are immovable.

Question 20. Why can our elbow not move backward?
Answer: Our elbow cannot move backward because the elbow has a hinge joint that allows movement in only one direction.

NCERT Class 6 Science Chapter 5 Body Movement Fill In The Blanks

1. Unscramble the jumbled words and write them in the blank spaces provided

  • Neosb-Bones
  • Tnemevom-Movement
  • Iontcaront-Contraction
  • Isecsum-Muscles
  • Arctigeal-Cartilage
  • Ephas-Shape
  • Inerlant sangro-Internal organs
  • laxaeriont-Relaxation.

2. Read the following paragraph and fill in the blanks using the word you unscrambled.

Bones and cartilage from the skeleton of the human body. they provide the framework, give shape to the body, and help in movement. They protect the internal organs. The bones are moved by alternate contraction and relaxation of two sets of muscles attached to them.

6. Contraction of muscles pulls the bone during movement.

7. The shoulder joint is an example of a ball and socket joint.

8. Joints of the bones help in the Movement of the body.

9. A combination of bones and cartilage forms the Skeleton of the body.

10. The bones at the elbow are joined by a Hinge joint.

11. The contraction of the Muscle pulls the bones during movement

NCERT Class 6 Science Chapter 5 Body Movement True Or False

1. All joints in our body are similar. False, joints are of mainly two types, i.e. fixed and
movable.

2. The upper jaw is movable. False, the lower jaw bone is the only movable bone in
the skull.

3. The ribcage protects vital internal organs like the heart, lungs, and liver. True

4. The joint of the upper arm and shoulder blade is a hinge joint. False, the joint of the upper arm and shoulder blade is a ball and socket joint.

5. Minute movable bristles in earthworms help in making a grip on the ground. True

6. An earthworm has a thick muscular foot. False

7. The body of a cockroach is covered with an exoskeleton. True

8. The body of a snail is streamlined. False

9. The movement and locomotion of all animals is the same. False, the movement and locomotion of animals differ.

10. The cartilage is harder than bones. False, the cartilage is not harder than bones

11. The finger bones do not have joints. False, the finger bones have joints.

12. The forearm has two bones. True

13. Cockroaches have an outer skeleton. True

NCERT Class 6 Science Chapter 5 Body Movement Match The Columns

Question 1. Match the column 1 with column 2 

Class 6 Science Chapter 5 Body Movements Match The Column 1 And 2

Answer: A-3, B-5, C-4, D-2, E-1

Question 2. Match the column 1 with column 2 

Class 6 Science Chapter 5 Body Movements Match The Column 1 And 2.

Answer: A-3, B-5; C-4, D-1, E-2

Question 3. Match the column 1 with column 2 

Class 6 Science Chapter 5 Body Movements Match The Column 1 And 2.3

Answer: A-4, B-1 or 7, C-5, D-6 or 2, E-2 or 3

NCERT Class 6 Science Chapter 3 Separation Of Substances Question And Answers

NCERT Class 6 Science Chapter 3 Separation Of Substances Long Question And Answers

Question 1. When separating sand and sawdust or powdered leaves, do both components fall in the same place?
Answer: When sand and sawdust are separated by winnowing, sawdust falls far away from the sand. This is because sand is heavy, while sawdust is light.

Question 2. Which method of separating tea leaves from prepared tea is better, decantation or filtration?
Answer: Tea leaves are better separated from prepared tea using the filtration method. It is because if the decantation method is used, it helps a little, but we still get a few leaves in our tea.

Read and Learn More Class 6 Science Question And Answers

Question 3. Many substances dissolve in water and form a solution. What happens if we go on adding more and more of these substances to a fixed quantity of water?
Answer: If we go on adding more and more of this substance to a fixed quantity of water, some of it remains undissolved in the solution and settles at the bottom of the vessel. This is called a saturated solution.

NCERT Class 6 Science Chapter 3 Separation Of Substance Questions And Answers

Question 4. What is sieving? Where is it used?
Answer: Sieving is a method by which fine particles are separated from bigger impurities by using a sieve. It is used in a flour mill, our homes, and at construction sites. In a flour mill, impurities like husk and stones are removed from wheat before grinding. At construction sites, pebbles and stones are removed from sand by sieving.

Question 5. Is it possible to separate sugar mixed with wheat flour? If yes, how will you do it?
Answer: Yes, it is possible to separate sugar mixed with wheat flour by sieving. The size of sugar particles is larger than that ofthe wheat flour. So, sugar will stay on a sieve and wheat flour will pass through the holes of the sieve.

Question 6. How will you separate sand and water from their mixture?
Answer: We will separate sand and water from their mixture by the process of sedimentation and decantation. Sedimentation During this process, we leave the mixture for some time. Heavier sand gets settled down at the bottom.

Decantation After the sedimentation process, water is poured into another vessel, and sand is left undisturbed in the first vessel. Thus, sand and water will be separated from their mixture.

Question 7. How would you obtain clear water from a sample of muddy water?
Answer: We will obtain clear water from a sample of muddy water by the process of filtration.

In this process, we use a piece of cloth to filter muddy water. Because in a piece of cloth, small holes or pores remain in between the woven threads. These pores in a cloth can be used as a filter.

If the water is still muddy, impurities can be separated by a filter paper that has even smaller pores.

A filter paper folded in the form of a cone is fixed onto a funnel. The mixture is then poured on the filter paper. Solid particles in the mixture do not pass through it and remain on the filter paper and in this way, clear water is obtained.

Question 8. Lemonade is prepared by mixing lemon juice and sugar in water. You wish to add ice to cool it. Should you add ice to the lemonade before or after dissolving sugar? In which case would it be possible to dissolve more sugar?
Answer: We should add ice to lemonade after dissolving sugar. Without ice, sugar dissolves easily in lemon juice because solubility depends on the temperature. When the temperature is high, then more sugar can be dissolved. After mixing ice, it gets cooled down so less sugar will dissolve in it.

Question 9. Both Sarika and Mohan were asked to make salt solutions. Sarika was given a teaspoonful of salt and half a glass of water, whereas, Mohan was given twenty teaspoonful of salt and half a glass of water.

  1. How would they make salt solutions?
  2. Who would be able to prepare a saturated solution

Answer:

They will mix salt with a certain amount of water to make a salt solution.

Saturation is a point at which no more substances can be dissolved in the solvent and the substance gets settled down. As Mohan has more quantity of salt than Sarika, hence, he would be able to prepare a saturated solution.

Question 10. Paheli was feeling thirsty, but there was only a pot of water at home which was muddy and unfit for drinking. How do you think Paheli would have made this water fit for drinking if the following materials were available to her? Tub, muslin cloth, gas stove, thread, alum, pan, and lid of the substance.
Answer: Paheli can make this water fit for drinking by working on the following sequence

At first, she needs to pour the water through a muslin cloth for filtration.

Next, the water has to be heated in a container on a gas stove until it vaporizes. The vapors are allowed to come in contact with a cold lid so that condense back into water which will be collected in a tub. Impuritiesifany will remain behind in the container.

Then she has to hang a piece of alum with a thread and submerge it in the water, keeping it undisturbed for some time. This allows the impurities to settle down as sediment.

Next, she has to pour off the undisturbed water in a tub by decantation.

Now, the water is fit for drinking.

Question 11. Read the story titled ‘Wise Farmer’ and tick the correct option from the underlined section to complete the story. A farmer was

  1. Sad/happy to see his healthy wheat crops ready for harvest. He harvested the crops and left them under the
  2. Sun/rain to dry the stalks. To separate the seeds from the bundles of the stalk, he
  3. Handpicked/threshed them. After gathering the seed grains, he wanted to separate the stones and husk from them. His wife
  4. Winnowed/ threshed them to separate the husk and later
  5. Sieved/ handpicked to remove stones from it. She grounded the wheat grains and
  6. Sieved/filtered the flour. The wise farmer and his wife got a good price for the flour. Can you tell me why?

Answer:

  1. Happy
  2. Sun
  3. Threshed
  4. Winnowed
  5. Handpicked
  6. Sieved

They got a good price as they used appropriate methods of separation to get good quality flour (atta).

Question 12. You are provided with a mixture of salt, sand, oil, and water. Write the steps involved in the separation of salt, sand, and oil from the mixture by giving an activity along with the diagram.
Answer: In the solution of salt, sand, oil, and water. Oil, being lighter will float on the solution forming a distinct layer, slowly oil is allowed to flow into another container and is separated from the water. This process is called decantation. Now, we are left with salt, sand, and water, to separate this.

Question 13. Pragati was helping her mother in the making of butter. She was amazed to see how easily the butter got separated from milk with the help of a churner.

Class 6 Science Chapter 3 Separation Of Substances Churching

1. What is the process that is being used above?

  1. Grinding
  2. Mashing
  3. Churning
  4. Threshing

Answer: 3. Churning

2. The principle behind the process of churning is

  1. Lighter particles of a solid get separated from liquid
  2. Heavier particles get settled at the bottom
  3. Heavier particles form heaps and lighter particles get swept away by the wind
  4. Particles of different sizes get separated

Answer: 1. Lighter particles of a solid get separated from liquid

Question 14. Saroj collected muddy water from a pond. He separated the water from the mud in three steps as shown in the pictures below.

Class 6 Science Chapter 3 Saroj collected muddy water from a pond. He Separated The Water From The Mud In Three Steps

1. Which separation method did Saroj use at each step? Select the correct row

Class 6 Science Chapter 3 Separation Of Substances Evaporation And Sedimentation Decantation

Answer: 1. Evaporation Sedimentation Decantation

2. Which of these properties did Saroj use to separate mud from water?

  1. It floats on water
  2. It dissolves in water
  3. Water turns cloudy due to mud
  4. It is heavier than water

Answer: 2. It dissolves in water

Question 15. Nidhi was doing an activity at home. She dissolved some sugar in a glass of water. Later she realised she had to add salt and not sugar. She wants to use the same water. Answer the following questions to help Nidhi with her activity.

  1. She can remove sugar from water by
  2. Filtering and then evaporating it
  3. Condensing and then evaporating it
  4. Evaporating and then condensing it
  5. Filtering and then condensing it

Answer: 1. Ice water

Water can be separated from sugar solution by first evaporating the water and then condensing it in another vessel to be used again with salt

2. In which of the following added sugar would be dissolved the least?

  1. Tap water
  2. Filtered water
  3. Ice water
  4. Boling water

Answer: 2. Filtered water

Ice water has the lowest temperature among all and hence would dissolve less amount of sugar as solubility decreases with a decrease in the temperature.

Question 16. Two beakers A and B are shown below. The components of each have been labeled.

Class 6 Science Chapter 3 Separation Of Substances Two beakers A And B

Paheli wants to mix the components of A and B beakers. However, she wants to first separate as many components as possible. Answer the following questions for her activity.

The order of separation of components from beaker A would be

  1. Oil layer > water = sand
  2. Sand = water > oil layer
  3. Sand > oil layer > water
  4. Water > oil layer > sand

Answer: 1. Oil layer > water = sand

The oil layer rests on the top of the solution hence it should be removed first. Sand is removed either by decantation or filtration. Hence, both are collected together separately. The correct order is Oil layer > water = sand.

Paheli can remove layers of oil by

  1. Filtration
  2. Decantation
  3. Handpicking
  4. Evaporation

Answer: 2. Decantation

Decantation is the process used to remove a layer of oil from the water-sulf solution as it forms a separate layer on top.

Question 17. Ashraf’s grandmother is a diabetic patient. Her doctor advised him to control her sugar level by taking less sugar and less fat content in her diet. So, she always prefers to take lassi instead of milk. Ashraf asked her mother when she explained the process of making lassi by removing cream from the curd. She told him this method is known as churning. Read this passage and answer the following questions.

  1. Explain the process of churning.
  2. What is the purpose of this method?
  3. What values are shown by Ashraf?

Answer:

  1. The process of separation of the lighter particles of a solid from a liquid is known as churning. It is also known as centrifugation.
  2. The purpose of churning is to remove butter or cream from the curd or milk. So in lassi, there is less fat content as compared to milk.
  3. Ashraf is a curious and intelligent boy.

Question 18. Rohit and Raman are best friends. They study in different classes, but in the same standard, Class 6. One day, they were playing in the school. They were throwing sand on each other. Suddenly, Raman asked Rohit if we mix salt in sand, then how will you separate the mixture of sand and salt? Rohit was quite intelligent, his science teacher had completed the chapter Separation of Substances last week. So, he immediately explained the method of separation. Rohit told him that his teacher had shown this activity in the class and the same method is applicable for the separation of the mixture of sugar and sand.

Read the passage carefully and answer the following questions.

  1. How is a mixture of sand and salt separated?
  2. Name the method(s) that are applicable in the separation of this mixture.
  3. What values are shown by Rohit?

Answer:

  1. This is done as follows
  2. Some water is added to the mixture of sand and salt in a beaker and stirred. Salt dissolves in water to form a salt solution, whereas sand remains undissolved. It is then filtered.
  3. On filtering, sand is obtained as a residue on the filter paper, and salt solution is obtained as a filtrate.
  4. The filtrate is evaporated and pure salt is left behind.
  5. The mixture of sand and salt has been separated into its components, i.e. sand and salt by doing two methods
  6. Filtration
  7. Evaporation
  8. Rohit is a knowledgeable and intelligent boy

Question 19. A construction worker poured a bag full of sand on the slanting sieve. All the sand particles filtered through the pores and the dust particles, stones, pebbles, and husk remain Name the process described above and state its principle.

  1. Why is it necessary to remove pebbles and dust particles from sand used for construction?
  2. You have been given a glass of muddy water, an empty glass, and a muslin cloth. Describe a method to obtain clean water.
  3. Identify the method of separation used to
  4. Separate chaff from the grain.
  5. Separate tea leaves from the water.

Class 6 Science Chapter 3 Separation Of Substances Pebbles and stones are removed from sand by sieving

Answer:

  1. The process used above is sieving. Principle – A mixture of components having different sizes gets separated with the help of a sieve
  2. A sieve has fine pores in it which lets the small-sized components pass through but holds back larger ones. Sand is sieved so that it can be used for construction work without any impurities.

Question 20.

Class 6 Science Chapter 3 Separation Of Substances Sepration of soild and liquid mixture A glass of muddy water An empty glass and a muslin cloth

  1. You have been given a glass of muddy water, an empty glass, and a muslin cloth. Describe a method to obtain clean water
  2. Identify the method of separation used to
  3. Separate chaff from the grain.
  4. Separate tea leaves from the water.

Answer:

  1. Filtration is done to obtain clean water from a glass of muddy water.
  2. Winnowing
  3. Filtration

Question 21. 

Class 6 Science Chapter 3 Separation Of Substances A group of salt solution The respective componets

  1. A group of students were given a salt solution and were asked to separate the respective components. They set up the apparatus as shown in the given figure.
  2. Name the solute and solvent for the given solution.
  3. Identify the processes for which the apparatus has been set up in the given figures.
  4. Identify the error in the set-up of the apparatus for the separation of a salt solution.

Answer:

  1. Solute-salt, solvent-water
  2. Filtration, condensation.
  3. Salt cannot be separated from water by filtration.

Question 22. Anu adds sugar and salt to two separate glasses of water. She stirs the mixture in each glass every time she adds salt or sugar to it.

Class 6 Science Chapter 3 Separation Of Substances Anu adds sugar and salt to two separate glasses of water

She notes her findings in a table

Class 6 Science Chapter 3 Separation Of Substances She Notes Her Findings

  1. What can Anu conclude from her activity?
  2. Sugar and salt dissolve equally in water
  3. Sugar is more soluble than salt in water
  4. Stirring helps in dissolving all substances in a liquid
  5. Sugar forms a saturated solution in water, but salt does not
  6. Anu used two spoons of the same size to add sugar and salt to the water. What else did Anu need to keep the same for her activity?
  7. Anu repeats the activity using hot water. Will the results in the table remain the same? Explain your answer.
  8. Which of these processes could be used to separate the sugar dissolved in the glass of water?
    1. Filtration
    2. Evaporation
    3. Condensation
    4. Sedimentation

Answer:

  1. Sugar is more soluble than salt in water
  2. Anu needs to keep the amount of water fixed to determine the solubility of a substance.
  3. No, results would not remain the same as temperature has a direct effect on solubility. With the increase in temperature, the solubility of both sugar and salt increases, but in different amounts.
  4. Evaporation

Question 23. A beaker contains a mixture of salt, sand, and water. The mixture is filtered using filter paper.

Class 6 Science Chapter 3 Separation Of Substances A beaker Contains a mixture of salt

  1. What is substance X?
  2. Write a single separation method by which water can be separated from a mixture of salt, sand, and water.

Answer:

  1. The substance ‘X sand as the process of filtration is used to separate insoluble solids, i.e. sand from its liquid mixture.
  2. By evaporating the mixture of salt, sand, and water, water will evaporate and separate. But sand and salt will remain in the vessel.

NCERT Class 6 Science Chapter 3 Separation Of Substances Short Question And Answer

Question 1. What is the principle of threshing?
Answer: Threshing is the method of separation based on the fact that the stalk of the crops and the chaff are soft materials, whereas the grains themselves are hard. Being soft, stalk and chaff can be broken into pieces on beating, but the grains remain unaffected.

Question 2. When a hot saturated solution is heated, crystals will be formed. Is It correct? If not, write the correct statement.
Answer: It is not correct as when a hot saturated solution is cooled, crystals will be formed.

Question 3. Name and describe briefly a method that can help separate a mixture of husk from grains. What is the principle of this method? NCERT Exemplar
Answer: Winnowing is used to separate grains from impurities like husk by dropping them vertically from a height. All light impurities are separated by a blowing air or wind and hence, husk from grains are separated easily.

Question 4. You have been given a solution of substance X. How will you test whether it is saturated or unsaturated? Give one factor affecting the solubility of a substance in solution.
Answer: When no more substance can be dissolved in the solution at a given temperature then the solution is saturated. Temperature affects the solubility of the substance.

Question 5. What happens when sand and powdered leaves are dropped at the same place in the wind?
Answer: Since, sand is heavier, while powdered leaves are lighter. When we drop these at the same place, blowing wind separates them as sand falls near that place vertically, while leaves are blown away. This process is known as winnowing.

Question 6. The best way to carry out the filtration of the mixture of water and sand is to use filter paper, not cloth, why?
Answer: Since, clothes may have small holes, but filter papers do not have any holes. It has very fine pores in it. Hence, sand particles settle down on the filter paper, while water after pouring reaches the bottom of the vessel.

NCERT Class 6 Science Chapter 3 Separation Of Substances Assertion-Reason Questions

Question 1. The following questions consist of two statements. Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below.

  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true, but R is not the correct explanation of A
  3. A is true, but R is false
  4. A is false, but R is true

1. Assertion (A) Husk and flour can be separated by the process of sieving.
Reason (R) Sieving is used when the particle sizes of the two components in the mixture differ.
Answer: Both A and R are true and R is the correct explanation of A.

2. Assertion (A) Evaporation can be used to separate a solid (like salt, or sugar) dissolved in a liquid. Reason (R) Evaporation is the process in which a solid gets converted to vapor.
Answer: A is true, but R is false. R can be corrected as Evaporation is the process in which liquid gets converted to vapor form and thus this process can be used to separate solids dissolved in a liquid.

3. Assertion (A) A substance gets dissolved more in hot water as compared to cold water. Reason (R) Solubility decreases with an increase in temperature.
Answer: A is true, but R is false. R can be corrected as Solubility increases with an increase in temperature

Question 2. Direction The following questions consist of two statements. Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below.

  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true, but R is not the correct explanation of
  3. A is true, but R is false
  4. A is false,

1. Assertion (A) The mixture of mud and water can be separated by using the process of filtration. Reason (R) The process of separating insoluble substances from a liquid using a filter is called filtration.
Answer: Both A and R are true and R is the correct explanation of A

2. Assertion (A) Filtration is used to separate chalk powder from water. Reason (R) Chalk powder is insoluble in water.
Answer: Both A and R are true and R is the correct explanation of A

NCERT Class 6 Science Chapter 3 Separation Of Substances Very Short Answer Type Questions

Question 1. Write one purpose of separating substances from a mixture.
Answer: To separate harmful or non-useful substances and to obtain the useful component.

Question 2. When is the process of handpicking used?
Answer: Handpicking is used to separate sightly large-sized undesirable components when present in small amounts.

Question 3. Which method is used to separate the pieces of stones from grains?
Answer: Handpicking is the method which is used to separate the pieces of stones from grains.

Question 4. Winnowing is based on which property?
Answer: Winnowing is based on the difference in the weight of particles of the different components.

Question 5. Give one example of sieving used in everyday life.
Answer: The separation of bran (choker) from flour is an example of sieving.

Question 6. A method is used to separate the components of a mixture that are of different sizes. Name it.
Answer: Sieving.

Question 7. What is the use of a strainer while preparing tea?
Answer: From prepared tea, we separate tea leaves by using a strainer as a filter. Tea leaves are bigger than the holes in the mesh.

Question 8. 1 kg wheat grains are mixed with 1 kg mustard seeds Write a suitable method to separate the mustard seeds from wheat grains.
Answer: The suitable method used to separate the mustard seeds from wheat grains is sieving.

Class 6 Science Chapter 3 Separation Of Substances Wheat Grains And Mustard Seeds

Question 9. Define the term condensation.
Answer: The process of conversion of water vapor into its liquid form by cooling is called condensation.

Question 10. Water is separated from rice and pulses after washing them. How?
Answer: By the process of decantation, water is separated from rice and pulses after washing them.

Question 11. How saturated solution can become an unsaturated solution?
Answer: When a saturated solution is heated, it becomes unsaturated.

Question 12. Temperature affects the solubility of solids in liquids. How?
Answer: The solubility of solids in liquids increases with increase in temperature.

Question 13. How chalk powder can be separated from a mixture of chalk powder and water?
Answer: A mixture of chalk powder and water can be separated by a filtration process.

Question 14. Sheela, Saima, and Ravi have to dissolve the maximum amount of sugar in the same amount of milk, to win in a game. Ravi took hot boiling milk, while Saima took ice-cold milk. Sheela managed to get milk at room temperature. Who do you think would win the game and why?
Answer: Milk at higher temperatures would dissolve more amount of sugar as solubility increases with temperature. Ravi took hot boiling milk so, he would win the game.

Question 15. Which method of separation is used to catch fish in the sea using a net?
Answer: A fishing net acts as a filter because it allows water to pass and holds fish in it. Thus, catching fishing using the net in the sea is an example of filtration.

Question 16. What will happen if a high quantity of substance is forced to dissolve in a fixed amount of water?
Answer: After the addition of a certain amount, substances remain undissolved in water as they become saturated.

Question 17. Does water dissolve equal amounts of different soluble substances?
Answer: No, water dissolves unequal amounts of different soluble substances.

Question 18. Write four materials that can be used for the process of filtration.
Answer: The materials that can be used for the process of filtration are a wire mesh, a piece of cotton, a piece of fine cloth, and a strainer.

NCERT Class 6 Science Chapter 3 Separation Of Substances Fill In The Blanks

1. The method of separating the seeds of paddy from its stalks is called Thereshing

2. When milk, cooled after boiling, is poured onto a piece of cloth, the cream (malai) is left behind on it. This process of separating cream from milk is an example of Filtration

3. Salt is obtained from seawater by the process of Evaporation

4. Impurities settled at the bottom when muddy water was kept overnight in a bucket. The clear water was then poured off from the top. The process of separation used in this example is called Sedimentation And Decantation

5. Insects can be separated from wheat by Handpicking

6. Crushing paddy under Cattle’s feet to separate grains is called Threshing

7. Husk from wheat flour is generally removed by Sieving

8. Slanting sieve Used at construction sites to separate pebbles and stones from sand.

9. The process of settling heavier particles in a solution is called Sedimentation.

NCERT Class 6 Science Chapter 3 Separation Of Substances True Or False.

1. A mixture of milk and water can be separated by filtration. False, we cannot separate the mixture of milk and water by filtration because milk is completely soluble in water.

2. A mixture of powdered salt and sugar can be separated by the process of winnowing. The false, winnowing process is used to separate heavier and lighter components of a mixture. Here, particles of size sugar and salt are almost of same size.

3. Separation of sugar from tea can be done by filtration. False, sugar cannot be separated from tea by filtration because sugar is completely soluble in tea.

4. Grain and husk can be separated by the process of decantation. False, grain and husk are separated by the process of winnowing.

5. Separation of components of a mixture is a useful process. True

6. A mixture of iron filings and rice flour can be separated by handpicking. False, iron filling can be separated from rice flour using a magnet

7. A mixture of wheat grains and rice flakes can be separated by winnowing. True

8. A mixture of oil and water can be separated by filtration. False, a mixture of oil and water can be separated by decantation

9. A mixture of tea leaves and milk can be separated by filtration. True

NCERT Class 6 Science Chapter 3 Separation Of Substances Match The Columns

Question 1. Match the following separation processes with their purposes and the way separated components are used.

Class 6 Science Chapter 3 Separation Of Substances The Following Separation process with their purpose

Answer: The correct match is 1.(b)-(i), 2.-(a)-(iii),3.-(c)-(ii)

Question 2. Match the Column 1 with Column 2

Class 6 Science Chapter 3 Separation Of Substances Match The Column 1 And Column 2

Answer: A-3, B-4,C-5,D-2, E-1

Question 3. Match The Column 1 And Column 2

Class 6 Science Chapter 3 Separation Of Substances Match The Column 1 And Column 2.

Answer: A-3,B-1,C-5,D-4, E-2

NCERT Class 6 Science Chapter 2 Sorting Materials Into Groups Question And Answers

NCERT Class 6 Science Chapter 2 Sorting Materials Into Groups Long Question And Answers

Question 1. Chalk, iron nails, wood, aluminium, candles, and cotton usually look different from each other. Give some properties by which we can prove that these materials are different.
Answer: We can differentiate these materials based on lustre, hardness, softness, roughness or smoothness.

Class 6 Science Sorting Materials Into Groups Chalk Iron nail wood aluminium, candle cotton usally look different from each other

Question 2. Sugar, salt, mustard oil, sand, sawdust, honey, chalk powder, petals of flowers, soil, copper sulphate crystals, glucose, and wheat flour are some substances given to Paheli. She wants to know whether these substances are soluble in water or not. Help her in identifying soluble and insoluble substances in water.
Answer: Substances soluble in water are sugar, salt, honey, copper sulphate crystals, and glucose. Substances insoluble in water are mustard oil, sand, sawdust, chalk powder, petals of flowers, soil, and wheat Hour.

Read and Learn More Class 6 Science Question And Answers

Question 4. Why do you think oxygen dissolved in water is important for the survival of aquatic animals and plants?
Answer: Oxygen gas dissolves in water, i.e. oxygen gas is soluble in water. The plants and animals which live in water use the oxygen dissolved in water for respiration. Thus, oxygen gas dissolved in water is very important for the survival of animals and plants that live in water.

NCERT Class 6 Science Chapter 2 Sorting Materials Into Gropus Questions And Answers

Question 5. Kavita bought new silver anklets. After a month, she found that the anklets had lost their lustre and turned black. She had doubts that they were not made up of pure silver. But her mother told her to wash it with lemon juice. She washed her silver anklets and found that the silver anklets had got the lustre back. Read the above passage and answer the following questions.
Answer:

1. Do you think that the anklets are not of pure silver?
Answer: No, the anklets were made up of pure silver.

2. Why silver anklets lose their lustre after some time?
Answer: Silver anklets lost their lustre and appear dull, because of the action of air and moisture On them

Question 6. Rohan and his friends planned to play a cricket match on Sunday morning. But, by chance on Saturday, Rohan got a high fever. Due to high fever, his parents did not allow him to play cricket. So, Rohan was very upset.  But his father suggested to him that he could see the cricket match that his friends were playing in the park through a glass window, although the window would be closed. Read the above passage and answer the following questions.
Answer:

1. How Rohan was able to see the cricket match, although the window was closed?
Answer: Rohan was able to see the cricket match through
glass window panes which were transparent

2. Write different types of materials based on their transparency.
Answer: Based on transparency different types of materials are

  • Transparent materials like glass, water air.
  • Opaque materials like wood, cardboard, and metal containers.
  • Translucent materials like oil paper, butter paper

Question 7. 

Class 6 Science Sorting Materials Into Groups Object To Be Seen
Answer:

1. Why is Sachin not able to see through the wall?

  1. The wall is transparent
  2. The wall is translucent
  3. The wall is opaque
  4. The wall is reflective

Answer: 3. The wall is opaque

2. Name the device Sachin is using to see the tree on the other side of the wall.

  1. Microscope
  2. Periscopes
  3. Telescope
  4. Electroscope

Answer: 2. Periscopes

3. Can we use cardboard instead of a plane mirror in this device? Give a reason for your answer. are often used in national defence. Describe two possible uses of this device by the military.
Answer: No, cardboard cannot be used as it does not have a shiny surface. Hence, it reflects only a small amount of light.

4. Periscopes are often used in national defence. Describe two possible uses of this device by the military.
Answer:

  • Perioscopes are used in submarines.
  • Soldiers use them to spot enemies in trenches.

Question 8. Namita has three boxes with the same thickness. She places a burning candle inside each box. The images show how Namita was able to view the candle through each box.

Class 6 Science Sorting Materials Into Groups Namita has three boxes with the same thickness. She places a burning candle inside each box.

Answer:

1. Which box is made of an opaque material?
Answer: Box 3 is made up of an opaque material as the candle is not visible

2. Namita replaced box 1 with a thicker box made of the same material and repeated the activity. Will there be any change in the results of the activity? Explain your answer.
Answer: With the increase in thickness, the candle will not be visible.

Question 9. Aqeel poured liquid 1 and liquid 2, each into a jar of water. He stirred the mixture in each jar with a spoon. He allowed the mixtures to rest for 10 mins.

Class 6 Science Sorting Materials Into Groups Aqeel poured liquid 1 and liquid 2, Each To A Jar Of Water.

Which of the following must Aqeel keep the same in his activity? Circle ‘Yes’ or ‘No’ for the correct response

Class 6 Science Sorting Materials Into Groups Should This Be Kept The Same In The Activity

Are liquids and 2 soluble in water? Write Yes or No in the space given in the table below.

Class 6 Science Sorting Materials Into Groups Are Liquid 1 and 2 soluble In water

Answer:

  1. Yes, Yes, No
  2. Yes, No

Question 10. Name five objects which can be made from wood.
Answer: Five objects which can be made from wood are

  1. Door
  2. Window
  3. Chair
  4. Table
  5. Boat

Question 12. Select those objects from the following that shine: Glass bowl, plastic toy, steel spoon, cotton shirt
Answer: Among the given objects, glass bowls and steel spoons are shiny objects which are also called lustrous materials.

Question 13. Match the objects given below with the materials from which they could be made. Remember, an object could be made from more than one material and a given material could be used for making many objects.

Class 6 Science Sorting Materials Into Groups The Objects given below With The Materials

Answer: The correct table is shown below

Class 6 Science Sorting Materials Into Groups The Objects given below With The Materials.

Question 14. State whether the statements given below are True or False.

  1. Stone is transparent, while glass is opaque.
  2. A notebook has lustre, while an eraser does not.
  3. Chalk dissolves in water.
  4. A piece of wood floats on water.
  5. Sugar does not dissolve in water.
  6. Oil mixes with water.
  7. Sand settles down in the water.
  8. Vinegar dissolves in water

Answer:

  1. False, glass is transparent because it allows light to pass through it while stone is opaque because it does not allow light to pass through it.
  2. False, notebook and eraser do not have any lustre.
  3. False, chalk is insoluble in water.
  4. True
  5. False, sugar dissolves in water.
  6. False, oil does not mix with water it floats on water.
  7. True
  8. True

Question 15. Below are the names of some objects and materials: Water, basketball, orange, sugar, globe, apple and earthen pitcher. Group them as

  1. Round shaped and other shapes.
  2. Eatables and non-eatables.

Answer: Objects or materials which are in the form of circles, are known as round shaped.

Class 6 Science Sorting Materials Into Groups Round Shape And Other Shape

The substances which we can eat, are known as eatables otherwise it is known as non-eatables

Class 6 Science Sorting Materials Into Groups The substance which we can eat, are known as eatables

Question 16. List all items known to you that float on water. Check and see if they will float on oil or kerosene.
Answer:

A list of some items that float on water are

  1. Wood
  2. Wax
  3. Thermocol
  4. Oil
  5. Ice
  6. Plastic

A list of items that float on oil or kerosene are

  1. Wood
  2. Paper
  3. Wax
  4. Plastic
  5. Thermocol

Question 17. Find the odd one out from the following

  1. Chair, Bed, Table, Baby, Cupboard
  2. Rose, Jasmine, Boat, Marigold, Lotus
  3. Aluminium, Iron, Copper, Silver, Sand
  4. Sugar, Salt, Sand, Copper sulphate

Answer:

  1. Except baby, all others are non-living things.
  2. Except for the boat, all others are flowers.
  3. Except for sand, all others are metals.
  4. Except for sand, all others are soluble in water.

NCERT Class 6 Science Chapter 2 Sorting Materials Into Groups Short Question And Answers

Question 1. Give three common characteristics of materials.
Answer: The three common characteristics of materials

  • All materials occupy space.
  • All materials possess mass.
  • All materials can be felt by one or more of our sense organs.

Question 2. Give the name of two objects for each which can be made from the following materials

  1. Glass
  2. Iron
  3. Gold
  4. Aluminium

Answer:

  1. Glass Window pane, tumbler.
  2. Iron Rods, iron nails
  3. Gold Jewellery, coins.
  4. Aluminium Aeroplane, window frames.

Question 3. List some materials that are used for making more than one type of object.
Answer: List of some materials that are used for making more than one type of an object as follows

Wood chair, table, plough, bullock cart and its wheels. Paper books, notebooks, newspapers, toys, calendars. Leather shoes, belts, wallets, bags and photo albums. Plastic buckets, lunch boxes, toys and pipes.

Question 4. Why paper-like materials are not used for cooking vessels?
Answer: Cooking vessels are made of a material with a high burning point but paper catches fire easily. Hence, it would not be wise to use paper-like materials for cooking vessels.

Question 5. Why are different materials used for making different objects?
Answer: Different materials have different properties. Materials are chosen to make objects depending on their properties and the purpose for which the object is to be used.

Question 6. Why are handles of utensils made from wood or plastic?
Answer: The handles of utensils are made from wood or plastics because these materials do not heat us easily unlike iron, aluminium, etc. and our hands do not burn while handling these utensils when they are hot.

Question 7. It was Paheli’s birthday. Her grandmother gave her two gifts made up of metals, one old dull silver spoon and a pair of lustrous gold earrings. She was surprised to see the difference in the appearance of the two metals. Can you explain the reason for this difference?
Answer: The silver spoon on long exposure to moist air has lost its shine and appears dull whereas gold does not tarnish.

Question 8. Which among the following materials would you identify as soft materials and why?
Ice, rubber band, leaf, eraser, pencil, pearl, a piece of wooden board, cooked rice, pulses and fresh chapati.
Answer: Materials which can be compressed or scratched easily – are called soft materials. Among the given materials rubber band, leaves, erasers, cooked rice and fresh chapati are soft materials because they can be compressed or scratched easily

Question 9. You are provided with the following materials: Turmeric, honey, mustard oil, water, glucose, rice flour, and groundnut oil. Make any three pairs of substances where one is soluble in the other and any three pairs of substances where one substance is insoluble in the other.
Answer: Soluble Substances

  • Honey in water
  • Groundnut oil in mustard oil
  • Insoluble substances
  • Turmeric in water
  • Mustard oil in water

Question 10. Give two examples of gas which dissolve in water.
Answer: Two examples of gas that dissolve in water are as follows

  • Oxygen gas dissolved in water is very important for the survival of animals and plants that live in water.
  • Carbon dioxide also dissolves in water.

Question 11. Give five examples of each of the objects that float and sink in water.
Answer: Objects that float in water are plastic, sawdust, dry leaves, ice, and kerosene. Objects that sink in water are copper, silver, gold, stone, and chalk.

Question 12. Our palm is translucent. Explain with an activity.
Answer: Cover the glass of a torch with your palm in a dark area. Switch on the torch and observe from the other side of the palm. We see that the light of the torch passes through the palm, but not clearly. This demonstration shows that our palm shows the property of translucent when a strong beam oflight passes through it.

NCERT Class 6 Science Chapter 2 Sorting Materials Into Groups Very Short Answer Type Questions

Question 1. Write two things which are made from glass.
Answer: Bowls and mirrors are made from glass.

Question 2. Name two materials which appear shiny.
Answer: Aluminium and iron appear to shine

Question 3. Why does iron appear different from copper and aluminium?
Answer: Iron appears different from copper and aluminium because it reacts with air and moisture loses its shape and appears dull, but in the case of copper and aluminium, these maintain their shine.

Question 4. Give the reason why some metals become dull and lose their shine (lustre).
Answer: Some metals lose their shine because when exposed to air react with moisture and gases present in them, thereby forming a dull layer of some other compounds.

Question 5. By which property, do steel and sponge differ from each other?
Answer: Hardness and softness are the properties by which steel and sponge differ from each other.

Question 6. Mixtures of red chilli powder in water, butter in water, petrol in water and honey in water were given to Radha, Sudha, Sofia and Raveena, respectively. Whose mixture is in solution form?
Answer: Raveena has a solution because honey will dissolve in water whereas red chilli powder, butter and petrol float on water.

Question 7. Are vinegar and lemon juice soluble or insoluble in water?
Answer: Vinegar and lemon juice are completely soluble in water.

Question 8. Write two names of liquids which are insoluble in water.
Answer: Mustard oil and kerosene oil are the two liquids which are insoluble in water.

Question 9. Materials which are denser than water sinks in water. Give two examples of such materials.
Answer: Honey and a steel spoon are examples of materials which are denser than water. These will sink.

Question 10. Some materials float on water. Name two such materials.
Answer: Wood and plastic are the materials which float on water.

Question 11. Take a small cotton ball and place it in a tumbler/bowl filled with water. Observe it for at least 10 minutes. Will it float or sink in water and why?
Answer: Cotton ball initially floats on water, but after a few minutes (i.e. 10 min), it sinks because it absorbs water.

Question 12. Do all liquids dissolve in water?
Answer: Some liquids get completely dissolved in water. Some others do not dissolve in water and form a separate layer when kept aside for some time.

Question 13. Do all gases also dissolve in water?
Answer: Some gases dissolve in water whereas others are not. Water usually has small quantities of some gases dissolved in it, e.g. oxygen gas is dissolved in the water.

Question 14. List the following substances as soluble and insoluble in water. Sand, sugar, vinegar, common salt, chilli, ghee.
Answer:

  1. Soluble sugar, vinegar, common salt.
  2. Insoluble sand, chilli, ghee.

Question 15. Why a shopkeeper prefers to keep biscuits and sweets in a glass or plastic container?
Answer: Due to the transparent nature of glass or plastic containers, biscuits, sweets, etc. can be seen easily by buyers. Hence, shopkeepers store them this way.

Question 16. Write two objects through which things can be seen clearly.
Answer: Polythene and glass containers are transparent objects through which things can be seen clearly.

Question 17. On a bright sunny day, Shikha was playing hide and seek with her brother. She hides behind a glass door. Do you think her brother will be able to locate her? If yes, why? If not, why not? Competency Based Que.
Answer: Yes, her brother can easily locate her because the glass door is a transparent material.

Question 18. Write the name of two objects which are made from opaque materials.
Answer: Blackboard and wooden doors are objects which are made from opaque materials.

Question 19. Why do we need to group materials in everyday life?
Answer: We need to group materials for our convenience. Grouping materials helps us to locate things easily.

Question 20. Do liquids also dissolve in water?
Answer: No, all liquids do not dissolve in water. Liquids like vinegar, and lemon juice, dissolve in water, while most oils do not mix and form a separate layer.

Question 21. What happens when dried leaves, a piece of stone and a drop of honey are dropped in water?
Answer: Dried leaves float on the surface of water, while a piece of stone and honey sink to the bottom.

Question 22. Give five examples of each object that floats on water and those that sink in water.
Answer: Examples of objects that float on water are balloons, sponges, ice, empty bottles, boat, etc. Examples of objects that sink in water are stone, iron blocks, marble, honey, jewellery, etc.

Question 23. Give some examples of translucent materials.
Answer: Some examples of translucent materials include frosted glass, thin plastic sheets, wax paper, some fabrics, etc.

Question 24.  Is the palm of our hand transparent, translucent or opaque?
Answer: The palm ofour hand is translucent. Hence, we can see some light coming through it

NCERT Class 6 Science Chapter 2 Sorting Materials Into Groups Fill In The Blanks

1. Materials through which we are not able to see are called Opaque materials.

2. Animals live underwater. It is because oxygen gas dissolves in it.

3. Rough Materials have uneven surfaces.

4. Rough materials have an uneven or irregular surface.

5. Materials which cannot be compressed are called Hard materials.

6. Soft materials can be cut with a knife.

7. Materials which do not disappear in water are called insoluble materials

8. Those materials through which we cannot see are called translucent materials.

NCERT Class 6 Science Chapter 2 Sorting Materials Into Groups Write True Or False

1. False Hard materials are easily compressible.

2. False Butter paper is a transparent material.

3. False Gold is not a lustrous material.

4. Diamond Is known as the hardest natural material. True

5. Ice Is lighter than water. True

6. Mustard oil Is soluble in water. False, mustard oil is insoluble in water

7. Carbon dioxide gas is soluble in water. False

8. Mirror is opaque. False

NCERT Class 6 Science Chapter 2 Sorting Materials Into Groups  Assertion-Reason Question And Answers

The following questions consist of two statements. Assertion and Reason (R). Answer these questions by selecting the appropriate option given below.

  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true, but, R is not the correct explanation of A
  3. A is true, but, R is false
  4. A is false, but, R is true

 

  1. Assertion Chalk powder floats on water.
    Reason (R) Chalk powder is insoluble in water.
  2. Assertion Animals can live underwater.
    Reason (R) O2 dissolves in water.
  3. Assertion Frying pan is translucent.
    Reason (R) Translucent objects allow some light to pass through them.
  4. Assertion Materials are grouped for convenience.
    Reason (R) Materials can be grouped based on their names.

Answers

  1. A is false, but R is true. This can be corrected as Chalk powder is insoluble in water. It settles down at the bottom.
  2. Both A and R are true and R is the correct explanation of A.
  3. A is false, but R is true. This can be corrected as the Frying pan is made up of metal. Hence, it is opaque, opaque materials allow no light to pass through, while translucent materials allow some light.
  4. A is true, but R is false. R can be corrected as Materials are grouped based on their properties and not names.

NCERT Class 6 Science Chapter 2 Sorting Materials Into Groups Match The Columns

Question 1. Match the following columns 

Class 6 Science Sorting Materials Into Groups Match The Following Columns

Answer: A-2, B-1, C-4, D-3

Question 2. Match Column 1 with Column 2.

Class 6 Science Sorting Materials Into Groups Match The Following Columns.2

Answer: A— 3, B— 5, C— 1, D— 2, E— 4.

NCERT Class 6 Science Chapter 6 The Living Organisms Characteristics And Habitats Questions And Answers

NCERT Class 6 Science Chapter 6 The Living Organisms Characteristics And Habitats Long Questions And Answers

Question 1. Like many animals, although a car also moves, It Is not considered a living organism. Give two or three reasons.
Answer:
A car though moving is not considered a living organism because Living organisms can move on their own, whereas a vehicle moves by the burning of fuel; like diesel and petrol. Like all living organisms, a car does not show any other living characteristics like respiration, digestion, reproduction or growth.

NCERT Class 6 Science Chapter 6 The Living Organisms Characteristics And Habitats Questions And Answers

Question 2. Ravi and Ritu went on a vacation to different places. They went to Rishikesh, Himalayas, Kerala, and Rajasthan. They saw many kinds of trees, plants and animals, e.g. pine trees in the Himalayas and cacti In Rajasthan. All the places were so different from each other. They clicked pictures, collected samples and showed Them to their teacher asking them to explain the reason behind these variations.

Read and Learn More Class 6 Science Question And Answers

  1. Why is there a difference In the type of plants growing In different places?
  2. What do you mean by the survival place of an organism?
  3. How many types of such places that give living organisms shelter, food, etc. are you aware of?
  4. Give an example of variation in plants from different regions.

Answer:

  1. Plants growing in different places appear different due to the presence of unique adaptive features which can be morphological or physiological.
  2. It is called a habitat. It provides food, water and shelter, on it. to all living organisms.
  3. Two types of habitats have been observed
  4. Terrestrial
  5. Aquatic.
  6. Leaves of cactus present in desert, are either absent or are very small and spiny to reduce water loss through transpiration.
  7. While leaves of trees like deodars are needle-shaped to let snow falling on them slide off easily.

Question 3. Riya lives in Delhi. During her vacations, she went to Leh with her parents. They all were fine on the way. When they reached Leh, all of them started feeling nauseous and complained of headaches and breathing difficulties. A man travelling with them told them to rest for a few days and everything would be fine. After a few days, Riya observed that the man was right and they were feeling fine.

  1. Why was Riya and her family suffering from these symptoms?
  2. After 2-3 days everybody was fine. Why?
  3. Explain what happened.

Answer: Riya and her family were suffering from these symptoms because they were present at high altitude regions where oxygen supply is limited.

After a few days, everybody’s body had adjusted to the conditions of the mountain region.

When we go from plains to mountains, our body faces difficulties due to the sudden changes it is exposed to. After a few days, however, they all get acclimatised to these sudden changes.

Question 4. A caterpillar coils when touched

Class 6 Science Chapter 6 The Living Organisms A Catepillar Coils When Toched

  • Which feature of the caterpillar Is shown In the picture?
  • It grows
  • It reproduces
  • It breathes through tiny pores on its bodies
  • It responds to changes in the environment
  • The green colour of a caterpillar helps It to blend with the green leaves. Explain how this helps the caterpillar.

Answer: It responds to changes in the environment.

The green colour of a caterpillar helps it get camouflaged with green leaves. This blending of colour protects the caterpillar from the predators like insects.

Question 5. Sea lions are mammals that live in the seas.

Class 6 Science Chapter 6 The Living Organisms Sea Lions Are Mammals That Live In The Seas

Some common features of sea lions are

  • A streamlined body
  • Whiskers on the face
  • Short fur on the body
  • Limbs shaped like flippers
  • A layer of fat under the belly

What does the picture show about sea lions?

  • They feed mainly on fish
  • They enjoy resting on rocks
  • They can see underwater
  • They produce babies that look similar to adults
  • Which two features of sea lions listed above help them swim in water?

Answer:

  • They produce babies that look similar to adults.
  • The two features of sea lions that help them to swim in water are as follows
  • A streamlined body
  • Limbs shaped like flippers

Question 6. The given picture is of a desert ecosystem. The biotic component of this ecosystem is the

  1. Date plan
  2. Sand Dune
  3. Sunlight
  4. Water

Class 6 Science Chapter 6 The Living Organisms A desert Ecosystem

Explain any two adaptations observed in a cactus that help it to survive in a desert.

Answer:

  1. Date palm
  2. The two adaptations observed in a cactus that help it to survive, in a desert are as follows (a) Roots are very deep for absorbing water from larger areas.
  3. Roots are very deep for absorbing water from larger areas.
  4. Leaves are either absent or very small and spiny to reduce water loss through transpiration.

Question 7. The picture shows a cactus. It Is found In hot deserts. Explain how the leaves of the cactus help it to survive In its habitat.

Class 6 Science Chapter 6 The Living Organisms The Leaves Of The Cactus Help it To Survie In Its Habitat

Answer: In cactus, leaves are reduced to spines. Which prevents loss of water (through transpiration). This adaptation helps them to survive in hot deserts

Question 8. State and explain any two features observed in a lotus plant that help it to adapt to an aquatic habitat.

Class 6 Science Chapter 6 The Living Organisms In A lotus Plant That helps it to adapt to an aquatic habitat

Answer: Two features observed in a lotus plant that help it to adapt to an aquatic habitat are

  • The stems are long, hollow and light. This feature makes their stem buoyant.
  • The stems grow up to the surface of water and the leaves and flowers float on the surface of water. This increases the exposure to light.

Question 9. The picture shows different animals

Class 6 Science Chapter 6 The Living Organisms Ant, Earthworm, snail, star fish

Which animals In the picture live In the same habitat?

  1. Starfish and snail
  2. Earthworm and snail
  3. Earthworms, ants and snail
  4. Snail, starfish and earthworm

What Is common to all the animals in the picture? Circle ‘Yes’ or ‘No’ for the correct response.

Class 6 Science Chapter 6 The Living Organisms Is This statement True Yes Or No

Answer:

  1. Earthworms, ants and snail
  2. No
  3. Yes
  4. Yes

NCERT Class 6 Science Chapter 6 The Living Organisms Characteristics And Habitats Short Questions And Answers

Question 1. What are the surroundings of plants and animals in the sea and desert?
Answer: In the sea, plants and animals are surrounded by saline (salty) water. Most of them use the air dissolved in water. In the desert, there is very little availability of water.

It is very hot during the daytime and very cold at night in the desert. The animals and plants of the desert live on the desert soil and breathe air from the surroundings.

Question 2. Are sunlight and heat biotic or abiotic components?
Answer: Sunlight and heat are abiotic components

Question 3. Which of the following are living? A forest, trees, creepers, small and big animals, birds, snakes, Insects, rocks, soil, water, air, dry leaves, dead animals, mushrooms and moss.
Answer: Living things are as follows A forest, trees, creepers, small and big animals, birds, snakes, insects, mushrooms and moss.

Question 4. How can you distinguish between living and non-living things?
Answer: The differences between living and non-living things are as follows

Class 6 Science Chapter 6 The Living Organisms Living Things And Non living Things

Question 5. Do non-living things show growth?
Answer: Yes, non-living things show growth. They increase in size and mass, for example; clouds.

Question 6. Can we live without breathing?
Answer: No, we cannot live without breathing. It is a part of a process called respiration. It is through respiration that the body finally obtains energy from the food it takes.

Question 7. How do we respond, if we suddenly step on a sharp object like a thorn, while walking barefoot?
Answer: If we suddenly step on a sharp object like a thorn, while walking barefoot we withdraw our foot.

Question 8. How do we feel when we see or think about our favourite food?
Answer: When we see or think about our favourite food, water starts filling our mouths.

Question 9. What happens when we suddenly move from a dark place to bright sunlight?
Answer: When we suddenly move from a dark place to bright sunlight our eyes shut themselves automatically for a moment till they adjust to the changed bright surroundings.

Question 10. Give some examples of responses of animals to stimuli.
Answer: Some examples of responses of animals to stimuli are as follows

  1. Wild animals run away when a bright light is flashed towards them.
  2. Cockroaches begin to move to their hiding places if the light in the kitchen is switched on at night.

Question 11. Do plants respond to stimuli? 
Answer: Yes, plants respond to stimuli. For example, in some plants, flowers close after sunset and in some plants like ‘touch-me-not’, leaves close when touched.

Question 12. What is a habitat?
Answer: The place where organisms live is called habitat. It means a dwelling place (a home).

Question 13. How are cactus adapted to survive in a desert?
Answer: To survive in a desert, a cactus has the following adaptations

  • Leaves are either modified into spines or are absent to prevent water loss through transpiration.
  • Stem is modified to perform photosynthesis and covered with a thick waxy layer to retain water.
  • The root system of the cactus is well-developed and grows deep to absorb water.

Question 14. Which of the things in the following list are non-living? Plough, mushroom, sewing machine, radio, boat, water hyacinth, earthworm
Answer: Non-living things do not possess life and do not need food, air or water. Thus, in the given list plough, sewing machines, radios and boats are non-living objects.

Question 15. Give an example of a non-living thing which shows any two characteristics of living things.
Answer: A cloud is an example of a non-living thing which shows the following two characteristics of living things

  1. It increases or decreases in size.
  2. It shows movements.
  3. (Another example can be a car which moves from one place to another and also requires energy, i.e. petrol).

Question 16. Which of the non-living things listed below, were once part of a living thing? Butter, leather, soil, wool, electric bulb, cooking oil, salt, apple, rubber
Answer: The following are the non-living things from the list which were once part of living things

  • Butter Obtained- from milk from dairy animals
  • leather- From animal skin
  • Wool- From the hair of sheep or goat
  • Cooking oil- Seeds Of plants
  • Apple- Fruit From Apple Tree
  • Rubber- Latex Of Rubber Tree.

Question 17. List the common characteristics of the living things.
Answer: Living things have the following common characteristics-

  1. Growth
  2. Movement
  3. Reproduction
  4. Respirator
  5. Respond to stimuli
  6. Excretion

Question 18. Explain why speed is important for survival in the grasslands for animals that live there. Hint There are few trees or places for animals to hide in grassland habitats.
Answer: In grasslands, there are a limited number of trees and places for animals to hide. When the enemy attacks an animal, it has to run fast, to reach a safe place. If it fails, it has to lose its life. So, speed in such places as grasslands is very important for the survival of animals.

Question 19. A habitat Is said to have two supporting components. What are they?
Answer: A habitat includes

  • A living component is called an abiotic.
  • Non-living components are called abiotic.

Question 20. Why do desert snakes burrow deep into the sand
Answer: Snakes in the desert burrow deep into the sand during the daytime because the deeper layers of sand are cooler which allows them to stay away from the heat of the desert that persists through the daytime.

Question 21. Some desert plants have very small leaves, whereas some others have only spines. How does this benefit you during the day? the plants?
Answer: Desert plants have very small leaves or spines as adaptations to dry conditions. Because of these modifications, the surface of the lamina is reduced thereby reducing water loss by transpiration as water is scarce in desert habitats.

Question 22. The presence of needle-like leaves on trees growing in extreme cold helps them to survive the cold conditions. Explain.
Answer: Many mountain trees have small, needle-like leaves due to which these leaves lose very little water in windy conditions. This helps the mountain trees to survive in winter when all the soil water is frozen in the form of ice and not available to their roots.

The needle-like leaves have a thick waxy layer (or cuticle) to reduce the loss of water through transpiration and to protect them from damage by rain and snow.

Question 23. An animal has a round body, small ears and thick fur on its body, feet and toes. This animal has a thick layer of fat beneath its skin to provide insulation.

  • The feet of this animal Are also bigger.
  • Identify this animal.
  • Name the possible habitat of this animal.

Answer: This animal is a snow leopard.

The habitat of this animal is the mountain where snow is present.

Question 24. Explain why many mountain trees are cone-shaped having sloping branches.
Answer: The trees in mountain habitats are usually cone-shaped having sloping branches because this shape of the mountain trees makes the rainwater and snow slide off easily without damaging the branches and leaves.

Question 25. What are the specific features present In a deer that help It to detect the presence of predators like lions?
Answer: A deer has the following specific features that help it to detect the presence of predators like lions.

  • Long ears to hear the movement of predators.
  • Eyes on the sides of its head which allows it to look in all directions.

Question 26. Aquatic plants undergo some modifications to adapt to their habitat. List any two of those.
Answer: Two adaptations of aquatic plants are

  • The roots of these plants are small just to keep the plant in place.
  • Stems are hollow with large spaces which help in floating.

Question 27. Write the adaptation in aquatic plants due to which

  • Submerged leaves can bend in the flowing water
  • Leaves can float on the surface of water

Answer: The adaptations are

  • Leaves ofsubmerged aquatic plants are narrow and ribbon-like which allows them to bend in the flowing water.
  • Stems/Stalks of leaves of aquatic plants are long, hollow and light which allows them to float on the surface of water.

Question 28. Read the features of different plants given below

  • Thick waxy stem
  • Short roots
  • Cone-shaped plants
  • Sloping branches
  • Small or Spine like leaves
  • Hollow stem

Choose the type of plant for every adaptive feature given in a, b, c, d, e and f from the list given below Aquatic plant, desert plant, mountainous plant.

Answer: Given below are the features of their corresponding type of plant,

  • Thick waxy stem – desert plant
  • Short roots – aquatic plant
  • Cone-shaped plants-mountainous plants
  • Sloping branches – mountainous plant
  • Small or Spine leaves- desert plant
  • Hollow stem – aquatic plant

Question 29. Montlonono adaptation present In the following

  • In camels to keep their bodies away from the heat of sand.
  • In frogs to enable them to swim.
  • Dolphins and whales breathe In air when they swim near the surface of the water.

Answer: Camels have long legs to keep their bodies away from the heat of sand.

  • Frogs have webbed feet that enable them to swim,
  • Dolphins and whales have blow holes which help them to breathe in air when they swim near the surface of water.

Question 30. List the types of movements seen In plants.
Answer: We sec the following movements in plants

  1. Opening and dosing off lower.
  2. Growth of stem and leaves.
  3. Growth of stems and twigs towards light and roots towards soil and water.

Question 31. Pahell has a rose plant In her garden. How can she Increase the number of rose plants In the garden?
Answer: Paheli can increase the number of rose plants in the garden by planting stem cuttings of the rose plant. This will lead to the reproduction of the rose plant.

Question 32. Plants are also called living beings. Mention any two characteristics that support this classification.
Answer: Plants are called living beings because

  • They carry out the respiratory process.
  • They respond to stimuli.
  • A car though moves is not considered a living organism because of the following reasons
  • Living organisms can move on their own, whereas a car moves by the burning of fuel; like diesel and petrol.
  • Like all living organisms, a car does not show any other living characteristics like respiration, digestion, reproduction or growth.

NCERT Class 6 Science Chapter 6 The Living Organisms Characteristics And Habitats Very Short Questions And Answers

Question 1. Camels live in desert habitats. How do they adapt to such a harsh climate?
Answer: Camels have long legs, a hump and padded feet. All of these are adaptive features that help camel to survive in desert habitats.

Question 2. Name the place where an organism lives and gets its food.
Answer: Habitat is a place where an animal lives and gets its food.

Question 3. Using the following words, write the habitat of each animal given in figures (a to d). Grassland, mountain, desert, pond, river

Class 6 Science Chapter 6 The Living Organisms Grassland, Moutain, Desert, Pond, River

Answer:

  • The diagram is of a deer which lives in grassland
  • The diagram is of a frog which lives in the pond
  • The diagram is of a yak which lives in the mountain
  • The diagram is of a camel which lives in the desert

Question 4. Classify the following habitats into terrestrial and aquatic types. Grassland, pond, ocean, rice

Answer: The habitats can be classified as

  • Terrestrial habitats
  • — Grassland and rice field
  • Aquatic habitats — Pond and ocean

Question 5. Adaptation to habitat does not take place in a short time. Explain.
Answer: It takes thousands of years for the abiotic factors to change and support the animals and plants present there,

Question 6. Animals like mountain goats and snow leopards do not slip or fall. Why?
Answer: These animals have big and padded feet and strong hooves to spread their weight which prevents them from slipping and falling on the mountains.

Question 7. Mention two ways by which animals cope with low temperatures.
Answer: Animals living in colder areas have hairy skin and a thick layer of fat under their skin to warm their bodies.

Question 8. Name the adaptation of hiding by merging with the surroundings.
Answer: Camouflage.

Question 9. Leaves of mountain trees prevent damage by snow or rain. How?
Answer: The needle-like leaves of these trees have a thick waxy layer called cuticle which protects the leaves from any kind of damage.

Question 10. Lions have very high agility. Why do they need this type of adaptation?
Answer: As a predator, they have to hunt for food which is not easy as prey like deer are very fast. Thus, they have high agility.

Question 11. A few animals can live in both places, i.e. land and water. Name at least two.
Answer: Few animals have adaptations to live both in water as well as land. These are called amphibians, e.g. frogs and toads.

Question 12. An adaptation requires a change In the body shape of the animals present. Name the habitat and the modification present.
Answer: Animals in aquatic habitats have streamlined bodies which help them survive underwater.

Question 13. Growth is an important characteristic of living beings. Support this statement.
Answer: Growth represents an increase in the weight and size of living beings as time passes. Non-living things do not grow over time.

Question 14. During respiration, gases are exchanged in our lungs. What are these gases? Answer: Carbon dioxide is given off and oxygen is taken in by our lungs.

Question 15. Why is reproduction important for organisms?
Answer: Reproduction is important for organisms as it leads to the production of more individuals of their kind. This helps in the continuity oflife on the Earth.

Question 16. Unscrnmblo tho words given below to got tho correct words using tho clues given against them.

  1. SATPADAOINT specific features or certain habits which enable a living being to live In Its surroundings.
  2. RUTRCQXNI waste products are removed by this process.
  3. LUMISIT all living things respond to these.
  4. ROUCDPRENTOI Because of this we find organisms of the same kind.

Answer:

  1. Adaptations
  2. Excretion
  3. Stimuli
  4. Reproduction.

NCERT Class 6 Science Chapter 6 The Living Organisms Characteristics And Habitats Fill In the blanks.

1. The presence of specific features which enable a plant or an animal to live in a particular habitat is called adaptation

2. The habitats of the plants and animals that live on land are called terrestrial

3. The habitats of plants and animals that live in water are called aquatic

4. Soil, water and air are the habitat abiotic

5. Changes in our surroundings that make us respond to them are called stimuli

6. Plants and animals living on land, are called terrestrial organisms.

7. Soil, water and air are the abiotic components of a habitat.

8. Desert plants have long roots for better absorption of minerals and water.

9. Some objects are characterised as living while others as non-living

10. The animals depend on it for their food plants.

11. Webbed feet is a feature found in frog

12. A habitat has biotic and Abiotic components that interact with each other.

13. Fishes have gills for respiration.

NCERT Class 6 Science Chapter 6 The Living Organisms Characteristics And Habitats True Or False

1. Fin is for aquatic life in fishes. True

2. Mountain is a special terrestrial habitat with low temperature. True

3. Almost all types of plants are found in the sea. False, Only Those Plants that are adapted to aquatic Habitats are found in the sea.

4. Frog has dry skin. False, the frog has wet and slimy skin

5. All living beings need water. True

6. The non-living objects carry out respiration. False, non-living things cannot carry out respiration. It is a feature of living things.

7. Leaves of touch-me-not plant curl up when touched. True

8. In cacti, leaves are reduced to spines. True

NCERT Class 6 Science Chapter 6 The Living Organisms Characteristics And Habitats Assertion-Reason Questions

The following questions consist of two statements. Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below.

  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true, but R is not the correct explanation of A
  3. A is true, but R is false
  4. A is false, but R is true

1. Assertion (A) Forests, grasslands, deserts and mountain regions are examples of terrestrial habitats.
Reason (R) The plants and animals that live on land are said to live in terrestrial habitats.

2. Assertion (A) The eyes of a lion are in front of the face.
Reason (R) They help to get the correct idea about the location of its prey.

3. Assertion (A) Non-living things react to changes in their surroundings.
Reason (R) Changes in surroundings that make us respond to them are called stimuli.

Answer:

  1. Both A and R are true, but R is not the correct explanation of A.
  2. Both A and R are true and R is the correct explanation of A.
  3. A is false, but R is true. A can be corrected as Living things react to changes in their surroundings, i.e. they respond to stimulus.

NCERT Class 6 Science Chapter 6 The Living Organisms Characteristics And Habitats Match The Columns

Question 1. Match the Column 1 with Column 2.

Class 6 Science Chapter 6 The Living Organisms Match The columns

Answer: A-2, B-4, C-5, D-3, E-1

Question 2. Match the Column 1 with Column 2.

Class 6 Science Chapter 6 The Living Organisms Match The columns.2

Answer: A-1,B-5, C-2, D-3.E-4

Question 3. Match the Column 1 with Column 2.

Class 6 Science Chapter 6 The Living Organisms Match The columns 3

Answer: A-4, B-5.C-2, D-1, E-3