NCERT Exemplar Solutions For Class 6 Maths Chapter 7 Fractions

Class 6 Maths Chapter 7 Fractions

Question 1. The fraction which is not equal to\(\frac{4}{5}\)

1. \(\frac{40}{50}\)
2. \(\frac{12}{15}\)
3. \(\frac{16}{20}\)
4. \(\frac{9}{15}\)

Solution:

4. Since, \(\frac{9}{15}=\frac{9 \div 3}{15 \div 3}=\frac{3}{5} \text {, }\) which is not equal to \(\frac{4}{5}\)

\(\frac{9}{15}\) is the required fraction.

Question 2. The two consecutive integers between which the fraction \(\frac{5}{7}\) lies are

1. 5 and 6
2. 0 and 1
3. 5 and 7
4. 6 and 7

Solution:

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2. Since we know that a proper fraction whose numerator is less than its denominator always lies between 0 and 1.

∴ \(\frac{5}{7},\) which is also a proper fraction, must
lie between 0 and 1.

Question 3. when \(\frac{1}{4}\) is written with denominator as 12, its numerator is

1. 3
2. 8
3. 24
4. 12

Solution:

1. Since, \(\frac{1}{4}=\frac{1}{4} \times \frac{3}{3}\) [By multiplying the numerator and denominator by 3 to get 12 as the denominator] \(=\frac{3}{12},\) which shows that the required numerator is 3.

Question 4. Which of the following is not in the lowest form?

1. \(\frac{7}{5}\)
2. \(\frac{15}{20}\)
3. \(\frac{13}{33}\)
4. \(\frac{27}{28}\)

Solution:

In the case of \(\frac{7}{5}\) the common factor of 7 and 5 is

i.e. already given in the simplest form.

Similarly,

In \(\frac{15}{20}\) the common factor of 15 and 20 is 4

\(\frac{13}{33}\) the common factor of 13 and 33 is 1

i.e., already given in the simplest form

\(\frac{27}{28}\) the common factor of 27 and 28 is 1

i.e., already given in the simplest form.

So, the above cases show that \(\frac{15}{20}\) is not in the lowest form.

Question 5. If \(\frac{5}{8}=\frac{20}{p}\) then value of P is

1. 23
2. 2
3. 32
4. 16

Solution:

3. We have given, \(\frac{5}{8}=\frac{20}{p}\)

To find the value of p, we must multiply the numerator and denominator of \(\frac{5}{8} \text { by } 4.\)

∴ \(\frac{5}{8}=\frac{5 \times 4}{8 \times 4}=\frac{20}{32}\)

∴ \(\frac{20}{32}=\frac{20}{p}\)

∴ The value of p is 32.

Question 6. Which of the following is not equal to the others?

1. \(\frac{6}{8}\)
2. \(\frac{12}{16}\)
3. \(\frac{15}{25}\)
4. \(\frac{18}{24}\)

Solution:

3. Firstly, we will simplify all the fractions into their lowest form, we get,

\(\frac{6}{8}=\frac{6 \div 2}{8 \div 2}=\frac{3}{4}, \frac{12}{16}=\frac{12 \div 4}{16 \div 4}=\frac{3}{4}, \frac{15}{25}=\frac{15 \div 5}{25 \div 5}=\frac{3}{5}\) and \(\frac{18}{24}=\frac{18 \div 6}{24 \div 6}=\frac{3}{4}\)

Thus, on comparing the above cases, observe that \(\frac{3}{5}=\frac{15}{25}\) is not equal to the above-given fractions.

Question 7. Which of the following fractions is the greatest?

1. \(\frac{5}{7}\)
2. \(\frac{5}{6}\)
3. \(\frac{5}{9}\)
4. \(\frac{5}{8}\)

Solution:

2. We have given, \(\frac{5}{7}, \frac{5}{6}, \frac{5}{9} \text { and } \frac{5}{8}\) Since we know that, among all the fractions with the same numerator, the one with a smaller denominator will be the greatest

\(\frac{5}{6}\) is the required fraction.

Question 8. Which of the following fractions is the smallest?

1. \(\frac{7}{8}\)
2. \(\frac{9}{8}\)
3. \(\frac{3}{8}\)
4. \(\frac{5}{8}\)

Solution:

3. We have given \(\frac{7}{8}, \frac{9}{8}, \frac{3}{8} \text { and } \frac{5}{8}\)

Since we know that, among all the fractions with the same denominator, the one with a smaller numerator will be the smallest.

∴ \(\frac{3}{8}\) is the required fraction.

QueSomeon 9. Sum of \(\frac{4}{17} \text { and } \frac{15}{17} \text { is }\)

1. \(\frac{19}{17}\)
2. \(\frac{11}{17}\)
3. \(\frac{19}{34}\)
4. \(\frac{2}{17}\)

Solution:

1.  We have, \(\frac{4}{17}+\frac{15}{17}\)

⇒ \(=\frac{4+15}{17}\)

⇒ \(\left[\begin{array}{l}
\text { Sum of like fractions } \\
=\frac{\text { Sum of their numerators }}{\text { common denominator }}
\end{array}\right]\)

⇒ \(=\frac{19}{17}\)

Question 10. On subtracting \(\frac{5}{9} \text { from } \frac{19}{9} \text {, }\) the result is

1. \(\frac{24}{9}\)
2. \(\frac{14}{9}\)
3. \(\frac{14}{18}\)
4. \(\frac{14}{0}\)

Solution:

According to the question, \(\frac{19}{9}-\frac{5}{9}\)

⇒ \(\left[\begin{array}{l}\text { Difference of like fraction } \\
=\frac{\text { Difference of their numerator }}{\text { Common denominator }}
\end{array}\right]\)

∴ \(=\frac{14}{9}\)

Question 11. \(\frac{11}{7}\) can be expressed in the form

1. \(\frac{33}{7}\)
2. \(\frac{39}{7}\)
3. \(\frac{33}{4}\)
4. \(\frac{39}{4}\)

Solution:

We have, \(5 \frac{4}{7}=\frac{(5 \times 7)+4}{7}=\frac{35+4}{7}=\frac{39}{7}\)

Question 13. A number representing a part of a ______ is called a fraction.
Solution: whole

Question 14. A fraction with a denominator greater than the numerator is called a______fraction.
Solution: Proper

Question 15. Fractions with the same denominator are called _____ fractions.
Solution: Like

Question 16. \(13 \frac{5}{18}\) is a _____ fraction.
Solution: Mixed

Question 17. \(\frac{18}{5}\) is an _____ fraction.
Solution: Improper

Question 18. \(\frac{7}{19}\) is an _____ fraction.
Solution: Proper

Question 19. \(\frac{5}{8} \text { and } \frac{3}{8} \text { are }\) proper _____ fraction.
Solution: Like

Question 20. \(\frac{6}{11} \text { and } \frac{6}{13}\) proper _____ fraction.
Solution: Unlike

Question 21. The fraction \(\frac{6}{15}\) in simplest form is_____.
Solution:

\(\frac{2}{5}: \text { We have, } \frac{6}{15}=\frac{6 \div 3}{15 \div 3}=\frac{2}{5}\)

Question 22. The fraction \(\frac{17}{34}\)  in simplest form is _____.
Solution:

\(\frac{1}{2} \text { : We have, } \frac{17}{34}=\frac{17 \div 17}{34 \div 17}=\frac{1}{2}\)

Question 23. \(\frac{18}{135} \text { and } \frac{90}{675}\)  are proper, unlike and _____ fractions.
Solution:  Equivalent

Question 24. \(8 \frac{2}{7}\)  is equal to the improper fraction______.
Solution:

\(\frac{58}{7}: 8 \frac{2}{7}=\frac{(8 \times 7)+2}{7}=\frac{56+2}{7}=\frac{58}{7}\)

Question 25. \(\frac{87}{7}\)  is equal to the mixed fraction______.
Solution:

\(12 \frac{3}{7}:\)

∴ \(\frac{87}{7}=12 \frac{3}{7}\)

Question 26. \(\frac{17}{9}+\frac{41}{9}=\)______.
Solution:

⇒ \(\frac{58}{9}: \frac{17}{9}+\frac{41}{9}=\frac{17+41}{9}=\frac{58}{9}\)

Question 27. \(\frac{67}{14}-\frac{24}{14}=\)_______.
Solution:

\(\frac{43}{14}: \frac{67}{14}-\frac{24}{14}=\frac{67-24}{14}=\frac{43}{14}\)

Question 28. \(\frac{17}{2}+3 \frac{1}{2}=\)______.
Solution:

\(\text { 12: } \frac{17}{2}+3 \frac{1}{2}=\frac{17}{2}+\frac{7}{2}=\frac{17+7}{2}=\frac{24}{2}=12\)

Question 29. \(9 \frac{1}{4}-\frac{5}{4}=\)______.
Solution:

\(8: 9 \frac{1}{4}-\frac{5}{4}=\frac{37}{4}-\frac{5}{4}=\frac{37-5}{4}=\frac{32}{4}=8\)

Question 30. Fractions with the same numerator are called fractions.
Solution: False

Because fractions with the same denominators are called fractions.

Question 31. Fraction \(\frac{18}{39}\) is in its lowest form
Solution: False

Since the common factor of 18 and 39 is 3 and its simplest form is

⇒ \(\frac{18 \div 3}{39 \div 3}=\frac{6}{13}\)

Question 32. Fractions  \(\frac{15}{39} \text { and } \frac{45}{117}\) are equivalent fractions.
Solution:  True

⇒\(\text { Since, } \frac{45 \div 3}{117 \div 3}=\frac{15}{39}\)

∴ Both are equivalent

Question 33. The sum of two fractions is always a fraction.
Solution: True

Let, \(\frac{2}{3} \text { and } \frac{4}{3}\) are two fractions.

When we add them, we get \(\frac{2}{3}+\frac{4}{3}=\frac{2+4}{3}\)

\(=\frac{6}{3}=\frac{2}{1} \text {, }\) which is a fraction.

Question 34. The result obtained by subtracting a fraction from another fraction is necessarily a fraction.
Solution:  False

Let \(\frac{1}{2} \text { and } \frac{2}{3}\) are two fractions.

When we subtract \(\frac{2}{3} \text { from } \frac{1}{2} \text {, }\)

⇒ \(\frac{1}{2}-\frac{2}{3}=\frac{3-4}{6}=\frac{-1}{6}\)

Which is not a fraction.

Question 35. If a whole or an object is divided into several equal parts, then each part represents a fraction.
Solution: True

Question 36. The fraction represented by the shaded portion in the adjoining figure is \(\frac{3}{8}.\)

Solution: True

The total number of parts in a given figure is 8 and the shaded parts are 3.

∴ The required fraction is \(\frac{3}{8}\)

Question 37. The fraction represented by the unshaded portion in the adjoining figure is\(\frac{5}{9}.\)

Solution: False

The total number of parts in a given figure is 9 of which unshaded parts are 4.

∴ The required fraction is \(\frac{4}{9}.\)

Question 38. \(\text { 38. } \frac{25}{19}+\frac{6}{19}=\frac{31}{38}\)
Solution: False

⇒ \(\frac{25}{19}+\frac{6}{19}=\frac{25+6}{19}=\frac{31}{19} \neq \frac{31}{38}\)

∴ The above condition is false.

Question 39. \(\frac{8}{18}-\frac{8}{15}=\frac{8}{3}\)
Solution: False

The L.C.M. of 18 and 15 is 2 * 3 x 3 x 5- 90

∴ \(\frac{8}{18}-\frac{8}{15}=\frac{8 \times 5}{90}-\frac{8 \times 6}{90}\)

⇒\(=\frac{40}{90}-\frac{48}{90}=-\frac{8}{90} \neq \frac{8}{3}\)

∴ The above condition is false.

Question 40. \(\frac{7}{12}+\frac{11}{12}=\frac{3}{2}\)
Solution: True

⇒ \(\frac{7}{12}+\frac{11}{12}=\frac{7+11}{12}=\frac{18}{12}=\frac{18 \div 6}{12 \div 6}=\frac{3}{2}\)

Question 41. \(\frac{16}{25}>\frac{13}{25}\)
Solution: True

The given fractions are like fractions. On comparing the numerators, we get

⇒ \(\frac{16}{25}>\frac{13}{25}\)

Question 42. \(\frac{11}{16} \ldots \frac{14}{15}\)
Solution:

< : The L.C.M. of 16 and 15 is 2x2x2x2x3x5 = 240

⇒ \(\text { Thus, } \frac{11}{16}=\frac{11 \times 15}{16 \times 15}=\frac{165}{240}\)

⇒ \(\text { and } \frac{14}{15}=\frac{14 \times 16}{15 \times 16}=\frac{224}{240}\)

On comparing, we observe that

∴ \(\frac{224}{240}>\frac{165}{240} \text {, i.e., } \frac{11}{16}<\frac{14}{15}\)

Question 43. \(\frac{8}{15} \ldots \frac{95}{14}\)
Solution:

< : The L.C.M. of 15 and 14 is 2 x 3 x 5 x 7 = 210

⇒ \(\text { Thus, } \frac{8}{15}=\frac{8 \times 14}{15 \times 14}=\frac{112}{210}\)

⇒ \(\text { and } \frac{95}{14}=\frac{95 \times 15}{14 \times 15}=\frac{1425}{210}\)

On comparing, we observed that

⇒ \(\frac{1425}{210}>\frac{112}{210} \text { i.e., } \frac{8}{15}<\frac{95}{14}\)

Question 44. \(\frac{12}{75} \ldots \frac{32}{200}\)
Solution:

= : The L.C.M. of 75 and 200 is 2x2x2x3x5x5 = 600

⇒ \(\text { Thus, } \frac{12}{75}=\frac{12 \times 8}{75 \times 8}=\frac{96}{600}\)

⇒ \(\text { and } \frac{32}{200}=\frac{32 \times 3}{200 \times 3}=\frac{96}{600}\)

On comparing, we observe that

⇒ \(\frac{12}{75}=\frac{32}{200}\)

Question 45. \(\frac{18}{15} \ldots 1.3\)
Solution:

< : \(1.3=\frac{13}{10}\)

The L.C.M. of 15 and 10 is 2 x 3 x 5 = 30

⇒ \(\text { Now, } \frac{18}{15}=\frac{18 \times 2}{15 \times 2}=\frac{36}{30} \text { and } \frac{13}{10}=\frac{13 \times 3}{10 \times 3}=\frac{39}{30}\)

⇒ \(\frac{36}{30}<\frac{39}{30}\)

∴ \(\text { Thus, } \frac{18}{15}<1.3\)

Question 46. Write the fraction represented by the shaded portion of the adjoining figure:

Solution:

In the given figure, the total parts in which the figure has been divided is 8 out of which 7 parts are shaded.

∴ The required fraction is \(\frac{7}{8}.\)

Question 47. Write the fraction represented by the unshaded portion of the adjoining figure:

Solution:

In the given figure, the total number of parts in which the 1444 figure has been divided is 15, out of which 4 are unshaded.

∴ The required fraction is \(\frac{4}{15}.\)

Question 48. Ali divided one fruit cake equally among six persons. What part of the cake he gave to each person?
Solution:  Since Ali has to divide one fruit cake equally among 6 persons

Each person will get \(\frac{1}{6}\) part.

Question 49. Express \(6 \frac{2}{3}\) as an improper fraction.
Solution:

We have \(6 \frac{2}{3}=\frac{(6 \times 3)+2}{3}=\frac{18+2}{3}=\frac{20}{3}\)

Question 50. Arrange the fractions \(\frac{2}{3}, \frac{3}{4}, \frac{1}{2} \text { and } \frac{5}{6} \text { in }\) in ascending order
Solution:

⇒ \(\text { We have, } 6 \frac{2}{3}=\frac{(6 \times 3)+2}{3}=\frac{18+2}{3}=\frac{20}{3}\)

Firstly find the L.C.M. of 3, 4, 2, and 6.

∴ The L.C.M. of 3, 4, 2 and 6 is 2 x 2 x 3 = 12

⇒ \(\text { Now, } \frac{2}{3}=\frac{2 \times 4}{3 \times 4}=\frac{8}{12}, \frac{3}{4}=\frac{3 \times 3}{4 \times 3}=\frac{9}{12} \text {, }\)

⇒ \(\frac{1}{2}=\frac{1 \times 6}{2 \times 6}=\frac{6}{12} \text { and } \frac{5}{6}=\frac{5 \times 2}{6 \times 2}=\frac{10}{12}\)

On comparing the above, we get

⇒ \(\frac{6}{12}<\frac{8}{12}<\frac{9}{12}<\frac{10}{12} \text { i.e., } \frac{1}{2}<\frac{2}{3}<\frac{3}{4}<\frac{5}{6}\)

Question 51. Arrange the fractions \(\frac{6}{7}, \frac{7}{8}, \frac{4}{5} \text { and } \frac{3}{4} \text { in }\) descending order.
Solution:

⇒ We have given, \(\frac{6}{7}, \frac{7}{8}, \frac{4}{5} \text { and } \frac{3}{4} \text {. }\)

The L.C.M. of 7, 8, 5 and 4 is 2x2x2x5x7 = 280

⇒ \(\text { Now, } \frac{6}{7}=\frac{6 \times 40}{7 \times 40}=\frac{240}{280}\)

⇒  \(\frac{7}{8}=\frac{7 \times 35}{8 \times 35}=\frac{245}{280}\)

⇒  \(\frac{4}{5}=\frac{4 \times 56}{5 \times 56}=\frac{224}{280} \text { and } \frac{3}{4}=\frac{3 \times 70}{4 \times 70}=\frac{210}{280}\)

On comparing the above, we get;

⇒  \(\frac{245}{280}>\frac{240}{280}>\frac{224}{280}>\frac{210}{280}\)

⇒ \(\text { i.e., } \frac{7}{8}>\frac{6}{7}>\frac{4}{5}>\frac{3}{4}\)

Question 52. Write \(\frac{3}{4}\) as a fraction with denominator 44.
Solution:

⇒ \(\text { Let } \frac{3}{4}=\frac{?}{44}\)

Then, we have to find the missing numeral. To get 44 in the denominator, we multiply by 4 by 11.

So, we multiply the numerator and denominator by 11.

∴ \(\frac{3}{4}=\frac{3 \times 11}{4 \times 11}=\frac{33}{44}\)

\(\text { Hence, } \frac{3}{4} \text { and } \frac{33}{44}\) are equivalent fractions.

Question 53. Write \(\frac{5}{6}\) as a fraction with the numerator 60.
Solution:

Let, \(\frac{5}{6}=\frac{60}{?}\)

Then, we have to find the missing numeral. To get 60 in the numerator, we multiply 5 by

So, we multiply the numerator and denominator by 12.

∴ \(\frac{5 \times 12}{6 \times 12}=\frac{60}{72}\)

⇒ \(\text { Hence, } \frac{5}{6} \text { and } \frac{60}{72}\) are equivalent fractions.

Question 54. Write \(\frac{129}{8}\) as a mixed fraction
Solution:

We have, \(\frac{129}{8}=16 \frac{1}{8}\)

Question 55. Add the fractions \(\frac{3}{8} \text { and } \frac{2}{3} \text {. }\)
Solution:

L.C.M. 0f 8 and 3 is 2x2x2x3 = 24

Now, \(\frac{3}{8}+\frac{2}{3}=\frac{3 \times 3}{8 \times 3}+\frac{2 \times 8}{3 \times 8}\)

∴  \(=\frac{9}{24}+\frac{16}{24}=\frac{9+16}{24}=\frac{25}{24}\)

Question 56. Add the fractions \(\frac{3}{8} \text { and } 6 \frac{3}{4} \text {. }\)
Solution:

L.C.M. of 8 and 4 is 2 x 2 x 2 = 8

Now, \(\frac{3}{8}+6 \frac{3}{4}=\frac{3}{8}+\frac{27}{4}\)

⇒  \(\frac{3}{8}+\frac{27 \times 2}{4 \times 2}\)

⇒  \(\frac{3}{8}+\frac{54}{8}=\frac{3+54}{8}=\frac{57}{8}=7 \frac{1}{8}\)

Question 57. Subtract \(\frac{1}{6} \text { from } \frac{1}{2} \text {. }\)
Solution:

The L.C.M. of 6 and 2 = 6

Now, \(\frac{1}{2}-\frac{1}{6}=\frac{1 \times 3}{2 \times 3}-\frac{1}{6}=\frac{3}{6}-\frac{1}{6}\)

⇒ \(\frac{3-1}{6}=\frac{2}{6}=\frac{2 \div 2}{6 \div 2}\)

[H.C.F. of 2 and 6 is 2]

⇒ \(\frac{1}{3}\)

Question 58. Subtract \(8 \frac{1}{3} \text { from } \frac{100}{9}\).
Solution:

The L.C.M. of 3 and 9 = 9

⇒ \(\frac{100}{9}-8 \frac{1}{3}=\frac{100}{9}-\frac{25}{3}\)

⇒ \(=\frac{100}{9}-\frac{25 \times 3}{3 \times 3}=\frac{100}{9}-\frac{75}{9}=\frac{100-75}{9}=\frac{25}{9}\)

⇒ \(2 \frac{7}{9}\)

Question 59. Subtract \(1 \frac{1}{4} \text { from } 6 \frac{1}{2}\)
Solution:

The L.C.M. of 4 and 2 = 4

Now, \(6 \frac{1}{2}-1 \frac{1}{4}=\frac{13}{2}-\frac{5}{4}\)

⇒ \(\frac{13 \times 2}{2 \times 2}-\frac{5}{4}=\frac{26}{4}-\frac{5}{4}=\frac{26-5}{4}\)

⇒ \(\frac{21}{4}=5 \frac{1}{4}\)

Question 60. Add \(1 \frac{1}{4} \text { and } 6 \frac{1}{2}\)
Solution:

The L.C.M. of 4 and 2 = 4

⇒ \(\text { Now, } 1 \frac{1}{4}+6 \frac{1}{2}=\frac{5}{4}+\frac{13}{2}\)

⇒ \(\frac{5}{4}+\frac{13 \times 2}{2 \times 2}=\frac{5}{4}+\frac{26}{4}=\frac{5+26}{4}=\frac{31}{4}\)

⇒ \(7 \frac{3}{4}\)

⇒ \(\begin{array}{l|l}
2 & 2,4 \\
\hline 2 & 1,2 \\
\hline & 1,1
\end{array}\)

Question 61. Katrina rode her bicycle \(6 \frac{1}{2}\) km in the morning and \(8 \frac{3}{4}\) km in the evening. Find the distance traveled by her altogether on that day.
Solution:

Given

Katrina rode her bicycle \(6 \frac{1}{2}\) km in the morning and \(8 \frac{3}{4}\) km in the evening.

The distance travelled by Katrina in the morning  \(=6 \frac{1}{2} \mathrm{~km}=\frac{13}{2} \mathrm{~km}\)

The distance traveled by Katrina in the evening \(=8 \frac{3}{4} \mathrm{~km}=\frac{35}{4} \mathrm{~km}\)

Total =distance traveled by her

⇒ \(\frac{13}{2} \mathrm{~km}+\frac{35}{4} \mathrm{~km}\)

⇒ \(\frac{13 \times 2}{2 \times 2} \mathrm{~km}+\frac{35}{4} \mathrm{~km}\)

∴ [ L.C.M. of 2 and 4 = 4]

⇒ \(\frac{26}{4} \mathrm{~km}+\frac{35}{4} \mathrm{~km}=\frac{(26+35) \mathrm{km}}{4}\)

⇒ \(\frac{61}{4} \mathrm{~km}=15 \frac{1}{4} \mathrm{~km}\)

Question 62. A rectangle is divided Into a certain number of equal parts. If 16 of the parts so formed represent the fraction \(\frac{1}{4}\) find the number of parts in which the rectangle has been divided.
Solution:

Given

A rectangle is divided Into a certain number of equal parts. If 16 of the parts so formed represent the fraction \(\frac{1}{4}\)

Let the number of parts in which the rectangle has been divided be x.

According to the question, \(\frac{1}{4}=\frac{16}{x}\)

By cross-multiplication, x = 16 Χ 4 = 64

∴ The required number of parts is 64.

Question 63. The grip size of a tennis racquet is \(11 \frac{9}{80} \mathrm{~cm}\). Express the size as an improper fraction.
Solution:

Given

The grip size of a tennis racquet is \(11 \frac{9}{80} \mathrm{~cm}\).

We have given, you the grip size of a tennis racquet \( =11 \frac{9}{80} \mathrm{~cm}=\frac{889}{80} \mathrm{~cm}\), which is the required improper fraction.

Question 64. On average \(\frac{1}{10}\) the food eaten is turned into an organism’s own body and is available for the next level of consumer in a food chain. What fraction of the food eaten is not available for the next level?
Solution:

Given

On average \(\frac{1}{10}\) the food eaten is turned into an organism’s own body and is available for the next level of consumer in a food chain.

We have given, \(\frac{1}{10}\) of the food eaten is turned into the organism’s own body.

The required fraction of the food eaten not available for the next level is

⇒ \(1-\frac{1}{10}=\frac{10-1}{10}\)

[ L.C.M. of1 and 10 is 10]

⇒ \(\frac{9}{10}\)

Question 65. Mr. Rajan got a job at the age of 24 years and he got retired from the job at the age of 60 years. What fraction of his age till retirement was he in the job?
Solution:

Given

Mr. Rajan got a job at the age of 24 years and he got retired from the job at the age of 60 years.

Mr. Rajan got a job at the age of 24 years.

He retired at the age of 60 years.

He worked for (60- 24) years = 36 years

∴ The required fraction \(=\frac{36}{60}=\frac{36 \div 12}{60 \div 12}=\frac{3}{5}\)

[ H.C.F. of 36 and 60 is 12]

Question 66. The food we eat remains in the stomach for a maximum of 4 hours. For what fraction of a day, does it remain there? 
Solution:

Given

The food we eat remains in the stomach for a maximum of 4 hours.

Total number of hours in a day 24 hours

The required fraction \(=\frac{4}{24}= \div {4+4}{24+4}= \div {1}{6}\)

[ H.C.F. of 4 and 24 is 4]

Question 67. It was estimated that because of people switching to Metro trains, about 33000 tonnes of CNG, 3300 tonnes of diesel, and 21000 tonnes of petrol were saved by the end of the year 2007. Find the fraction of:

1. The quantity of diesel saved to the quantity of petrol saved.
2. The quantity of diesel saved to the quantity of CNG saved.

Solution:

Given

It was estimated that because of people switching to Metro trains, about 33000 tonnes of CNG, 3300 tonnes of diesel, and 21000 tonnes of petrol were saved by the end of the year 2007.

By the end of the year 2007,

The quantity of CNG saved 33000 tonnes

The quantity of diesel saved 3300 tonnes and

The quantity of petrol saved 21000 tonnes

⇒ \(\frac{\text { The quantity of diesel saved }}{\text { The quantity of petrol saved }}=\frac{3300}{21000}\)

⇒ \(\frac{3300 \div 300}{21000 \div 300}\)

[H.C.F. of 3300 and 21000 is 300]

\(=\frac{11}{70}\)

⇒ \(\frac{\text { The quantity of diesel saved }}{\text { The quantity of CNG saved }}=\frac{3300}{33000}\)

⇒ \(\frac{3300 \div 3300}{33000 \div 3300}\)

[H.C.F. of 3300 and 33000 is 3300]

∴ \(=\frac{1}{10}\)

Question 68. A cup is \(\frac{1}{3}\) full of milk. What part of the cup is still to be filled with milk to make it full?
Solution:

Given

A cup is \(\frac{1}{3}\) full of milk.

The remaining part of the cup which is still to be filled with milk  \(=1-\frac{1}{3}\)

⇒ \(\frac{3-1}{3}\)

⇒  \(\frac{2}{3}\)

Question 69. Mary bought \(3 \frac{1}{2} \mathrm{~m}\) of lace. She used \(1 \frac{3}{4} \mathrm{~m}\) of Hires lace for her new dress. How much lace is left with her?
Solution:

Given

Mary bought the lace = \(3 \frac{1}{2} \mathrm{~m}\) = \(\frac{7}{2} \mathrm{~m}\)

Lace used by Mary \(=1 \frac{3}{4} \mathrm{~m}=\frac{7}{4} \mathrm{~m}\)

She is left with \(\frac{7}{2} m-\frac{7}{4} m\)

⇒\(\frac{7 \times 2}{2 \times 2} m-\frac{7}{4} m=\frac{14 m-7 m}{4}\)

[L.C.M. of 2 and 4 is 4]

⇒ \(\frac{7}{4} m=1 \frac{3}{4} m \text { of lace }\)

Question 70. When Sunita weighed herself on Monday, she found that she had gained \( \frac{1}{4} \mathrm{~kg}\) Earlier her weight was \(46 \frac{3}{8} \mathrm{~kg}\)  What was her weight on Monday?
Solution:

Given

Sunita weighed herself on Monday, she found that she had gained \( \frac{1}{4} \mathrm{~kg}\) Earlier her weight was \(46 \frac{3}{8} \mathrm{~kg}\)

Sunita had gained \(=1 \frac{1}{4} \mathrm{~kg}=\frac{5}{4} \mathrm{~kg}\)

Earlier her weight was \(46 \frac{3}{8} \mathrm{~kg}=\frac{371}{8} \mathrm{~kg}\)

Her total weight on Monday

⇒ \(\frac{371}{8} \mathrm{~kg}+\frac{5}{4} \mathrm{~kg}=\frac{371}{8} \mathrm{~kg}+\frac{5 \times 2}{4 \times 2} \mathrm{~kg}\)

[L.C.M. of 8 and 4 is 8]

⇒ \(\frac{371}{8} \mathrm{~kg}+\frac{10}{8} \mathrm{~kg}\)

⇒ \(\frac{(371+10)}{8} \mathrm{~kg}=\frac{381}{8} \mathrm{~kg}=47 \frac{5}{8} \mathrm{~kg}\)

Question 71. Sunil purchased \(12 \frac{1}{2}\)liters of juice on Monday and \(14 \frac{3}{4}\) liters of juice on Tuesday. How many liters of juice did he purchase together in two days?
Solution:

Given

Sunil purchased \(12 \frac{1}{2}\)liters of juice on Monday and \(14 \frac{3}{4}\) liters of juice on Tuesday.

Sunil purchased juice on Monday = \(12 \frac{1}{2}\)liters \(=\frac{25}{2} \text { litres }\)

and on Tuesday \(14 \frac{3}{4}\)liters  \(=\frac{59}{4} \text { litres }\)

The total quantity of juice Sunil purchased in two days  \(=\left(\frac{25}{2}+\frac{59}{4}\right) \text {liters}\)

⇒ \(=\left(\frac{25 \times 2}{2 \times 2}+\frac{59}{4}\right) \text { litres }=\left(\frac{50}{4}+\frac{59}{4}\right) \text { litres }\)

⇒ \(=\left(\frac{50+59}{4}\right) \text { litres }=\frac{109}{4} \text { litres }=27 \frac{1}{4} \text { litres }\)

Question 72. Nazima gave \(2 \frac{3}{4}\) liters out of the \(5 \frac{1}{2}\) liters of juice she purchased for her friends. How many liters of juice is left with her?
Solution:

Given

Nazima gave \(2 \frac{3}{4}\) liters out of the \(5 \frac{1}{2}\) liters of juice she purchased for her friends.

Total quantity of juice = \(5 \frac{1}{2}\)liters

= \(\frac{11}{2} \text { litres }\)

Nazima gave to her friends \(=2 \frac{3}{4} \text { litres }\)

⇒ \(\frac{11}{4} \text { litres }\)

The required quantity of juice she is left \(=\left(\frac{11}{2}-\frac{11}{4}\right) \text { litres }=\left(\frac{11 \times 2}{2 \times 2}-\frac{11}{4}\right) \text { litres }\)

⇒ \(\left(\frac{22-11}{4}\right) \text { litres }=\frac{11}{4} \text { litres }=2 \frac{3}{4} \text { litres }\)

Question 73. Roma gave a wooden board of length \(150 \frac{1}{4} \mathrm{~cm}\) to a carpenter for making a shelf. The Carpenter sawed off a piece of \(40 \frac{1}{5} \mathrm{~cm}\) cm from it. What is the length of the remaining piece?
Solution:

Given

Roma gave a wooden board of length \(150 \frac{1}{4} \mathrm{~cm}\) to a carpenter for making a shelf. The Carpenter sawed off a piece of \(40 \frac{1}{5} \mathrm{~cm}\) cm from it.

Total length of a wooden board \(=150 \frac{1}{4} \mathrm{~cm}=\frac{601}{4} \mathrm{~cm}\)

The carpenter sawed off a piece of length \(=40 \frac{1}{5} \mathrm{~cm}=\frac{201}{5} \mathrm{~cm}\)

The length of the remaining piece \(=\left(\frac{601}{4}-\frac{201}{5}\right) \mathrm{cm}=\left(\frac{601 \times 5}{4 \times 5}-\frac{201 \times 4}{5 \times 4}\right) \mathrm{cm}\)

⇒\(\left(\frac{3005}{20}-\frac{804}{20}\right) \mathrm{cm}=\frac{2201}{20} \mathrm{~cm} \text { or } 110 \frac{1}{20} \mathrm{~cm}\)

Question 74. Naslr travelled \(3 \frac{1}{2} \mathrm{~km}\) in a bus and then walked \(1 \frac{1}{8} \mathrm{~km}\) to reach a town. How much did he travel to reach the town?
Answer:

Given

Naslr travelled \(3 \frac{1}{2} \mathrm{~km}\) in a bus and then walked \(1 \frac{1}{8} \mathrm{~km}\) to reach a town.

Nasir traveled by a bus = \(3 \frac{1}{2} \mathrm{~km}\)

⇒ \(\frac{7}{2} \mathrm{~km}\)

Nasir walked \(=1 \frac{1}{8} \mathrm{~km}=\frac{9}{8} \mathrm{~km}\)

Total distance traveled by him latex]=\frac{7}{2} \mathrm{~km}+\frac{9}{8} \mathrm{~km}=\frac{7 \times 4}{2 \times 4} \mathrm{~km}+\frac{9}{8} \mathrm{~km}[/latex]

⇒ \(\left(\frac{28}{8}+\frac{9}{8}\right) \mathrm{km}\)

∴ \(\left(\frac{28+9}{8}\right) \mathrm{km}=\frac{37}{8} \mathrm{~km} \text { or } 4 \frac{5}{8} \mathrm{~km}\)

Question 75. The fish caught by Neetu was of weight \(3 \frac{3}{4} \mathrm{~kg}\) kg and the fish caught by Narendra was of weight 2 \frac{1}{2} \mathrm{~kg} kg. How much more did Neetu’s fish weigh than that of Narendra?
Solution:

Given

The fish caught by Neetu was of weight \(3 \frac{3}{4} \mathrm{~kg}\) kg and the fish caught by Narendra was of weight 2 \frac{1}{2} \mathrm{~kg} kg.

The weight of fish caught by Neetu \(=3 \frac{3}{4} \mathrm{~kg}=\frac{15}{4} \mathrm{~kg}\)

The weight of fish caught by Narendra \(=2 \frac{1}{2} \mathrm{~kg}=\frac{5}{2} \mathrm{~kg}\)

∴ Neetu’s fish weigh more than that of Narendra by \(\frac{15}{4} \mathrm{~kg}-\frac{5}{2} \mathrm{~kg}=\frac{15}{4} \mathrm{~kg}-\frac{5 \times 2}{2 \times 2} \mathrm{~kg}=\frac{15}{4} \mathrm{~kg}-\frac{10}{4} \mathrm{~kg}\)

⇒ \(\frac{(15-10)}{4} \mathrm{~kg}=\frac{5}{4} \mathrm{~kg} \text { or } 1 \frac{1}{4} \mathrm{~kg}\)

Question 76. Neelam’s father needs \(1 \frac{3}{4} \mathrm{~m}\) of cloth for the skirt of Neelam’s new dress and \(\frac{1}{2} m\) for the scarf. How much cloth must he buy in all?
Solution:

Given

Neelam’s father needs \(1 \frac{3}{4} \mathrm{~m}\) of cloth for the skirt of Neelam’s new dress and \(\frac{1}{2} m\) for the scarf.

Neelam’s father purchased the length of cloth for the skirt \(=1 \frac{3}{4} \mathrm{~m}=\frac{7}{4} \mathrm{~m}\) and for the scarf \(=\frac{1}{2} \mathrm{~m}\)

Total length he buys in all = \(=\frac{7}{4} m+\frac{1}{2} m\)

⇒ \(\frac{7}{4} m+\frac{1 \times 2}{2 \times 2} m=\frac{7}{4} m+\frac{2}{4} m\)

⇒ \(\frac{(7+2)}{4} m=\frac{9}{4} m \text { or } 2 \frac{1}{4} m\)

Question 77. What is wrong with the following additions?

Solution:

1. Equal denominators are added.
2. Numerators and denominators are added

Question 78. Match the fractions of Column I with the shaded or marked portion of figures of Column II:

Solution:

1→(D); 2→(A); 3→(E); 4→(B);

Marked portion in (A) \(=\frac{6}{10}\)

Shaded fraction in (B) \(=\frac{6}{16}\)

Shaded fraction in (C) \(=\frac{6}{7}\)

Shaded fraction in (D) \(=\frac{4}{4}+\frac{2}{4}=\frac{6}{4}\)

Shaded fraction in (E) \(=\frac{6}{6}\)

Question 79. Find the fraction that represents the number of natural numbers to total numbers in the collection 0, 1, 2, 3, 4, 5. What fraction will it be for whole numbers?
Solution:

Out of 0, 1, 2, 3, 4, 5 → 1, 2, 3, 4 and 5 are the natural numbers.

The fraction that represents the number of natural numbers to the total numbers \(=\frac{5}{6}\) and the whole numbers are 0, 1, 2, 3, 4, and 5.

∴ The fraction that represents the number of whole numbers to the total numbers \(=\frac{6}{6}\)

Question 80. Write the fraction representing the total number of natural numbers in the collection of numbers -3, -2, -1, 0, 1, 2, 3. What fraction will it be for whole numbers? What fraction will it be for integers?
Solution:

Out of -3, -2, -1, 0, 1, 2, 3 → 1, 2, and 3 are the natural numbers, 0, 1, 2, and 3 are the whole numbers and -3, -2, -l, 0, 1, 2, 3 are integers.

∴ The fraction representing the natural numbers to the total numbers \(=\frac{3}{7}\)

The fraction representing the whole numbers to the total numbers \(=\frac{4}{7}\)

The required fraction representing the integers to the total numbers  \(=\frac{7}{7}\)

Question 81. Write a pair of fractions whose sum is \(\frac{7}{11}\) and the difference is \(\frac{2}{11}\)
Solution: Let one fraction be x

Another fraction be \(\frac{7}{11}-x\)

Now, according to the question,

⇒ \(x-\left(\frac{7}{11}-x\right)=\frac{2}{11}\)

⇒ \(\Rightarrow x-\frac{7}{11}+x=\frac{2}{11}\)

⇒ \(\frac{11 x}{11}-\frac{7}{11}+\frac{11 x}{11}=\frac{2}{11}\)

⇒ \(\frac{11 x-7+11 x}{11}=\frac{2}{11}\)

⇒ \(22 x-7=\frac{2}{11} \times 11\)

⇒ \(22 x-7=2 \Rightarrow 22 x=2+7=9\)

⇒ \(x=\frac{9}{22}\)

Thus, one fraction is \(\frac{9}{22}\) and another fraction is \(\frac{7}{11}-\frac{9}{22}=\frac{7 \times 2}{11 \times 2}-\frac{9}{22}=\frac{14-9}{22}=\frac{5}{22}\)

Question 82. What fraction of a straight angle is a right angle?
Solution: Since we know that the measurement of a straight angle is 180° and a right angle is 90°

∴ The required fraction is \(\frac{90^{\circ}}{180^{\circ}}=\frac{1}{2} .\)

Question 83. Put the right card in the right bag.

Solution:

We know that if the numerator is less than the denominator, then the fraction is less than 1.

If the numerator is equal to the denominator, then the fraction is equal to 1 and if the numerator is greater than the denominator, then the fraction is greater than 1.

The cards in Bag 1 are

1.  \(\frac{3}{7},\)

4. \(\frac{8}{9}\)

5.  \(\frac{5}{6},\)

6. \(\frac{6}{11},\)

8. \(\frac{19}{25}\)

9. \(\frac{2}{3}\) and

10. \(\frac{13}{17}\)

The cards in Bag 2 are

2.  \(\frac{4}{4}\)

7. \(\frac{18}{18}\)

And cards in B ag 3 are

3. \(\frac{9}{8} .\)