NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Recalling Ratios And Percentages Important Points
1. We usually compare two quantities by division, i.e., by using fractions. Comparison by division is called ratio.
Note that two quantities can be compared only when they have the same units. A ratio has no unit.
However, if the two quantities are not in the same units, we convert them into the same units before comparing.
2. Two quantities can also be compared using percentages. By percentage, we mean a fraction where the denominator is 100. The numerator of the fraction is called the rate percent.
For example: \(\frac{1}{5}\) means 5%. The symbol % is often used to express ‘percent’ (p. c.).
3. To convert the ratio into a percentage, we convert it into a fraction whose denominator is 100. [or we multiply by 100 and employ the % sign.]
4. To convert percentages in \(\frac{5}{100}\) to fractions, we divide the numerator by 100 and express it in the lowest form.
For example : 5% =\(\frac{1}{20}\)
Read and Learn More NCERT Solutions For Class 8 Maths
.
5. In the unitary method, we find the value of one unit from the given value of some units and then we find the value of the required number of units.
Question 1. In primary school, the parents were asked about the number of hours they spend per day helping their children to do homework. 90 parents helped for hour to \(\frac{1}{2}\) hours to 1 \(\frac{1}{2}\) . The distribution of parents according to the time for which, they said they helped is given in the adjoining figure; 20% helped for more than 1\(\frac{1}{2}\) ho per day; 30% helped for \(\frac{1}{2}\) an hour to 1 \(\frac{1}{2}\) hours, 50% did not help at all.
Using this, answer the following:
- How many parents were surveyed?
- How many said that they did not help
- How many said that they helped for more than 1\(\frac{1}{2}\) hour?
Solution:
1. According to the question
⇒ \(\frac{30}{100}\) of number of parents surveyed = 90
Number of parents surveyed = 90
Number of parents surveyed
= \(\frac{90 \times 100}{30}\)
= 300
Hence, 300 parents were surveyed.
2. A number of those parents said that they did not help
= 50% of 300
= \(300 \times \frac{50}{100}\)
= 150
Hence, 150 parents said that they did 30% not help.
3. A number of those parents said that they helped for more than 1\(\frac{1}{2}\) hour
= 20% of 300 = \(300 \times \frac{20}{100}\)
= 60
Hence, 60 parents said that they helped for more than 1/[latxe]\frac{1}{2}[/latex] hours
Note: ‘Of’ means multiplication
NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Exercise 7.1
Question 1. Find the ratio of the following:
- The speed of a cycle 15 km per hour to the speed of a scooter 30 km per hour.
- 5 m to 10 km
- 50 paise to 5
Solution:
1. The speed ofa a cycle 15 km per hour to the speed of a scooter 30 km per hour
= 15 km per hour: 30 km per hour
= 15: 30
= \(\frac{15}{30}=\frac{1}{2}\) Or 1:2
2. 5 m to 10 km
10 km = 10 × 1000 m = 10000 m
∴Ratio of 5 m to 10 km
= 5 m: 10000 m
= 5: 10000
⇒ \(\frac{5}{10000}=\frac{1}{2000}\)
= 1:20000
3. 50 paise to ₹ 5
₹ 5 = 5× 100 = 500 paise
A Ratio of 50 paise to ₹ 5
= 50 paise : 500 paise
= 50: 500
= \(\frac{50}{500}=\frac{1}{10}\) Or 1:10
Question 2. Convert the following ratios to percentages
- 3:4
- 2: 3.
Solution:
1. 3: 4
= \(\frac{3}{4}=\frac{3}{4} \times \frac{25}{25}\)
Making denominator 100
= \(\frac{75}{100}=75 \%\)
Aliter:
⇒ \(3: 4=\frac{3}{4} \times 100 \%\)
= 75 %
2. 2:3
= \(\frac{2}{3}=\frac{2}{3} \times \frac{100}{100}\)
= \(66 \frac{2}{3} \%\)
Aliter:
2: 3 = \(\frac{2}{3} \times 100 \%\)
= \(\frac{200}{3} \%=66 \frac{2}{3} \%\)
Question 3. 72% of 25 students are interested in mathematics. How many are not interested in mathematics?
Solution:
Total number of students = 25
Percentage of students interested in Mathematics s = 72%
Percentage of students who are not interested in Mathematics
= (100 – 72)% = 28%
Number of those students who are not interested in Mathematics
= 28% of 25
= \(\frac{28}{100} \times 25\)
= 7
Hence, 7 students are not interested in or 1: 10 Mathematics.
Question 4. A football team won 10 matches out ofthe total number of matches they played. Iftheir win percentage was 40, then how many matches did they play in all?
Solution:
If 40 matches were won, then the total number of matches played = 100
∴ If 1 match was won, then the total number of matches played
=\(\frac{100}{40}\)
If 10 matches were won, then the total number of matches played
= \(\frac{100}{40} \times 10\)
= 25
Hence, they played 25 matches in all.
Aliter:
According to the question,
40% of (total number of matches) = 10
⇒ \(\frac{40}{100} \times \text { (total number of matches) }\)=10
Total number of matches = \(\frac{10 \times 100}{40}\) = 25
Hence, they played 25 matches in all.
Question 5. If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning ?’
Solution:
Percentage of money spent = 75%
Percentage of money left = (100- 75)% = 25%
∴ If Chameli had ₹ 25 left, then the money she had in the beginning
= 100
∴ If Chameli had ₹1 left, then the money she had in the beginning
= \(₹ \frac{100}{25}\)
∴ If Chameli has v600 left, then the money she had in the beginning
= \(₹ \frac{100}{25} \times 600\)
= ₹ 24000
Hence, the money she had in the beginning was ₹ 2400.
Aliter :
According to this question,
25% of total money = ₹ 600
⇒ \(\frac{25}{100}\) of total money = ₹ 600
⇒ \(₹ \frac{600 \times 100}{25}\)
= ₹ 2400.
Hence, the money she had in the beginning was? 2400
Question 6. If 60% of people in a city like cricket, 30% like football and, the remaining like other games, then what percent of the people like other games? If the total number of people is 50 lakh, find the exact number who like each type of game.
Solution:
Percentage of people who like cricket = 60%
Percentage of people who like football = 30%
∴Percentage of people who like other games
= [100 -(60 +30)]%
= (100-90)% = 10%.
Total number of people = 50 lakh
= 50,00,000
∴ Number of people who like cricket
= 60% of 50,00,000
= \(50,00,000 \times \frac{60}{100}\)
= 30,00,000 = 30 lakh
Number of people who like football
= 30% of 50,00,000
= \(50,00,000 \times \frac{30}{100}\)
Number of people who like the other games
= 10% of 50,00,000
= \(50,00,000 \times \frac{10}{100}\)
= 5,00,000 = 5 lakh
Finding Discounts
Discount = Marked Price – Sale Price
Discount per cent = \(\frac{\text { Discount }}{\text { Marked Price }} \times 100 \%\)
Question 1. 4 shop gives 20% discount. What Would the sale price ofeach of these be?
- A dress marked at.₹ 120
- A pair of shoes marked at ₹ 750
- A bag marked at₹ 250.
Solution:
1. A dress marked at ₹120
Marked price ofthe dress = ₹120
Discount rate= 20%
,
∴ Discount = 20% of ₹ 120
20 MarkedDiscountprice × 100
=\(₹ \frac{20}{100} \times 120\)
= ₹ 24
∴ Sale price ofthe dress x 100 % = 4%.
= Marked price – Discount
= ₹ 120 – ₹ 24 = ₹ 96
2. A pair of shoes marked at ₹ 750
Marked price ofthe pair ofshoes= ₹ 750
Discount rate = 20%
Discounts 20% of ₹ 750
= \(₹ \frac{20}{100} \times 750\)
= ₹150
∴ Sale price ofthe pair of shoes
= Marked price – Discount
= ₹ 750 -₹ 150
= ₹600
3. A bag marked at ₹ 250
The marked price of the bag =₹ 250
Discount rate = 20%
Discount = 20% of ₹ 250
= \(₹ \frac{20}{100} \times 250\)
= ₹50
∴ The sale price of the bag
= Marked price – Discount
=₹250 – ₹ 50 =₹200.
Question 2. If the table marked at f 15,000 is * available for f14, 400. Find the discount given £ and the discount percent.
Solution:
Marked price ofthe table =₹ 15,000
Sale price ofthe table = ₹ 14,400
∴ Discount given
= Marked price – Sale price
= ₹ 15,000 – ₹14,400
= ₹ 600
∴ Discount per cent
= \(\left(\frac{\text { Discount }}{\text { Marked price }} \times 100\right) \% \)
= \(\left(\frac{600}{15000} \times 100\right) \%=4 \%\)
Question 3. An almirah is sold at ₹5,225 after allowing a discount of 5%. Find its marked price.
Solution:
Let the marked price bet x.
Discount rate = 5%
Discount = 5% of ₹ x
=\(₹ \frac{5}{100} \times x=₹ \frac{x}{20}\)
∴ Sale price
= Marked price -Discount
= \(₹ x-₹ \frac{x}{20}=₹ \frac{19 x}{20}\)
According to the question,
∴ \(\frac{19 x}{20}=5225\)
x = \(\frac{5225 \times 20}{19}\)
x = ₹ 5500
Hence, the marked price of the almirah is ₹5500
Question 4. Try estimating 20% of the same bill amount.
- Round of the bill to the nearest tens.
- Find the amount of discount.
- Reduce the bill amount by the discount amount
Solution:
Your bill in a shop is ₹ 577.80.
1. Round off the bill to the nearest tens of ₹ 577.80, i.e., to ₹ 580.
2. Find 20% of this,
⇒ \(₹ \frac{20}{100} \times 580\)
= ₹ 116
3. Therefore estimated 20% of the same bill amount is ₹116 or ₹120 (rounded off to the nearest tens).
Question 5. Try finding 15% of ₹ 375.
Solution:
Find 10% of ₹ 375, i.e.
⇒ ₹\(₹ \frac{10}{100} \times 375\)
= ₹ 37.5
= ₹ 40
(rounded off to the nearest tens)
2. Take half of this, i.e \(\frac{1}{2} \times ₹ 40\)
= ₹ 20
(3) Add (2) and (1) to get ₹ 60.
Therefore, an estimated 15% of 7 375 is
= 60
Sales Tax/Value Added Tax/Goods And Services Tax
Sales tax was charged at a specified rate on the sale price of an item by the state government and was added to the bill amount. It is different for different items and also for different states.
- Amount of Sales Tax = Tax% of bill amount
- These days, the prices include the tax known as Value Added Tax (VAT).
- From July 1, 2017, the Government of India introduced GST which stands for Goods and Services Tax which is levied on the supply of goods or services or both
Question 1. Find the buying price of each of the following when 5%STis added to the purchase
- A towel at ₹ 50
- Two bars of soap at ₹ 35 each
- 5 kg of flour at ₹15 per kg.
Solution:
1. Cost of the towel
Rate of ST = 5%
∴ ST =5% of 7 50
=₹ \(\frac{5}{100} \times 50\)
= ₹ 2.50
∴ Buying price of the towel
= Cost of the towel + ST
= ₹ 50 + ₹ 2.50 =₹ 52.50.
2. The cost of two bars of soap
= ₹35 × 2 = ₹ 70
Rate of ST= 5%
ST = 5% of ₹ 70
= \(₹ \frac{5}{100} \times 70\)
= ₹ 3.50
∴ Buying price of two bars of soap
= Cost of two bars of soap + ST
= ₹ 70 +₹ 3.50 = ₹ 73.50
3. Cost of 5 kg of flour
= ₹ 15 × 5 = ₹ 75
Rate of ST = 5%
ST = 5% of ₹ 75
= \(₹ \frac{5}{100} \times 75\)
= ₹ 3.75
∴ The buying price of 5 kg of flour + ST
= Cost of 5 kg of flour + ST
= ₹ 75 +₹ 3.75
=₹ 78.75.
Question 2. If 8% is included in the prices, find the original price of
- A TV bought for ₹13,500
- A shampoo bottle was bought for ₹ 180.
Solution:
1. When the price of a TV including
VAT is ₹108,
original price = ₹ 100
When the price of the TV including VAT is
₹ 1, original price = ₹ \( \frac{100}{108}\)
The price of a TV including VAT is ₹ 13500,
Original price = \(₹ \frac{100}{108} .\)
= ₹ 12500
Hence, the original price of the TV Is ₹ 12500.
Aliter:
Price of TV including VAT
= ₹ 13500
Rate of VAT = 8%
Let, the original price of the TV be ₹x
Then,
x+8 % of x =₹ 13500
⇒ \(x+\frac{8}{100} x =₹ 13500 \)
⇒ 1+ \(\frac{8}{100}\)x =₹ 13500
⇒ \(\frac{108}{100} x =₹ 13500\)
= ₹ \(\frac{13500 \times 100}{108}\)
Hence, the original price of the T.V. is ₹12500.
2. When the price of a shampoo bottle including VAT is ₹ 108,
Original price = ₹ 100
When the price of a shampoo bottle including VAT is ₹1, the original price
= \(₹ \frac{100}{108} .\)
∴ The price of a shampoo bottle including VAT is? 180,
Original price = \(=₹ \frac{100}{108} \times 180 \)
⇒ \(₹ \frac{100 \times 5}{3}=₹ \frac{500}{3}\)
=\(₹ 166 \frac{2}{3}\)
Hence, the original price ofthe shampoo Hence, the original price ofthe shampoo bottle is \(166 \frac{2}{3}\)
Alitor: Price of shampoo bot t lo including
VAT = ₹ 180
Rate of VAT = 8%
Let the original price of shampoo bottle be ₹x
Then,
x+ 8%of x = ₹180
⇒ \(x+\frac{8}{100} x=₹ 180\)
⇒ \(\left(1+\frac{8}{100}\right) x=₹ 180\)
⇒ \(\frac{108}{100} x=₹ 180\)
x =\(₹ \frac{180 \times 100}{108}\)
x = \(₹ \frac{500}{3}=₹ 166 \frac{2}{3}\)
Hence, the original price ofthe shampoo bottle is \(166 \frac{2}{3}\)
NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Exercise 7.2
Question 1. During a sale, a shop offered a discount of 10% on the marked prices of all the items. W7iat would a customer have to pay for a pair of jeans marked at ₹ 1450 and two shirts marked at f850 each
Solution:
For a pair of jeans
Marked price = ₹ 1450
Discount rate = 10%
Discount = 10% of marked price
= 10% of ₹ 1450
= \(₹ \frac{10}{100} \times 1450\)
= ₹ 145
Sale price – Marked price – Discount
= ₹1450-₹ 145
= ₹ 1305
For two shirts
Marked price =₹ 850 × 2 = ₹ 1700
Discount rate = 10%
Discount = 10% of marked price
= 10% of ₹ 1700
=₹ \(\frac{10}{100} \times 1700=₹ 170\)
Sale price
= Marked price – Discount
= ₹1700 – ₹70 =₹1530
Total payment made by the customer
= ₹ 1305 + ₹ 1530 =₹ 2835
Hence, the customer will have to pay. 2835 for a pair of jeans and two shirts.
Question 2. The price ofa W is f13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it
Solution:
Price of TV = ₹ 13,000
Sales tax charged on it
= 12% of ₹ 13,000
= \(₹ \frac{12}{100} \times 13,000\)
= ₹ 1560
Amount to be paid
= Price + Sales Tax
= ₹ 13,000 + ₹ 1,560
= ₹14,560.
Hence, the amount that Vinod will haveto pay for the TV if he buys it is ₹14,560
Question 3. Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is ₹ 1600, find the marked price
Solution:
Let the marked price of the pair of 6 shoes be ₹ 100.
Kate of discount = 20
Discount = 20% of ₹ 100
= ₹ \(\frac{12}{100} \times 100\)
=₹ 20
Amount paid = Marked price -Discount
= ₹ 100 – ₹20 – = ₹ 80
If the amount paid is ₹80,
Then the marked price – \(₹ \frac{100}{80}\)
If the amount paid is ₹ 1, then the marked price = \(₹ \frac{100}{80} \times 1600\)
= ₹ 2000
Hence, the marked price of the pair of shoes is ₹ 2000.
Question 4. I purchased a hair dryer for f 5,400 including 8% VAT. Find the price before it was added,
Solution:
Price of hair-dryer including VAT
= ₹ 5400
VAT rate = 8%
Let the original price be ₹ 100
∴ VAT = ₹ 8
Price including VAT = Original price + VAT
= ₹100 + ₹ 8
= ₹ 108
If the price including VAT is ₹108, then
Original price = ₹ 100
If the price including VAT is ₹ 1, then
Orginalprice = \(₹ \frac{100}{108}\)
If the price including VAT is ₹5,400, then
Original price = ₹ \(\frac{100}{108} \times 5,400\)
= ₹ 5000
Hence, ‘the price before VAT was added is ₹5,000
Question 5. An article was purchased for ₹1239 including a GST of 18%. Find the price of the article before GST was added,
Solution:
Let the original price of the article be 100. GST Rate = 18% ,
Price after GST is included
= ₹(100 + 18) = ₹118
If the selling price is? 118 then original price = ₹100
If the selling price is ₹ 1, then
Original Price = \(₹ \frac{100}{108} \times 5,400\)
If the selling price is ₹ 1239, then
original price = \(₹ \frac{100}{118} \times 1239\)
= 100 × 10.5
= ₹ 1050
Hence, the price ofthe article before GST was added is? 1050
Compound Interest
1. Interest:
Interest is the extra money paid by institutions like banks or post offices on money that is deposited (kept) with them. Interest is also paid by people when they borrow money. The money deposited or borrowed is called the principal. Interest is generally given in percent for one year.
2. Simple interest (SI):
The interest is called simple when the principal does not change.
3. The formula for simple interest:
Simple interest on a principal of 1 P at R% rate of interest per year for T years is given by
Simple Interest = \(\frac{\text { Principal } \times \text { Rate } \times \text { Time }}{100}\)
SI = \(\frac{\text { P R T }}{100}\)
4. Amount :
Amount (A)= Principal (P) + Simple Interest(SI)
Question 1. Find interest and amount to be paid on 115,000 at 5% per annum after 2 years
Solution:
Interest on₹ 100 for 1 year is = 5
Interest on ₹ for 1 year is
= \(₹ \frac{5}{100}\)
Interest On ₹ for 1 year is
= \(₹ \frac{5}{100} \times 15,000\)
Interest on ₹ 15000 for 2 years is
= \(₹ \frac{5 \times 15,000 \times 2}{100}\)
= ₹ 1,500
∴ Amount = Principal + Interest
= ₹ 15,000 + ₹ 1,500
= ₹ 16,500
Aliter:
Here, P = ₹ 15,000
R = 5% per annum
T = 2 years
∴ \(\text { S.I. }=\frac{\text { PRT }}{1000}\)
= \(₹ \frac{15000 \times 5 \times 2}{100}\)
₹ 1500
Amount = Principal (P) + SimpleInterest (SI)
= ₹ 15000 + ₹ 1500
= ₹ 1,500 = ₹ 16500
Deducing A Formula For Compound Interest
A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n\)
where P = Principal
R = Rate of interest per annum compounded annually
n = Number of years
A = Amount
CI = A-P
Question 1. Find Cl on a sum of ₹ 8000 for 2 years at 5% per annum compounded annually.
Solution:
Here,
P =₹ 8000
R = 5%p.a.
n = 2 years
A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n\)
= \(8000\left(1+\frac{5}{100}\right)^2\)
= \(8000\left(1+\frac{1}{20}\right)^2\)
= \(8000\left(\frac{21}{20}\right)^2 \)
= \(8000 \times \frac{21}{20} \times \frac{21}{20}\)
= 20 × 21 × 21
=₹ 8820
Cl = A – P
= ₹ 8820 – ₹ 8000
= ₹ 820.
∴ Hence, the compound interest is ₹ 820
Applications Of Compound Interest Formula
We use the compound interest formula to find
- Increase (or decrease) in population.
- The growth of a bacteria if the rate of growth is known.
- The value of an item, if its price increases or decreases in the intermediate years.
Note: For increase, R is positive and for decrease, R is negative.
Question 1. A machinery worth f 10,500 depreciated by 5%. Find its value after one year.
Solution:
P = ₹ 10,500
R = -5% per annum
There is depreciation
n = year
A = \(P\left(1-\frac{R}{100}\right)^n \)
= \(10,500\left(1-\frac{5}{100}\right)\)
= \(10,500\left(1-\frac{1}{20}\right) \)
= \(10,500 \times \frac{19}{20}\)
= 9,975
Hence value after 1 year is ₹ 9,975.
Question 2. Find the population ofa city after 2 years, which is at present 12 lacks, if the rate of increase is 4%.
Solution:
P = 12,00,000
R = 4% per annum
n = 2years
A = \(P\left(1+\frac{R}{100}\right)^n\)
= \(12,00,000\left(1+\frac{4}{100}\right)^2 \)
= \(12,00,000\left(1+\frac{1}{25}\right)^2\)
= \(12,00,000\left(\frac{26}{25}\right)^2 \)
= \(12,00,000 \times \frac{26}{25} \times \frac{26}{25} \)
= 12,97,920
Hence, the population of the city after 2 years is 12,97,920
Question 3. Calculate the amount and compound interest on
- 10,800 for 3 years at \(2 \frac{1}{2} \)% perannum compounded annually.
- ₹ 18,000 for \(2 \frac{1}{2} \)% years at perannum compounded annually.
- ₹ 62,500 for \(1 \frac{1}{2} \)% years at perannum compounded half-yearly.
- ₹ 8,000 for 1 year at 9% per annum compounded half yearly.You could use the year-by-year calculation to verify).
- ₹ 10,000 for a year at 8% per annum compounded halfyearly
Solution:
1. By using year-by-year calculation
Here, P =₹ 10800
R = ₹ 12% per annum
T = ₹ year
SI on ₹ 10,800 at \(12 \frac{1}{2}\) per annum for
= \(10,800 \times \frac{25}{2} \times \frac{1}{100}\)
= ₹ 1,350
∴ Amount at the end of 1st year
= ₹ 10,800 + ₹ 1,350 (A = P + SI)
= ₹ 12,150
= Principal for 2nd year.
SI on ₹ 12,150 at \(12 \frac{1}{2} \% \)per annum
For 1 year = \(\frac{\text { PRT }}{100}\)
=’\(12,150 \times \frac{25}{2} \times \frac{1}{100}\)
= ₹ 1,518.75
∴Amount at the end of the 2nd year
= ₹ 12,150+ ₹ 1,518.75
= ₹ 13,608.75
= Principal for 3rd year
SI on ₹ 13,668.75 at 12\(\frac{1}{2}\)% per annum
For 1year =\(\frac{P R T}{100}\)
=\(13,668.75 \times \frac{25}{2} \times \frac{1}{100} \)
=\(13,668.75 \times \frac{1}{8}\)
= ₹1,708.59
Amount at the end of the year
= ₹13,668.75 + ₹ 1,708.59
=₹ 15,377.34
This is the required amount. Now, CI = ₹15,377.34 – ₹ 10,800
= ₹ 4,577.34
By using the compound interest formula
P = ₹ 10,800
R = 12\(\frac{1}{2}\) per annum
= \(\frac{25}{2}\)% per annum
n = 3 years
A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n \)
= \(10,800\left(1+\frac{25}{2 \times 100}\right)^3 \)
= \(10,800\left(1+\frac{1}{8}\right)^3\)
= \(10,800\left(\frac{9}{8}\right)^3\)
= \(10,800 \times \frac{9}{8} \times \frac{9}{8} \times \frac{9}{8}\)
= ₹ 15,377.34
2. By using year-by-year calculation
Here
P = ₹ 18000
R = ₹ 10% per annum
T = ₹ year
SI on ₹ 18,000 at 10% p.a. for 1 year
= \(\frac{\mathrm{PRT}}{100}\)
= \(\frac{18,000 \times 10 \times 1}{100}\)
= ₹ 1,800
∴ Amount at the end of the year
= ₹ 18,000 + ₹ 1,800
= ₹ 19,800
= Principal for 2nd year
SI on ₹ 19,800 at 10% p.a. for 1 year
= \(\frac{19,800 \times 10 \times 1}{100}\)
= ₹ 1,980
Amount at the end of 2nd year
= ₹ 19,800 + ₹1,980
= ₹ 21,780
= Principal for 3rd year
SI on ₹ 21,780 at 10% p.a. for \(\frac{1}{2}\)year
⇒ \(\frac{\mathrm{PRT}}{100}=\frac{21,780 \times 10 \times 1}{2 \times 100}\)
= ₹1089
∴ Amount at the end of \(2 \frac{1}{2}\) year
= ₹ 21,780 +₹1,089
= ₹ 22,869
This is the required amount.
Now, CI = ₹ 22,869 – ₹ 18,000
= ₹ 4,869
By using the compound interest formula
P = ₹ 18,000
R = 10% p.a.
n = 2 years
A = \(P\left(1+\frac{R}{100}\right)^n \)
= \(18,000\left(1+\frac{10}{100}\right)^2 \)
= \(18,000\left(1+\frac{1}{10}\right)^2 \)
= \(18,000\left(\frac{11}{10}\right)^2\)
=18,000 × \(\frac{11}{10} \times \frac{11}{10}\)
=₹ 21,780
SI on? 21,780 at 10% p.a. for \(\frac{1}{2}\) year
=\(\frac{21,780 \times 10 \times 1}{2 \times 100}\)
= ₹ 1089
Amount at the end of \(\frac{1}{2}\) years
= ₹21,780 + ₹ 1,089
= ₹ 22,869
Cl = A- P
= ₹22,869- ₹ 18,000
= ₹ 4,869
3. By using half-year by half-year calculation
Here, P = ₹ 62500
R = 8% per annum
= \(\frac{8}{2}\) per half-year
T = 1 year
∴ SI on ₹ 62,500 at 8% p.a. for 1st half year
= \(\frac{P R T}{100} \)
= \(\frac{62,500 \times 8 \times 1}{2 \times 100}=\frac{62500}{25}\)
= ₹ 2,500
∴ Amount at the end of 1st half-year
= ₹ 62,500 +₹ 2,500
= ₹65,000
= Principal for 2nd half year ₹ 65,000 at 8% p.a. for 2nd half-year
= \(\frac{P R T}{100}\)
= \(\frac{65,000 \times 8 \times 1}{2 \times 100}\)
= ₹2,600
Amount at the end of 2nd halfyear
= ₹ 65,000 + ₹ 2,600
= ₹ 67,600
= Principal for 3rd halfyear
SI on ₹ 67,600 at 8% p.a. for 3rd half-year
= \(\frac{P R T}{100} \)
⇒ \(\frac{67,600 \times 8 \times 1}{2 \times 100}\)
= ₹ 24704
∴ Amount at the end of 3rd half-year
= ₹ 67,600 + ₹ 2,704
= ₹ 70,304
This is the required amount.
Now, Cl = Amount – Principal
= ₹ 70,304 – ₹ 62,500
= ₹7,804
Cl = ₹ 2,500 + ₹ 2,600 +₹ 2,704
= ₹ 7,804
By using the compound interest 2 formula
P =₹ 62,500
= \(\frac{1}{2} \times 8 \% \text { per half year } \)
= 4% per half-year
n = \(1 \frac{1}{2} \text { year } \)
= \(1 \frac{1}{2} \times 2 \text { half years }\)
3 half years
A =\(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n \)
= \(62,500\left(1+\frac{4}{100}\right)^3 \)
= \(62,500\left(1+\frac{1}{25}\right)^3 \)
= \(62,500\left(\frac{26}{25}\right)^3 \)
= \(62,500 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25}\)
= \(4\times 26 \times 676\)
= 4 × 26 × 676
= ₹ 70,304
Cl = A- P = ₹ 70,304- ₹ 62,500
=₹ 7804.
By using the compound interest formula
P =₹ 62,500
R = 8%p.a
= \(\frac{1}{2} \times 8 \% \text { per half year }\)
1% per half-year
n =\(\frac{1}{2}\)
= 1 \(\frac{1}{2} \times 2\)
= 3 half years
A = \(P\left(1+\frac{R}{100}\right)^n\)
= \(62,500\left(1+\frac{4}{100}\right)^3 \)
= \(62,500\left(1+\frac{1}{25}\right)^3\)
= \(62,500\left(\frac{26}{25}\right)^3\)
=\(62,500 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25}\)
4. By using the compound interest formula
P = ₹ 8,000
R = 9% p.a.
= \(\frac{9}{2}\) % per half-year
= \(8,000\left(1+\frac{9}{2 \times 100}\right)^2\)
= \(8,000\left(1+\frac{9}{200}\right)^2\)
=\(8,000\left(\frac{209}{200}\right)^2 \)
= \(8,000 \times \frac{209}{200} \times \frac{209}{200} \)
\(=\frac{209 \times 209}{5}\)= 4 × 26× 676
= ₹ 70,304
Cl = A- P = ₹ 70,304 – ₹ 62,500
= ₹ 7804
By using compound interest formula
P=8000
R=9%p.a
=\(\frac{9}{2} \%\)per halfyear
n=1 year = 1 × 2 half-year
= 2 half years
∴ A =\(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n\)
= \(8,000\left(1+\frac{9}{2 \times 100}\right)^2\)
= \(8,000\left(1+\frac{9}{200}\right)^2 \)
= \(8,000\left(\frac{209}{200}\right)^2\)
= \(8,000 \times \frac{209}{200} \times \frac{209}{200} \)
=\(\frac{209 \times 209}{5}\)
= ₹8,736.20
Cl = A-P
= ₹ 8,736.20 – ₹8,000
=₹ 736.20
5. By using the compound interest formula
P = ₹ 10,000
= 8% per annum
= \(\frac{8}{2}\) per half-year
= 4 % per half-year
n = year = 1 × 2 half years
= 2 half years
= \(P\left(1+\frac{R}{100}\right)^n \)
= \(10,000\left(1+\frac{4}{100}\right)^2 \)
=\(10,000\left(1+\frac{1}{25}\right)^2 \)
= \(10,000 \times \frac{26}{25} \times \frac{26}{25}\)
16 × 26 × 26
= ₹ 10,816
∴ Cl = A-P
= ₹ 10,816 – ₹ 10,000
= ₹ 816
Question 4. Kamala borrowed ₹26,400from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of the year and 4 months to clear the loan
(Hint: Find A for 2 years, interest is compounded yearly and then find SI on the 2nd year amount for \(\frac{4}{12}\)
Solution:
P = ₹ 26,400
R = 15% p.a.
n = 2 years
A = \(P\left(1+\frac{R}{100}\right)^n \)
= \(26,400\left(1+\frac{15}{100}\right)^2 \)
= \(26,400\left(1+\frac{3}{20}\right)^2 \)
= \(26,400\left(\frac{23}{20}\right)^2\)
= \(26,400 \times \frac{23}{20} \times \frac{23}{20}\)
= \(66 \times 23 \times 23 \)
= ₹ 34,914
S.I. on X 34,914 at 15% p.a. for 4 months
=\(\text { (i.e., } \left.\frac{4}{12} \text { year, i.e., } \frac{1}{3} \text { year }\right) \)
= \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100} \)
= \(34,914 \times \frac{15}{100} \times \frac{1}{3} \)
= \(\frac{34,914 \times 15 \times 1}{3 \times 100}\)
= 1,745.70
Required amount
= ₹ 34,914 +₹ 1,745.70
= ₹ 36,659.70
Hence, the amount that Kavita will pay is ₹ 36,659.70.
NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Exercise 7.3
Question 1. The population of the place increased to 54,000 in 2003 at a rate of 5% per annum. Find the population in 2001. What, would be its population in 2005?
Solution:
1. Let the population in 2001 be P.
R = 5% p.a.
n = 2 years
A = \(P\left(1+\frac{R}{100}\right)^n=P\left(1+\frac{5}{100}\right)^2 \)
= \(P\left(1+\frac{1}{20}\right)^2=P\left(\frac{21}{20}\right)^2\)
According to the question
⇒ \(\mathrm{P}\left(\frac{21}{20}\right)^2=54,000\)
P = \(54,000\left(\frac{20}{21}\right)^2\)
= \(54,000 \times \frac{20}{21} \times \frac{20}{21} \)
= 48,980(approx.)
Hence, the population in 2001 was 48,980 (approx.)
Initial population in 2003
(P) = 54,000
R = 5% p.a.
n = 2 years (2005 – 2003 = 2)
A=\(\left(1+\frac{R}{100}\right)^n\)
= \(54,000 \times \frac{20}{21} \times \frac{20}{21}\)
= \(54,000\left(1+\frac{1}{20}\right)^2\)
= \(54,000\left(\frac{21}{20}\right)^2\)
= \(54,000 \times \frac{21}{20} \times \frac{21}{20}\)
=\(135 \times 21 \times 21=59,535\)
Hence, the population in 2005 would be 59,535. .1
Question 2. In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of2 hours, if the count was initially 5,06,000
Solution:
The initial count of bacteria
(P) = 5,06,000
R = 2.5% per hour
n = 2 hours
=\(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n \)
=\(5,06,000\left(1+\frac{2.5}{100}\right)^2 \)
= \(5,06,000\left(1+\frac{1}{40}\right)^2 \)
= \(5,06,000\left(\frac{41}{40}\right)^2 \)
= \(5,06,000 \times \frac{41}{40} \times \frac{41}{40} \)
= \(\frac{1265}{4} \times 41 \times 41\)
= 531616.25
= 5,31,616 (approx.)
Hence, the bacteria count at the end UJ of 2 hours is 5,31,616 (approx.).
Question 3. A scooter was bought at ₹42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:
Initial value of the scooter
P =₹ 42,000
R = -8% per annum
I v There is depreciation
A = \(P\left(1-\frac{R}{100}\right)^n \)
=\(42,000\left(1-\frac{8}{100}\right)^1\)
= \(42,000\left(1-\frac{2}{25}\right)\)
= \(42,000 \times \frac{23}{25}\)
= \(1680 \times 23=₹ 38,640\)
= 1680 × 23 = ₹ 38,640
Hence, its value after 1 year is ₹ 38,640.
NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Multiple Choice Questions
Question 1. The ratio of 50 paise to ₹ 1 is
- 1:2
- 2:1
- 1:1
- 1:5.
Solution: 1. 1:2
1 = 100 paise
50paise : 100 paise or 50 : 100 or 1:2
Question 2. The ratio of 10 m to 1 km is
- 1:10
- 10:1
- 1:100
- 100 :1.
Solution: 3.1:100
1 km = 1000 m
10 m: 1000 m or 10: 1000 or 1: 100
Question 3. The ratio of 10 km per hour to 30 km per hour is
- 3:1
- 1: 2
- 1: 3
- 2:1.
Solution: 3. 1: 3
10 : 30 = 1:3
Question 4. The ratio 1: 4 converted to percentage is
- 50%
- 25%
- 75%
- 4%.
Solution: 2. 25%
1: 4 = \(\frac{1}{4} \times 100 \%=25 \% \text {. }\)
Question 5. The ratio 4:25 converted to percentage is
- 8%
- 4%
- 16%
- 25%.
Solution: 3. 16%
4: 25 = \(\frac{4}{25} \times 100 \%=16 \%\)
Question 6. The fraction ^ converted to a percentage is
- 20%
- 30%
- 40%
- 50%.
Solution: 3.40%
⇒ \(\frac{2}{5}\) = \(\frac{2}{5} \times 100 \%=40 \% \text {. }\)
Question 7. The fraction\(\frac{1}{8}\) converted to percentage
- 8
- 12%
- 25%
- 8%
- 16%.
Solution: 1. 8
⇒ \(\frac{1}{8}=\frac{1}{8} \times 100 \%=12 \frac{1}{2} \%\)
Question 8. Out of 40 students in a class, 25% passed. How many students passed?
- 20
- 10
- 30
- 40.
Solution: 2.10
Out of 100, passed = 25
Out of40, passed =\(\frac{25}{100} \times 40\)
⇒ \(25 \% \text { of } 40=\frac{25}{100} \times 40\)
= 10
Question 9. Out of 100 students in a class, 30% like to watch T.V. How many students like to watch T.V.?
- 70
- 50
- 60
- 30.
Solution: 4.30
Number of students who like to watch
TV =30% of\(100=\frac{30}{100} \times 100\)
= 30
Question 10. There are 50 students in a class of which 40 are boys and the rest are girls. The ratio of the number of boys and the number of girls is
- 2: 3
- 1:5
- 4:1
- 2:5.
Solution: 3. 4:1
Number of Girls = 50 – 40 – 10
Required ratio = 40 : 10 = 4 : 1
Question 11. 40% of 50 students in a class are good at Science. How many students are not good at Science?
- 20
- 30
- 10
- 40.
Solution: 2. 30
Out of 100, good at Science = 40
Out of 50, good at Science
Number of students good at science
= \(40 \% \text { of } 50=\frac{40}{100} \times 50=20\)
Or
Number of students good at science
= 40% of 50
= \(40 \% \text { of } 50=\frac{40}{100} \times 50=20\)
Number of students not good at science
= 50-20 = 30
Question 12. Apala has ₹ 200 with her. She spent 80% amount she had. How much money is left with her?
- 10
- 20
- 30
- 40.
Solution: 4.40
Out of 100, money spent =? 80
Out of? 200, money spent
= \(\frac{80}{100} \times 200\)
Money spent = 80% of₹ 200
= \(200 \times \frac{80}{100}\)
= ₹ 160
Money left = ₹ 200 – ₹ 160
= ₹ 40.
Question 13. A toy marked at t 40 is available for 132. What percent discount is given on the marked price?
- 10%
- 20%
- 25%
- 40%.
Solution: 2.20%
Discount = ₹ 40 – ₹ 32 = ?8
Discount on ₹ 40 = ? 8
Discount on ₹ 100
⇒\(\frac{80}{40}\) x100
= ₹ 20
Question 14. Find the simple interest on? 1000 for 2 years at 8% per annum.
- ₹ 80
- ₹ 40
- ₹ 120
- ₹ 60.
Solution: 4. ₹ 60
S.I = \(\frac{1000 \times 2 \times 8}{100}\)
= ₹ 160
Question 15. A sofa-set was bought for l 10000. Its value depreciated at the rate of 10% per annum. Find its value after one year.
- ₹11000
- ₹19000
- ₹10000
- ₹1000.
Solution: 2. ₹19000
Depreciation in 1 year =\(10000 \times \frac{10}{100}\)
= 1000
Value after 1 year
= 10000 – 1000 =₹ 9000.
Question 16. There are 1275 trees in Chaudhary Farm. Out of these 36% trees are of fruits. How many trees of fruits are there in Chaudhary Farm?
- 459
- 549
- 945
- 954.
Solution: 1.459
Required number = \(\frac{1275 \times 36}{100}\)
= 459
Question 17. The quantity of protein in a particular variety of pulse is 25%. Find the amount of protein in 4 kg of pulse.
- 1 kg
- 2 kg
- 3 kg
- kg.
Solution: 1. 1 kg
Amount of protein =\(4 \times \frac{25}{100}\)
= 1kg
Question 18. In a mixture, the amount of zinc is 45%. Find the amount of zinc in the 400 g mixture.
- 60 g
- 120 g
- 180 g
- 200 g.
Solution: 3. 180 g
Amount of zinc =\(400\times \frac{45}{100}\)
= 80kg
Question 19. In a school out of 340 students, 55% of students are of Science. The remaining students are of Commerce. Find the number of students of Commerce.
- 135
- 153
- 315
- 140.
Solution: 2. 153
Number of Science students
= \(340 \times \frac{55}{100}\) = 187
Number of students of Commerce
= 340 – 187 = 153
Question 20. The salary of Manish is ₹ 10000. His salary gets increased by 10%. Find his increased salary.
- ₹ 9000
- ₹ 11000
- ₹ 8000
- ₹ 12000.
Solution: 2.₹ 11000
Increase in salary
= ₹ \(10000 \times \frac{10}{100}=₹ 1000\)
Question 21. A shopkeeper purchased 2 refrigerators for ₹ 9800 and ₹ 8200 respectively. He sold them for ₹ 16920. Find loss%.
- 2%
- 4%
- 5%
- 6%.
Solution: 4.6%
Total C.P. =₹ 9800 + ₹ 8200 = ₹18000
S.P. = ₹ 16920
Loss = ₹18000 – ₹16920 = ₹1080
=\(\frac{1080}{18000} \times 100 \%=6 \%\)
Question 22. In selling a plot of land for ₹ 61200, a profit of 20% is made. The cost price of the plot is
- ₹ 51000
- ₹ 50000
- ₹ 49000
- ₹ 52000.
Solution: 1. ₹ 51000
If S.P. is ₹120(100 + 20),
C.P. =₹ 100
If S.P. is ₹ 61200,
C.P =\(\frac{100}{120} \times 61200\)
= ₹ 51000
Question 23. The simple interest on X 2000 for 4 years is ? 400. The rate percent of interest is
1. \(\frac{2000 \times 100}{400 \times 4}\)
2. \(\frac{400 \times 4}{2000 \times 10} \)
3. \(\frac{400 \times 100}{2000 \times 4}\)
4. \( \frac{2000 \times 4}{400 \times 100} \)
Solution: 3.\(\frac{400 \times 100}{2000 \times 4}\)
⇒ \(\frac{2000 \times 4 \times \mathrm{R}}{100}=400\)
R= \(\frac{400 \times 100}{2000 \times 4}\)
Question 24. The simple interest of ₹ 500 at the rate of 5% is ₹ 100. This interest is of the time
- 1 year
- 4 years
- 10 years
- 20 years.
Solution: 2. 4 years
⇒ \(\frac{500 \times 5 \times \mathrm{T}}{100}=100\)
⇒ \(\mathrm{T}=\frac{100 \times 100}{500 \times 5}=4\)
Question 25. The S.I. of ₹ 100 for 1 year at the rate of 3 paise per rupee per month is
- ₹ 30
- ₹ 36
- ₹ 24
- ₹ 8.
Solution: 2.₹ 36
S.I. = 100 × 1 × (3 × 12)
= 3600 paise
= ₹ 36
NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities True-False
1. VAT is always calculated on the selling price – True
2. A machinery worth .₹ P is depreciated by R% per annum. Its value after 1 year will be \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)\)– False
3. Comparison of two quantities by division is called ratio-True
NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Fill In the Blanks
1. The overhead changes are added to → CP
2. If the conversion period is not specified, then it is taken as → Oneyear
3. If a discount of ₹ 4y is available on the market price of ₹ 8y, then the discount percent – 50%
4. If — % of a number is 154, then the number is → 4900
5. Find 10% of [100- 20% of 300] → 4
6. Explain 0.1234 as a percentage → 12.34
7 A jacket work for? 4000 is offered for sale at? 3600. What per cent discount is offered during the sale → 10%
8. Apala got 100 marks out of 200 and Meenu got 120 works out of 300. Whose performance is better → Apalas.