NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities

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NCERT SOLUTIONS FOR CLASS 8 Maths Chapter 8

Comparing Quantities

NCERT SOLUTIONS FOR CLASS 8 Maths Chapter 8
ncert class 8 maths solutions pdf
Ex 8.1

1. Find the ratio of the following.

(a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
(b) 5 m to 10 km (c) 50 paise to ₹ 5

Answer – 
(a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
Sol – Speed of a cycle : Speed of Scooter
= 15 : 30
= 1 : 2

(b) 5 m to 10 km
Sol – By converting km into m
∵ 1 km = 1000 m
∴ 10 km = 10 × 1000
= 10000 m
So, the ratio is
5 : 10000
= 1 : 2000

(c) 50 paise to ₹ 5
Sol – By converting ₹ in to paise –
∵ 1 ₹ = 100 Paise
∴ 5 ₹ = 5 × 100
= 500 Paise

So, the ratio is 
50 : 500
= 1 : 10

2. Convert the following ratios to percentages.

(a) 3 : 4 (b) 2 : 3

Answer – 
(a) 3 : 4
Sol – Multiplying by 100 to convert to percentage –
= \(\displaystyle \frac{3}{4}\)
= 3 × 25
= 75%

(b) 2 : 3
Sol – Multiplying by 100 to convert to percentage –
= \(\displaystyle \frac{2}{3}\times 100\)
\(\displaystyle \begin{array}{l}=\frac{{200}}{3}\\=66\frac{2}{3}\end{array}\)

3. 72% of 25 students are interested in mathematics. How many are not interested in mathematics?
Sol – Number of total students = 25
Students who interested in mathematics = 72%
Percentage of students who are not good in mathematics = 100 – 72 = 28%
So, number of students who are not good in mathematics = \(\displaystyle \frac{{25\times 28}}{{100}}\)
\(\displaystyle \begin{array}{l}=\frac{{28}}{4}\\=7\end{array}\)
Thus, 7 students are not good in mathematics

4. A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?
Sol – Let the total number of matches played by the team be x.
Winning percentage of the team was 40%.
Number of winning match = 10 
Therefore, 
40% of x = 10
\(\displaystyle x\times \frac{{40}}{{100}}=10\)
40x = 1000 (cross multiplication)
\(\displaystyle x=\frac{{1000}}{{40}}\)
𝑥 = 25 matches

5. If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?
Sol – Let’s total money Chameli had = x
Percentage of spending money = 75%
So, 25% of x = 600
\(\displaystyle x\times \frac{{25}}{{100}}=600\)
25𝑥 = 60000
\(\displaystyle x=\frac{{60000}}{{25}}\)𝑥 = 2400 ₹.

6. If 60% people in a city like cricket, 30% like football and the remaining like other games, then what per cent 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.
Sol – Percentage who like cricket = 60%
Percentage who like football = 30%
Percentage who like other games = 100 – (60 + 30)
= 100 – 90
= 10%
If total number of people = 50 lakh
Then, number of people who like cricket = \(\displaystyle \frac{{50L\times 60}}{{100}}\)
= 5 × 6
= 30 lakh
number of people who like football = \(\displaystyle \frac{{50L\times 30}}{{100}}\)
= 5 × 3
= 15 lakh
number of people who like other games = \(\displaystyle \frac{{50L\times 10}}{{100}}\)
= 5 lakh

NCERT SOLUTIONS FOR CLASS 8 Maths Chapter 8
ncert class 8 maths solutions pdf
Ex 8.2

1. A man got a 10% increase in his salary. If his new salary is ₹ 1,54,000, find his original salary.
Sol – Let’s original salary = 100 ₹
Increasing 10% or 10 ₹
Then new salary = 100 + 10 = 110 ₹
∵ New salary is 110 ₹ then original salary is = 100 ₹
∴ New salary is 1 ₹ then original salary is = \(\displaystyle \frac{{100}}{{110}}\)
∴ New salary is 1,54,000 ₹ then original salary is = \(\displaystyle \frac{{100}}{{110}}\times 15400\)
= 100 × 1400
= 140000 ₹

2. On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the per cent decrease in the people visiting the Zoo on Monday?
Sol – Number of people who went zoo on Sunday = 845
Number of people who went zoo on Monday = 169
decrease = 845 – 169
= 676
Percentage of decrease = \(\displaystyle \frac{{676}}{{845}}\times 100\)= 80%

3. A shopkeeper buys 80 articles for ₹ 2,400 and sells them for a profit of 16%. Find the selling price of one article.
Sol –
∵ Cost price of 80 articles is = 2400 ₹
∴ Cost price of 1 articles is = \(\displaystyle \frac{{2400}}{{80}}\)= 30 ₹
Let’s cost price of a article = 100 ₹
Profit 16% or 16 ₹
then, selling price = 100 + 16 = 116 ₹
∵ cost price 100 ₹ then selling price is = 116 ₹
∴ cost price 1 ₹ then selling price is = \(\displaystyle \frac{{116}}{{100}}\)∴ cost price 30 ₹ then selling price is = \(\displaystyle \frac{{116}}{{100}}\times 30\)= \(\displaystyle \frac{{3480}}{{100}}\)= 34.80 ₹

4. The cost of an article was ₹ 15,500. ₹ 450 were spent on its repai₹. If it is sold for a profit of 15%, find the selling price of the article.
Sol – Item purchased = ₹ 15,500
Repair cost = ₹ 450
Total cost price of article = 15500 + 450
= ₹ 15950
Let the cost price of the article = ₹ 100
Profit = 15% or ₹ 15
So selling price = 100 + 15 = 115 ₹
∵ Cost price ₹ 100 then selling price = ₹ 115
∴ Cost price ₹ 1 then selling price = \(\displaystyle \frac{{115}}{{100}}\)∴ Cost price ₹ 15950 then selling price = \(\displaystyle \frac{{115}}{{100}}\times 15950\)

= \(\displaystyle \frac{{115}}{{10}}\times 1595\)
= \(\displaystyle \frac{{183425}}{{10}}\)
= 18342.50 ₹

5. A VCR and TV were bought for ₹ 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.
Sol – Cost price of VCR = ₹.8000
Let cost price of VCR = ₹.100
Loss loss = 4% or ₹ 4
So selling price = 100 – 4 = ₹ 96
∵ Cost price of VCR is ₹ 100 then selling price = ₹ 96
∴ Cost price of VCR is ₹ 1 then selling price = \(\displaystyle \frac{{96}}{{100}}\)
∴ Cost price of VCR is ₹ 8000 then selling price = \(\displaystyle \frac{{96}}{{100}}\times 8000\)
= 96 × 80
= ₹ 7680
Similarly, cost price of TV = ₹ 8000
Let the cost price of TV = ₹ 100
Profit = 8% or ₹.8
So selling price = 100 + 8 = ₹ 108
∵ Cost price ₹ 100 then selling price = ₹ 108
∴ Cost price ₹ 1 then selling price = \(\displaystyle \frac{{108}}{{100}}\)
∴ Cost price ₹ 8000 then selling price = \(\displaystyle \frac{{108}}{{100}}\times 8000\)
= 108 × 80
= ₹ 8640
Total cost price = cost price of VCR, cost price of TV
= 8000 + 8000
= ₹ 16000
Total selling price = cost price of VCR, cost price of TV
= 7680 + 8640
= ₹ 16320
Total Profit = Total Selling Price – Total Cost Price
= 16320 – 16000
= ₹ 320
Total profit percentage = (Total profit / Total cost price) × 100
= \(\displaystyle \frac{{320}}{{16000}}\times 100\)
= \(\displaystyle \frac{{320}}{{160}}\)
= 2% profit

6. During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at ₹ 1450 and two shirts marked at ₹ 850 each?
Sol – Price of a Jeans = 1450 ₹ 
Price of two shirts = 2 × 850 = 1700 ₹
Total cost price of one jeans and two shirts = 1450 + 1700 = ₹ 3150
Discount percentage = 10%
Discount value = \(\displaystyle \frac{{3150\times 10}}{{100}}\)
= ₹ 315
Hence, after deducting the discount, the value of the articles = 3150 – 315
= ₹ 2835
So the customer has to pay ₹ 2835.

7. A milkman sold two of his buffaloes for ₹ 20,000 each. On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. (Hint: Find CP of each)
Sol – Let the cost price of the first buffalo = ₹ 100
Profit = 5% or ₹ 5
So selling price = 100 + 5 = ₹ 105
So by the unitary rule –
∵ Selling price is ₹ 105 then cost price = ₹ 100
∴ Selling price ₹ 1 then cost price = \(\displaystyle \frac{{100}}{{105}}\)
∴ Selling price ₹ 20,000 then cost price = \(\displaystyle \frac{{100}}{{105}}\times 20000\)
\(\displaystyle \begin{array}{l}=\frac{{20}}{{21}}\times 20000\\=\frac{{400000}}{{21}}\end{array}\)
= ₹ 19047.61
Similary,
Let cost price of second buffalo = 100
Loss = 10% or ₹ 10
So selling price = 100 – 10 = ₹ 90
So by the unitary rule –
∵ Selling price ₹ 90 then cost price = ₹ 100
∴ Selling price ₹ 1 then cost price =\(\displaystyle \frac{{100}}{{90}}\)
∴ Selling price ₹ 20,000 then cost price = \(\displaystyle \frac{{100}}{{90}}\times 20000\)
\(\displaystyle \begin{array}{l}=\frac{{10}}{9}\times 20000\\=\frac{{200000}}{{21}}\end{array}\)
= ₹ 22,222.22
So total selling price = 20,000 + 20,000 = ₹ 40,000
Total cost price = cost price of first buffalo + cost price of second buffalo
= 19,047.61 + 22,222.22 = ₹ 41,269.83
Since the cost price is more than the selling price, the total will be on the deal.
Loss = Cost Price – Selling Price
= 41,269.83 – 40,000 = ₹ 1269.83

8. The price of a TV is ₹ 13,000. 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.
Sol – Cost of television = ₹ 13,000
Sales Tax = 12%
13,000 sales tax = \(\displaystyle \frac{{13000\times 12}}{{100}}\)
= 130 × 12
= ₹ 1560
Total cost price of television = cost of television sales tax
= 13,000 + 1560
=  ₹ 14,560
Hence Vinod will buy that television for ₹ 14,560.

9. Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is ₹ 1,600, find the marked price.
Sol – Let the marked price of the shoe = ₹ 100
Discount = 20% or ₹ 20
Hence cost price = 100 – 20 = ₹ 80
By unitary rule –
∵ Cost price ₹ 80 then marked price = ₹ 100
∴ Cost price ₹ 1 then marked price = \(\displaystyle \frac{{100}}{{80}}\)
∴ Cost price ₹ 1600 then marked price = \(\displaystyle \frac{{100}}{{80}}\times 1600\)
= 100 × 20
= ₹ 2000
Hence the marked price of the shoe will be ₹ 2000.

10. I purchased a hair-dryer for ₹ 5,400 including 8% VAT. Find the price before VAT as added.
Sol – Let cost of hair dryer = ₹ 100
VAT = 8% or ₹ 8
So cost price = 100 + 8 = ₹ 108
By unitary rule –
∵ Cost price of hair dryer is ₹ 108  then earlier price = ₹ 100
∴ Cost price of hair dryer is ₹ 1 then earlier price = \(\displaystyle \frac{{100}}{{108}}\)
∴ Cost price of hair dryer is ₹ 5400 then earlier price = \(\displaystyle \frac{{100}}{{108}}\times 5400\)
= 100 × 50
= ₹ 5000
Hence, the earlier cost of hair dryer = ₹ 5000.

11. An article was purchased for ₹ 1239 including GST of 18%. Find the price of the article before GST was added?
Sol – Assuming GST Cost price of article before ₹ 100
GST = 18% or ₹ 18
Hence, later cost price of article = 100 18 = ₹ 118
By unitary rule –
Later price is ₹ 118 then first price = ₹ 100
The latter value is ₹ 1 then the first = \(\displaystyle \frac{{100}}{{118}}\)
Subsequent value ₹ 1239 then price = \(\displaystyle \frac{{100}}{{108}}\times 1239\)
= \(\displaystyle \frac{{123900}}{{118}}\)
= ₹ 1050

NCERT SOLUTIONS FOR CLASS 8 Maths Chapter 8
ncert class 8 maths solutions pdf
Ex 8.3

1. Calculate the amount and compound interest on
(a) ₹ 10,800 for 3 years at \(\displaystyle 12\frac{1}{2}\)% per annum compounded annually.
(b) ₹ 18,000 for \(\displaystyle 2\frac{1}{2}\) years at 10% per annum compounded annually.
(c) ₹ 62,500 for \(\displaystyle 1\frac{1}{2}\) years at 8% per annum compounded half yearly.
(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly.
(You could use the year by year calculation using SI formula to verify).
(e) ₹ 10,000 for 1 year at 8% per annum compounded half yearly.
Answer – 
(a) ₹ 10,800 for 3 years at \(\displaystyle 12\frac{1}{2}\)% per annum compounded annually.
Sol – \(\displaystyle A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\)
where A (Amount) is to be found.
P (Principal) = ₹ 10,800
R (Rate) = \(\displaystyle 12\frac{1}{2}\)%
T (Time) = 3 years
\(\displaystyle \begin{array}{l}A=10800{{\left( {1+\frac{{25}}{2}\times \frac{1}{{100}}} \right)}^{3}}\\A=10800{{\left( {1+\frac{1}{8}} \right)}^{3}}\\A=10800{{\left( {\frac{{8+1}}{8}} \right)}^{3}}\\A=10800{{\left( {\frac{9}{8}} \right)}^{3}}\\A=10800\times \frac{9}{8}\times \frac{9}{8}\times \frac{9}{8}\\A=1350\times 9\times \frac{9}{8}\times \frac{9}{8}\\A=\frac{{984150}}{{64}}\end{array}\)
A (Amaunt) = RU 15377.34
Compound Interest (CI) = Amount (A) – Principal (P)
CI = 15377.34 – 10800
CI = ₹ 4577.34

(b) ₹ 18,000 for \(\displaystyle 2\frac{1}{2}\) years at 10% per annum compounded annually.
Sol – \(\displaystyle A=P{{\left( {1 \frac{R}{{100}}} \right)}^{T}}\)
where A (Amount) is to be found.
P (Principal) = ₹ 18,000
R (Rate) = 10% p.a.
T (Time) = \(\displaystyle 2\frac{1}{2}\) year
\(\displaystyle \begin{array}{l}A=18000{{\left( {1+\frac{{10}}{{100}}} \right)}^{{2\frac{1}{2}}}}\\A=18000{{\left( {\frac{{110}}{{100}}} \right)}^{{2\frac{1}{2}}}}\\A=18000\times \frac{{110}}{{100}}\times \frac{{110}}{{100}}\times \frac{{105}}{{100}}\end{array}\)
A = 9 × 11 × 11 × 21
A (Amount) = ₹ 22869
Compound Interest (CI) = Amount (A) – Principal (P)
CI = 22869 – 18000
CI = ₹ 4869

(c) ₹ 62,500 for \(\displaystyle 1\frac{1}{2}\) years at 8% per annum compounded half yearly.
Sol – Since the rate is compounded half yearly, the time will be doubled and the rate will be halved.
\(\displaystyle A=P{{\left( {1 \frac{R}{{100}}} \right)}^{T}}\)
where A (Amount) is to be found.
P (Principal) = ₹ 62,500
R (rate) = 10% half-yearly (rate will become 4%)
T (time) = \(\displaystyle 1\frac{1}{2}\) year (time will be 3 years)
\(\displaystyle \begin{array}{l}A=62500{{\left( {1+\frac{4}{{100}}} \right)}^{3}}\\A=62500{{\left( {\frac{{104}}{{100}}} \right)}^{3}}\\A=62500\times \frac{{104}}{{100}}\times \frac{{104}}{{100}}\times \frac{{104}}{{100}}\end{array}\)
A = 4 × 26 × 26 × 26
A = 70304
A (Amount) = ₹ 70304
Compound Interest (CI) = Amount (A) – Principal (P)
CI = 70304 – 62500
CI = ₹ 7804

(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly.
(You could use the year by year calculation using SI formula to verify).
Sol – Since the rate is compounded half yearly, the time will be doubled and the rate will be halved.
\(\displaystyle A=P{{\left( {1 \frac{R}{{100}}} \right)}^{T}}\)
where A (mixture) is to be found.
P (Principal) = ₹ 8,000
R (rate) = 9% half-yearly (rate will become 4.5%)
T (time) = 1 year (time will become 2 years)
\(\displaystyle \begin{array}{l}A=8000{{\left( {1+\frac{{4.5}}{{100}}} \right)}^{2}}\\A=8000{{\left( {\frac{{104.5}}{{100}}} \right)}^{2}}\\A=8000\times \frac{{104.5}}{{100}}\times \frac{{104.5}}{{100}}\times \frac{{104.5}}{{100}}\end{array}\)
A = 8736.20
A (Amount) = ₹ 8736.20
Compound Interest (CI) = Amount (A) – Principal (P)
CI = 8736.20 – 8000
CI = ₹ 736.20

(e) ₹ 10,000 for 1 year at 8% per annum compounded half yearly.
Sol – Since the rate is compounded half yearly, the time will be doubled and the rate will be halved.
\(\displaystyle A=P{{\left( {1 \frac{R}{{100}}} \right)}^{T}}\)
where A (Amount) is to be found.
P (Principal) = ₹ 10,000
R (Rate) = 8% half-yearly (rate will become 4%)
T (Time) = 1 year (time will become 2 years)
\(\displaystyle \begin{array}{l}A=10000{{\left( {1+\frac{4}{{100}}} \right)}^{2}}\\A=10000{{\left( {\frac{{100+4}}{{100}}} \right)}^{2}}\\A=10000{{\left( {\frac{{104}}{{100}}} \right)}^{2}}\\A=10000\times \frac{{104}}{{100}}\times \frac{{104}}{{100}}\end{array}\)
A = 104 × 104
A = 10816
A (Principal) = 10816
Compound Interest (CI) = Amount (A) – Principal (P)
CI = 10816 – 10000
CI = ₹ 816

2. Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for \(\displaystyle \frac{4}{{12}}\) years).
Sol – \(\displaystyle A=P{{\left( {1 \frac{R}{{100}}} \right)}^{T}}\)
where A (Amount) is to be found.
P (Principal) = ₹ 26,400
R (Rate) = 15%
T (Time) = 2 years 4 months or \(\displaystyle 2\frac{1}{4}\) years
\(\displaystyle \begin{array}{l}A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\\A=26400{{\left( {1+\frac{{15}}{{100}}} \right)}^{{2\frac{1}{4}}}}\\A=26400{{\left( {\frac{{100+15}}{{100}}} \right)}^{{2\frac{1}{4}}}}\\A=26400{{\left( {\frac{{115}}{{100}}} \right)}^{{2\frac{1}{4}}}}\\A=26400\times \frac{{115}}{{100}}\times \frac{{115}}{{100}}\times \frac{{105}}{{100}}\\A=\frac{{33\times 23\times 23\times 21}}{{10}}\end{array}\)
A = 36659.70
A (Amount) = ₹ 36659.70

3. Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
Sol – \(\displaystyle SI=\frac{{PRT}}{{100}}\)
\(\displaystyle S.I.=\frac{{PRT}}{{100}}\)
\(\displaystyle S.I.=\frac{{12500\times 12\times 3}}{{100}}\)
S.I. = 125 × 12 × 3
S.I = ₹ 4500  
Amount borrowed by Radha = Principal = Rs 12,500
Rate = 10%
Time = 3 years
\(\displaystyle CI=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}-P\)
\(\displaystyle CI=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}-P\)
4500 – 4137.50
\(\displaystyle \begin{array}{l}CI=12500{{\left( {1+\frac{{10}}{{100}}} \right)}^{3}}-12500\\CI=12500{{\left( {\frac{{100+10}}{{100}}} \right)}^{3}}-12500\\CI=12500{{\left( {\frac{{110}}{{100}}} \right)}^{3}}-12500\\CI=12500{{\left( {\frac{{11}}{{10}}} \right)}^{3}}-12500\\CI=12500\times \frac{{11}}{{10}}\times \frac{{11}}{{10}}\times \frac{{11}}{{10}}-12500\\CI=25\times \frac{{1331}}{2}-12500\\CI=\frac{{33275}}{2}-12500\end{array}\)
CI = 16637.50 – 12500
CI = 4137.50
So Fabina had to pay more interest.
Difference in interest = 4500 – 4137.50
= ₹ 362.50 more.

4. I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?
Sol – Principal = ₹ 12,000
Time = 2 years
Rate = 6% p.a.
When money is borrowed at simple interest –
\(\displaystyle SI=\frac{{PRT}}{{100}}\)\(\displaystyle S.I.=\frac{{PRT}}{{100}}\)\(\displaystyle SI=\frac{{12000\times 2\times 6}}{{100}}\)SI = 120 × 2 × 6
SI = ₹ 1440 
When money is borrowed at compound interest –
\(\displaystyle CI=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}-P\)\(\displaystyle \begin{array}{l}CI=12000{{\left( {1+\frac{6}{{100}}} \right)}^{2}}-12000\\CI=12000{{\left( {\frac{{100+6}}{{100}}} \right)}^{2}}-12000\\CI=12000{{\left( {\frac{{106}}{{100}}} \right)}^{2}}-12000\\CI=12000\times \frac{{106}}{{100}}\times \frac{{106}}{{100}}-12000\\CI=12\times \frac{{106\times 106}}{{10}}-12000\\CI=\frac{{134832}}{{10}}-12000\end{array}\)CI = 13483.20 – 12000
CI = ₹ 1483.20
Hence the difference between the two interest on the amount taken by Jamshed = 1483.20 – 1440
= ₹ 43.20

5. Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months?
(ii) after 1 year?
Sol – Principal = ₹ 60,000
Rate = 12% (half yearly)
When the rate is compounded half yearly, the rate is halved and the time is doubled.

(i) So when the time is 6 months,
Principal = ₹ 60,000
rate = 6%
time = 1 year
\(\displaystyle A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\)\(\displaystyle \begin{array}{l}A=60000{{\left( {1+\frac{6}{{100}}} \right)}^{1}}\\A=60000\left( {\frac{{100+6}}{{100}}} \right)\\A=60000\times \frac{{106}}{{100}}\end{array}\)A = 600 × 106
A = ₹ 63600  
So when the time is 6 months, then the amount = ₹ 63,600.

(ii) So when the time is 1 year,
Principal = ₹ 60,000
rate = 6%
time = 2 years
\(\displaystyle \begin{array}{l}A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\\A=60000{{\left( {1+\frac{6}{{100}}} \right)}^{2}}\\A=60000{{\left( {\frac{{100+6}}{{100}}} \right)}^{2}}\\A=60000\times {{\left( {\frac{{106}}{{100}}} \right)}^{2}}\\A=60000\times \frac{{106}}{{100}}\times \frac{{106}}{{100}}\end{array}\)A = 6 × 106 × 106
A = 67,416
So when the time is 1 year, then the amount = ₹ 67,416.

6. Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after \(\displaystyle 1\frac{1}{2}\) years if the interest is
(i) compounded annually.
(ii) compounded half yearly.
Sol –
(i) compounded annually

Principal = ₹ 80,000
Rate of interest = 10%
time = \(\displaystyle 1\frac{1}{2}\)\(\displaystyle A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\)\(\displaystyle \begin{array}{l}A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\\A=80000{{\left( {1+\frac{{10}}{{100}}} \right)}^{{1\frac{1}{2}}}}\\A=80000{{\left( {\frac{{100+10}}{{100}}} \right)}^{{1\frac{1}{2}}}}\\A=80000\times {{\left( {\frac{{110}}{{100}}} \right)}^{{1\frac{1}{2}}}}\\A=80000\times \frac{{110}}{{100}}\times \frac{{105}}{{100}}\end{array}\)\(\displaystyle \begin{array}{l}A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\\A=80000{{\left( {1+\frac{{10}}{{100}}} \right)}^{{1\frac{1}{2}}}}\\A=80000{{\left( {\frac{{100+10}}{{100}}} \right)}^{{1\frac{1}{2}}}}\\A=80000\times {{\left( {\frac{{110}}{{100}}} \right)}^{{1\frac{1}{2}}}}\\A=80000\times \frac{{110}}{{100}}\times \frac{{105}}{{100}}\end{array}\)

(ii) is compounded half yearly.
Principal = ₹ 80,000
(When the rate is half yearly, the time is doubled and the rate is halved.)
Rate of interest = 5%
time = 3 years
\(\displaystyle \begin{array}{l}A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\\A=80000{{\left( {1+\frac{5}{{100}}} \right)}^{3}}\\A=80000{{\left( {\frac{{100+5}}{{100}}} \right)}^{3}}\\A=80000\times {{\left( {\frac{{105}}{{100}}} \right)}^{3}}\\A=80000\times \frac{{105}}{{100}}\times \frac{{105}}{{100}}\times \frac{{105}}{{100}}\end{array}\)A = 2 × 21 × 21 × 105
A = Rs 92610

7. Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the 3rd year.
Sol –
(i) Find the amount deposited in his name at the end of two years.

Principal = Rs 8000
Rate = 5% p.a.
time = 2 years
\(\displaystyle A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\)
\(\displaystyle \begin{array}{l}A=8000{{\left( {1+\frac{5}{{100}}} \right)}^{2}}\\A=8000{{\left( {\frac{{100+5}}{{100}}} \right)}^{2}}\\A=8000\times {{\left( {\frac{{105}}{{100}}} \right)}^{2}}\\A=8000\times \frac{{105}}{{100}}\times \frac{{105}}{{100}}\\A=8\times 105\times \frac{{105}}{{10}}\end{array}\)
A = 4 × 21 × 105
A = Rs 8820 

(ii) Find the interest for the third year.
\(\displaystyle \begin{array}{l}A=8000{{\left( {1 \frac{5}{{100}}} \right)}^{3}}\\A=8000{{\ left( {\frac{{100 5}}{{100}}} \right)}^{3}}\\A=8000\times {{\left( {\frac{{105}}{{100} }} \right)}^{3}}\\A=8000\times \frac{{105}}{{100}}\times \frac{{105}}{{100}}\times \frac{{ 105}}{{100}}\\A=\frac{{8\times 105\times 105\times 105}}{{1000}}\end{array}\)
A = 21 × 21 × 21
A = Rs 9261
Hence the difference in compound interest for 3 years and 2 years = 9261 – 8820
= Rs 441

8. Find the amount and the compound interest on ₹ 10,000 for \(\displaystyle 1\frac{1}{2}\) years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?
Sol – (When the rate is half yearly, the time is doubled and the rate is halved.)
Principal = ₹ 10,000
time = 3 years
rate = 5%
\(\displaystyle A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\)
\(\displaystyle \begin{array}{l}A=10000{{\left( {1+\frac{5}{{100}}} \right)}^{3}}\\A=10000{{\left( {\frac{{100+5}}{{100}}} \right)}^{3}}\\A=10000\times {{\left( {\frac{{105}}{{100}}} \right)}^{3}}\\A=10000\times \frac{{105}}{{100}}\times \frac{{105}}{{100}}\times \frac{{105}}{{100}}\\A=\frac{{105\times 105\times 105}}{{100}}\\A=\frac{{21\times 21\times 105}}{4}\\A=\frac{{46305}}{4}\end{array}\)
A = ₹ 11576.25
CI = A – P
CI = 11576.25 – 10000
CI = ₹ 1576.25 

If the principal is drawn at the rate of interest per annum then –
Principal = ₹ 10,000
rate = \(\displaystyle 1\frac{1}{2}\) year
time = 10%
\(\displaystyle \begin{array}{l}A=10000{{\left( {1+\frac{{10}}{{100}}} \right)}^{{1\frac{1}{2}}}}\\A=10000{{\left( {\frac{{100+10}}{{100}}} \right)}^{{1\frac{1}{2}}}}\\A=10000\times {{\left( {\frac{{110}}{{100}}} \right)}^{{1\frac{1}{2}}}}\\A=10000\times \frac{{110}}{{100}}\times \frac{{105}}{{100}}\end{array}\)
A = 110 × 105
A = ₹ 11550
Yes. The half yearly rate of interest is higher than the annual rate of interest.

9. Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at 2 % per annum, interest being compounded half yearly.
Sol – Principal = ₹ 4096
Since when the rate is half yearly, the time is doubled and the rate is halved.
So time = 18 months or \(\displaystyle 1\frac{1}{2}\) = 3 years
rate = \(\displaystyle 12\frac{1}{2}\%\) = \(\displaystyle 1\frac{25}{4}\) %
\(\displaystyle A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\)
\(\displaystyle \begin{array}{l}A=4096{{\left( {1+\frac{{25}}{4}\times \frac{1}{{100}}} \right)}^{3}}\\A=4096{{\left( {1+\frac{1}{{16}}} \right)}^{3}}\\A=4096{{\left( {\frac{{16+1}}{{16}}} \right)}^{3}}\\A=4096{{\left( {\frac{{17}}{{16}}} \right)}^{3}}\\A=4096\times \frac{{17}}{{16}}\times \frac{{17}}{{16}}\times \frac{{17}}{{16}}\end{array}\)
A = 17 × 17 × 17
A = ₹ 4913  

10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) find the population in 2001.
(ii) what would be its population in 2005?
Sol –
(i) Population in the year 2001
Population of 2003 (post population) = 54,000
rate = 5%
time = 2 years
2001 population = x
\(\displaystyle A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\)
\(\displaystyle \begin{array}{l}54000=x{{\left( {1+\frac{5}{{100}}} \right)}^{2}}\\54000=x{{\left( {\frac{{100+5}}{{100}}} \right)}^{2}}\\54000=x{{\left( {\frac{{105}}{{100}}} \right)}^{2}}\\54000=x{{\left( {\frac{{21}}{{20}}} \right)}^{2}}\\54000=x\times \frac{{21}}{{20}}\times \frac{{21}}{{20}}\end{array}\)
x × 21 × 21 = 54000 × 20 × 20  (By cross multiplying)
\(\displaystyle \begin{array}{l}x=\frac{{54000\times 20\times 20}}{{21\times 21}}\\\x=\frac{{6000\times 20\times 20}\\x=\frac{{6000\times 20\times 20 }}{{7\times 7}}\\x=\frac{{2400000}}{{49}}\end{array}\)
x = 48,980 (approx)

(ii) What will be the population in the year 2005?
Population of 2003 (East population) = 54,000
rate = 5%
time = 2 years
Population of 2005 = x (post population)
\(\displaystyle A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\)
\(\displaystyle \begin{array}{l}x=54000{{\left( {1+\frac{5}{{100}}} \right)}^{2}}\\x=54000{{\left( {\frac{{100+5}}{{100}}} \right)}^{2}}\\x=54000{{\left( {\frac{{105}}{{100}}} \right)}^{2}}\\x=54000\times \frac{{105}}{{100}}\times \frac{{105}}{{100}}\end{array}\)
x = 27 × 21 × 105
x = 59535

11. 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 of 2 hours if the count was initially 5, 06,000.
Sol – Prior number of bacteria = 5,06,000
time = 2 hours
rate = 2.5%
Subsequent number = x
\(\displaystyle A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\)
\(\displaystyle \begin{array}{l}x=506000{{\left( {1+\frac{{2.5}}{{100}}} \right)}^{2}}\\x=506000{{\left( {\frac{{100+2.5}}{{100}}} \right)}^{2}}\\x=506000{{\left( {\frac{{102.5}}{{100}}} \right)}^{2}}\\x=506000\times \frac{{102.5}}{{100}}\times \frac{{102.5}}{{100}}\\x=506\times 102.5\times \frac{{102.5}}{{10}}\\x=\frac{{5316162.5}}{{10}}\end{array}\)
x = 531616 (Aproximataly)

12. A scooter was bought at Rs 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Sol – Cost of scooter = Rs.42,000
Rate = – 8% (Since the annual rate is decreasing, there will be a negative sign)
time = 1 year
\(\displaystyle A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}\)
\(\displaystyle \begin{array}{l}x=42000{{\left( {1-\frac{8}{{100}}} \right)}^{1}}\\x=42000{{\left( {\frac{{100-8}}{{100}}} \right)}^{1}}\\x=42000{{\left( {\frac{{92}}{{100}}} \right)}^{1}}\\x=42000\times \frac{{92}}{{100}}\end{array}\)
x = 420 × 92
x = 38,640

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