NCERT Solutions for Class 8 Maths Part 2 Chapter 1 - Fractions in Disguise

Class 8 Maths Part 2 Chapter 1 Solutions PDF Free Download - Ganita Prakash Book 2

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NCERT Solutions for Class 8 Maths Part 2 Chapter 1 - Exercise Questions and Answers

Here are the NCERT Solutions for Class 8 Maths Ganita Prakash Part 2 Chapter 1 Fractions in Disguise question answers with clear and detailed solutions.

Question 1. Express the following fractions as percentages.
(i) $\frac{3}{5}$
(ii) $\frac{7}{14}$
(iii) $\frac{9}{20}$
(iv) $\frac{72}{150}$
(v) $\frac{1}{3}$
(vi) $\frac{5}{11}$

$\textbf{Answer:}$

1 (i): 60%
$\textbf{Explanation:}$
$\frac{3}{5}\times100=60\%$

1 (ii): 50%
$\textbf{Explanation:}$
$\frac{7}{14}\times100=50\%$

1(iii): 45%
$\textbf{Explanation:}$
$\frac{9}{20}\times100=45\%$

1 (iv): 48%
$\textbf{Explanation:}$
$\frac{72}{150}\times100=48\%$
1 (v): 33.33%
Explanation:
$\frac{1}{3}\times100=33.33\%$

1 (vi): 45.45%
Explanation:
$\frac{5}{11}\times100=45.45\%$

Question 2. Nandini has 25 marbles, of which 15 are white. What percentage of her marbles are white?
(i) $10 \%$
(ii) $15 \%$
(iii) $25 \%$
(iv) $60 \%$
(v) $40 \%$
(vi) None of these

$\textbf{Answer:}$ Option (iv)

$\textbf{Explanation:}$

Nandini has 25 marbles, of which 15 are white.
$\therefore$ Required percentage $=\frac{15}{25}\times100=60\%$

Hence, the correct answer is option (iv).

Question 3. In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking?

$\textbf{Answer:}$ 18.75%

$\textbf{Explanation:}$

In a school, 15 of the 80 students come to school by walking.
$\therefore$ Required percentage $=\frac{15}{80}\times100=18.75\%$

Question 4. A group of friends is participating in a long-distance run. The positions of each of them after 15 minutes are shown in the following picture. Match (among the given options) what percentage of the race each of them has approximately completed.

$\textbf{Answer:}$

A is about one-third of the way → approximately 38%

B is a little more than halfway → approximately 55%

C is about three-fourths of the way → approximately 72%

D is very close to the finish → approximately 93%

Question 5: Pairs of quantities are shown below. Identify and write appropriate symbols ' $>$ ', ' $<$ ', ' $=$ ' in the blanks. Try to do it without calculations.

(i) $50 \%$ $\_\_\_\_ 5\%$

(ii) $\frac{5}{10} \_\_\_\_50 \%$

(iii) $\frac{3}{11}$ $\_\_\_\_ 61\%$

(iv) $30 \%$ $\_\_\_\_$ $\frac{1}{3}$

$\textbf{Answer:}$

5 (i): >

$\textbf{Explanation:}$

$50\%=\frac{50}{100}=0.5$
$5\%=\frac5{100}=0.05$
So, $50\%>5\%$

5 (ii): =

$\textbf{Explanation:}$

$\frac5{10}=0.5$
$50\%=\frac{50}{100}=0.5$
So, $\frac5{10}=50\%$

5 (iii): <

$\textbf{Explanation:}$

$\frac3{11}=0.27$
$61\%=\frac{61}{100}=0.61$
So, $\frac3{11}<61\%$

5 (iv): <

$\textbf{Explanation:}$

$30\%=\frac{30}{100}=0.3$
$\frac{1}{3}=0.33$
So, $30\%<\frac13$

Estimate first before making any computations to solve the following questions. Try different methods including mental computations.

Question 1: Find the missing numbers. The first problem has been worked out.

$\textbf{Answer:}$

1 (ii):
Here, there are 10 parts.
So, 1 part = $\frac{100}{10}=10\%$

Total = 90
So, sum of 6 part = $\frac{6}{10}\times90=54$

1 (iii):

Here, there are 4 parts.
So, 1 part = $\frac{100}{4}=25\%$

Total = 140
So, sum of 3 parts = $\frac{3}{4}\times140=105$

Question 2. Find the value of the following and also draw their bar models.
(i) $25 \%$ of 160
(ii) $16 \%$ of 250
(iii) $62 \%$ of 360
(iv) $140 \%$ of 40
(v) $1 \%$ of 1 hour
(vi) $7 \%$ of 10 kg

$\textbf{Answer:}$

2 (i): 40

$\textbf{Explanation:}$
$25 \%$ of 160 = $\frac{25}{100}\times160=40$
2 (ii): 40

$\textbf{Explanation:}$
$16 \%$ of 250 = $\frac{16}{100}\times250=40$

2 (iii): 223.2
$\textbf{Explanation:}$
$62 \%$ of 360 = $\frac{62}{100}\times360=223.2$

2 (iv): 56

$\textbf{Explanation:}$
$140 \%$ of 40 = $\frac{140}{100}\times40=56$

2 (v): 36

$\textbf{Explanation:}$
$1 \% $ of 1 hour
1 hour = 60 minutes
= $\frac{1}{100}\times60=0.6$ minutes = $0.6\times60=36$ seconds

2 (vi): 700

$\textbf{Explanation:}$
$7 \%$ of 10 kg = $\frac{7}{100}\times10=0.7$ kg $=0.7\times1000=700$ gm

Question 3. Surya made 60 ml of deep orange paint. How much red paint did he use if red paint made up $\frac{3}{4}$ of the deep orange paint?

$\textbf{Answer:}$ 45 ml

$\textbf{Explanation:}$

Total deep orange paint = 60 ml
So, the amount of red paint he uses is $60\times\frac34=45$ ml

Question 4. Pairs of quantities are shown below. Identify and write appropriate symbols '>', '<', '=' in the boxes. Visualising or estimating can help. Compute only if necessary or for verification.
(i) $50 \%$ of 510 □ $50 \%$ of 515
(ii) $37 \%$ of 148 □ $73\%$ of 148
(iii) $29 \%$ of 43 □ $92\%$ of 110
(iv) $30 \%$ of 40 □ $40\%$ of 50
(v) $45 \%$ of 200 □ $10\%$ of 490
(vi) $30 \%$ of 80 □ $24\%$ of 64

$\textbf{Answer:}$

4 (i): <

$\textbf{Explanation:}$
$50 \% \text { of } 510=255 $
$ 50 \% \text { of } 515=257.5$
So, $50 \%$ of 510 $<$ $50 \%$ of 515

4 (ii): <

$\textbf{Explanation:}$
Both are percentages of the same number, and 37 < 73
So, $37 \%$ of 148 $<$ $73\%$ of 148

4 (iii): <

$\textbf{Explanation:}$
$29 \% \text { of } 43=12.47 $
$ 92 \% \text { of } 110=101.2$
So, $29 \%$ of 43 $<$ $92\%$ of 110

4 (iv) <

$\textbf{Explanation:}$
$\begin{aligned} & 30 \% \text { of } 40=12 \\ & 40 \% \text { of } 50=20\end{aligned}$
So, $30 \%$ of 40 $<$ $40\%$ of 50

4 (v): >

$\textbf{Explanation:}$
$\begin{aligned} & 45 \% \text { of } 200=90 \\ & 10 \% \text { of } 490=49\end{aligned}$
So, $45 \%$ of 200 $>$ $10\%$ of 490

4 (vi): >

$\textbf{Explanation:}$
$30 \% \text { of } 80=24 $
$ 24 \% \text { of } 64=15.36$
So, $30 \%$ of 80 $>$ $24\%$ of 64

Question 5. Fill in the blanks appropriately:
(i) $30 \%$ of $k$ is $70,60 \%$ of $k$ is $\_\_\_\_$ , $90 \%$ of $k$ is $\_\_\_\_$ , $120 \%$ of $k$ is $\_\_\_\_$ .
(ii) $100 \%$ of $m$ is $215,10 \%$ of $m$ is $\_\_\_\_$ , $1 \%$ of $m$ is $\_\_\_\_,$ $6\%$ of $m$ is $\_\_\_\_$ .
(iii) $90 \%$ of $n$ is $270,9 \%$ of $n$ is $\_\_\_\_$ , $18 \%$ of $n$ is $\_\_\_\_$ , $100 \%$ of $n$ is $\_\_\_\_$ .
(iv) Make 2 more such questions and challenge your peers.

$\textbf{Answer:}$

5 (i):
Since, $30 \%$ of $k$ is 70,
Then, $30\%\times2=60\%$ of $k$ is 70 × 2 = 140
$30\%\times3=90\%$ of $k$ is 70 × 3 = 210
$30\%\times4=120\%$ of $k$ is 70 × 4 = 280

5 (ii):
Since, $100 \%$ of $m$ is 215,
Then, $\frac{100\%}{10}=10\%$ of $m$ is $\frac{215}{10}=21.5$
$\frac{100\%}{100}=1\%$ of $m$ is $\frac{215}{100}=2.15$
Now, $6\%$ of $m$ $=2.15\times6=12.9$

5 (iii):
Since, $90 \%$ of $n$ is 270,
Then, $\frac{90\%}{10}=9\%$ of $n$ is $\frac{270}{10}=27$
Now, $9\times2=18\%$ of $n$ is $27\times2=54$

100% of $n=\frac{100}{90} \times 270=300$

5 (iv):

• If $20 \%$ of $p$ is 50, what is $80 \%$ of $p$?

• If $75 \%$ of $q$ is 150, what is $25 \%$ of $q$?

Question 6. Fill in the blanks:
(i) 3 is $\_\_\_\_ \%$ of 300.
(ii) $\_\_\_\_$ is $40 \%$ of 4 .
(iii) 40 is $80 \%$ of $\_\_\_\_$ .

$\textbf{Answer:}$

6 (i): 1%

$\textbf{Explanation:}$
$\frac3{300}\times100=1\%$
Hence, 3 is 1% of 300.

6 (ii): 1.6

$\textbf{Explanation:}$
$\frac{40}{100}\times4=1.6$
Hence, 1.6 is 40% of 4.

6 (iii): 50

$\textbf{Explanation:}$
Let the number be $x$.
So, $x\times\frac{80}{100}=40$
$\therefore x= 50$
Hence, 40 is 80% of 50.

Question 7. Is $10 \%$ of a day longer than $1 \%$ of a week? Create such questions and challenge your peers.

$\textbf{Answer:}$

1 day = 24 hours
10% of a day $=\frac{10}{100}\times24=2.4$ hours
1 week = 24 × 7 = 168 hours
1% of 1 week $=\frac1{100}\times168=1.68$ hours

Hence, $10 \%$ of a day is longer than $1 \%$ of a week.

Also, discuss:

• Is $5 \%$ of a month longer than $20 \%$ of a week?

• Is $25 \%$ of 80 m greater than $40 \%$ of 45 m?

Question 8. Mariam's farm has a peculiar bull. One day she gave the bull 2 units of fodder and the bull ate 1 unit. The next day, she gave the bull 3 units of fodder and the bull ate 2 units. The day after, she gave the bull 4 units and the bull ate 3 units. This continued, and on the 99th day she gave the bull 100 units and the bull ate 99 units. Represent these quantities as percentages. This task can be distributed among the class. What do you observe?

$\textbf{Answer:}$

On each day:
Day 1: Ate $\frac{1}{2} \times 100=50 \%$
Day 2: Ate $\frac{2}{3} \times 100=66.67 \%$
Day 3: Ate $\frac{3}{4} \times 100=75 \%$
Day 4: Ate $\frac{4}{5} \times 100=80 \%$
Day 5: Ate $\frac{5}{6} \times 100=83.33 \%$
.....
......
.....
Day 99: Ate $\frac{99}{100} \times 100=99 \%$

So, on day $n$, $\text { Percentage eaten }=\frac{n}{n+1} \times 100 \%$

As the days increase, the percentage eaten becomes larger and gets closer and closer to 100%, though it never actually reaches 100%.

Question 9. Workers in a coffee plantation take 18 days to pick coffee berries in 20% of the plantation. How many days will they take to complete the picking work for the entire plantation, assuming the rate of work stays the same? Why is this assumption necessary?

$\textbf{Answer:}$

20% of the plantation is completed in 18 days.
Rate of work stays the same.
So, 100% of the plantation is completed in $\frac{18}{20}\times100=90$ days

This assumption is necessary because this calculation is valid only if the workers continue picking berries at the same rate throughout. If the number of workers, weather conditions, or work speed changes, the total time will also change.

Question 10. The badminton coach has planned the training sessions such that the ratio of warm up : play : cool down is 10% : 80% : 10%. If he wants to conduct a training of 90 minutes. How long should each activity be done?

$\textbf{Answer:}$

Given:
warm up : play : cool down = 10% : 80% : 10%
Total = 10 + 80 + 10 = 100%
He wants to conduct a 90-minute training session.
So, Warm up time = 10% of 90 = $\frac{10}{100}\times90=9$ minutes
Play time = 80% of 90 = $\frac{80}{100}\times90=72$ minutes
Cool down time = 10% of 90 = $\frac{10}{100}\times90=9$ minutes

Question 11. An estimated $90 \%$ of the world's population lives in the Northern Hemisphere. Find the (approximate) number of people living in the Northern Hemisphere based on this year's worldwide population.

$\textbf{Answer:}$ 7.4 billion people approximately

$\textbf{Explanation:}$

The current world population is approximately 8.2 billion.
So, 90% of 8.2 billion = $\frac{90}{100}\times8.2=7.38$ billion

Hence, approximately 7.4 billion people live in the Northern Hemisphere.

Question 12. A recipe for the dish, halwa, for 4 people has the following ingredients in the given proportions: Rava: 40%, Sugar: 40%, and Ghee: 20%.
(i) If you want to make halwa for 8 people, what is the proportion of each of the above ingredients?
(ii) If the total weight of the ingredients is 2 kg, how much rava, sugar and ghee are present?

$\textbf{Answer:}$

12 (i):
Doubling the number of people doubles the quantities, but the proportions remain the same.
So, Rava = 40%, Sugar = 40%, and Ghee = 20%

12 (ii):
Total weight of the ingredients is 2 kg
So, total weight of Rava = 40% of 2 kg = $\frac{40}{100}\times2=0.8$ kg
Total weight of Sugar = 40% of 2 kg = $\frac{40}{100}\times2=0.8$ kg
Total weight of Ghee = 20% of 2 kg = $\frac{20}{100}\times2=0.4$ kg

Question 1. If a shopkeeper buys a geometry box for ₹75 and sells it for ₹110, what is his profit margin with respect to the cost?

$\textbf{Answer:}$ 46.67%

$\textbf{Explanation:}$

Cost price = ₹75
Selling price = ₹110
Profit = ₹110 − ₹75 = ₹35
$\therefore$ Profit percentage = $\frac{\text{Profit}}{\text{Cost price}}\times100=\frac{35}{75}\times100=46.67\%$

Hence, his profit percentage is 46.67%.

Question 2. I am a carpenter and I make chairs. The cost of materials for a chair is ₹475 and I want to have a profit margin of $50 \%$. At what price should I sell a chair?

$\textbf{Answer:}$ 712.5

$\textbf{Explanation:}$

Cost price = ₹475
Profit margin = 50%
So, profit = 50% of 475 = $\frac{50}{100}\times475=237.5$
Therefore, selling price = 475 + 237.5 = ₹712.5

Question 3. The total sales of a company (also called revenue) was ₹ 2.5 crore last year. They had a healthy profit margin of $25 \%$. What was the total expenditure (costs) of the company last year?

$\textbf{Answer:}$ 1.875 crore

$\textbf{Explanation:}$

Revenue = ₹2.5 crore
Profit margin = 25%
So, Profit = 25% of 2.5 crore = $\frac{25}{100}\times2.5=0.625$ crore
Now, Expenditure = Revenue − Profit = 2.5 − 0.625 = 1.875 crore

Hence, the total expenditure of the company is 1.875 crore.

Question 4. A clothing shop offers a $25 \%$ discount on all shirts. If the original price of a shirt is ₹300, how much will Anwar have to pay to buy this shirt?

$\textbf{Answer:}$ 225

$\textbf{Explanation:}$

Original price = ₹300
Discount = 25%
Discount amount = 25% of 300 = $\frac{25}{100}\times300=75$
Therefore, Amount to be paid = 300 – 75 = 225

Question 5. The petrol price in 2015 was ₹60 and ₹100 in 2025. What is the percentage increase in the price of petrol?
(i) $50 \%$
(ii) $40 \%$
(iii) $60 \%$
(iv) $66.66 \%$
(v) $140 \%$
(vi) $160.66 \%$

$\textbf{Answer:}$ Option (iv)

$\textbf{Explanation:}$

Old price = ₹60
New price = ₹100
Difference between petrol prices in 2025 and 2015 = 100 - 60 = 40
So, percentage increase = $\frac{\text{Price difference}}{\text{Old price}}\times100=66.66\%$

Hence, the correct answer is option (iv).

Question 6. Samson bought a car for ₹ $4,40,000$ after getting a $15 \%$ discount from the car dealer. What was the original price of the car?

$\textbf{Answer:}$ ₹517647.06

$\textbf{Explanation:}$

Samson paid ₹4,40,000 after a 15% discount.
Therefore, 85% of the original price is ₹4,40,000.
So, 100% of the original price $=\frac{440000}{85}\times100\approx517647.06$

Hence, the original price of the car is ₹517647.06.

Question 7: 1600 people voted in an election and the winner got 500 votes. What percent of the total votes did the winner get?
Can you guess the minimum number of candidates who stood for the election?

$\textbf{Answer:}$ 31.25% and 4

$\textbf{Explanation:}$

Total votes = 1600
Winner's votes = 500
$\therefore$ Percentage of Winner's votes $=\frac{500}{1600} \times 100=31.25 \%$

If there were only 3 candidates, the remaining 1100 votes could be split as 550 and 550, and the winner with 500 votes would not actually win.
With 4 candidates, the remaining 1100 votes can be distributed among three candidates so that each gets fewer than 500 votes.
So, the minimum number of candidates is 4.

Question 8: The price of 1 kg of rice was ₹38 in 2024. It is ₹42 in 2025. What is the rate of inflation? (Inflation is the percentage increase in prices.)

$\textbf{Answer:}$ 10.53%

$\textbf{Explanation:}$

Price in 2024 = ₹38
Price in 2025 = ₹42
Increase = ₹42 − ₹38 = ₹4
$\therefore$ Increase percentage $=\frac{4}{38}\times100\approx10.53\%$

Hence, the rate of inflation is approximately 10.53%.

Question 9. A number increased by $20 \%$ becomes 90. What is the number?

$\textbf{Answer:}$ 75

$\textbf{Explanation:}$

After increasing by 20%, the number becomes 90.
So, (100 + 20)% = 120% of the number is 90.
Therefore, required number = $\frac{90}{120}\times100=75$

Question 10. A milkman sold two buffaloes for ₹ 80,000 each. On one of them, he made a profit of $5 \%$ and on the other a loss of $10 \%$. Find his overall profit or loss.

$\textbf{Answer:}$ Loss = 5079.37

$\textbf{Explanation:}$

Selling price of each buffalo = ₹80,000
Total selling price = 80000 + 80000 = 160000
5% profit on the first buffalo.
Cost price of first buffalo $=\frac{80000 \times 100}{105}= 76190.48$
10% loss on the 2nd buffalo.
Cost price of the 2nd buffalo $=\frac{80000 \times 100}{90}= 88888.89$
So, total cost price = 76190.48 + 88888.89 = 165079.37
$\therefore$ Overall loss
= Total cost price – Total selling price $= 165079.37 - 160000 = 5079.37$

Question 11. The population of elephants in a national park increased by $5 \%$ in the last decade. If the population of the elephants last decade is $p$, the population now is
(i) $p \times 0.5$
(ii) $p \times 0.05$
(iii) $p \times 1.5$
(iv) $p \times 1.05$
(v) $p+1.50$

$\textbf{Answer:}$ option (iv)

$\textbf{Explanation:}$

Population of the last decade = $p$
Increase = 5% of $p$
New population $=p+\frac{5}{100}\times p=p(1+\frac5{100})=p\times1.05$

Hence, the correct answer is option (iv).

Question 12. Which of the following statement(s) mean the same as - "The demand for cameras has fallen by $85 \%$ in the last decade"?
(i) The demand now is $85 \%$ of the demand a decade ago.
(ii) The demand a decade ago was $85 \%$ of the demand now.
(iii) The demand now is $15\%$ of the demand a decade ago.
(iv) The demand a decade ago was $15 \%$ of the demand now.
(v) The demand a decade ago was $185 \%$ of the demand now.
(vi) The demand now is $185 \%$ of the demand a decade ago.

$\textbf{Answer:}$ option (iii)

$\textbf{Explanation:}$

A fall of 85% means only 15% remains.
Let the demand a decade ago be 100 units.
So, current demand = 100 - 85 = 15 units
Therefore, the current demand is 15% of the previous demand.

Hence, the correct answer is option (iii).

Question 1. Bank of Yahapur offers an interest of $10 \%$ p.a. Compare how much one gets if they deposit ₹20,000 for a period of 2 years with compounding and without compounding annually.

$\textbf{Answer:}$

Let the principal amount be $P= 20000$, the rate of interest be $R=10 \%$, and the time period be $T=2$ years
Without compounding:
Simple interest = $\frac{\text{Principal × Time × Rate}}{100}=\frac{20000 \times 10 \times 2}{100}=4000$
Total amount = 20000 + 4000 = 24000

With annual compounding:
Total amount
= Principal × $(1+\frac{\text{Rate}}{100})^{\text{Time}}$
$=20000\left(1+\frac{10}{100}\right)^2=20000\times 1.21=24200$

Question 2. Bank of Wahapur offers an interest of $5 \%$ p.a. Compare how much one gets if one deposits ₹20,000 for a period of 4 years with compounding and without compounding annually.

$\textbf{Answer:}$

Let the principal amount be $P= 20000$, the rate of interest be $R=5 \%$, and the time period be $T=4$ years

Without compounding:
Simple interest = $\frac{\text{Principal × Time × Rate}}{100}=\frac{20000 \times 4 \times 5}{100}=4000$
Total amount = 20000 + 4000 = 24000

With annual compounding:
Total amount
= Principal × $(1+\frac{\text{Rate}}{100})^{\text{Time}}$
$=20000\left(1+\frac{5}{100}\right)^4=20000\times 1.21=24200\times1.21550625=24310.13$

Question 3. Do you observe anything interesting in the solutions of the two questions above? Share and discuss.

$\textbf{Answer:}$

Main observations are:

• In both cases, compound interest gives a larger amount than simple interest.

• The difference becomes larger when the time period increases.

• Compound interest earns interest on interest, whereas simple interest increases uniformly every year.

Question 4. Jasmine invests amount '$p$' for 4 years at an interest of $6 \%$ p.a. Which of the following expression(s) describe the total amount she will get after 4 years when compounding is not done?
(i) $p \times 6 \times 4$
(ii) $p \times 0.6 \times 4$
(iii) $p \times \frac{0.6}{100} \times 4$
(iv) $p \times \frac{0.06}{100} \times 4$
(v) $p \times 1.6 \times 4$
(vi) $p \times 1.06 \times 4$
(vii) $p+(p \times 0.06 \times 4)$

$\textbf{Answer:}$ option (vii)

$\textbf{Explanation:}$

Jasmine invests a principal amount $p$ for $T=4$ years at a simple interest rate of $R=6 \%$ p.a.
Simple interest
= $\frac{\text{Principal × Time × Rate}}{100}=\frac{p \times 6 \times 4}{100}=p \times 0.06 \times 4$
Total amount = Principal + Simple interest = $p+(p \times 0.06 \times 4)$

Hence, the correct answer is option (vii).

Question 5. The post office offers an interest of $7 \%$ p.a. How much interest would one get if one invests ₹50,000 for 3 years without compounding?
How much more would one get if it was compounded?

$\textbf{Answer:}$

Let the principal amount be $P= 50000$, the rate of interest be $R=7 \%$, and the time period be $T=3$ years.

Without compounding:
Simple interest = $\frac{\text{Principal × Time × Rate}}{100}=\frac{50000 \times 7 \times 3}{100}=10500$

With annual compounding:
Compound interest
= Principal × $(1+\frac{\text{Rate}}{100})^{\text{Time}}$ – Principal
= $50000\left(1+\frac{7}{100}\right)^3-50000$
= $61252.15-50000$
= $11252.15$
Difference between the two returns
= $11252.15-10500=752.15$

Hence, one gets an interest of ₹10,500 without compounding, and would receive ₹752.15 more if it were compounded.

Question 6. Giridhar borrows a loan of ₹ 12,500 at $12 \%$ per annum for 3 years without compounding and Raghava borrows the same amount for the same time period at $10 \%$ per annum, compounded annually. Who pays more interest and by how much?

$\textbf{Answer:}$ Giridhar, ₹362.5

$\textbf{Explanation:}$

Giridhar's interest:
Without compounding:
Simple interest = $\frac{\text{Principal × Time × Rate}}{100}=\frac{12500 \times 3 \times 12}{100}=4500$

Raghava's interest:
With compounding:
Compound interest
= Principal × $(1+\frac{\text{Rate}}{100})^{\text{Time}}$ – Principal
= $12500\left(1+\frac{10}{100}\right)^3-12500$
= $16637.5-12500$
= $4137.5$
Difference between the two interests
= $4500-4137.5=362.5$

Hence, Giridhar pays more interest of ₹362.5.

Question 7. Consider an amount ₹1000. If this grows at $10 \%$ p.a., how long will it take to double when compounding is done vs. when compounding is not done?
Is compounding an example of exponential growth and not-compounding an example of linear growth?

$\textbf{Answer:}$ 8 years

$\textbf{Explanation:}$

Let the time be $t$ when the amount will be double.
Principal, $p=1000$ and rate of interest, $r=10$
For without compounding:
According to the question,
$2000-1000=\frac{1000\times t\times10}{100}$
$⇒1000=100t$
$\therefore t=10$ years
Hence, it takes 10 years without compounding to double.

For with compounding:
According to the question,
$1000=1000 \times (1+\frac{10}{100})^t-1000$
$⇒2000=1000 \times(1.1)^t$
If we take $t=7$, we get, $1000\times 1.1^7\approx1949$
If we take $t=8$, we get $1000(1.10)^8\approx2144$
Hence, the amount doubles in about 8 years with compounding.

Compound growth is exponential growth, and Simple interest growth is linear growth.

Question 8. The population of a city is rising by about $3 \%$ every year. If the current population is 1.5 crore, what is the expected population after 3 years?

$\textbf{Answer:}$ 1.64 crore

$\textbf{Explanation:}$

Let the initial population be $P = 1.5\text{ crore}$, the annual growth rate be $R = 3\%$, and the duration be $T = 3\text{ years}$.
Population after 3 years
= Initial population × $(1+\frac{\text{Growth rate}}{100})^{\text{Time}}$
= $1.5 \times\left(1+\frac{3}{100}\right)^3$
$\approx$ 1.64 crore

Hence, the expected population after 3 years is 1.64 crore.

Question 9. In a laboratory, the number of bacteria in a certain experiment increases at the rate of $2.5 \%$ per hour. Find the number of bacteria at the end of 2 hours if the initial count is $5,06,000$.

$\textbf{Answer:}$ 5,31,616

$\textbf{Explanation:}$

Initial number of bacteria = 506000
Growth rate = 2.5% per hour
Time = 2 hours
Final Bacteria count
= Initial Bacteria count × $(1+\frac{\text{Rate}}{100})^{\text{Time}}$
= $506000 \times\left(1+\frac{2.5}{100}\right)^2$
$\approx$ 531616

Hence, the number of bacteria at the end of 2 hours is approximately 5,31,616.

Question 1. The population of Bengaluru in 2025 is about $250 \%$ of its population in 2000. If the population in 2000 was 50 lakhs, what is the population in 2025?

$\textbf{Answer:}$ 1.25 crore

$\textbf{Explanation:}$

Population in 2025
= 250% of the population in 2000 = $\frac{250}{100}\times50=125$ lakhs or 1.25 Crore

Hence, the population of Bengaluru in 2025 is 125 lakhs or 1.25 crore.

Question 2. The population of the world in 2025 is about 8.2 billion. The populations of some countries in 2025 are given. Match them with their approximate percentage share of the worldwide population.
[Hint: Writing these numbers in the standard form and estimating can help].

$\textbf{Answer:}$

The population of the world in 2025 is about 8.2 billion. i.e, 8200 million

Population of Germany is 83 million.
$\therefore$ Percentage share of Germany $= \frac{83}{8200} \times 100 \approx 1\%$

Population of India is 1.46 billion, i.e, 1460 million.
$\therefore$ Percentage share of Germany $= \frac{1460}{8200} \times 100 \approx 18\%$

Population of Bangladesh is 175 million.
$\therefore$ Percentage share of Germany $= \frac{175}{8200} \times 100 \approx 2\%$

Population of USA is 347 million.
$\therefore$ Percentage share of Germany $= \frac{347}{8200} \times 100 \approx 4\%$

Question 3. The price of a mobile phone is ₹ 8,250. A GST of $18 \%$ is added to the price. Which of the following gives the final price of the phone including the GST?
(i) $8250+18$
(ii) $8250+1800$
(iii) $8250+\frac{18}{100}$
(iv) $8250 \times 18$
(v) $8250 \times 1.18$
(vi) $8250+8250 \times 0.18$
(vii) $1.8 \times 8250$

$\textbf{Answer:}$ Option (v) and (vi)

$\textbf{Explanation:}$

The base price of the mobile phone is $P=8,250$, and a GST of $18 \%$ is added to this amount. The GST amount is calculated as 18% of the base price:
$\text { GST Amount }=8250 \times \frac{18}{100}=8250 \times 0.18$
The final price of the phone, including GST, is the sum of the base price and the GST amount:
$\text { Final Price }=8250+(8250 \times 0.18)$
$⇒\text { Final Price }=8250 \times(1+0.18)=8250 \times 1.18$

Both $(v)$ and $(v i)$ are mathematically identical to these expressions.

Hence, the correct expressions are $(\mathrm{v}) 8250 \times 1.18$ and $(\mathrm{vi}) 8250+8250 \times 0.18$.

Question 4. The monthly percentage change in population (compared to the previous month) of mice in a lab is given: Month 1 change was $+5 \%$, Month 2 change was $-2 \%$, and Month 3 change was $-3 \%$. Which of the following statement(s) are true?
The initial population is $p$.
(i) The population after three months was $p \times 0.05 \times 0.02 \times 0.03$.
(ii) The population after three months was $p \times 1.05 \times 0.98 \times 0.97$.
(iii) The population after three months was $p+0.05-0.02-0.03$.
(iv) The population after three months was $p$.
(v) The population after three months was more than $p$.
(vi) The population after three months was less than $p$.

$\textbf{Answer:}$ Option (ii) and (vi)

$\textbf{Explanation:}$

The initial population of mice is $p$.
After Month 1($+5\%$ change):
The population becomes $p \times(1+0.05)=p \times 1.05$
After Month 2 ($-2\%$ change):
The population becomes $(p \times 1.05) \times(1-0.02)=p \times 1.05 \times 0.98$
After Month 3 (- $3 \%$ change):
The population becomes $(p \times 1.05 \times 0.98) \times(1-$ 0.03) $=p \times 1.05 \times 0.98 \times 0.97$
This calculation directly matches statement (ii).
$\text { Total Multiplier }=1.05 \times 0.98 \times 0.97$
⇒ Total Multiplier $=1.029 \times 0.97=0.99813$
Since the final multiplier, 0.99813, is strictly less than 1, the population after three months is less than the initial population $p$. This matches statement (vi).

Hence, the correct statements are (ii) and (vi).

Question 5. A shopkeeper initially set the price of a product with a $35 \%$ profit margin. Due to poor sales, he decided to offer a $30 \%$ discount on the selling price. Will he make a profit or a loss? Give reasons for your answer.

$\textbf{Answer:}$ Loss, 5.5%

$\textbf{Explanation:}$
Let the cost price of the product be 100.
To have a 35% profit, the selling price = 100 + 35 = 135
After 30% on the selling price, the new selling price
$=\frac{100-30}{100}\times135=94.5$
So, he will have a loss of $100-94.5=5.5\%$

Question 6. What percentage of area is occupied by the region marked ' $E$ ' in the figure?

$\textbf{Answer:}$ 12.5%

$\textbf{Explanation:}$

Let the total area of the square be 100 square units.

It has been halved into two equal parts.

So, area of both sides = $\frac{100}2=50$ square units

Again, it has been halved.

So, area of each half = $\frac{50}2=25$ square units

Again, there are two halves, D and E.

$\therefore$ Area of E = $\frac{25}2=12.5$ square units

Therefore, required percentage $=\frac{12.5}{100}\times100=12.5\%$

Question 7. What is $5 \%$ of 40?
What is $40 \%$ of 5?
What is $25 \%$ of 12?
What is $12 \%$ of 25?
What is $15 \%$ of 60?
What is $60 \%$ of 15?
What do you notice?
Can you make a general statement and justify it using algebra, comparing $x \%$ of $y$ and $y \%$ of $x$?

$\textbf{Answer:}$

For any numbers $x$ and $y$,

$x\%$ of $y=\frac{x}{100} \times y=\frac{xy}{100}$

Similarly,

$y\%$ of $x=\frac{y}{100} \times x=\frac{xy}{100}$

Hence, $x\%$ of $y=y\%$ of $x$

Question 8. A school is organising an excursion for its students. 40% of them are Grade 8 students and the rest are Grade 9 students. Among these Grade 8 students, 60% are girls.
[Hint: Drawing a rough diagram can help].

(i) What percentage of the students going to the excursion are Grade 8 girls?
(ii) If the total number of students going to the excursion is 160, how many of them are Grade 8 girls?

$\textbf{Answer:}$

Let the total number of students be 100
40% grade 8 students. i.e., 40 students
So, 100 - 40 = 60% students are in grade 9, i.e., 60 students
Among these Grade 8 students, 60% are girls. i.e., $40\times\frac{60}{100}=24$
So, 40 - 24 = 16 of the grade 8 students are boys.

(i)
Total students = 100
Grade 8 girls = 24
$\therefore$ Required percentage = $\frac{24}{100}\times100=24\%$

(ii)
If the total number of students = 160
Number of grade 8 girls = $\frac{24}{100}\times160=38.5$
Person value must be an integer.
So, either 38 or 39 girls from grade 8 are going to the excursion.

Question 9. A shopkeeper sells pencils at a price such that the selling price of 3 pencils is equal to the cost of 5 pencils. Does he make a profit or a loss? What is his profit or loss percentage?

$\textbf{Answer:}$ Profit, $66 \frac{2}{3} \%$

$\textbf{Explanation:}$

Let the cost price of one pencil be ₹1.
So, the cost price of 5 pencils = ₹5
Given: Selling price of 3 pencils = Cost price of 5 pencils
So, the selling price of 3 pencils = ₹5
Then the selling price of 1 pencil = $\frac53$
Profit on 1 pencil = $\frac53-1=\frac23$
$\therefore$ Profit percentage $=\frac{\frac{2}{3}}{1} \times 100$

Hence, the shopkeeper makes a profit of $66 \frac{2}{3} \%$.

Question 10. The bus fares were increased by $3 \%$ last year and by $4 \%$ this year. What is the overall percentage price increase in the last 2 years?

$\textbf{Answer:}$ 7.12%

$\textbf{Explanation:}$

Single equivalent price increase
$= (a + b+ \frac{ab}{100})$, where $a$ and $b$ are successive price increase percentages
$=3+4+\frac{3\times4}{100}$
$=7+0.12$
$=7.12\%$

Hence, the overall percentage increase in bus fares over the two years is 7.12%.

Question 11. If the length of a rectangle is increased by $10 \%$ and the area is unchanged, by what percentage (exactly) does the breadth decrease by?

$\textbf{Answer:}$ Decrease, 9.1%

$\textbf{Explanation:}$

Let the initial length and breadth of the rectangle be 100 units each.
Then the area of the rectangle = Length × Breadth = 100 × 100 = 10000 square units
New length = 100 + 10 = 110
Area remains the same.
So, Breadth = $\frac{\text{Total area}}{\text{Length}}=\frac{10000}{100}\approx90.9$ unit
Hence, decrease percentage $=\frac{100-90.9}{100}\times100=9.1\%$

Question 12. The percentage of ingredients in a 65 g chips packet is shown in the picture. Find out the weight each ingredient makes up in this packet.

$\textbf{Answer:}$

Potato = 70% of 65g = $\frac{70}{100}\times65=45.5$ g

Vegetable oil = 24% of 65g = $\frac{24}{100}\times65=16.8$ g

Salt = 3% of 65g = $\frac{3}{100}\times65=2.1$ g

Spices = 3% of 65g = $\frac{3}{100}\times65=2.1$ g

Question 13. Three shops sell the same items at the same price. The shops offer deals as follows:
Shop A: "Buy 1 and get 1 free"
Shop B: "Buy 2 and get 1 free"
Shop C: "Buy 3 and get 1 free"
Answer the following:

(i) If the price of one item is ₹100, what is the effective price per item in each shop? Arrange the shops from cheapest to costliest.
(ii) For each shop, calculate the percentage discount on the items.
[Hint: Compare the free items to the total items you receive.]
(iii) Suppose you need 4 items. Which shop would you choose? Why?

$\textbf{Answer:}$

13 (i) Shop A < Shop B < Shop C

$\textbf{Explanation:}$
Price of one item = ₹100
Shop A:
"Buy 1 and get 1 free" = 2 items.
So, price of 1 item = $\frac{100}2=50$

Shop B:
"Buy 2 and get 1 free" = 3 items
Total price = 2 × 100 = 200
So, price of 1 item = $\frac{200}3 \approx 66\frac23$

Shop C:
"Buy 3 and get 1 free" = 4 items
Total price = 3 × 100 = 300
So, price of 1 item = $\frac{300}4 =75$

So, the shops from cheapest to costliest: Shop A < Shop B < Shop C

13 (ii): 25%

$\textbf{Explanation:}$
Discount percentage = $\frac{\text { Number of free items }}{\text { Total items received }} \times 100$

Shop A: Free items = 1 and Total items received = 2
So, Discount percentage = $\frac12\times100=50\%$

Shop B: Free items = 1 and Total items received = 3
So, Discount percentage = $\frac13\times100\approx 33\frac13\%$

Shop C: Free items = 1 and Total items received = 4
So, Discount percentage = $\frac14\times100=25\%$

13 (iii): Shop A

$\textbf{Explanation:}$
Shop A: Buy 2 items (₹200) and get 2 free → 4 items for ₹200.
Shop B: Buy 3 items (₹300) and get 1 free → 4 items for ₹300.
Shop C: Buy 3 items (₹300) and get 1 free → 4 items for ₹300.
Therefore, Shop A is the best choice because it provides 4 items for only ₹200, which is the lowest cost.

Question 14. In a room of 100 people, $99 \%$ are left-handed. How many left-handed people have to leave the room to bring that percentage down to $98 \%$?

$\textbf{Answer:}$ 50

$\textbf{Explanation:}$

Total people in the room = 100
99% are left-handed, i.e., 99 people.
Let the number of left-handed people left be $x$.
According ot the question,
$\frac{99-x}{100-x}=\frac{98}{100}$
$⇒9900-100 x=9800-98 x$
$⇒2x=100$
$\therefore x=50$

Hence, 50 left-handed people must leave the room to bring that percentage down to 98%.

Question 15. Look at the following graph.

Based on the graph, which of the following statement(s) are valid?

(i) People in their twenties are the most computer-literate among all age groups.

(ii) Women lag behind in the ability to use computers across age groups.

(iii) There are more people in their twenties than teenagers.

(iv) More than a quarter of people in their thirties can use computers.

(v) Less than 1 in 10 aged 60 and above can use computers.

(vi) Half of the people in their twenties can use computers.

$\textbf{Answer:}$

Let's review each statement individually.
Statement (i): Valid
People in their twenties are the most computer-literate among all age groups.
It is true as Male percentage of 37 and the female percentage of 26 are computer literate.

Statement (ii): Valid
Women lag behind in the ability to use computers across age groups.
It is true that, in all age groups except children, the percentage of women who are computer literate is lower.

Statement (iii): Not Valid
There are more people in their twenties than teenagers.
This statement is not valid. The chart displays the percentage (proportion) of people within each age group who can use a computer, not the absolute population size of each demographic group.

Statement (iv): Not Valid
More than a quarter of people in their thirties can use computers.
This statement is not valid. For people in their thirties, 14% of females and 25% of males can use computers. Even if we look at the highest performing gender in this group (males), it hits exactly a quarter (25%), not more than a quarter. The overall average for the entire thirties age group would sit somewhere between 14% and 25%, which is well below a quarter.

Statement (v): Valid
Less than 1 in 10 aged 60 and above can use computers.
"Seniors" represents those aged 60 and above. The data shows 2% for females and 4% for males. Since both values (and their combined average) are significantly under 10% (which is 1 in 10), this statement holds true.

Statement (vi): Not Valid
Half of the people in their twenties can use computers.
This statement is not valid. In the twenties bracket, 26% of females and 37% of males can use computers. Neither individual group reaches 50% (half), nor would their combined average reach 50%.

Key Concepts Explained in Fractions in Disguise - Ganita Prakash Book 2

Topics you will learn in NCERT Solutions for Class 8 Maths Part 2 Chapter 1 Fractions in Disguise include:

• 1.1 Fractions as Percentages

• 1.2 Percentage of Some Quantity

• 1.3 Using Percentages

Extra Questions for Class 8 Maths Part 2 Chapter 1 Fractions in Disguise

Question 1:

There are 500 guests at an event, among whom 28% are adult males, 64% are adult females, and the rest are children. Could you find the total number of children at the event?

$\textbf{Answer:}$

According to the question
⇒ Percentage of children = 100% − (28% + 64%) = 100% − 92% = 8%
Now,
Number of children = $\frac{8}{100}$ × 500 = 0.08 × 500 = 40
Hence, the correct answer is 40.

Question 2:

A is 25% less than B and B is 30% more than 120. What is the value of A?

$\textbf{Answer:}$

A is 25% less than B.
B is 30% more than 120.
Value of B = 120 + 30% of 120
$= 120 + \frac{30}{100} \times 120 = 120 + 36 = 156$
Value of A = B – 25% of B
$= 156 - \frac{25}{100} \times 156 = 156 - 39 = 117$
$\therefore$ The required value of A is 117.
Hence, the correct answer is 117.

Question 3:

The compound interest on INR 18,000 at 7% per annum, compounded annually, is INR 1,260. What is the period?

$\textbf{Answer:}$

Use :
A = P × $[1+ \frac{R}{100}]^{t}$
A = CI + P, where
A = Amount, CI = Compound interest, P = Principal, R = Rate of interest, t = time
According to the question
⇒ A = CI + P = 1260 + 18000 = Rs.19,260
⇒ 19260 = 18000 × $[1+ \frac{7}{100}]^{t}$
⇒ 19260 = 18000 × $[1.07]^{t}$
⇒ [1.07]$^{t}$ = $\frac{19260}{18000}$
⇒ [1.07]$^{t}$ = 1.07
⇒ t = 1
Hence, the correct answer is 1 year.

Question 4:

Ritika bought each set of 10 shirts at Rs 8,000 and sold them at Rs 9,000 for each set of 12 shirts. Find her profit or loss percentage.

$\textbf{Answer:}$

Cost price of 10 shirts = Rs. 8000
Cost price of 1 shirt $= \frac{8000}{10}=800$
Selling price of 12 shirts = Rs. 9000
Selling price of 1 shirt $=\frac{9000}{12}=750$
Loss = cost price – selling price = 800 – 750 = 50
Loss percentage = $\frac{\text{Loss}}{\text{Cost price}}\times 100$
= $\frac{50}{800}\times 100$
= 6.25%
Hence, the correct answer is a 6.25% loss.

Question 5:

At what time does a sum of money amount to twice itself at the simple interest of 25% per annum?

$\textbf{Answer:}$

Given: A sum of money amounts to twice itself at the simple interest of 25% per annum.
Let the principal be INR $x$.
According to the question,
Total amount $=2x$
$\therefore$ Simple interest = $2x-x=x$
Also, We know, Simple interest = $\frac{\text{Principal × Rate × Time}}{100}$
So, $x=\frac{x\times 25\times \text{Time}}{100}$
⇒ Time $=4$ years
Hence, the correct answer is 4 years.

NCERT Solutions for Class 8 Maths Chapter Wise

We at Careers360 compiled all the NCERT Class 8 Maths solutions based on latest textbook in one place for easy student reference. The following links will allow you to access them.

NCERT Books and NCERT Syllabus

Before planning a study schedule, always analyse the latest syllabus. Here are the links to the latest NCERT syllabus and some of the important books that will help students in this cause.

• NCERT Books Class 8 Maths

• NCERT Syllabus Class 8 Maths

• NCERT Books Class 8

• NCERT Syllabus Class 8

Chapter Summary: NCERT Solutions for Class 8 Maths Ganita Prakash Part 2 Chapter 1

Number of Exercises: 5

Total number of questions: 53

Expert Review of NCERT Solutions for Class 8 Maths Part 2 Chapter 1

NCERT Solutions for Class 8 Maths Ganita Prakash Part 2 Chapter 1 are fundamental for students in class 8, as it prepares them for some important concepts in Class 9 and beyond. So, let's check which key concepts are strengthened by this chapter.

Class 9-10

• Fractions and rational numbers

• Conversion between fractions, decimals, and percentages

• Percentage calculations

• Profit and loss

• Discount and marked price

• Simple Interest

• Compound Interest

• Applications of percentages in daily life

• Ratio and proportion

Class 11-12

• Percentage-based algebraic manipulation

• Growth and depreciation concepts

• Exponential growth models

• Sequences and series involving percentage increase/decrease

• Logarithms

• Financial mathematics basics

• Data interpretation involving percentages and ratios

JEE (Main & Advanced)

• Percentage and ratio concepts

• Fraction and decimal manipulation

• Exponential growth and decay

• Logarithms and compound growth

• Sequences and series applications

• Quantitative aptitude and data interpretation

• Algebraic simplification involving ratios and percentages

• Mathematical modelling using growth and depreciation concepts