Fractions in Disguise Class 8 Solutions Maths Ganita Prakash Part 2 Chapter 1

Solving questions with the help of Class 8 Ganita Prakash Solutions and NCERT Class 8 Maths Part 2 Chapter 1 Fractions in Disguise Question Answer Solutions improves confidence.

Class 8 Maths Ganita Prakash Part 2 Chapter 1 Solutions

Ganita Prakash Class 8 Chapter 1 Solutions Fractions in Disguise

Class 8 Maths Ganita Prakash Part 2 Chapter 1 Fractions in Disguise Solutions Question Answer

1.1 Fractions As Percentages

Figure It Out (Pages 3-4)

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) \(\frac {5}{11}\)
Solution:

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
Solution:
(iv) 60%
Percentage of white marbles = \(\frac {15}{25}\) × 100% = 60%

Question 3.
In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking?
Solution:
Percentage of students coming by walking = \(\frac {15}{80}\) × 100% = 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.

Solution:
A = 20%
B = 38%
C = 72%
D = 93%

Question 5.
A pair of quantities is 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% ____ \(\frac {1}{3}\)
Solution:
(i) 50% > 5%
(ii) \(\frac {5}{10}\) = 50%
(iii) \(\frac {3}{11}\) < 61%
(iv) 30% < \(\frac {1}{3}\)

1.2 Percentage of Some Quantity

Figure It Out (Pages 12-14)

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.


Solution:
(i) 20%; 60
(ii) 10%; 54
(iii) 25%; 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
Solution:
(i) 25% of 160 = \(\frac {25}{100}\) × 160 = 40

(ii) 16% of 250 = \(\frac {16}{100}\) × 250 = 40

(iii) 62% of 360 = \(\frac {62}{100}\) × 360 = 223.2
(iv) 140% of 40 = \(\frac {140}{100}\) × 40 = 56
(v) 1% of 1 hour = 1% of 60 min
= 1% of 3600 sec.
= \(\frac {1}{100}\) × 3600 sec.
= 36 sec.
(vi) 7% of 10 kg = \(\frac {7}{100}\) × 10 kg = 0.7 kg.

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?
Solution:
Quantity of red paint = \(\frac {3}{4}\) × 60 ml = 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
Solution:
(i) 50% of 510 < 50% of 515 (510 < 515)
(ii) 37% of 148 < 73% of 148 (37 < 73)
(iii) 29% of 43 < 92% of 110 (29 < 92; 43 < 110)
(iv) 30% of 40 < 40% of 50 (30 < 40; 40 < 50)
(v) 45% of 200 > 10% of 490 (90 > 49)
(vi) 30% of 80 > 24% of 64 (24 > 15.36)

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.
Solution:
(i) (30% of k is 70) × 2 = 60% of k = 140
(30% of k is 70) × 3 = 90% of k = 210
(30% of k is 70) × 4 = 120% of k = 280

(ii) (100% of m is 215) ÷ 10
10% of m is 21.5
(100% of m is 215) ÷ 100
1% of m is 2.15
(1% of m is 2.15) × 6 = 6% of m is 12.9

(iii) (90% of n is 270) ÷ 10
9% of n is 27
(90% of n is 210) ÷ 9
(10% of n is 30) × 10
100% of n is 300
(9% of n is 27) × 2
18% of n is 54.

(iv) Do yourself.

Question 6.
Fill in the blanks:
(i) 3 is ____ % of 300.
(ii) ____ is 40% of 4.
(iii) 40 is 80% of ____
Solution:
(i) 1 (\(\frac {3}{300}\) × 100% = 1%)
(ii) 1.6 (\(\frac {4}{100}\) × 4 = 1.6)
(iii) 50 (\(\frac {40}{80}\) × 100 = 50)

Question 7.
Is 10% of a day longer than 1% of a week? Create such questions and challenge your peers.
Solution:
Yes, 10% of a day = \(\frac {10}{100}\) × 24 hrs = 2.4 hr
1% of week = \(\frac {1}{100}\) × (24 × 7) hrs = 1.68 hrs
10% of a day > 1 % of a week.
Is 10% of a month (30 days) < 50% of a week?
Is 50% of a dozen (12) > 10% of a score (20)?
Is 80% of a century < 45% of a double century?

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?
Solution:
\(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{98}{99}, \frac{99}{100},\)
50%, 66\(\frac {2}{3}\)%, 75%, 80%,… 98\(\frac {98}{99}\)%, 99%.
It follows the pattern \(\left(\frac{n}{n+1}\right)\) × 100 where n is the day number.
As n increases, the percentage gets closer and closer to 100%, but never quite reaches it.

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?
Solution:
(20% work in 18 days) × 5 = 100% work in 90 days.
The work will be completed in 90 days.
Necessary Assumptions

• Weather conditions might change.
• Workers might get tired over time. This reduces their efficiency.
• Some workers might take leave or breaks.

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 last?

Solution:
Warm-up time = 10% of 90 min
= \(\frac {10}{100}\) × 90
= 9 min
Play time = 80% of 90 min
= \(\frac {80}{100}\) × 90
= 72 min
Cool down time = 10% of 90 min = 9 min.

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.
Solution:
World population as of Jan 1, 2026 = 8.3 billion
90% of 8.3 billion = \(\frac {90}{100}\) × 8.3 billion = 7.47
Hence, 7.47 billion people live in northern hemisphere.

Question 12.
A recipe for the dish, ha I wa, 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?
Solution:
(i) Proportion remains the same.
(ii) Rava = \(\frac {40}{100}\) × 2 kg = 0.8 kg
Sugar = \(\frac {40}{100}\) × 2 kg = 0.8 kg
Ghee = \(\frac {20}{100}\) × 2 kg = 0.4 kg

1.3 Using Percentages

Figure It Out (Pages 19-20)

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?
Solution:
Profit = ₹ 110 – ₹ 75 = ₹ 35
Profit % = \(\frac {35}{75}\) × 100
= 0.4667 × 100
= 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?
Solution:
Cost of material = ₹ 475
Profit = 50% of 475
= \(\frac {50}{100}\) × 475
= ₹ 237.50
Sale price = ₹ 475 + ₹ 237.50 = ₹ 712.50

Question 3.
The total sales of a company (also called revenue) were ₹ 2.5 crore last year. They had a healthy profit margin of 25%. What was the total expenditure (costs) of the company last year?
Solution:
Let the cost be ₹ x
Then profit = ₹ \(\frac {25x}{100}\) or ₹ 0.25x
∴ Revenue = x + 0.25x = 2.5 crores
⇒ 1.25x = 2.5
⇒ x = \(\frac{2.5}{1.25}=\frac{250}{125}\) = 2
∴ Cost is ₹ 2 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?
Solution:
Marked price = ₹ 300
Discount = ₹ \(\frac {25}{100}\) × 300 = ₹ 75
Sale price = ₹ 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%
Solution:
Increase in price = ₹ 100 – ₹ 60 = ₹ 40
Increase % = \(\frac {40}{60}\) × 100% = 66\(\frac {2}{3}\)% or 66.66%

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? (NCERT Q. No. 3)
Solution:
Let the marked price of the car be ₹ x
Discount = \(\frac {15}{100}\)x = 0.15x
Sale price = x – 0.15x = 0.85x
Now 0.85x = 4,40,000
x = \(\frac {4,40,000}{0.85}\) = 517647
Marked price of the car is ₹ 5,17,647.

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 stoodfor the election? (NCERTQ. No. 4)
Solution:
Vote% (winner) = \(\frac {500}{1600}\) × 100% = 31.25%
and 100 ÷ 31.25 = 3.2
∴ In all, there were at least 4 candidates.
This means at least 3 more candidates.

Question 8.
The price of l 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.) (NCERT Q. No. 5)
Solution:
Increase in price = ₹ 42 – ₹ 38 = ₹ 4
Rate of inflation = \(\frac {4}{38}\) × 100% = 10.52%

Question 9.
A number increased by 20% becomes 90. What is the number? (NCERT Q. No. 6)
Solution:
120% of a number is 90
∴ 1% of the number is \(\frac {90}{120}\)
100% of the number = \(\frac {90}{120}\) × 100 = 75
or the number was 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. (NCERT Q. No. 7)
Solution:
SP of 1 st buffalo = ₹ 80,000
Profit% = 5%
∴ CP = ₹ \(\frac{100}{100+5} \times 80,000\) = ₹ 76,190
SP of 2nd buffalo = ₹ 80,000
Loss % = 10%
∴ CP = ₹ \(\frac{100}{100-10} \times 80,000\) = ₹ 88,889
Total CP = ₹ 76,190 + ₹ 88,889 = ₹ 1,65,079
Total SP = ₹ 80,000 + ₹ 80,000 = ₹ 1,60,000
Loss = ₹ 1,65,079 – ₹ 1,60,000 = ₹ 5,079
Loss% = \(\frac {5079}{165079}\) × 100% = 3%

Question 11.
The population of elephants in a national park increased by 5% in the last decade. If the population of the elephants over the last decade is p, the population now is (NCERT Q. No. 8)
(i) p × 0.5
(ii) p × 0.05
(iii) p × 1.5
(iv) p × 1.05
(v) p + 1.50
Solution:
(iv) p × 1.05
Population 10 years ago = p
Increase in population = \(\frac {5}{100}\)p = 0.05p
∴ Current population = p + 0.05p = 1.05p

Question 12.
Which of the following statements means the same as — “The demand for cameras has fallen by 85% in the last decade”? (NCERT Q. No. 9)
(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.
Solution:
Statement: The demand for cameras has fallen by 85% in last decade.
Only (iii) means the same.

1.4 Growth and Compounding

Figure It Out (Pages 22-24)

Question 1.
The Bank of Yahapur offers an interest rate of 10% p.a. Compare the amount one earns by depositing ₹20,000 for a period of 2 years with and without compounding annually.
Solution:
Without compounding
Amount = \(P\left(1+\frac{r t}{100}\right)\)
= 20,000 × \(P\left(1+\frac{10 \times 2}{100}\right)\)
= 20,000(1 + 0.20)
= 20,000 × 1.20
= ₹ 24, 000
With compounding
Amount = P(1 + r)^t
= 20,000 × \(\left(1+\frac{10}{100}\right)^2\)
= 20,000 × 1.21
= 24,200
Comparison:
Without compounding = ₹ 24,000
With compounding = ₹ 24,200
Difference = ₹ 24,200 – ₹ 24,000 = ₹ 200
Hence, with compounding, one gets ₹ 200 more than without compounding.

Question 2.
The Bank of Wahapur offers an interest rate of 5% p.a. Compare the amount one earns by depositing ₹20,000 for a period of 4 years with and without compounding annually.
Solution:
P = ₹ 20,000; t = 4, r = 5
SI = \(\frac{20,000 \times 5 \times 4}{100}\) = ₹ 4,000
A = ₹ 20,000 × \(\left(1+\frac{5}{100}\right)^4\)
= ₹ 20,000 × \(\left(\frac{21}{20}\right)^4\)
= ₹ 20,000 × \(\frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}\)
= ₹ 24310.13
CI = ₹ 24,310.13 – ₹ 20,000 = ₹ 4,310.13
SI (without compounding) < CI (with compounding)

Question 3.
Do you observe anything interesting in the solutions of the two questions above? Share and discuss.
Solution:
If the rate percent and time is same, the interest received with compounding is more than the interest received without compounding.

Question 4.
Jasmine invests an 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 × 6 × 4
(ii) p × 0.6 × 4
(iii) p × \(\frac {0.6}{100}\) × 4
(iv) p × \(\frac {0.06}{100}\) × 4
(v) p × 1.6 × 4
(vi) p × 1.06 × 4
(vii) p + (p × 0.06 × 4)
Solution:
P = p, R = 6, T = 4
Amount = p + I
= p + \(\frac{p \times 6 \times 4}{100}\)
= p + p × 0.06 × 4
= p + 0.24p
= 1.24 p
= p × 1.06 × 4
Hence (vi) and (vii) are correct.

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 were compounded?
Solution:
Without compounding
P = ₹ 50,000; R = 7% pa; T = 3 years
I = ₹ \(\frac{50,000 \times 7 \times 3}{100}\) = ₹ 10,500
Amount = ₹ 50,000 + ₹ 10, 500 = ₹ 60,500
With compounding
A = 50,000(1.07)³ = 61252.15
Difference = 61252.15 – 50,000 = 11252.15
Extra interest = 11252.15 – 10500 = ₹ 752.15

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?
Solution:
I (Giridhar) = \(\frac{12,500 \times 12 \times 3}{100}\) = ₹ 4,500
For Raghava
A = 12,500\(\left(1+\frac{10}{100}\right)^3\)
= 12,500 × \(\frac {1331}{1000}\)
= ₹ 16637.5
I (Raghava) = ₹ 16,637.5 – ₹ 12,500 = ₹ 4137.50
₹ 4500 – ₹ 4137.50 = ₹ 362.50
Giridhar pays ₹ 362.5 more than Raghava.

Question 7.
Consider an amount of ₹ 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?
Solution:
₹ 1000 becomes ₹ 2,000
Interest = ₹ 1000
Without compounding
1000 = \(\frac{1000 \times 10 \times t}{100}\)
t = 10 years
With compounding
\(1000\left(1+\frac{10}{100}\right)^n\) = 2000
(1.1)ⁿ = 2
This can be done by hit and trial
1.1² = 1.21
1.1³ = 1.331
1.1⁴ = 1.4641
1.1⁵ = 1.6
1.1⁶ = 1.77
1.1⁷ = 1.94
1.1⁸ = 2.14
1.94 < 2 < 2.14
Time would be between 7 and 8 years (The nearest answer is 7.2 years)

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?
Solution:
Population after 3 years = 1.5 × \(\left(1+\frac{3}{100}\right)^3\) crores
= 1.5 × (1.03)³ crores
= 1.639 crores

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.
Solution:
No of bacteria after 2 hours = 5,06,000\(\left(1+\frac{2.5}{100}\right)^2\)
= 506000 × (1.025)²
= 5,31,616

1.5 Decline

Figure It Out (Pages 28-30)

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?
Solution:
Population in 2000 = 50 Lakhs
Population in 2025 = \(\frac {250}{100}\) × 50 L = 125 L or 1 crore 25 L

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.

Solution:
Germany = \(\frac {83}{8200}\) × 100% ~ 1%
India = \(\frac {1.46}{8.20}\) × 100% ~ 18%
Bangladesh = \(\frac {175}{8200}\) × 100% ~ 2%
USA = \(\frac {347}{8200}\) × 100% ~ 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 × 18
(v) 8250 × 1.18
(vi) 8250 + 8250 × 0.18
(vii) 1.8 × 8250
Solution:
Price of mobile phone = ₹ 8,250
GST @ 18% = ₹ \(\frac{8250 \times 18}{100}\) = ₹ 8250 × 0.18
Total cost = ₹ (8250 + 8250 × 0.18)
Options (v) and (vi) are correct.

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 statements are true? The initial population is p.
(i) The population after three months was p × 0.05 × 0.02 × 0.03.
(ii) The population after three months was p × 1.05 × 0.98 × 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.
Solution:
Population after 3 months = \(p\left(1+\frac{5}{100}\right)\left(1-\frac{2}{100}\right)\left(1-\frac{3}{100}\right)\)
= p × 1.05 × 0.98 × 0.97
= 0.99813p
Options (ii) and (vi) are correct.

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.
Solution:
Let CP be ₹ 100.
P% = 35%
P = ₹ 35
MP = ₹ 135
Discount = 30% of 135 = ₹ 40.50
New MP = ₹ 135 – ₹ 40.50 = ₹ 94.50
New MP < CP
∴ Loss
Reason: Although he initially added a 35% profit margin, the 30% discount is calculated on the increased selling price (not the cost price), which results in a larger absolute discount amount that exceeds the original profit.

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

Solution:
Total area = 8 × 8 = 64 sq. units
And area of E = 8 sq. units
∴ Required % = \(\frac {8}{64}\) × 100% = 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?
Solution:
5% of 40 = \(\frac {5}{100}\) × 40 =
40% of 5 = \(\frac {40}{100}\) × 5 = 2
25% of 12 = \(\frac {25}{100}\) × 12 = 3
12% of 25 = \(\frac {12}{100}\) × 25 = 3
15% of 60 = \(\frac {15}{100}\) × 60 = 9
60% of 15 = \(\frac {60}{100}\) × 15 = 9
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?
Solution:
Let no. of students be 100.
Then, no. of students of grade 8 = \(\frac {40}{100}\) × 100 = 40
No. of students of grade 9 = 100 – 40 = 60
(i) No.of grade 8 girls = \(\frac {60}{100}\) × 40 = 24
(ii) 100 : 24 :: 160 : x
100x = 24 × 160
⇒ x = 38.4
No. of grade 8 girls is 38.4

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?
Solution:
SP of 3 pencils = CP of 5 pencils
Let SP of 3 pencils = CP of 5 pencils
= 3 × 5
= 15
Then SP = ₹ 5; CP = ₹ 3
Profit = ₹ 2
Profit% = \(\frac {2}{3}\) × 100% = 66\(\frac {2}{3}\)% (~ 67%)

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?
Solution:
Let the bus fare 2 years ago be ₹ 100
Present bus fare = ₹ 100 × 1.03 × 1.04 = ₹ 107.12
Increase = ₹ 7.12
Increase% = \(\frac {7.12}{100}\) × 100% = 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?
Solution:


Breadth is decreased by 9\(\frac {1}{11}\)%.

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.

Solution:
Potato = \(\frac {70}{100}\) × 65 g = 45.5 g
Veg oil = \(\frac {24}{100}\) × 65 g = 15.6 g
Salt = \(\frac {3}{100}\) × 65 g = 1.95 g
Spice = \(\frac {3}{100}\) × 65 g = 1.95 g
Verification: 45.5 + 15.6 + 1.95 + 1.95 = 65.05 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?
Solution:
(i) Effective price per item at shop A = ₹ \(\frac {100}{2}\) = ₹ 50
Effective price per item at shop B = ₹ \(\frac {200}{3}\) = ₹ 66\(\frac {2}{3}\)
Effective price per item at shop C = ₹ \(\frac {300}{4}\) = ₹ 75
Cheapest to costliest: A; B; C.
(ii) Discount at shop A = \(\frac {1}{2}\) × 100% = 50%
Discount at shop B = \(\frac {1}{3}\) × 100% = 66\(\frac {2}{3}\)%
Discount at shop C = \(\frac {1}{4}\) × 100% = 25%
(iii) To buy 4 items, we choose shop A. Pay for 2, and the other 2 are free.

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%?
Solution:
No. of people in room = 100
No. of left handers = 99% of 100 = 99
No. of right-handers = 1
Let x people leave.
No. of left-handers is 98%.
and no. of right-handers is 2%.
Then \(\frac{99-x}{100-x}=\frac{98}{100}\)
⇒ 100(99 – x) = 98(100 – x)
⇒ 9900 – 100x = 9800 – 98x
⇒ 9900 – 9800 = 100x – 98x
⇒ 2x = 100
⇒ x = 50
∴ 50 left-handers must leave the room.

Question 15.
Look at the following graph.

Based on the graph, which of the following statements are valid?
(i) People in their twenties are the most computer-literate among all age groups.
(ii) Women lag 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.
Solution:
(i) True
(ii) True
(iii) True
(iv) False
(v) True
(vi) False

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