I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Fractions in Disguise Exercise 1.7 Class 8 Maths – Chapter 1 NCERT Solutions

NCERT Class 8 Maths Term 1 Chapter 1 Fractions in Disguise Exercise 1.6 detailed solutions are provided below. Prepared by the ALLEN expert team, these solutions explain the exercise questions clearly for practice and revision.Exercise 1.6 uses the concept of percentages in several real life situations. It covers population comparisons, GST, successive percentage changes, profit and loss, area percentages, ingredient quantities and buy one get one free offers

1.0Download Class 8 Maths Chapter 1 Ex 1.7 NCERT Solutions PDF

Download the Exercise 1.6 NCERT Solutions PDF to refer to the solved percentage questions and review the methods used in the exercise.

Chapter 1 : Fractions in Disguise Exercise 1.6

2.0Detailed NCERT Solutions for Fractions in Disguise Exercise 1.6 Class 8 Maths

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?

Sol. Population in $2000 = 50$ lakhs Population in 2025 $= 50$ lakhs $+ 250%$ of 50 lakhs $= 50$ lakhs $+ 2.5 \times 50$ lakhs $= 50$ lakhs + 125 lakhs $= 175$ lakhs or 1.75 crores

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 Germany 83 million $13%8%$ $13%8%$ 18% 18% 10% 1% 35% 2% 2% Sol. Germany $= \frac{83}{8200} \times 100% \sim 1%$ India $= \frac{1.46}{8.20} \times 100% \sim 18%$ Bangladesh $= \frac{175}{8200} \times 100% \sim 2%$ USA $= \frac{347}{8200} \times 100% \sim 4%$

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? of the worldwide population.

Hint: Writing these numbers in the standard form and estimating can help. (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) $1.8 \times 8250$

Sol. Price of mobile phone $= ₹8, 250$ GST @ $18% = \frac{8250 \times 18}{100}$ $= ₹8250 \times 0.18$ Total cost $= ₹ (8250 + 8250 \times 0.18)$ Options (v) and (vi) are correct.

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 $p \times 0.05 \times 0.02 \times 0.03$ . (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. Sol. Population after 3 months $= p (1 + \frac{5}{100}) (1 - \frac{2}{100}) (1 - \frac{3}{100})$ $= p \times 1.05 \times 0.98 \times 0.97 = 0.99813 p$ Option (ii) and (vi) are correct.

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 loss? Give reasons for your answer. Sol. 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 ∴ the shopkeeper makes a 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.

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

Sol. Total area $= 8 \times 8 = 64$ sq. units And area of $E = 8$ sq. units ∴ Required $% = \frac{8}{64} \times 100% = 12.5%$

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 ?

Sol. $5%$ of $40 = \frac{5}{100} \times 40 = 2$ $40%$ of $5 = \frac{40}{100} \times 5 = 2$ $25%$ of $12 = \frac{25}{100} \times 12 = 3$ $12%$ of $25 = \frac{12}{100} \times 25 = 3$ $15%$ of $60 = \frac{15}{100} \times 60 = 9$ $60%$ of $15 = \frac{60}{100} \times 15 = 9$ Justification $x %$ of $y = y %$ of x Since Multiplication is commutative $\frac{x}{100} \times y = \frac{y}{100} \times x$

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? Sol. Let no. of students be 100. Then, no. of students of grade 8 $= \frac{40}{100} \times 100 = 40$ No. of students of grade $9 = 100 - 40 = 60$ (i) No. of grade 8 girls $= \frac{60}{100} \times 40 = 24%$ (ii) $100 : 24 :: 160 : x$ $100 x = 24 \times 160$ $x = 38.4$ No. of grade 8 girls is 38.4

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?

Sol. SP of 3 pencils = CP of 5 pencils Let SP of 3 pencils = CP of 5 pencils $= 3 \times 5 = 15$ Then SP = ₹5; CP = ₹3 Profit $= ₹2$ Profit $% = \frac{2}{3} \times 100% = 66 \frac{2}{3} % (\sim 67%)$

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?

Sol. Let the bus fare 2 years ago be ₹ 100 Present bus fare $= ₹100 \times 1.03 \times 1.04 = ₹107.12$ Increase = ₹7.12 Increase $% = \frac{7.12}{100} \times 100% = 7.12%$

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?

Sol.

$l (l + \frac{10}{100} l) (b - \frac{x}{100} b) = A$

$1.1 l \times \frac{( 100 - x ) b}{100} = A$ $1.1 l \times \frac{100 - x}{100} b = l \times b$ $1.1 l \times (\frac{100 - x}{100}) = 1 \Rightarrow 100 - x = \frac{1 \times 100}{1.1}$ $x = 100 - \frac{1000}{11} \Rightarrow x = \frac{1100 - 1000}{11}$ $\frac{100}{11} = 9 \frac{1}{11}$ Breadth is decreased by $9 \frac{1}{11} %$ .

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.

Sol. Potato $= \frac{70}{100} \times 65 g = 45.5 g$ Veg Oil $= \frac{24}{100} \times 65 g = 15.6 g$ Salt $= 390$ of $65 g = 1.95 g$ Spice $= \frac{3}{100} \times 65 g = 1.95 g$ Verification: $45.5 + 15.6 + 1.95 + 1.95 = 65.05 g$

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?

Sol. (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: Shop A< Shop B < Shop C (ii) Discount at shop $A = \frac{1}{2} \times 100% = 50%$

Discount at shop B $= \frac{1}{3} \times 100%$ $= \frac{100}{3} % = 33.3%$ Discount at shop C $= \frac{1}{4} \times 100% = 25%$ (iii) To buy 4 items, we choose shop A. Pay for 2 , and the other 2 are free.

14. In a room of 100 people, $99%$ are lefthanded. How many left-handed people have to leave the room to bring that percentage down to $98%$ ? Sol. 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 $\frac{99 - x}{100 - x} = \frac{98}{100}$ $\Rightarrow 100 (99 - x) = 98 (100 - x)$ $\Rightarrow 9900 - 100 x = 9800 - 98 x$ $\Rightarrow 9900 - 9800 = 100 x - 98 x$ $\Rightarrow 2 x = 100$ $\Rightarrow x = 50$ $∴ 50$ left-handers must leave the room.

15. Look at the following graph.

Ability to use computer by age and gender (2023)

The ability to use computers is highest among those is their twenties and teenagers.

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.

Sol. (i) True (ii) True (iii) True (iv) False (v) True (vi) False

3.0Key Concepts of Chapter 1 Fractions in Disguise Exercise 1.6

Finding a quantity when a percentage of the original quantity is given.
Calculating the percentage share of a population.
Finding the final price of a product after adding GST.
Applying successive percentage increases and decreases to a quantity.
Working with profit, loss, discount, and their percentages.
Finding the percentage of an area and comparing x% of y with y% of x.
Applying percentages to dimensions, ingredient quantities, and buy-one-get-one-free offers.

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 1 Exercise 1.6

Finds population values from a given percentage increase.
Works out the percentage share of different populations.
Calculates the final price after GST is added.
Applies successive percentage changes to find the updated value.
Works out profit or loss from the given cost and selling prices.
Calculates the percentage of an area occupied by a particular region.
Uses percentage relationships to work with ingredient quantities and promotional offers.

NCERT Class 8 Maths Ch. 1 Fractions in Disguise Other Exercises:-

Exercise 1.1

Exercise 1.2