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Fractions in Disguise - Exercise 1.5 Class 8 Maths - Chapter 1 NCERT Solutions

This page provides detailed NCERT Solutions for Class 8 Maths Ganita Prakash Term 2 Chapter 1, Fractions in Disguise Exercise 1.5. Prepared by the experienced faculty at ALLEN, the solutions provide clear, question-wise explanations based on the updated syllabus, making the concepts and calculations easier to follow.

As the final exercise of the chapter, Exercise 1.5 focuses on percentage decrease and how a quantity changes when its value declines over successive periods. It covers calculating the reduced value after a percentage decrease, repeated percentage decrease on the updated amount, depreciation over successive periods, comparing the original and remaining values, and finding the final value after repeated decreases.

1.0Download Class 8 Maths Chapter 1 Ex 1.5 NCERT Solutions PDF

Download the detailed NCERT Solutions PDF for Exercise 1.5 for free and refer to the step-by-step solutions whenever required.

Chapter 1 : Fractions in Disguise Exercise 1.5

2.0Detailed NCERT Solutions on Fractions in Disguise Exercise 1.5 Class 8 Maths (Ganita Prakashan Term 2)

1. 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) $p \times 6 \times 4$ (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) $p + (p \times 0.06 \times 4)$

Sol. $P = p, R = 6, T = 4$ Amount $= p + I = p + \frac{p \times 6 \times 4}{100} = p + (p \times 0.06 \times 4)$ $= p + 0.24 p = 1.24 p$ Hence (vii) is correct.

2. 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? Sol. 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

3. 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? Sol. Interest (Giridhar) $= (1 + \frac{10}{100})^{3} = ₹4, 500$ For Raghava with compounding $A = 12, 500 \times (1 + \frac{10}{100})^{3} = 12, 500 \times \frac{1331}{1000}$ = ₹16637.5 (Raghava) Interest $=$ ₹16,637.5 - ₹12,500 $=$ ₹4137.50 Difference $=$ ₹4500 - ₹4137.50 = ₹362.50 Giridhar pays ₹362.5 more than Raghava.

4. 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?

Sol. ₹1000 becomes ₹2,000 Interest = ₹1000 Without compounding $1000 = \frac{1000 \times 10 \times t}{100}$ $t = 10$ years With compounding $1000 (1 + \frac{10}{100})^{n} = 2000$ $(1.1)^{n} = 2$ This can be done by hit and trial $1. 1^{2} = 1.21$ $1. 1^{3} = 1.331$ $1. 1^{4} = 1.4641$ $1. 1^{5} = 1.6$ $1. 1^{6} = 1.77$ $1. 1^{7} = 1.94$ $1. 1^{8} = 2.14$ 1.94 < 2 < 2.14

Time would be between 7 and 8 years (The nearest answer is 7.2 years)

5. 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? Sol. Population after 3 years $= 1.5 \times (1 + \frac{3}{100})^{3}$ crores $= 1.5 \times (1.03)^{3}$ crores $= 1.639$ crores

6. 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. Sol. Total bacteria after 2 hours $= 5, 06, 000 \times (1 + \frac{25}{100})^{2}$ $= 506000 \times (1.025)^{2} = 5, 31, 616$

3.0Key Concepts of Chapter 1- Fractions in Disguise Exercise 1.5

Understanding percentage decrease when a quantity declines over a period.
Calculating the reduced amount after applying a given percentage decrease.
Understanding repeated decline when the percentage decrease is applied to the updated amount.
Calculating depreciation of a quantity or value over successive periods.
Comparing the original amount with the amount remaining after one or more periods of decline.
Finding the final value after repeated percentage decreases.
Understanding how successive percentage decreases affect the overall change in a quantity.

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 1 Exercise 1.5

Explains how to find the new value after a given percent decrease.
Demonstrates the calculation of depreciation during successive periods.
Explains how the successive percentage decreases are to be calculated to the new value.
Deconstructs questions involving a quantity or value decreasing over time.
Shows how to find the final value after several percentage reductions.
Helps compare the original value to the value remaining after each successive drop.
Explains how the net change in a quantity is affected by successive percentage decreases.

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

Exercise 1.1

Exercise 1.2