Class 8 Maths Exercise 7.3 Comparing Quantities: NCERT Solutions

Presenting the NCERT solutions exercise 7.3 from chapter Comparing Quantities. You will learn few new concepts in this exercise like compound interest, calculating compound interest, deducing a formula for compound interest, and applications of compound interest formula. Lets have a look at some of the basic concepts you would require while solving the questions.

Compound Interest

Compound Interest is the interest on a loan or deposit calculated based on both the initial principal and the accumulated interest from previous periods. It is common in bank accounts and investments.

Calculating Compound Interest

For example, if ₹10,000 is invested at an annual interest rate of 5% compounded yearly for 2 years, the compound interest is calculated as follows:
First Year Interest: 5% of ₹10,000 = ₹500.
Second Year Interest: 5% of (₹10,000 + ₹500) = ₹525.
Total Interest: ₹500 + ₹525 = ₹1025.
Total Amount: ₹10,000 + ₹1025 = ₹11,025.

Deducing a Formula for Compound Interest

The formula for compound interest is A = P(1 + r/n)ⁿᵗ, where:
A = the future value of the investment/loan, including interest,
P = principal amount (initial investment),
r = annual interest rate (decimal),
n = number of times interest is compounded per year,
t = number of years.

Applications of Compound Interest Formula

Compound interest is used in various financial products and investments, like savings accounts, fixed deposits, and mutual funds. Understanding the compound interest formula is going to help in calculating the future value of investments and comparing different financial products.

NCERT Solutions for Class 8 Maths Exercise 7.3 Chapter 7 Comparing Quantities

Question 1. The population of a place increased to 54000 in 2003 at a rate of 5% per annum. (i) find the population in 2001. (ii) what would be its population in 2005?

Solution

(i) find the population in 2001

The population in 2003 is the compounded amount of the population in 2001.
Given:
Population in 2003 = 54000
Rate of increase = 5%
Time Period = 2 year

Let the population of year 2001 be P
Population in 2003 = x(1 + 5/100)²
=) 54000 = P((100 + 5)/100)²
=) 54000 = P(105/100)²
=) 54000 = P × (21/20) × (21/20)
=) 54000 × 20 × 20 = P × 21 × 21
=) P = (54000 × 20 × 20)/(21 × 21) = 21600000/(21 × 21) = 48980
Population in 2001 is 48980

Method 2:
Population in 2003 = Population in 2001 × (1 + 5/100)².

54000 = Population in 2001 × 1.1025.

Population in 2001 = 54000 / 1.1025 = 48980

(ii) what would be its population in 2005

Year 2005 is 2 years after the given 2003
Given:
Population in 2003 = 54000
Rate of increase = 5%
Time Period = 2 year

Let the population of 2005 be X

X = P(1 + R/100)ⁿ
= 54000 (1 + 5/100)²
= 54000 ((100 + 5)/100)²
= 54000 (105/100)²
= 54000 × 105/100 × 105/100
= 59535

The population in 2005 is 59535

Question 2. 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 506000.

Solution

Given:
Initial count of bacteria = 506000
Rate of increase = 2.5% per hour
Time Period = 2 hours

Let the count of bacteria at the end of 2 hours be B

B = Initial count × (1 + R/100)ⁿ
= 506000 × (1 + 2.5/100)²
= 506000 × (102.5/100)²
= 506000 × 1.025 × 1.025
= 506000 × 1.050625
= 531616.25

The count of bacteria at the end of 2 hours is approximately 531616

Question 3. A scooter was bought at ₹ 42000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Solution

Given:
Initial value of scooter = ₹ 42000
Rate of depreciation = 8% per annum
Time Period = 1 year

Let the value of the scooter after one year be V

V = Initial value × (1 – R/100)ⁿ
= 42000 × (1 – 8/100)
= 42000 × (92/100)
= 42000 × 0.92
= 38640

The value of the scooter after one year is ₹ 38640

Additional Challenging Questions like EXERCISE 7.3

Since there are only 3 questions in EXERCISE 7.3, we are presenting additional 5 questions for you to practice.

Question 1. The price of a commodity increases by 15% annually. If its current price is ₹ 5000, what will be its price after two years?

Solution

Price after 1st year = ₹ 5000 × (1 + 15/100) = ₹ 5000 × 1.15 = ₹ 5750.

Price after 2nd year = ₹ 5750 × (1 + 15/100) = ₹ 5750 × 1.15 ≈ ₹ 6612.50.

Question 2. A car’s value depreciates by 10% every year. If the current value of the car is ₹ 800,000, what will be its value after 3 years?

Solution

Value after 1st year = ₹ 800,000 × (1 – 10/100) = ₹ 800,000 × 0.90 = ₹ 720,000.

Value after 2nd year = ₹ 720,000 × 0.90 = ₹ 648,000.

Value after 3rd year = ₹ 648,000 × 0.90 = ₹ 583,200.

Question 3. A population of bacteria increases by 20% every 3 hours. If initially there are 10,000 bacteria, how many will there be in 15 hours?

Solution

After 3 hours = 10,000 × (1 + 20/100) = 10,000 × 1.20 = 12,000.

The pattern repeats every 3 hours for 15 hours (5 cycles).

After 15 hours = 10,000 × 1.20⁵ ≈ 24,883.2 ≈ 24,883 bacteria.

Question 4. A student scores 70% in the first exam and wishes to improve his score by 30% in the next exam. What percentage should he aim for in the second exam?

Solution

Improvement required = 70% × 30% = 21%.

Target percentage in the second exam = 70% + 21% = 91%.

Question 5. An investor receives an annual interest of 5% on a certain amount of fixed deposit. If he received ₹ 2500 as interest, find the amount deposited.

Solution

Let the deposited amount be ₹ x.

Annual interest = 5% of x = 0.05x.

Since the interest received is ₹ 2500, we have 0.05x = ₹ 2500.

Therefore, x = ₹ 2500 / 0.05 = ₹ 50,000.

Practice Worksheet with Challenging Questions For Class 8 Ex. 7.3 Comparing Quantities

Questions

1. If a sum of money doubles in 5 years on compound interest, what is the annual interest rate?
2. Find the compound interest on ₹10,000 for 1 year at 7% per annum, compounded half-yearly.
3. If the price of a commodity decreases by 5% each year, what will be its price after 3 years, if the current price is ₹8000?
4. Calculate the amount when ₹15,000 is invested for 2 years at an annual interest rate of 10%, compounded annually.
5. If the population of a town increases by 2% per annum and its current population is 50000, what will it be after two years?
6. A car’s value depreciates at the rate of 10% per annum. Find its value after two years if it is currently worth ₹400,000.
7. A student’s marks increased by 20% from the first test to the second. If the marks in the second test were 72, what were they in the first test?
8. If an investor gets 5% more on his investment every year, how much more will he get on an investment of ₹20000 after 2 years?
9. The diameter of a wheel increases by 10% due to wear and tear. By what percent does the area increase?
10. A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits ₹2000 each on 1st January and 1st July of a year. At the end of the year, the amount he would get is?
11. A shop offers a discount of 20% on marked price. If the sale price after discount is ₹1200, what is the marked price?
12. If the price of gold increases by 15% in a year, how much gold can now be bought for the same amount that could buy 10 grams a year ago?
13. Find the effective percentage increase in an area of a square if each of its sides is increased by 10%.
14. If the side of a square is increased by 30%, by what percent does the area of the square increase?
15. Find the compound interest on ₹5000 for 2 years at 8% per annum, compounded quarterly.

Hints

1. Use the formula A = P(1 + r/n)^(nt) to find the rate.
2. Calculate the interest for two six-month periods.
3. Apply the percentage decrease successively for three years.
4. Calculate the compound amount for 2 years.
5. Apply the percentage increase successively for two years.
6. Use the formula for depreciation to find the value after two years.
7. Calculate the original marks before the 20% increase.
8. Calculate 5% of ₹20000 and then increase it by 5% for the second year.
9. Use the formula for the area of a circle and calculate the percentage increase.
10. Find the compound interest for both the investments separately and add them.
11. Find the original price before the 20% discount.
12. Calculate the decreased quantity that can be bought with the same amount of money.
13. Calculate the new area and compare it with the original area.
14. Use the formula for the area of a square with the increased side.
15. Calculate the compound interest for 8 quarters.

Answers

1. 14.87%
2. ₹714
3. ₹6840
4. ₹18150
5. 52020
6. ₹324000
7. 60 marks
8. ₹2100
9. 21% increase
10. ₹4410
11. ₹1500
12. 8.7 grams
13. 21% increase
14. 69% increase
15. ₹979.04