NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.3

Class VIII Mathematics NCERT chapter 7 Ex. 7.3 Comparing Quantities updated for CBSE and State board students appearing in 2024-25 exams. All the exercises are in Hindi and English Medium free to use online or download in PDF file format. Videos related to exercise 7.3 is also given here, so that students can easily understand each questions. If someone is facing problem to access NCERT textbook solutions or other contents on Tiwari Academy, please contact us for help.

Interest on capital is the interest paid to owners for providing a firm with the required capital to start a business. It is similar to obtaining a loan from any financial institution. The partners are paid interest on the capital that remains outstanding.

When money is borrowed on simple interest then the interest is calculated uniformly on the original principal throughout the loan period.
S. I. = P X R X T 100

In such cases where interest is compounded yearly, the interest accrued during the first year is added to the principal and the amount so obtained becomes the principal for the second year. The amount at the end of the second year becomes the principal for the third year, and so on.

Find the compound interest on 25000 for 3 years at 10% per annum, compounded annually.
Principal for the first year = 25000.
S. I. = (P X R X T)/100
Interest for the first year = (25000 x 10 x 1)/100 = Rs. 2500
Amount at the end of the first year = (25000 + 2500) = Rs. 27500.
Principal for the second year = 27500.
Interest for the second year = (27500 x 10 x 1)/100 = Rs. 2750
Amount at the end of the second year = (27500 + 2750) = 30250.
Principal for the third year = 30250.
Interest for the third year = (30250 x 10 x 1)/100 = Rs. 3025
Amount at the end of the third year = (30250 + 3025) = 33275.
Compound interest = (33275 – 25000) = 8275.

In such cases, if the rate of interest is R% per annum then it is clearly % per half-year. The amount alter the first half-year becomes the principal for the next half-year, and so on. The method for calculating the compound interest in such cases is shown in the example given below.

Find the compound interest on 5000 for 1 year at 8% per annum, compounded half-yearly.
Rate of interest = 8% per annum = 4% per half-year.
Time = 1 year = 2 half-years.
Original principal = 5000.
Interest for the first half-year = Rs. (5000 x 4 x 1)/100 = Rs. 200
Amount at the end of the first half-year = (5000 + 200) = 5200.
Principal for the second half-year = 5200.
Interest for the second half-year = Rs. (5200 x 4 x 1)/100 = Rs. 208.
Amount at the end of the second half-year = (5200 + 208) = 5408
Compound interest = (5408 – 5000) = 408.

If the rate of interest is R% per annum and the interest is compounded quarterly, then the rate is R/4% per quarter.

Find the compound interest on 8000 for 1 year at 20% per annum, interest being payable quarterly.
Principal = 8000
Rate of Interest = 20% per annum = 20/4% = 5% per quarter
Time = 1 year = 4 quarters
Interest for 1st quarter = Rs. (8000 x 5 x 1)/100 = Rs. 400
Amount at the end of 1st quarter = (8000 + 400) = 8400.
Principal for 2nd quarter = 8400.
Interest for 2nd quarter = Rs. (8400 x 4 x 1)/100 = 420.
Amount at the end of 2nd quarter = (8400 + 420) = 8820.
Principal for 3rd quarter = 8820
Interest for 3rd quarter = Rs. (8820 x 5 x 1)/100 = 441.
Amount at the end of 3rd quarter = (8820 + 441) = 9261.
Principal for 4th quarter = 9261
Interest for 4th quarter = Rs. (9261 x 5 x 1)/100 = 463.05.
Amount at the end of 4th quarter = (9261 + 463.05) = 9724.05.
So, Compound Interest = (9724.05 – 8000) = 1724.05

The method of calculating the compound interest discussed previously is quite long and cumbersome, especially when the time period is large.
We can derive general formulae for calculating compound interest in various cases, as given below. It is very easy to calculate compound interest by using these formulae.
When the interest is compounded annually
Let principal = P, rate = R% per annum and time = n years. Then, the amount A is given by the formula.
A = Rs. P (1 + r/100)ⁿ

Find the amount of 8000 for 3 years, compounded annually at 10% per annum. Also, find the compound interest.
Here, p = 8000, R = 10% per annum and n = 3 years
amount after 3 years = Rs. 8000 (1 + 10/100)3
= 8000 x 11/10 x 11/10 x 11/10 = 10648
Thus, amount after 3 years = 10648. And, compound interest = (10648 – 8000) = 2648.

In each of the following situations we use the CI formula.

1. Increase (or decrease) in population.
2. The growth of bacteria when the rate of growth is known.
3. Depreciation in the values of machines, etc., at a given rate.

Note: For decrease (or depreciated value), we use the formula:
Decreased value (or depreciated value) = P {1 – R/100}ⁿ

Let principal = P, rate = R% per annum, time = n years.
Suppose that the interest is compounded half-yearly. Then,
Rate = R/2% per half-year, time = (2n) half-years, and amount = Rs. P x (1 + R/200)2n
Compound interest = (amount) – (principal)

Let principal = P, rate = R% per annum, time = n years.
Suppose that the interest is compounded quarterly, Then,
Rate = R/4% per half-year, time = (4n) half-years, and amount = Rs. P x (1 + R/400)4n
Compound interest = (amount) – (principal)

(i) Suppose A can finish a piece of work in 8 days.
Then, work done by A in 1 day = [by unitary method].
(ii) Suppose that the work done by A is 1 day is.
Then, time taken by A to finish the whole work = 6 days.

(i) Suppose A can finish a work in n days.
Then, work done by A in 1 day = 1/n
(ii) Suppose that the done by A in 1 day is 1/n.
Then, time taken by A to finish the whole work = n days.

A cistern or a water tank is connected with two types of pipes.
(i) Inlet: The pipe which fills the tank is called on inlet.
(ii) Outlet: The pipe which empties the tank is called an outlet.
Rule 1. Suppose an outlet fills a tank in n hours.
Then, part of the tank filled in 1 hour = 1/n,
i.e., work done by the inlet in 1 hour = 1/n.
Rule 2. Suppose an outlet empties a full tank in m hours.
Then, part of the tank emptied in 1 hour = 1/m
i.e., work done by the outlet in 1 hour = (-1/m)