Maharashtra Board Class 8 Maths Chapter 14 Compound Interest Set 14.1 Solutions

Question 1. Find the amount and the compound interest.

Answer:
(i) Solution:
Here P = Rs 2000, R = 5 p.c.p.a. and N = 2 years
\(A=P \left[1+\frac{R}{100}\right]^N\)
\( = 2000 \left[1+\frac{5}{100}\right]^2 \)
\( = 2000 \left[\frac{100+5}{100}\right]^2 \)
\( = 2000 \left[\frac{105}{100}\right]^2 \)
\( = 2000 \left[\frac{21 \times 5}{20 \times 5}\right]^2 \)
\( = 2000 \left[\frac{21}{20}\right]^2 \)
\( = 2000 \left[\frac{441}{400}\right] \)
\( = 5 \times 441 \)
\( \therefore \) A = Rs 2205
I = Amount (A) - Principal (P)
= 2205 - 2000
= Rs 205
\( \therefore \) The amount is Rs 2205 and the compound interest is Rs 205.
(ii) Solution:
Here, P = Rs 5000, R = 8 p.c.p.a. and N = 3 years
\(A=P \left[1+\frac{R}{100}\right]^N\)
\( = 5000 \left[1+\frac{8}{100}\right]^3 \)
\( = 5000 \left[\frac{100+8}{100}\right]^3 \)
\( = 5000 \left[\frac{108}{100}\right]^3 \)
\( = 5000 \left[\frac{27 \times 4}{25 \times 4}\right]^3 \)
\( = 5000 \left[\frac{27}{25}\right]^3 \)
\( = 5000 \times \frac{27}{25} \times \frac{27}{25} \times \frac{27}{25} \)
\( = \frac{8}{25} \times 27 \times 27 \times 27 \)
\( = 0.32 \times 19683 \)
\( \therefore \) A = Rs 6298.56
I = Amount (A) - Principal (P)
= 6298.56 - 5000
= Rs 1298.56
\( \therefore \) The amount is Rs 6298.56 and the compound interest is Rs 1298.56.
(iii) Solution:
Here, P = Rs 4000, R = 7.5 p.c.p.a. and N = 2 years
\(A=P \left[1+\frac{R}{100}\right]^N\)
\( = 4000 \left[1+\frac{7.5}{100}\right]^2 \)
\( = 4000 \left[1+\frac{75}{1000}\right]^2 \)
\( = 4000 \left[\frac{1000+75}{1000}\right]^2 \)
\( = 4000 \left[\frac{1075}{1000}\right]^2 \)
\( = 4000 \left[\frac{43 \times 25}{40 \times 25}\right]^2 \)
\( = 4000 \left[\frac{43}{40}\right]^2 \)
\( = 4000 \times \frac{43}{40} \times \frac{43}{40} \)
\( = 2.5 \times 43 \times 43 \)
\( \therefore \) A = Rs 4622.50
I = Amount (A) - Principal (P)
= 4622.50 - 4000
= Rs 622.50
\( \therefore \) The amount is Rs 4622.50 and the compound interest is Rs 622.50.
In simple words: Compound interest calculations involve finding the future value (amount) by adding interest to the principal, and then calculating interest on the new total. The compound interest is the difference between the final amount and the initial principal.

🎯 Exam Tip: Always state the given values (Principal, Rate, Time) clearly. Ensure correct application of the compound interest formula \(A = P(1 + \frac{R}{100})^N\) and perform calculations accurately to avoid errors in the final amount and interest.

Question 2. Sameerrao has taken a loan of Rs 12500 at the rate of 12 p.c.p.a. for 3 years. If the interest is compounded annually then how many rupees should he pay to clear his loan?
Answer:
Solution:
Here, P = Rs 12,500, R = 12 p.c.p.a. and
N = 3 years
\(A=P \left[1+\frac{R}{100}\right]^N\)
\( = 12500 \left[1+\frac{12}{100}\right]^3 \)
\( = 12500 \left[\frac{100+12}{100}\right]^3 \)
\( = 12500 \left[\frac{112}{100}\right]^3 \)
\( = 12500 \left[\frac{28 \times 4}{25 \times 4}\right]^3 \)
\( = 12500 \left[\frac{28}{25}\right]^3 \)
\( = 12500 \times \frac{28}{25} \times \frac{28}{25} \times \frac{28}{25} \)
\( = \frac{200}{25} \times 28 \times 28 \times 28 \)
\( = 0.8 \times 28 \times 28 \times 28 \)
\( = \) Rs 17,561.60
Sameerrao should pay Rs 17,561.60 to clear his loan.
In simple words: Sameerrao needs to pay back the accumulated amount after 3 years, which is calculated using the compound interest formula. This amount covers both the principal loan and the interest accrued over the period.

🎯 Exam Tip: When calculating loan repayment with compound interest, the final answer is always the 'Amount' (A), which includes both the principal and the compound interest. Carefully multiply fractions or decimals to get the precise value.

Question 3. To start a business Shalaka has taken a loan of Rs 8000 at a rate of \(10\frac{1}{2}\) p.c.p.a. After two years how much compound interest will she have to pay?
Answer:
Solution:
Here, P = Rs 8000, N = 2 years and
\(R = 10\frac{1}{2}\% = \frac{21}{2}\% = 10.5\) p.c.p.a.
\(A=P \left[1+\frac{R}{100}\right]^N\)
\( = 8000 \left[1+\frac{10.5}{100}\right]^2 \)
\( = 8000 \left[1+\frac{105}{1000}\right]^2 \)
\( = 8000 \left[\frac{1000+105}{1000}\right]^2 \)
\( = 8000 \left[\frac{1105}{1000}\right]^2 \)
\( = 8000 \left[\frac{221 \times 5}{200 \times 5}\right]^2 \)
\( = 8000 \left[\frac{221}{200}\right]^2 \)
\( = 8000 \times \frac{221}{200} \times \frac{221}{200} \)
\( = \frac{1}{5} \times 221 \times 221 \)
\( = 0.2 \times 48,841 \)
\( = \) Rs 9768.20
I = Amount (A) - Principal (P)
= 9768.20 - 8000
= Rs 1768.20
\( \therefore \) After two years Shalaka will have to pay Rs 1768.20 as compound interest.
In simple words: Shalaka's compound interest is calculated by first finding the total amount accumulated after two years, using the principal loan and the given interest rate. The interest she has to pay is the difference between this total amount and her original loan.

🎯 Exam Tip: Pay close attention to the specific question asked - whether it's the total 'Amount' or just the 'Compound Interest'. Mixed fractions in rates should be converted to decimals or improper fractions for easier calculation, and intermediate steps should be clearly shown for partial credit.