## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 8 Simple and Compound Interest Ex 8.3

Question 1.

Calculate the amount and compound interest on

(i) ₹15000 for 2 years at 10% per annum compounded annually.

(ii) ₹156250 for \(1 \frac{1}{2}\) years at 8% per annum compounded half yearly.

(iii) ₹ 100000 for 9 months at 4% per annum compounded quarterly.

Solution:

(i) Principal (P) = ₹15000

Rate (R) = 10% p.a.

Period (n) = 2 years

∴ C.I. = A – P = ₹18150 – 15000 = ₹3150

(ii) Principal (P) = ₹156250

Rate (R) = 8% p.a. or 4% half-yearly

Period (n) = \(1 \frac{1}{2}\) years = 3 half-year

∴ C.I. = A – P = ₹175760 – ₹156250 = ₹19510

Question 2.

Find the difference between the simple interest and compound interest on ₹4800 for 2 years at 5% per annum, compound interest being reckoned annually.

Solution:

Principal (P) = ₹4800

Rate (R) = 5% p.a.

Period (n) = 2 years

Question 3.

Find the compound interest on ₹3125 for 3 years if the rates of interest for the first, second and third year are respectively 4%, 5% and 6% per annum.

Solution:

Principal (P) = ₹3125

Rate of interest for continuous 3 years = 4%, 5%, 6%

Period (n) = 3 years

∴ C.I. = A – P = ₹3617·25 – ₹3125 = ₹492·25

Question 4.

Kamla borrowed ₹26400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?

Solution:

Sum borrowed (P) = ₹26400

Rate (R) = 15% p.a.

Period (n) = 2 years 2 months

Question 5.

Anil borrowed ₹18000 from Rakesh at 8% per annum simple interest for 2 years. If Anil had borrowed this sum at 8% per annum compound interest, what extra amount would he have to pay?

Solution:

Sum borrowed (P) = ₹18000

Rate (R) = 8% p.a.

Time (n) = 2 years

∴ C.I. = ₹20995.20 – 18000 = ₹2995.20

Now difference in both interest = ₹2995.20 – ₹2880 = ₹115.20

Question 6.

Mukesh borrowed 775000 from a bank. If the rate of interest is 12% per annum, find the amount he would be paying after \(1 \frac{1}{2}\) years if the interest is

(i) compounded annually

(ii) compounded half-yearly.

Solution:

Sum borrowed (P) = ₹75000

Rate (R) = 12% p.a. or 6% half-yearly

Period (n) = \(1 \frac{1}{2}\) years or 3 half-years

(i) When the interest compounded yearly

(ii) When the interest compounded half-yearly, then

Question 7.

Aryaman invested ₹10000 in a company, he would be paid interest at 7% per annum compounded annually. Find

(i) the amount received by him at the end of 2 years.

(ii) the interest for the 3rd year.

Solution:

Investment to a company (P) = ₹10000

Rate of interest (R) = 7% p.a.

Period (n) = 2 years

Question 8.

What sum of money will amount to ₹9261 in 3 years at 5% per annum compound interest?

Solution:

Amount (A) = ₹9261

r = 5%, Time (t) = 3

P = ?

Hence the sum of money = ₹8000

Question 9.

What sum invested for \(1 \frac{1}{2}\) years compounded half-yearly at the rate 8% p.a. will amount to ₹ 140608?

Solution:

Amount (A) = ₹140608

Rate (R) = 8% p.a. = 4% half-yearly

Period (n) = \(1 \frac{1}{2}\) years = 3 half-year

∴ Principal = ₹ 125000

Question 10.

At what rate percent will ₹2000 amount to ₹2315·25 in 3 years at compound interest?

Solution:

Principal (P) = ₹2000

Amount (A) = ₹2315·25

Period (n) = 3 years

Let rate of interest = r % p.a.

∴ r = \(\frac{100}{20}\) = 5

∴ Rate of interest = 5% p.a.

Question 11.

If ₹40000 amounts to ₹46305 in \(1 \frac{1}{2}\) years, compound interest payable half yearly, find the rate of interest per annum.

Solution:

Amount (A) = ₹17576

Principal (P) = ₹15625

Rate (R) = 4% p.a.

Comparing, we get n = 3

∴ Time 3 years

Question 12.

In what time will ₹15625 amount to ₹ 17576 at 4% per annum compound interest?

Solution:

Amount (A) = ₹ 17576

Principal (P) = ₹15625

Rate = 4% p.a.

Let Period = n years

We know that

∴ n = 3

Hence time = 3 years

Question 13.

₹ 16000 invested at 10% p.a. compounded semi-annually, amounts to ₹18522. Find the time period of investment.

Solution:

We know thatPrincipal (P) = ₹16000

Amount (A) = ₹18522

Rate = 10% p.a. or 5% semi-annually

Let period = n half-years

We know that

Hence time = \(\frac{3}{2}=1 \frac{1}{2}\) years