ML Aggarwal Solutions Class 9 Mathematics Solutions for Compound Interest Exercise 2.1 in Chapter 2 - Compound Interest
Question 5 Compound Interest Exercise 2.1
A man invests ₹ 5000 for three years at a certain rate of interest, compounded annually. At the end of
one year it amounts to ₹ 5600. Calculate:
(i) the rate of interest per annum
(ii) the interest accrued in the second year.
(iii) the amount at the end of the third year.
Answer:
Compound interest is computed on both the principal and the interest earned over a period of time. It differs from simple interest in that interest is not added to the principle when the interest for the next month is calculated.
It is given that
Principal = ₹ 5000
Consider r% p.a. as the rate of interest
(i) We know that
At the end of one year
Interest = Prt/100
Substituting the values
= (5000 × r × 1)/ 100
= 50r
Here
Amount = 5000 + 50r
We can write it as
5000 + 50r = 5600
By further calculation
50r = 5600 – 5000 = 600
So we get
r = 600/50 = 12
Hence, the rate of interest is 12% p.a.
(ii) We know that
Interest for the second year = (5600 × 12 × 1)/ 100
= ₹ 672
So the amount at the end of the second year = 5600 + 672
= ₹ 6272
(iii) We know that
Interest for the third year = (6272 × 12 × 1)/ 100
= ₹ 752.64
So the amount after the third year = 6272 + 752.64
= ₹ 7024.64
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Q.
A sum of Rs. 44,200 is divided between John and Smith, 12 years and 14 years old respectively, in such a way that if their portions be invested at 10 percent per annum compound interest, they will receive equal amounts on reaching 16 years of age.
(i) What is the share of each out of Rs. 44,200 ?
(ii) What will each receive, when 16 years old ?
Given: P = Rs. 5,000; A = Rs. 6,272 and n = 2 years.
(i) ∴ `"A" = "P"( 1 + r/100)^n`
⇒ `6,272 = 5,000( 1 + r/100 )^2`
⇒ `6,272/5,000 = ( 1 + r/100 )^2`
⇒ `784/625 = ( 1 + r/100 )^2`
⇒ `(28/25)^2 = ( 1 + r/100 )^2`
On comparing,
`28/25 = 1 + r/100`
On solving, we get
r = 12%
(ii) Amount at the third year
= `5,000( 1 + 12/100 )^3`
= `5000(28/25)^3`
= Rs. 7,024.64