# Suppose a certain sum doubles in 2 years at r% rate of simple interest per annum or at R% rate of interest per annum compounded annually. We have

(a) r < R

(b) R < r

(c) R = r

(d) can’t be decided

**Solution:**

Suppose for the principal P, rate r% and time T, then simple interest

S is given as:

S = [P × r × T]/100 --- (1)

Given that the sum doubles in two years at simple rate of interest of r% we can state:

S = P(In other words the simple interest after two years will be the principal itself)

Therefore,

P = [P × r × 2]/100

P/P = [r × 2]/100

1 = [r × 2]/100

r = 100/2 = 50%

The compound interest relation is given as :

\(A = P(1 + R/100)^{T}\) --- (2)

Since the amount doubles in 1 year @ R% compounded annually we can write equation (2) as:

2P = P(1 + R/100)

2 = 1 + R/100

2 - 1 = R/100

1 = R/100

R = 100%

Therefore we can conclude that

R = 2r

Therefore r < R

**✦ ****Try This:** Suppose a certain sum doubles in 1 years at r% rate of simple interest per annum or at R% rate of interest per annum compounded annually. We have (a) r < R, (b) R < r, (c) R = r, (d) can’t be decided

Since a certain sum doubles in one year at r% simple interest, means that the simple

rate of interest is 100%. Since the amount doubles at R% compounded annually means that R is also equal to 100%.

**☛ Also Check: **NCERT Solutions for Class 8 Maths Chapter 8

**NCERT Exemplar Class 8 Maths Chapter 9 Problem 2**

## Suppose a certain sum doubles in 2 years at r% rate of simple interest per annum or at R% rate of interest per annum compounded annually. We have (a) r < R, (b) R < r, (c) R = r, (d) can’t be decided

**Summary:**

When a certain sum doubles in 2 years at r % rate of simple interest per annum or at R% rate of interest per annum compounded annually, then it can be concluded that r < R

**☛ Related Questions:**

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