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
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