Simple interest
 Mr. Deepak invested an amount of ₹ 21250 for 6 yr. At what rate of simple interest, will he obtain the total amount of ₹ 26350 at the end of 6 yr ?

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Rate = (SI x 100) / (Principal x Time)
Correct Option: E
SI = 26350  21250 = ₹ 5100
∴ Rate = (SI x 100) / (Principal x Time)
= (5100 x 100) / (21250 x 6)
= 4%
 At what rate per annum will the simple interest on a certain sum of money be 1/5 of the amount in 10 yr ?

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Let sum = P, then SI = P/5 time = 10 yr.
∴ Rate = (100 x P)/(P x 5 x 10)Correct Option: B
Let sum = P, then SI = P/5 time = 10 yr.
∴ Rate = (100 x P)/(P x 5 x 10) = 2%
 Simple interest for the sum of ₹ 1500 is ₹ 30 in 4 yr and ₹ 60 in 8 yr. Find the rate of simple interest.

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Given, SI_{2} = 60, SI_{1} = 30, T_{1} = 4 yr, T_{2} = 8 yr
According to the question,
[(1500 x R x 8)/100]  [(1500 x R x 4)/100] = 60  30Correct Option: C
Given, SI_{2} = 60, SI_{1} = 30, T_{1} = 4 yr, T_{2} = 8 yr
According to the question,
[(1500 x R x 8)/100]  [(1500 x R x 4)/100] = 60  30
⇒ (6000 x R)/100 = 30
∴ R = 30/60 = 1/2 = 0.5%
 At simple interest, a sum becomes 3 times in 20 yr. Find the time, in which the sum will be double at the same rate of interest.

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Here, n = 3, m = 2, T_{1} = 20 yr
∴ T_{2} = [(m 1) / (n  1)] x T_{1}Correct Option: B
Here, n = 3, m = 2, T_{1} = 20 yr
∴ T_{2} = [(m 1) / (n  1)] x T_{1}
= (2  1) / (3  1) x 20 = 10 yr
 At a simple interest, a sum amounts to ₹ 1012 in 2^{1}/_{2} and becomes ₹ 1067.20 in 4 yr. What is the rate of interest

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Given, T_{1} = 2^{1}/_{2} yr, T_{2} = 4 Yr
According to the question.
[P + (P x R x 4)/100]  [P + (P x R x 2.5)/100] = 1067.20  1012 = 55.2
For 4 yr.
SI = PRT/100 = (3680 x 4)/100 = ₹ 147.2Correct Option: C
Given, T_{1} = 2^{1}/_{2} yr, T_{2} = 4 Yr
According to the question.
[P + (P x R x 4)/100]  [P + (P x R x 2.5)/100] = 1067.20  1012 = 55.2
⇒ (1.5 x P x R)/100 = 55.2
⇒ PR = (552 x 100) / 15 = 3680
For 4 yr.
SI = PRT/100 = (3680 x 4)/100 = ₹ 147.2
∴ Sum (P) = 1067.2  147.2 = ₹ 920
We have, PR = 3680
∴ R = 3680/P = 3680/920 = 4%