Simple interest
- A sum becomes its double in 10 yr. Find the annual rate of simple interest.
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Here, n = 2, T = 10 yr
∴ R = 100 (n - 1)/TCorrect Option: C
Here, n = 2, T = 10 yr
∴ R = 100 (n - 1)/T
= 100(2 - 1)/10
= 100/10
= 10%
- The simple interest on a sum of money at 8% per annum for 6 yr is half the sum. What is the sum ?
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Let sum = P
Then, according to the question.
SI = P/2
∴ P/2 = (P x 8 x 6)/100Correct Option: D
Let sum = P
Then, according to the question.
SI = P/2
∴ P/2 = (P x 8 x 6)/100
∴ It is clear that data is inadequate.
- The simple interest on a sum of money at 9% per annum for 5 yr is half the sum. What is the sum ?
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Let the sum be P.
∵ SI = P/2
∴ P/2 = (P x 9 x 5)/100Correct Option: D
Let the sum be P.
∵ SI = P/2
∴ P/2 = (P x 9 x 5)/100
Clearly data is inadequate.
- A sum of money amounts to ₹ 2240 at 4% per annum simple interest in 3 yr. The interest on the same sum for 6 months at 3.5% per annum is
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If the sum be ₹ P, then
(2240 - P) = (P x 4 x 3)/100
⇒ 2240 = 12P/100 + P
⇒ 2240 = 112P/100
∴ P = (2240 x 100)/112 = ₹ 2000
Now, required interest,
SI = PRT/100 = [{2000 x (7/2) x (1/2)} /100]Correct Option: C
If the sum be ₹ P, then
(2240 - P) = (P x 4 x 3)/100
⇒ 2240 = 12P/100 + P
⇒ 2240 = 112P/100
∴ P = (2240 x 100)/112 = ₹ 2000
Now, required interest,
SI = PRT/100 = [{2000 x (7/2) x (1/2)} /100]
= ₹ 35
- The simple interest on ₹ 4000 in 3 yr at the rate of R % per annum equals equals to the simple interest on ₹ 5000 at the rate of 12% per annum in 2 yr The value of R is
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Since, the two simple interest are equal.
Then, (4000 x 3 x R)/100 = (5000 x 12 x 2)/100Correct Option: D
Since, the two simple interest are equal.
Then, (4000 x 3 x R)/100 = (5000 x 12 x 2)/100
∴ R = 10%