Compound Interest
- The compound interest on ₹ 30000 at 7% pa for a certain time is ₹4347. The time is
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Given, CI = ₹ 4347, P = ₹ 30000 and R = 7%
By formula, CI = P[(1 + R/100)n - 1 ]
⇒ 4347 = 30000 [(1+7/100)n - 1]
⇒ (107/100)n = (4347/30000) + 1Correct Option: A
Given, CI = ₹ 4347, P = ₹ 30000 and R = 7%
By formula, CI = P[(1 + R/100)n - 1 ]
⇒ 4347 = 30000 [(1+7/100)n - 1]
⇒ (107/100)n = (4347/30000) + 1
⇒ (107/100)n = 34347/30000 = 11449/10000
⇒ (107/100)n = (107/100)2
∴ n = 2
- What is the difference between compound interest and simple interest for 2 yr on the sum of ₹ 1250 at 4% pa?
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Given, P = ₹ 1250, n = 2 yr and R = 4%
According to the formula,
Difference between compound interest and simple interest = PR2/1002Correct Option: C
Given, P = ₹ 1250, n = 2 yr and R = 4%
According to the formula,
Difference between compound interest and simple interest = PR2/1002
∴ Required difference = (1250 x 4 x 4)/(100 x 100) = ₹ 2
- The difference between compound and simple rates of interest on ₹ 10000 for 3 yr at 5% per annum is
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Required difference = P(R/100)2 x [(300 + R)/100]
= 10000(5/100)2 (305/100)Correct Option: A
Required difference = P(R/100)2 x [(300 + R)/100]
= 10000(5/100)2 (305/100)
= 76.25
- A borrowed sum was paid in the two annual installments of ₹ 121 each. If the rate of compound interest is 10% pa, what sum was borrowed?
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According to the question,
Borrowed sum = [121/{1/(1 + 10/100)}] + 121/(1 + 10/100)2Correct Option: B
According to the question,
Borrowed sum = [121/{1/(1 + 10/100)}] + 121/(1 + 10/100)2
= 121/(11/10) + 121/{(11/10) x (11/10)]
=110 + 100
= ₹ 210
