Question : Find the Rate percent per annum, if Rs. 2000 amounts to Rs. 2315.25 in a year and a half, with interest being compounded half-yearly.
Option 1: 11.5%
Option 2: 10%
Option 3: 5%
Option 4: 20%
Correct Answer: 10%
Solution : Amount(A) = Rs. 2315.25 Principal(P) = Rs. 2000 Time = 1.5 years ($n$ = 3) By applying the formula: Amount = P[$(1+\frac{r}{100})^n$] where P is principal, $r$ is the rate of interest per term for $n$ terms. Since CI is compounded half-yearly. Let the rate be $R\%$ per annum. So, $r=\frac{R}{2}$ Putting the value in the formula of CI, $2315.25=2000(1+\frac{\frac{R}{2}}{100})^{3}$ ⇒ $\frac{2315.25}{2000}=(1+\frac{\frac{R}{2}}{100})^{3}$ ⇒ $\frac{231525}{200000}=(1+\frac{\frac{R}{2}}{100})^{3}$ ⇒ $\frac{9261}{8000}=(1+\frac{\frac{R}{2}}{100})^{3}$ ⇒ $(\frac{21}{20})^{3}=(1+\frac{R}{100})^{3}$ Comparing both sides, we get, ⇒ $1+\frac{\frac{R}{2}}{100}=\frac{21}{20}$ ⇒ $1+\frac{\frac{R}{2}}{100}=1+\frac{1}{20}$ ⇒ $\frac{R}{200}=\frac{1}{20}$ ⇒ $R=\frac{200}{20}$ = 10% Hence, the correct answer is 10%
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Question : A sum amounts to INR 7,562 in 4 years and to INR 8,469.44 in 5 years at a certain rate per annum, when the interest is compounded yearly. The rate of interest is:
Option 1: 12%
Option 2: 15%
Option 3: 20%
Option 4: 8%
Question : What is the Compound Interest (in Rs.) on Rs. 12500 at the rate of 12% per annum compounded yearly for 2 years?
Option 1: Rs. 3000
Option 2: Rs. 2980
Option 3: Rs. 3050
Option 4: Rs. 3180
Question : What will be the amount received on Rs. 25000 at the rate of 20% per annum compounded yearly for 4 years?
Option 1: Rs. 51840
Option 2: Rs. 50350
Option 3: Rs. 53550
Option 4: Rs. 48750
Question : What will be the amount of Rs. 12500 at the rate of 20% per annum compounded yearly for 3 years?
Option 1: Rs. 21080
Option 2: Rs. 21560
Option 3: Rs. 20600
Option 4: Rs. 21600
Question : At what time will Rs. 64000 amount to Rs. 68921 at 5% per annum, with interest being compounded half-yearly?
Option 1: $3$ years
Option 2: $2\frac{1}{2}$ years
Option 3: $2$ years
Option 4: $1\frac{1}{2}$ years
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