Simple Interest Questions and Answers

Answer: D

\[ \text{SI} = \frac{P \times R \times T}{100} \]

The rate is 8% per annum and the period two and a half years.

\[ \text{SI} = \frac{6500 \times 8 \times 2.5}{100} = \text{Rs. } 1300 \]

\[ \text{amount} = 6500 + 1300 = \text{Rs. } 7800 \]

The question asks what the sum will yield in total, so the principal is included; the interest alone is Rs. 1300.

Answer: B

\[ \text{SI} = \frac{P \times R \times T}{100} \]

Equal interests give the ratio of the two parts.

\[ \frac{p_{1} \times 6.25 \times 8}{100} = \frac{p_{2} \times 7 \times 5}{100} \]

\[ 50\,p_{1} = 35\,p_{2} \quad\Longrightarrow\quad \frac{p_{1}}{p_{2}} = \frac{7}{10} \]

So the parts are 7 and 10 of 17 shares:

\[ p_{1} = \frac{7}{17} \times 20400 = \text{Rs. } 8400, \qquad p_{2} = \text{Rs. } 12{,}000 \]

The smaller part is Rs. 8400 - the one at the higher \( R \times T \) product, since it needs less capital for the same return.

Answer: A

\[ \text{SI} = \frac{P \times R \times T}{100} \]

The four parts must account for the whole, so the last fraction is

\[ 1 - \frac{1}{5} - \frac{1}{3} - \frac{2}{5} = \frac{15 - 3 - 5 - 6}{15} = \frac{1}{15} \]

Adding the four interests for one year:

\[ \frac{P}{100}\left(\frac{3}{5} + \frac{5}{3} + \frac{18}{5} + \frac{11}{15}\right) = 297 \]

\[ \frac{P}{100} \times \frac{9 + 25 + 54 + 11}{15} = \frac{P}{100} \times \frac{99}{15} = 297 \]

\[ P = \frac{297 \times 1500}{99} = \text{Rs. } 4500 \]

The three named fractions come to \( \tfrac{14}{15} \), so the unnamed remainder is \( \tfrac{1}{15} \) - easy to overlook.

Answer: B

\[ \text{SI} = \frac{P \times R \times T}{100} \]

The two investments share the rate and the four years, so the Rs. 130 gap comes entirely from the Rs. 250 difference in principal.

\[ \frac{(2250 - 2000) \times R \times 4}{100} = 130 \]

\[ \frac{250 \times 4R}{100} = 10R = 130 \quad\Longrightarrow\quad R = 13\% \]

Working out each interest separately gives the same answer but needs two calculations instead of one.

Answer: D

Simple interest of Rs. 20 over two years is Rs. 10 a year. Compound interest exceeds it by Re. 1, which is one year's interest on that first Rs. 10:

\[ 10 \times \frac{R}{100} = 1 \quad\Longrightarrow\quad R = 10\% \]

Now use the simple interest to find the principal:

\[ 20 = \frac{P \times 10 \times 2}{100} = 0.2P \quad\Longrightarrow\quad P = \text{Rs. } 100 \]

Checking: at 10%, Rs. 100 gives compound interest of \( 100(1.21 - 1) = 21 \).

Answer: C

\[ \text{SI} = \frac{P \times R \times T}{100} \]

Find the original rate first.

\[ \text{SI} = 4600 - 4000 = 600 \quad\Longrightarrow\quad 600 = \frac{4000 \times R \times 5}{100} = 200R \]

\[ R = 3\% \]

At \( 3 + 3 = 6\% \):

\[ \text{SI} = \frac{4000 \times 6 \times 5}{100} = 1200 \quad\Longrightarrow\quad \text{amount} = \text{Rs. } 5200 \]

The extra 3% over 5 years adds \( 4000 \times 0.15 = 600 \), doubling the interest.

Answer: A

\[ \text{SI} = \frac{P \times R \times T}{100} \]

Beyond five years leaves \( 9 - 5 = 4 \) years at 14%, so the three slabs are:

\[ P\left(\frac{6 \times 2 + 9 \times 3 + 14 \times 4}{100}\right) = 8550 \]

\[ P\left(\frac{12 + 27 + 56}{100}\right) = 0.95P = 8550 \]

\[ P = \text{Rs. } 9000 \]

Answer: D

Let us assume Mr. Karan borrowed amount is Rs. A. (the principal)
By formula of simple interest, S.I. = PRT / 100
Where P = The principal, R = Rate of interest as a %, T = Time in years
S.I. = (P * 6 * 9)/100 = 54 P/100
Amount = Principal + S.I.
8110 = P + (54P/100)
8110 = (100P + 54P)/100
8110 = 154P/100
P = (8110 * 100)/154 = Rs. 5266.233

Answer: A

P = Rs 900, R = 3 ½ % = 7/2 %, T = 2 years.
Therefore,
S.I. = PRT/100
S.I. = Rs. (900 x 7/2 x 2/100) = Rs. 63
Now, P = Rs. 200, S.I. = Rs. 63, R = 5 %
Time = ((100 x 63) / (200 x 5) ) years = 6.3 years

Answer: C

Simple interest of Rs. 800 over two years is Rs. 400 a year. Compound interest exceeds it by

\[ 840 - 800 = \text{Rs. } 40 \]

which is one year's interest on that first Rs. 400:

\[ 400 \times \frac{R}{100} = 40 \quad\Longrightarrow\quad R = 10\% \text{ p.a.} \]

The sum works out at Rs. 4000, and \( 4000(1.21 - 1) = 840 \) confirms it - but the principal is not needed to answer.