Simple Interest Questions and Answers

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: B

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

Equal interests give the ratio of the parts.

\[ \frac{p_{1} \times 3 \times 8}{100} = \frac{p_{2} \times 5 \times 3}{100} \]

\[ 24\,p_{1} = 15\,p_{2} \quad\Longrightarrow\quad \frac{p_{1}}{p_{2}} = \frac{15}{24} = \frac{5}{8} \]

So the sum splits into 13 shares of \( 2665 \div 13 = 205 \):

\[ p_{2} = 8 \times 205 = \text{Rs. } 1640 \]

The first part is Rs. 1025, and both yield Rs. 246 of interest.

Answer: D

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

Tripling means the interest is twice the principal.

\[ 2P = \frac{P \times R \times 12}{100} \quad\Longrightarrow\quad 200 = 12R \]

\[ R = \frac{50}{3} = 16\tfrac{2}{3}\% \text{ p.a.} \]

That rate is not among the choices, so the answer is none of these. The multiple counts the amount, so the interest is one less than it - using 3 instead of 2 would give 25%.

Answer: D

Simple interest is proportional to time, so over eight years instead of five the interest rises by a fixed factor no matter what the rate is.

\[ \frac{\text{SI}_{8}}{\text{SI}_{5}} = \frac{8}{5} = 1.6 \]

That is a 60% increase for every rate and every sum, so the stated condition carries no information about \( R \).

The rate therefore cannot be determined. A condition that holds identically is not a constraint.

Answer: D

Given that Rs.10,000 is invested in S.I for two years and Rs.8,000 in C.I for two years 

=>  C.I - S.I

=> 8,000 { [1 + r/100]2   - 1} = (10,000)2r /100

=> 8{ 1 + 2r/100 + r2 / (100)2 - 1} = r/5 

=> 16r/100 + 8r2/(100)2 = 20r/100

 => 4r/10 = 8r2/(100)2 

=> 8[r/100]2 - 4r/100 = 0 

=> r/100 {8r/100 -4} = 0 

=> r = 0% of 50% 

Since r!= 0%, r =50%

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: C

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

The sum and the period are common, so only the gap in rates produces the difference.

\[ \frac{1500 \times \Delta R \times 3}{100} = 13.50 \]

\[ 45\,\Delta R = 13.50 \quad\Longrightarrow\quad \Delta R = 0.3\% \]

Neither rate needs to be known individually - only their difference enters the calculation.

Answer: A

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

\[ 0.125P = \frac{P \times 10 \times T}{100} = 0.1PT \]

\[ T = \frac{0.125}{0.1} = 1.25 = 1\tfrac{1}{4} \text{ years} \]

At 10% a year, interest of 12.5% of the sum takes a year and a quarter.