Simple Interest Questions and Answers

Answer: A

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

Set the two interests equal and see what it forces.

\[ \frac{p_{1} \times 15 \times 4}{100} = \frac{p_{2} \times 20 \times 3}{100} \]

\[ 0.6\,p_{1} = 0.6\,p_{2} \quad\Longrightarrow\quad p_{1} = p_{2} \]

\[ \text{difference} = \text{Rs. } 0 \]

Both products \( R \times T \) come to 60, so the two schemes are identical and the salary splits exactly in half - Rs. 4300 each.

Answer: A

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

Doubling means the interest equals the principal.

\[ P = \frac{P \times R \times 5}{100} \quad\Longrightarrow\quad R = 20\% \]

\[ \text{Raj's interest} = \frac{6250 \times 20 \times 3}{100} = \text{Rs. } 3750 \]

Aman's Rs. 8000 never enters Raj's calculation; it serves only to fix the rate, and even that cancels out of the doubling condition.

Answer: B

Simple interest of Rs. 600 over two years means Rs. 300 in each year. Compound interest exceeds it by

\[ 645 - 600 = \text{Rs. } 45 \]

and that excess is one year's interest on the first year's Rs. 300:

\[ 300 \times \frac{R}{100} = 45 \quad\Longrightarrow\quad R = 15\% \]

The principal is never needed. This shortcut works for any two-year simple-versus-compound comparison.

Answer: C

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

Write the interest as a fraction of the principal.

\[ \frac{9}{25}P = \frac{P \times R \times 4}{100} \]

The \( P \) cancels:

\[ \frac{9}{25} = \frac{4R}{100} \quad\Longrightarrow\quad R = \frac{9 \times 100}{25 \times 4} = 9\% \]

Because the interest is quoted as a fraction of the principal, no actual sum is needed.

Answer: C

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

Convert both periods to years: 15 months is \( \tfrac{5}{4} \) and 8 months is \( \tfrac{2}{3} \).

\[ \frac{P \times 7.5 \times \tfrac{5}{4}}{100} - \frac{P \times 12.5 \times \tfrac{2}{3}}{100} = 32.50 \]

\[ \frac{P(9.375 - 8.3333)}{100} = 32.50 \]

\[ \frac{1.04167P}{100} = 32.50 \quad\Longrightarrow\quad P = \text{Rs. } 3120 \]

The two \( R \times T \) products are close (9.375 against 8.333), so the small difference needs a large principal to reach Rs. 32.50.

Answer: D

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

Write both amounts for the same \( P \) and \( T \).

\[ P + \frac{12PT}{100} = 3000, \qquad P + \frac{15PT}{100} = 3300 \]

Subtracting:

\[ \frac{3PT}{100} = 300 \quad\Longrightarrow\quad PT = 10{,}000 \]

Substituting back:

\[ P + 0.12 \times 10000 = 3000 \quad\Longrightarrow\quad P = \text{Rs. } 1800 \]

\[ T = \frac{10000}{1800} = 5.56 \approx 5.5 \text{ years} \]

Subtracting the two equations removes the principal and leaves the product \( PT \) - the key step.

Answer: C

Under simple interest the growth is proportional to time, so work with the interest rather than the amount.

Becoming 7 times means the interest is \( 6P \) in 15 years:

\[ 6P = \frac{P \times R \times 15}{100} \quad\Longrightarrow\quad R = 40\% \]

Becoming 10 times needs interest of \( 9P \):

\[ 9P = \frac{P \times 40 \times T}{100} \quad\Longrightarrow\quad T = 22.5 \text{ years} \]

Equivalently the times are in the ratio \( 9 : 6 \), so \( 15 \times \tfrac{9}{6} = 22.5 \).

Answer: A

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

Becoming 6 times means the interest is 5 times the principal.

\[ 5P = \frac{P \times R \times 40}{100} \]

\[ 5 = \frac{40R}{100} \quad\Longrightarrow\quad R = 12.5\% \]

The multiple counts the amount, so the interest is one less than it - using 6 instead of 5 would give 15%.

Answer: D

After the down payment the balance is \( 20000 - 4000 = \text{Rs. } 16{,}000 \), cleared by four monthly installments of \( x \).

Carry everything forward to the end of month 4. The debt grows for 4 months, while the installments earn interest for 3, 2, 1 and 0 months:

\[ 16000\left(1 + \frac{8}{100} \times \frac{4}{12}\right) = x\left[\left(1 + \tfrac{8}{100}\cdot\tfrac{3}{12}\right) + \left(1 + \tfrac{8}{100}\cdot\tfrac{2}{12}\right) + \left(1 + \tfrac{8}{100}\cdot\tfrac{1}{12}\right) + 1\right] \]

\[ 16000 \times 1.026667 = x(1.02 + 1.013333 + 1.006667 + 1) \]

\[ 16426.67 = 4.04\,x \quad\Longrightarrow\quad x = \text{Rs. } 4066.01 \]

Dividing the balance by four gives Rs. 4000, which ignores the interest entirely - hence that distractor.

Answer: A

Carry the debt and the installments forward to the end of year 6. The debt accrues 6 years of interest; the installments earn interest for 5, 4, 3, 2, 1 and 0 years.

\[ x\left[6 + \frac{9}{100}(5+4+3+2+1+0)\right] = 1078\left(1 + \frac{9 \times 6}{100}\right) \]

\[ x(6 + 1.35) = 1078 \times 1.54 \]

\[ 7.35\,x = 1660.12 \quad\Longrightarrow\quad x = \text{Rs. } 146.67 \]

The same result comes from the standard formula

\[ x = \frac{100 \times \text{Debt}}{100n + R \cdot \frac{n(n-1)}{2}} = \frac{107800}{600 + 135} = 146.67 \]