Simple Interest Questions and Answers

Answer: C

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

\[ 5400 = \frac{P \times 12 \times 3}{100} = 0.36P \]

\[ P = \frac{5400}{0.36} = \text{Rs. } 15{,}000 \]

Three years at 12% comes to 36% of the principal, so the principal is a little under three times the interest.

Answer: A

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

Beyond eight years leaves \( 11 - 8 = 3 \) years at 13%, so the three slabs are:

\[ P\left(\frac{6 \times 3 + 9 \times 5 + 13 \times 3}{100}\right) = 8160 \]

\[ P\left(\frac{18 + 45 + 39}{100}\right) = 1.02P = 8160 \]

\[ P = \text{Rs. } 8000 \]

The final slab covers only the last three years, not all eleven.

Answer: B

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

Let \( p \) be the part at 8%; the rest, \( 10000 - p \), earns 10%.

\[ 0.08p + 0.10(10000 - p) = 890 \]

\[ 0.08p + 1000 - 0.10p = 890 \quad\Longrightarrow\quad -0.02p = -110 \]

\[ p = \text{Rs. } 5500 \]

Checking: Rs. 440 from the 8% part and Rs. 450 from the Rs. 4500 at 10%.

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: 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 rate from the four-year figure.

\[ \text{SI} = 540 - 450 = 90 \quad\Longrightarrow\quad 90 = \frac{450 \times R \times 4}{100} = 18R \]

\[ R = 5\% \]

\[ \text{SI over 6 years} = \frac{450 \times 5 \times 6}{100} = 135 \quad\Longrightarrow\quad \text{amount} = \text{Rs. } 585 \]

Since simple interest is linear in time, the shortcut is \( 90 \times \tfrac{6}{4} = 135 \).

Answer: C

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

Under simple interest each year adds the same amount, so the gap between the two figures is one year's interest.

\[ 854 - 815 = \text{Rs. } 39 \text{ per year} \]

\[ P = 815 - 3 \times 39 = 815 - 117 = \text{Rs. } 698 \]

Working back three years from the three-year amount recovers the principal.

Answer: C

Write both statements with the same \( R \) and \( T \), and let \( k = \tfrac{RT}{100} \).

\[ x = yk, \qquad y = zk \]

Dividing one by the other:

\[ \frac{x}{y} = \frac{y}{z} \quad\Longrightarrow\quad y^{2} = xz \]

So \( y \) is the geometric mean of \( x \) and \( z \). Substituting directly gives the same thing: \( x = zk^{2} \) and \( y^{2} = z^{2}k^{2} = xz \).

Answer: D

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

\[ \text{SI} = 920 - 800 = 120 \quad\Longrightarrow\quad 120 = \frac{800 \times R \times 3}{100} = 24R \]

\[ R = 5\% \]

At \( 5 + 3 = 8\% \):

\[ \text{SI} = \frac{800 \times 8 \times 3}{100} = 192 \quad\Longrightarrow\quad \text{amount} = \text{Rs. } 992 \]

The extra 3% over three years adds \( 800 \times 0.09 = 72 \) - the same answer in one step.

Answer: B

Let the sum be Rs. 100. Then,

S.I. for first 6 months = (100 * 10 *1) / (100 * 2) = Rs. 5

S.I. for last 6 months = (105 * 10 * 1) / (100 * 2) = Rs. 5.25

So, amount at the end of 1 year = (100 + 5 + 5.25) = Rs. 110.25

Effective rate = (110.25 - 100) = 10.25%.