Simple Interest Questions and Answers

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: 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: D

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

Growing from Re. 1 to Rs. 9 means interest of Rs. 8.

\[ 8 = \frac{1 \times R \times 20}{100} = 0.2R \]

\[ R = 40\% \]

The amount is 9 times the principal, so the interest is 8 times it - using 9 here would give 45%.

Answer: A

100 ---- 200 ---- 5

             200 ---- 5                                         

             100 ---- 2 1/2

             ------------------

              600 ---- 12 ½ years

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

The extra interest from two more percentage points over seven years would be

\[ \frac{P \times 2 \times 7}{100} = 0.14P \]

which needs the principal. All that is known is

\[ \frac{P \times R \times 7}{100} = 1750 \quad\Longrightarrow\quad PR = 25{,}000 \]

one equation in two unknowns. A sum of Rs. 5000 at 5% and a sum of Rs. 2500 at 10% both fit, yet they would give Rs. 700 and Rs. 350 of extra interest respectively.

So the answer cannot be determined.

Answer: C

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

\[ \text{SI} = 3600 - 3000 = 600 \quad\Longrightarrow\quad 600 = \frac{3000 \times R \times 4}{100} = 120R \]

\[ R = 5\% \]

At 6%:

\[ \text{SI} = \frac{3000 \times 6 \times 4}{100} = 720 \quad\Longrightarrow\quad \text{amount} = \text{Rs. } 3720 \]

The data are complete, so "cannot be determined" is a trap: one extra percentage point over four years adds \( 3000 \times 0.04 = 120 \).

Answer: C

S.I. = (956 - 800) = Rs. 156.

Rate = (100 * 156) / (800 * 3) = 6 1/2 %

Now rate = (6 1/2 + 4) = 10 1/2 %

New S.I. = (800 * 21/2 * 3/100) = Rs. 252

New amount = (800 + 252) = Rs. 1052.

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: 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

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