Simple Interest Questions and Answers

Answer: C

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

\[ \text{SI} = 15600 - 12000 = \text{Rs. } 3600 \]

\[ 3600 = \frac{12000 \times 6 \times T}{100} = 720\,T \]

\[ T = 5 \text{ years} \]

The Rs. 15,600 is the amount, so the principal must be subtracted before the formula can be used.

Answer: D

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

Both loans share one rate \( R \), so add the two interests.

\[ \frac{5000 \times R \times 2}{100} + \frac{3000 \times R \times 4}{100} = 2200 \]

\[ 100R + 120R = 2200 \quad\Longrightarrow\quad 220R = 2200 \]

\[ R = 10\% \]

Checking: Rs. 1000 from B and Rs. 1200 from C.

Answer: B

Let the interest for one year be x.

As amount = Principal + Interest, we have 

P + 4x = 20720 --- (1)  ; P + 6x = 24080 --- (2)

Solving the equations (1) and (2), we can get 

P = Rs. 14000 and x = Rs. 1680

Interest for one year on Rs. 14000 is Rs. 1680

So, R = (100 * 1680)/(14000 * 1) = 12% p.a.

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 \( 3000 \times 0.04 = 120 \).