Quantitative Aptitude - MCQ Practice Questions

We use the simple interest formula $I=\frac{P\times R\times T}{100}$ to find interest on equal principal amounts invested at different rates and periods, then set up an equation using the given difference.

Step 1: Write the interest formula for each investment

Let the principal be $P$ (same for both).

For Investment 1 (12% p.a. for 4 years):

For Investment 2 (15% p.a. for 3 years):

Step 2: Find the difference in interests

Since $I_1>I_2$ (higher rate × longer time):

Step 3: Use the given difference to find P

We're told the difference is ₹540:

Step 4: Verify the answer

$I_1=0.48\times 18000=8640$

$I_2=0.45\times 18000=8100$

Difference: $8640-8100=540$ ✓

Let each principal amount be P.

Using Simple Interest formula:

SI=

100

P×R×T

​

First investment

SI

1

​

=

100

P×12×4

​

=

100

48P

​

Second investment

SI

2

​

=

100

P×15×3

​

=

100

45P

​

Difference in interests:

100

48P

​

−

100

45P

​

=540

100

3P

​

=540

3P=54000

P=18000

Therefore, each principal amount is ₹18,000.

Answer: Each principal amount is ₹18,000 (Option B)

Step 1: From first investment, 2800 = (P × 14 × 5) / 100 = 0.7P.

Step 2: P = $\frac{2800}{0}$.7 = ₹4000.

Step 3: At 10% for 5 years, SI = (4000 × 10 × 5) / 100 = $\frac{200000}{100}$ = ₹2000.

Step 1: If amount doubles, SI = P.

So P = (P × R × 8) / 100, giving R = $\frac{100}{8}$ = 12.5% per annum.

Step 2: For amount to become 3 times, SI = 2P.

Step 3: 2P = (P × 12.5 × T) / 100, so 2 = 0.125T, therefore T = $\frac{2}{0}$.125 = 16 years.

SI for 2 years = 6000 - 5500 = 500. SI per year = 250. SI for 3 years = 750. Principal = 5500 - 750 = 4000

SI = 8640 - 7200 = 1440. Using SI = (P × R × T)/100; 1440 = (7200 × R × 2)/100; R = 10%

SI = 11700 - 9000 = 2700. Using T = (SI × 100)/(P × R) = (2700 × 100)/(9000 × 15) = 2 years

SI = (24000 × 8.5 × 4)/100 = 8160

If amount = 2P, then SI = P. Using P = (P × R × 10)/100; 100 = 10R; R = 10% p.a.

SI = (5000 × 12 × 2.5)/100 = 1500. Amount = 5000 + 1500 = 6500

Amount = P + (P × R × T)/100; 9200 = P + (P × 8 × 4)/100; 9200 = P(1 + 0.32); P = 7000

Let sum = P. Difference in SI = (P × 12 × 3)/100 - (P × 10 × 3)/100 = (6P)/100 = 600; P = 10000

SI = (20000 × 8 × 5)/100 = 8000. Amount = 20000 + 8000 = 28000

17600 = P(1 + (11 × 4)/100); 17600 = P(1.44); P = 12222.22 ≈ 12000 (approx)

Using SI = (P × R × T)/100; 800 = (4000 × R × 2)/100; R = 10% p.a.

SI = 33250 - 25000 = 8250. Using T = (SI × 100)/(P × R) = (8250 × 100)/(25000 × 9) = 3.67 ≈ 3.5 years

We use the simple interest formula $SI=\frac{P\times R\times T}{100}$ to equate the interests earned on both parts.

Step 1: Set up variables

Let the first part be $x$ and the second part be $(12000-x)$.

Step 2: Write the simple interest formula for each part

Simple interest on first part at 10% for 2 years:

Simple interest on second part at 12% for 2 years:

Step 3: Equate the interests

Given that both simple interests are equal:

Step 4: Solve for x

Answer: The first part is ₹6,545.45 (Option B)

(6000 × R × 3)/100 = (9000 × R' × 2)/100. If same rate: 18000R = 18000R, but comparing different principals/times: (6000 × R × 3) = (9000 × R × 2) doesn't work. Recalc: If they want same SI, 18R = 18R (same). Rate = 10% works as standard

SI = 10240 - 8000 = 2240. Rate = (2240 × 100)/(8000 × 4) = 7% p.a.

Simple interest is calculated as a percentage of the principal amount and remains constant each year.

Step 1: Calculate Interest for Friend 1 (8% rate)

Friend 1 invested ₹10,000 at 8% simple interest for 3 years.

Step 2: Calculate Interest for Friend 2 (10% rate)

Friend 2 invested ₹10,000 at 10% simple interest for 3 years.

Step 3: Calculate Interest for Friend 3 (12% rate)

Friend 3 invested ₹10,000 at 12% simple interest for 3 years.

Step 4: Calculate Total Interest

Add the interest earned by all three friends.

The total interest earned by all three friends is ₹9,000.

Answer: (A) ₹9,000

Q = P[1 + (RT/100)]; Therefore P = Q/[1 + (RT/100)]