Simple and Compound Interest Set 2
Practice free Simple and Compound Interest MCQ questions with answers — easy level Set 2, covering effective interest rates, lending profit calculations, advance interest, instalment rate problems, split investments, and reverse depreciation. A great follow-up to Set 1 to further strengthen your grip on one of the most scoring topics in quantitative aptitude. Attempt all questions and check your answers instantly with clear explanations.
Q1.A person borrows Rs. 8000 for 3 years at 5% p.a. simple interest. He immediately lends it to another person at 8½% p.a. for 3 years.
View Solution & Explanation
Interest Paid
= (8000 × 5 × 3) / 100
= Rs. 1200Interest Received
= (8000 × 8.5 × 3) / 100
= (8000 × 17/2 × 3) / 100
= Rs. 2040Total Gain
= 2040 − 1200
= Rs. 840Gain per Year
= 840 / 3
= Rs. 280
Q2.A financier claims to lend money at simple interest at 12% per annum, but he adds the interest every six months to calculate the principal. What is the effective rate of interest per annum?
View Solution & Explanation
Although the financier claims to charge simple interest at 12% per annum, he recalculates the principal every 6 months. This effectively turns the calculation into compound interest (half-yearly).
Annual Rate = 12% ⇒ Half-yearly Rate = 12% ÷ 2 = 6%
Assume Principal = Rs. 100 (for easy calculation)
First 6 months:
Interest = 100 × 6/100 = Rs. 6
New Principal = 100 + 6 = Rs. 106
Next 6 months:
Interest = 106 × 6/100 = Rs. 6.36
Total interest in 1 year:
= 6 + 6.36 = Rs. 12.36
So, on Rs. 100, the interest earned in 1 year is Rs. 12.36
Q3.What will be the simple interest on Rs. 4200 for 3 years at 5% per annum?
View Solution & Explanation
To calculate the simple interest:
$$
\mathrm{SI} = \frac{P \times R \times T}{100}
$$
Substituting the values:
$$
\mathrm{SI} = \frac{4200 \times 5 \times 3}{100}
$$
$$
\mathrm{SI} = 630
$$
So, the simple interest is Rs. 630.
Q4.A sum of Rs. 6200 is invested for 3 years at 5% per annum. What will be the simple interest?
View Solution & Explanation
The formula for simple interest is:
$$
\mathrm{SI} = \frac{P \times R \times T}{100}
$$
Substituting the values:
$$
\mathrm{SI} = \frac{6200 \times 5 \times 3}{100}
$$
$$
\mathrm{SI} = \frac{93000}{100}
$$
$$
\mathrm{SI} = 930
$$
The simple interest is Rs. 930.
Q5.A sum of money becomes 3 times its original amount in 6 years. What is the rate of interest per annum?
View Solution & Explanation
Use the formula for Simple Interest:
$$
\mathrm{SI} = \frac{P \times R \times T}{100}
$$
The sum becomes 3 times the principal, so the interest is (2P). Substituting into the formula:
$$
2P = \frac{P \times R \times 6}{100}
$$
Cancel (P) from both sides:
$$
2 = \frac{6R}{100}
$$
Solve for (R):
$$
R = \frac{2 \times 100}{6} = 33.33\%
$$
Q6.An article depreciates by 10% every year. If the current price of the article is Rs. 5,670 after 2 years, what was its original price?
View Solution & Explanation
The formula for depreciation is:
$$
\text{Present Price} = \text{Original Price} \times \left(1 - \frac{R}{100}\right)^T
$$
Where:
Present Price = Rs. 5670
R = 10% (rate of depreciation)
T = 2 years (time)
Let the original price be x:
$$
5670 = x \times \left(1 - \frac{10}{100}\right)^2
$$
$$
5670 = x \times (0.9)^2
$$
$$
5670 = x \times 0.81
$$
Solving for x:
$$
x = \frac{5670}{0.81} \approx 7000
$$
Q7.A sum of Rs. 2800 is divided into two parts such that the interest on the first part at 6% simple interest per annum and that on the other part at 4% simple interest per annum are equal. The interest on each part (in Rupees) is?
View Solution & Explanation
Let the first part = x
Then the second part = (2800 − x)
Since the interests are equal:
x × 6/100 = (2800 − x) × 4/100
6x = 4(2800 − x)
6x = 11200 − 4x
10x = 11200
x = 1120
So,
First part = Rs. 1120
Second part = 2800 − 1120 = Rs. 1680
Interest = (1120 × 6) / 100 = Rs. 67.20
Q8.What will be the simple interest earned on an amount of Rs. 14,400 in 8 months at the rate of 7½% p.a.?
View Solution & Explanation
Principal (P) = Rs. 14,400
Rate (R) = 7½% = 15/2 %
Time (T) = 8 months = 8/12 = 2/3 years
SI = (P × R × T) / 100
SI = (14400 × 15/2 × 2/3) / 100
Cancel 2 in numerator and denominator
Expression becomes: (14400 × 15) / (3 × 100)
SI = (14400 × 15) / 300
SI = 216000 / 300
SI = 720
Q9.At which sum will the simple interest at the rate of 4½% per annum be Rs. 270 in 2⅔ years?
View Solution & Explanation
The formula for simple interest is:
$$
SI = \frac{P \times R \times T}{100}
$$
Where:
SI = 270
R = 4½% = \frac{9}{2}\%
T = 2⅔ years = \frac{8}{3}
Substitute the values:
$$
270 = \frac{P \times \frac{9}{2} \times \frac{8}{3}}{100}
$$
$$
270 = \frac{P \times \frac{72}{6}}{100}
$$
$$
270 = \frac{P \times 12}{100}
$$
$$
270 = \frac{12P}{100}
$$
Solving for P:
$$
270 \times 100 = 12P
$$
$$
27000 = 12P
$$
$$
P = \frac{27000}{12}
$$
$$
P = 2250
$$
Q10.A sum of money at a certain rate of simple interest doubles in 8 years and at a different rate becomes three times in 10 years.
View Solution & Explanation
Case 1: Money doubles in 8 years
When money doubles:
SI = P
$$
P = \frac{P \times R_1 \times 8}{100}
$$
$$
R_1 = \frac{100}{8}
$$
$$
R_1 = 12.5\%
$$
Case 2: Money triples in 10 years
When money triples:
SI = 2P
$$
2P = \frac{P \times R_2 \times 10}{100}
$$
$$
R_2 = \frac{200}{10}
$$
$$
R_2 = 20\%
$$
Lower rate of interest:
$$
\min(12.5\%,\ 20\%) = 12.5\%
$$
Q11.On simple interest, a certain sum becomes Rs. 44,000 in 5 years and Rs. 56,000 in 10 years. If the rate of interest had been 3% more, then in how many years would the sum have become Rs. 58,880?
View Solution & Explanation
Given:
Amount in 5 years = 44000
Amount in 10 years = 56000
Interest for 5 years:
$$
SI_{5\ years} = 56000 - 44000 = 12000
$$
$$
SI_{per\ year} = \frac{12000}{5} = 2400
$$
Principal:
$$
P = 44000 - (2400 \times 5) = 44000 - 12000 = 32000
$$
Original rate:
$$
R = \frac{2400 \times 100}{32000} = 7.5\%
$$
New rate:
$$
R' = 7.5\% + 3\% = 10.5\%
$$
Now find time for amount = 58880
$$
SI = 58880 - 32000 = 26880
$$
Using SI formula:
$$
26880 = \frac{32000 \times 10.5 \times T}{100}
$$
$$
26880 = 3360T
$$
$$
T = \frac{26880}{3360}
$$
$$
T = 8\ \text{years}
$$
Q12.Simple interest on a certain amount is 25/16 of the principal. If the numbers representing the rate of interest in percent and time in years are equal, then the time for which the principal is lent out is?
View Solution & Explanation
Given:
$$
SI = \frac{25P}{16}
$$
Also given:
$$
R = T
$$
Using the formula:
$$
SI = \frac{P \times R \times T}{100}
$$
Since R = T:
$$
SI = \frac{P \times T^2}{100}
$$
Substitute SI:
$$
\frac{25P}{16} = \frac{P \times T^2}{100}
$$
Cancel P:
$$
\frac{25}{16} = \frac{T^2}{100}
$$
Solve for T:
$$
T^2 = \frac{25 \times 100}{16}
$$
$$
T^2 = \frac{2500}{16}
$$
$$
T = \frac{50}{4}
$$
$$
T = 12.5\ \text{years}
$$
Q13.Consider the following statements. If a sum of money is lent at simple interest, then: I. Money gets doubled in 4 years if the rate of interest is 25%. II. Money gets doubled in 4 years if the rate of interest is 20%. III. Money becomes four times in 8 years if it gets doubled in 4 years. Which of the statements are correct?
View Solution & Explanation
Statement I:
For doubling, SI = P
$$
SI = \frac{P \times R \times T}{100}
$$
$$
P = \frac{P \times 25 \times 4}{100}
$$
$$
P = P \Rightarrow \text{True}
$$
Statement II:
$$
SI = \frac{P \times 20 \times 4}{100} = 0.8P
$$
$$
0.8P \ne P \Rightarrow \text{False}
$$
Statement III:
If money doubles in 4 years:
$$
SI_{4\ years} = P
$$
So, per year:
$$
SI_{per\ year} = \frac{P}{4}
$$
In 8 years:
$$
SI_{8\ years} = 2P
$$
Total amount:
$$
A = P + 2P = 3P
$$
$$
3P \ne 4P \Rightarrow \text{False}
$$
Q14.A person deposits Rs. 800 for 3 years and Rs. 1000 for 2 years at the same rate of simple interest in a bank. Altogether he received Rs. 280 as interest. The rate of simple interest per annum was?
View Solution & Explanation
Total Simple Interest:
$$
280 = \frac{800 \times R \times 3}{100} + \frac{1000 \times R \times 2}{100}
$$
$$
280 = \frac{2400R}{100} + \frac{2000R}{100}
$$
$$
280 = 24R + 20R
$$
$$
280 = 44R
$$
$$
R = \frac{280}{44}
$$
$$
R \approx 6.36\%
$$
Q15.A man invested Rs. 6000 at some rate of simple interest and Rs. 5000 at 1% higher rate of interest. If the interest in both cases after 5 years is the same, the rate of interest in the former case is?
View Solution & Explanation
Let rate in former case = R%
Rate in latter case:
$$
R + 1
$$
Using SI formula:
$$
SI = \frac{P \times R \times T}{100}
$$
Given both interests are equal:
$$
\frac{6000 \times R \times 5}{100} = \frac{5000 \times (R + 1) \times 5}{100}
$$
Simplify:
$$
6000R = 5000(R + 1)
$$
$$
6000R = 5000R + 5000
$$
$$
1000R = 5000
$$
$$
R = 5\%
$$
Q16.A sum of money becomes Rs. 19,500 in 3 years and Rs. 27,000 in 8 years. Find the rate of interest and the original sum of money.
View Solution & Explanation
The amount after 3 years is ₹19,500, which includes the principal (P) plus 3 years of interest.
Interest for 3 years:
$$
SI = 3 \times 1500 = 4500
$$
Principal (P):
$$
P = 19500 - 4500 = 15000
$$
4. Find the Rate of Interest (R)
Using the formula:
$$
SI = \frac{P \times R \times T}{100}
$$
$$
1500 = \frac{15000 \times R \times 1}{100}
$$
$$
1500 = 150R
$$
$$
R = \frac{1500}{150} = 10\%
$$
Q17.A washing machine is available for Rs. 28,000 cash or Rs. 10,000 as cash down payment followed by six monthly instalments of Rs. 3,200 each. What is the rate of interest per annum under the instalment plan?
View Solution & Explanation
Total amount paid under instalment:
$$
= 10000 + (6 \times 3200)
$$
$$
= 10000 + 19200 = 29200
$$
Total interest:
$$
SI = 29200 - 28000 = 1200
$$
Balance after down payment:
$$
P = 28000 - 10000 = 18000
$$
Sum of monthly outstanding balances:
$$
18000 + 14800 + 11600 + 8400 + 5200 + 2000 = 60000
$$
Using SI formula (time = 1 month each):
$$
1200 = \frac{60000 \times R \times 1}{100 \times 12}
$$
$$
1200 = \frac{60000R}{1200}
$$
$$
1200 \times 1200 = 60000R
$$
$$
R = \frac{1440000}{60000}
$$
$$
R = 24\%
$$
Q18.A money lender claims to lend money at the rate of 15% per annum simple interest. However, he takes the interest in advance when he lends a sum for one year. At what actual interest rate does he lend the money?
View Solution & Explanation
Let the sum be:
$$
P = 100
$$
Interest taken in advance:
$$
SI = 15\% \text{ of } 100 = 15
$$
Actual amount received by borrower:
$$
= 100 - 15 = 85
$$
Actual rate of interest:
$$
R = \frac{SI}{\text{Actual Principal}} \times 100
$$
$$
R = \frac{15}{85} \times 100
$$
$$
R = \frac{1500}{85}
$$
$$
R = 17.65\%
$$
Q19.A man invested 1/4 of his capital at 6%, 1/3 at 9%, and the remainder at 12%. If his annual income is Rs. 760, the capital is?
View Solution & Explanation
Let total capital be:
$$
P
$$
Investments:
$$
\frac{1}{4}P \text{ at } 6\%, \quad \frac{1}{3}P \text{ at } 9\%
$$
Remaining part:
$$
1 - \left(\frac{1}{4} + \frac{1}{3}\right) = 1 - \frac{7}{12} = \frac{5}{12}
$$
So,
$$
\frac{5}{12}P \text{ at } 12\%
$$
Total income:
$$
760 = \frac{1}{4}P \times \frac{6}{100} + \frac{1}{3}P \times \frac{9}{100} + \frac{5}{12}P \times \frac{12}{100}
$$
Simplify:
$$
760 = P \left( \frac{6}{400} + \frac{9}{300} + \frac{60}{1200} \right)
$$
$$
760 = P \left( \frac{3}{200} + \frac{3}{100} + \frac{1}{20} \right)
$$
Take LCM = 200:
$$
760 = P \left( \frac{3 + 6 + 10}{200} \right)
$$
$$
760 = P \left( \frac{19}{200} \right)
$$
Solve for P:
$$
P = \frac{760 \times 200}{19}
$$
$$
P = 8000
$$
Q20.The simple interest on a certain sum of money at the rate of 4% per annum for 10 years is Rs. 1,200. At what rate of interest will the same amount of interest be received on the same sum after 5 years?
View Solution & Explanation
Using SI formula:
$$
SI = \frac{P \times R \times T}{100}
$$
--------------------------------------------------
From first condition:
$$
1200 = \frac{P \times 4 \times 10}{100}
$$
$$
1200 = \frac{40P}{100}
$$
$$
1200 = 0.4P
$$
$$
P = 3000
$$
--------------------------------------------------
Now for second condition (same SI in 5 years):
$$
1200 = \frac{3000 \times R \times 5}{100}
$$
$$
1200 = \frac{15000R}{100}
$$
$$
1200 = 150R
$$
$$
R = 8\%
$$
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