Quantitative Aptitude
Simple and Compound Interest
3,394 Questions
Simple and Compound Interest Questions
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$10\%$
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$15\%$
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$50\%$
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$30\%$
C
Correct answer
Explanation
A = P(1 + r/100)^n. 5000 = 1000(1 + r/100)^4. 5 = (1 + r/100)^4. 5^(1/4) = 1 + r/100. 5^(0.25) is approximately 1.495. So r/100 = 0.495, r = 49.5%, which is approximately 50%.
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$ $ 20,000$
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$ $ 15,000$
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$ $ 12,000$
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$ $ 10,000$
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$ $ 9,000$
A
Correct answer
Explanation
The rule of 70 states that money doubles in 70/8 = 8.75 years. In 18 years, the money doubles approximately twice (18 / 8.75 is slightly more than 2). Doubling \$5000 twice results in \$5000 * 2 * 2 = $20,000.
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Rs. 1050
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Rs. 2050
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Rs. 3000
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None of these
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$Rs. 1750$
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$Rs. 1680$
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$Rs. 1840$
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$Rs. 1443.75$
B
Correct answer
Explanation
Present Worth = Amount / (1 + (R*T)/100). PW = 2310 / (1 + (15 * 2.5)/100) = 2310 / (1 + 37.5/100) = 2310 / 1.375 = 1680.
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Rs. 116.64
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Rs. 116.00
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Rs. 120
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Rs. 108
A
Correct answer
Explanation
Principal P. Interest for 1st year = P * 0.08 = 100, so P = 1250. Amount after 1st year = 1350. Interest for 2nd year = 1350 * 0.08 = 108. Amount after 2nd year = 1350 + 108 = 1458. Interest for 3rd year = 1458 * 0.08 = 116.64.
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$\displaystyle \frac { w }{ 1+1.08 } $
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$\displaystyle \frac { w }{ 1.08+1.16 } $
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$\displaystyle \frac { w }{ 1.16+1.24 } $
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$\displaystyle \frac { w }{ 1.08+{ \left( 1.08 \right) }^{ 2 } } $
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$\displaystyle \frac { w }{ { \left( 1.08 \right) }^{ 2 }+{ \left( 1.08 \right) }^{ 3 } } $
D
Correct answer
Explanation
After 1 year, the account has x * 1.08. Then another x is added, so the balance is x * 1.08 + x. After the second year, this amount earns 8% interest: (x * 1.08 + x) * 1.08 = w. So, w = x * (1.08^2 + 1.08). Therefore, x = w / (1.08^2 + 1.08).
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$\displaystyle 1,000 \left(1 + \frac{5-3}{1,200}\right)^{12}$
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$\displaystyle 1,000 \left(1 + \frac{\displaystyle \frac{5}{3}}{1,200}\right)^{12}$
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$\displaystyle \frac{1,000 \left(1 + \displaystyle \frac{5}{1,200}\right)^{12}}{1,000 \left(1 + \displaystyle \frac{3}{1,200}\right)^{12}}$
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$\displaystyle 1,000 \left(1 + \displaystyle \frac{5}{1,200}\right)^{12} - \displaystyle 1,000 \left(1 + \displaystyle \frac{3}{1,200}\right)^{12}$
D
Correct answer
Explanation
The amount generated is A(r) = 1000 * (1 + r/1200)^12. Additional money at 5% vs 3% is A(5) - A(3).
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Rs. $13000 $ in scheme $A$, Rs. $12000$ in scheme $B$
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Rs. $12000$ in scheme $A$, Rs. $10000$ in scheme $B$
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Rs. $ 11000$ in scheme $A$, Rs. $10000$ in scheme $B$
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Rs. $10000$ in scheme $A$, Rs. $13000$ in scheme $B$
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Both (i) and (iii)
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Both (ii) and (iii)
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Both (i) and (ii)
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Any two of the three
D
Correct answer
Explanation
Simple Interest (SI) = P * R * T / 100. (i) 690 = P + P * R * 3 / 100. (ii) 750 = P + P * R * 5 / 100. (iii) R = 5. With any two statements, we have two variables (P and R) or can solve for P using the difference in interest over time.
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x = £8000, y = £18000
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x = £12000, y = £7000
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x = £9000, y = £15000
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x = £12000, y = £8000
D
Correct answer
Explanation
Let x be amount at 6%, y at 7%. x + y = 20000. 0.06x + 0.07y = 1280. 6x + 7y = 128000. 6x + 6y = 120000. y = 8000. x = 12000.
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4 years
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3 years
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2 years
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1 year
B
Correct answer
Explanation
Using the compound interest formula A = P(1 + r/100)^n, we have 6655 = 5000(1 + 0.10)^n. Dividing by 5000 gives 1.331 = (1.1)^n, and since 1.1^3 = 1.331, n = 3.
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£1264.29
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£1324.82
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£1391.85
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£1529.21
C
Correct answer
Explanation
Amount = P * (1 + r/100)^n = 1100 * (1.04)^6. (1.04)^6 is approximately 1.2653. 1100 * 1.2653 = 1391.85.
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1 year
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2 years
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3 years
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4 years
C
Correct answer
Explanation
Compound interest formula: A = P(1 + r/100)^n. 665.50 = 500(1 + 0.10)^n. 1.331 = (1.1)^n. Since 1.1^3 = 1.331, n = 3 years.
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6 years
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5 years
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4 years
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3 years
B
Correct answer
Explanation
Using the compound interest formula, we have 608 = 500 * (1.04)^n, which simplifies to 1.04^n = 1.216. Calculating the powers of 1.04, we find that 1.04^5 is approximately 1.2167, making 5 years the closest integer duration for the investment to reach 608.
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1.5 years
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2 years
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2.5 years
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3 years