Question 1

A sum of money lent at compound interest for 2 years at 20% per annum would fetch ₹ 482 more, if the interest was payable half-yearly than if it was payable annually. The sum is :

Accepted Answer

Concept & Compound Interest Frequency: The effective interest earned changes based on compounding frequency. Compounding more frequently yields higher total interest. Amount = P(1 + (R/n × 100))^{n × T} where n is the number of compounding periods per year. Step-by-Step Solution: 1. Let the principal sum be P. 2. Case 1: Annually Amount after 2 years = P(1 + (20/100))^2 = P(1.2)^2 = 1.44P Compound Interest (CIA) = 1.44P - P = 0.44P 3. Case 2: Half-Yearly Rate per half-year = (20%/2) = 10%. Number of periods = 2 × 2 = 4. Amount after 2 years = P(1 + (10/100))^4 = P(1.1)^4 = 1.4641P Compound Interest (CIH) = 1.4641P - P = 0.4641P 4. Difference Given difference = ₹ 482 0.4641P - 0.44P = 482 0.0241P = 482 P = (482/0.0241) = 20,000 Exam Strategy & Shortcut: For 20% over 2 years: Annual effective CI is 44%. For 10% over 4 periods: Effective CI is 46.41% (memorizing Pascal's triangle or successive percentage changes helps: 10% twice is 21%, 21% twice is 21 + 21 + 4.41 = 46.41%). Difference = 2.41%. 2.41% of P = 482 \implies 1% of P = 200 \implies 100% = 20,000. Common Pitfall: A common mistake is forgetting to adjust both the rate and the time period when switching from annual to half-yearly compounding. Final Answer: Therefore, the correct answer is ₹ 20,000.

Question 2

A person lent out a certain sum on simple interest and the same sum on compound interest at a certain rate of interest per annum. He noticed that the ratio between the difference of compound interest and simple interest of 3 years and that of 2 year…

Accepted Answer

Concept & Ratio of 3-Year to 2-Year CI-SI Differences: For a principal P at rate R%, the difference between CI and SI for 2 years (D_2) and 3 years (D_3) follows a direct algebraic ratio. D_2 = P((R/100))^2 D_3 = P((R/100))^2 × (3 + (R/100)) (D_3/D_2) = 3 + (R/100) Step-by-Step Solution: 1. Equate the formula to the given ratio: The ratio is given as 25 : 8. (D_3/D_2) = (25/8) 2. Substitute into the standard formula: 3 + (R/100) = (25/8) 3. Solve for R: (R/100) = (25/8) - 3 (R/100) = (25 - 24/8) (R/100) = (1/8) R = (100/8) = 12.5 4. Format the answer: 12.5% = 12(1/2)% Exam Strategy & Shortcut: Memorize the relationship (D_3/D_2) = 3 + (R/100) for CI-SI difference problems. When you see a ratio like (25/8), immediately split it as 3 + (1/8). This immediately tells you that (R/100) = (1/8) = 12.5%. Common Pitfall: Trying to solve this by expanding the full (1+R/100)^3 binomial formula manually from scratch will consume minutes. Recognize the D_3/D_2 shortcut pattern. Final Answer: Therefore, the correct answer is 12(1/2)%.

Question 3

A father left a will of ₹ 16400 for his two sons aged 17 and 18 years. They must get equal amounts when they are 20 years, at 5% compound interest. Find the present share of the younger son.

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Concept & Equating Future Amounts: When a total sum is divided such that future amounts are equal, the present shares must be inversely proportional to their respective compound interest growth factors. Amount = P(1 + (R/100))^T Step-by-Step Solution: 1. Identify the time periods: Younger son (age 17) gets interest for 20 - 17 = 3 years. Let his share be A. Elder son (age 18) gets interest for 20 - 18 = 2 years. Let his share be B. 2. Set up the equality: Both must have equal amounts at age 20. The rate R = 5%. A × (1 + 0.05)^3 = B × (1 + 0.05)^2 A × (1.05)^3 = B × (1.05)^2 3. Find the ratio of their shares: Dividing both sides by (1.05)^2: A × 1.05 = B B = A × (105/100) = A × (21/20) So, the ratio A : B = 20 : 21. 4. Calculate the younger son's share (A): Total sum = A + B = 16400 Total ratio units = 20 + 21 = 41 1 unit = (16400/41) = 400 Younger son's share (A) = 20 × 400 = 8000. Exam Strategy & Shortcut: Difference in time = 1 year. This means the elder son's share is just the younger son's share multiplied by (1 + R/100) once. Ratio of younger to elder = 100 : (100 + R) = 100 : 105 = 20 : 21. Younger share = (20/(20+21)) × 16400 = (20/41) × 16400 = 8000. Common Pitfall: A common error is confusing which son gets the larger principal. The younger son has *more* time to earn interest, so his initial principal (A) must be *smaller* to reach the same final amount. Final Answer: Therefore, the correct answer is ₹ 8000.

Question 4

A taperecorder is sold for ₹ 3500 cash, or ₹ 1000 cash down payment and the balance in three equal easy instalments. If $12 \frac{1}{2}\%$ is the rate of interest compounded annually, find the amount of instalment.

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Concept & Instalments with Down Payment: When a down payment is made, the principal amount to be amortized through instalments is the total cash price minus the down payment. We then apply the standard present value formula for instalments on this balance. Balance = \sum_{t=1}^{n} (x/(1 + \frac{R){100})^t} Step-by-Step Solution: • Given: Cash price = ₹ 3500, Down payment = ₹ 1000. • Principal balance to be paid in instalments P = 3500 - 1000 = 2500. • Rate of interest R = 12 (1/2)% = 12.5% = (1/8). • Number of instalments n = 3. Let the instalment be x. • The discounting factor 1 + (R/100) = 1 + (1/8) = (9/8). • 2500 = (x/\frac{9){8}} + (x/(\frac{9){8})^2} + (x/(\frac{9){8})^3} • 2500 = x [ (8/9) + (64/81) + (512/729) ] • Take LCM of 9, 81, and 729, which is 729. • 2500 = x [ (8 × 81 + 64 × 9 + 512/729) ] • 2500 = x [ (648 + 576 + 512/729) ] • 2500 = x [ (1736/729) ] • Solve for x: x = (2500 × 729/1736) • x = (1822500/1736) \approx 1049.827 • Rounding to two decimal places, we get approximately ₹ 1049.83, which makes ₹ 1049.85 the closest matching option. Exam Strategy & Shortcut: Memorize fraction equivalents: 12.5% = 1/8. This immediately converts the ratio to 9/8, making the powers much simpler to compute mentally. Common Pitfall: A major mistake is calculating instalments on the full ₹ 3500 without first deducting the ₹ 1000 down payment. Final Answer: Therefore, the correct answer is ₹ 1049.85.

Question 5

What annual payment will discharge a debt of ₹ 1025 due in 2 years at the rate of 5% compound interest?

Accepted Answer

Concept & Instalments for Compound Interest: The present value of a debt paid in equal instalments is calculated using the formula for compound interest present worth. P = (x/(1 + \frac{R){100})} + (x/(1 + \frac{R){100})^2} where P is the principal borrowed, x is the annual instalment, and R is the rate of interest. Step-by-Step Solution: • Given: Debt (Principal) P = 1025, Rate R = 5%, Time T = 2 years. • Let the annual instalment be x. • 1025 = (x/1 + \frac{5){100}} + (x/(1 + \frac{5){100})^2} • 1025 = (x/1.05) + (x/(1.05)^2) = (x/\frac{21){20}} + (x/(\frac{21){20})^2} • 1025 = x((20/21) + (400/441)) • 1025 = x((420 + 400/441)) = x((820/441)) • x = (1025 × 441/820) • x = (205 × 441/164) = (5 × 441/4) = (2205/4) = 551.25 Exam Strategy & Shortcut: Recognize that the sum of the series for 2 years at 5% is (820/441). The instalment is simply P × (441/820). Notice that 820 and 1025 are both multiples of 205 (which is 41 × 5), simplifying the calculation quickly to 5 × (441/4). Common Pitfall: A common mistake is treating the "debt due in 2 years" as the future amount rather than the present principal, leading to incorrect applications of the instalment formula. Final Answer: Therefore, the correct answer is ₹ 551.25.

Question 6

A sum of money is borrowed and paid back in two annual instalments of ₹ 882 each allowing 5% compound interest. The sum borrowed was

Accepted Answer

Concept & Present Value of Instalments: The sum borrowed (Principal) is the total present value of all future instalments discounted at the given interest rate. P = (x/(1 + \frac{R){100})} + (x/(1 + \frac{R){100})^2} Step-by-Step Solution: • Given: Instalment x = 882, Rate R = 5%. • The fractional rate is 1 + (5/100) = 1 + (1/20) = (21/20). • P = (882/\frac{21){20}} + (882/(\frac{21){20})^2} • P = 882 × (20/21) + 882 × (400/441) • Calculate the first term: 882 \div 21 = 42, so 42 × 20 = 840. • Calculate the second term: 882 \div 441 = 2, so 2 × 400 = 800. • Sum the present values: P = 840 + 800 = 1640. Exam Strategy & Shortcut: Recognize that 882 is exactly 2 × 441, which is a strong hint since 21^2 = 441. This makes the cancellation immediate: 2 × 21 × 20 + 2 × 400 = 840 + 800 = 1640. Common Pitfall: Students sometimes incorrectly multiply the instalment amount by the number of years and then subtract interest, effectively confusing simple interest arithmetic with compound discounting. Final Answer: Therefore, the correct answer is ₹ 1640.

Question 7

Directions: Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.  What is the rate of i…

Accepted Answer

Concept & Evaluating Multiple Statements: When given three statements in Data Sufficiency, test each independently first, then in pairs. The goal is to find the minimum information required to find the rate of interest (R). Step-by-Step Solution: 1. Evaluate III alone: An amount doubles in 5 years at simple interest. This means SI = P. Formula: P = (P × R × 5/100) \Rightarrow R = 20%. Statement III alone is sufficient. 2. Evaluate I and II together: Statement I: SI per annum = ₹ 5300. Thus, (P × R/100) = 5300. Statement II: CI - SI for 2 yrs = ₹ 1060. Thus, P((R/100))^2 = 1060. Divide II by I: (P(\frac{R/100))^2}{(P × R/100)} = (1060/5300) \Rightarrow (R/100) = 0.2 \Rightarrow R = 20%. Statements I and II together are also sufficient. 3. Conclusion: The rate can be found using either III alone, or I and II together. Exam Strategy & Shortcut: Always check the simplest statement first. Statement III is a classic standard property (doubling in 5 years always means R = (100/5) = 20%). Finding this immediately narrows down the options. Common Pitfall: Stopping after finding one sufficient statement (like III) and choosing "Only III" without checking if combinations of the remaining statements also work. Final Answer: Therefore, the correct answer is Only III or I and II.

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