Question 1

A man invests ₹ 5000 for 3 years at 5% p.a. compound interest reckoned yearly. Income tax at the rate of 20% on the interest earned is deducted at the end of each year. Find the amount at the end of the third year. (M.A.T., 2004)

Accepted Answer

Concept & Strategy: When income tax is deducted on the interest earned each year, it effectively reduces the rate of interest. If 20% of the interest goes to tax, the investor keeps 80% of the interest. Effective Rate of Interest = Nominal Rate × (1 - Tax Rate). Step-by-Step Solution: • Nominal Rate = 5% p.a. • Tax Rate on interest = 20% • The effective rate of interest the man earns is 80% of 5%. Effective Rate = 5% × ((100 - 20/100)) = 5 × 0.8 = 4% • Now, calculate the compound interest amount with P = 5000, R = 4%, and n = 3 years: Amount A = P × (1 + (R/100))^n A = 5000 × (1 + (4/100))^3 A = 5000 × (1.04)^3 • Calculate (1.04)^3: 1.04^2 = 1.0816 1.0816 × 1.04 = 1.124864 • A = 5000 × 1.124864 = 5624.32 Exam Strategy & Shortcut: By finding the effective interest rate upfront (4%), you transform a complex taxation problem into a standard, straightforward compound interest calculation. Common Pitfall: Calculating the interest year by year, deducting 20%, and adding it to the principal for three iterations is extremely tedious and prone to calculation errors. Always find the effective rate first. Final Answer: Therefore, the correct answer is ₹ 5624.32.

Question 2

The compound interest on ₹ 30,000 at 7% per annum is ₹ 4347. The period (in years) is (L.I.C.A.A.O., 2003)

Accepted Answer

Concept & Formula: The relationship between Amount (A), Principal (P), Rate (R), and Time (n) in compound interest is given by: A = P (1 + (R/100))^n The Amount is also equal to Principal + Compound Interest. Step-by-Step Solution: • Given Principal P = 30000 • Compound Interest C.I. = 4347 • Rate R = 7% p.a. • Amount A = P + C.I. = 30000 + 4347 = 34347 • Using the formula: 34347 = 30000 × (1 + (7/100))^n (34347/30000) = ((107/100))^n (11449/10000) = ((107/100))^n *(Dividing numerator and denominator by 3)* • Notice that 10000 = 100^2. Let us check if 11449 = 107^2. 107^2 = (100 + 7)^2 = 10000 + 1400 + 49 = 11449. • Therefore, ((107/100))^2 = ((107/100))^n • Comparing the powers, we get n = 2. Exam Strategy & Shortcut: Look at the denominator after isolating the ratio A/P. The fraction is (34347/30000). Dividing by 3 yields (11449/10000). The denominator 10000 is perfectly 100^2. This strongly hints that the power n is 2, as the base rate fraction is (107/100). Common Pitfall: Getting stuck trying to solve for n using logarithms when simple fraction simplification and recognizing perfect squares is much faster. Final Answer: Therefore, the correct answer is 2 years.

Question 3

₹ 2000 amounts to ₹ 2226.05 in 2 years at compound interest. What will be the rate of interest? (E.S.I.C., 2006)

Accepted Answer

Concept & Formula: The fundamental formula for calculating the amount under compound interest is: A = P (1 + (R/100))^n To find the rate, isolate the (1 + R/100) term and take the n-th root of the A/P ratio. Step-by-Step Solution: • Given Principal P = 2000 • Amount A = 2226.05 • Time n = 2 years • Plugging these into the formula: 2226.05 = 2000 × (1 + (R/100))^2 (1 + (R/100))^2 = (2226.05/2000) • Multiply numerator and denominator by 100 to remove the decimal: (1 + (R/100))^2 = (222605/200000) • Simplify the fraction by dividing by 5: (1 + (R/100))^2 = (44521/40000) • The denominator 40000 is a perfect square: 200^2. We need to find the square root of 44521. Since it ends in 1, the root ends in 1 or 9. It is larger than 200^2 (40000) and smaller than 210^2 (44100). Wait, 210^2 = 44100. So it must be 211^2. Let us check 211^2: 211 × 211 = 44521. • So, (1 + (R/100))^2 = ((211/200))^2 1 + (R/100) = (211/200) = 1.055 (R/100) = 0.055 \implies R = 5.5% Exam Strategy & Shortcut: Once you have the ratio (A/P) = (2226.05/2000) = 1.113025, you are looking for \sqrt{1.113025}. Check the options: (a) 5% \implies 1.05^2 = 1.1025 (d) 6% \implies 1.06^2 = 1.1236 Our value 1.113025 is squarely between them. Let us test 5.5% \implies 1.055^2. Using the rule for squaring numbers ending in 5: 105 × 106 = 11130, append 25 \implies 1113025. Perfect match. Common Pitfall: Not eliminating the decimal points before trying to simplify the fraction (2226.05/2000). Multiplying by 100 first makes the perfect squares much easier to identify. Final Answer: Therefore, the correct answer is 5.5%.

Question 4

A man invests ₹ 4000 for 3 years at compound interest. After one year the money amounts to ₹ 4320. What will be the amount (to the nearest rupee) due at the end of 3 years? (R.R.B., 2006)

Accepted Answer

Concept & Formula: The amount under compound interest is calculated using the formula: A = P (1 + (R/100))^n When the amount after 1 year is given, we can directly find the rate of interest, which remains constant for subsequent years. Step-by-Step Solution: • Given Principal P = 4000 • Amount after 1 year A_1 = 4320 • Interest for the first year = 4320 - 4000 = 320 • Rate R = (\frac{Interest × 100}{P × 1}) = (320 × 100/4000) = 8% p.a. • Now, find the amount due at the end of 3 years using R = 8% and n = 3: A_3 = 4000 × (1 + (8/100))^3 A_3 = 4000 × (1.08)^3 • Calculate (1.08)^3: 1.08 × 1.08 = 1.1664 1.1664 × 1.08 = 1.259712 • A_3 = 4000 × 1.259712 = 5038.848 • Rounding to the nearest rupee gives ₹ 5039. Exam Strategy & Shortcut: Instead of recalculating from the base P = 4000 for 3 years, you can treat the amount after 1 year (4320) as the principal for the remaining 2 years: 4320 × (1.08)^2 = 4320 × 1.1664 = 5038.848. This saves one multiplication step. Common Pitfall: A frequent mistake is applying the interest for 3 years simply as simple interest (320 × 3 = 960 \implies 4960), completely ignoring the compounding effect on the interest itself. Final Answer: Therefore, the correct answer is ₹ 5039.

Question 5

The present worth of ₹ 169 due in 2 years at 4% per annum compound interest is

Accepted Answer

Concept & Logic: "Present worth" is simply the Principal (P) that needs to be invested today to yield a specific future Amount (A) under compound interest. P = (A/(1 + \frac{R){100})^n} Step-by-Step Solution: • Given Future Amount A = 169 • Time n = 2 years • Rate R = 4% p.a. • Convert 4% to a fraction: 4% = (4/100) = (1/25). • The multiplier is (1 + (1/25)) = (26/25). • Substitute into the formula: P = (169/(\frac{26){25})^2} P = 169 × ((25/26))^2 P = 169 × (625/676) • Simplify the expression. Notice that 169 and 676 are related. 169 × 4 = 676. (Alternatively, 13^2 = 169 and 26^2 = 676). So, P = (169 × 625/169 × 4) P = (625/4) • Evaluate the final fraction: P = 156.25 Exam Strategy & Shortcut: Instead of dealing with large multiplications, break numbers into prime factors or recognizable squares early. Seeing 169 as 13^2 and 26 as 13 × 2 makes the cancellation obvious without doing heavy division. Common Pitfall: Confusing "Present Worth" in a compound interest context with formulas for "True Discount", which are primarily used in simple interest or specific commercial discounting scenarios. Final Answer: Therefore, the correct answer is ₹ 156.25.

Question 6

In how many years will a sum of ₹ 800 at 10% per annum compounded semi-annually become ₹ 926.10? (S.S.C., 2010)

Accepted Answer

Concept & Conversion: For semi-annual (half-yearly) compounding, the annual interest rate is divided by 2, and the compounding periods are in half-years. A = P (1 + (R/2/100))^n where n is the number of half-years. Step-by-Step Solution: • Given Principal P = 800 • Amount A = 926.10 • Annual Rate = 10% \implies Semi-annual Rate r = 5% • Substitute into the formula: 926.10 = 800 (1 + (5/100))^n (926.10/800) = (1 + (1/20))^n (926.10/800) = ((21/20))^n • Remove the decimal by multiplying numerator and denominator by 10: (9261/8000) = ((21/20))^n • Observe the denominator: 8000 = 20^3. Let us check if 9261 is 21^3. 21^2 = 441. 441 × 21 = 9261. Yes. • Therefore, ((21/20))^3 = ((21/20))^n n = 3 half-years. • Convert half-years to years: 3 half-years = (3/2) years = 1(1/2) years. Exam Strategy & Shortcut: Always look for perfect squares or cubes when you have the fraction (A/P). Once you see 8000, recognize immediately that it is 20^3. This instantly tells you n = 3 periods without calculating 21^3. Common Pitfall: A classic mistake is finding n = 3 and looking for "3 years" in the options, completely forgetting that n represents the number of semi-annual periods, not years. Final Answer: Therefore, the correct answer is 1(1/2) years.

Question 7

The compound interest on a certain sum for 2 years at 10% per annum is ₹ 525. The simple interest on the same sum for double the time at half the rate percent per annum is

Accepted Answer

Concept & Formula: Compound interest for 2 years compounded annually is given by: CI = P [(1 + (R/100))^T - 1] Simple interest is calculated as: SI = (P × R × T/100) Step-by-Step Solution: 1. Find the Principal (P): Given CI = ₹ 525, T = 2 years, and R = 10%. CI = P [(1 + (10/100))^2 - 1] = P [((11/10))^2 - 1] 525 = P [(121/100) - 1] = P ((21/100)) P = (525 × 100/21) = 25 × 100 = ₹ 2500 2. Calculate the Simple Interest (SI): • Double the time: T' = 2 × 2 = 4 years • Half the rate: R' = (10%/2) = 5% SI = (2500 × 5 × 4/100) = 25 × 20 = ₹ 500 Exam Strategy & Shortcut: • Effective CI rate for 2 years at 10% = 10 + 10 + (10 × 10/100) = 21%. • Effective SI rate for double the time (4 years) at half the rate (5%) is 4 × 5% = 20%. • Therefore, SI = 525 × (20/21) = 25 × 20 = ₹ 500. Common Pitfall: Students often forget to scale the rate and time for the simple interest part, mistakenly using the original time and rate values. Final Answer: Therefore, the correct answer is ₹ 500.

Question 8

There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of ₹ 12,000 after 3 years at the same rate?

Accepted Answer

Concept & Formula: Rate of Simple Interest: R = (SI × 100/P × T) Compound Interest formula: CI = P [(1 + (R/100))^T - 1] Step-by-Step Solution: 1. Find the Rate of Interest (R): • Let the original sum be P = 100. • Increase in 6 years (SI) = 60. R = (60 × 100/100 × 6) = 10% per annum 2. Calculate Compound Interest for 3 years: • P = ₹ 12000, R = 10%, T = 3 years A = 12000 × (1 + (10/100))^3 = 12000 × ((11/10))^3 A = 12000 × (1331/1000) = 12 × 1331 = ₹ 15972 CI = A - P = 15972 - 12000 = ₹ 3972 Exam Strategy & Shortcut: For R = 10% over 3 years, the effective compound interest rate is 33.1%. CI = 33.1% of 12000 = 33.1 × 120 = ₹ 3972 Common Pitfall: Confusing the 6-year period from the first part of the problem with the 3-year period required for calculating compound interest. Final Answer: Therefore, the correct answer is ₹ 3972.

Question 9

The compound interest accrued on an amount of ₹ 25500 at the end of 3 years is ₹ 8440.50. What would be the simple interest accrued on the same amount at the same rate in the same period?

Accepted Answer

Concept & Formula: Total Amount under compound interest: A = P + CI = P (1 + (R/100))^T Simple Interest formula: SI = (P × R × T/100) Step-by-Step Solution: 1. Find the Rate of Interest (R): • P = ₹ 25500, CI = ₹ 8440.50, T = 3 years A = 25500 + 8440.50 = ₹ 33940.50 (A/P) = (1 + (R/100))^3 (1 + (R/100))^3 = (33940.50/25500) = (339405/255000) = (1331/1000) = ((11/10))^3 1 + (R/100) = (11/10) \implies (R/100) = (1/10) \implies R = 10% 2. Calculate the Simple Interest (SI): SI = (25500 × 10 × 3/100) = 255 × 30 = ₹ 7650 Exam Strategy & Shortcut: Recognize the ratio (A/P) = (33940.5/25500) = 1.331 = (1.1)^3, which immediately indicates R = 10%. Then: SI = 30% of 25500 = 0.30 × 25500 = ₹ 7650 Common Pitfall: Lengthy division without simplifying fractions to detect standard cube powers like 1331/1000. Final Answer: Therefore, the correct answer is ₹ 7650.

Question 10

The difference between the amount the amount of compound interest and simple interest accrued on an amount of ₹ 26000 at the end of 3 years is ₹ 2994.134. What is the rate of interest p.c.p.a.?  (Bank P.O., 2005)

Accepted Answer

Concept & Formula: The difference between Compound Interest (CI) and Simple Interest (SI) for 3 years on a principal P at a rate R% per annum is given by the formula: D_3 = P ((R/100))^2 (3 + (R/100)) Step-by-Step Solution: 1. Extract Given Values: Principal (P) = ₹ 26000 Time (T) = 3 years Difference (D_3) = ₹ 2994.134 Let r = (R/100) 2. Set up the Equation: 2994.134 = 26000 × r^2 × (3 + r) r^2(3 + r) = (2994.134/26000) r^2(3 + r) = 0.115159 3. Test the Given Options: Instead of solving the cubic equation, substitute the options into r = (R/100): • Option (a) 17%: r = 0.17 \implies (0.17)^2 × (3 + 0.17) = 0.0289 × 3.17 = 0.091613 (Incorrect) • Option (b) 19%: r = 0.19 \implies (0.19)^2 × (3 + 0.19) = 0.0361 × 3.19 = 0.115159 (Correct) • Option (c) 22%: r = 0.22 \implies (0.22)^2 × (3 + 0.22) = 0.0484 × 3.22 = 0.155848 (Incorrect) Exam Strategy & Shortcut: Solving cubic equations in a competitive exam is too time-consuming. Always rely on option checking. Look at the unit digits or the approximate scale to quickly narrow down the choices. Since 10% yields roughly 0.03, and 20% yields 0.128, the answer must be very close to 20%. Thus, 19% is the immediate logical candidate to test first. Common Pitfall: Attempting to algebraically solve the cubic polynomial r^3 + 3r^2 - 0.115159 = 0 manually, which leads to massive time loss. Final Answer: Therefore, the correct answer is 19%.

Question 11

The difference between compound interest and simple interest on a sum for 2 years at 8 per cent is ₹ 768. The sum is

Accepted Answer

Concept & Formula: The difference between Compound Interest (CI) and Simple Interest (SI) for exactly 2 years on a principal P at a rate of R% per annum is given by the formula: D_2 = P ((R/100))^2 Step-by-Step Solution: 1. Extract Given Values: Difference for 2 years (D_2) = ₹ 768 Rate of interest (R) = 8% Time (T) = 2 years 2. Apply the 2-Year Difference Formula: 768 = P ((8/100))^2 768 = P ((64/10000)) 3. Solve for Principal (P): P = (768 × 10000/64) P = 12 × 10000 P = ₹ 120000 Exam Strategy & Shortcut: The difference essentially arises from the interest calculated on the first year's interest. First year's interest rate = 8%. Interest on that 8% in the second year = 8% of 8% = 0.64%. So, 0.64% of the Principal = 768. P = (768/0.64) × 100 = 1200 × 100 = ₹ 120000 Common Pitfall: Using the 3-year difference formula or mistakenly computing D = P × (R^2/100) instead of dividing by 10000. Final Answer: Therefore, the correct answer is ₹ 120000.

Question 12

The difference between compound interest and simple interest on a sum for 2 years at 10% per annum, when the interest is compounded annually is ₹ 16. If the interest were compounded half-yearly, the difference in two interests would be

Accepted Answer

Concept & Logic: First, use the annual difference to find the Principal (P). D_2 = P ((R/100))^2 Then, find the new difference using the half-yearly compounding rate over the same duration. Step-by-Step Solution: 1. Find the Principal (P): Given D_2 = ₹ 16, R = 10%, T = 2 years. 16 = P ((10/100))^2 16 = P ((1/100)) \implies P = ₹ 1600 2. Calculate SI for 2 years at 10%: SI = (1600 × 10 × 2/100) = ₹ 320 3. Calculate CI compounded half-yearly: Rate per half-year = (10%/2) = 5%. Time periods = 2 × 2 = 4 half-years. A = 1600 (1 + (5/100))^4 = 1600 × (1.05)^4 (1.05)^4 = 1.21550625 A = 1600 × 1.21550625 = 1944.81 CI = 1944.81 - 1600 = ₹ 344.81 4. Find the New Difference: New Difference = CI - SI = 344.81 - 320 = ₹ 24.81 Exam Strategy & Shortcut: From 10% annually for 2 years, the difference is exactly 1% of P. Thus, P = 1600. For half-yearly compounding, effective CI rate is 21.55%. SI rate is 20%. Difference in effective rates = 1.55%. 1.55% of 1600 = 16 × 1.55 = 24.8. Using exact decimal yields 24.81. Common Pitfall: Forgetting that the new difference requested is still between the new CI (half-yearly) and the original SI, not between the two different CIs. Final Answer: Therefore, the correct answer is ₹ 24.81.

Question 13

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 14

On a sum of money, the simple interest for 2 years is ₹ 660, while the compound interest is ₹ 696.30, the rate of interest being the same in both the cases. The rate of interest is

Accepted Answer

Concept & Difference between CI and SI: For the first year, Simple Interest (SI) and Compound Interest (CI) are identical. The difference in the second year is purely the interest earned on the first year's interest. Step-by-Step Solution: 1. Find SI per year: Total SI for 2 years = ₹ 660 SI for 1 year = (660/2) = ₹ 330 2. Find the difference between CI and SI for 2 years: Difference = CI - SI = 696.30 - 660 = 36.30 3. Determine the rate of interest (R): This ₹ 36.30 is exactly the interest generated in the second year on the first year's interest of ₹ 330. R = ((36.30/330)) × 100 R = ((363/3300)) × 100 = 11% 4. Compare with options: The calculated rate is 11%, which is not present in options (a), (b), or (c). Exam Strategy & Shortcut: Always remember: for 2 years, Difference = (SI × R/2 × 100). 36.30 = (660 × R/200) \implies R = (36.30 × 200/660) = (7260/660) = 11%. Scanning the options immediately points to "None of these". Common Pitfall: Students often try to set up complex quadratic equations involving the principal P, which wastes significant time. Use the direct relationship between SI and CI difference. Final Answer: Therefore, the correct answer is None of these.

Question 15

The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly is

Accepted Answer

Concept & Effective Annual Rate (EAR): When interest is compounded more than once a year, the effective annual rate (EAR) is higher than the nominal annual rate due to compounding. EAR = (1 + (R/n × 100))^n - 1 Step-by-Step Solution: 1. Identify the given values: Nominal annual rate = 6% Compounding frequency = half-yearly (n = 2) 2. Determine half-yearly rate: Rate per half-year = (6%/2) = 3% 3. Calculate effective rate for two periods (1 year): Equivalent multiplier = (1 + (3/100))^2 = (1.03)^2 (1.03)^2 = 1.0609 4. Convert to percentage: Effective rate = (1.0609 - 1) × 100% = 0.0609 × 100% = 6.09% Exam Strategy & Shortcut: Use the successive percentage formula for 2 periods: A + B + (A × B/100). Here, A = 3%, B = 3%. Effective Rate = 3 + 3 + (3 × 3/100) = 6 + 0.09 = 6.09%. Common Pitfall: Misunderstanding "nominal rate". Some students mistakenly apply the 6% rate twice, yielding 6 + 6 + 0.36 = 12.36%, failing to divide the annual rate by the number of compounding periods first. Final Answer: Therefore, the correct answer is 6.09%.

Question 16

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 17

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.

Accepted Answer

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 18

A sum of money invested at compound interest amounts to ₹ 4624 in 2 years and to ₹ 4913 in 3 years. The sum of money is :

Accepted Answer

Concept & Deriving Principal from Consecutive Amounts: First, find the rate of interest using the difference between consecutive years' amounts. Then, trace backward using the compound interest formula or multipliers to find the original principal. Amount = P(1 + (R/100))^T Step-by-Step Solution: 1. Find the interest for the 3rd year: Amount after 2 years (A_2) = ₹ 4624 Amount after 3 years (A_3) = ₹ 4913 Interest = 4913 - 4624 = 289 2. Calculate the rate of interest (R): R = ((289/4624)) × 100 Notice that 289 × 16 = 4624. R = ((1/16)) × 100 = 6.25% 3. Find the principal (P): The multiplier for each year is (1 + (1/16)) = (17/16). P × ((17/16))^2 = A_2 P × ((17/16))^2 = 4624 P × ((289/256)) = 4624 P = (4624 × 256/289) P = 16 × 256 = 4096 Exam Strategy & Shortcut: Recognize squares and cubes. 4913 is 17^3. 4624 is 68^2, but more usefully, the multiplier is (4913/4624) = (17/16). Since A_2 = P × ((17/16))^2, then P = A_2 × ((16/17))^2 = 4624 × (256/289) = 4096. Common Pitfall: A common pitfall is stopping after finding the rate of interest, or making calculation errors when squaring fractions like (17/16). Always carry out the final reverse calculation step carefully. Final Answer: Therefore, the correct answer is ₹ 4096.

Question 19

The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is

Accepted Answer

Concept & Exponential Growth: To find when a sum more than doubles, we need to find the smallest integer value of n for which the amount is greater than 2P. P(1 + (R/100))^n > 2P Step-by-Step Solution: • Given: Rate of interest R = 20%. • We need: P(1 + (20/100))^n > 2P • Divide both sides by P: (1 + (1/5))^n > 2 • ((6/5))^n > 2, which means (1.2)^n > 2. • Let's evaluate powers of 1.2: • For n = 1: (1.2)^1 = 1.2 • For n = 2: (1.2)^2 = 1.44 • For n = 3: (1.2)^3 = 1.44 × 1.2 = 1.728 • For n = 4: (1.2)^4 = 1.728 × 1.2 = 2.0736 • Since 2.0736 > 2, the sum will be more than doubled in 4 years. Exam Strategy & Shortcut: For 20%, you can use the Rule of 72 as a quick approximation for doubling time. 72 / 20 = 3.6 years. Since the interest is compounded annually, we round up to the next complete year, which is 4. Common Pitfall: Miscalculating the decimal multiplications (e.g., assuming 1.2^3 is greater than 2) or confusing it with simple interest where it would take exactly 5 years to double. Final Answer: Therefore, the correct answer is 4.

Question 20

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.

Accepted Answer

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.

Compound Interest Questions

Questions

Practice smarter