Question 1

Cash Discount versus Credit Terms: A merchant receives an invoice for a motor boat for $20,000 with credit terms 4/30, n/100 (a 4% discount if paid within 30 days, otherwise the full amount is due in 100 days). What is the highest simple annual inte…

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Introduction / Context: This question compares cash discount terms with delayed payment terms and asks you to find the highest simple annual interest rate at which it is still beneficial to borrow money to pay early. It is a classic time value of money and trade credit problem, where you relate the discount saved to the time period over which payment is advanced. Given Data / Assumptions: Invoice amount = $20,000. Terms 4/30, n/100: 4% discount if paid within 30 days; otherwise pay full in 100 days. If paying early, the merchant pays within 30 days instead of 100 days. Therefore, the merchant pays 70 days earlier (100 − 30 = 70 days). Interest rate sought is a simple annual rate. We use simple interest and a 360 day year (commonly used in such problems) to match the given options. Concept / Approach: If the merchant takes the discount, he pays 96% of the invoice now instead of 100% later. The discount saved is 4% of 20,000, which is $800. Effectively, he is earning $800 by paying $19,200 earlier than otherwise required. This is like earning interest on $19,200 for 70 days. The implied simple interest rate is (interest / principal) * (year length / days), converted to a percentage. Step-by-Step Solution: Step 1: Compute the discounted price: 96% of $20,000 = 0.96 * 20,000 = $19,200. Step 2: Discount saved = 20,000 − 19,200 = $800. Step 3: The merchant advances payment by 70 days (100 − 30). Step 4: Think of this as earning $800 interest on a "principal" of $19,200 over 70 days. Step 5: Simple rate r (per year) is given by r = (I / P) * (year length / time in days) * 100. Step 6: Using a 360 day year, r = (800 / 19,200) * (360 / 70) * 100. Step 7: First compute 800 / 19,200 ≈ 0.0416667. Step 8: Next compute 360 / 70 ≈ 5.142857. Step 9: Multiply: 0.0416667 * 5.142857 ≈ 0.2142857. Step 10: Convert to percent: r ≈ 0.2142857 * 100 ≈ 21.43% per annum. Verification / Alternative check: Because the discount is 4% for 70 days, a quick rule of thumb is to annualise the discount rate: approximate annual rate ≈ (discount / discounted price) * (360 / days). Here, 4 / 96 ≈ 4.17%, and 4.17% * (360 / 70) ≈ 21.43%. This matches the detailed calculation above. Taking any borrowing rate lower than this is beneficial, since the cost of borrowing would be less than the effective return from taking the discount. Why Other Options Are Wrong: 18.00% and 15.00%: These are below the implied rate; if the merchant can borrow at these rates, he certainly should take the discount, but they are not the highest break even rate. 24.00% and 30.00%: These are higher than the calculated effective rate; borrowing at such rates may make taking the discount uneconomical. Common Pitfalls: Common mistakes include forgetting to use the discounted price as the principal, using 70 / 360 instead of 360 / 70, or computing the discount as a percentage of the full amount without adjusting for the shorter time period. Some students also confuse the 4% discount with a 4% annual rate, which is incorrect. Always annualise the effective short term return to compare with an annual borrowing rate. Final Answer: The highest simple annual interest rate at which taking the discount remains worthwhile is approximately 21.43% per annum.

Question 2

Present Value under Compound Interest: Find the principal (in $) that will amount to $25,000 if invested at 3% per annum for 20 years, compounded annually. You are given the final amount and must determine the present value.

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Introduction / Context: This question asks you to work backward from a future amount under compound interest to find the present value or principal. Such problems appear frequently in time value of money topics in finance and aptitude tests. The investment grows at a fixed rate with annual compounding, and you must find how much must be invested today to reach a target amount after a given number of years. Given Data / Assumptions: Future amount A = $25,000. Annual interest rate r = 3% per annum. Time t = 20 years. Compounding is annual. Principal P (present value) is unknown. Concept / Approach: For compound interest with annual compounding, the amount A after t years is given by A = P * (1 + r / 100)^t. To find P when A, r, and t are known, rearrange this formula: P = A / (1 + r / 100)^t. Then compute the power term accurately and divide the future amount by this factor to get the present value. Step-by-Step Solution: Step 1: Write the compound interest amount formula: A = P * (1 + r / 100)^t. Step 2: Rearrange for P: P = A / (1 + r / 100)^t. Step 3: Substitute A = 25,000, r = 3, t = 20. Step 4: Compute the factor (1 + 3 / 100)^20 = (1.03)^20. Step 5: (1.03)^20 ≈ 1.8061 (using standard compound factor tables or calculator). Step 6: Now compute P = 25,000 / 1.8061 ≈ $13,841.89. Step 7: Therefore, the required principal is approximately $13,841.89. Verification / Alternative check: Multiply the computed principal back by the compound factor: 13,841.89 * (1.03)^20 ≈ 13,841.89 * 1.8061 ≈ 25,000 (within rounding tolerance). This confirms that investing about $13,841.89 at 3% per annum for 20 years will grow to $25,000 with annual compounding. Why Other Options Are Wrong: $14,821.00 and $15,625.00: These are larger than the correct present value; invested at 3% for 20 years, they would grow to more than $25,000. $13,468.00: Slightly smaller than the correct value; after compounding it would not reach the full $25,000. $57,389.00: Much larger than needed and clearly unrealistic as a present value for the given future amount and rate. Common Pitfalls: Some candidates mistakenly use the simple interest formula when the question clearly specifies compound interest and annual compounding. Others try to multiply by (1 + r)^t instead of dividing by it when moving from future value back to present value. Always remember that compounding grows a present amount into a future amount; to find the starting amount, you must discount the future amount by the same factor. Final Answer: The principal that will grow to $25,000 in 20 years at 3% compounded annually is approximately $13,841.89.

Question 3

Data Sufficiency – Principal Sum from Simple Interest and Doubling: What is the principal sum (in ₹) that earned the interest? Statement A: The total simple interest is ₹9,000 after 9 years. Statement B: The total of principal and simple interest be…

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Introduction / Context: This data sufficiency question is similar in structure to earlier ones: you must decide whether each statement alone or both together are sufficient to determine the principal sum on which simple interest is calculated. The problem combines information about the total simple interest over 9 years with an amount doubling condition over 6 years at the same rate. Given Data / Assumptions: Simple interest at a constant annual rate r%. Principal = P (unknown). Statement A: Total SI = ₹9,000 after 9 years. Statement B: After 6 years, the amount (principal + SI) becomes double of the principal. Concept / Approach: Simple interest follows I = (P * r * t) / 100. From Statement A we get one equation linking P and r. From Statement B we get another equation for the same P and r based on the doubling condition. The key is to see whether these two equations together are sufficient to solve uniquely for P, while each alone is not. Step-by-Step Solution: Step 1: Analyse Statement A. It gives I = 9,000 after t = 9 years. Step 2: Apply SI formula: 9,000 = (P * r * 9) / 100. Step 3: Rearranging, P * r = (9,000 * 100) / 9 = 100,000. Step 4: This is one equation with two unknowns (P and r), so P cannot be determined uniquely from Statement A alone. Step 5: Analyse Statement B. Amount doubles in 6 years, so P + I = 2P, which gives I = P. Step 6: Under SI for 6 years, I = (P * r * 6) / 100 = P. Step 7: Cancelling P, r * 6 / 100 = 1, so r = 100 / 6 ≈ 16.67% per annum. Step 8: Statement B alone determines r but not P, so P is still unknown. Step 9: Combine both statements. From A, P * r = 100,000, and from B, r = 100 / 6. Step 10: Substitute r into P * r = 100,000: P * (100 / 6) = 100,000. Step 11: Therefore, P = 100,000 * (6 / 100) = 6,000. Step 12: Thus, only by using both Statement A and Statement B together can we uniquely find the principal. Verification / Alternative check: With P = ₹6,000 and r = 100 / 6% per annum, check Statement A: SI in 9 years = (6,000 * (100 / 6) * 9) / 100 = 6,000 * (9 / 6) = 6,000 * 1.5 = ₹9,000, which matches. For Statement B, SI in 6 years = (6,000 * (100 / 6) * 6) / 100 = 6,000, so amount becomes P + I = 6,000 + 6,000 = 12,000 = 2P, so doubling holds. Both statements are consistent with these values. Why Other Options Are Wrong: Only Statement A is sufficient: Incorrect, as A alone leaves two unknowns (P and r). Only Statement B is sufficient: Incorrect, as B alone gives r but not P. Either statement alone is sufficient: Incorrect; each alone is insufficient. Neither statement is sufficient: Incorrect, because together they yield a unique P. Common Pitfalls: Test takers sometimes stop after finding r from Statement B and incorrectly claim B is sufficient, forgetting that the question asks for the principal. Others misinterpret the doubling condition and apply it to 9 years instead of 6. Always connect each statement correctly to the formula and check whether all required variables can be uniquely found. Final Answer: The correct choice is Both Statement A and Statement B are sufficient together to determine the principal sum.

Question 4

Simple Interest Rate for Doubling in a Given Time: An amount becomes double of itself in 15 years at simple interest. What is the annual rate of simple interest (in % per annum)?

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Introduction / Context: This question tests your understanding of how simple interest causes an amount to grow over time. You are told that an investment doubles in 15 years under simple interest and asked to find the annual rate of simple interest. Doubling-time questions are very common in aptitude exams and often have neat fractional answers. Given Data / Assumptions: Initial principal = P. Amount after 15 years = 2P (doubles). Time t = 15 years. Simple interest is used at a constant annual rate r%. Concept / Approach: Under simple interest, total interest I over t years is I = (P * r * t) / 100. If the amount doubles in t years, then the total interest equals the original principal: P + I = 2P, so I = P. This gives a direct equation relating r and t, which you can solve quickly. Then convert the resulting fraction into a percentage and match it to the options. Step-by-Step Solution: Step 1: Amount doubles in 15 years, so final amount A = 2P. Step 2: Total interest I = A − P = 2P − P = P. Step 3: Use the SI formula: I = (P * r * t) / 100. Step 4: Substitute I = P and t = 15: P = (P * r * 15) / 100. Step 5: Cancel P on both sides (P ≠ 0): 1 = (r * 15) / 100. Step 6: Rearrange to solve for r: r = 100 / 15. Step 7: Compute r: 100 / 15 = 6.666..., which is approximately 6.66% per annum. Step 8: Therefore, the annual simple interest rate is about 6.66%. Verification / Alternative check: Check by substituting r ≈ 6.66% back into the SI formula. SI for 15 years = (P * 6.66 * 15) / 100 ≈ P * (99.9 / 100) ≈ P. So the total amount is P + I ≈ P + P = 2P, confirming the doubling condition. The slight approximation arises from rounding 6.666... to 6.66%, but for aptitude purposes, this is accurate enough. Why Other Options Are Wrong: 7.85%, 9.41%: Both are higher than needed and would cause the amount to more than double in 15 years. 4.21% and 5.00%: These are lower than required and would not be enough to double the amount in 15 years under simple interest. Common Pitfalls: Some students mistakenly use the compound interest doubling rule or apply logarithms unnecessarily. Others divide time by 2 instead of linking interest to the principal. Under simple interest, doubling simply means interest equals the principal over the given time, leading directly to the relation r = 100 / t. Keep this shortcut in mind for quick mental calculations. Final Answer: The annual rate of simple interest is approximately 6.66% per annum.

Question 5

Althaf borrows a total of Rs 1500 from two moneylenders under simple interest. One part of the loan carries 12% per annum and the remaining part carries 14% per annum for one full year. If the total interest he pays for that year is Rs 186, how much…

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Introduction / Context: This aptitude question tests understanding of the simple interest formula when a total principal is split into two parts that earn different rates of interest. The learner needs to express the total interest as the sum of interests on each part and then form a linear equation in one variable. Such problems are common in bank exam quantitative aptitude sections and help build confidence with algebraic representation of real life borrowing situations. Given Data / Assumptions: Total amount borrowed by Althaf = Rs 1500. One part of the money is borrowed at 12% simple interest per annum. The remaining part is borrowed at 14% simple interest per annum. Time period for both loans = 1 year. Total interest paid after 1 year = Rs 186. Interest is calculated using simple interest throughout. Concept / Approach: For simple interest, the formula is: Simple interest = (Principal * Rate * Time) / 100 Here the total principal is split into two parts, say x and 1500 − x. Each part earns interest at its own rate for the same time. The total interest is the sum of the two simple interests. By equating this total to the given interest (Rs 186), we obtain a linear equation in x. Solving that equation gives the amount borrowed at 12% per annum. Step-by-Step Solution: Let the amount borrowed at 12% per annum be x rupees. Then the amount borrowed at 14% per annum is 1500 − x rupees. Simple interest on the first part for 1 year = (x * 12 * 1) / 100 = 0.12x. Simple interest on the second part for 1 year = ((1500 − x) * 14 * 1) / 100 = 0.14 * (1500 − x). Total interest for 1 year = 0.12x + 0.14 * (1500 − x). Given total interest = Rs 186, so 0.12x + 0.14 * (1500 − x) = 186. 0.12x + 210 − 0.14x = 186. −0.02x + 210 = 186. −0.02x = 186 − 210 = −24. x = (−24) / (−0.02) = 1200. Therefore, the amount borrowed at 12% per annum is Rs 1200. Verification / Alternative check: If Rs 1200 is borrowed at 12% for 1 year, interest = (1200 * 12 * 1) / 100 = Rs 144. The remaining Rs 300 is borrowed at 14%, so interest = (300 * 14 * 1) / 100 = Rs 42. Total interest = 144 + 42 = Rs 186, which matches the given information. Hence the solution is consistent and correct. Why Other Options Are Wrong: Rs 300 would give very small interest at 12% and would not produce the required total of Rs 186 when combined with the remaining part at 14%. Rs 400 also fails to satisfy the total interest equation when calculated precisely. Rs 1100 produces a total interest that is less than or greater than Rs 186 when checked with both rates. Rs 900 similarly does not give a total interest of Rs 186 for the given rates and period. Common Pitfalls: Students sometimes assume equal division of the principal between the two rates, which is not given in this question. Another common mistake is to forget that the time period is the same for both loans and accidentally multiply by different time values. Some learners also directly divide the difference in rates into the total interest without forming the correct equation, which leads to wrong values. Careful setup of the equation and systematic simplification prevent such errors. Final Answer: Thus, the amount of the loan that was borrowed at 12% simple interest per annum is Rs 1200.

Question 6

Raghu invests Rs 1000 at a simple interest rate of 6% per annum and receives a total amount of Rs 1300 after X years. Using the standard simple interest formula, determine the value of X, that is, the number of years for which the money was invested…

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Introduction / Context: This problem checks basic understanding of simple interest and how to work backward from the final amount to find the time period. Instead of asking for interest directly, the question gives the principal, rate, and final amount and asks for the duration of the investment. Such questions appear frequently in bank exams, SSC exams, and other aptitude tests where candidates must be very comfortable with the simple interest formula and rearranging it to isolate any required variable. Given Data / Assumptions: Principal invested by Raghu, P = Rs 1000. Rate of simple interest, R = 6% per annum. Number of years of investment, T = X years (unknown). Total amount received after X years, A = Rs 1300. Interest is calculated using simple interest for the entire period. Concept / Approach: The simple interest formula is: Simple interest (SI) = (P * R * T) / 100 The final amount with simple interest is: Amount A = P + SI In this problem, A and P are known, so SI can first be found as A − P. Once SI is known, we can substitute SI, P, and R into the simple interest formula and solve for T. This involves only basic algebra, rearranging a linear equation in one variable. Step-by-Step Solution: Given principal P = Rs 1000. Given rate R = 6% per annum. Amount after X years A = Rs 1300. First, find the simple interest earned: SI = A − P = 1300 − 1000 = Rs 300. Use the simple interest formula: SI = (P * R * T) / 100. So, 300 = (1000 * 6 * T) / 100. Simplify the right side: (1000 * 6) / 100 = 60, so 300 = 60T. Therefore, T = 300 / 60 = 5 years. Hence, Raghu invested the money for 5 years. Verification / Alternative check: We can verify the result by computing the amount for 5 years directly. Simple interest for 5 years at 6% per annum on Rs 1000 is SI = (1000 * 6 * 5) / 100 = 300. The final amount after 5 years is P + SI = 1000 + 300 = Rs 1300, which matches the given amount. Therefore, the solution is verified as correct. Why Other Options Are Wrong: If X were 4 years, interest would be (1000 * 6 * 4) / 100 = Rs 240, giving an amount of only Rs 1240. If X were 3 years, interest would be Rs 180, so the amount would be Rs 1180, which is too small. If X were 2 years, interest would be Rs 120, so the amount would be Rs 1120. If X were 6 years, interest would be Rs 360, leading to an amount of Rs 1360, which is larger than Rs 1300. Hence those options do not match the problem data. Common Pitfalls: A common mistake is to misread the given amount as interest rather than the total amount, which leads to an incorrect equation. Another error is to forget to subtract principal from amount to obtain the interest. Some students also mishandle the division when solving for T, especially when simplifying (P * R) / 100. Writing each step clearly prevents calculation errors and builds accuracy. Final Answer: Therefore, the money was invested for 5 years in order to grow from Rs 1000 to Rs 1300 at 6% simple interest per annum.

Question 7

A man lends a sum of money at a simple interest rate of 4% per annum. The total interest earned in 8 years is Rs 340 less than the original principal amount that he lent. Using the simple interest relation and the given difference between principal …

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Introduction / Context: This question explores how to connect the simple interest earned over a period of time with the principal when a relationship between the two is given. Instead of providing the interest directly, the problem states that the total interest earned over 8 years is Rs 340 less than the principal. The student must translate this verbal condition into an algebraic equation involving the simple interest formula. Such problems strengthen the ability to interpret words into equations, a key skill in quantitative aptitude. Given Data / Assumptions: Rate of simple interest, R = 4% per annum. Time period, T = 8 years. Principal lent, P rupees (unknown). Total simple interest earned over 8 years is Rs 340 less than the original principal. Interest is calculated using simple interest. Concept / Approach: The simple interest formula is: SI = (P * R * T) / 100 The problem states that the simple interest is less than the principal by Rs 340. That means: P − SI = 340 or equivalently: SI = P − 340 By substituting SI from the simple interest formula into this relationship, we get an equation with one unknown P. Solving this linear equation yields the required principal. Step-by-Step Solution: Let the principal be P rupees. Rate R = 4% per annum, time T = 8 years. Simple interest SI = (P * 4 * 8) / 100 = (32P) / 100 = 0.32P. Given that the interest is Rs 340 less than the principal, so SI = P − 340. Therefore, 0.32P = P − 340. Bring terms together: P − 0.32P = 340. 0.68P = 340. P = 340 / 0.68. P = 500. Hence, the principal lent by the man is Rs 500. Verification / Alternative check: For P = Rs 500, simple interest for 8 years at 4% per annum is SI = (500 * 4 * 8) / 100 = 160. Now compare principal and interest: P − SI = 500 − 160 = 340, which matches the condition given in the question. Therefore, the value P = Rs 500 satisfies both the simple interest formula and the relationship between principal and interest. Why Other Options Are Wrong: If P = Rs 520, then SI = (520 * 4 * 8) / 100 = 166.4, and P − SI = 353.6, not 340. If P = Rs 540, then SI = 172.8, and P − SI = 367.2, which does not match 340. If P = Rs 560, then SI = 179.2, so P − SI = 380.8, again not equal to 340. If P = Rs 600, then SI = 192, and P − SI = 408, which is different from 340. Thus those options do not satisfy the given condition. Common Pitfalls: Learners sometimes misunderstand the phrase “interest is Rs 340 less than the principal” and mistakenly set SI = 340 instead of P − 340. Another frequent error is to compute the amount instead of comparing principal and interest. Algebraic mistakes in combining terms, such as miscomputing 1 − 0.32, can also lead to wrong answers. Writing the relationship clearly before substituting into the simple interest formula helps to avoid confusion. Final Answer: The value of the original principal that the man lent is Rs 500.

Question 8

A sum of money is invested for 3 years at a certain rate of simple interest. If instead it had been invested at a rate that is 4% higher per annum, the total simple interest earned in the same 3 years would have been Rs 600 more than before. Using t…

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Introduction / Context: This problem tests the ability to work with changes in interest rate under simple interest. When the interest rate is increased by a fixed percentage, the extra interest earned over a given time can be related directly to the principal. Instead of giving the actual rate, the question provides the increase in rate (4% per annum) and the resulting increase in interest (Rs 600 in 3 years). The task is to use the simple interest formula to relate these quantities and solve for the principal or original sum invested. Given Data / Assumptions: Time period T = 3 years. Original rate of interest = R% per annum (unknown). New rate of interest = (R + 4)% per annum. Increase in total simple interest over 3 years when rate increases by 4% = Rs 600. Original principal or sum invested = P rupees (unknown). Interest is calculated using simple interest for both scenarios. Concept / Approach: Simple interest is given by: SI = (P * R * T) / 100 When the rate increases from R to R + 4, the difference in interest for the same principal and time is due only to the change in rate. The extra interest is: Extra SI = (P * (R + 4) * T) / 100 − (P * R * T) / 100 This simplifies neatly because the R terms cancel out, leaving an expression in terms of P, T, and the rate change 4. By equating that extra interest to Rs 600, we can solve for P directly. Step-by-Step Solution: Let the original principal be P rupees. Original simple interest for 3 years at R% = (P * R * 3) / 100. New simple interest for 3 years at (R + 4)% = (P * (R + 4) * 3) / 100. Increase in interest = New SI − Old SI. Increase in interest = (P * (R + 4) * 3) / 100 − (P * R * 3) / 100. Factor out P * 3 / 100: Increase = (P * 3 / 100) * ((R + 4) − R) = (P * 3 / 100) * 4. So increase in interest = (P * 12) / 100 = 0.12P. Given that this increase equals Rs 600, we have 0.12P = 600. Therefore, P = 600 / 0.12 = 5000. Thus, the original sum invested is Rs 5000. Verification / Alternative check: To verify, choose any convenient original rate, for example R = 10% per annum. At 10% for 3 years on Rs 5000, SI = (5000 * 10 * 3) / 100 = 1500. At the increased rate of 14%, SI = (5000 * 14 * 3) / 100 = 2100. Difference in interest = 2100 − 1500 = Rs 600, which matches the question. The choice of R is arbitrary because the difference depends only on the 4% increase, confirming that P = Rs 5000 is correct. Why Other Options Are Wrong: If P were Rs 4000, then extra interest at 4% for 3 years would be (4000 * 4 * 3) / 100 = Rs 480, not 600. If P were Rs 4950, the extra interest would be (4950 * 4 * 3) / 100 = Rs 594, slightly less than 600. If P were Rs 5150, the extra interest would be Rs 618, larger than 600. If P were Rs 6000, the extra interest would be Rs 720. Therefore these options do not satisfy the condition that the increase in interest is Rs 600. Common Pitfalls: A common mistake is to attempt to find the original rate R first, even though it is not needed. Another error is to use 4 as the total interest rather than as the change in rate. Some students also forget to multiply by time when computing the extra interest, using P * 4 / 100 instead of P * 4 * 3 / 100. Focusing on the structure of the formula and carefully substituting the change in rate over the full time period avoids these errors. Final Answer: The original principal or sum of money invested is Rs 5000.

Question 9

How much simple interest will an amount of Rs 4000 earn in a period of 18 months at the rate of 12% per annum, assuming that the interest is calculated only using the simple interest formula for the entire time and that 18 months is treated as 1.5 y…

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Introduction / Context: This question focuses on straightforward application of the simple interest formula, with an additional twist that the time period is given in months rather than years. The candidate needs to convert the time from months to years correctly and then substitute the values of principal, rate, and time into the simple interest formula. Correct handling of time units is very important in all interest problems, especially in competitive exams where small mistakes in conversion can lead to wrong answers. Given Data / Assumptions: Principal P = Rs 4000. Rate of simple interest R = 12% per annum. Time period given = 18 months. Simple interest is used for the entire period. Assume 12 months = 1 year, so 18 months = 1.5 years. Concept / Approach: The simple interest formula is: SI = (P * R * T) / 100 Here P is in rupees, R is in percent per annum, and T must be in years. Because the time is given in months, we must first convert 18 months to years by dividing by 12. Then we can substitute P = 4000, R = 12, and T = 1.5 into the formula. The calculation is a direct multiplication followed by division by 100. Step-by-Step Solution: Principal P = Rs 4000. Rate R = 12% per annum. Time = 18 months = 18 / 12 years = 1.5 years. Use the formula SI = (P * R * T) / 100. SI = (4000 * 12 * 1.5) / 100. First compute 12 * 1.5 = 18. So SI = (4000 * 18) / 100. 4000 * 18 = 72000. SI = 72000 / 100 = 720. Therefore, the simple interest earned in 18 months is Rs 720. Verification / Alternative check: We can also view the problem in two parts. For 1 year, simple interest on Rs 4000 at 12% is SI = (4000 * 12 * 1) / 100 = Rs 480. For 0.5 year (6 months), the interest will be half of the yearly interest, that is Rs 240. Total interest for 1.5 years = 480 + 240 = Rs 720. This matches the earlier calculation and confirms that the answer is correct. Why Other Options Are Wrong: Rs 216 and Rs 360 correspond to much smaller times or lower rates and do not match 18 months at 12% per annum on Rs 4000. Rs 960 would arise from higher time or rate than given, so it overestimates the interest. Rs 540 is also less than the correct value and does not satisfy the exact substitution into the simple interest formula for 1.5 years. Common Pitfalls: A frequent mistake is to treat 18 months as 18 years in the formula or to use 18 directly as T, which gives a huge interest value and is clearly inconsistent. Some students forget to convert months to years and instead convert the rate incorrectly. Another common error is mishandling the multiplication 4000 * 12 * 1.5. Writing every intermediate step clearly and performing a quick reasonableness check prevents such issues. Final Answer: The simple interest earned in 18 months at 12% per annum on Rs 4000 is Rs 720.

Question 10

Simple interest on a certain unknown sum for 3 years at 10% per annum is equal to half of the compound interest earned on Rs 6000 for 2 years at 10% per annum when interest is compounded annually. Using this relationship between simple interest and …

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Introduction / Context: This question combines both simple interest and compound interest concepts in a single scenario. The main idea is that the simple interest on an unknown principal over a given time equals half the compound interest on a known principal over a different time, both at the same rate of interest. Such mixed interest questions are important in competitive exams because they check a student’s ability to compute compound interest accurately and then relate it to simple interest through an algebraic equation. Given Data / Assumptions: Rate of interest for both cases, R = 10% per annum. Simple interest case: time T1 = 3 years, principal P1 is unknown. Compound interest case: time T2 = 2 years, principal P2 = Rs 6000. Interest in the compound interest case is compounded annually. Simple interest for 3 years on P1 equals half of the compound interest for 2 years on Rs 6000. Concept / Approach: For simple interest, we use: SI = (P1 * R * T1) / 100 For compound interest compounded annually, we use: Amount A2 = P2 * (1 + R / 100) ^ T2 Compound interest CI = A2 − P2 The relation given is: SI = (1 / 2) * CI First we compute the compound interest on Rs 6000 for 2 years at 10% and then take half of that value. That value is equal to the simple interest on P1 for 3 years, which can be expressed using the simple interest formula. Solving this equation gives P1. Step-by-Step Solution: P2 = Rs 6000, R = 10% per annum, T2 = 2 years. Amount after 2 years with compound interest: A2 = 6000 * (1 + 10 / 100) ^ 2. A2 = 6000 * (1.1) ^ 2 = 6000 * 1.21 = 7260. Compound interest CI = A2 − P2 = 7260 − 6000 = Rs 1260. Half of this compound interest = 1260 / 2 = Rs 630. Let P1 be the principal on simple interest for 3 years at 10%. Simple interest SI = (P1 * 10 * 3) / 100 = 0.3P1. Given that SI = 630, so 0.3P1 = 630. P1 = 630 / 0.3 = 2100. Therefore, the sum placed on simple interest is Rs 2100. Verification / Alternative check: Check the simple interest on Rs 2100 for 3 years at 10%. SI = (2100 * 10 * 3) / 100 = 630. This matches half of the compound interest computed earlier, which was 1260. Since both conditions are satisfied, P1 = Rs 2100 is consistent with the relationship stated in the question. Why Other Options Are Wrong: If P1 were Rs 4200, then simple interest for 3 years at 10% would be SI = 1260, which equals the full compound interest, not half of it. If P1 were Rs 1680, then SI = (1680 * 10 * 3) / 100 = 504, which is less than 630. If P1 were Rs 1050, then SI = 315, also not equal to 630. If P1 were Rs 3000, then SI = 900, too large compared with 630. Hence only Rs 2100 satisfies the condition that simple interest equals half of the compound interest. Common Pitfalls: A typical mistake is to confuse simple interest and compound interest formulas or to treat compound interest as simple interest over 2 years. Some learners forget to subtract the principal when finding compound interest and accidentally use the amount directly. Another error is misinterpreting “half of the compound interest” as “compound interest at half the rate”, which is incorrect. Using the standard formulas carefully and reading the condition precisely avoids these misunderstandings. Final Answer: The sum of money that was placed on simple interest is Rs 2100.

Question 11

Each question contains Quantity I and Quantity II. Read carefully and compare the two values. Quantity I: The simple interest on a certain sum of money for 3 years at 5% per annum is Rs 4800, so find the principal in this simple interest case. Quant…

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Introduction / Context: This is a comparison based aptitude question where the candidate must compute two separate principal amounts using different interest models and then compare the results. Quantity I uses simple interest over 3 years, while Quantity II uses compound interest over 2 years. Such problems are common in banking and insurance exams, where candidates must quickly evaluate which of the two quantities is greater without necessarily focusing on the absolute amounts beyond what is needed for comparison. Given Data / Assumptions: Quantity I: Simple interest SI1 = Rs 4800. Simple interest rate for Quantity I, R1 = 5% per annum. Time for Quantity I, T1 = 3 years. Quantity II: Compound interest CI2 = Rs 3708. Compound interest rate for Quantity II, R2 = 6% per annum. Time for Quantity II, T2 = 2 years with annual compounding. Principals in the two cases may be different and must be found separately. Concept / Approach: For Quantity I with simple interest, the formula is: SI1 = (P1 * R1 * T1) / 100 We can rearrange this to find P1. For Quantity II with compound interest, we use: Amount A2 = P2 * (1 + R2 / 100) ^ T2 Compound interest CI2 = A2 − P2 By expressing CI2 in terms of P2 and equating to Rs 3708, we can solve for P2. Once P1 and P2 are obtained, they are compared to determine the correct option. Step-by-Step Solution: For Quantity I (simple interest): SI1 = Rs 4800, R1 = 5% per annum, T1 = 3 years. 4800 = (P1 * 5 * 3) / 100. 4800 = (15P1) / 100. 4800 = 0.15P1. P1 = 4800 / 0.15 = 32000. So, Quantity I principal = Rs 32000. For Quantity II (compound interest): CI2 = Rs 3708, R2 = 6% per annum, T2 = 2 years. Let the principal be P2. Amount A2 = P2 * (1 + 6 / 100) ^ 2 = P2 * (1.06) ^ 2. (1.06) ^ 2 = 1.1236, so A2 = 1.1236P2. Compound interest CI2 = A2 − P2 = 1.1236P2 − P2 = 0.1236P2. Given CI2 = 3708, so 0.1236P2 = 3708. P2 = 3708 / 0.1236 = 30000. Thus Quantity II principal = Rs 30000. Verification / Alternative check: We can quickly recheck both values. For Quantity I, 5% of 32000 is 1600 per year, and for 3 years the interest is 1600 * 3 = 4800, which matches. For Quantity II, 6% of 30000 is 1800 for the first year, so amount becomes 31800. Six percent of 31800 is 1908 for the second year, making the total amount 33708. Compound interest is 33708 − 30000 = 3708, which agrees with the given data. Thus both principal values are correct. Why Other Options Are Wrong: Since P1 = 32000 and P2 = 30000, clearly Quantity I is greater than Quantity II. Therefore, options suggesting equality or the reverse inequality, such as Quantity I ≤ Quantity II, Quantity I < Quantity II, or Quantity I = Quantity II, are incorrect. Quantity I ≥ Quantity II is technically correct but less precise than the strict inequality, so the best answer is Quantity I > Quantity II. Common Pitfalls: Some students may miscalculate (1.06) ^ 2 or may take compound interest simply as 2 times 6% of the principal, which would be wrong. Others might mix up the interest and amount or forget to subtract the principal when calculating compound interest. Careful arithmetic and a clear distinction between amount and interest are essential. Also, one must remember that the simplest correct comparison is Quantity I > Quantity II, not just a weak inequality. Final Answer: Since the principal in Quantity I is Rs 32000 and the principal in Quantity II is Rs 30000, the correct relationship is Quantity I > Quantity II.

Question 12

A sum of Rs 8000 is divided into two parts. The simple interest on the first part at the rate of 21% per annum for one year is equal to the simple interest on the second part at the rate of 35% per annum for the same one year. Using this condition a…

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Introduction / Context: This question involves dividing a total sum into two parts such that the simple interests on the two parts, at different rates, are equal over the same time period. These types of problems test a candidate’s skill in translating a verbal condition about equal interests into an equation involving the two parts of the principal. Once the relation is established, the problem reduces to solving a simple linear equation and then computing the interest on either part using the standard simple interest formula. Given Data / Assumptions: Total sum of money = Rs 8000. First part is invested at 21% per annum simple interest for 1 year. Second part is invested at 35% per annum simple interest for 1 year. Simple interest on the first part equals simple interest on the second part. Time period is one year for both parts. Concept / Approach: For simple interest, we use: SI = (P * R * T) / 100 Let the first part be x rupees, then the second part is 8000 − x rupees. The interests on both parts for one year can be computed in terms of x. The condition that these interests are equal gives an equation that can be solved for x. After finding x, we can compute the simple interest on either part, which will be the same value and is what the question asks for. Step-by-Step Solution: Let the first part be x rupees. Then the second part is 8000 − x rupees. Interest on the first part for 1 year at 21%: SI1 = (x * 21 * 1) / 100 = 0.21x. Interest on the second part for 1 year at 35%: SI2 = ((8000 − x) * 35 * 1) / 100 = 0.35 * (8000 − x). Given SI1 = SI2, so 0.21x = 0.35 * (8000 − x). 0.21x = 0.35 * 8000 − 0.35x = 2800 − 0.35x. Bring like terms together: 0.21x + 0.35x = 2800. 0.56x = 2800. x = 2800 / 0.56 = 5000. So the first part is Rs 5000 and the second part is 8000 − 5000 = Rs 3000. Simple interest on the first part: SI1 = 0.21 * 5000 = Rs 1050. Simple interest on the second part: SI2 = 0.35 * 3000 = Rs 1050. Therefore, the interest earned on each part is Rs 1050. Verification / Alternative check: We can verify the arithmetic quickly. For Rs 5000 at 21% for 1 year, 21% of 5000 is 0.21 * 5000 = 1050. For Rs 3000 at 35% for 1 year, 35% of 3000 is 0.35 * 3000 = 1050. Since both interests are equal and match the condition, and the total of the two parts is still Rs 8000, our solution is consistent and correct. Why Other Options Are Wrong: Rs 840, Rs 1400, Rs 1220, and Rs 700 do not match the equal interest condition when we check with correct principal values. If the interest were Rs 840, the parts would not sum to Rs 8000 under the given rates. Similar inconsistencies arise for the other incorrect options. Only Rs 1050 satisfies both the simple interest formula for each part and the equality condition. Common Pitfalls: One common mistake is to assume the two parts are equal (each Rs 4000), which does not satisfy the equal interest condition at different rates. Another error is misapplying the simple interest formula without dividing by 100 or mishandling decimals. Some learners also incorrectly treat the rates as if they apply to the same principal. Writing the equation systematically and double checking each algebraic step prevents such errors. Final Answer: The simple interest earned on each part of the investment for that year is Rs 1050.

Question 13

The simple interest earned on a principal of Rs 1600 at the rate of 5% per annum for a period of 6 years is to be calculated. Using the standard simple interest formula, what will be the amount of simple interest earned over these 6 years?

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Introduction / Context: This is a direct computation problem on simple interest. The principal, rate, and time are all given explicitly, and the task is simply to apply the simple interest formula correctly. Such problems are common in the early part of quantitative aptitude sections and serve to check the candidate’s accuracy with basic percentage calculations and multiplication. Even though the calculation is simple, exam pressure can cause mistakes, so a clear step-by-step approach is recommended. Given Data / Assumptions: Principal P = Rs 1600. Rate of interest R = 5% per annum. Time period T = 6 years. Interest is calculated using simple interest throughout. Concept / Approach: The formula for simple interest is: SI = (P * R * T) / 100 Here P is the amount of money invested or borrowed, R is the annual rate of interest in percent, and T is the time in years for which the money is invested or borrowed. After computing the simple interest, one can also find the final amount if needed by adding the simple interest to the principal, but the question here asks only for the interest amount. Step-by-Step Solution: Principal P = Rs 1600. Rate R = 5% per annum. Time T = 6 years. Use the formula SI = (P * R * T) / 100. SI = (1600 * 5 * 6) / 100. First compute 5 * 6 = 30. So SI = (1600 * 30) / 100. 1600 * 30 = 48000. SI = 48000 / 100 = 480. Therefore, the simple interest earned over 6 years is Rs 480. Verification / Alternative check: We can check the result by reasoning yearly. At 5% per annum, yearly interest on Rs 1600 is 5% of 1600, which is (1600 * 5) / 100 = 80. Over 6 years, simple interest equals 80 * 6 = 480. This matches the earlier calculation, confirming that Rs 480 is correct. Why Other Options Are Wrong: Rs 2080 is much larger than the principal and would require a higher rate or longer time period. Rs 450, Rs 496, and Rs 520 are all different from the exact calculated value of Rs 480 and do not correspond to correct application of the formula with the given inputs. Common Pitfalls: Common mistakes include forgetting to divide by 100 after multiplying P, R, and T, or miscalculating the product 1600 * 30. Some candidates might also incorrectly use compound interest logic for a simple interest question. Carefully writing down the formula and substituting the values step by step helps to avoid these errors and ensures accuracy, even under exam conditions. Final Answer: The amount of simple interest earned over 6 years is Rs 480.

Question 14

At a simple interest rate of 9.5% per annum, a sum of money grows to Rs 942 in 6 years. Using the simple interest relationship between principal, rate, time, and amount, find the initial principal amount that was invested so that it becomes Rs 942 a…

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Introduction / Context: This problem requires working backwards from the final amount to determine the principal under simple interest. The amount after a given time at a stated rate is known, and the principal is unknown. Such reverse simple interest problems are very common in quantitative aptitude, where candidates must be comfortable rearranging the simple interest formulas and performing accurate calculations with decimal interest rates like 9.5% per annum. Given Data / Assumptions: Final amount after 6 years, A = Rs 942. Rate of simple interest, R = 9.5% per annum. Time period, T = 6 years. Principal P is unknown. Interest is calculated using simple interest only. Concept / Approach: For simple interest, the relationship between amount and principal is: Amount A = Principal P + Simple interest SI and SI = (P * R * T) / 100 Combining, we get: A = P + (P * R * T) / 100 = P * (1 + (R * T) / 100) Here, A, R, and T are known, so we can rearrange this equation to solve for P. The factor multiplying P is computed first, and then P is found by dividing A by that factor. Step-by-Step Solution: Given A = Rs 942, R = 9.5% per annum, T = 6 years. Compute the effective multiplier: 1 + (R * T) / 100. R * T = 9.5 * 6 = 57. (R * T) / 100 = 57 / 100 = 0.57. So the amount factor is 1 + 0.57 = 1.57. Thus A = P * 1.57. Given A = 942, so 942 = 1.57P. P = 942 / 1.57. P = 600. Therefore, the initial principal invested is Rs 600. Verification / Alternative check: We can verify by computing the simple interest on Rs 600 at 9.5% per annum for 6 years. Simple interest SI = (600 * 9.5 * 6) / 100. First compute 9.5 * 6 = 57, so SI = (600 * 57) / 100 = 34200 / 100 = 342. The final amount is P + SI = 600 + 342 = 942, which matches the given amount. This confirms that P = Rs 600 is correct. Why Other Options Are Wrong: If P were Rs 616, multiplying by 1.57 gives approximately 967.12, not 942. If P were Rs 626, 626 * 1.57 is about 982.82, again not 942. If P were Rs 636, 636 * 1.57 is greater than 942, and if P were Rs 650, 650 * 1.57 = 1020.5, which is also incorrect. Only Rs 600 gives the exact final amount of Rs 942 when the simple interest is applied. Common Pitfalls: Mistakes often occur when multiplying a decimal rate by time or when handling the decimal factor for the amount. Some students forget to convert 9.5% properly or try to use compound interest formulas instead of simple interest. Others may compute A = P + SI but then not factor P correctly. Writing the combined formula for A directly in terms of P and carefully computing the factor 1.57 helps to avoid such errors. Final Answer: The initial principal amount that was invested so that it becomes Rs 942 in 6 years is Rs 600.

Question 15

Simple interest on a certain unknown sum of money for 3 years at 8% per annum is equal to half of the compound interest on Rs 1200 for 2 years at 10% per annum, assuming the compound interest is calculated annually. Using this relationship, find the…

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Introduction / Context: This question is another mixed-interest problem that links simple interest to compound interest. The simple interest on an unknown principal is given to be half of the compound interest on a known principal. Students must compute the compound interest on Rs 1200 for 2 years at 10% per annum, take half of that value, and then relate it to the simple interest formula over 3 years at 8% per annum to find the unknown principal. Such questions are excellent practice for combining different interest concepts in a single problem. Given Data / Assumptions: Rate for simple interest case = 8% per annum. Time for simple interest case = 3 years. Principal for simple interest case = P rupees (unknown). Rate for compound interest case = 10% per annum. Time for compound interest case = 2 years, compounded annually. Principal for compound interest case = Rs 1200. Simple interest for 3 years equals half of the compound interest for 2 years on Rs 1200. Concept / Approach: For compound interest with annual compounding: Amount A = P * (1 + R / 100) ^ T Compound interest CI = A − P For simple interest: SI = (P * R * T) / 100 The problem gives: SI (on unknown principal) = 0.5 * CI (on Rs 1200) So we first compute CI on Rs 1200 at 10% for 2 years, then take half, and finally set that equal to the simple interest expression for P at 8% for 3 years. Solving this equation yields P. Step-by-Step Solution: Compute compound interest on Rs 1200 at 10% for 2 years. Amount A = 1200 * (1 + 10 / 100) ^ 2 = 1200 * (1.1) ^ 2 = 1200 * 1.21. A = 1452. Compound interest CI = A − Principal = 1452 − 1200 = Rs 252. Half of this compound interest = 252 / 2 = Rs 126. Now consider simple interest case: Rate = 8% per annum, time = 3 years, principal P rupees. Simple interest SI = (P * 8 * 3) / 100 = 0.24P. Given that SI = 126, so 0.24P = 126. P = 126 / 0.24 = 525. Therefore, the principal placed on simple interest is Rs 525. Verification / Alternative check: Check the simple interest on Rs 525 at 8% for 3 years. SI = (525 * 8 * 3) / 100. First compute 8 * 3 = 24, so SI = (525 * 24) / 100 = 12600 / 100 = Rs 126. This matches half of the compound interest on Rs 1200, which was 252. Hence, the relationship is satisfied and P = Rs 525 is correct. Why Other Options Are Wrong: If P were Rs 1050, simple interest would be 0.24 * 1050 = 252, which equals the full compound interest, not half of it. If P were Rs 260, SI would be 0.24 * 260 = 62.4, which is less than 126. If P were Rs 420, SI would be 100.8, still not equal to 126. If P were Rs 630, SI would be 151.2, larger than 126. Only Rs 525 satisfies the required equality condition. Common Pitfalls: Students sometimes confuse half of compound interest with compound interest at half the rate, which is incorrect. Another mistake is to miscalculate the compound amount by using simple interest instead of compounding, or by forgetting to subtract the principal when finding CI. Rounding errors may appear if intermediate steps are approximated. Using exact values and carefully following the formula helps to avoid these common issues. Final Answer: The principal that was placed on simple interest is Rs 525.

Question 16

The difference between the simple interest obtained by investing an amount X rupees at 8% per annum for one year and the simple interest obtained by investing X plus 1400 rupees at 8% per annum for two years is 240 rupees. Using the simple interest …

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Introduction / Context: This problem compares simple interest earned on two different principals over different time periods, both at the same rate of interest. The interest on a smaller amount X for one year is compared with interest on a larger amount (X + 1400) for two years. The difference between these two simple interests is given as Rs 240. The task is to translate this relationship into an algebraic equation using the simple interest formula and then solve for X. Such questions build comfort with forming equations involving unknowns from verbal descriptions. Given Data / Assumptions: Rate of interest for both cases, R = 8% per annum. Case 1: Principal = X rupees, time T1 = 1 year. Case 2: Principal = X + 1400 rupees, time T2 = 2 years. Difference between simple interest in Case 2 and Case 1 is Rs 240. Interest is calculated using simple interest only. Concept / Approach: For simple interest, the formula is: SI = (P * R * T) / 100 We compute the interest for each case in terms of X. Let SI1 be the interest in the first case and SI2 be the interest in the second case. The problem states that the difference between these two interests is Rs 240. Since the second investment has larger principal and time, SI2 is greater than SI1, so we set SI2 − SI1 = 240. Substituting the expressions for SI1 and SI2 leads to a linear equation in X that can be solved easily. Step-by-Step Solution: Rate R = 8% per annum. Case 1: Principal = X, time T1 = 1 year. SI1 = (X * 8 * 1) / 100 = 0.08X. Case 2: Principal = X + 1400, time T2 = 2 years. SI2 = ((X + 1400) * 8 * 2) / 100. Simplify: SI2 = (X + 1400) * 16 / 100 = 0.16(X + 1400). Given that SI2 − SI1 = 240. So 0.16(X + 1400) − 0.08X = 240. Expand: 0.16X + 0.16 * 1400 − 0.08X = 240. 0.16X − 0.08X + 224 = 240. 0.08X + 224 = 240. 0.08X = 240 − 224 = 16. X = 16 / 0.08 = 200. Therefore, the required value of X is Rs 200. Verification / Alternative check: Compute the two interests with X = 200. For Case 1, principal = 200, so SI1 = (200 * 8 * 1) / 100 = 16. For Case 2, principal = 200 + 1400 = 1600, time = 2 years, so SI2 = (1600 * 8 * 2) / 100 = (1600 * 16) / 100 = 256. The difference SI2 − SI1 = 256 − 16 = 240, which matches the given condition. This confirms that X = Rs 200 is correct. Why Other Options Are Wrong: If X were Rs 100, SI1 would be 8, and SI2 would be (1500 * 16) / 100 = 240, giving a difference of 232, not 240. If X were Rs 400, SI1 would be 32, and SI2 would be (1800 * 16) / 100 = 288, giving a difference of 256. If X were Rs 300, or Rs 250, similar calculations show that the difference between SI2 and SI1 does not equal 240. Only X = Rs 200 satisfies the exact relationship. Common Pitfalls: A common error is to set up the equation as SI1 − SI2 = 240 instead of SI2 − SI1 = 240, which can lead to negative values and confusion. Another mistake is to forget that the time in the second case is 2 years and use T = 1 year for both cases. Some students also mishandle the decimal 0.08 and 0.16 calculations. Careful substitution and stepwise simplification reduce the risk of such errors. Final Answer: The value of the original amount X is Rs 200.

Question 17

The difference between the compound interest and the simple interest on the same principal at the rate of 12% per annum for 3 years is Rs 112.32, when interest is compounded annually. Using this information about the excess of compound interest over…

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Introduction / Context: This question compares compound interest with simple interest on the same principal, time, and rate. For a given principal, compound interest over multiple years is always higher than simple interest because interest is earned on both principal and accumulated interest. The problem tells us that this excess of compound interest over simple interest is Rs 112.32 when the rate is 12% per annum for 3 years. The candidate must use the standard formulas for simple and compound interest and set up an equation to determine the principal. This is a typical higher level interest problem in many aptitude exams. Given Data / Assumptions: Rate of interest R = 12% per annum. Time period T = 3 years. Principal amount P rupees (unknown). Interest is compounded annually in the compound interest case. Difference (CI − SI) over 3 years = Rs 112.32. Concept / Approach: Simple interest for 3 years is: SI = (P * R * T) / 100 = (P * 12 * 3) / 100 = 0.36P For compound interest compounded annually, amount is: Amount A = P * (1 + R / 100) ^ T With R = 12% and T = 3 years: A = P * (1.12) ^ 3 Then compound interest is: CI = A − P = P * ((1.12) ^ 3 − 1) The problem gives: CI − SI = P * ((1.12) ^ 3 − 1) − 0.36P = 112.32 Factoring P allows us to compute P from this equation once we evaluate (1.12) ^ 3 accurately. Step-by-Step Solution: Compute (1.12) ^ 3. (1.12) ^ 2 = 1.2544. (1.12) ^ 3 = 1.2544 * 1.12 = 1.404928. Thus CI = P * (1.404928 − 1) = P * 0.404928. Simple interest for 3 years at 12%: SI = 0.36P. Difference CI − SI = 0.404928P − 0.36P = 0.044928P. Given CI − SI = Rs 112.32. So 0.044928P = 112.32. P = 112.32 / 0.044928. P = 2500. Therefore, the principal is Rs 2500. Verification / Alternative check: Check simple interest on Rs 2500 for 3 years at 12%. SI = (2500 * 12 * 3) / 100 = 900. For compound interest, amount A = 2500 * (1.12) ^ 3 = 2500 * 1.404928 = 3512.32. Compound interest CI = A − P = 3512.32 − 2500 = 1012.32. The difference CI − SI = 1012.32 − 900 = 112.32, which matches the given value. This confirms that P = Rs 2500 is correct. Why Other Options Are Wrong: If P were Rs 25000 or Rs 50000, the difference CI − SI would be ten or twenty times larger than 112.32, which is clearly inconsistent. If P were Rs 5000 or Rs 3000, recalculating CI and SI would give differences that do not equal 112.32. Only P = Rs 2500 produces exactly the given difference between compound interest and simple interest over 3 years at 12% per annum. Common Pitfalls: Many students miscalculate the power (1.12) ^ 3 or mistakenly use 3 * 12% for compound interest as if it were simple interest. Another error is to subtract incorrectly when finding CI − SI, or to round intermediate steps too aggressively, losing precision. Keeping a few decimal places until the end and systematically applying the formulas helps prevent these mistakes. It is also important to remember that the difference increases with time and rate, so unrealistic principal guesses can be ruled out by estimation. Final Answer: The principal amount for which the difference between compound interest and simple interest is Rs 112.32 is Rs 2500.

Question 18

What principal amount will yield simple interest of Rs 120 at a rate of 6% per annum over a period of 10 years, assuming that interest is calculated using the standard simple interest formula for the entire duration?

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Introduction / Context: This is a simple reverse calculation question on simple interest. The simple interest earned, the rate of interest, and the time period are given. The task is to find the principal that would generate this interest. Such problems are very common in the early stages of quantitative aptitude practice and build familiarity with rearranging the simple interest formula to make the principal the subject of the formula. Given Data / Assumptions: Simple interest SI = Rs 120. Rate of interest R = 6% per annum. Time period T = 10 years. Principal P is unknown. Interest is calculated using simple interest for the full duration. Concept / Approach: The simple interest formula is: SI = (P * R * T) / 100 To find the principal P, we rearrange the formula as: P = (SI * 100) / (R * T) By substituting SI = 120, R = 6, and T = 10, we can compute the value of P directly through straightforward arithmetic. It is useful to simplify R * T first, then divide SI * 100 by this product. Step-by-Step Solution: Given SI = Rs 120, R = 6% per annum, T = 10 years. Use the rearranged formula: P = (SI * 100) / (R * T). Compute R * T = 6 * 10 = 60. Compute SI * 100 = 120 * 100 = 12000. Now divide: P = 12000 / 60. P = 200. Therefore, the required principal amount is Rs 200. Verification / Alternative check: We can verify by calculating the simple interest on Rs 200 at 6% per annum for 10 years. SI = (200 * 6 * 10) / 100. Compute 6 * 10 = 60, so SI = (200 * 60) / 100 = 12000 / 100 = 120. This matches the given interest, confirming that Rs 200 is the correct principal. Why Other Options Are Wrong: If P were Rs 120, simple interest would be (120 * 6 * 10) / 100 = 72, which is less than 120. If P were Rs 180, SI would be 108, still not 120. If P were Rs 210, SI would be 126, which is more than 120. If P were Rs 240, SI would be 144. Therefore these options do not match the given interest at the specified rate and time. Common Pitfalls: A common error is to forget to divide by 100 in the simple interest formula or to miscompute R * T. Some students invert the formula incorrectly and multiply by R * T instead of dividing by it. Others may confuse amount and interest, treating Rs 120 as the final amount instead of the interest. Writing out the formula clearly and substituting carefully helps to avoid these mistakes. Final Answer: The principal amount that will yield simple interest of Rs 120 at 6% per annum for 10 years is Rs 200.

Question 19

A person repays Rs 22500 after 10 years of borrowing a loan at 10% per annum simple interest. Using the relationship between amount, principal, rate, and time under simple interest, find the original loan amount that was borrowed.

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Introduction / Context: This problem asks us to find the original loan amount when the final amount repaid after a certain time at simple interest is known. The loan runs for 10 years at 10% per annum simple interest, and the borrower repays Rs 22500 in total. The candidate must express the amount in terms of the principal and simple interest, relate the interest to the principal using the simple interest formula, and then solve for the principal. This kind of question is standard in loan and banking based aptitude topics. Given Data / Assumptions: Final amount repaid A = Rs 22500. Rate of simple interest R = 10% per annum. Time period T = 10 years. Original loan amount or principal P is unknown. Interest is calculated using simple interest throughout the 10 years. Concept / Approach: For simple interest, the interest earned or paid is: SI = (P * R * T) / 100 The final amount A is: A = P + SI Substituting the expression for SI, we have: A = P + (P * R * T) / 100 = P * (1 + (R * T) / 100) Since A, R, and T are known, we can compute the factor multiplying P and then solve for P by dividing the amount by this factor. Step-by-Step Solution: Given A = Rs 22500, R = 10% per annum, T = 10 years. Compute the multiplier: 1 + (R * T) / 100. R * T = 10 * 10 = 100. (R * T) / 100 = 100 / 100 = 1. So the multiplier is 1 + 1 = 2. Therefore, A = P * 2. 22500 = 2P. P = 22500 / 2. P = 11250. Thus, the original loan amount borrowed was Rs 11250. Verification / Alternative check: We can verify by calculating the simple interest on Rs 11250 at 10% per annum for 10 years. SI = (11250 * 10 * 10) / 100. First compute 10 * 10 = 100, so SI = (11250 * 100) / 100 = 11250. The final amount A = P + SI = 11250 + 11250 = 22500, which matches the given amount repaid. This confirms that the original principal is Rs 11250. Why Other Options Are Wrong: If P were Rs 11225, the amount at 10% for 10 years would be 11225 * 2 = 22450, which is less than 22500. If P were Rs 10000, the final amount would be 20000, too small. If P were Rs 7500, the amount would be 15000, much smaller than 22500. If P were Rs 12000, the amount would be 24000, which is more than 22500. Only Rs 11250 produces the correct final repayment amount under simple interest at 10% for 10 years. Common Pitfalls: Students sometimes confuse amount with interest and might try to treat Rs 22500 as interest instead of the total amount. Another error is to miscalculate the multiplier 1 + (R * T) / 100, especially for large time periods. Some candidates also forget that with 10% for 10 years, the simple interest equals the principal, doubling the amount. Recognizing that pattern can speed up calculations and reduce mistakes. Final Answer: The original loan amount borrowed by the person is Rs 11,250.

Question 20

A sum of money invested at simple interest triples itself in 8 years, meaning that the final amount after 8 years is three times the original principal. At the same rate of simple interest, how many times of the original principal will the amount be…

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Introduction / Context: This question tests conceptual understanding of simple interest and how the amount grows linearly with time. If a sum of money triples in 8 years under simple interest, we can use this information to find the rate of interest and then predict the amount after any other time, such as 20 years. The key property of simple interest is that interest increases in direct proportion to time. This allows us to use ratios or to compute the rate explicitly and apply it to a different time period. Given Data / Assumptions: The sum triples in 8 years under simple interest. Final amount after 8 years = 3P, where P is the principal. Time for first scenario T1 = 8 years. Same rate of simple interest is applied for 20 years in the second scenario. We must find the multiple of P that the amount becomes after 20 years. Concept / Approach: Under simple interest, the amount after time T is: A = P * (1 + (R * T) / 100) If the amount after 8 years is 3P, we have: 3P = P * (1 + (R * 8) / 100) Dividing both sides by P, we get: 3 = 1 + (8R / 100) This allows us to find R, the rate of interest per annum. Then, using the same formula for T = 20 years, we can determine how many times the principal the amount becomes by computing 1 + (R * 20) / 100 and interpreting the result as a multiple of P. Step-by-Step Solution: From the given condition for 8 years: 3P = P * (1 + 8R / 100). Divide both sides by P: 3 = 1 + 8R / 100. 8R / 100 = 3 − 1 = 2. 8R = 200. R = 200 / 8 = 25% per annum. Now consider 20 years at this rate. Amount after 20 years: A20 = P * (1 + (R * 20) / 100). Substitute R = 25: A20 = P * (1 + (25 * 20) / 100). 25 * 20 = 500, so (R * T) / 100 = 500 / 100 = 5. So A20 = P * (1 + 5) = P * 6. Therefore, after 20 years the amount is 6 times the original principal. Verification / Alternative check: We can also reason directly in terms of interest. In 8 years, interest earned is 3P − P = 2P. Thus, in 8 years, simple interest is 2P, so in 1 year, interest is (2P / 8) = P / 4. In 20 years, total interest is (P / 4) * 20 = 5P. Therefore, the amount after 20 years is P + 5P = 6P, which matches the result obtained using the formula approach. Why Other Options Are Wrong: 8 times, 7 times, 9 times, and 5 times correspond to interest totals of 7P, 6P, 8P, and 4P respectively, which do not align with the rate deduced from the initial 8 year tripling condition. Only 6 times is consistent with the linear accumulation of simple interest at 25% per annum over 20 years. Common Pitfalls: Some students confuse simple interest with compound interest and try to raise factors to powers rather than using a linear relationship. Another common error is to misinterpret “triples in 8 years” as meaning the interest is 3P rather than the amount being 3P. Mismanaging ratios or incorrectly solving for R can also lead to wrong results. Focusing on the formula and carefully isolating R first ensures a correct and systematic solution. Final Answer: At the same rate of simple interest, the amount will become 6 times the original principal in 20 years.

Simple Interest Questions