Question 1

A person invests ₹12,000 at 7.5% simple interest per annum. How much total amount will he receive after 2.5 years?

Accepted Answer

Step 1: Calculate Simple Interest using SI = (P × R × T) / 100. Step 2: SI = (12000 × 7.5 × 2.5) / 100 = (12000 × 18.75) / 100 = 225000 / 100 = ₹2,250. Step 3: Total Amount = Principal + SI = 12000 + 2250 = ₹14,250.

Question 2

A sum of ₹5,000 is invested at a simple interest rate of 8% per annum. What will be the interest earned after 3 years?

Accepted Answer

Step 1: Use the Simple Interest formula: SI = (P × R × T) / 100. Step 2: Substitute values: SI = (5000 × 8 × 3) / 100 = 120000 / 100 = ₹1,200. Therefore, the interest earned is ₹1,200.

Question 3

At what rate of simple interest per annum will ₹2,400 amount to ₹3,120 in 5 years?

Accepted Answer

Step 1: Calculate Simple Interest: SI = Amount - Principal = 3120 - 2400 = ₹720. Step 2: Use SI = (P × R × T) / 100 to find R. Step 3: 720 = (2400 × R × 5) / 100. Step 4: 720 = 120R, so R = 720 / 120 = 6% per annum.

Question 4

A principal of ₹8,000 earns ₹1,600 as simple interest in 4 years. What is the rate of interest per annum?

Accepted Answer

Step 1: Use the formula R = (SI × 100) / (P × T). Step 2: Substitute: R = (1600 × 100) / (8000 × 4) = 160000 / 32000 = 5% per annum.

Question 5

In how many years will ₹6,000 become ₹7,800 at a simple interest rate of 6% per annum?

Accepted Answer

Step 1: Calculate Simple Interest: SI = 7800 - 6000 = ₹1,800. Step 2: Use T = (SI × 100) / (P × R). Step 3: T = (1800 × 100) / (6000 × 6) = 180000 / 36000 = 5 years.

Question 6

Priya lent ₹20,000 to Raj at 5% per annum simple interest for 3 years. After 3 years, Raj returned the amount with interest. Priya then invested this entire amount at 8% per annum simple interest for 2 years. What is the total interest earned by Priya over the entire period?

Accepted Answer

Step 1: SI earned in first 3 years = (20,000 × 5 × 3) / 100 = ₹3,000. Step 2: Amount after 3 years = 20,000 + 3,000 = ₹23,000. Step 3: This ₹23,000 is invested at 8% for 2 years. Step 4: SI earned in next 2 years = (23,000 × 8 × 2) / 100 = ₹3,680. Step 5: Total interest earned = 3,000 + 3,680 = ₹6,680.

Question 7

A sum of money becomes ₹8,400 after 5 years at a simple interest rate of 8% per annum. What was the principal amount?

Accepted Answer

Step 1: Use SI formula: SI = (P × R × T) / 100. Step 2: Amount = P + SI, so 8400 = P + (P × 8 × 5) / 100. Step 3: 8400 = P + 0.4P = 1.4P. Step 4: P = 8400 / 1.4 = ₹6,000.

Question 8

Rahul borrowed ₹12,000 at 9% per annum simple interest. After how many years will the total amount (principal + interest) become ₹16,320?

Accepted Answer

Step 1: Amount = P + SI, so 16,320 = 12,000 + SI. Step 2: SI = 16,320 - 12,000 = ₹4,320. Step 3: Use SI = (P × R × T) / 100: 4,320 = (12,000 × 9 × T) / 100. Step 4: 4,320 = 1,080T. Step 5: T = 4,320 / 1,080 = 4 years.

Question 9

A person invests ₹15,000 at 10% per annum simple interest. How much additional principal must be invested at 12% per annum to make the average rate of interest 11% per annum?

Accepted Answer

Step 1: Let additional principal = y. Step 2: Total SI from both = (15,000 × 10 × 1)/100 + (y × 12 × 1)/100 = 1,500 + 0.12y. Step 3: For average rate 11%: [1,500 + 0.12y] / (15,000 + y) = 0.11. Step 4: 1,500 + 0.12y = 0.11(15,000 + y) = 1,650 + 0.11y. Step 5: 0.12y - 0.11y = 1,650 - 1,500. Step 6: 0.01y = 150. Step 7: y = ₹15,000.

Question 10

A sum of ₹8,000 is divided into two parts. One part is invested at 6% per annum and the other at 9% per annum simple interest. If the total interest earned in 2 years is ₹1,200, what is the amount invested at 9% per annum?

Accepted Answer

Step 1: Let amount at 6% = x, amount at 9% = (8,000 - x). Step 2: SI from 6% part = (x × 6 × 2) / 100 = 12x/100. Step 3: SI from 9% part = [(8,000 - x) × 9 × 2] / 100 = 18(8,000 - x)/100. Step 4: Total SI = [12x + 18(8,000 - x)] / 100 = 1,200. Step 5: 12x + 144,000 - 18x = 120,000. Step 6: -6x = -24,000. Step 7: x = 4,000. Step 8: Amount at 9% = 8,000 - 4,000 = ₹4,000.

Question 11

A sum of money becomes ₹5,400 in 3 years and ₹6,480 in 5 years at simple interest. What is the principal amount?

Accepted Answer

Step 1: Let P be principal and R be rate. Using SI formula: A = P + SI, we have 5,400 = P + (P × R × 3)/100 and 6,480 = P + (P × R × 5)/100. Step 2: Subtract the first equation from the second: 6,480 - 5,400 = (P × R × 5)/100 - (P × R × 3)/100, so 1,080 = (P × R × 2)/100, which gives P × R = 54,000. Step 3: From first equation: 5,400 = P + (54,000 × 3)/100 = P + 1,620, so P = 3,780.

Question 12

Rajesh borrowed ₹8,000 at a simple interest rate of 12% per annum. After some time, he repaid ₹10,240. If he had borrowed the same amount at 15% per annum for the same period, how much more interest would he have paid?

Accepted Answer

Step 1: Find the time period using SI formula. SI = A - P = 10,240 - 8,000 = ₹2,240. Using SI = (P × R × T)/100, we get 2,240 = (8,000 × 12 × T)/100, so T = 2,240 × 100 / (8,000 × 12) = 2.33 years (or 28/12 years = 7/3 years). Step 2: Calculate SI at 15% for the same period: SI = (8,000 × 15 × 7/3)/100 = (8,000 × 15 × 7)/(100 × 3) = ₹2,800. Step 3: Additional interest = 2,800 - 2,240 = ₹560.

Question 13

A person invests ₹15,000 at 10% per annum simple interest. Every year, he withdraws ₹1,500 (principal + interest combined). After how many complete years will his investment reduce to zero or below?

Accepted Answer

Step 1: Year 1: SI = (15,000 × 10 × 1)/100 = ₹1,500. Amount = 15,000 + 1,500 = ₹16,500. After withdrawal: 16,500 - 1,500 = ₹15,000. Step 2: Year 2: SI = (15,000 × 10 × 1)/100 = ₹1,500. Amount = 15,000 + 1,500 = ₹16,500. After withdrawal: 16,500 - 1,500 = ₹15,000. Step 3: The principal remains constant at ₹15,000 because SI earned equals the withdrawal amount. This continues indefinitely. However, the question asks when investment reduces to zero or below. Since it never does under these conditions, we need to reconsider: the withdrawal is fixed at ₹1,500, but SI is calculated on the remaining balance. Year 1: SI = 1,500, balance after withdrawal = 15,000. Year 2: SI = 1,500, balance = 15,000. This pattern continues forever. But if the question implies the withdrawal is from the total (principal + interest), then the principal never decreases. Re-reading: the investment will never reduce to zero under simple interest with these parameters. However, if this is a trick question testing understanding, the answer is that it never reduces to zero. But for SSC purposes, let's assume the question means: when does the balance first become less than the withdrawal amount? This never happens. The most reasonable interpretation for an SSC question: the investment never reduces to zero, but if forced to choose, after infinite years. For practical purposes, this question may have an error, but the intended answer is likely that the balance stabilizes and never reaches zero.

Question 14

Arun borrowed ₹20,000 at 9% per annum simple interest from a bank. He repaid ₹8,000 after 1 year and ₹8,000 after 2 years. How much does he still owe at the end of 3 years?

Accepted Answer

Step 1: Calculate total SI for 3 years on ₹20,000: SI = (20,000 × 9 × 3)/100 = ₹5,400. Total amount due = 20,000 + 5,400 = ₹25,400. Step 2: Total repaid = 8,000 + 8,000 = ₹16,000. Step 3: Amount still owed = 25,400 - 16,000 = ₹9,400.

Question 15

Two sums of money are in the ratio 4:5. They are lent at simple interest rates of 5% and 4% per annum respectively. After 4 years, the difference in the interests earned is ₹160. What is the larger sum?

Accepted Answer

Step 1: Let the two sums be 4x and 5x. Step 2: SI on first sum = (4x × 5 × 4)/100 = 0.8x. SI on second sum = (5x × 4 × 4)/100 = 0.8x. Step 3: Difference in SI = 0.8x - 0.8x = 0. This doesn't match ₹160. Let me recalculate: SI₁ = (4x × 5 × 4)/100 = 80x/100 = 0.8x. SI₂ = (5x × 4 × 4)/100 = 80x/100 = 0.8x. The interests are equal, so the difference is 0, which contradicts the problem. Let me reinterpret: perhaps the rates are different. Assuming SI₁ = (4x × 5 × 4)/100 = 0.8x and SI₂ = (5x × 4 × 4)/100 = 0.8x. Since both equal 0.8x, the difference is 0. However, if the problem states the difference is 160, there's an inconsistency. Let me assume the first sum earns more: |0.8x - 0.8x| = 0, which is impossible. Re-reading the problem: if the difference in interests is ₹160 and we need to find which is larger, let's assume SI₁ > SI₂. Then 0.8x - 0.8x = 0 ≠ 160. This suggests an error in the problem setup. However, for SSC purposes, let me assume the rates are 5% and 3% (not 4%): SI₁ = (4x × 5 × 4)/100 = 0.8x. SI₂ = (5x × 3 × 4)/100 = 0.6x. Difference = 0.8x - 0.6x = 0.2x = 160, so x = 800. Larger sum = 5x = ₹4,000. Alternatively, if rates are 6% and 4%: SI₁ = (4x × 6 × 4)/100 = 0.96x. SI₂ = (5x × 4 × 4)/100 = 0.8x. Difference = 0.96x - 0.8x = 0.16x = 160, so x = 1,000. Larger sum = 5x = ₹5,000. Let me use the second interpretation as it's more standard.

Question 16

Priya lent ₹12,000 to Neha at 8% per annum simple interest. After 2 years, Neha returned ₹14,000 and the remaining amount after 1 more year. What amount did Neha return in the 3rd year?

Accepted Answer

Step 1: Calculate total SI for 3 years: SI = (12,000 × 8 × 3)/100 = ₹2,880. Total amount due = 12,000 + 2,880 = ₹14,880. Step 2: After 2 years, Neha returned ₹14,000. Remaining amount = 14,880 - 14,000 = ₹880. Note: In simple interest, the remaining principal still accrues interest on the original principal amount, not on the returned amount. Step 3: SI for the 3rd year on original principal = (12,000 × 8 × 1)/100 = ₹960. However, since ₹14,000 was already paid, the remaining ₹880 is what Neha owes, which already includes the SI for year 3. Therefore, Neha returns ₹880 in the 3rd year.

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