Title: M6.0 Finance, Simple & Compound Interest Test
Course: Technical Math for Industry > Section: Module 6.0 Finance & Simple and Compound Interest

Q: A sum of money at simple interest amounts to 815 in 3 years and 854 in 4 years. The sum is?
- 698 (Correct. Simple interest for 1 year = (854 - 815) = 39. Simple interest for 3 years = (39 x 3) = 117. Principal = (815 - 117) = 698.)
- 650 (Incorrect)
- 690 (Incorrect)
- 700 (Incorrect)
A: 698

Q: Mr. Thomas invested an amount of 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be 3508, what was the amount invested in Scheme B?
- 6400 (Correct. Explanation: Let the sum invested in Scheme A be x and that in Scheme B be (13900 - x). Then, (x x 14 x 2)/100 + ((13900 - x) x 11 x 2)/100 = 3508 28x - 22x = 350800 - (13900 x 22) 6x = 45000 x = 7500. So, sum invested in Scheme B = (13900 - 7500) = 6400.)
- 6500 (Incorrect)
- 7200 (Incorrect)
- 7500 (Incorrect)
A: 6400

Q: A sum fetched a total simple interest of 4016.25 at the rate of 9 %.p.a. in 5 years. What is the sum?
- 8925 (Correct. Explanation: Principal = (100 x 4016.25)/(9 x 5) = 401625/45 = 8925. - See more at: http://www.theonlinetestcentre.com/simple-interest.html#3)
- 4460.50 (Incorrect)
- 8032.50 (Incorrect)
- None of the answers (Incorrect)
A: 8925

Q: How much time will it take for an amount of 450 to yield 81 as interest at 4.5% per annum of simple interest?
- 4 years (Correct Time = (100 x 81)/(450 x 4.5) years = 4 years.)
- 3.5 years (Incorrect)
- 4.5 years (Incorrect)
- 5 years (Incorrect)
A: 4 years

Q: Reena took a loan of 1200 with simple interest for as many years as the rate of interest. If she paid 432 as interest at the end of the loan period, what was the rate of interest?
- 6 (Explanation: Let rate = R% and time = R years. Then, (1200 x R x R)/100 = 432 12R² = 432 R² = 36 R = 6.)
- 18 (Incorrect)
- 3.6 (Incorrect)
- 12 (Incorrect)
A: 6

Q: The difference between the compound interest, compounded annually and the simple interest on a certain sum for 2 years at 6% per annum is $ 18. Find the sum.
- $5000 (Correct. Let the sum be $ P. Then, SI = $ (P × 6/100 × 2) = $ 3P/25 And, CI = $ {P × (1 + 6/100)^2 - P} = $ {(P × 53/50 × 53/50) - P} = $ (^2809/_2500 P - P) = $ (309P/2500) (CI) - (SI) = $ (309P/2500 – 3P/25) = $ (9P/2500) Therefore, 9P/2500 = 18 P = 2500 × 18/9 P = 5000. Hence, the required sum is $ 5000.)
- $4000 (Incorrect)
- $4500 (Incorrect)
- $5500 (Incorrect)
A: $5000

Q: A certain sum amounts to $ 72900 in 2 years at 8% per annum compound interest, compounded annually. Find the sum.
- $62500 (Correct Let the sum be $ P. Then, amount = $ {P × (1 + 8/100)^2} = $ {P × 27/25 × 27/25} = $ (729P/625) Therefore, 729P/625 = 72900 P = (72900 × 625)/729 P = 62500. Hence, the required sum is $ 62500.)
- $62000 (Incorrect)
- $52000 (Incorrect)
- $52500 (Incorrect)
A: $62500

Q: At what rate per cent per annum will Ron lends a sum of $2000 to Ben. Ben returned after 2 years $2205, compounded annually?
- 5% (Correct Let the required rate be R% per annum. Here, A = $ 2205, P = $ 2000 and n = 2 years. Using the formula A = P(1 + R/100)^n, 2205 = 2000 × ( 1 + R/100)^2 ⇒ (1 + R/100)^2 = 2205/2000 = 441/400 = (21/20)^2 ⇒ ( 1 + R/100) = 21/20 R/100 = (21/20 – 1) = 1/20 R = (100 × 1/20) = 5 Hence, the required rate of interest is 5% per annum.)
- 4% (Incorrect)
- 8% (Incorrect)
- 10% (Incorrect)
A: 5%

Q: A man deposited $1000 in a bank. In return he got $1331. Bank gave interest 10% per annum. How long did he kept the money in the bank?
- 3 (Correct Let the required time be n years. Then, amount = $ {1000 × (1 + 10/100)^n} = $ {1000 × (11/10)^n} Therefore, 1000 × (11/10)^n = 1331 [since, amount = $ 1331 (given)] (11/10)^n = 1331/1000 = 11 × 11 × 11/ 10 × 10 × 10 = (11/10)^3 (11/10)^n = (11/10)^3 n = 3. Thus, n = 3. Hence, the required time is 3 years.)
- 3.5 (Incorrect)
- 6 (Incorrect)
- 2.5 (Incorrect)
A: 3