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Aptitude Questions: Simple Interest Set 9

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Hello Aspirants. Welcome to Online Quants Quiz in AffairsCloud.com. Here we are creating question sample in Simple Interest, which is common for all the IBPS, RRB exam and other competitive exams. We have included Some questions that are repeatedly asked in exams !!

Questions penned by Yogit

  1. A certain sum of money amounts to rupees 2900 at 4% per annum in 4 years. In how many years will it amount to rupees 5000 at the same rate?
    a) 20
    b) 22
    c) 24
    d) 25
    e) None of these
    Answer – d) 25
    Explanation :
    2900 = p + p*(4/100)*4, p = 2500
    5000 = 2500 + 2500*(4/100)*t

  2. A sum of money becomes four times at simple interest rate of 5%. At what rate it becomes seven times
    a) 8%
    b) 10%
    c) 12%
    d) 14%
    e) None of these
    Answer – b) 10%
    Explanation :
    3p = p*(5/100)*t, t = 60
    6p = p*(r/100)*60

  3. Arya borrows rupees 6000 from a bank at SI. After 4 years she paid Rs 2500 to the bank and at the end of 5 years from the date of borrowing he paid Rs 4560 to settle the account. Find the rate of interest (approx)
    a) 3.25%
    b) 3.50%
    c) 3.85%
    d) 4%
    e) None of these
    Answer – c) 3.85%
    Explanation :
    Total interest she paid = 6000*(r/100)*4 + 3500*(r/100)*1 = 275r
    total interest = 2500 + 4560 – 6000 = 1060
    so 1060 = 275r, r = 3.85% approx

  4. A sum of money lent out at simple interest amounts to rupees 840 in 3 years and to rupees 1200 in 8 years. Find the rate of interest
    a) 11.8/13%
    b) 11.7/13%
    c) 12.8/13%
    d) 12.7/13%
    e) None of these
    Answer – b) 11.7/13%
    Explanation :
    1200 = p + p*(r/100)*8
    840 = p + p*(r/100)*3
    Solve both equations to get r

  5. The simple interest on a certain sum of money at 4% per annum for 5 years is 100 more than the interest on the same sum for 3 years at 5% per annum. Find the sum
    a) 1000
    b) 1500
    c) 2000
    d) 2500
    e) None of these
    Answer – c) 2000
    Explanation :
    p*(4/100)*5 – p*(5/100)*3 = 100

  6. A sum of rupees 4800 is lent out in two parts in such a way that the interest on one part at 10% for 4 years is equal to that on another part at 8% for 7 years. Find the two sums
    a) 2800, 2000
    b) 2400, 2400
    c) 2600, 2200
    d) 2700, 2100
    e) None of these
    Answer – a) 2800, 2000
    Explanation :
    Let first part is A then second will be 4800 – A
    A*(10/100)*4 = (4800 – A)*(8/100)*7

  7. The simple interest on a sum of money will be rupees 400 after 5 years. In the next 5 years the principal is doubled, what will be the total interest at the end of the 10th year
    a) 800
    b) 1000
    c) 1200
    d) 1600
    e) None of these
    Answer – c) 1200
    Explanation :
    400 = p*(r/100)*5, pr = 8000
    SI = 2p*(r/100)*5 = pr/10 = 8000/10 = 800
    Total interest = 400 + 800 = 1200

  8. What annual instalment will discharge a debt of rupees 1060 due in 4 years at 4 % simple interest?
    a) 200
    b) 250
    c) 300
    d) 400
    e) None of these
    Answer – b) 250
    Explanation :
    Let each instalment be P,
    [P + P*(4/100)*3] + [P + P*(4/100)*2] + [P + P*(4/100)*1] + P = 1060

  9. Anil borrowed some money at the rate of 5 percent per annum for the first 2 years, 3 percent for the next 4 years and 4 percent per annum beyond 6 years. If the total interest paid by him at the end of 9 years is rupees 3400, how much money did anil borrow.
    a) 5000
    b) 8000
    c) 10000
    d) 12000
    e) None of these
    Answer – c) 10000
    Explanation :
    3400 = p*(5/100)*2 + p*(3/100)*4 + p*(4/100)*3 = 3400

  10. Saroj invested 20000 rupee in fixed deposit at the rate of 10% simple interest. After every 3 rd year he added interest to principal. Find the interest earned at the end of 6th year.
    a) 7800
    b) 8000
    c) 7600
    d) 8200.
    e) None of these
    Answer – a)7800
    Explanation :
    For the first 3 years SI will be = 20000*10/100*3 = 6000
    Now he add 6000 to the principal i.e = 20000+6000 = 26000
    Now interest earned at end of 6th year = 26000*10/100*3 = 7800

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