simple and compound interest Model Questions & Answers, Practice Test for ibps clerk prelims 2023
ibps clerk prelims 2023 SYLLABUS WISE SUBJECTS MCQs
Ratio & Proportion
Percentages
Profit & Loss
Time & Work
Time & Distance
Simple Interest & Compound Interest
Mensuration: Area & Volumes
Algebraic Expressions
Trigonometric Ratios & Identity
Linear Equations
Quadratic Equations
Logarithm
Surya borrowed Rs. 25,000 @ 15% p.a. S.I. for 5 years. After 5 years he repaid Rs. 15,000 and promised to pay the balance amount after 3 years. Find the amount repayable as final settlement.
Answer: (d)
SI = ${25,000×15×5}/100=18750$
Amt = 25000 + 18750 = 43750
Balance = 43750 – 15000 = 28750
SI = ${28750×15×3}/100=12937.5$
Amt = 28750 + 12937.5 = 41687.5
Ms. Neelam deposits an amount of Rs. 16420 at simple interest and obtained Rs. 25451 at the end of 5 years. What was the rate of interest per year?
Answer: (d)
Interest = (25451 – 16420) = Rs.9031
Rate =${\text"Interest"×100}/\text"Principal×Time"$
${9031×100}/{16420×5}$ = 11%
Mohan borrowed Rs.18,000 at 10% p.a. simple interest and then lend it to Sohan at 10% C.I. After 3 years he will earn a profit of
Answer: (c)
SI=${18000×10×3}/100=5400$
Amt=$18000(1+10/100)^3=23958$
CI = 23958 – 18000 = 5958.
Profit = 5958 – 5400 = 558
Asmita invests an amount of Rs. 9535 at the rate of 4 per cent per annum to obtain a total amount of Rs. 11442 on simple interest after a certain period. For how many year did she invest the amount to obtain the total sum?
Answer: (a)
Let the required time = t years
Simple interest = (11442 – 9535) = Rs.1907
Simple = ${P×T×R}/100$
1907 =${9535×4×t}/100$
∴ t=${1907×100}/{9535×4}$ = 5 years
What would be the compound interest accrued on an amount of Rs. 7850 at the rate of 14% per annum in two years ?
Answer: (c)
CI = P$[(1 + r/{100})^t -1]$
=7850 $[(1 + {{114}/{100}})^2 -1] = 7850[({1.14}/{100})^2 -1]$
= 7850 [$(1.14)^2$ – 1] = 7850 [1.2996 – 1]
= 7850 [0.2996] = Rs. 2351.86
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