# Compound Interest Quiz 6

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# Compound Interest Quiz 6

### Introduction

What is compound interest? Compound interest is the addition of interest to the principal sum of a loan or deposit, or in other words, interest on interest. It is the result of reinvesting interest, rather than paying it out, so that interest in the next period is then earned on the principal sum plus previously accumulated interest.
Compound Interest is one of important topic in Quantitative Aptitude Section. In Compound Interest Quiz 6 article candidates can find questions with answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Compound Interest Quiz 6 questions are very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.

### Q1

What sum of money will amount to rupees 1124.76 in 3 years, if the rate of interest is 5% for the first year, 4% for the second year and 3% for the third year?
A. 1500 B. 1200 C. 1000 D. 1900

C
1124.76 = $p \times (\frac {105}{100}) \times (\frac {104}{100}) \times (\frac {103}{100})$ = 1000

### Q2

Riya saves an amount of 500 every year and then lent that amount at an interest of 10 percent compounded annually. Find the amount after 3 years.
A. 1820.5 B. 1840.5 C. 1920.5 D. 1940.5

A
Total amount = $500 \times {(1 + \frac {10}{100})} ^ {3} + 500 \times {(1 + \frac {10}{100})} ^ {2} + 500 \times (1 + \frac {10}{100}) = 1820.5$

### Q3

A sum of 3000 becomes 3600 in 3 years at 15 percent per annum. What will be the sum at the same rate after 9 years?
A. 5124 B. 5184 C. 5186 D. 5192

B
$3600 = 3000 \times {(1 + \frac {15}{100})} ^ {3}$
${(1 + \frac {15}{100})} ^ {3} = \frac {6}{5}$
Amount = $3000 \times{ [{(1 + \frac {15}{100})}^{3}]} ^ {3}$
Amount = $3000 \times {(\frac {6}{5})}^{3} = 5184$

### Q4

On a certain sum of money, after 2 years the simple interest and compound interest obtained are Rs 400 and Rs 600 respectively. What is the sum of money invested?
A. 100 B. 200 C. 300 D. 400

B
$400 = P \times (\frac {R}{100}) \times 2$
600 = $P \times {(1 + \frac {R}{100})}^{2} – P$
Solve both equations to get P
i.e, P = 200

### Q5

A sum of money becomes Rs 35,280 after 2 years and Rs 37,044 after 3 years when lent on compound interest. Find the principal amount.
A. 2800 B. 3000 C. 3200 D. 4000

C
$37044 = p \times {(1 + \frac {r}{100})}^{3}$
$35280 = p \times {(1 + \frac {r}{100})}^{2}$
Divide both equations to get the value of r and then substitute in any equation to get P
i.e, P = 3200

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