# Simple Interest Quiz 9

Home > > Tutorial
5 Steps - 3 Clicks

# Simple Interest Quiz 9

### Introduction

What is simple interest? Simple interest is a quick and easy method of calculating the interest charge on a loan. Simple interest is determined by multiplying the daily interest rate by the principal by the number of days that elapse between payments.
Important Formulae :
1. Simple interest(S.I) = $\frac{principal(P) \times rate(R) \times time(T)}{100}$
2. Principal(P) = $\frac{100 \times Simple interest}{Rate \times time}$
3. Rate = $\frac{100 \times Simple interest}{principal \times time}$
4. Time = $\frac{100 \times Simple interest}{principal \times rate}$
5. Amount = Principal+ simple interest (or) Amount = P(1 + $\frac{R \times T}{100}$)
Simple Interest is one of important topic in Quantitative Aptitude Section. In Simple Interest Quiz 9 article candidates can find questions with answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Simple Interest Quiz 9 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

A sum of 5000 is invested in which the investor gets 800 as SI at the @ 4 %.p.a. To get an interest of 2000 rupees on the same sum in the same number of years, what will be the rate of interest?
A. 5% B. 10% C. 15% D. 20%

B
$800 = 5000 \times (\frac {4}{100}) \times t$, we got t = 4 years
Now, $2000 = 5000 \times (\frac {r}{100}) \times 4,$ we get r = 10%

### Q2

Some amount of money out of 8000 rupees is lent at 8 % and remaining at 6 %.p.a. If the total interest obtained from both sum in 4 years is 2400, then find the sum lent at 8% rate (approx)
A. 6000 B. 3000 C. 5000 D. 4000

A
$2400 = p \times (\frac {8}{100}) \times 4 + (8000 \times p) \times 4 \times (\frac {6}{100})$
p = 6000

### Q3

Two equal amount of sum are deposited in banks at the @ 5 %.p.a. The amount deposited for 3 and 5 years respectively. If the difference between the SI obtained is 120 rupees, then find the sum.
A. 1100 B. 1200 C. 1300 D. 1400

B
$120 = p \times (\frac {5}{100}) \times 5 - p \times (\frac {5}{100}) \times 3$ = 1200

### Q4

Mohan invested 20000 rupee in fixed deposit at the rate of 10% SI. After every 3rd year he added interest to principal. Find the interest earned at the end of 6 year.
A. 7800 B. 8000 C. 7600 D. 8200

A
For the first 3 years SI will be $= 20000 \times \frac {10}{100} \times 3 = 6000$
Now he add 3000 to the principal i.e, = 20000 + 6000 = 26000
Now interest earned at the end of 6th year $= 26000 \times (\frac {10}{100}) \times 3 = 7800$

### Q5

The simple interest on a certain sum is 4/9 of the principal and the numbers of years is equal to the rate of interest. The rate of interest is.
A. 6.66 B. 3.33 C. 4.33 D. None of these

A
$(\frac {4}{9}) \times P = \frac {P \times R}{100 \times R}$ = 6.66