# Simple Interest Quiz 4

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# Simple Interest Quiz 4

### 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 4 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 4 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

Suresh lends 40% of his money at 15% p.a, 50% of the rest at 10% per annum and the rest at 18% p.a. rate of interest. What would be the annual rate of interest, if the interest is calculated on the whole sum?
A. 18.5% B. 14.4% C. 16.5% D. 19.5%

B
x – $(\frac {40}{100}) × x = \frac {60x}{100}$
$\frac {40}{100} at 15% p.a = \frac {40}{100} × \frac {15}{100} = \frac {60x}{1000}$
$\frac {50}{100} × \frac {40x}{100} × \frac {30x}{100} at 10% p.a = \frac {30x}{100} × \frac {10}{100} = \frac {30x}{1000}$
Balance amount = $\frac {40x}{100} - \frac {30x}{100} = \frac {30x}{100}$at 18% p.a = $\frac {18}{100} × \frac {30x}{100} = \frac {54x}{1000}$
R = $[\frac {\frac {144x}{1000}}{x}] × 100 = 14.4$%

### Q2

Ajay borrows Rs 1000 at the rate of 12% per annum S.I. and Babu borrows Rs 1050 at the rate of 10% per annum simple interest. In how many years will their amounts of debts be equal?
A. $\frac {18}{5}$ B. $\frac {10}{3}$ C. $\frac {22}{3}$ D. $\frac {10}{5}$

B
Let Time = x years Then,
$[1000 + \frac {(1000 × 12 × x)}{100}] = [1050 + \frac {(1050 × 10 × x)}{100}] \Rightarrow 1000 + 120x = 1050 + 105x$
$\Rightarrow 15x = 50 \Rightarrow x = \frac {10}{3}$ years

### Q3

A sum of Rs. 8800 is to be divided among three brothers A, D and R in such a way that S.I. on each part at 5% p.a. after 1, 2 and 3 year respectively remains equal. The share of A is more than that of R by?
A. Rs. 3200 B. Rs. 2500 C. Rs. 3000 D. Rs. 2700

A
$x × 5 × \frac {1}{100} = y × 5 × \frac {2}{100} = z × 5 × \frac {3}{100}$
x:y:z = 6:3:2
The share of A is more than that of R by = $\frac {4}{11}$ × 8800 = 3200

### Q4

Mayank invested a certain sum of money in a S.I. bond, that value grew to Rs. 300 at the end of 3 year and to Rs. 400 at the end of another 5 year. Then what was the rate of interest in which he invested his sum?
A. 12% B. 12.5% C. 6.67% D. 8.33%

D
$\frac {P × R × 3}{100}$ + P = 300 –(i)
$\frac {P × R × 8}{100}$ + P = 400 –(ii)
From (i) and (ii)
$\frac {P × R × 5}{100}$ = 100
P × R = 2000 —(iii)
Sub (iii) in (i)
$\frac {6000}{100}$ + P = 300
P = 240
240 × R = 2000 $\Rightarrow$ R = 8.33%

### Q5

A borrows Rs.7000 at S.I. from a lender. At the end of 3 years, A again borrows Rs.3000 and settled that amount after paying Rs.4615 as interest after 8 years from the time A made the first borrowing. What is the rate of interest?
A. 5.5% B. 9.5% C. 7.5% D. 6.5%

D
SI for Rs.7000 for 8 years = $\frac {(7000 × r × 8)}{100}$
Again borrowed = 3000
SI = $\frac{(3000 × r × 5)}{100}$
Total interest= $[\frac{(7000 × r × 8)}{100}] + [\frac {(3000 × r × 5)}{100}] = 4615$
560r + 150r = 4615
710r = 4615
r = 6.5%

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