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) = [latex]\frac{principal(P) \times rate(R) \times time(T)}{100}[/latex]
2. Principal(P) = [latex]\frac{100 \times Simple interest}{Rate \times time}[/latex]
3. Rate = [latex]\frac{100 \times Simple interest}{principal \times time}[/latex]
4. Time = [latex]\frac{100 \times Simple interest}{principal \times rate}[/latex]
5. Amount = Principal+ simple interest (or) Amount = P(1 + [latex]\frac{R \times T}{100} [/latex])
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%
[latex] 800 = 5000 \times (\frac {4}{100}) \times t [/latex], we got t = 4 years
Now, [latex] 2000 = 5000 \times (\frac {r}{100}) \times 4, [/latex] 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
[latex]2400 = p \times (\frac {8}{100}) \times 4 + (8000 \times p) \times 4 \times (\frac {6}{100})[/latex]
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
[latex]120 = p \times (\frac {5}{100}) \times 5 - p \times (\frac {5}{100}) \times 3[/latex] = 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
For the first 3 years SI will be [latex]= 20000 \times \frac {10}{100} \times 3 = 6000 [/latex]
Now he add 3000 to the principal i.e, = 20000 + 6000 = 26000
Now interest earned at the end of 6th year [latex] = 26000 \times (\frac {10}{100}) \times 3 = 7800[/latex]
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
[latex](\frac {4}{9}) \times P = \frac {P \times R}{100 \times R}[/latex] = 6.66