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Simple Interest Quiz 8

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Simple Interest Quiz 8

shape 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 8 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 8 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.

shape Q1

What annual instalment will discharge a debt of rupees 1060 due in 4 years at 4 % simple interest?
    A. 200 B. 250 C. 300 D. 400

SPLessons

Let each instalment be P,
[latex][P + P \times (\frac {4}{100}) \times 3] + [P + P \times (\frac {4}{100}) \times 2] + [P + P \times (\frac {4}{100}) \times 1] + P = 1060[/latex]

shape Q2

Anil borrowed some money at the rate of 5% p.a for the first 2 years, 3% the total interest paid by him at the end of 9 years is rupees 3400, how much money did anil borrow.
    A. 5000 B. 8000 C. 10000 D. 12000

SPLessons

3400 = [latex]p \times (\frac {5}{100}) \times 2 + p \times (\frac {3}{100}) \times 4 + p \times (\frac {4}{100}) \times 3 = 3400 [/latex]

shape Q3

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

SPLessons

For the first 3 years SI will be = [latex]20000 \times \frac {10}{100} \times 3 = 6000[/latex]
Now he add 6000 to the principal i.e = [latex]20000 + 6000 = 26000[/latex]
Now interest earned at end of 6th year = [latex]26000 \times \frac {10}{100} \times 3 = 7800[/latex]

shape Q4

The simple interest on a sum of money will be rupees 210 after 3 years. In the next 3 years, principal become 4 times, then the total interest at the end of 6 years.
    A. 1020 B. 1050 C. 1080 D. 1100

SPLessons

210 = [latex]P \times (\frac {r}{100}) \times 3[/latex]
now, SI = [latex]4 \times p \times (\frac {r}{100}) \times 3[/latex]
SI = [latex]4 \times 210 = 840[/latex]. So total SI for 6 years = 840 + 210 = 1050.

shape Q5

A certain sum of money is borrowed by ankit at 5 percent per annum for 10 years. If he pays an interest of rupees 200, then the total amount paid by ankit.
    A. 450 B. 500 C. 600 D. 650

SPLessons

[latex]200 = p \times (\frac {5}{100}) \times 10, [/latex] we get P = 400
So total amount = 400 + 200 = 600
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