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 7 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 7 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 money becomes four times at simple interest rate of 5%. At what rate it becomes seven times
- A. 8%
B. 10%
C. 12%
D. 14%
3p = [latex] p \times (\frac {5}{100}) \times t[/latex], t = 60
6p = [latex] p \times (\frac {r}{100}) \times 60[/latex]
= 10%
Q2
Arya borrows rupees 6000 from a bank at SI. After 4 years she paid Rs 2500 to the bank and at the end of 5 years from the date of borrowing he paid Rs 4560 to settle the account. Find the rate of interest
- A. 3.25%
B. 3.50%
C. 3.85%
D. 4%
Total interest she paid = [latex] 6000 \times (\frac {r}{100}) \times 4 + 3500 \times (\frac{r}{100}) \times 1[/latex] = 275r
Total interest = 2500 + 4560 – 6000 = 1060
So 1060 = 275r, r = 3.85% approx
Q3
The simple interest on a certain sum of money at 4% per annum for 5 years is 100 more than the interest on the same sum for 3 years at 5% per annum. Find the sum
- A. 1000
B. 1500
C. 2000
D. 2500
[latex] p \times (\frac {4}{100}) \times 5 - p \times (\frac{5}{100}) \times 3 = 100[/latex]
= 2000
Q4
A sum of rupees 4800 is lent out in two parts in such a way that the interest on one part at 10% for 4 years is equal to that on another part at 8% for 7 years. Find the two sums
- A. 2800, 2000
B. 2400, 2400
C. 2600, 2200
D. 2700, 2100
Let first part is A then second will be 4800 – A
[latex] A \times (\frac {10}{100}) \times 4 = (4800 - A) \times (\frac{8}{100}) \times 7[/latex]
Q5
The simple interest on a sum of money will be rupees 400 after 5 years. In the next 5 years the principal is doubled, what will be the total interest at the end of the 10th year
- A. 800
B. 1000
C. 1200
D. 1600
[latex] 400 = p \times (\frac {r}{100}) \times 5, pr = 8000[/latex]
[latex] SI = 2p \times (\frac {r}{100}) \times 5 = \frac {pr}{10} = \frac {8000}{10} = 800[/latex]
Total interest = 400 + 800 = 1200