Simple Interest Formulas, Shortcuts and Study Material

Interest is the cost of borrowing money.An interest rate is the cost stated as a percent of the amount borrowed per period of time, usually one year.

Simple Interest: Simple interest is calculated on the original principal only.Accumulated interest from prior periods is not used in calculations for the following periods. Simple interest is normally used for a single period of less than a year, such as 30 or 60 days.
Simple interest = $(\frac{P\times R\times T}{100})$

P = principal (original amount borrowed or loaned)
R = interest rate for one period
T = number of periods

Principal: The money borrowed or lent out for a certain period is called the principal or the sum.

Interest: Extra money paid for using other's money is called interest.

Simple Interest (S.I.): If the interest on a sum borrowed for certain period is reckoned uniformly, then it is called simple interest.

Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then

(i) Simple interest= $(\frac{P\times R\times T}{100 })$

(ii) P= $\left [ \frac{100\times S.I}{R\times T} \right ] ;R=\left [ \frac{100\times S.I}{T\times R} \right ] and\, T=\left [ \frac{100\times S.I}{P\times R} \right ]$

Simple Interest Formulas

Example 1. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

Answer: Rs.698.
Explanation:
S.I. for 1 year = Rs. (854-815) = Rs. 39.
S.I. for 3 years = Rs. $(39\times 3)$ = Rs. 117.
Principal = Rs. (815-117) = Rs. 698.

Example 2.Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?

Answer: Rs.6400.
Explanation:
Let the sum invested in Scheme A be Rs.x and that in Scheme B be Rs. (13900-x).
Then, $\left ( \frac{x\times 14\times 2}{100} \right )+\left ( \frac{(13900-X)\times 11\times 2}{100} \right )$ =3508.
$\Rightarrow 28x-22x=350800 -(13900\times 22)$
$\Rightarrow 6x=45000$
$\Rightarrow x=7500$
So, sum invested in Scheme B = Rs. (13900-7500) = Rs. 6400.

Example 3. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years.What is the sum?

Answer: Rs.8925.
Explanation:
Principal= Rs. $\left [ \frac{100\times 4016.25}{9\times 5} \right ]$
= Rs. $\left [ \frac{401625}{45} \right ]$ = Rs. 8925

Example 4. How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

Answer: 4 Years
Explanation:
Time = $\left [ \frac{100\times 81}{450\times 4.5} \right ]$ years = 4 years.

Example 5. A person takes a loan of Rs. 200 at 5% simple interest. He returns Rs. 100 at the end of 1 year. In order to clear his dues at the end of 2 years, he would pay:

Answer: Rs.115.
Explanation:
Amount to be paid
= Rs. $\left ( 100+\frac{200\times 5\times 1}{100}+\frac{100\times 5\times 1}{100} \right )$ = Rs.115.

Simple Interest Questions from Previous Year Exams

IBPS questions related to Simple and Compound Interests

Question 1 of 18

1. Question
1 points
Mahesh invested an amount of Rs. 12050 at simple interest. He got an amount of Rs. 13496 at the end of 2 yr. At what rate of interest did he invest?
- 12% per annum
- 8% per annum
- 6% per annum
- 4.5% per annum
- None of these
Correct

Let the rate of interest be r% per annum.

$\therefore$ (13496 - 12050) = $\frac{12050\times 2}{100}$

$\Rightarrow r=\frac{1446 \times100}{12050 \times 2}$ = 6 % per annum

Incorrect

Let the rate of interest be r% per annum.

$\therefore$ (13496 - 12050) = $\frac{12050\times 2}{100}$

$\Rightarrow r=\frac{1446 \times100}{12050 \times 2}$ = 6 % per annum
Question 2 of 18

2. Question
5 points
If the difference between the simple and the compound interest earned on a sum of money at the rate of 5% per annum for 2 yr is Rs. 21, find the principal.
- Rs. 8000
- Rs. 8250
- Rs. 8400
- Cannot be deterrnined
- None of these
Correct

Let the principal be Rs. x

$\therefore\, \, \,21=x\left ( \frac{5}{100}\right )^{2}$

$\Rightarrow x = \frac{21 \times100 \times100}{5 \times5} = Rs. 8400$

Incorrect

Let the principal be Rs. x

$\therefore\, \, \,21=x\left ( \frac{5}{100}\right )^{2}$

$\Rightarrow x = \frac{21 \times100 \times100}{5 \times5} = Rs. 8400$
Question 3 of 18

3. Question
1 points
Sitaram invested an amount of Rs. 7450 @ 6% per annum rate of simple interest. After how many years will he obtain the total amount of Rs. 8791?
- 5 yr
- 3 yr
- 2 yr
- 6 yr
- None of these
Correct

$(8791 - 7450) = \frac{7450 \times6}{100}$

$\Rightarrow t = \frac{1341 \times100}{7450 \times6} = 3 yr$

Incorrect

$(8791 - 7450) = \frac{7450 \times6}{100}$

$\Rightarrow t = \frac{1341 \times100}{7450 \times6} = 3 yr$
Question 4 of 18

4. Question
1 points
Asmita invests an amount of Rs. 9535 @ 4% per annum to obtain a total amount of Rs. 11442 on simple interest after a certain period. For how many years did she invest the amount to obtain the total sum?
- 10 yr
- 2 yr
- 5 yr
- 4 yr
- None of these
Correct

$(11442 - 9535) =\frac{9535 \times4}{100}$

$\Rightarrow \, \,t = \frac{1907 \times100}{9535 \times4}=5\,yr$

Incorrect

$(11442 - 9535) =\frac{9535 \times4}{100}$

$\Rightarrow \, \,t = \frac{1907 \times100}{9535 \times4}=5\,yr$
Question 5 of 18

5. Question
1 points
Ms Suchi deposits an amount of Rs. 24000 to obtain a simple interest at the rate of 14 % per annum for 8 yr. What total amount will Ms Suchi get at the end of 8 yr?
- Rs. 52080
- Rs. 28000
- Rs. 50880
- Rs. 26880
- None of these
Correct

The amount which Suchi will get

$= 24000 +\frac{24000 \times14 \times8}{100}$

$= 24000 + 26880 =Rs. 50880$

Incorrect

The amount which Suchi will get

$= 24000 +\frac{24000 \times14 \times8}{100}$

$= 24000 + 26880 =Rs. 50880$
Question 6 of 18

6. Question
1 points
What amount of compound interest can be obtained on an amount of Rs. 4500 at the rate of 4% per annum at the and end of 2 yr ?
- Rs. 360
- Rs.358.40
- Rs. 367.20
- None of these
Correct

Incorrect

$Compund \, \, Interest=4500\left [ \left ( 1+\frac{4}{100} \right )^{2}-1 \right ]$

$=4500\left [ \left ( \frac{26}{25} \right )^{2}-1 \right ]$

$=4500\left [ \frac{26^{2}-25^{2}}{25^{2}} \right ]$

$=4500\left [ \frac{\left ( 26-25 \right )\left ( 26+25 \right )}{625} \right ]$

$=4500\left [ \frac{\left ( 1 \right )\left ( 51 \right )}{625} \right ]=Rs \, \, 367.20$
Question 7 of 18

7. Question
1 points
What amount of compound interest can be obtained on an amount of Rs. 4800 @ 6 % per annum at the end of 2 yr?
- Rs. 544.96
- Rs. 576
- Rs. 593.28
- Rs. 588
- None of these
Correct

$Compund \, \, Interest=4800\left [ \left ( 1+\frac{6}{100} \right )^{2}-1 \right ]$

$=4800\left [ \left ( \frac{52}{50} \right )^{2}-1 \right ]$

$=4800\left [ \frac{53^{2}-50^{2}}{50^{2}} \right ]$

$=4800\left [ \frac{\left ( 53-50 \right )\left ( 53+50 \right )}{2500} \right ]$

$=4800\left [ \frac{\left ( 3 \right )\left ( 103 \right )}{2500} \right ]=\, \,Rs \, \,593.28$

Incorrect

$Compund \, \, Interest=4800\left [ \left ( 1+\frac{6}{100} \right )^{2}-1 \right ]$

$=4800\left [ \left ( \frac{52}{50} \right )^{2}-1 \right ]$

$=4800\left [ \frac{53^{2}-50^{2}}{50^{2}} \right ]$

$=4800\left [ \frac{\left ( 53-50 \right )\left ( 53+50 \right )}{2500} \right ]$

$=4800\left [ \frac{\left ( 3 \right )\left ( 103 \right )}{2500} \right ]=\, \,Rs \, \,593.28$
Question 8 of 18

8. Question
1 points
What would be the compound interest obtained on an amount of Rs. 2000 at the rate of 15% per annum after 3 yr?
- Rs. 1141.75
- Rs. 1209.75
- Rs. 1041.75
- Rs. 1248.75
- None of these
Correct

$Compund\,Interest=2000\left [ \left ( 1+\frac{15}{100} \right )^{3} -1\right ]$

$=2000\left [ \left ( \frac{23}{20} \right )^{3}-1 \right ]$

$=2000\left [ \frac{23^{3}-20^{3}}{20^{3}} \right ]$

$=2000\left [ \frac{12167-8000}{8000} \right ]$

$=2000\left [ \frac{4167}{8000} \right ]=\,Rs\,1041.75$

Incorrect

$Compund\,Interest=2000\left [ \left ( 1+\frac{15}{100} \right )^{3} -1\right ]$

$=2000\left [ \left ( \frac{23}{20} \right )^{3}-1 \right ]$

$=2000\left [ \frac{23^{3}-20^{3}}{20^{3}} \right ]$

$=2000\left [ \frac{12167-8000}{8000} \right ]$

$=2000\left [ \frac{4167}{8000} \right ]=\,Rs\,1041.75$
Question 9 of 18

9. Question
1 points
A sum of money doubles itself in 4 yr at compound interest. It will amount to 8 times itself at the same rate of interest in
- 18 yr
- 12 yr
- 16 yr
- 24 yr
- None of these
Correct

Incorrect

The sum doubles in 4 yr.
$\therefore$ It will be 8 times i.e.,2³, times in 3 x 4 = 12 yr
Question 10 of 18

10. Question
1 points
If the compound interest on a certain sum for 2yr at 3% per annum be Rs. 101.50, then the simple interest on the same sum at the same rate and for the same time will be
- Rs. 90
- Rs. 95.50
- Rs. 100
- Rs. 98.25
- None of these
Correct

Let the sum be Rs. x .

$\therefore 101.50=x\left [ \left ( 1+\frac{3}{100} \right )^{2}-1 \right ]$

$\Rightarrow \frac{1015}{10}=x\left [ \frac{103^{2}-100^{2}}{10000} \right ]$

$\Rightarrow \frac{1015}{10}=x.\frac{\left ( 3 \right )\left ( 203 \right )}{10000}$

$x=\frac{1015\times 1000}{3\times 203}=\frac{5000}{3}$

$\therefore\, Simple\,Interest=\frac{5000\times3\times 2}{3\times 1000}=Rs \, \, 100$

Incorrect

Let the sum be Rs. x .

$\therefore 101.50=x\left [ \left ( 1+\frac{3}{100} \right )^{2}-1 \right ]$

$\Rightarrow \frac{1015}{10}=x\left [ \frac{103^{2}-100^{2}}{10000} \right ]$

$\Rightarrow \frac{1015}{10}=x.\frac{\left ( 3 \right )\left ( 203 \right )}{10000}$

$x=\frac{1015\times 1000}{3\times 203}=\frac{5000}{3}$

$\therefore\, Simple\,Interest=\frac{5000\times3\times 2}{3\times 1000}=Rs \, \, 100$
Question 11 of 18

11. Question
1 points
Shivangi invested Rs. 15000 at the rate of interest of 10% per annum for 1 yr. If the interest is compounded every six months, what amount wilt shivang get,at the end of the year?
- Rs. 16537.50
- Rs. 16500
- Rs. 16525.50
- Rs. 18150
- None of these
Correct

$The\,amount=15000\left (1+\frac{\frac{10}{2}}{100} \right )^{1\times 2}$

$=15000\left ( 1+\frac{5}{100} \right )^{2}$

$=15000\left ( \frac{21}{20} \right )^{2}=15000\times \frac{441}{400}=\, \, Rs\, \, 16537.50$

Incorrect

$The\,amount=15000\left (1+\frac{\frac{10}{2}}{100} \right )^{1\times 2}$

$=15000\left ( 1+\frac{5}{100} \right )^{2}$

$=15000\left ( \frac{21}{20} \right )^{2}=15000\times \frac{441}{400}=\, \, Rs\, \, 16537.50$
Question 12 of 18

12. Question
1 points
The difference between CI and SI on a certain,sum at 4% for 3 yr is Rs. 608. The sum is
- Rs. 25000
- Rs.75000
- Rs. 100000
- Rs. 125000
- None of these
Correct

$608=P\left ( \frac{4+300}{100} \right )\left ( \frac{4}{100} \right )^{2}$

$\Rightarrow P=\frac{608\times 100\times 100\times 100}{304\times 16}=Rs\, 125000$

Incorrect

$608=P\left ( \frac{4+300}{100} \right )\left ( \frac{4}{100} \right )^{2}$

$\Rightarrow P=\frac{608\times 100\times 100\times 100}{304\times 16}=Rs\, 125000$
Question 13 of 18

13. Question
1 points
Ram invests Rs. 5000 in a bond which gives interest at 4%
per annum during the first year, 5% during the second year and 10% during the third year. How much does he get at the third year?

get at the end of the third Year?
- Rs. 5060
- Rs. 5900
- Rs. 6000
- Rs. 6006
- None of these
Correct

$He\, will \,get=5000\left ( 1+\frac{4}{100} \right )\left ( 1+\frac{5}{100} \right )\left ( 1+\frac{10}{100} \right )$

$=5000\left ( \frac{26}{25} \right )\left ( \frac{21}{20} \right )\left ( \frac{11}{10} \right ) = Rs \,6006$

Incorrect

$He\, will \,get=5000\left ( 1+\frac{4}{100} \right )\left ( 1+\frac{5}{100} \right )\left ( 1+\frac{10}{100} \right )$

$=5000\left ( \frac{26}{25} \right )\left ( \frac{21}{20} \right )\left ( \frac{11}{10} \right ) = Rs \,6006$
Question 14 of 18

14. Question
1 points
An amount of Rs. 25220 borrowed at 5% per annum, compounded yearly, is to be repaid in 3 equal instalments. The amount of each instalment is
- Rs. 9261
- Rs. 9100
- Rs. 9050
- Rs. 8930
- None of these
Correct

Let the amount of each installment be Rs x.

$\therefore \, \, \, 25220=\frac{x}{\left ( 1+\frac{5}{100} \right )}+\frac{x}{\left ( 1+\frac{5}{100} \right) ^{2}}+\frac{x}{\left ( 1+\frac{5}{100} \right ) ^{3}}$

$\Rightarrow \, \, 25220=x\left [ \frac{20}{21} +\left ( \frac{20}{21} \right )^{2}+\left ( \frac{20}{21} \right )^{3}\right]$

$\Rightarrow \, \, \, 25220=x\left ( \frac{20}{21} \right )\left [ 1+\frac{20}{21}+\frac{400}{441} \right ]$

$\Rightarrow \, \, \, 25220\times \frac{21}{20}=x\left [ \frac{441+420+400}{441} \right ]$

$\Rightarrow \, \, \, x=25220\times \frac{21}{20}\times \frac{441}{1261}=Rs\, 9261$

Incorrect

Let the amount of each installment be Rs x.

$\therefore \, \, \, 25220=\frac{x}{\left ( 1+\frac{5}{100} \right )}+\frac{x}{\left ( 1+\frac{5}{100} \right) ^{2}}+\frac{x}{\left ( 1+\frac{5}{100} \right ) ^{3}}$

$\Rightarrow \, \, 25220=x\left [ \frac{20}{21} +\left ( \frac{20}{21} \right )^{2}+\left ( \frac{20}{21} \right )^{3}\right]$

$\Rightarrow \, \, \, 25220=x\left ( \frac{20}{21} \right )\left [ 1+\frac{20}{21}+\frac{400}{441} \right ]$

$\Rightarrow \, \, \, 25220\times \frac{21}{20}=x\left [ \frac{441+420+400}{441} \right ]$

$\Rightarrow \, \, \, x=25220\times \frac{21}{20}\times \frac{441}{1261}=Rs\, 9261$
Question 15 of 18

15. Question
1 points
Aniket deposited two parts of a sum of Rs. 25000 in different banks at the rate of 15% per annum and 18% per annum respectively. In one year, he got Rs. 4050 as the total interest. What was the amount deposited at the rate of 18% Per annum?
- Rs. 9000
- Rs. 12000
- Rs. 15000
- Data inadequate
- None of these
Correct

Let Aniket invests Rs. x at the rate of 18 % per annum.

$\therefore$ $\frac{x\times18 \times1}{100} + \frac{(25000 - x) \times15 \times1 }{100} = 4050$

$\Rightarrow\, \,18x + 375000 - 15x = 405000$

$\Rightarrow \, \,3x = 405000 - 375000$

$\Rightarrow\, \,x = \frac{30000}{3} = Rs.10000$

Incorrect

Let Aniket invests Rs. x at the rate of 18 % per annum.

$\therefore$ $\frac{x\times18 \times1}{100} + \frac{(25000 - x) \times15 \times1 }{100} = 4050$

$\Rightarrow\, \,18x + 375000 - 15x = 405000$

$\Rightarrow \, \,3x = 405000 - 375000$

$\Rightarrow\, \,x = \frac{30000}{3} = Rs.10000$
Question 16 of 18

16. Question
1 points
The simple interest on a sum of money is 4/9 of the principal and the number of years is equal to the rate per cent per annum. The rate per annum is
- 5%
- 6%
Correct

Let the rate of interest and time be r% per annum and r yr respectively.

$\therefore \frac{4}{9}\times p=\frac{p\times r}{100}$

$\Rightarrow \, \, r^{2}=\frac{400}{9}$

$\Rightarrow\, \,r = \frac{20}{3} = 6\frac{2}{3}$ %

Incorrect

Let the rate of interest and time be r% per annum and r yr respectively.

$\therefore \frac{4}{9}\times p=\frac{p\times r}{100}$

$\Rightarrow \, \, r^{2}=\frac{400}{9}$

$\Rightarrow\, \,r = \frac{20}{3} = 6\frac{2}{3}$ %
Question 17 of 18

17. Question
1 points
A certain sum of money invested at compound interest, compounded annually, becomes Rs. 8820 in 2 yr and Rs. 9261 in 3 yr. The rate of interest is
- 5 % per annum
- 10% per annum
- 11 % per annum
- 12 % per annum
- None of these
Correct

Incorrect

Let the principal and rate of interest be P and r % per annum respectively.

$\therefore \, \, 8820=P\left ( 1+\frac{r}{100} \right )^{2}$

$\therefore \, \, 9261=P\left ( 1+\frac{r}{100} \right )^{3}$

From Equation (i) and (ii), we get

$\frac{9261}{8820}=1+\frac{r}{100}$

$\Rightarrow \, \, \, \frac{r}{100}=\frac{9261-8820}{8820}=\frac{441}{8820}$

$\Rightarrow \, \, \, r=\frac{441}{8820}\times 100=5$ % $per\, \,annum$
Question 18 of 18

18. Question
1 points
In what time will Rs. 1000 becomes Rs. 1331 at 10% per annul compounded annually?
- o years
- 3 years
- 8 years
- 12 years
Correct

Principal = Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a ,
Let the time be n years. Then,

$1000(1+\frac{10}{100})^{n}$ =1331

$(\frac{11}{100})^{n}$ = $\frac{1331}{1000}$ = $(\frac{11}{10})^{3}$

So, n = 3 years.

Incorrect

Principal = Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a ,
Let the time be n years. Then,

$1000(1+\frac{10}{100})^{n}$ =1331

$(\frac{11}{100})^{n}$ = $\frac{1331}{1000}$ = $(\frac{11}{10})^{3}$

So, n = 3 years.