Simple interest is a fast and easy method of calculating the interest charge on an amount . Simple interest is determined by multiplying the daily interest rate by the principal amount by the number of days that elapse between payments.It is the interest calculated on the principal amount of a loan or the original contribution to a savings account. Simple interest does not compound, meaning that an account holder will only gain interest on the principal, and a borrower will never have to pay interest on interest already accrued.
Example – What is simple interest of Rs 5000/- for 5 years at 5% interest per annum.
Solution – SI= $\mathbf{\frac{P * R * T}{100}}$
$=\frac{5000 *5 *5}{100}$
=Rs 1250.
1. Find the simple interest on Rs. 60,000 at 13/5 % p.a. for a period of 9 months?
Options:
A. 1170
B. 1710
C. 11700
D. 1017
Solution:
We know that, SI = $\mathbf{\frac{P * R * T}{100}}$
P = 60000
R = $\frac{13}{5}$
T = 9 months = $\frac{ 3}{4}_{th}$ year
SI =$\frac{60000 * 13 * 3}{5 * 4 * 100}$
SI = $\frac{117000}{100}$
SI = 1170
Correct option: A
2. What will be the ratio of simple interest earned on a certain amount at the same rate of interest for 6 years and that for 24 years?
Options:
A. 1:2
B. 3:5
C. 1:3
D. 1:4
Solution:
Required Ratio = Simple Interest for 6 years/Simple Interest for 24 years
=$\frac{T_{1}}{T_{2}}$
$=\frac{6}{24}$
$=\frac{1}{4}$
=1:4
Correct option: D
3.Find the simple interest on Rs.500 for 10 months at 5 paisa per month?
Options:
A. Rs. 2500
B. Rs. 250
C. Rs. 25
D. Rs. 25.5
Solution:
SI = $\mathbf{\frac{P * R * T}{100}}$
Downstream speed = 2x
SI = $\frac{500 * 5 * 10}{100}$
= Rs. 250
Correct option: B
1. Nisha borrowed some money at the rate of 5% p.a. for the first two years. She again borrowed at the rate of 10% p.a. for the next three years. Later at the rate of 15% p.a. for the rest of the years. Total interest paid by her was Rs. 15000 at the end of 10 years. Calculate the amount of money she borrowed?
Options:
A. Rs. 15005
B. Rs. 20000
C. Rs. 15000
D. Rs. 10000
Solution:
According to the question,
r1 = 5% , T1 = 2 years
r2 = 10% , T2 = 3 years
r3 = 15% , T3 = 5 years
(since, beyond 5 years rate is 14%)
(since, beyond 5 years rate is 14%)
Simple interest = 15000
Therefore, P =$ \frac{15000 * 100}{5*2 +10*3 +15*4 }$
$=\frac{1500000}{10 + 30 + 60 }$
= $\frac{1500000}{100}$
= Rs. 15000
Correct option: C
2. Rahul invests some amount of money in three different schemes for 5 years, 10 years and 15 years at 10%, 12% and 15% Simple Interest respectively. At the completion of each scheme, he gets the same interest. Find out the ratio of his investment?
Options:
A. 3: 9: 15
B. 10: 24: 45
C. 10: 24: 40
D. 9: 24: 45
Solution:
If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are 10 %, 12 %,15% respectively and time periods are 5 years , 10 years , 15 years respectively.
Let the three amounts be Rs. x, Rs. y and Rs. z,
Then , According to question
$\frac{x × 10 × 5}{100} = \frac{y× 12 × 10}{100} = \frac{z × 15 × 15}{100}$
50 x = 120 y = 225 z = k(say)
10 x = 24 y = 45 z = k
$\frac{k}{10}:\frac{k}{24}:\frac{k}{24}$
Hence, the ratio of his investment will be
$\frac{k}{10}:\frac{k}{24}:\frac{k}{45}$
10 : 24 : 45
Correct option: B
Ram tool a loan for Rs.10000 from bank for a period of 3 years. The bank charged the interest rates as 5% for first year, 7% for second year and 9% for the third year. Find the amount he has to pay back to the bank after three years?
Options:
A. Rs. 18500
B. Rs. 145000
C. Rs. 15100
D. Rs 12100.
Solution:
Simple Interest = 5% + 7% + 9% = 21%.
Amount = Principal + Rate of Interest.
Principal is 100% of the amount.
Therefore, Amount = 100% + 21%.= 121%
According to the question, total amount to be paid
A =$\frac{121 * 10000}{100}$
A = 12100
Therefore, the amount that Ram has to pay back to the bank after three years Rs 12100.
Correct option: D
1.A sum of money becomes six times in 30 years. Calculate the rate of interest.
Options:
A. 16.66%
B. 4.5%
C. 15.45%
D. 15%
Solution:
We know that, if sum of money becomes x times in n years at some rate of interest, then rate of interest is calculated as,
R =$ 100 ( \frac{x-1}{n})%$
R = $100 ( \frac{6-1}{30})%$
R = $100 ( \frac{6-1}{30})%$
R = $100 ( \frac{6-1}{30})$
R = 16.66%
Correct option: A
2. How much time will it take for an amount of Rs. 630 to yield Rs. 72 as interest at 5.4 % p.a. of simple interest?
Options:
A. 2 years and 10 months
B. 1 years and 11 months
C. 2 years and 11 months
D. 1 years and 11 months
Solution:
We know that, Time =$ \frac{100 * SI }{R * P }$
We know that, Time = $\frac{100 * SI }{R * P }$
T = $7200/3402 = \frac{7200}{3402}$
T = 2 years and 11 months
Correct option: C
3.Mr. Tata invested Rs. 13,900 in two different schemes I and II. The rate of interest for both the schemes were 14% 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 II?
Options:
A. Rs. 2400
B. Rs. 6200
C. Rs. 6400
D. Rs. 4600
Solution:
Let the amount invested in schemes I = x
Therefore, in schemes II = (13900 – x)
SI (scheme I) = $\frac{P * R * T}{100}$
Then For scheme I,
SI = $\frac{x * 14 * 2}{100}$
Then For scheme II,
SI (scheme II) =$ \frac{(13900-x) * 14 * 2}{100}$
SI (scheme I) + SI (scheme II) = 3508
$\frac{( x * 14 * 2)}{100}+\frac{((13900-x) * 14 * 2)}{100}$ = 3508
$\frac{28x}{100}+\frac{(13900 – x) * 22}{100}$
6x = 45000
x = 45000/6
x = 7500
Hence, the sum invested in Scheme II = 13900 – 7500 = Rs. 6400
Correct option: C
Go through the page completely to know How To Solve Compound Interest Questions Quickly .Compound interest is the multiplication of the initial principal amount and one plus the annual interest rate raised to the number of compound periods minus one i.e Compound Interest , to know How To Solve Compound Interest Questions Quickly, we can use the formula. $CI =P((1+ \mathbf{\frac{r}{100n})^{nT} – 1 })$
Basic Formula for the Compound Interest ,
$A= P(1 +\mathbf{\frac{r}{n}}nr)nt$
Here,
A = Amount
P = Principa
r = Interest rate(decimal)
n = number of times interest is compounded per unit ‘t’
t = total time
1 . If a sum of Rs.100 is invested for 10% p.a at CI then the sum of the amount will be Rs.121 in
Options:
A. 2 years
B. 1 year
C. 1.5 years
D. 3 years
Solution:
We know that,
$Amount =P(1+ \mathbf{\frac{r}{100}})^{t}$
$121 = 100 (1 + \frac{10}{100})^t$ =
$(\frac{121}{100})$ = $(\frac{11}{10})^t$
$(\frac{11}{10})^2$ = $(\frac{11}{10})^t$
t = 2 years
Correct option: A
2. Find the CI on Rs. 10,000 in 2 years at 2 % per annum, the interest being compounded half-yearly.
Options:
A. 408
B. 406.04
C. 409.03
D. 405.50
Solution:
According to the question
P =10000, r = 2%, t = 2
We know that
Amount = $P \mathbf{(1+ \frac{{\frac{r}{2}}}{100})^{2t}}$
$A = 10000 \mathbf{(1+ \frac{{\frac{2}{2}}}{100})^{2*2}}$
$A = 10000 \mathbf{(1+ \frac{1}{100})^{4}}$
A = 10000 (1+ 0.01)4
A = 10000 (1.01)4
A = 10000 (1.04)
A = 10406.04
Now, CI = A – P
CI = 10406.04 – 10000
CI = 406.04
Correct option: B
3. Find the CI on Rs. 2000 in 9 months at 12% p.a. if the interest is calculated quarterly.
Options:
A. 251.01
B. 251
C. 304
D. 185.45
Solution:
According to the question,
P = 2000, r = 6%, t = 9 months (3 quarter)
We know that,
$Amount = P (1+ \frac{{\frac{r}{4}}}{100})^{4t}$
$A = 2000 (1+ \frac{{\frac{12}{4}}}{100})^{4*\frac{9}{12}}$
$A = 2000 (1+ \frac{{\frac{12}{4}}}{100})^{4*\frac{3}{4}}$
$A = 2000 (1+ \frac{3}{100})^{3}$
A = 2000 (1+0.03)3
A = 2000 (1.03) 3
A = 2000 (1.09)
A = 2185.45
Now, CI = A – P
x = 7500
CI = 2185.45– 2000 = 185.45
Correct option: D
1. The CI on Rs. 20,000 at 6% per annum is Rs. 2472. Find the period (in years):
Options:
A. 2
B. 4
C. 5
D. 3
Solution:
Amount = 20000 + 2472 = Rs. 22472
Let the time = n years
So,$ 20000 (1+ \frac{6}{100})^{n}$ = 22472
$(\frac{106}{100})^{n} $ = $(\frac{22472}{20000})$
$(\frac{11296}{10000})$ = $(\frac{106}{100})^{2}$
Therefore, n = 2 years
Correct option: A
2. The principal amount is put on CI for two years at 40%. It gets 964 more if the interest is payable half yearly. Calculate the sum.
Options:
A. Rs 8485
B. Rs 8485.91
C. Rs 8480
D. Rs 8455.91
Solution:
Let us assume the Principal as Rs. 100
When compounded annually
$A = 100 (1+ \frac{40}{100})^{2}$
A= 196
When compounded half yearly
A = $100 (1+ \frac{\frac{40}{2}}{100})^{4}$
$A = 100 (1+ \frac{20}{100})^{4}$
A = 207.36
Difference, 207.36 – 196 = 11.36
If difference is 11.36, then Principal = Rs 100
If difference is 964, then Principal = ( \frac{100}{11.36})
P = Rs 8485.91
Correct option: B
3. In what time the CI on Rs 800 at 30% pa will amount to Rs.1352 if calculated annually?
Options:
A. 3 years
B. 1.6 years
C. 2 years
D. 5 years
Solution:
We know that,
$Amount = P (1+ \frac{r}{100})^{t}$
$1352 = 800 (1+ \frac{30}{100})^{t}$
$(\frac{1352}{800})$
1.69 = (1.3)^{t}(1.3)t
$1.69 = (1.3)^{2}$
Therefore, t = 2 years
Correct option: C
1. If the difference between compound interest and simple interest on a certain principal amount is Rs 500 at 10% p.a. for 2 years. Then calculate the sum.
Options:
A. 45000
B. 50000
C. 55000
D. 5000
Solution:
Given the difference between SI and CI = 500
Time = 2 years
When the difference between SI and CI is of two years then,
Difference = $P \mathbf{ \frac{(R)^{2}}{(100)^{2}}}$
Where P = principal amount, R = rate of interest
According to the question:
500 = P \frac{(10)^{2}}{(100)^{2}}
P = 50,000
Correct option: B
2. The difference between CI and SI on an amount of Rs. 15,000 for 2 years is Rs. 96. Find the rate of interest?
Options:
A. 8%
B. 5%
C. 4%
D. 3%
Solution:
Difference =$ P × \mathbf{ \frac{(R)^{2}}{(100)^{2}}}$
$96 =15000 × \frac{(R)^{2}}{10000}$
$\frac{96 ×10000 }{15000} = R^{2}
R^{2} = 64
R = 8%
Correct option: A
3. The difference between CI and SI calculated annually on a certain amount of money for two years at 4% pa is Re. 1. Find the sum:
Options:
A. 600
B. 700
C. 650
D. 625
Solution:
When the difference between SI and CI is of two years then,
$Difference =P × \mathbf{ \frac{(R)^{2}}{(100)^2}}$
Where P = principal amount, R = rate of interest
$1 = P × \frac{(4)^{2}}{(100)^2}$
P = 625
Correct option: D
1. A sum of money borrowed under CI gets double in 5 years. When will it become eight times of itself if the rate of interest remains same?
Options:
A. 15 years
B. 20 years
C. 10 years
D. 5 years
Solution:
We know that,
$(x)^{\frac{1}{a}} = (y)^{\frac{1}{b}}$
$(2)^{\frac{1}{5}}$ = $(8)^{\frac{1}{b}}$
$(2)^{\frac{1}{5}}$ = $(2)^{\frac{3}{x}}$
$\frac{1}{5} = \frac{3}{x}$
x = 15 years
Correct option: A
2. . If a certain amount of sum becomes 9 times in 2 years at compound interest, then find out the rate of interest?
Options:
A. 250%
B. 100%
C. 200%
D. 20%
Solution:
We know that,
$r = 100 ((\frac{A}{P})^\frac{1}{t} – 1)$
Therefore,
$r = 100((\frac{9P}{P})^\frac{1}{2} – 1)$
$r =100 (9^\frac{1}{2} – 1)$
r = 100 (3-1)= 100*2
r = 200%
Correct option: C
3. At what rate percentage will certain amount of money become 8 times in three years?
Options:
A. 113. 79%
B. 100%
C. 110. 79%
D. 130%
Solution:
We know that,
$r = 100 ((\frac{A}{P})^\frac{1}{t} – 1)$
$r = 100 ((8)^\frac{1}{3} – 1)$
$r = 100 * (2-1)$
$r = 100%$
Correct option: B
1. Ravi took an amount of Rs.20000 as loan at CI charging 5%, 10% and 20% for the 1st year, 2nd year, and 3rd year respectively. Find out the total interest to be paid by Ravi at the end of the 3rd year?
Options:
A. Rs. 7270
B. Rs. 2270
C. Rs. 7720
D. Rs. 7027
Solution:
According to the question,
$P = 20000$
$R = 5%, 10%, and 20%$
T= 1, 2, and 3 years
$Amount = 20000 (1+\frac{5}{100})( 1+\frac{10}{100})(1+\frac{20}{100}$
Amount = 20000 (\frac{105}{100})(\frac{110}{100})(\frac{120}{100})$
Amount = (20000) (1.05) (1.1) ( 1.2)
Amount = 27720
We know that, CI = A – P
We know that, CI = A – P
We know that, CI = A – P
Correct option: C
2. If the rate of CI for the 1st year is 8%, 2nd year is 10%, and for 3rd year is 15% then find the amount and the CI on Rs 10000 in three years.
Options:
A. 3626
B. 6236
C. 3666
D. 3662
Solution:
We know that, Amount = $P (1 + \frac{r_{1}}{100}) (1 + \frac{r_{2}}{100})(1 + \frac{r_{3}}{100})(1+\frac{10}{100}) (1+\frac{15}{100})$
Amount = 10000 (1.08) (1.1) (1.15)
Amount = 13662
Now, CI = amount –principal
CI = 13662 – 10000
CI = 3662
Correct option: D
3.Karan bought a bike of Rs. 60000 by paying cash. He borrowed the cash from his friend at rate of interest 5% for the 1st year and 15% for the 2nd year. Find out the total amount he has to pay after 2 years to his friend.
Options:
A. Rs. 72450
B. Rs. 74250
C. Rs. 72540
D. Rs. 75420
Solution:
We know that, Amount = $P (1+\frac{r_{1}}{100})(1+\frac{r_{2}}{100})$
Amount after two years = $60000 (1+\frac{5}{100})(1+\frac{15}{100})$
Amount after two years = (60000)(1.05)(1.15)
Amount after two years = Rs. 72450
Correct Option: A