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# IBPS PO Prelims Simple & Compound Interest Questions

Question 1: Find the difference between the Compound and Simple Interest for a sum of Rs 1000 at a 10% rate of interest per annum for 2 years.

a) Rs 10

b) Rs 20

c) Rs 100

d) None of the above

Solution:

The difference between the Compound and Simple Interest for 2 years is given as =
$P\times (\dfrac{R}{100})^2$, where P is Principle amount and R is the rate of interest.
Here P = Rs 1000 and R = 10%
From the above formula, we get the difference between Ci and SI for two years
= $1000\times (\dfrac{10}{100})^2$
Rs 10.

Question 2: A sum of money doubles itself in 15 years. By how much time it will become 8 times at Compound Interest?

a) 50 years

b) 40 years

c) 45 years

d) None of the above

Solution:

If money will become m times in n years.
So, $m^n$ times in n x number of years
2 times will become in 15 years
So, $2^3$ will become double in 3 x 15 years = 45 years.

Question 3: What will be the approximate compound interest on a sum of Rs. 16500 invested for 3 years at the rate of 15% per annum compounded annually?

a) Rs.8418

b) Rs.8594

c) Rs.8680

d) Rs.8742

Solution:

Compound interest = $P(1+\frac{R}{100})^{n} – P$
= $16500(1+\frac{15}{100})^{3} – 16500$
= $16500(\frac{115}{100})^{3} – 16500$

= $16500\times1.15^{3} – 16500$

= $16500\times 1.520875 – 16500$
= $16500( 1.520875 – 1)$
= $16500 \times0.520875$
= 8594.4375

= Rs.8594 (approx.)
Hence, option b is the correct answer.

Question 4: Find the compound interest on Rs 7540 at 8% per annum for 1 year, when compounded half yearly.

a) Rs. 454.66

b) Rs. 254.66

c) Rs. 615.264

d) Rs. 554.664

Solution:

According to the question,
Rate = 8%,
Principal = Rs 7540
Time = 2 half years
Compound Interest = $P[(1+\frac{r}{200})^2 – 1]$
CI = $7540[(1+\frac{8}{200})^2 – 1]$
CI = $7540[(\frac{208}{200})^2 – 1]$
CI = $7540 \times (\frac{43264}{40000}-1)$, solving this we get,
CI = Rs. 615.264.

Question 5: If the sum of money kept in compound interest triple itself in 3 years, then find the time taken by it to be 9 times of itself.

a) 9 years

b) 6 years

c) 5 years

d) 12 years

Solution:

According to the question,
Amount = $Principal(1 + \frac{rate}{100})^{time}$
Amount gets double in three years so,
3 x Amount = $Principal(1 + \frac{rate}{100})^{3}$
Amount gets 9 times in,
3 x 3 x Amount = $Principal(1 + \frac{rate}{100})^{n}$
We know,
$(Principal(1 + \frac{rate}{100})^{3})^2$ = $Principal(1 + \frac{rate}{100})^{n}$
So
N = 6 years

Question 6: The rate of interest at which a certain sum of money at simple interest amounts to Rs.18174 in 2 years and Rs.22035 in 5 years is:

a) 8.25%

b) 6.75%

c) 6.67%

d) 15%

Solution:

Given, Amount for 2 years = Rs.18174.
Amount for 5 years = Rs.22035.
Then, Simple Interest for 3 years = Rs.22035 – Rs.18174 = Rs.3861.
Simple Interest for 2 years = $\dfrac{2\times3861}{3} = Rs.2574$.
Principal = Amount for 2 years – Simple Interest for 2 years = Rs.18174 – Rs.2574 = Rs.15600.
Let the rate of interest be R%.
$\dfrac{15600\times R\times2}{100} = 2574$
$312R = 2574$
$R = 8.25\%$

Question 7: A certain sum of money amounts to Rs.14725 at 6.25% per annum simple interest in 3 years. What approximate interest would have been obtained if the same sum is invested at the same rate of compound interest in the same time period?

a) Rs.2583

b) Rs.2473

c) Rs.2613

d) Rs.2437

Solution:

Let the principal be Rs.P.
Rate of interest = 6.25%.
Time period = 3 years.
Interest = Rs.14725 – P.
$\dfrac{P\times6.25\times3}{100} = 14725 – P$

$0.1875P = 14725-P$
$1.1875P = 14725$
$P = Rs.12400$
If Rs.12400 is invested at compound interest,
Amount after three years = $12400\times\dfrac{106.25}{100}\times\dfrac{106.25}{100}\times\dfrac{106.25}{100}$

$\approx 14873$

Hence, Interest = 14873 – 12400 = Rs.2473.

Question 8: A loan was repaid in two equal instalments of Rs.51577.5 each in two years. If the rate of interest is 15% compounded annually, then the difference between the amount of loan taken and the total interest paid is:

a) Rs.64895

b) Rs.64545

c) Rs.68750

d) Rs.65800

Solution:

Let Principal = Rs.x.
Total amount repaid = 51577.5 $\times$2 = Rs.103155
Loan amount = $\dfrac{51577.5}{1.15}+\dfrac{51577.5}{1.15\times1.15} = 44850+39000 = Rs.83850$.

Total interest paid = 103155 – 83850 = 19305.
Hence, Required difference = 83850 – 19305 = Rs.64545.

Question 9: A certain sum of money amounts to Rs.101167.5 in 4 years 6 months at 7.69% per annum simple interest. Find the interest obtained by the same sum of money at 15.38% per annum simple interest for 6.8 years.

a) Rs.78621.6

b) Rs.72186.6

c) Rs.76218.6

d) Rs.71626.8

Solution:

7.69% = $\dfrac{1}{13}$
Let the principal be Rs.65P.
Given, $\dfrac{65P\times4.5}{13} + 65P = 101167.5$

$22.5P+65P = 101167.5$
$87.5P = 101167.5$
P = 1156.2.
Then, Principal = 65P = Rs.75153.
Rate of interest = 15.38% = $\dfrac{2}{13}$
Time period = 6.8 years.
Interest = $\dfrac{75153\times2\times6.8}{13} = Rs.78621.6$.

Question 10: Which of the following yields the highest interest of all when invested for 2 years?
I. Rs.24000 invested at 12.5% per annum Simple Interest.
II. Rs.19440 invested at 11.11% per annum Compound interest compounded annually.
III. Rs.28750 invested at 10% per annum compounded half-yearly.

a) Only I

b) Both II and III

c) Only III

d) Both I and III

Solution:

I. Principal = Rs.24000
Rate of interest = 12.5%
Time period = 2 years.
Interest = $\dfrac{24000\times12.5\times2}{100} = Rs.6000$.

II. Principal = Rs.19440
Rate of interest = 11.11% = $\dfrac{1}{9}$
Amount after two years = $19440\times\dfrac{10}{9}\times\dfrac{10}{9} = Rs.24000$

Interest = Rs.24000-19440 = Rs.4560.

III. Principal = Rs.28750.
Rate of interest = 10% per annum compounded half-yearly.
Amount after two years = $28750\times1.05\times1.05\times1.05\times1.05 = Rs.34945.8$.
Interest = Rs.34945.8 – Rs.28750 = Rs.6195.8
Hence, Option C is the correct answer.

Question 11: In approximately how many years will a certain sum of money become 21 times of itself at 7% per annum Compound interest compounded annually?

a) 45 years

b) 84 years

c) 68 years

d) 39 years

Solution:

Let the sum of money be Rs.x
Let the time period be ‘T’ years.
Amount after T years = Rs.343x

$x\times\dfrac{107}{100}^T = 21x$

$\dfrac{107}{100}^T = 21$

$1.07^T = 21$

From the options,
A) $1.07^\text{45} \approx 21$
B) $1.07^\text{84} \approx 294$
C) $1.07^\text{68} \approx 100$
D) $1.07^\text{39} \approx 14$

Hence, Option A is the correct answer.

Question 12: A sum of money of Rs.15000 is lent in two parts such that one part is lent at Simple Interest at 11.11% per annum for two years and the other part is lent at Compound Interest at 12.5% per annum compounded annually for two years. If the difference between the interests earned from both the schemes is Rs.507, then find the interest earned on the sum of money which was invested at simple interest is now invested at 8.33% per annum Compound interest compounded annually for 2 years.

a) Rs.1677.77

b) Rs.1432.67

c) Rs.1756

d) Rs.1237.5

Solution:

Given, Principal = Rs.15000.

Let the sum of money invested at Simple interest be Rs.9x. (Since, 11.11% = $\dfrac{1}{9}$)

Interest earned in two years = $\dfrac{9x\times2}{9} = Rs.2x.$

Sum of money invested at Compound Interest = Rs.(15000-9x).

Amount earned in two years = $(15000-9x)\times\dfrac{9}{8}\times\dfrac{9}{8}$

Interest earned through Compound Interest = $(15000-9x)\times\dfrac{81}{64}-(15000-9x)$

Given, $(15000-9x)\times\dfrac{81}{64}-(15000-9x) – 2x = 507$

$\dfrac{151875}{8}-\dfrac{729x}{64}-15000+9x-2x=507$

$9x-2x-\dfrac{729x}{64}+\dfrac{151875}{8}-15000=507$

$\dfrac{31875}{8}-\dfrac{281x}{64}=507$

$\dfrac{281x}{64}=\dfrac{31875}{8}-507$

$\dfrac{281x}{64}=\dfrac{27819}{8}$

$281x=222552$

$x=792$.

Therefore, Sum of money invested at Simple Interest = 9x = Rs.7128.

Hence, Required Interest = $7128\times\dfrac{13}{12}\times\dfrac{13}{12} = Rs.8365.5$.

Hence, Interest earned = 8365.5 – 7128 = Rs.1237.5

Question 13: Simple interest on a certain sum of money is Rs 500 for two years and the compound interest on the same sum of money is Rs 560 for two years on the same rate of interest per year. So find the rate of interest for which they invested?

a) 20%

b) 24%

c) 22%

d) 15%

Solution:

According to the question,
Let the principle be p and rate be r
Given
$\frac{prt}{100} = 500$ where t = 2 solving this we get pr = 25,000
$p(r/100)^{2}$ = 560-500 = 60

$p(r/100)^{2}$ = 60

$25000\times (r/10000) = 60$

solving this we get r = 24%

Question 14: Two friends invested in the ratio 7:6 in compound interest at 20% and 10% respectively, then what will be the ratio of their interest after one year.

a) 3:7

b) 77:60

c) 7:3

d) 6:7

Solution:

According to the question,
Let principal = 7x and 6x
We know compound interest =  P(1+r/100) -P
Therefore ratio of interest =P(1+r/100) -P : p(1+R/100) -p
I.e  7x(1+20/100) -7x : 6x(1+10/100) -6x solving this we get,
The ratio of interest = 7:3

Question 15: If a sum of money after 4 years amounts to Rs 6600 and after 7 years it amounts to 7800 at Simple interest, then find the principal and interest.

a) 5000,8%

b) 4500,12%

c) 5000,10%

d) 4500,8%

Solution:

So interest for 3 years=7800-6600
=1200
Interest for 1 year=1200/3
=Rs 400
So for 4 years interest=4*400
=1600
A=P+I
P=6600-1600
P=Rs 5000
For 4 years interest =1600
So 5000*r*4/100 =1600
r=8%

Question 16: A person took a loan of certain amount at simple interest of 7% per annum for 2 years and paid an interest of Rs.280. If he took the same amount at compound interest at the same rate of interest and for the same time period, then the interest would be?

a) Rs.289.8

b) Rs.324.6

c) Rs.312.8

d) Rs.329.8

Solution:

Let the sum be Rs. x
Given, $\dfrac{x \times 7 \times 2}{100} = 280$

⇒ $x = 2000$
Therefore, Principal $= Rs.2000$

Compound Interest for Rs.2000 at 7% per annum for 2 years:
Effective compound interest percentage for 2 years $= 7 + 7 + \dfrac{7\times7}{100} = 14.49$ %

Therefore, Compound Interest = 14.49% of 2000 = Rs.289.8

Question 17: A sum of money amounts to Rs.680 in 3 years and to Rs.750 in 5 years. Then find the principal.

a) Rs.560

b) Rs.575

c) Rs.450

d) Rs.350

Solution:

Given that the amount after 3 years = Rs.680
Amount after 5 years = Rs.750
Then, S.I. for 2 years = Rs.750-Rs.680 = Rs.70
Then, S.I. for 3 years = 70$\times\frac{3}{2}$ = Rs.105
Then Principal = Rs.680-Rs.105 = Rs.575

Question 18: A certain sum of money doubles itself in 20 years at simple interest. Then in how many years, will it become 8 times itself?

a) 120 years

b) 140 years

c) 180 years

d) 200 years

Solution:

Let Principal be Rs.100
Then, Amount after 20 years = Rs.200
Interest = Rs.200-Rs.100 = Rs.100
Let Rate of interest = R%
$\Large\frac{100\times20\times R}{100}$ $= 100$

=> $R = 5$%

Now, Amount should be 8 times of Principal
=> Amount $= 8*100 = Rs.800$
Interest $= Rs.800 – Rs.100 = Rs.700$
Let time period = T years

=> $\Large\frac{100\times5\times T}{100}$ $= 700$

=> T $= 140$ years

Hence, Required time period $= 140$ years

Question 19: A sum of money doubles itself in 20 years at simple interest. Find the rate of interest.

a) 4 %

b) 7 %

c) 10 %

d) 5 %

Solution:

Let the principal be Rs.100
Then, Final amount = Rs.100*2 = Rs.200
Interest = Rs.200-Rs.100 = Rs.100
Time period = 20 years
Let rate of interest = R %
Then, $\frac{100\times20\times R}{100} = 100$

=> $20x = 100$
=> $x = 5$
Therefore, Rate of interest $= 5$ %

Question 20: The difference between Simple Interest and Compound Interest of a certain amount of money for 2 years at 5% compound interest is Rs.5. Then find the sum.

a) Rs.400

b) Rs.800

c) Rs.1200

d) Rs.2000

Solution:

Let the principal be Rs.10000
Simple Interest $= \Large\frac{10000\times2\times5}{100}$ $= Rs.1000$
Amount at the end of 1st year at 5% compound interest $= 105$% of $10000 = Rs.10500$
Amount at the end of 2 years $= 105$% of $10500 = Rs.11025$
Compound interest $= 11025-10000 = Rs.1025$
Difference between SI and CI = Rs.1025-1000 = Rs.25
Given difference = Rs.5
25 → 5
10000 → ?
$\Large\frac{10000\times5}{25}$ $= Rs.2000$
Therefore, Principal $= Rs.2000$