**Introduction:**

If the interest at the end of a year or fixed period is
added to the sum lent, and the amount thus obtained becomes the principal for
the next period, then sum of the money is said to be lent at compound interest.

If P is the principal, T is the number of years and R
is rate of interest per annum.

Amount= P [ 1 +

R
100

]^{T}**Note 1 :**When the interest is compounded half yearly

Amount= P[1 +

R/2
100

]^{2T}= p[1+
R
200

]^{2T}**Note 2:**When the interest is compounded quarterly

Amount= P[1 +

R/4
100

]^{4T}= p[1+
R
400

]^{4T}
Compound interest is obtained by when principal is
subtracted from Amount.

Solved Problems on Compound Interest

*1. Find the compound interest on Rs 80000 for 3 years at 5% per annum rate of interest?*

*a)Rs 12000 b)Rs 12610 c)Rs 14800 c)Rs 15300***Answer:B**

**Explanation:**Here the given principal is Rs 80000. Time period is 3 years and rate of interest is 5%.

A =P [ 1 +

R
100

]^{T }=> A= 80000 [1+
5
100

]^{3}
Ã° A
= 80000 x

21
20

x
21
20

x
21
20

Ã° A=
Rs 92610

Compound Interest = Amount – Principal

=92610 – 80000 = Rs 12610

=92610 – 80000 = Rs 12610

*2. Find the compound interest on Rs 24000 at 10% per annum for 2 years 6 months*

*a)*

*Rs 6492 b)Rs 6200 c)Rs 6000 d)Rs 5825***Answer :A**

**Explanation:**Principal = Rs 24000 Rate of interest = 10% p.a

Time Period = 2 ½ years

For 2 years , the rate of interest is 10% and for next
6 months rate of interest will be 10%/2= 5%

Amount => A = Rs 24000 ( 1 +

10
100

)^{2}( 1 +
5
100

)
A= 24000 x

110
100

x
110
100

x
105
100

A
= 24000 x

11
10

x
11
10

x
21
20

=30492
Compound interest
= Amount – Principal

= Rs 30492 – Rs 24000 = Rs 6492

= Rs 30492 – Rs 24000 = Rs 6492

*3. Find the Compound interest on Rs 5000 in 2 years, the rate of interest being 5% for the first year and 10% for the second year ?*

*a)Rs 775 b) Rs 3875 c)Rs 1650 d)Rs 2150***Answer: A**

**Explanation:**Given Principal = Rs 5000 T= 2 years and Rate of interest for 1

^{st}year is 5% and 2

^{nd}year is 10% => R1= 5% and R2= 10%

A=
[1+

R1
100

][1+
R2
100

]
A = 5000 x [ 1+

5
100

[1+
10
100

]
=>A=5000 x

=> A= Rs 5775

21
20

x
11
10

==> A= Rs 5775

∴ Compound interest= A – P => 5775 – 5000 = Rs 775

**4. At what rate percent per annum compound interest will Rs 12500 amount to Rs 13520 in 2 years ?**

**a)13% b)10% c)8% d)4%**

**Answer: D**

**Explanation:**Amount A= Rs 13520 Principal = Rs 12500 Rate of Interest= R and Time period T = 2 years

A =P [ 1 +

R
100

]^{T}=> 13520 = 12500 [ 1 +
R
100

]^{2}
=>

13520
12500

13 = [ 1 +
R
100

]^{2}
=>

676
675

= [
1 +
R
100

]^{2}
=> (

26
25

)^{2}= [ 1 +
R
100

]^{2}
=> [ 1 +

R
100

] =
26
25

=>
R
100

=
1
25

Therefore,
rate of interest is 4%.

*5. What sum will amount to Rs 30000 in 3 years at 25% p.a compound interest?*

*a)*

*Rs 15000 b) Rs 15360 c)Rs 20000 d)Rs 24000***Answer :B**

**Explanation:**Given Amount= Rs 30000 Time period T= 3 years and Rate of interest= 25%

We have to find the princpali P

A =P [ 1 +

R
100

]^{T}=>30000= P[ 1 +
25
100

]^{3}
=>30000=
P [

5
4

]^{3}=> P= 30000 x
4
5

x
4
5

x
4
5

= Rs 15,360
On a sum of Rs 15,360, the amount we get in 3 years 25%
rate of compound interest is Rs 30000

*6. At what rate percent compound interest, will Rs 20000 amount to Rs 22050 in 2 years?*

*a)*

*12% b)8% c)5% d)2%***Answer: C**

**Explanation:**Given the Amount is Rs 22050 And Principal = Rs 20000

Time period T= 2 years

A =P [ 1 +

R
100

]^{T}=> 22050 = 20000[ 1 +
R
100

]^{2}
Ã°

22050
20000

= [ 1 +$\frac{R}{100}$
]^{2}
Ã° $\frac{441}{400}$
=[ 1 +$\frac{R}{100}$
]

^{2}
Ã° ($\frac{21}{20}$
)

^{2}= [ 1 +$\frac{R}{100}$ ]^{2}
Ã° $\frac{21}{20}$
= 1 +$\frac{R}{100}$

Ã° R=
5%

At 5% p.a compound interest, Rs 20000 becomes Rs 22050 in 2 years

*7. Find the compound interest on Rs 32000 at 10% p.a for one and half years, the interest being compounded half yearly?*

*a)Rs 5044 b)Rs 6000 c)Rs 3822 d)Rs 4000***Answer:A**

**Explanation:**Given Principal P= Rs 32000 Rate of Interest =10% and Time period= one and half years = 1

1
2

years =
3
2

years
Amount= P[1 +

R/2
100

]^{2T}=> A = 32000 [ 1 +
10
200

]^{2 x 3/2}
Ã° A=
32000 x $\left[\frac{21}{20}\right]
^{3} $

Ã° A=
32000 x $ \frac{21}{20}$
x $ \frac{21}{20}$ x $ \frac{21}{20}$

Ã° A
= 4 x21x21x21 =37044

Compound Interest = Rs 37044 – Rs 32000 = 5044

*8. What is the difference between Compound interest and simple interest on Rs 12800 for 2 years at 10% p.a rate of interest?*

*a)*

*Rs 120 b)Rs 128 c)Rs 150 d)Rs 172***Answer:B**

**Explanation :**Simple Interest =

PTR
100

= 12800 X 2 X
10
100

= Rs 2560
To find compound interest, first we find Amount

Amount= P[1 +

R
100

]^{T}= > 12800 x [ 1 +
10
100

]^{2}=> 12800x
11
10

x
11
10

=15488
Compound interest = A – P => Rs 15488 – 12800 = Rs
2688

The difference between CI and SI = Rs 2688 – Rsd 2560 =
Rs 128

Short cut : The difference between CI and SI for 2
years at R% p.a is D =

PR

^{2}100^{2}
Therefore, D =

12800 X 10

= ^{2 }100^{2 }
12800 X 10 X 10
100 X 100

= Rs 128^{2 }

*9. The difference between CI and SI on Rs 4000 for 2 years is Rs 10. What is the rate of interest per annum?*

*a)*

*2% b)7% c)9% d)5%***Answer: D**

**Explanation:**The difference between CI and SI for 2 years at R% p.a is D =

PR

^{2}100^{2}
Ã° R

^{2}=
D x 100

^{2}P
Ã° R

^{2}= $\frac{10\times100\times100}{100}$
Ã° R=
5%

*10. The difference between CI and SI on a certain sum of money for 3 years at 10% p.a rate of interest is Rs 1550. Find the principal?*

*a)*

*Rs 50000 b)Rs 34000 c)Rs 40000 d)Rs 42000***Answer : A**

**Explanation:**

Shortcut method :When the difference between the simple
interest and compound interest on P for 3 years at R% rate of interest, then P
=

100

^{3}D R^{2}(300+R)
P=

100 x 100 x 100 x 1550
10

^{2}x (300+10)
P=

100x100x100x1550
100

x 310 = Rs 50000

*11. The value of a land increases by 15% annually. If its present value is 1058000. What was its value 2 years ago?*

*a)*

*Rs 400000 b)Rs 500000 c)Rs 800000 d)Rs 1000000***Answer:C**

**Explanation:**Its present value is Rs 1058000 means that is the amount.

A= P [ 1+

R
100

]^{T}=> Rs 1058000 = P [1+
15
100

]^{2}
Ã° 1058000=
P [ $\frac{23}{20}$
]

^{2}
Ã° 1058000=
P X $\frac{529}{400}$

Ã° P
= 1058000 X $\frac{400}{529}$ = Rs 800000

*12. At Compound interest, a sum of money becomes 2 times itself in 4 years, In how many years will it become 8 times?*

*a)*

*10 years b)12 years c)14 years d)15 years***Answer: B**

**Explanation:**At compound interest, a principal is always multiplied.

A sum
of money becomes 2 times in 4 years. Means in every 4 years, the principal
becomes 2 times.

To become 8 times => 2

^{3}times (for each 2 times, it takes 4 years) , it takes 3 x 4 = 12 years

*13. The compound interest on a certain sum of money for 2 years at 10% pa. Is Rs 2520. Find the simple interest on the same sum of money at the same rate for 2 years?*

*a)*

*Rs 2000 b)Rs 2200 c)Rs 2400 d)Rs 2540***Answer:C**

**Explanation:**Given CI= Rs 2520 and Rate of intest R = 10% and T= 2 years

Let Principal P
be Rs 100

=> Amount at
compound interest

=> A = 100 x (

110
100

)^{2}=> A= Rs 121
Theefore CI for 2 years is Rs 21

If P is Rs 100
---------CI is Rs 21

? ---------CI is Rs 2520

P =

2520
21

x 100 = Rs 12000
Now SI on Rs
12000 for 2 years at 10% rate of interest per annum => SI =

PTR
100

=> 12000 x 2 x

10
100

= Rs 2400

*14. The SI on a certain sum of money for 3 years at 8% per annum is Rs 1200. What is the compound interest on the same sum of money at the same rate of interest for 2 years?*

*a)*

*Rs 840 b)Rs 800 c)Rs 1000 d)Rs 832***Answer:D**

**Explanation**: Given SI= Rs 1200 T= 3 Years R= 8% T= 3 years

SI =

PTR
100

=> 1200 = P x
3 x
8
100

=> P = Rs 5000
To find CI, first we find Amount =>
A = P[ 1+

R
100

]^{T}
Ã° A
= Rs 5000 x $\frac{108}{100}$
x $\frac{108}{100}$
= Rs 5832

Therefore, CI = A –P= Rs 5832 – Rs
5000 = Rs 832

**Shortcut Method**:

Given SI for 3 years is Rs 1200 . SI for 1 years is

1200
3

= Rs 400
SI and CI are same for 1

^{st}year.
Now CI for 2

^{nd}year = 400 + 8% on 1^{st}year interest 400 = 432^{ }CI for 2 years = Rs 400+ Rs 432 = Rs 832

*15. A sum of money amounts to Rs 2400 in 2 years and Rs 2640 in 3 years at CI. Find the rate of percent per annum?*

*a)10 % B)12$ C)12.5% D)14%***Answer:A**

**Explanation:**A1 = Rs 2400 and A2= 2640

P + CI for 3 years – PI + CI for 2 years = Rs 2640 – 2400
= Rs 240

This Rs 240 is the simple interest obtained on Rs 2400
in the 3

^{rd}year.
Ã° SI=
$\frac{PTR}{100}$

Ã° 240
= 2400 x 1 x $\frac{R}{100}$

Ã° R=
10%