Compound Interest


Compound interest differs from Simple interest as the interests earned in the previous years considered as principal for the next years.  Observe the table
compound interest concept
From the above table, we can see that in Simple interest case, Interests earned in each year is same, but in the compound interest, except for the first year, interests are different.  This is due to the fact that in the CI case interest earned in the 1st year is considered as principal and interest will be calculated on this too.  In the first box 10+1 can be explained as 10 is the interest on the principal, 1 is the interest on the first year interest.  For the 2nd box, 10 is the interest on principal, 1 is interest on first year interest, another 1 is interest on second year 10, and 0.1 is the interest on 1.

Compound Interest formula:

There is no direct formula for Compound interest but $P{\left( {1 + \displaystyle\frac{R}{{100}}} \right)^n}$ gives us the Amount which includes principal as well as interest.  So to find the interest we need to substract principal from the amount. 
CI = A - P = \(P{\left( {1 + \dfrac{R}{{100}}} \right)^n} - P\)

Compound interest through Pascal Triangle:

Find the compound interest on Rs.6000 at the rate 10% per annum for 3 years. 
Formula: A = $P{\left( {1 + \displaystyle\frac{R}{{100}}} \right)^n}$ = $6000{\left( {1 + \displaystyle\frac{{10}}{{100}}} \right)^3}$
$ \Rightarrow 6000 \times {\left( {\displaystyle\frac{{11}}{{10}}} \right)^3}$ = 7986
Compound Interest = 7986 - 6000 = 1986

Alternate method:
We can solve compound interest problems easily with the use of pascal triangle. 
compound interest shortcut
To solve above problem, we need to identify the coefficients where 3 appears first time in a row.  We know that in 4th row 3 appears.  We need to take coefficients excluding 1.  i.e., 3, 3, 1. Now
compound interest shortcut 2

CI = 3 × (10% of 6000) + 3 × (10% of 600) + 1 × (10% of 60) = 1986


Finding compound interest when periods (\(n\)) is not an integer:

For example, we have to find compound interest on Rs.2000 at 15% interest for 2 years 4 months.
There is no difference in calculation for the 2 years. But for 4 months, take 4 months interest.
4 months =\(\dfrac{4}{{12}} = \dfrac{1}{3}\)rd of an year. So interest rate for the last 4 months = \(\dfrac{1}{3}\left( {15} \right)\)
In this case, A = \(2000{\left( {1 + \dfrac{{15}}{{100}}} \right)^2}\left( {1 + \dfrac{{{\textstyle{1 \over 3}}\left( {15} \right)}}{{100}}} \right)\)
= \(2000{\left( {1 + \dfrac{3}{{20}}} \right)^2}\left( {1 + \dfrac{5}{{100}}} \right)\)
= \(2000{\left( {\dfrac{{23}}{{20}}} \right)^2}\left( {1 + \dfrac{1}{{20}}} \right)\)
= \(2000{\left( {\dfrac{{23}}{{20}}} \right)^2}\left( {\dfrac{{21}}{{20}}} \right)\)
= \(\dfrac{{11109}}{4}\) = \(2777.25\)
So compound interest = 2777.25 - 2000 = 777.25

To find the difference between compound interest and simple interest for two years :

Compound interest for two years = \(p{\left( {1 + \dfrac{R}{{100}}} \right)^2} - p\)
Simple interest for two years = \(p \times 2 \times \dfrac{R}{{100}}\)

=\(\left[ {p{{\left( {1 + \dfrac{R}{{100}}} \right)}^2} - p} \right] - \left( {p \times 2 \times \dfrac{R}{{100}}} \right)\)
= \(p\left[ {{{\left( {1 + \dfrac{R}{{100}}} \right)}^2} - 1 - \left( {\dfrac{{2R}}{{100}}} \right)} \right]\)
= \(p\left[ {1 + \dfrac{{2R}}{{100}} + {{\left( {\dfrac{R}{{100}}} \right)}^2} - 1 - \dfrac{{2R}}{{100}}} \right]\)
= \(p{\left( {\dfrac{R}{{100}}} \right)^2}\)
Hence, difference between compound interest and simple interest @R % p.a. for two years = \(p{\left( {\dfrac{R}{{100}}} \right)^2}\)

To find the difference between compound interest and simple interest for three years :

Use the following formula. 
Difference = \(p{\left( {\dfrac{R}{{100}}} \right)^2}\left( {3 + \dfrac{R}{{100}}} \right)\)

Practice problems


1. Find the amount for Rs. 6000 at 10% per annum, compounded semi-annually for 2 years.
Solution:
Semi annually means we have to compound the interest for every six months.  So we have to calculate interest for 6 months. So R =$\dfrac{{10}}{2}$= 5%  (for half year)
Also, n = 2 years x 2 = 4 periods
Therefore, P = 6000, n = 4, r = 5%
A = 6,000${\left( {1 + \dfrac{5}{{100}}} \right)^4}$  = Rs. 7,293
Interest = Rs.7293 - 6000 = 1293

Alternate method:
Using pascal triangle: the coefficients are 4, 6, 4, 1 for 4 years. 
compound interest shortcut 3
= 4 × (300) + 6 × (15) + 4 × ( .75) + ( ) = 1293
We can stop after using 3 coefficients as the 4th coefficient multiple is too small and may not cause any big difference in our answer.

2. The difference between the CI and  SI on a certain amount at 10% per annum for 2 years, compounded annually is Rs. 372. Find the principal.

Solution:
Let the principal be a.
SI = $\dfrac{{a \times 2 \times 10}}{{100}} = \dfrac{a}{5}$  and CI = Amount – a = a${\left( {1 + \displaystyle\frac{{10}}{{100}}} \right)^2}$ – a = $\displaystyle\frac{{21}}{{100}} \times a$

CI – SI = Rs. 372
 $\displaystyle\frac{{21}}{{100}} \times a - \displaystyle\frac{a}{5}$ = Rs. 372 

a = Rs. 37,200

Alternate method:
We know that the difference in interest comes from second year.  Assume principal is Rs.100 then Interests are calculated as below.
simple and compound interest difference
Now the difference is Rs.1 but actually it was given as Rs.372. If principal is Rs.100 difference is Rs.1, what is the principal if difference has to be Rs.372
$ \Rightarrow \displaystyle\frac{{372}}{1} \times 100 = 37200$

Alternate method:
The above problem has an alternate method. You need to understand the fact that for 1st period, SI = CI.
The difference between the values of CI and SI is because of accumulated interest building on interest which is reinvested. Therefore, for period 2, the difference between CI and SI is the interest for 1 period on the interest of period 1.
In the above example, the difference being 372 is the interest generated on interest for period 1 on the principal.
Interest for period 1 = Rs. 372 × $\displaystyle\frac{{100}}{{10}}$ = Rs. 3,720
Therefore, principal = Rs. 3,720 × $\displaystyle\frac{{100}}{{10}}$ = Rs. 37,200.

3. Find compound interest on Rs. 10000 at 10% p.a. for 4 years, if interest is compounded annually.
Solution:
Amount = \(p{\left( {1 + \dfrac{r}{{100}}} \right)^n}\)
Amount = \(10000 \times {\left( {1 + \dfrac{{10}}{{100}}} \right)^4}\) = \(10000 \times {\left( {\dfrac{{11}}{{10}}} \right)^4}\) = 14641
Therefore, Compound interest = Rs. 14641 - Rs. 10000 = Rs. 4641
Note: Steps for calculation of \({11^4}\)
11 x 11 = 121; 121 x 11 = 1331; 1331 x 11 = 14641

Short cut Method:
Use coefficients of pascal triangle, 4, 6, 4, 1
Successive interests = 10% × 10000 = 1000
10% × 1000 = 100
10% × 100 = 10
10% × 10 = 1
Therefore, Interest = (4 × 1000) + (6 × 100) + (4 × 10) + (1 × 1) = 4641

4. If a certain sum of money invested at a certain rate of compound interest doubles in 5 years. In how many years will it become 4 times?.

Solution:
In compound interest, sum of principal and interests earned equal to principal for next year.  If \(p\) becomes \(2p\) in 5 years, then \(2p\) becomes \(4p\) in another 5 years. So answer is 10 years.
Shortcut:
If a certain sum of amount becomes $a$ times in $t$ years, then time take to become $b$ times = \(t \times \dfrac{{{\mathop{\rm log~b}\nolimits} }}{{{\mathop{\rm log~a}\nolimits} }}\)
Therefore,  \(5 \times \dfrac{{\log 4}}{{\log 2}}\) = \(5 \times \dfrac{{\log {2^2}}}{{\log 2}}\) = \(5 \times \dfrac{{2 \times \log 2}}{{\log 2}}\) = \(10\) years
( \( \because \) \({{\mathop{\rm log~p}\nolimits} ^q} = q \times \log~p\) )

5. If a certain sum of money invested at a certain rate of compound interest doubles in 6 years. In how many years will it become 8 times?
Solution:
If "p" becomes "2p" in 6 years, it will become "4p" in 12 years and "8p" in 18 years. 
Shortcut:
By using the formula \(t \times \dfrac{{{\mathop{\rm log~b}\nolimits} }}{{{\mathop{\rm log~a}\nolimits} }}\) = \(6 \times \dfrac{{{\rm{log}}8}}{{{\rm{log}}2}}\) = \(6 \times \dfrac{{\log {2^3}}}{{\log 2}} = 18\)

6. At what rate per cent of compound interest, a sum of Rs. 2000 will amount to Rs. 2662 in 3 years?
Solution:
Note: To simplify the calculation, we take \(\dfrac{R}{{100}} = r\) and we multiply the answer by 100 in the end. 
We know that, $\left( {{\rm{1 + }}\dfrac{{{{R}}}}{{{\rm{100}}}}} \right)^{{{n}}} {\rm{ = }}\dfrac{{{\rm{Amount}}}}{{{\rm{Principal}}}}$
Let \(\dfrac{R}{{100}} = r\)
Therefore, \({\left( {1 + r} \right)^3} = \dfrac{{2662}}{{2000}}\) = \(\dfrac{{1331}}{{1000}} = {\left( {\frac{{11}}{{10}}} \right)^3}\)

\( \Rightarrow \) \(1 + r = \dfrac{{11}}{{10}}\)
\( \Rightarrow \) \(r = \dfrac{{11}}{{10}} - 1 = \dfrac{1}{{10}}\)

Rate of interest = \(\dfrac{1}{{10}} \times 100 = 10\% \)

7. A man invested Rs. 16000 at compound interest for 3 years, interest compounded annually. If he got Rs. 18522 at the end of 3 years, what is rate of interest?
Solution:
Here, $\left( {{\rm{1 + r}}} \right)^{\rm{3}} {\rm{ = }}\displaystyle\frac{{{\rm{18522}}}}{{{\rm{16000}}}}{\rm{ = }}\displaystyle\frac{{{\rm{9261}}}}{{{\rm{8000}}}}{\rm{ = }}\left( {\displaystyle\frac{{{\rm{21}}}}{{{\rm{20}}}}} \right)^{\rm{3}} {\rm{ = }}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{20}}}}} \right)^{\rm{3}} $
Therefore, Rate of interest = $\displaystyle\frac{{\rm{1}}}{{{\rm{20}}}}$ = 5%

8. A sum of money amounts to Rs. 2880 in 2 years and 3456 in 3 years at compound interest. Find the sum.
Solution:
Given that after two years amount is Rs.2880.  This amount becomes principal for the third year.  So interest earned in third year = 3456 – 2880 = 576
Therefore rate of interest = \(\dfrac{{576}}{{2880}} \times 100 = 20\% \)
\( \Rightarrow Sum \times {\left( {1 + \dfrac{{20}}{{100}}} \right)^2} = 2880\)
\( \Rightarrow Sum \times {\left( {\dfrac{6}{5}} \right)^2} = 2880\)
\( \Rightarrow Sum{\left( {\dfrac{6}{5}} \right)^2} = 2880 \times {\left( {\dfrac{5}{6}} \right)^2}\)
\( \Rightarrow Sum = 2000\)

9. A certain sum is to be divided between A and B so that after 5 years the amount received by A is equal to the amount received by B after 7 years. The rate of interest is 10%, interest compounded annually. Find the ratio of amounts invested by them.
Solution:
Let the sum (principal) received by A and B are m and n respectively.
\( \Rightarrow m{\left( {1 + \dfrac{{10}}{{100}}} \right)^5} = n{\left( {1 + \frac{{10}}{{100}}} \right)^7}\)
\( \Rightarrow m = n{\left( {1 + \dfrac{{10}}{{100}}} \right)^2}\)
\( \Rightarrow \dfrac{m}{n} = {\left( {1 + \dfrac{{10}}{{100}}} \right)^2}\) = \({\left( {\dfrac{{11}}{{10}}} \right)^2}\) = \(\dfrac{{121}}{{100}}\)

10. A father wants to divide Rs. 5100 between his two sons, Mohan and Sohan who are 23 and 24 at present. Divide the amount in such a way that if their shares are invested at compound interest @ 4% p.a. they will receive equal amount on attaining the age of 26 years. Find Mohan's share.
Solution:
Let, Mohan and Sohan receives Rs. m and Rs. n respectively at present.  The amount that they receive 3 years and 2 years after should be equal.
\( \Rightarrow m{\left( {1 + \dfrac{4}{{100}}} \right)^3} = n{\left( {1 + \dfrac{4}{{100}}} \right)^2}\)
\( \Rightarrow m\left( {1 + \dfrac{4}{{100}}} \right) = n\)
\( \Rightarrow m\left( {\dfrac{{26}}{{25}}} \right) = n\)
\( \Rightarrow \dfrac{m}{n} = \dfrac{{25}}{{26}}\)
Therefore, Rs.5100 must be distribued in the ratio 25 : 26
So Mohan's share = \(5100 \times \dfrac{{25}}{{\left( {25 + 26} \right)}} = 2500\)

11. Find the difference between Compound Interest and Simple Interest on Rs. 4000 for 1 year at 10% p.a., if the interest is compounded half-yearly.
Solution:
Simple interest for 1 year = \(\dfrac{{p \times t \times r}}{{100}}\) = \(\dfrac{{4000 \times 1 \times 10}}{{100}} = 400\)
Since, Interest is compounded half-yearly, Rate of interest is halved and time is doubled. Therefore, Rate = $\dfrac{{10}}{2}$% = 5%
And, n = 2 x 1 = 2.
Compound interest = \(p{\left( {1 + \dfrac{r}{{100}}} \right)^n} - p\)
= \(4000{\left( {1 + \dfrac{{5}}{{100}}} \right)^2} - 4000\)
= \(4000{\left( {\dfrac{{21}}{{20}}} \right)^2} - 4000\)
= \(4000\left( {\frac{{441}}{{400}}} \right) - 4000\)
= 410
Therefore, Difference between Compound Interest and Simple Interest = 410 - 400 = Rs.10

12. Find the difference between Compound Interest and Simple Interest on Rs. 10000 for 4 years at 10% p.a., if interest is compounded annually. 
Solution:
Difference between Compound Interest and Simple Interest for 4 years = P${\rm{r}}^{\rm{2}} $ (6 + 4r + ${\rm{r}}^{\rm{2}} $) = 10000 x $\displaystyle\frac{1}{{10}}$ x $\displaystyle\frac{1}{{10}}$ x $\left( {6 + \displaystyle\frac{4}{{10}} + \displaystyle\frac{1}{{100}}} \right)$
= 10000 x $\displaystyle\frac{1}{{100}} \times \displaystyle\frac{{641}}{{100}}$ = Rs. 641

13. If Compound Interest on a certain sum for 2 years at 5% p.a. is Rs.328, the Simple interest will be ?
Solution:
Suppose, Compound Interest for first year = Rs. 100
Then, Compound Interest for second year =100 + 5% (100) =  Rs. 105
Total Compound Interest for two years = (Rs. 100 + Rs. 105) = Rs. 205
And Simple Interest for two years = 2 x Rs. 100 = Rs. 200
If Compound Interest is Rs. 205, Simple Interest = Rs. 200
If compound interest is Rs.1, then simple interest = $\dfrac{{200}}{{205}}$
If Compound Interest is Rs. 328, Simple Interest = Rs.\(328 \times \dfrac{{200}}{{205}}\) = Rs. 320

MCQ's

1. A sum of money becomes Rs.6690 after three years and Rs.10,035 after 6 years on compound interest. The sum is :
a. Rs.4400
b. Rs.4445
c. Rs.4460
d. Rs.4520
Correct Option: C
Explanation:
Let the sum be P.
Then, P ${\left[ {1 + \displaystyle\frac{R}{{100}}} \right]^3} = 6690$.........(i)
and P ${\left[ {1 + \displaystyle\frac{R}{{100}}} \right]^6} = 10,035$ ..... (ii)
Dividing (ii) by (i), we get
${\left( {1 + \displaystyle\frac{R}{{100}}} \right)^3} = \displaystyle\frac{{10035}}{{6690}} = \displaystyle\frac{3}{2}$
P=$\left( {6690 \times \displaystyle\frac{2}{3}} \right)$=Rs.4460

2. Rs.1600 at 10% per annum compound interest compound half-yearly amount to Rs.1944.81  in 
a. 2 years
b. 3 years
c..$1\displaystyle\frac{1}{2}$ years
d. $2\displaystyle\frac{1}{2}$ years
Correct Option: A
Explanation:
As interest is compounded semi-annually, we have to take half of interest for 6 months. And finally we have to half the "n" to get time. 
\(1600{\left( {1 + \dfrac{5}{{100}}} \right)^n} = 1944.81\)
\( \Rightarrow 1600{\left( {\dfrac{{21}}{{20}}} \right)^n} = 1944.81\)
\( \Rightarrow {\left( {\dfrac{{21}}{{20}}} \right)^n} = \dfrac{{1944.81}}{{1600}}\)
\( \Rightarrow {\left( {\dfrac{{21}}{{20}}} \right)^n} = \dfrac{{194481}}{{160000}} = {\left( {\dfrac{{441}}{{400}}} \right)^2}\)
 \( \Rightarrow {\left( {\dfrac{{21}}{{20}}} \right)^n} = {\left( {\dfrac{{{{21}^2}}}{{{{20}^2}}}} \right)^2} = {\left( {\dfrac{{21}}{{20}}} \right)^4}\)
\(\therefore \) \(n = 4\)
As the interest is compounded semi annually, Time = \(\dfrac{n}{2} = \dfrac{4}{2} = 2\)

3. The difference between simple interest and compound interest on a sum for 2 years at 8%, when the interest is compounded annually Rs.16. If the interest was compounded half-yearly, the difference in two interests would be nearly :
a. Rs.16
b. Rs.16.80
c. Rs.21.85
d. Rs.24.64
Correct Option: D
Explanation:
For 1st year, S.I = C.I
Thus, Rs.16 difference is interest on interest
So 8% (SI) = 16  or SI = 200
Now we use simple interest formula to calculate principal. 
 \( \Rightarrow SI = \dfrac{{p \times t \times r}}{{100}}\)
 \( \Rightarrow 200 = \dfrac{{p \times 1 \times 8}}{{100}}\)
 \( \Rightarrow p = 2500\)
To calculate amount for 2 years, compounded half-yearly,we need to take n = 4 and r =4% ( \(\because\) double the time and half the interest )
= \(p{\left( {1 + \dfrac{r}{{100}}} \right)^n} = 2500 \times {\left( {1 + \dfrac{4}{{100}}} \right)^4}\)  = Rs.2924.64
C.I = Rs.424.64
Also, S.I = \( \dfrac{{p \times t \times r}}{{100}} = \dfrac{{2500 \times 8 \times 2}}{{100}}\) = Rs.400
Hence, CI - SI = 424.64 - 400 = Rs.24.64

4. The difference in C.I and S.I for 2 years on a sum of money is Rs.160. If the S.I for 2 years be Rs.2880, the rate percent is :
a. $5\displaystyle\frac{5}{9}$%
b. $12\displaystyle\frac{1}{2}$%
c. $11\displaystyle\frac{1}{9}$%
d. 9%
Correct Option: C
Explanation:
S.I for 1 year = Rs.1440
S.I on Rs.1440 for 1 year = Rs.160
Hence, rate percent = $\left( {\displaystyle\frac{{100 \times 160}}{{1440 \times 1}}} \right)$ = $11\displaystyle\frac{1}{9}\% $

5. The value k of a machine depreciates every year at the rate of 10% on its value at the beginning of that year. If the present value of the machine is Rs.729, its worth 3 years ago was :
a. Rs.947.10
b. Rs.800
c. Rs.1000
d. Rs.750.87
Correct Option: C
Explanation:
The value of the machine after \(n\) years when depreciates at  \(r\% \) = \(p{\left( {1 - \dfrac{r}{{100}}} \right)^n}\)
Here  \(p\) is present value of the machine
\( \Rightarrow p{\left( {1 - \dfrac{{10}}{{100}}} \right)^3} = 729\)
\( \Rightarrow p{\left( {\dfrac{9}{{10}}} \right)^3} = 729\) 
\( \Rightarrow p \times \dfrac{{729}}{{1000}} = 729\)
\( \Rightarrow p = 1000\)

6. The least number of complete years in which a sum of money put out at 20% C.I. will be more than doubled is :
a. 3
b. 4
c. 5
d. 6
Correct Option: B
Explanation:
Let the sum be  \(p\).
\( \Rightarrow p{\left( {1 + \dfrac{{20}}{{100}}} \right)^n} = 2p\)
\( \Rightarrow {\left( {\dfrac{6}{5}} \right)^n} = 2\)
 Here it is difficult to solve.  So to simplify it, we take log on both sides
\( \Rightarrow \log {\left( {\dfrac{6}{5}} \right)^n} = \log 2\)
\( \Rightarrow n\left( {\log 6 - \log 5} \right) = \log 2\)
\( \Rightarrow n = \dfrac{{\log 2}}{{\left( {\log 6 - \log 5} \right)}} = 3.8017\)
So it will take nearly 4 years to become double.
Shortcut Technique:
The number of years to take a certain amount of money to become double is \(t\) at \(r%\) then \(r \times t = 72\)
This technique you can find in any MBA finance text book.  This formula gives you only approximate value. 
So \(t = \dfrac{{72}}{{20}} = 3.6\) years

7. A sum of Rs.550 was taken a loan. This is to be repaid in two equal annual installments. If the rate of interest be 20% compounded annually, then the value of each installment is :
a. Rs.421
b. Rs.396
c. Rs.360
d. Rs.350
Correct Option: C
Explanation:
Let the value of each instalment be Rs.x. Then,
$\displaystyle\frac{x}{{\left( {1 + \displaystyle\frac{{20}}{{100}}} \right)}} + \displaystyle\frac{x}{{{{\left( {1 + \displaystyle\frac{{20}}{{100}}} \right)}^2}}} = 550$
or $\displaystyle\frac{{5x}}{6} + \displaystyle\frac{{25x}}{{36}} = 550$ or x = 360

8. A loan was repaid in two annual instalments of Rs.112 each. If the rate of interest be 10% per annum compounded annually, the sum borrowed was :
a. Rs.200
b. Rs.210
c. Rs.217.80
d. Rs.280
Correct Option: B
Explanation:
Principal = (Present value of Rs.121 due 1 year hence ) + (Present value of Rs.121 due 2 years hence )
= Rs. $\displaystyle\frac{{121}}{{\left( {1 + \displaystyle\frac{{10}}{{100}}} \right)}} + \displaystyle\frac{{121}}{{{{\left( {1 + \displaystyle\frac{{10}}{{100}}} \right)}^2}}}$ =Rs.210

9. A sum amounts to Rs.2916 in 2 years and to Rs.3149.28 in 3 years at compound interest. The sum is :
a. Rs.1500
b. Rs.2000
c. Rs.2500
d. Rs.3000
Correct Option: C
Explanation:
Let P be the principal and R%  per annum be rate.
Then P ${{{\left( {1 + \displaystyle\frac{R}{{100}}} \right)}^3}}$=3149.28 ........ (i)
and P ${{{\left( {1 + \displaystyle\frac{R}{{100}}} \right)}^2}}$=2916  ..........(ii)
On dividing (i) and (ii) we get
${\left( {1 + \displaystyle\frac{R}{{100}}} \right)}$ = $\displaystyle\frac{{3149.28}}{{2916}}$
or ${100 = \displaystyle\frac{{233.28}}{{2916}}}$ or R = $\displaystyle\frac{{233.28}}{{2916}} \times 100 = 8\% $
Now P ${\left( {1 + \displaystyle\frac{8}{{100}}} \right)^2} = 2916$
or P $ \times \displaystyle\frac{{27}}{{25}} \times \displaystyle\frac{{27}}{{25}} = 2916$
or P = $\displaystyle\frac{{2916 \times 25 \times 25}}{{27 \times 27}}$=Rs. 2500

10. A sum of money amounts to Rs.10648 in 3 years and Rs.9680 in 2 years. The rate of interest is :
a. 5% 
b. 10%
c. 15%
d. 20%
Correct Option: B
Explanation:
Let P be the principal and R% annum be the rate. Then.
P ${\left( {1 + \displaystyle\frac{R}{{100}}} \right)^3} = 10648$...(i)
and P ${\left( {1 + \displaystyle\frac{R}{{100}}} \right)^2} = 9680$ ....(ii)
On dividing (i) by (ii), we have
$\left( {1 + \displaystyle\frac{R}{{100}}} \right) = \displaystyle\frac{{10648}}{{9680}}$
or $\displaystyle\frac{R}{{100}} = \displaystyle\frac{{968}}{{9680}} = \displaystyle\frac{1}{{10}}$
or R = $\displaystyle\frac{1}{{10}} \times 100 = 10\% $

11. The difference between simple interest and compound interest at the same rate for Rs.5000 for 2 years is Rs.72. The rate of interest is :
a. 10%
b. 12%
c. 6%
d. 8%
Correct Option: B
Explanation:
$\left[ {5000 \times \left( {1 + {{\displaystyle\frac{R}{{100}}}^2}} \right) - 5000} \right] - \displaystyle\frac{{5000 \times 2 \times R}}{{100}} = 72$
$ \Rightarrow 5000\left[ {\left( {1 + {{\displaystyle\frac{R}{{100}}}^2}} \right) - 1 - \displaystyle\frac{R}{{50}}} \right] = 72$
$ \Rightarrow 1 + \displaystyle\frac{{{R^2}}}{{100}} + \displaystyle\frac{{2R}}{{100}} - 1 - \displaystyle\frac{R}{{50}} = \displaystyle\frac{{72}}{{5000}}$
$ \Rightarrow {R^2} = \left( {\displaystyle\frac{{72}}{{5000}} \times 10000} \right) = 144$ or R = 12%

12. The compound interest on a certain sum of money for 2 years at 10% per annum is Rs.420. The simple interest on the same sum at the same rate and for the same time will be :
a. Rs.350
b. Rs.375
c. Rs.380
d. Rs.400
Correct Option: D
Explanation:
Let principal be P. Then, $P\left( {1 + {{\displaystyle\frac{P}{{100}}}^2}} \right) - P = 420 \Rightarrow P $=Rs.2000
S.I = Rs.$\displaystyle\frac{{2000 \times 2 \times 10}}{{100}}$= Rs.400

13.The difference between the compound interest and simple interest on a certain sum at 5% per annum for 2 years is Rs.1.50. The sum is :
a. Rs.600
b. Rs.500
c. Rs.400
d. Rs.300
Correct Option: A
Explanation:
Let the sum be Rs. 100. Then.
S.I = Rs. $\left( {\displaystyle\frac{{100 \times 5 \times 2}}{{100}}} \right)$ = Rs.10
C.I = Rs.$\left[ {\left\{ {100 \times {{\left( {1 + \displaystyle\frac{5}{{100}}} \right)}^2}} \right\} - 100} \right]$ = Rs. $\displaystyle\frac{{41}}{4}$
Difference between C.I and S.I. = Rs. $\left( {\displaystyle\frac{{41}}{4} - 10} \right)$=Rs.0.25
0.25 : 1.50 : : 100 : x
x = $\displaystyle\frac{{1.50 \times 100}}{{0.25}}$= Rs.600

14. A sum of money placed at C.I doubles itself in 5 years. It will amount to eight times itself in :
a. 15 years
b. 20 years
c. 12 years
d. 10 years
Correct Option: A
Explanation:
Let the principal P and rate be r% . Then, 2P = P ${\left( {1 + \displaystyle\frac{r}{{100}}} \right)^5}$ or
${\left( {1 + \displaystyle\frac{r}{{100}}} \right)^5}$ = 2
Let it be 8 times in t years . Then, 8p = p ${\left( {1 + \displaystyle\frac{r}{{100}}} \right)^t}$
or ${\left( {1 + \displaystyle\frac{r}{{100}}} \right)^t}$=8=${(2)^3}$ = ${\left( {{{\left( {1 + \displaystyle\frac{r}{{100}}} \right)}^5}} \right)^3} = {\left( {1 + \displaystyle\frac{r}{{100}}} \right)^{15}}$
t = 15 years

15. The simple interest on a certain sum for 2 years at 10% per annum is Rs.90. The corresponding compound interest is :
a. Rs.99
b. Rs.95.60
c. Rs.94.50
d. Rs.108
Correct Option: C
Explanation:
Sum = Rs. $\left( {\displaystyle\frac{{100 \times 190}}{{2 \times 10}}} \right)$ = Rs.450

C.I = Rs. $\left[ {450 \times {{\left( {1 + \displaystyle\frac{{10}}{{100}}} \right)}^2} - 450} \right]$ = Rs.94.50

16. What is the principal amount which earns Rs.132 as compound interest for the second year at 10% per annum ?
a. Rs.1000
b. Rs.1200
c. Rs.1320
d. None of these
Correct Option: B
Explanation:
Let x be the principal at the end of first year.  
So interest on first year ending amount = $\displaystyle\frac{{x \times 10 \times 1}}{{100}} = 132 \Rightarrow x = 1320$
Let y be the original principal.  Then one year interest + principal = 1320
So, y + $\displaystyle\frac{{y \times 10 \times 1}}{{100}} = 1320 \Rightarrow y = 1200$

17. A sum amounts to Rs.1352 in 2 years at 4% compound interest. The sum is :
a. Rs.1300
b. Rs.1250
c. Rs.1260
d. Rs.1200
Correct Option: B
Explanation:
Let the sum be P . Then, 1352= P ${\left( {1 + \displaystyle\frac{4}{{100}}} \right)^2}$
$ \Rightarrow 1352 = P \times \displaystyle\frac{{26}}{{25}} \times \displaystyle\frac{{26}}{{25}}$
$ \Rightarrow P = \displaystyle\frac{{1352 \times 25 \times 25}}{{26 \times 26}} = 1250$

18. The compound interest on Rs.30000 at 7% per annum for a certain time is Rs.4347. The time is :
a. 2 years
b. $2\displaystyle\frac{1}{2}$ years
c. 3 years
d. 4 years
Correct Option: A
Explanation:
$30000 \times {\left( {1 + \displaystyle\frac{7}{{100}}} \right)^t} = 30000 + 4347$
or ${\left( {\displaystyle\frac{{107}}{{100}}} \right)^t} = \displaystyle\frac{{34347}}{{30000}} = \displaystyle\frac{{11449}}{{10000}} = {\left( {\displaystyle\frac{{107}}{{100}}} \right)^2}$
Time = 2 years

19. Rs.800 at 5% per annum compound interest will amount to Rs.882 in :
a. 1 year
b. 2 years
c. 3 years
d. 4 years
Correct Option: B
Explanation:
Let time be t years
$882 = 800{\left( {1 + \displaystyle\frac{5}{{100}}} \right)^t} = \displaystyle\frac{{882}}{{800}} = {\left( {\displaystyle\frac{{21}}{{20}}} \right)^t}$
= ${\left( {\displaystyle\frac{{21}}{{20}}} \right)^2} = {\left( {\displaystyle\frac{{21}}{{20}}} \right)^t} \Rightarrow t = 2$
time = 2 years

20. Simple interest on a sum at 4% per annum for 2 years is Rs.80.The compound interest on the same sum for the same period is :
a. Rs.81.60
b. Rs.160
c. Rs.1081.60
d. None of these
Correct Option: A
Explanation:
Principal = Rs. $\left( {\displaystyle\frac{{100 \times 80}}{{4 \times 2}}} \right)$ = Rs.1000
C.I = Rs. $\left[ {\left\{ {1000 \times {{\left( {1 + \displaystyle\frac{4}{{100}}} \right)}^2} - 1000} \right\}} \right]$ = Rs.81.60

21. The difference of compound interest on Rs.800 for 1 year at 20% per annum when compounded half-yearly and quarterly is :
a. Nil
b. Rs.2.50
c. Rs.4.40
d. Rs.6.60
Correct Option: C
Explanation:
C.I when reckoned half-yearly
= Rs. $\left[ {800 \times {{\left( {1 + \displaystyle\frac{{10}}{{100}}} \right)}^4} - 800} \right]$ = Rs. 172.40
Difference = Rs.(172.40-168) Rs.4.40

22. The difference between simple interest and the compound interest on Rs.600 for 1 year at 10% per annum, reckoned half-yearly is :
a. Nil
b. Rs.6.60
c. Rs.4.40
d. Rs.1.50
Correct Option: D
Explanation:
S.I = Rs. $\left( {\displaystyle\frac{{600 \times 10 \times 1}}{{100}}} \right)$ = Rs.60
C.I = Rs. $\left[ {600 \times {{\left( {1 + \displaystyle\frac{5}{{100}}} \right)}^2} - 600} \right]$ = Rs.61.50
Difference = Rs.(61.50-60) = Rs.1.50

23. The compound interest of Rs.20480 at $6\displaystyle\frac{1}{4}$% per annum for 2 years 73 days is :
a. Rs.3000
b. Rs.3131
c. Rs.2929
d. Rs.3636
Correct Option: C
Explanation:
73 days is 1/5 th of an year.  So interest would be 1/5th of the rate of interest for the 73 days.
A = \({20480 \times {{\left( {1 + \dfrac{{{\textstyle{{25} \over 4}}}}{{100}}} \right)}^2}\left( {1 + \dfrac{{\left( {{\textstyle{1 \over 5}} \times {\textstyle{{25} \over 4}}} \right)}}{{100}}} \right)}\)
= \({20480 \times {{\left( {1 + \dfrac{1}{{16}}} \right)}^2}\left( {1 + \dfrac{1}{{80}}} \right)}\)
= \({20480 \times {{\left( {\dfrac{{17}}{{16}}} \right)}^2}\left( {\dfrac{{81}}{{80}}} \right)}\)
= 23409
Compound interest = 23409 - 20480 = 2929

24. The compound interest on Rs.2800 for $1\displaystyle\frac{1}{2}$years at 10% per annum is :
a. Rs.441.35
b. Rs.436.75
c. Rs.434
d. Rs.420
Correct Option: C
Explanation:
Amount = Rs.$\left[ {\left[ {2800 \times \left( {1 + \displaystyle\frac{{10}}{{100}}} \right)} \right]\left( {1 + \displaystyle\frac{5}{{100}}} \right)} \right]$
= Rs.$\left[ {2800 \times \displaystyle\frac{{11}}{{100}} \times \displaystyle\frac{{21}}{{20}}} \right]$ = Rs.3234
C.I = Rs.(3234-2800)= Rs.434

25. If Rs.7500 are borrowed at C.I at the rate of 4% per annum, then after 2 years the amount to be paid is :
a. Rs.8082
b. Rs.7800
c. Rs.8100
d. Rs.8112
Correct Option: D
Explanation:
Amount = Rs. $\left[ {7500{{\left( {1 + \displaystyle\frac{4}{{100}}} \right)}^2}} \right]$ = Rs. $\left[ {7500 \times \displaystyle\frac{{26}}{{25}} \times \displaystyle\frac{{26}}{{25}}} \right]$=Rs.8112.