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?
a. 5%
b. 6%
c. 7%
d. 8%

8. A sum of money amounts to Rs. 2880 in 2 years and 3456 in 3 years at compound interest. Find the sum.
a. Rs.2000
b. Rs.2200
c. Rs.2255
d. Rs.2400

Answer: A

Explanation:
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.
a. 100 : 121
b. 121 : 100
c. 110 : 121
d. 110 : 131

Answer: B

Explanation:
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.
a. Rs.2460
b. Rs.2600
c. Rs.2500
d. Rs.2720

Answer: C

Explanation:
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.
a. Rs.40
b. Rs.35
c. Rs.25
d. Rs.10

Answer: D

Explanation:
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.
a. Rs.541
b. RS.578
c. Rs.641
d. Rs.666

Answer: C

Explanation:
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