13. If Compound Interest on a certain sum for 2 years at 5% p.a. is Rs.328, the Simple interest will be ?
a. Rs.320
b. Rs.340
c. Rs.360
d. Rs.380

Answer: A

Explanation:
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

14. 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

Answer: 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

15. 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

Answer: 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 = \(\dfrac{n}{2} = \dfrac{4}{2} = 2\)

16. 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

Answer: 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

17. 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%
d. Data inadequate

Answer: 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}\% $

18. The value 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

Answer: 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.
\( \Rightarrow p{\left( {1 - \dfrac{{10}}{{100}}} \right)^3} = 729\)
\( \Rightarrow p{\left( {\dfrac{9}{{10}}} \right)^3} = 729\)
\( \Rightarrow p \times \dfrac{{729}}{{1000}} = 729\)
\( \Rightarrow p = 1000\)