1. What will be the simple interest earned on an amount of Rs. 16,800 in 9 months at the rate of $$6\frac{1}{4}$$ % p.a. ?

a) Rs. 787.50

b) Rs. 812.50

c) Rs. 860

d) Rs. 887.50

Explanation:

$$\eqalign{ & {\text{P}} = {\text{Rs}}{\text{. }}16800 \cr & {\text{R}} = 6\frac{1}{4}\% = \frac{{25}}{4}\% \cr & {\text{T}} = 9\,{\text{months}} = \frac{3}{4}{\text{yr}}{\text{.}} \cr & {\text{S}}{\text{.I}}{\text{.}} = {\text{Rs}}{\text{.}}\left( {16800 \times \frac{{25}}{4} \times \frac{3}{4} \times \frac{1}{{100}}} \right) \cr & = {\text{Rs}}{\text{.}}\,787.50 \cr} $$

2. The simple interest on a sum of money is $$\frac{4}{9}$$ of the principal and the number of years is equal to the rate percent per annum. The rate per annum is =

a) 5%

b) $$6\frac{2}{3}$$ %

c) 6%

d) $$7\frac{1}{5}$$ %

Explanation:

$$\eqalign{ & {\text{Let principal}} \cr & {\text{ = 9 units }} \cr & {\text{Hence simple interest}} \cr & {\text{ = }}\frac{4}{9} \times {\text{9 = 4 units}} \cr & {\text{Let, Rate of interest = R% }} \cr & {\text{R = T (given)}} \cr & {\text{Using formula}} \cr & {{\text{SI = }}\frac{{{\text{P}} \times {\text{T}} \times {\text{R}}}}{{100}}} \cr & 4 = \frac{{9 \times {\text{R}} \times {\text{R}}}}{{100}} \cr & {{\text{R}}^2} = \frac{{400}}{9} \cr & {\text{R = }}\frac{{20}}{3} = 6\frac{2}{3}\% \cr} $$

3. At what rate percent per annum will the simple interest on a sum of money be $$\frac{2}{5}$$ of the principal amount in 10 years ?

a) 4%

b) 6%

c) $$5\frac{2}{3}$$ %

d) $$6\frac{2}{3}$$ %

Explanation:

$$\eqalign{ & {\text{Let principal = 5 units}} \cr & {\text{Hence interest}} \cr & {\text{ = 5}} \times \frac{2}{5} \cr & {\text{ = 2 units}} \cr & {\text{Time = 10 years}} \cr & {\text{Using formula, }} \cr & {\text{Rate% = }}\frac{2}{5} \times \frac{{100}}{{10}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = 4% }} \cr} $$

4. A sum of Rs. 1750 is divided into two parts such that the interests on the first part at 8% simple interest per annum and that on the other part at 6% simple interest per annum are equal. The interest on each part ( in Rupees) is ?

a) Rs. 60

b) Rs. 65

c) Rs. 70

d) Rs. 40

Explanation:

$$\eqalign{ & {\text{Principal = Rs}}{\text{. 1750}} \cr & {\text{Let the first part = }}x \cr & {\text{Hence second part}} \cr & {\text{ = }}\left( {1750 - x} \right) \cr & \Rightarrow x \times \frac{8}{{100}} \times 1 = \left( {1750 - x} \right) \times \frac{6}{{100}} \times 1 \cr & 4x = 5250 - 3x \cr & 7x = 5250 \cr & x = 750 \cr & {\text{First part = Rs}}{\text{. 750}} \cr & {\text{ Second part}} \cr & {\text{ = Rs}}{\text{. }}\left( {1750 - 750} \right) \cr & {\text{ = Rs}}{\text{. 1000}} \cr & {\text{Required interest}} \cr & {\text{ = 750}} \times \frac{8}{{100}} \cr & {\text{ = Rs}}{\text{. 60}} \cr} $$

5. A person borrows Rs. 5000 for 2 year at 4% p.a. simple interest. He immediately lends it to another person at $$6\frac{1}{4}$$ % p.a. for 2 years. Find his gain in the transaction per year.

a) Rs. 112.50

b) Rs. 125

c) Rs. 150

d) Rs. 167.50

Explanation:

$$\eqalign{ & {\text{Gain in 2 years}} \cr & = {\left( {5000 \times \frac{{25}}{4} \times \frac{2}{{100}}} \right) - \left( {\frac{{5000 \times 4 \times 2}}{{100}}} \right)} \cr & = {\text{Rs}}{\text{.}}\left( {625 - 400} \right) \cr & = {\text{Rs}}{\text{. }}225 \cr & {\text{Gain 1 year}} = {\text{Rs}}{\text{.}}\left( {\frac{{225}}{2}} \right) \cr & = {\text{Rs}}{\text{. }}112.50 \cr} $$

6. A sum of Rs. 210 was taken as a loan. This is to be paid back in two equal installments. If the rate of the interest be 10% compounded annually, then the value of each installment is:

a) Rs. 121

b) Rs. 127

c) Rs. 210

d) Rs. 225

Explanation: Let X = equal installment at the end of one year( rate% annually) .

Now 1st year,

P =210,

Interest = $$\frac{{{\text{PTR}}}}{{100}}$$ = 210 * 0.1 = 21.

Let X is to be paid as an equal installment.

At the beginning of 2nd year,

P = 210 + 21 - X,

Interest at the end of 2nd year,

= (231 - X) * 0.1 = 23.1 - 0.1X.

Total installment,

2X = 210 + 21 + 23.1 - 0.1X,

X = $$\frac{{{\text{254}}{\text{.1}}}}{{2.1}}$$ = 121.

7. There is a decrease of 10% yearly on an article. If this article was bought 3 years ago and present cost is Rs. 5,832 then what was the cost of article at buying time?

a) Rs. 7,200

b) Rs. 7,862

c) Rs. 8,000

d) Rs. 8,500

Explanation: $$A = P {\left( {1 - \frac{R}{{100}}} \right)^n}$$

Where A = Value of goods after n years

P = Initial Price

R = Rate of depriciation

$$\eqalign{ & P = \frac{{5832}}{{{{\left( {1 - \frac{{10}}{{100}}} \right)}^3}}} \cr & P = \frac{{5832}}{{{{\left( {1 - \frac{1}{{10}}} \right)}^3}}} \cr & P = \frac{{5832}}{{{{\left( {\frac{9}{{10}}} \right)}^3}}} \cr & P = 5832 \times \frac{{10}}{9} \times \frac{{10}}{9} \times \frac{{10}}{9} \cr & P = 8000 \cr} $$

8. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

a) Rs. 650

b) Rs. 690

c) Rs. 698

d) Rs. 700

Explanation: S.I. for 1 year = Rs. (854 - 815) = Rs. 39

S.I. for 3 years = Rs.(39 x 3) = Rs. 117

Principal = Rs. (815 - 117) = Rs. 698

9. Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?

a) Rs. 6400

b) Rs. 6500

c) Rs. 7200

d) Rs. 7500

Explanation: Let the sum invested in Scheme A be Rs.

*x*and that in Scheme B be Rs. (13900 -

*x*)

$$ {\frac{{x \times 14 \times 2}}{{100}}} + $$ $$ {\frac{{\left( {13900 - x} \right) \times 11 \times 2}}{{100}}} $$ $$ = 3508$$

28x - 22x = 350800 - (13900 x 22)

6x = 45000

x = 7500

Sum invested in Scheme B

= Rs. (13900 - 7500)

= Rs. 6400

10. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?

a) Rs. 4462.50

b) Rs. 8032.50

c) Rs. 8900

d) Rs. 8925

Explanation:

$$\eqalign{ & {\text{Principal}} = Rs.\,\left( {\frac{{100 \times 4016.25}}{{9 \times 5}}} \right) \cr & = Rs.\,\left( {\frac{{401625}}{{45}}} \right) \cr & = Rs.\,8925 \cr} $$