## Simple Interest Questions and Answers Part-1

1. What is the simple interest on a sum of Rs. 700 if the rate of interest for the first 3 years is 8% per annum and for the last 2 years is 7.5% per annum?
a) Rs. 280
b) Rs. 269.5
c) Rs. 273
d) Rs. 283

Explanation:
\eqalign{ & {1^{{\text{st}}}}{\kern 1pt} {\text{case}}: \cr & {I_1} = \frac{{700 \times 3 \times 8}}{{100}} = {\text{Rs}}{\text{. }}168 \cr & {2^{{\text{nd}}}}{\kern 1pt} {\text{case}}: \cr & {I_2} = \frac{{700 \times 2 \times 7.5}}{{100}} = {\text{Rs}}{\text{. }}105 \cr}
Then total interest for five years
$$= \left( {{I_1} + {I_2}} \right) = {\text{Rs}}{\text{. }}273$$

2. The difference between simple and compound interest on a sum of money at 20% per annum for 3 years is Rs. 48. What is the sum?
a) Rs. 550
b) Rs. 500
c) Rs. 375
d) Rs. 400

Explanation:
\eqalign{ & {\text{Let}}\,{\text{sum}}\,{\text{is}}\,P. \cr & {\text{The}}\,{\text{difference}}\,{\text{between}}\,{\text{compound}}\,{\text{interest}}\,{\text{and}} \cr & \,{\text{simple}}\,{\text{interest}}\,{\text{over}}\,{\text{three}}\,{\text{years}}\,{\text{is}}\,{\text{given}}\,{\text{by}} \cr & = P\left( {\frac{r}{{100}}} \right)2 \times \left\{ {\left( {\frac{r}{{100}}} \right) + 3} \right\} \cr & 48 = P \times \left( {\frac{{20}}{{100}}} \right)2 \times \left\{ {\left( {\frac{{20}}{{100}}} \right) + 3} \right\} \cr & 48 = P \times \frac{4}{{100}} \times \frac{{16}}{5} \cr & 48 = P \times \frac{{64}}{{500}} \cr & \,64P = 48 \times 500 \cr & \,P = Rs.\,375 \cr}

3. What will be the simple interest on Rs. 700 at 9% per annum for the period from February 5, 1994 to April 18, 1994?
a) Rs. 13
b) Rs. 12.60
c) Rs. 11.30
d) Rs. 15

Explanation:
\eqalign{ & {\text{Here,}}\,{\text{time}}\,{\text{interval}}\,{\text{is}}\,{\text{given}}\,{\text{as}}\, \cr & {\text{February}}\,5,\,1994\,{\text{to}}\,{\text{April}}\,18,\,1994 \cr & = 73\,{\text{days}} = \frac{{73}}{{365}} = 0.2\,{\text{years}}. \cr & {\text{Now}}\,{\text{interest}} = \frac{{PTR}}{{100}} \cr & = \frac{{ {700 \times 9 \times 0.2} }}{{100}} \cr & = Rs.\,12.60 \cr}

4. In what time will the simple interest on Rs. 1750 at 9% per annum be the same as that on Rs. 2500 at 10.5% per annum in 4 years?
a) 6 years and 8 months
b) 7 years and 3 months
c) 6 years
d) 7 years and 6 months

Explanation:
\eqalign{ & {\text{Let}}\,{\text{time}}\,{\text{is}}\,T\,{\text{years}}. \cr & {\text{According}}\,{\text{to}}\,{\text{question}}, \cr & \frac{{1750 \times 9 \times T}}{{100}} = \frac{{\left( {2500 \times 10.5 \times 4} \right)}}{{100}} \cr & \,T = \frac{{ {2500 \times 10.5 \times 4} }}{{1750 \times 9}} \cr & \,T = 6.66 = 6\,{\text{years}}\,{\text{and}}\,8\,{\text{months}} \cr}

5. A sum of money becomes $$\frac{7}{4}$$ of itself in 6 years at a certain rate of simple interest. Find the rate of interest.
a) 12%
b) $$12\frac{1}{2}$$ %
c) 14%
d) 8%

Explanation:
\eqalign{ & {\text{Let}}\,{\text{sum}}\,is\,{\text{Rs}}.\,100, \cr & {\text{Then}}\,{\text{it}}\,{\text{become}}\,\frac{7}{4}\,{\text{times}} \cr & i.e.\,{\text{Rs}}.\,\frac{{700}}{4}\,{\text{in}}\,6\,{\text{years}}. \cr & {\text{Interest}} = {\frac{{700}}{4}} - 100 = Rs.\,\frac{{300}}{4} \cr & {\text{Rate}} = \frac{{{\text{Total}}\,{\text{interest}}}}{{{\text{Given}}\,{\text{time}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{300}}{4} \times 6 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 12 {\frac{1}{2}} \% \cr}

6. What is the rate of simple interest for the first 4 years if the sum of Rs. 360 becomes Rs. 540 in 9 years and the rate of interest for the last 5 years is 6%?
a) 3%
b) 4%
c) 6%
d) 5%

Explanation:
\eqalign{ & {\text{Interest}}\,{\text{for}}\,{\text{the}}\,{\text{last}}\,{\text{5}}\,{\text{years}} \cr & = \frac{{PTR}}{{100}} \cr & = \frac{{360 \times 5 \times 6}}{{100}} = Rs.\,108 \cr & {\text{Interest}}\,{\text{for}}\,{\text{year}} = 540 - 360 = 180 \cr & {\text{So,}}\,{\text{interest}}\,{\text{for}}\,{\text{first}}\,{\text{four}}\,{\text{years}} \cr & = 180 - 108 = Rs.\,72 \cr & {\text{Now,}}\,{\text{rate}}\,{\text{for}}\,{\text{first}}\,{\text{four}}\,{\text{years}} \cr & = \frac{{ {72 \times 100} }}{{360 \times 4}} \cr & = 5\% \cr}

7. Find the rate of interest if the amount after 2 years of simple interest on a capital of Rs. 1200 is Rs. 1440.
a) 8%
b) 9%
c) 10%
d) 11%

Explanation:
\eqalign{ & {\text{Amount}},\,A = Rs.\,1440 \cr & {\text{Principal}},\,P = Rs.\,1200 \cr & {\text{Interest}},\,I = Rs.\,\left( {1440 - 1200} \right) = 240 \cr & R = \frac{{ {240 \times 100} }}{{ {1200 \times 2} }} = 10\% \cr & \cr }

8. A sum was invested at simple interest at a certain interest for 2 years. It would have fetched Rs. 60 more had it been invested at 2% higher rate. What was the sum?
a) Rs. 1500
b) Rs. 1300
c) Rs. 2500
d) Rs. 1000

Explanation: Let the rate be R at which Principal P is invested for 2 years.
According to question,
{Interest at Rate (R + 2)}% - (interest at rate R%) = Rs. 60
$$\frac{{\left( {P \times 2 \times \left( {R + 2} \right)} \right)}}{{100}} -$$     $$\frac{{\left( {P \times 2 \times R} \right)}}{{100}}$$   $$= 60$$
\eqalign{ & \frac{{ {2PR + 4P - 2PR} }}{{100}} = 60 \cr & 4P = 60 \times 100 \cr & P = \frac{{60 \times 100}}{4} \cr & P = {\text{Rs}}{\text{.}}\,1500 \cr}

9. If a sum of Rs. 13040 is to be repaid in two equal installments at $$3\frac{3}{4}$$ % per annum, what is the amount of each installment?
a) 7045
b) 8000
c) 65067
d) 6889

\eqalign{ & {\text{Let}}\,{\text{each}}\,{\text{installment}}\,{\text{be}}\,P \cr & {\text{Hence}}, \cr & {\frac{x}{{ {\left( {\frac{{100}}{{100 + r}}} \right) + {{\left( {\frac{{100}}{{100 + r}}} \right)}^2}} }}} \cr & \,\frac{x}{{ {1 + {\frac{{15}}{{400}}} } }} + \frac{x}{{ {1 + {{\left( {\frac{{15}}{{400}}} \right)}^2}} }} = Rs.\,13040 \cr & x = Rs.\,6889 \cr}
d) $$1\frac{1}{2}$$ years
\eqalign{ & P = Rs.\,3300 \cr & A = Rs.\,3399 \cr & R = 6\% \,{\text{per}}\,{\text{annum}} \cr & {\text{Let}}\,{\text{the}}\,{\text{time}}\,{\text{be}}\,{\text{n}}\,{\text{years}}{\text{.}} \cr & {\text{Compound}}\,{\text{interest}}\,{\text{is}}\,{\text{taken}}\,{\text{half - yearly}}. \cr & A = P \times {\left[ {1 + \left( {\frac{R}{2} \times 100} \right)} \right]^{2n}} \cr & 3399 = 3300{\left( {1 + \frac{3}{{100}}} \right)^{2n}} \cr & {\left( {1.03} \right)^{2n}} = \frac{{3399}}{{3300}} \cr & {\left( {1.03} \right)^{2n}} = {\left( {1.03} \right)^1} \cr & Thus,\,2n = 1\,year \cr & n = \frac{1}{2}{\text{year}} = 6\,{\text{months}} \cr}