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Quantitative Aptitude > Interest

SIMPLE INTEREST MCQs

Total Questions : 234 | Page 16 of 24 pages
Question 151. A money lender claims to lend money at the rate of 10% per annum simple interest. However, he takes the interest in advance when he lends a sum for one year. At what interest rate does he lend the money actually ?
  1.    10%
  2.    $$10\frac{1}{9}$$ %
  3.    11%
  4.    $$11\frac{1}{9}$$ %
 Discuss Question
Answer: Option D. -> $$11\frac{1}{9}$$ %
$$\eqalign{
& {\text{Amount}} \to {\text{10}} \cr
& \,\,\,\,\, \downarrow \cr
& \,\,\,\,\,90 \to \frac{{10}}{{90}} \times 100 \cr
& \,\,\,\,\,\,\,\,\,\, = 11\frac{1}{9}\% \cr} $$
Question 152. A sum of Rs. 1550 was lent partly at 5% and partly at 8% p.a. simple interest. The total interest received after 3 years was Rs. 300. The ratio of the money lent at 5% to that lent at 8% is:
  1.    5 : 8
  2.    8 : 5
  3.    16 : 15
  4.    31 : 6
 Discuss Question
Answer: Option C. -> 16 : 15
Let the sum lent at 5% be Rs. x and that lent 8% be Rs. (1550 - x).
Then,
$$\eqalign{
& \left( {\frac{{x \times 5 \times 3}}{{100}}} \right) + \left[ {\frac{{\left( {1550 - x} \right) \times 8 \times 3}}{{100}}} \right] = 300 \cr
& \Leftrightarrow 15x - 24x + \left( {1550 \times 24} \right) = 30000 \cr
& \Leftrightarrow 9x = 7200 \cr
& \Leftrightarrow x = 800. \cr
& \therefore {\text{Required ratio}} = 800:750 \cr
& = 16:15 \cr} $$
Question 153. An amount of Rs. 1,00,000 is invested in two types of shares. The first yields an interest of 9% p.a. and second, 11% p.a. If the total interest at the end of one year is $$9\frac{3}{4}$$ %, then the amount invested in each share was -
  1.    Rs. 52, 500; Rs. 47, 500
  2.    Rs. 62, 500; Rs. 37, 500
  3.    Rs. 72, 500; Rs. 27, 500
  4.    Rs. 82, 500; Rs. 17, 500
 Discuss Question
Answer: Option B. -> Rs. 62, 500; Rs. 37, 500
Let the sum invested at 9% be Rs. x and that invested at 11% be Rs. (100000 - x).
Then,
$$\eqalign{
& = \left( {\frac{{x \times 9 \times 1}}{{100}}} \right) + \left[ {\frac{{\left( {100000 - x} \right) \times 11 \times 1}}{{100}}} \right] \cr
& = \left( {100000 \times \frac{{39}}{4} \times \frac{1}{{100}}} \right) \cr
& \Leftrightarrow \frac{{9x + 1100000 - 11x}}{{100}} = \frac{{39000}}{4} = 9750 \cr
& \Leftrightarrow 2x = \left( {1100000 - 975000} \right) = 125000 \cr
& \Leftrightarrow x = 62500 \cr
& \therefore {\text{Sum invested at 9}}\% \cr
& = {\text{Rs}}{\text{. }}62500 \cr
& {\text{Sum invested at 11% }} \cr
& {\text{ = Rs}}{\text{.}}\left( {100000 - 62500} \right) \cr
& = {\text{Rs}}{\text{. }}37500 \cr} $$
Question 154. A sum of Rs. 1440 is lent out in three parts in such away that the interests on first part at 2% for 3 years, second part at 3% for 4 years and third part at 4% for 5 years are equal. Then the difference between the largest and the smallest sum is -
  1.    Rs. 200
  2.    Rs. 400
  3.    Rs. 460
  4.    Rs. 560
 Discuss Question
Answer: Option D. -> Rs. 560
Let the parts be Rs. x, Rs. y and Rs. [1440 - (x + y)].
Then,
$$\eqalign{
& = \frac{{x \times 2 \times 3}}{{100}} = \frac{{y \times 3 \times 4}}{{100}} \cr
& = \frac{{\left[ {1440 - \left( {x + y} \right)} \right] \times 4 \times 5}}{{100}} \cr
& \therefore 6x = 12y\,or\,x = 2y. \cr
& So,\frac{{x \times 2 \times 3}}{{100}} = \frac{{\left[ {1440 - \left( {x + y} \right)} \right] \times 4 \times 5}}{{100}} \cr
& \Leftrightarrow 12y = \left( {1440 - 3y} \right) \times 20 \cr
& \Rightarrow 72y = 28800 \cr
& \Rightarrow y = 400 \cr
& {\text{First part}} = x = 2y = {\text{Rs}}{\text{. }}800, \cr
& {\text{Second part}} = {\text{Rs}}{\text{. }}400 \cr
& {\text{Third part}} \cr
& = {\text{Rs}}.\left[ {1440 - \left( {800 + 400} \right)} \right] \cr
& = {\text{Rs}}{\text{. }}240. \cr
& \therefore {\text{Required difference}} \cr
& {\text{ = Rs}}{\text{.}}\left( {800 - 240} \right) \cr
& = {\text{Rs}}.560 \cr} $$
Question 155. The simple interest at x% for x years will be Rs. x on a sum of:
  1.    Rs. x
  2.    Rs. $$\frac{{100}}{x}$$
  3.    Rs. 100x
  4.    Rs. $$\frac{{100}}{{{x^2}}}$$
 Discuss Question
Answer: Option B. -> Rs. $$\frac{{100}}{x}$$
$$\eqalign{
& {\text{Sum}} = {\frac{{100 \times S.I.}}{{R \times T}}} \cr
& = {\text{Rs}}{\text{.}}\,\, {\frac{{100 \times x}}{{x \times x}}} \cr
& = {\text{Rs}}{\text{.}}\,\, {\frac{{100}}{x}} \cr} $$
Question 156. Rs. 6200 amounts to Rs. 9176 in 4 years at simple interest. If the interest rate is increased by 3% it would amount to how much?
  1.    Rs. 8432
  2.    Rs. 9820
  3.    Rs. 9920
  4.    Rs. 10920
 Discuss Question
Answer: Option C. -> Rs. 9920
$$\eqalign{
& {\text{P}} = {\text{Rs}}{\text{. 6200}} \cr
& {\text{S}}{\text{.I}}{\text{.}} \cr
& = {\text{Rs}}{\text{.}}\left( {{\text{9176}} - {\text{6200}}} \right) \cr
& = {\text{Rs}}{\text{. }}2976 \cr
& {\text{T}} = 4\,{\text{years}} \cr
& \therefore {\text{Rate}} \cr
& = \left( {\frac{{100 \times 2976}}{{6200 \times 4}}} \right)\% \cr
& = 12\% \cr
& {\text{New }}{\text{rate}} \cr
& = \left( {12 + 3} \right)\% \cr
& = 15\% \cr
& {\text{New}}\,{\text{S}}{\text{.I}}{\text{.}} \cr
& = {\text{Rs}}{\text{.}}\left( {\frac{{{\text{6200}} \times 15 \times 4}}{{100}}} \right) \cr
& = \text{Rs.} \,3720 \cr
& {\text{New}}\,{\text{amount}} \cr
& = {\text{Rs}}{\text{.}}\left( {{\text{6200}} + 3720} \right) \cr
& = {\text{Rs}}{\text{.}}\,9920 \cr} $$
Question 157. The simple interest on a certain sum of money at the rate of 5% p.a. for 8 years is Rs. 840. At what rate of interest the same account of interest can be received on the same sum after 5 years?
  1.    6%
  2.    8%
  3.    9%
  4.    10%
 Discuss Question
Answer: Option B. -> 8%
$$\eqalign{
& {\text{S}}{\text{.I}}{\text{.}} = {\text{Rs}}{\text{. 840}} \cr
& {\text{R}} = {\text{5}}\% \cr
& {\text{T}} = 8\,{\text{years}} \cr
& {\text{Principal}} \cr
& = {\text{Rs}}{\text{. }}\left( {\frac{{100 \times 840}}{{5 \times 8}}} \right) \cr
& = {\text{Rs}}{\text{.}}\,2100 \cr
& {\text{Now,}} \cr
& {\text{P}} = {\text{Rs}}{\text{.}}\,2100 \cr
& {\text{S}}{\text{.I}}{\text{.}} = {\text{Rs}}{\text{.}}\,{\text{840}} \cr
& {\text{T}} = {\text{5 years}}{\text{}} \cr
& \therefore {\text{Rate}} \cr
& = \left( {\frac{{100 \times 840}}{{2100 \times 5}}} \right)\% \cr
& = {\text{8}}\% \cr} $$
Question 158. The simple interest accrued on a certain principal in 5 years at the rate of 12 p.c.p.a. is Rs.1536. what amount of simple interest would one get if one invests Rs.1000 more than the previous principal for 2 years and at the same rate p.c.p.a. ?
  1.    Rs. 614.40
  2.    Rs. 845.40
  3.    Rs. 1536
  4.    Rs. 2136
  5.    None of these
 Discuss Question
Answer: Option E. -> None of these
$$\eqalign{
& {\text{Sum}} = {\text{Rs}}{\text{.}}\left( {\frac{{100 \times 1536}}{{12 \times 5}}} \right) \cr
& = {\text{Rs}}{\text{. }}2560 \cr
& Now, \cr
& P = {\text{Rs}}{\text{.}}\left( {2560 + 1000} \right) \cr
& \,\,\,\,\,\, = {\text{Rs}}{\text{. }}3560 \cr
& {\text{T}} = {\text{2years}} \cr
& {\text{R}} = {\text{12}}\% \cr
& \therefore S.I. \cr
& = {\text{Rs}}{\text{.}}\left( {\frac{{3560 \times 12 \times 2}}{{100}}} \right) \cr
& = {\text{Rs}}{\text{.}}\,854.40 \cr} $$
Question 159. Rs. 6000 becomes Rs. 7200 in 4 years. If the rate becomes 1.5 times of itself, the amount of the same principal in 5 years will be = ?
  1.    Rs. 8000
  2.    Rs. 8250
  3.    Rs. 9250
  4.    Rs. 9000
 Discuss Question
Answer: Option B. -> Rs. 8250
$$\eqalign{
& {\text{Principal}}\,\,\,\,\,\,\,{\text{Amount}} \cr
& \underbrace {\,\,\,\,\,{\text{6000}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{7200}}\,\,\,\,\,}_{ + 1200} \cr
& {\text{By using formula,}} \cr
& {\text{Rate }}\% \cr
& = \frac{{1200}}{{6000}} \times \frac{{100}}{4} \cr
& {\text{ = 5}}\% \cr
& {\text{New rate}}\% \cr
& = {\text{5}} \times \frac{3}{2} = {\text{7}}{\text{.5}}\% \cr
& {\text{Interest after 5 years}} \cr
& = \frac{{6000 \times 7.5 \times 5}}{{100}} \cr
& {\text{ = Rs}}{\text{. 2250}} \cr
& {\text{Hence,}} \cr
& {\text{amount }} \cr
& {\text{ = Rs}}{\text{. }}\left( {6000 + 2250} \right) \cr
& {\text{ = Rs 8250}} \cr} $$
Question 160. Simple interest on Rs. 500 for 4 years at 6.25% per annum is equal to the simple interest on Rs. 400 at 5% per annum for a certain period of time. The period of time is = ?
  1.    4 years
  2.    5 years
  3.    $$6\frac{1}{4}$$ years
  4.    $$8\frac{2}{3}$$ years
 Discuss Question
Answer: Option C. -> $$6\frac{1}{4}$$ years
$$\eqalign{
& {\text{Let the required time = t years }} \cr
& {\text{According to the question,}} \cr
& \Leftrightarrow \frac{{500 \times 4 \times 6.25}}{{100}} = \frac{{400 \times 5 \times {\text{t}}}}{{100}} \cr
& \Leftrightarrow 5 \times 4 \times 625 = 400 \times 5 \times {\text{t}} \cr
& \Leftrightarrow {\text{t = }}\frac{{625}}{{100}} = \frac{{25}}{4} \cr
& \Leftrightarrow {\text{t}} = 6\frac{1}{4}{\text{years}} \cr} $$

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