Quantitative Aptitude > Interest
COMPOUND INTEREST MCQs
Total Questions : 262
| Page 16 of 27 pages
Answer: Option B. -> Rs. 675
$$\eqalign{
& {\text{Amount = Rs}}{\text{. 2916}} \cr
& {\text{Time = 2 years }} \cr
& {\text{Rate = 8}}\% \cr
& {\text{Effective rate }}\% {\text{ CI for 2 years}} \cr
& {\text{ = 8 + 8 + }}\frac{{8 \times 8}}{{100}} = 16.64\% \cr
& {\text{Required sum}} \cr
& {\text{ = }}\frac{{2916}}{{\left( {100 + 16.64} \right)}} \times 100 \cr
& = {\text{Rs}}{\text{. }}2500 \cr
& {\text{Required simple interest}} \cr
& {\text{ = }}\frac{{2500 \times 9 \times 3}}{{100}} \cr
& = {\text{Rs}}{\text{. }}675 \cr} $$
$$\eqalign{
& {\text{Amount = Rs}}{\text{. 2916}} \cr
& {\text{Time = 2 years }} \cr
& {\text{Rate = 8}}\% \cr
& {\text{Effective rate }}\% {\text{ CI for 2 years}} \cr
& {\text{ = 8 + 8 + }}\frac{{8 \times 8}}{{100}} = 16.64\% \cr
& {\text{Required sum}} \cr
& {\text{ = }}\frac{{2916}}{{\left( {100 + 16.64} \right)}} \times 100 \cr
& = {\text{Rs}}{\text{. }}2500 \cr
& {\text{Required simple interest}} \cr
& {\text{ = }}\frac{{2500 \times 9 \times 3}}{{100}} \cr
& = {\text{Rs}}{\text{. }}675 \cr} $$
Answer: Option A. -> 4 years
$$\eqalign{
& P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^3} = 8P \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^3} = 8 \cr
& {\text{Let }}P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 16P \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 16 = {2^4} = {\left( {{2^3}} \right)^{\frac{4}{3}}} \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = {\left( 8 \right)^{\frac{4}{3}}} \cr
& \Rightarrow {\left\{ {{{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^3}} \right\}^{\frac{4}{3}}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^4} \cr
& \Rightarrow n = 4 \cr
& \therefore {\text{ Required time 4 years}} \cr} $$
$$\eqalign{
& P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^3} = 8P \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^3} = 8 \cr
& {\text{Let }}P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 16P \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 16 = {2^4} = {\left( {{2^3}} \right)^{\frac{4}{3}}} \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = {\left( 8 \right)^{\frac{4}{3}}} \cr
& \Rightarrow {\left\{ {{{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^3}} \right\}^{\frac{4}{3}}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^4} \cr
& \Rightarrow n = 4 \cr
& \therefore {\text{ Required time 4 years}} \cr} $$
Answer: Option B. -> Rs. 121
$$\eqalign{
& {\text{Rate of interest}} \Rightarrow {\text{ 10% = }}\frac{1}{{10}} \cr
& {\text{Each installment of 2 years}} \cr
& \Rightarrow \frac{{10}}{{11}} \times \frac{{\left( {10 + 11} \right)}}{{11}} \times {\text{ Installment = P}}{\text{.A}} \cr
& \Rightarrow \frac{{10}}{{11}} \times \frac{{\left( {10 + 11} \right)}}{{11}} \times {\text{ Installment = 210}} \cr
& \Rightarrow {\text{Installment = 121}} \cr} $$
$$\eqalign{
& {\text{Rate of interest}} \Rightarrow {\text{ 10% = }}\frac{1}{{10}} \cr
& {\text{Each installment of 2 years}} \cr
& \Rightarrow \frac{{10}}{{11}} \times \frac{{\left( {10 + 11} \right)}}{{11}} \times {\text{ Installment = P}}{\text{.A}} \cr
& \Rightarrow \frac{{10}}{{11}} \times \frac{{\left( {10 + 11} \right)}}{{11}} \times {\text{ Installment = 210}} \cr
& \Rightarrow {\text{Installment = 121}} \cr} $$
Answer: Option D. -> Rs. 10000
Given,
Amount = 12,100; r = 10%, t = 2 yrs
$$\eqalign{
& {\text{Amount}} = P{\left[ {1 + \frac{r}{{100}}} \right]^t} \cr
& 12100 = P{\left[ {1 + \frac{{10}}{{100}}} \right]^2} \cr
& \Rightarrow 12100 = P{\left[ {\frac{{11}}{{10}}} \right]^2} \cr
& \Rightarrow 12100 = P \times \frac{{11}}{{10}} \times \frac{{11}}{{10}} \cr
& \Rightarrow P = \frac{{12100 \times 10 \times 10}}{{11 \times 11}} \cr
& \Rightarrow P = 10000 \cr} $$
Given,
Amount = 12,100; r = 10%, t = 2 yrs
$$\eqalign{
& {\text{Amount}} = P{\left[ {1 + \frac{r}{{100}}} \right]^t} \cr
& 12100 = P{\left[ {1 + \frac{{10}}{{100}}} \right]^2} \cr
& \Rightarrow 12100 = P{\left[ {\frac{{11}}{{10}}} \right]^2} \cr
& \Rightarrow 12100 = P \times \frac{{11}}{{10}} \times \frac{{11}}{{10}} \cr
& \Rightarrow P = \frac{{12100 \times 10 \times 10}}{{11 \times 11}} \cr
& \Rightarrow P = 10000 \cr} $$
Question 155. One can purchase a flat from a house building society for Rs. 55000 cash or on the terms that he should pay Rs. 4275 as cash down payment and get the rest in three equal installments. The society charges interest at the rate of 16% per annum compounded half-yearly. If the flat is purchased under installment plan, find the value of each installment ?
Answer: Option C. -> Rs. 19683
Total cost of the flat = Rs. 55000
Down payment = Rs. 4275
Balance = Rs. (55000 - 4275) = Rs. 50725
Rate of interest = 8% per half year
Let the value of each instalment be Rs. x
P.W. of Rs. x due 6 months hence + P.W. of Rs. x due 1 year hence + P.W. of Rs. x due $$1\frac{1}{2}$$ years hence = 50725
$$ \Rightarrow \frac{x}{{\left( {1 + \frac{8}{{100}}} \right)}} + $$ $$\frac{x}{{{{\left( {1 + \frac{8}{{100}}} \right)}^2}}} + $$ $$\frac{x}{{{{\left( {1 + \frac{8}{{100}}} \right)}^3}}} = $$ $$50725$$
$$\eqalign{
& \Rightarrow \frac{{25x}}{{27}} + \frac{{625x}}{{729}} + \frac{{15625x}}{{19683}} = 50725 \cr
& \Rightarrow \frac{{50725x}}{{19683}} = 50725 \cr
& \Rightarrow x = \left( {\frac{{50725 \times 19683}}{{50725}}} \right) = 19683 \cr} $$
Total cost of the flat = Rs. 55000
Down payment = Rs. 4275
Balance = Rs. (55000 - 4275) = Rs. 50725
Rate of interest = 8% per half year
Let the value of each instalment be Rs. x
P.W. of Rs. x due 6 months hence + P.W. of Rs. x due 1 year hence + P.W. of Rs. x due $$1\frac{1}{2}$$ years hence = 50725
$$ \Rightarrow \frac{x}{{\left( {1 + \frac{8}{{100}}} \right)}} + $$ $$\frac{x}{{{{\left( {1 + \frac{8}{{100}}} \right)}^2}}} + $$ $$\frac{x}{{{{\left( {1 + \frac{8}{{100}}} \right)}^3}}} = $$ $$50725$$
$$\eqalign{
& \Rightarrow \frac{{25x}}{{27}} + \frac{{625x}}{{729}} + \frac{{15625x}}{{19683}} = 50725 \cr
& \Rightarrow \frac{{50725x}}{{19683}} = 50725 \cr
& \Rightarrow x = \left( {\frac{{50725 \times 19683}}{{50725}}} \right) = 19683 \cr} $$
Answer: Option D. -> Rs. 50000
Rate of interest = 18%
Time = 2 year
When the interest is payable half yearly
Then, rate of interest = 9%
Time = 4 half - years
Let the principal be Rs. x
$$\eqalign{
& {\text{C}}{\text{.I}}{\text{. = }}x\left[ {{{\left( {1 + \frac{R}{{100}}} \right)}^T} - 1} \right]{\text{ }} \cr
& = x\left[ {{{\left( {1 + \frac{9}{{100}}} \right)}^4} - 1} \right] \cr
& = x\left[ {{{\left( {\frac{{109}}{{100}}} \right)}^4} - 1} \right] \cr
& = x\left[ {1.4116 - 1} \right] \cr
& = Rs.\,0.4116x \cr
& {\text{According to question}} \cr
& = x\left[ {{{\left( {1 + \frac{{18}}{{100}}} \right)}^2} - 1} \right] \cr
& = x\left[ {{{\left( {\frac{{118}}{{100}}} \right)}^2} - 1} \right] \cr
& = x\left[ {{{\left( {1.18} \right)}^2} - 1} \right] \cr
& = x\left[ {1.3924 - 1} \right] \cr
& = Rs.\,0.3924x \cr
& {\text{According to question,}} \cr
& 0.4116x - 0.3924x = 960 \cr
& \Rightarrow x = \frac{{960}}{{0.0192}} \cr
& \Rightarrow x = \frac{{960 \times 10000}}{{192}} \cr
& \Rightarrow x = 50000 \cr
& {\text{Hence, sum of money}} \cr
& {\text{ = Rs. 50000}} \cr} $$
Rate of interest = 18%
Time = 2 year
When the interest is payable half yearly
Then, rate of interest = 9%
Time = 4 half - years
Let the principal be Rs. x
$$\eqalign{
& {\text{C}}{\text{.I}}{\text{. = }}x\left[ {{{\left( {1 + \frac{R}{{100}}} \right)}^T} - 1} \right]{\text{ }} \cr
& = x\left[ {{{\left( {1 + \frac{9}{{100}}} \right)}^4} - 1} \right] \cr
& = x\left[ {{{\left( {\frac{{109}}{{100}}} \right)}^4} - 1} \right] \cr
& = x\left[ {1.4116 - 1} \right] \cr
& = Rs.\,0.4116x \cr
& {\text{According to question}} \cr
& = x\left[ {{{\left( {1 + \frac{{18}}{{100}}} \right)}^2} - 1} \right] \cr
& = x\left[ {{{\left( {\frac{{118}}{{100}}} \right)}^2} - 1} \right] \cr
& = x\left[ {{{\left( {1.18} \right)}^2} - 1} \right] \cr
& = x\left[ {1.3924 - 1} \right] \cr
& = Rs.\,0.3924x \cr
& {\text{According to question,}} \cr
& 0.4116x - 0.3924x = 960 \cr
& \Rightarrow x = \frac{{960}}{{0.0192}} \cr
& \Rightarrow x = \frac{{960 \times 10000}}{{192}} \cr
& \Rightarrow x = 50000 \cr
& {\text{Hence, sum of money}} \cr
& {\text{ = Rs. 50000}} \cr} $$
Answer: Option A. -> 10%
$$\eqalign{
& {\text{According to the question,}} \cr
& {\text{compounded half yearly}} \cr
& {\text{Rate = }}\frac{{\text{R}}}{2} \cr
& {\text{Time = }}\frac{{{\text{2T}}}}{3} \cr
& {\text{Amount = P}}{\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr
& \Rightarrow 2315.25 = 2000{\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr
& \Rightarrow \frac{{2315.25}}{{2000}} = {\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr
& \Rightarrow \frac{{231525}}{{200000}} = {\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr
& \Rightarrow \frac{{9261}}{{8000}} = {\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr
& \Rightarrow {\left( {\frac{{21}}{{20}}} \right)^3} = {\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr
& \Rightarrow 1 + \frac{{\text{R}}}{{200}} = \frac{{21}}{{20}} \cr
& \Rightarrow {\text{R = 10}}\% \cr} $$
$$\eqalign{
& {\text{According to the question,}} \cr
& {\text{compounded half yearly}} \cr
& {\text{Rate = }}\frac{{\text{R}}}{2} \cr
& {\text{Time = }}\frac{{{\text{2T}}}}{3} \cr
& {\text{Amount = P}}{\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr
& \Rightarrow 2315.25 = 2000{\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr
& \Rightarrow \frac{{2315.25}}{{2000}} = {\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr
& \Rightarrow \frac{{231525}}{{200000}} = {\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr
& \Rightarrow \frac{{9261}}{{8000}} = {\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr
& \Rightarrow {\left( {\frac{{21}}{{20}}} \right)^3} = {\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr
& \Rightarrow 1 + \frac{{\text{R}}}{{200}} = \frac{{21}}{{20}} \cr
& \Rightarrow {\text{R = 10}}\% \cr} $$
Answer: Option B. -> $${\text{Rs}}{\text{. S}}{\left( {1 + \frac{{\text{r}}}{{50}}} \right)^3}$$
$$\eqalign{
& {\text{According to the question}} \cr
& {\text{Principal = Rs S}} \cr
& {\text{Rate }}\% {\text{ = 2r}}\,\% {\text{ p}}{\text{.a}}{\text{.}} \cr
& {\text{Time = 3 years}} \cr
& \therefore {\text{A = P}}{\left( {1 + \frac{{\text{r}}}{{100}}} \right)^T} \cr
& \Leftrightarrow {\text{A = S}}{\left( {1 + \frac{{{\text{2r}}}}{{100}}} \right)^3} \cr
& \Leftrightarrow {\text{A = S}}{\left( {1 + \frac{{\text{r}}}{{50}}} \right)^3} \cr} $$
$$\eqalign{
& {\text{According to the question}} \cr
& {\text{Principal = Rs S}} \cr
& {\text{Rate }}\% {\text{ = 2r}}\,\% {\text{ p}}{\text{.a}}{\text{.}} \cr
& {\text{Time = 3 years}} \cr
& \therefore {\text{A = P}}{\left( {1 + \frac{{\text{r}}}{{100}}} \right)^T} \cr
& \Leftrightarrow {\text{A = S}}{\left( {1 + \frac{{{\text{2r}}}}{{100}}} \right)^3} \cr
& \Leftrightarrow {\text{A = S}}{\left( {1 + \frac{{\text{r}}}{{50}}} \right)^3} \cr} $$
Answer: Option A. -> $$1\frac{1}{2}$$ years
$$\eqalign{
& {\text{According to the question,}} \cr
& {\text{Amount}} = {\text{ }}{\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^{2 \times {\text{t}}}} \cr
& \Rightarrow 68921 = 64000{\left( {1 + \frac{5}{{2 \times 100}}} \right)^{2 \times {\text{t}}}} \cr
& \Rightarrow \frac{{68921}}{{64000}} = {\left( {1 + \frac{1}{{40}}} \right)^{2 \times {\text{t}}}} \cr
& \Rightarrow {\left( {\frac{{41}}{{40}}} \right)^3} = {\left( {\frac{{41}}{{40}}} \right)^{2 \times {\text{t}}}} \cr
& \Rightarrow 2{\text{t = 3}} \cr
& \Rightarrow {\text{t = }}\frac{3}{2} \cr
& \Rightarrow {\text{t = 1}}\frac{1}{2}{\text{ years}} \cr} $$
$$\eqalign{
& {\text{According to the question,}} \cr
& {\text{Amount}} = {\text{ }}{\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^{2 \times {\text{t}}}} \cr
& \Rightarrow 68921 = 64000{\left( {1 + \frac{5}{{2 \times 100}}} \right)^{2 \times {\text{t}}}} \cr
& \Rightarrow \frac{{68921}}{{64000}} = {\left( {1 + \frac{1}{{40}}} \right)^{2 \times {\text{t}}}} \cr
& \Rightarrow {\left( {\frac{{41}}{{40}}} \right)^3} = {\left( {\frac{{41}}{{40}}} \right)^{2 \times {\text{t}}}} \cr
& \Rightarrow 2{\text{t = 3}} \cr
& \Rightarrow {\text{t = }}\frac{3}{2} \cr
& \Rightarrow {\text{t = 1}}\frac{1}{2}{\text{ years}} \cr} $$
Answer: Option C. -> 6%
$$\eqalign{
& {\text{A = P }}{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} \cr
& \Rightarrow 1348.32 = 1200{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{134832}}{{120000}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{231525}}{{200000}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{2809}}{{2500}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow {\left( {\frac{{53}}{{50}}} \right)^2} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{53}}{{50}} = 1 + \frac{{\text{R}}}{{100}} \cr
& \Rightarrow {\text{R}} = {\text{ 6% }} \cr} $$
$$\eqalign{
& {\text{A = P }}{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} \cr
& \Rightarrow 1348.32 = 1200{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{134832}}{{120000}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{231525}}{{200000}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{2809}}{{2500}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow {\left( {\frac{{53}}{{50}}} \right)^2} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{53}}{{50}} = 1 + \frac{{\text{R}}}{{100}} \cr
& \Rightarrow {\text{R}} = {\text{ 6% }} \cr} $$
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