Interest - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Let the required time = t years }} \cr & {\text{According to the question,}} \cr & \Leftrightarrow \frac{{500 \times 4 \times 6.25}}{{100}} = \frac{{400 \times 5 \times {\text{t}}}}{{100}} \cr & \Leftrightarrow 5 \times 4 \times 625 = 400 \times 5 \times {\text{t}} \cr & \Leftrightarrow {\text{t = }}\frac{{625}}{{100}} = \frac{{25}}{4} \cr & \Leftrightarrow {\text{t}} = 6\frac{1}{4}{\text{years}} \cr} $$

$$\eqalign{ & \frac{{{\text{Principal}}}}{{{\text{Amount}}}} = \frac{{4 \times 5}}{{5 \times 5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20}}{{25}} \cr & {\text{After three year}} \cr & \frac{{\text{P}}}{{\text{A}}} = \frac{{5 \times 4}}{{7 \times 4}} \cr & \,\,\,\,\,\,\,\, = \frac{{20}}{{28}} \cr & {\text{In three year S}}{\text{.I}}{\text{.}} \cr & = 28x - 25x \cr & = 3x \cr & \text{So, the required interest will be} \cr & 3x = \frac{{20x \times {\text{R}} \times 3}}{{100}} \cr & {\text{R}} = 5\% \cr} $$

Let time be T years and rate be R% p.a.
$$\eqalign{ & {\text{Then, }}y{\text{ is the S}}{\text{.I}}{\text{. on x}} \cr & \Rightarrow \frac{{x{\text{RT}}}}{{100}} = y......(i) \cr & {\text{And, }}z{\text{ is the S}}{\text{.I}}{\text{. on y}} \cr & \Rightarrow \frac{{y{\text{RT}}}}{{100}} = z \cr & \Rightarrow y = \frac{{100z}}{{RT}}......(ii) \cr & {\text{From (i) and (ii) we have:}} \cr & \frac{{x{\text{RT}}}}{{100}} = \frac{{100z}}{{{\text{RT}}}} \cr & \Rightarrow \frac{{x{{\text{R}}^2}{{\text{T}}^2}}}{{{{\left( {100} \right)}^2}}} = z \cr & \Rightarrow \frac{{{y^2}}}{x} = z \cr & \Rightarrow {y^2}= xz \cr }$$

$$\eqalign{ & {\text{Let the sum be Rs}}{\text{. }}x \cr & {\text{Then,}} \cr} $$
$$ {\frac{{x \times 8 \times 4}}{{100}}} + {\frac{{x \times 10 \times 6}}{{100}}} \,+ $$     $$ {\frac{{x \times 12 \times 5}}{{100}}} $$   $$ = 12160$$
$$\eqalign{ & \Rightarrow 32x + 60x + 60x = 1216000 \cr & \Rightarrow 152x = 1216000 \cr & \Rightarrow x = 8000 \cr} $$

Let the loan amount be Rs. x
$$\eqalign{ & {\text{Then,}} \cr & \Rightarrow \frac{{6x}}{{100}} + \frac{{7.5x}}{{100}} + \frac{{9x}}{{100}} = 8190 \cr & \Rightarrow 22.5x = 819000 \cr & \Rightarrow x = 36400 \cr} $$

Note : In such type of questions to save your valuable time follow the given below method.
$$\eqalign{ & {\text{Value of installment}} \cr & {\text{ = }}\frac{{{\text{Principal}} \times {\text{100}}}}{{{\text{Time}} \times {\text{100}} + \left( {{{\text{t}}_{{\text{n - 1}}}} + {{\text{t}}_{{\text{n - 2}}}}...1} \right) \times {\text{Rate}}\% }} \cr & {\text{Principal = Rs}}{\text{. 848}} \cr & {\text{Rate = 4}}\% \cr & {\text{Time = 4 year}} \cr & {\text{Installment}} \cr & {\text{ = }}\frac{{848 \times 100}}{{4 \times 100 + \left( {3 + 2 + 1} \right) \times 4}}{\text{ }} \cr & {\text{ = }}\frac{{848 \times 100}}{{\left( {400 + 24} \right)}}{\text{ }} \cr & = \frac{{848 \times 100}}{{424}} \cr & {\text{ = Rs}}{\text{. 200}} \cr} $$

Rate of interest
$$ = \frac{{100(x - 1)}}{t}\% $$
Where x is the no. of times the sum becomes of itself. Here the sum is getting 3 times.
Therefore, x = 3
Where t is the time taken by sum to become x times of itself. Here, t = 12 years
By the short trick approach, we get
$$\eqalign{ & = \frac{{100(3 - 1)}}{{12}} \cr & = \frac{{200}}{{12}} \cr & = \frac{{50}}{3} = 16\frac{2}{3}\% \cr} $$

Alternative Method :
Let the principal be P and in the 2nd scenario, SI = 2P
$$\eqalign{ & {\text{Rate}} = \frac{{{\text{SI}} \times 100}}{{{\text{Principal}} \times {\text{Time}}}} \cr & = \frac{{2{\text{P}} \times 100}}{{{\text{P}} \times 12}} \cr & = \frac{{50}}{3} \cr & = 16\frac{2}{3}\% \cr} $$

$$\eqalign{ & {\text{Let the first part = Rs}}{\text{. }}x \cr & \therefore {\text{Hence second part}} \cr & {\text{ = Rs}}{\text{. }}\left( {12000 - x} \right) \cr & {\text{According to the question,}} \cr & \Rightarrow \frac{{x \times 12 \times 3}}{{100}} = \frac{{\left( {12000 - x} \right) \times 9 \times 16}}{{2 \times 100}} \cr & \Rightarrow 36x = 72\left( {12000 - x} \right) \cr & \Rightarrow x = 24000 - 2x \cr & \Rightarrow 3x = 24000 \cr & \Rightarrow x = {\text{Rs}}{\text{. 8000}} \cr & {{\text{1}}^{{\text{st}}}}{\text{ part = Rs}}{\text{. 8000}} \cr & {{\text{2}}^{{\text{nd}}}}{\text{ part}} \cr & {\text{ = Rs}}{\text{. }}\left( {12000 - 8000} \right) \cr & = {\text{Rs}}{\text{. 4000 }} \cr & {\text{Hence maximum part}} \cr & {\text{ = Rs}}{\text{. 8000}} \cr} $$
Alternate
Note : In such type of questions to save your valuable time follow the given below method.
Let two parts P₁ and P₂ respectively
According to the question,
$$\eqalign{ & \Rightarrow {{\text{P}}_1} \times \frac{{36}}{{100}} \times 1 = {{\text{P}}_2} \times \frac{9}{2} \times \frac{{16}}{{100}} \times 1 \cr & \Rightarrow {{\text{P}}_1} \times 4 = 8{{\text{P}}_2} \cr & \Rightarrow \frac{{{{\text{P}}_1}}}{{{{\text{P}}_2}}} = \frac{2}{1} \cr & \Rightarrow {{\text{P}}_1}{\text{:}}{{\text{P}}_2} = 2:1 \cr & {\text{Hence greater part}} \cr & {\text{ = }}\frac{{12000}}{{\left( {2 + 1} \right)}} \times 2 \cr & = {\text{Rs}}{\text{. }}8000 \cr} $$

$$\eqalign{ & {\text{Let the sum be Rs}}{\text{.100}}{\text{}} \cr & {\text{Then,}} \cr & {\text{S}}{\text{.I}}{\text{.for first 6 months}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{100 \times 10 \times 1}}{{100 \times 2}}} \right) \cr & = {\text{Rs}}{\text{. }}5 \cr & {\text{S}}{\text{.I}}{\text{.for last 6 months}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{105 \times 10 \times 1}}{{100 \times 2}}} \right) \cr & = {\text{Rs}}{\text{. }}5.25 \cr & So, \cr & {\text{Amount at the end of 1year}} \cr & = {\text{Rs}}{\text{.}}\left( {100 + 5 + 5.25} \right) \cr & = {\text{Rs}}{\text{.}}\,110.25 \cr & \therefore {\text{Effective rate}} \cr & = \left( {110.25 - {\text{100}}} \right) \cr & = 10.25\% \cr} $$

$$\eqalign{ & {\text{S}}{\text{.I}}{\text{. for 1 year}} \cr & = {\text{Rs}}.\left( {854 - 815} \right) \cr & = {\text{Rs}}.39 \cr & {\text{S}}{\text{.I}}{\text{. for 3 years}} \cr & = {\text{Rs}}{\text{.}}\left( {39 \times 3} \right) \cr & = {\text{Rs}}{\text{. }}117 \cr & \therefore \text{Principal} \cr & = {\text{Rs}}{\text{.}}\left( {854 - 117} \right) \cr & = {\text{Rs}}{\text{. }}698 \cr} $$