Chain Rule - Aptitude MCQ Questions and Solutions with Explanations

Let the required quantity of coal be x metric tonnes
More engines, More coal (Direct proportion)
More hours per day, More coal (Direct proportion)
More rate, More coal (Direct proportion)
\[\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Engines 5}}:8 \hfill \\ {\text{Hours per day 9}}:10 \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Rate }}\frac{1}{3}:\frac{1}{4} \hfill \\ \end{gathered} \right\}::6:x\]
$$\eqalign{ & \therefore \,\,\left( {5 \times 9 \times \frac{1}{3} \times x} \right) = \left( {8 \times 10 \times \frac{1}{4} \times 6} \right) \cr & \Leftrightarrow 15x = 120 \cr & \Leftrightarrow x = 8 \cr} $$

Let the required number of days be x
2 men of latter type = 3 men of former type
12 men of latter type
= $$\left( {\frac{3}{2} \times 12} \right)$$
= 18 men of former type
More men, Less days (Indirect proportion)
Less working hours, More days (Indirect proportion)
\[\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Men 18}}:9 \hfill \\ {\text{Working hrs 6}}:\frac{{15}}{2} \hfill \\ \end{gathered} \right\}::20:x\]
$$\eqalign{ & \therefore \,18 \times 6 \times x = 9 \times \frac{{15}}{2} \times 20 \cr & \Leftrightarrow 108x = 1350 \cr & \Leftrightarrow x = \frac{{25}}{2} \cr & \Leftrightarrow x = 12\frac{1}{2} \cr} $$

3 women ≡ 2 men
So, 21 women ≡ 14 men
Less men, More days (Indirect proportion)
Less hours per day, More days (Indirect proportion)
\[\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Men 14}}:15 \hfill \\ {\text{Hours per day 6}}:8 \hfill \\ \end{gathered} \right\}::21:x\]
$$\eqalign{ & \therefore \,\left( {14 \times 6 \times x} \right) = \left( {15 \times 8 \times 21} \right) \cr & \Leftrightarrow x = \frac{{\left( {15 \times 8 \times 21} \right)}}{{\left( {14 \times 6} \right)}} \cr & \Leftrightarrow x = 30 \cr} $$
∴ Required number of days = 30

Let the remaining food last for x days
4000 soldiers had provision for 160 days
3200 soldiers had provision for x days
Less men, More days (Indirect proportion)
$$\eqalign{ & \therefore \,3200:4000::160:x \cr & \Leftrightarrow 3200x = 4000 \times 160 \cr & \Leftrightarrow x = \frac{{\left( {4000 \times 160} \right)}}{{3200}} \cr & \Leftrightarrow x = 200 \cr} $$

Let the remaining food last for x days
500 men had provision for = (27 - 3) = 24 days
(500 + 300) men had provision for x days
More men, Less days (Indirect proportion)
$$\eqalign{ & \therefore \,800:500::24:x \cr & \Leftrightarrow \left( {800 \times x} \right) = \left( {500 \times 24} \right) \cr & \Leftrightarrow x = \frac{{\left( {500 \times 24} \right)}}{{800}} \cr & \Leftrightarrow x = 15 \cr} $$

Initially, Let there be x men having food for y days
After 10 days, x men had food for days (y - 10)
Also, $$\left( {x - \frac{x}{5}} \right)$$   men had food for y days
$$\eqalign{ & \therefore \,x\left( {y - 10} \right) = \frac{{4x}}{5} \times y \cr & \Leftrightarrow 5xy - 50x = 4xy \cr & \Leftrightarrow xy - 50x = 0 \cr & \Leftrightarrow x\left( {y - 50} \right) = 0 \cr & \Leftrightarrow y - 50 = 0 \cr & \Leftrightarrow y = 50 \cr} $$

Let there be x men originally
So, x men had provisions 40 days whereas (x + 500) men consumed it in 35 days
More men, Less days (Indirect proportion)
$$\eqalign{ & \therefore \,\left( {x + 500} \right):x::40:35 \cr & \Leftrightarrow 35 \times \left( {x + 500} \right) = 40x \cr & \Leftrightarrow 5x = 35 \times 500 \cr & \Leftrightarrow x = \left( {\frac{{35 \times 500}}{5}} \right) \cr & \Leftrightarrow x = 3500 \cr} $$

Let the team take x days to finish 360 pieces
Then, number of pieces made each day = $$\frac{{360}}{x}$$
More number of pieces per day, Less days (Indirect proportion)
$$\eqalign{ & \therefore \,\left( {\frac{{360}}{x} + 4} \right):\frac{{360}}{x}::x:\left( {x - 1} \right) \cr & \Leftrightarrow \left( {\frac{{360}}{x} + 4} \right) \left( {x - 1} \right) = \frac{{360}}{x} \times x \cr & \Leftrightarrow 360 - \frac{{360}}{x} + 4x - 4 = 360 \cr & \Leftrightarrow 4x - \frac{{360}}{x} - 4 = 0 \cr & \Leftrightarrow x - \frac{{90}}{x} - 1 = 0 \cr & \Leftrightarrow {x^2} - x - 90 = 0 \cr & \Leftrightarrow \left( {x - 10} \right)\left( {x + 9} \right) = 0 \cr & \Leftrightarrow x = 10 \cr} $$

Ratio of time taken by a woman, a man and a boy
$$\eqalign{ & = 8:6:12 \cr & = 4:3:6 \cr} $$
So, 4 women ≡ 3 men ≡ 6 boy
(12 mens + 12 womens + 12 boys)
$$\eqalign{ & = \left[ {12 + \left( {\frac{3}{4} \times 12} \right) + \left( {\frac{3}{6} \times 12} \right)} \right]{\text{men}} \cr & {\text{ = }}\left( {12 + 9 + 6} \right){\text{men}} \cr & = 27{\text{ men}} \cr} $$
Let the required number of days be x
More men, Less days (Indirect proportion)
More working hours, Less days (Indirect proportion)
\[\left. \begin{gathered} {\text{Working hours 8}}:6 \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Men 27}}:9 \hfill \\ \end{gathered} \right\}::6:x\]
$$\eqalign{ & \therefore \,27 \times 8 \times x = 9 \times 6 \times 6 \cr & \Leftrightarrow x = \frac{{\left( {9 \times 6 \times 6} \right)}}{{\left( {27 \times 8} \right)}} \cr & \Leftrightarrow x = \frac{3}{2} \cr & \Leftrightarrow x = 1\frac{1}{2} \cr} $$

1 man ≡ 2 boys ⇔ (12 men + 18 boys)
          ≡ (12 × 2 ×18) boys = 42 boys
Let required number of boys = x
⇒ (21 men + x boys) ≡ (21 × 2 × x) boys = (42 + x) boys
Less days, More boys (Indirect proportion)
More hours per day, Less boys (Indirect proportion)
More work, More boys (Direct proportion)
\[\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Days 50}}:60 \hfill \\ {\text{Hours per day 9}}:\frac{{15}}{2} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Work }}1:2 \hfill \\ \end{gathered} \right\}::42:\left( {42 + x} \right)\]
$$\therefore \left[ {50 \times 9 \times 1 \times \left( {42 + x} \right)} \right] = $$     $$\left( {60 \times \frac{{15}}{2} \times 2 \times 42} \right)$$
$$\eqalign{ & \Leftrightarrow \left( {42 + x} \right) = \frac{{37800}}{{450}} \cr & \Leftrightarrow 42 + x = 84 \cr & \Leftrightarrow x = 42 \cr} $$