Practice MCQ Questions and Answer on Chain Rule

Solution:

 Let the required number of chairs be x Then, More carpenters, More chairs (Direct proportion) More hours per day, More chairs (Direct proportion) More days, More chairs (Direct proportion) \[\left. \begin{gathered} \,\,\,\,{\text{Carpenters 12}}:18 \hfill \\ {\text{Hours per day 6}}:8 \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Days 24}}:{\text{36}} \hfill \\ \end{gathered} \right\}::460:x\] $$\eqalign{ & \therefore {\text{ }}12 \times 6 \times 24 \times x = 18 \times 8 \times 36 \times 460 \cr & \Leftrightarrow x = \frac{{\left( {18 \times 8 \times 36 \times 460} \right)}}{{\left( {12 \times 6 \times 24} \right)}} \cr & \Leftrightarrow x = 1380 \cr} $$ ∴ Required number of chairs = 1380

Solution:

 Let the required number of labourers be x Less working hour/day, More labours (Indirect proportion) More days, Less labourers (Indirect proportion) \[\left. \begin{gathered} {\text{Working hours/day 6}}:7 \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Days 30}}:18 \hfill \\ \end{gathered} \right\}::30:x\] $$\eqalign{ & \therefore 6 \times 30 \times x = 7 \times 18 \times 30 \cr & \Leftrightarrow 6x = 126 \cr & \Leftrightarrow x = 21 \cr} $$

Solution:

 Let the required time be x seconds Then, More metres, More time (direct proportion) $$\eqalign{ & \therefore 0.128:25::1:x \cr & \Leftrightarrow x = \frac{{25}}{{0.128}} \cr & \Leftrightarrow x = \frac{{25 \times 1000}}{{128}} \cr & \Leftrightarrow x = 195.31 \cr} $$` ∴ Required time =195 seconds (approx)

Solution:

 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) \[\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Engines 5}}:8 \hfill \\ {\text{Hours per day 9}}:10 \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Rate }}\frac{1}{3}:\frac{1}{4} \hfill \\ \end{gathered} \right\}::6:x\] $$\eqalign{ & \therefore \,\,\left( {5 \times 9 \times \frac{1}{3} \times x} \right) = \left( {8 \times 10 \times \frac{1}{4} \times 6} \right) \cr & \Leftrightarrow 15x = 120 \cr & \Leftrightarrow x = 8 \cr} $$

Solution:

 Let, L = large ships, M = medium ships, S = small ships Now from the question, 4L ≡ 7S $$\eqalign{ & {\text{15L}} = \left( {\frac{7}{4} \times 15} \right){\text{S}} = \frac{{105}}{4}{\text{S}} \cr & {\text{Also, 2L}} = \left( {\frac{7}{4} \times 2} \right){\text{S}} = \frac{7}{2}{\text{S}} \cr & {\text{3M}} \equiv 2{\text{L}} + {\text{1S }} \equiv \left( {\frac{7}{2} + 1} \right){\text{S}} = \frac{9}{2}{\text{S}} \cr & \Leftrightarrow {\text{ 7M}} = \left( {\frac{9}{2} \times \frac{1}{3} \times 7} \right)\,\,{\text{S}} = \frac{{21}}{2}{\text{S}} \cr & \therefore \,\left( {{\text{15L}} + {\text{7M}} + {\text{14S}}} \right){\text{ ships}} \cr & \equiv \left( {\frac{{105}}{4} + \frac{{21}}{2} + 14} \right){\text{S}} \cr & = \frac{{203}}{4}{\text{S}} \cr & \left( {{\text{12 large}} + 14{\text{ medium}} + 2{\text{1 small}}} \right){\text{ ships}} \cr & \equiv \left[ {\left( {\frac{7}{4} \times 12} \right) + \left( {\frac{{21}}{2} \times 2} \right) + 21} \right]{\text{S}} \cr & = \left( {21 + 21 + 21} \right){\text{S}} \cr & = 63\,{\text{S}} \cr} $$ Let the required number of journeys be x More ships, Less journeys (Indirect proportion) $$\eqalign{ & \therefore \,63:\frac{{203}}{4}::36:x \cr & \Leftrightarrow 63x = \frac{{203}}{4} \times 36 = 1827 \cr & \Leftrightarrow x = \frac{{1827}}{{63}} \cr & \Leftrightarrow x = 29 \cr} $$ Alternate: Now from the question, $$\eqalign{ & 4L = 7S \cr & \Rightarrow \frac{L}{S} = \frac{7}{4} \cr & \Rightarrow L = 7x;S = 4x \cr & 3M = 2L + S \cr & \Rightarrow M = \frac{{2 \times 7x + 4x}}{3} \cr & \Rightarrow M = 6x \cr} $$ Thus, ratio of large, medium and small = 7 : 6 : 4 So, numbers of journeys: $$\eqalign{ & = \frac{{(15 \times 7 + 7 \times 6 + 14 \times 4)36}}{{12 \times 7 + 14 \times 6 + 21 \times 4}} \cr & = \frac{{7308}}{{252}} \cr & = 29 \cr} $$

Solution:

 Let the weekly earning be Rs. xMore persons , More earning (Direct proportion) Less working hours, Less earning (Direct proportion) \[\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Persons 6}}:9 \hfill \\ {\text{Working hours 8}}:6 \hfill \\ \end{gathered} \right\}::8400:x\] $$\eqalign{ & \therefore {\text{ }}6 \times 8 \times x = 9 \times 6 \times 8400 \cr & \Leftrightarrow x = \frac{{\left( {9 \times 6 \times 8400} \right)}}{{\left( {6 \times 8} \right)}} \cr & \Leftrightarrow x = 9450 \cr} $$

Solution:

 Let the required number days be x. Less spiders, More days (Indirect Proportion) Less webs, Less days (Direct Proportion) \[\left. \begin{gathered} {\text{Spiders}}\,\,\,\,\,1:7 \hfill \\ \,\,\,{\text{Webs}}\,\,\,\,\,7:1 \hfill \\ \end{gathered} \right\}::7:x\] $$\eqalign{ & \therefore 1 \times 7 \times x = 7 \times 1 \times 7 \cr & \Rightarrow x = 7 \cr} $$

Solution:

 $$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{required}}\,{\text{weight}}\,{\text{be}}\,x\,{\text{kg}}. \cr & {\text{Less}}\,{\text{weight,}}\,{\text{less}}\,{\text{cost}}\,\left( {{\text{Direct}}\,{\text{Proportion}}} \right) \cr & \therefore 250:200::60:x \cr & \Rightarrow 250 \times x = {200 \times 60} \cr & \Rightarrow x = \frac{{ {200 \times 60} }}{{250}} \cr & \Rightarrow x = 48 \cr} $$

Solution:

 Let the required number of machine be x More persons, More machines (Direct proportion) More working working hours, More machines (Direct proportion) More days, More machines (Direct proportion) \[\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Persons 2}}:6 \hfill \\ \,\,\,\, {\text{Working hours 2}}:6 \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Days 2}}:{\text{6}} \hfill \\ \end{gathered} \right\}::2:x\] $$\eqalign{ & \therefore {\text{ }}2 \times 2 \times 2 \times x = 6 \times 6 \times 6 \times 2 \cr & \Leftrightarrow x = \frac{{\left( {6 \times 6 \times 6 \times 2} \right)}}{{\left( {2 \times 2 \times 2} \right)}} \cr & \Leftrightarrow x = 54 \cr} $$

Solution:

 Let the height of the building be x metres Lees lengthy shadow, Less is the height (Direct proportion) $$\eqalign{ & \therefore 40.25:28.75::17.5:x \cr & \Leftrightarrow 40.25 \times x = 28.75 \times 17.5 \cr & \Leftrightarrow x = \frac{{\left( {28.75 \times 17.5} \right)}}{{40.25}} \cr & \Leftrightarrow x = 12.5 \cr} $$