Practice MCQ Questions and Answer on Chain Rule

Solution:

 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) $$\eqalign{ & \therefore \,\left( {\frac{{360}}{x} + 4} \right):\frac{{360}}{x}::x:\left( {x - 1} \right) \cr & \Leftrightarrow \left( {\frac{{360}}{x} + 4} \right) \left( {x - 1} \right) = \frac{{360}}{x} \times x \cr & \Leftrightarrow 360 - \frac{{360}}{x} + 4x - 4 = 360 \cr & \Leftrightarrow 4x - \frac{{360}}{x} - 4 = 0 \cr & \Leftrightarrow x - \frac{{90}}{x} - 1 = 0 \cr & \Leftrightarrow {x^2} - x - 90 = 0 \cr & \Leftrightarrow \left( {x - 10} \right)\left( {x + 9} \right) = 0 \cr & \Leftrightarrow x = 10 \cr} $$

Solution:

 Let the required number of days be x 7 men = 10 women (14 men and 20 women) = (20 + 20) women = 40 women More length, More days (Direct proportion) More women, Less days (Indirect proportion) \[\left. \begin{gathered} {\text{Length 100}}:600 \hfill \\ \,\,{\text{Women 40}}:10 \hfill \\ \end{gathered} \right\}::10:x\] $$\eqalign{ & \therefore {\text{ }}100 \times 40 \times x = 600 \times 10 \times 10 \cr & \Leftrightarrow x = \frac{{\left( {600 \times 10 \times 10} \right)}}{{\left( {100 \times 40} \right)}} \cr & \Leftrightarrow x = 15 \cr} $$

Solution:

 Let the required number of days be x. Less men, More days (Indirect Proportion) $$\eqalign{ & \therefore 27:36::18:x \cr & \Rightarrow 27 \times x = 36 \times 18 \cr & \Rightarrow x = \frac{{36 \times 18}}{{27}} \cr & \Rightarrow x = 24 \cr} $$

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, 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 total number of men be x Work done = $$\frac{1}{3}$$ Remaining work = $$\left( {1 - \frac{1}{3}} \right) = \frac{2}{3}$$ More work, More men (Direct proportion) More days, Less men (Indirect proportion) \[\left. \begin{gathered} \,{\text{Work }}\frac{1}{3}:\frac{2}{3} \hfill \\ {\text{Days 25}}:20 \hfill \\ \end{gathered} \right\}::20:x\] $$\eqalign{ & \therefore {\text{ }}\left( {\frac{1}{3} \times 25 \times x} \right) = \left( {\frac{2}{3} \times 20 \times 20} \right) \cr & \Leftrightarrow x = \frac{{800}}{{25}} \cr & \Leftrightarrow x = 32 \cr} $$ ∴ More men to be employed = (32 - 20) = 12

Solution:

 Remaining part $$\eqalign{ & = \left( {1 - \frac{4}{9}} \right) \cr & = \frac{5}{9}{\text{ }} \cr} $$ Let the required time be x minutesMore volume to be filled , More time taken (Direct proportion) $$\eqalign{ & \therefore {\text{ }}\frac{4}{9}:\frac{5}{9}::1:x \cr & \Rightarrow \frac{4}{9}x = \frac{5}{9} \cr & \Rightarrow x = \left( {\frac{5}{9} \times \frac{9}{4}} \right) \cr & \Rightarrow x = \frac{5}{4}{\text{ }} \cr} $$

Solution:

 Let the required number of days be x Less men, More days (Indirect proportion) $$\eqalign{ & \therefore 42:56::24:x \cr & \Leftrightarrow 42 \times x = 56 \times 24 \cr & \Leftrightarrow x = \left( {\frac{{56 \times 24}}{{42}}} \right) \cr & \Leftrightarrow x = 32 \cr} $$

Solution:

 Let the required number of days bex. Less persons, More days (Indirect Proportion) More working hours per day, Less days (Indirect Proportion) \[\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Persons }}\,30:39 \hfill \\ {\text{Working hour/day }}\,{\text{6}}:{\text{5}} \hfill \\ \end{gathered} \right\}::12:x\] $$\eqalign{ & \therefore 30 \times 6 \times x = 39 \times 5 \times 12 \cr & \Rightarrow x = \frac{{ {39 \times 5 \times 12} }}{{ {30 \times 6} }} \cr & \Rightarrow x = 13 \cr} $$

Solution:

 Let the number of working hours per day be x More persons, Less working hours ( Indirect proportion) Less days, More working hours (Indirect proportion) \[\left. \begin{gathered} {\text{Persons 7}}:5 \hfill \\ {\text{Quantity 4}}:8 \hfill \\ \end{gathered} \right\}::7:x\] $$\eqalign{ & \therefore {\text{ }}7 \times 4 \times x = 5 \times 8 \times 7 \cr & \Leftrightarrow x = \frac{{\left( {5 \times 8 \times 7} \right)}}{{\left( {7 \times 4} \right)}} \cr & \Leftrightarrow x = 10 \cr} $$