Chain Rule — Part 2 MCQs with Explanations

Let the required number of binders be x Less books, Less binders (Direct proportion) More days, Less binders (Indirect proportion) \[\left. \begin{gathered} {\text{Books 1400}}:800 \hfill \\ \,\,\,\,\,\,\,\,\,{\text{Days 20}}:15 \hfill \\ \end{gathered} \right\}::21:x\] $$\eqalign{ & \therefore {\text{ }}1400 \times 20 \times x = 800 \times 15 \times 21 \cr & \Leftrightarrow x = \frac{{\left( {800 \times 15 \times 21} \right)}}{{\left( {1400 \times 20} \right)}} \cr & \Leftrightarrow x = 9 \cr} $$

Let the total number of men to be engaged be $$x$$ More length, More labourers (Direct proportion) Less days, More labourers (Indirect proportion) \[\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,{\text{Lentgh 26}}:39 \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Days 6}}:18 \hfill \\ {\text{Hours per day }}9:8 \hfill \\ \end{gathered} \right\}::17:x\] $$\eqalign{ & \therefore {\text{ }}26 \times 6 \times 9 \times x = 39 \times 18 \times 8 \times 17 \cr & \Leftrightarrow x = \frac{{\left( {39 \times 18 \times 8 \times 17} \right)}}{{\left( {26 \times 6 \times 9} \right)}} \cr & \Leftrightarrow x = 68 \cr} $$ ∴ Number of more labourers = (68 -17) = 51

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

Let the required number of rounds be x More radius, Less rounds (Indirect proportion) $$\eqalign{ & \therefore {\text{ }}20:14::70:x \cr & \Leftrightarrow \left( {20 \times x} \right) = \left( {14 \times 70} \right) \cr & \Leftrightarrow x = \frac{{\left( {14 \times 70} \right)}}{{20}} \cr & \Leftrightarrow x = 49 \cr} $$

Remaining work = $$\left( {1 - \frac{4}{7}} \right){\text{ = }}\frac{3}{7}$$ Remaining period = (46 - 33) = 13 days Let the total men working at it be x Less work, Less men (Direct proportion) Less days, More men (Indirect proportion) More hour/day, Less men (Indirect proportion) \[\left. \begin{gathered} \,\,\,\,\,{\text{Work }}\frac{4}{7}:\frac{3}{7} \hfill \\ \,\,\,\,\,\,\,{\text{Men 13}}:33 \hfill \\ {\text{Hour/day }}9:8 \hfill \\ \end{gathered} \right\}::117:x\] $$\eqalign{ & \therefore {\text{ }}\frac{4}{7} \times 13 \times 9 \times x = \frac{3}{7} \times 33 \times 8 \times 117 \cr & \Leftrightarrow x = \frac{{\left( {3 \times 33 \times 8 \times 117} \right)}}{{\left( {4 \times 13 \times 9} \right)}} \cr & \Leftrightarrow x = 198 \cr} $$ ∴ Addition men to be employed = (198 - 117) = 81

Let the required number of persons be x Less days , More persons (Indirect proportion) More work , More persons (Direct proportion) \[\left. \begin{gathered} \,{\text{Days 2}}:4 \hfill \\ {\text{Work 1}}:8 \hfill \\ \end{gathered} \right\}::12:x\] $$\eqalign{ & \therefore \,2 \times 1 \times x = 4 \times 8 \times 12 \cr & \Leftrightarrow x = \frac{{\left( {4 \times 8 \times 12} \right)}}{{\left( 2 \right)}} \cr & \Leftrightarrow x = 192 \cr} $$

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) \[\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Days 50}}:60 \hfill \\ {\text{Hours per day 9}}:\frac{{15}}{2} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Work }}1:2 \hfill \\ \end{gathered} \right\}::42:\left( {42 + x} \right)\] $$\therefore \left[ {50 \times 9 \times 1 \times \left( {42 + x} \right)} \right] = $$ $$\left( {60 \times \frac{{15}}{2} \times 2 \times 42} \right)$$ $$\eqalign{ & \Leftrightarrow \left( {42 + x} \right) = \frac{{37800}}{{450}} \cr & \Leftrightarrow 42 + x = 84 \cr & \Leftrightarrow x = 42 \cr} $$

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: $$\eqalign{ & = \frac{{(15 \times 7 + 7 \times 6 + 14 \times 4)36}}{{12 \times 7 + 14 \times 6 + 21 \times 4}} \cr & = \frac{{7308}}{{252}} \cr & = 29 \cr} $$

Cloth is required for 1 shirt = 2 m, 60 cm or 260 cm Cloth is required for 7 shirt = 260 × 7 = 1820 cm or 18 m 20 cm

1 score of papers = 20 papers Cost of 4 dozen papers = Rs.24 Cost of 20 papers $$\eqalign{ & = \frac{{24}}{{4 \times 12}} \times 20 \cr & = {\text{Rs}}{\text{.10}} \cr} $$

Cost of 8 fan's and 14 oven's is Rs. 36520 Cost of 4 fan's and 7 oven's is Rs. $$\frac{{{\text{36520}}}}{2}$$ = Rs.18260 Cost of 12 fan's and 21 oven's is = 18260 × 3 = Rs. 54780

Cost of 5 kgs apples Rs. 450 Cost of 1 kg apples Rs. $$\frac{{{\text{450}}}}{5}$$ Cost of 8 kgs apples = $$\frac{{450}}{5} \times 8$$ = Rs.720 Cost of 12 dozens mangoes Rs. 4320 Cost of 1 dozen mangoes Rs. $$\frac{{{\text{4320}}}}{{12}}$$ Cost of 8 dozen mangoes = $$\frac{{4320}}{{12}} \times 8$$ = Rs. 2880 Cost of 4 kgs oranges = Rs. 240 Cost of 1 kg orange = Rs. $$\frac{{{\text{240}}}}{4}$$ Cost of 8 kgs oranges = $$\frac{{240}}{4} \times 8$$ = Rs. 480 Total cost = 720 + 2880 + 480 = Rs. 4080

Cost of 21 pencils and 9 clippers = Rs. 819 Cost of 7 pencils and 3 clippers = $$\frac{{819}}{3}$$ = Rs. 273