Practice MCQ Questions and Answer on Chain Rule

31.

On a certain map of India the actual distance of 1450 km between two cities Delhi and Kolkata is shown as 5 cm. What scale is used to draw the map ?

• (A) $$1:15\times {10}^6$$
• (B) $$1:20\times {10}^6$$
• (C) $$1:25\times {10}^6$$
• (D) $$1:29\times {10}^6$$

Show Answer

 5 cm on the map represents 1450 km ∴ 1 cm on the map represents $$\left( {\frac{{1450}}{5}} \right)$$  km = 290 km Hence, the scale is 1 cm : 290 km i.e., $$1:29 \times {10^6}{\text{ }}$$ $$\left[ {\because 290{\text{km}} = \left( {29 \times {{10}^6}} \right){\text{cm}}} \right]$$

32.

A garrison had provision for a certain number of days. After 10 days, $$\frac{1}{5}$$ of the men desert and it is found that the provisions will now last just as long as before. How long was that ?

• (A) 15 days
• (B) 25 days
• (C) 35 days
• (D) 50 days

Show Answer

 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 $$\eqalign{ & \therefore \,x\left( {y - 10} \right) = \frac{{4x}}{5} \times y \cr & \Leftrightarrow 5xy - 50x = 4xy \cr & \Leftrightarrow xy - 50x = 0 \cr & \Leftrightarrow x\left( {y - 50} \right) = 0 \cr & \Leftrightarrow y - 50 = 0 \cr & \Leftrightarrow y = 50 \cr} $$

33.

In a barrack of soldiers there was stock of food for 190 days for 4000 soldiers. After 30 days 800 soldiers left the barrack. For how many days shall the left over food last for the remaining soldiers ?

• (A) 175 days
• (B) 200 days
• (C) 225 days
• (D) 250 days

Show Answer

 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) $$\eqalign{ & \therefore \,3200:4000::160:x \cr & \Leftrightarrow 3200x = 4000 \times 160 \cr & \Leftrightarrow x = \frac{{\left( {4000 \times 160} \right)}}{{3200}} \cr & \Leftrightarrow x = 200 \cr} $$

34.

3 pumps , working 8 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work to empty the tank in 1 day ?

• (A) 9
• (B) 10
• (C) 11
• (D) 12

Show Answer

 Let the required number of working hours per day be x More pumps, Less working hours per day (Indirect proportion) Less days, More working hours per day (Indirect proportion) \[\left. \begin{gathered} {\text{Pumps 4}}:3 \hfill \\ \,\,\,\,\,\,{\text{Days 1}}:2 \hfill \\ \end{gathered} \right\}::8:x\] $$\eqalign{ & \therefore {\text{ }}4 \times 1 \times x = 3 \times 2 \times 8 \cr & \Leftrightarrow x = \frac{{\left( {3 \times 2 \times 8} \right)}}{{\left( 4 \right)}} \cr & \Leftrightarrow x = 12 \cr} $$

35.

The work done by a women in 8 hours is equal to the work done by a man in 6 hours and by a boy in 12 hours. If working 6 hours per day 9 men can complete a work in 6 days, then in how many days can 12 men, 12 women and 12 boys together finish the same work, working 8 hours per day ?

• (A) $$\text{1}\frac{1}{2}\text{ days}$$
• (B) $$\text{3 days}$$
• (C) $$\text{3}\frac{2}{3}\text{ days}$$
• (D) $$\text{4}\frac{1}{2}\text{ days}$$

Show Answer

 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) \[\left. \begin{gathered} {\text{Working hours 8}}:6 \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Men 27}}:9 \hfill \\ \end{gathered} \right\}::6:x\] $$\eqalign{ & \therefore \,27 \times 8 \times x = 9 \times 6 \times 6 \cr & \Leftrightarrow x = \frac{{\left( {9 \times 6 \times 6} \right)}}{{\left( {27 \times 8} \right)}} \cr & \Leftrightarrow x = \frac{3}{2} \cr & \Leftrightarrow x = 1\frac{1}{2} \cr} $$

36.

5 persons can prepare an admission list in 8 days working 7 hours a day. If 2 persons join them so as to complete the work in 4 days, they need to work per day for ?

• (A) 8 hours
• (B) 9 hours
• (C) 10 hours
• (D) 12 hours

Show Answer

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

37.

The cost of 5 kgs of apples is Rs. 450. The cost of 12 dozen mangoes is Rs. 4320 and the cost of 4 kgs of oranges is Rs. 240. What is the total cost of 8 kg of apples, 8 dozen of mangoes and 8 kg of oranges ?

• (A) Rs. 4020
• (B) Rs. 4080
• (C) Rs. 4050
• (D) Other than those given as option

Show Answer

 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

38.

A fort has provisions for 50 days. If after 10 days they are strengthened by 500 men and the food lasts for 35 days longer, the number of men originally in the fort were ?

• (A) 2500
• (B) 3000
• (C) 3500
• (D) 4000

Show Answer

 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) $$\eqalign{ & \therefore \,\left( {x + 500} \right):x::40:35 \cr & \Leftrightarrow 35 \times \left( {x + 500} \right) = 40x \cr & \Leftrightarrow 5x = 35 \times 500 \cr & \Leftrightarrow x = \left( {\frac{{35 \times 500}}{5}} \right) \cr & \Leftrightarrow x = 3500 \cr} $$

39.

A flagstaff 17.5 m high casts a shadow of length 40.25 m. The height of the building, which casts a shadow of length 28.75 m under similar conditions will be:

• (A) 10 m
• (B) 12.5 m
• (C) 17.5 m
• (D) 21.25 m

Show Answer

 Let the height of the building x metres. Less lengthy shadow, Less in the height (Direct Proportion) $$\eqalign{ & \therefore 40.25:28.75::17.5:x \cr & \Rightarrow 40.25 \times x = 28.75 \times 17.5 \cr & \Rightarrow x = \frac{{28.75 \times 17.5}}{{40.25}} \cr & \Rightarrow x = 12.5 \cr} $$

40.

49 pumps can empty a reservoir in $$6\frac{1}{2}$$ days, working 8 hours a day. If 196 pumps are used for 5 hours each day, then the same work will be complete in ?

• (A) 2 days
• (B) $$\text{2}\frac{1}{2}$$ days
• (C) $$\text{2}\frac{3}{5}$$ days
• (D) 3 days

Show Answer

 Let the required number of days be x Then, More pumps, Less days (Indirect proportion) Less working hour/day, More days (Indirect proportion) \[\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Pumps 196}}:49 \hfill \\ {\text{Working hour/day 5}}:8 \hfill \\ \end{gathered} \right\}::\frac{{13}}{2}:x\] $$\eqalign{ & \therefore {\text{ }}196 \times 5 \times x = 49 \times 8 \times \frac{{13}}{2} \cr & \Leftrightarrow x = \left( {49 \times 8 \times \frac{{13}}{2} \times \frac{1}{{196 \times 5}}} \right) \cr & \Leftrightarrow x = \frac{{13}}{5} \cr & \Leftrightarrow x = 2\frac{3}{5} \cr} $$