Practice MCQ Questions and Answer on Time and Work
321.
A contractor undertake to do a piece of work in 40 days. He engages 100 men at the beginning and 100 more after 35 days and completes the work in stipulated time. If he had not engaged the work in men, how many days behind the scheduled the work should have been finished ?
(A) 5
(B) 6
(C) 7
(D) 8
Solution:
100 men's 40 day's work + 100 men's 5 day's work = 1 ⇒ 100 men's 45 day's work = 1 So, if the contractor had not engaged additional men, 100 men would have finished the work in 45 days. Difference in time = (45 - 40) = 5 days
322.
Some carpenters promised to do a job in 9 days but 5 of them were absent and remaining men did the job in 12 day. The original number of carpenters was ?
(A) 24
(B) 20
(C) 16
(D) 18
Solution:
Let there were N carpenters in the beginning According to the question, $$\frac{{{{\text{N}}_{{\text{men}}}} \times {9_{{\text{days}}}}}}{{{{\text{1}}_{{\text{work}}}}}} = $$ $$\frac{{{{\left( {{\text{N}} - 5} \right)}_{{\text{men}}}} \times {{12}_{{\text{days}}}}}}{{{{\text{1}}_{{\text{work}}}}}}$$ $$\eqalign{ & {\text{3N}} = {\text{4N}} - 20 \cr & {\text{N}} = {\text{20 men}} \cr} $$
323.
Tapas works twice as fast as Mihir. If both of them together complete a work in 12 days, Tapas alone can complete it in ?
Working together B and C take 50% more number of days than A, B and C together take and A and B working together, take $$\frac{8}{3}$$ more number of days than A, B and C take together. If A, B and C all have worked together till the completion of the work and B has received Rs. 120 out of total earnings of Rs. 450, then in how many days did A, B and C together complete the whole work?
(A) 2 days
(B) 4 days
(C) 6 days
(D) 8 days
Solution:
Ratio of efficiencies of A, B and C, = 5x : 4x : 6x Number of days required by A and B = $$\frac{{100}}{{9{\text{x}}}}$$ ------ (1) Number of days required by A, B and C = $$\frac{{100}}{{15{\text{x}}}}$$ ------ (2) $$\eqalign{ & \frac{{100}}{{9{\text{x}}}} - \frac{{100}}{{15{\text{x}}}} = \frac{8}{3} \cr & \Rightarrow {\text{x}} = \frac{5}{3} \cr} $$ Number of days required by A, B and C = $$\frac{{100}}{{15{\text{x}}}}$$ = $$\frac{{100}}{{15 \times \frac{5}{3}}}$$ = 4 days
325.
There are three boats A, B and C, working together they carry 60 people in each trip. One day an early morning A carried 50 people in few trips alone. When it stopped carrying the passengers B and C started carrying the people together. It took a total of 10 trips to carry 300 people by A, B and C. It is known that each day on an average 300 people cross the river using only one of the 3 boats A, B and C. How many trips it would take to A to carry 150 passengers alone?
(A) 15
(B) 30
(C) 25
(D) 10
Solution:
Combined efficiency of all the three boats = 60 passengers /trip Now, consider option (A) 15 trips and 150 passengers means efficiency of A = 10 passengers per trip A's efficiency = 10 passengers per trip Then, (B + C) combined efficiency = 50 passengers per trip Since, combined efficiency is 60 so option (A) is correct
326.
Kiran, Vishal and Dinesh work in a juice factory. Kiran takes 2 hours to extract as much juice as Vishal can in 3 hours. Dinesh takes 5 hours to extract as much juice that Kiran extracts in 4 hours. A tank can be filled with juice in 48 hours, if all of them work together. How long will it take to fill the tank, if Dinesh alone is trying to fill the tank?
20 men can do a piece of work in 18 days. They worked together for 3 days, then 5 men joined. In how many days is the remaining work completed ?
(A) 12 days
(B) 14 days
(C) 13 days
(D) 15 days
Solution:
20 men → 18 days ⇒ Work done by 20 men working Together = 1 work ⇒ Work done by them in 3 days working Together = 1 × 3 = 3 work ⇒ Remaining work = 18 - 3 = 15 work ⇒ 15 work is to be done by (20 + 5) = 25 men $$\eqalign{ & \therefore {\text{Efficiency of 1 man}} = \frac{1}{{20}} \cr & \Rightarrow {\text{Efficincy of 5 men}} \cr & = \frac{5}{{20}} \cr & = \frac{1}{4} \cr & \Rightarrow {\text{So, efficiency of }}\left( {20 + 5} \right) \cr & \Rightarrow 25{\text{ men}} = 1 + \frac{1}{4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{4}{\text{ working days}} \cr & {\text{Required time}} \cr & = \frac{{{\text{Work}}}}{{{\text{Efficiency}}}} \cr & = \frac{{15}}{{\frac{5}{4}}} \cr & = 12{\text{ days}} \cr} $$ Therefore, 12 more days will be taken to finish the remaining work Alternate : 20 men can do 18 days So, total work = 18 × 20 = 360 20 men 3 days work = 20 × 3 = 60 Remaining work = 360 - 60 = 300 After joining 5 men total men = 20 + 5 = 25 So, finish the remaining work in = $$\frac{{300}}{{25}} = 12{\text{ days}}$$
328.
Two pipes A and B can fill a cistern in 12 min and 16 min respectively. Both the pipes are opened together for a certain time but due to some obstruction the flow of water was restricted to $$\frac{7}{8}$$ of full flow in pipe A and $$\frac{5}{6}$$ of full in pipe B. This obstruction is removed after some time and tank is now filled in 3 min from that moment. How long was it before the full flow.
(A) 8 min
(B) 3 min
(C) 5.6 min
(D) 4.5 min
Solution:
Let the obstruction remain for x min. Hence, Part of cistern filled in X min + part of cistern filled in 3 min = full cistern $$\left[ {\frac{{7{\text{x}}}}{{8 \times 12}} + \frac{{5{\text{x}}}}{{6 \times 16}}} \right]$$ $$ + $$ $$\left[ {\frac{3}{{12}} + \frac{3}{{16}}} \right]$$ = 1 $$\frac{{12{\text{x}}}}{{96}} + \frac{7}{{16}} = 1$$ Thus, X = 4.5 min.
329.
A can do a work in 36 days, B in 18 days and C in 12 days. Every 2nd day B and every 3rd day C, helps A .Then in how many days the work will be completed?
(A) 12
(B) 14
(C) 10
(D) 8
Solution:
Let to work Total Work = 36 One day work of A = $$\frac{{36}}{{36}}$$ = 1 unit/day One day work of B = $$\frac{{36}}{{18}}$$ = 2 unit/day One day work of C = $$\frac{{36}}{{12}}$$ = 3 unit/day In 3 days cycle total work done is A, A + B, A + C = 1 + (1 + 2) + (1 + 3) = 8 unit/Cycle ∴ 32 units of the work completed in 4 cycle and reminder 4 units of works in next two days. 1 cycle = 3 days ∴ 4 cycle = 12 days And remaining 4 unit work done in next two days. In days 13, A will work 1 unit and in day 14, A and B will work 3 units. Total number the day required to complete the work in the given condition is 14 days
330.
A and B can together finish a work in 30 days. They worked together for 20 days and B left. After another 20 days, A finished the remaining work. In how many days A alone can finish the job ?
(A) 40 days
(B) 50 days
(C) 54 days
(D) 60 days
Solution:
$$\eqalign{ & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 20 day's work}}{\text{.}} \cr & = \left( {\frac{1}{{30}} \times 20} \right) \cr & = \frac{2}{3} \cr & {\text{Remaining work }} \cr & = \left( {1 - \frac{2}{3}} \right) \cr & = \frac{1}{3}{\text{ }} \cr} $$ Now, $$\frac{1}{3}$$ work is done by A in 20 days Whole work will be done by A in (20 × 3) = 60 days.