Practice MCQ Questions and Answer on Time and Work

31.

X can do a piece of work in 40 days. He works at it for 8 days and then Y finished it in 16 days. How long will they together take to complete the work?

(A) $13 \frac{1}{3}$ days
(B) 15 days
(C) 20 days
(D) 26 days

Solution: Work done by X in 8 days = $\frac{1}{40} \times 8$ = $\frac{1}{5}$ Remaining work = $1 - \frac{1}{5}$ = $\frac{4}{5}$ Now, $\frac{4}{5}$ work is done by Y in 16 days Whole work will be done by Y in = $16 \times \frac{5}{4}$ = 20 days ∴ X's 1 day's work = $\frac{1}{40}$ ∴ Y's 1 day's work = $\frac{1}{20}$ (X + Y)'s 1 day's work $\begin{aligned} = \frac{1}{40} + \frac{1}{20} \\ = \frac{3}{40} \end{aligned}$ Hence, X and Y will together complete the work in $\begin{aligned} = \frac{40}{3} \\ = 13 \frac{1}{3} days \end{aligned}$

32.

A does 20% less work than B. If A can complete a piece of work in $7 \frac{1}{2}$ hours, then B can do it in ?

(A) $6 \frac{1}{2} hours$
(B) 6 hours
(C) $5 \frac{1}{2} hours$
(D) 5 hours

Solution: Let time taken by B = x $Efficiency \propto \frac{1}{Time taken}$ So, if B is 100% efficient. then A is 80% efficient So, $\begin{aligned} \Rightarrow \frac{80}{100} = \frac{x}{15 / 2} \\ \Rightarrow x = \frac{80 \times 15}{100 \times 2} \\ \Rightarrow x = 6 hours \end{aligned}$ Alternate: Number of hours taken by A to finish the work = $7 \frac{1}{2}$ hours = $\frac{15}{2}$ hours Work done by A in one hour : $\frac{2}{15}$ Let number of hours taken by B to finish the work : $\frac{1}{x}$ A can work 20% less than B that is $\frac{20}{100}$ = $\frac{4}{5}$ times of B’s work. $\begin{aligned} Here, \frac{4}{5} : 1 = \frac{2}{15} : \frac{1}{x} \\ \frac{4}{5} = \frac{2 x}{15} \\ x = \frac{15 \times 4}{5 \times 2} = 6 hours \end{aligned}$

33.

To do a certain work, the ratio of the efficiencies of X and Y is 5 : 4. Working together, they can complete the same work in 10 days. Y alone starts the work and leaves after 5 days. The remaining work will be completed by X alone in:

(A) 14 days
(B) 12 days
(C) 15 days
(D) 10 days

Solution: $\begin{matrix} X & Y \\ Efficiency & 5 & : & 4 \end{matrix}$ Total work = (5 + 4) × 10 = 90 Y × 5 + X × a = 90 4 × 5 + 5 × a = 90 20 + 5a = 90 5a = 70 a = 14

34.

A can lay railway track between two given stations in 16 days and B can do the same job in 12 days. With help of C, they did the job in 4 days only. Then, C alone can do the job in:

(A) $9 \frac{1}{5}$ days
(B) $9 \frac{2}{7}$ days
(C) $9 \frac{3}{5}$ days
(D) 10

Solution: $\begin{aligned} (A + B + C)'s 1 day's work = \frac{1}{4} \\ A's 1 day's work = \frac{1}{16} \\ B's 1 day's work = \frac{1}{12} \\ ∴ C's 1 day's work \\ = \frac{1}{4} - (\frac{1}{16} + \frac{1}{12}) = \frac{1}{4} - \frac{7}{48} = \frac{5}{48} \\ So, C alone can do the work in \\ \frac{48}{5} = 9 \frac{3}{5} days \end{aligned}$

35.

A, B and C can do a piece of work in 11 days, 20 days and 55 days respectively. If B works daily and is supported by A and C on alternate days beginning with A, then in how many days will the work be finished?

(A) 10
(B) $9 \frac{1}{3}$
(C) 12
(D) 8

36.

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

37.

16 women take 12 days to complete a work which can be completed by 12 men in 8 days. 16 men started working and after 3 days 10 men left and 4 women joined them. How many days will they take to complete the remaining work ?

(A) 4 days
(B) 6 days
(C) 8 days
(D) 10 days

Solution: $\begin{aligned} 16 women \times 12 = 12 men \times 8 \\ 2 women = 1 man \\ Total work \\ = 12 men \times 8 days \\ = 96 units \\ 16 men do work in 3 days \\ = 16 \times 3 \\ = 48 units \\ Work left 96 - 48 = 48 units \end{aligned}$ 16 men - 10 men left = 6 men + 4 women join $\begin{aligned} 6men + 2 men = 8 men \\ 8 men will do 48 units in \\ = \frac{48}{8} \\ = 6 days \end{aligned}$

38.

A alone can complete a piece of work in 18 days and B alone in 15 days, B alone worked at it for 10 days and then left the work. In how many more days, will A alone complete the remaining work ?

(A) 5 days
(B) $5 \frac{1}{2} days$
(C) 6 days
(D) 8 days

Solution: $\begin{aligned} B's 10 day's work \\ = (\frac{1}{15} \times 10) = \frac{2}{3} \\ Remaining work \\ = (1 - \frac{2}{3}) = \frac{1}{3} \\ Now, \frac{1}{18} work is done by A in 1 day \\ ∴ \frac{1}{3} work is done by A in \\ (18 \times \frac{1}{3}) = 6 days \end{aligned}$

39.

Three men A, B, C working together can do a job in 6 hours less time than A alone, in one hour less time than B alone and in one half the time needed by C when working alone. Then A and B together can do the job in:

(A) $\frac{2}{3}$ hours
(B) $\frac{3}{4}$ hours
(C) $\frac{3}{2}$ hours
(D) $\frac{4}{3}$ hours

Solution: Time taken by A =x hours. Therefore taken by A, B and C together = (x - 6) Time taken by B = (x - 5) Time taken by C = 2(x - 6) Now, rate of work of A + Rate of work of B + Rate of work of C = Rate of work of ABC. $\Rightarrow \frac{1}{x} + \frac{1}{x - 5} + \frac{1}{2 (x - 6)} = \frac{1}{x - 6}$ On solving above equation, we get x = 3, $\frac{40}{6}$ When x = 3, the expression (x - 6) becomes negative, thus it's not possible. $\Rightarrow x = \frac{40}{6}$ Time taken by A & B together = $\frac{1}{\frac{3}{20} + \frac{3}{5}}$ = $\frac{4}{3}$ hours

40.

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 $[\frac{7 x}{8 \times 12} + \frac{5 x}{6 \times 16}]$ $+$ $[\frac{3}{12} + \frac{3}{16}]$ = 1 $\frac{12 x}{96} + \frac{7}{16} = 1$ Thus, X = 4.5 min.

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Practice MCQ Questions and Answer on Time and Work

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