Practice MCQ Questions and Answer on Time and Work

61.

If the work done by (x - 1) men in (x + 1) days and the work done by (x + 2) men in (x - 1) days are in the ratio 9 : 10, then the value of x is equal to ?

(A) 5
(B) 6
(C) 7
(D) 8

Solution: Put values in formula $\frac{{(x - 1)}_{men} \times {(x + 1)}_{days}}{9_{work}} =$ $\frac{{(x + 2)}_{men} \times {(x - 1)}_{days}}{10_{work}}$ $\begin{aligned} \Rightarrow \frac{x + 1}{9} = \frac{x + 2}{10} \\ \Rightarrow 10 x + 10 = 9 x + 18 \\ \Rightarrow x = 8 \end{aligned}$

62.

X can do a piece of work in 24 days. When he had worked for 4 days, Y joined him. If complete work was finished in 16 days, Y can alone finish that work in ?

(A) 27 days
(B) 36 days
(C) 42 days
(D) 18 days

Solution: $\begin{aligned} According to the question, \\ X \to 24 days \\ \Rightarrow Work done X in 4 days alone \\ = 4 \times 1 \\ = 4 units \\ \Rightarrow Remaining work \\ = 24 - 4 = 20 units \\ \Rightarrow 20 units done by both together in \\ = (16 - 4 days) \\ = 12 days \\ Then efficiencies of (X + Y) \\ = \frac{Work}{Days} \\ = \frac{20}{12} \\ = \frac{5}{3} \\ = 1 + \frac{2}{3} \\ \Rightarrow Efficiency of Y = \frac{2}{3} \end{aligned}$ Time taken by Y alone to complete the total work $\begin{aligned} = \frac{24}{\frac{2}{3}} \\ = 36 days \end{aligned}$ Alternate : $\begin{aligned} X \times 20 = (X + Y) \times 12 \\ \frac{X}{X + Y} \\ = \frac{12}{15} \\ = \frac{3 \to Efficiency of X}{5 \to Efficiency of (X + Y)} \\ Efficiency of Y \\ = 5 - 3 \\ = 2 units/day \\ Total work \\ = 24 \times 3 \\ = 72 units \\ Total time taken by Y \\ = \frac{72}{2} \\ = 36 days \end{aligned}$

63.

x does $\frac{1}{4}$ of a job in 6 days. y completes rest of the job in 12 days. Then x and y could complete the job together in = ?

(A) $9 days$
(B) $8 \frac{1}{8} days$
(C) $9 \frac{3}{5} days$
(D) $7 \frac{1}{3} days$

Solution: x does $\frac{1}{4}$ work in 6 days. ∴ x does complete work in 6 × 4 = 24 days y does complete the $\frac{3}{4}$ work in 12 days. ∴ y does complete work in 12 × $\frac{4}{3}$ = 16 days x and y together can complete a work in $\begin{aligned} = \frac{16 \times 24}{16 + 24} \\ = \frac{48}{5} \\ = 9 \frac{3}{5} days \end{aligned}$

64.

P can complete $\frac{1}{4}$ of a work in 10 days, Q can complete 40% of the same work in 145 days. R, complete $\frac{1}{3}$ of the work in 13 days and S, $\frac{1}{6}$ of the work in 7 days. Who will be able complete the work first ?

(A) P
(B) Q
(C) R
(D) S

Solution: P completes $\frac{1}{4}$ of work in 10 days P completes full of work in $\begin{aligned} = \frac{10}{1} \times 4 \\ = 40 days \end{aligned}$ Q completes 40% of work in 145 days Q completes full 100% of work in $\begin{aligned} = \frac{145}{40} \times 100 \\ = 362.5 days \end{aligned}$ R completes $\frac{1}{3}$ of work in 13 days R completes full of work in $\begin{aligned} = \frac{13}{1} \times 3 \\ = 39 days \end{aligned}$ S completes $\frac{1}{6}$ of work in 7 days S completes full of work in $\begin{aligned} = \frac{7}{1} \times 6 \\ = 42 days \end{aligned}$ Clearly, we can see R completes the work first

65.

If 3 men or 6 women can do a piece of work in 16 days, in how many days can 12 men and 8 women do the same piece of work ?

(A) 4 days
(B) 5 days
(C) 3 days
(D) 2 days

Solution: $\begin{aligned} According to the question, \\ 3m \times 16 = 6w \times 16 \end{aligned}$ $\frac{m}{w} =$ $\frac{2 \to Efficiency of men}{1 \to Efficiency of women}$ $\begin{aligned} Total work = 3 \times 2 \times 16 \\ = 96 units \end{aligned}$ $One day work of$ $(12m + 8w)$ $\begin{aligned} = 12 \times 2 + 8 \times 1 \\ = 32 units \end{aligned}$ $Total time taken by$ $(12m + 8w)$ $\begin{aligned} = \frac{96}{32} \\ = 3 days \end{aligned}$

66.

A and B can together do a piece of work it in 6 days and A alone can do it in 9 days. The number of days B will take to do it alone is ?

(A) 18 days
(B) 24 days
(C) 9 days
(D) 12 days

Solution: L.C.M. of Total Work = 36 One day work of A + B = $\frac{36}{6}$ = 6 unit/day One day work of A = $\frac{36}{9}$ = 4 unit/day $\begin{aligned} B's efficiency \\ = 6 - 4 \\ = 2 \\ Time taken by B \\ = \frac{36}{2} \\ = 18 days \end{aligned}$

67.

A and B together can complete a work in 12 days. B and C together can complete the same work in 8 days and A and C together can complete it in 16 days. In total, how many days do A, B and C together take to complete the same work ?

(A) $3 \frac{5}{12}$
(B) $3 \frac{9}{13}$
(C) $7 \frac{5}{12}$
(D) $7 \frac{5}{13}$

Solution: $\begin{aligned} (A + B)'s 1 day's work = \frac{1}{12} \\ (B + C)'s 1 day's work = \frac{1}{8} \\ (A + C)'s 1 day's work = \frac{1}{16} \end{aligned}$ Adding, we get 2(A + B + C)'s 1 day's work $\begin{aligned} = (\frac{1}{12} + \frac{1}{8} + \frac{1}{16}) \\ = \frac{13}{96} \end{aligned}$ So, A, B and C together can complete the work in $\begin{aligned} = \frac{96}{13} \\ = 7 \frac{5}{13} days \end{aligned}$

68.

A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?

(A) 12 days
(B) 15 days
(C) 16 days
(D) 18 days

Solution: $\begin{aligned} A's 2 day's work = \frac{1}{20} \times 2 = \frac{1}{10} \\ (A + B + C)'s 1 day's work \\ = \frac{1}{20} + \frac{1}{30} + \frac{1}{60} = \frac{6}{60} = \frac{1}{10} \\ Work done in 3 days = \frac{1}{10} + \frac{1}{10} = \frac{1}{5} \\ Now, \frac{1}{5} work is done in 3 days \\ ∴ Whole work will be done in \\ 3 \times 5 = 15 days \end{aligned}$

69.

If 2 men or 3 women or 4 boys can do a piece of work in 52 days, then the same piece of work will be done by 1 man, 1 woman and 1 boy in:

(A) 48 days
(B) 36 days
(C) 45 days
(D) 50 days

Solution: Work done by 2 men = 3 women = 4 boys 1 man = 2 boys 1 woman = $\frac{4}{3}$ boys Thus, Boys × days = 4 × 52 boys × days Again, 1 man + 1 women + 1 boys, $= 2 + \frac{4}{3} + 1$ $= \frac{13}{3}$ boys Using work equivalent method, boys × day = BOYS × DAYS $4 \times 52 = \frac{13}{3} \times x (let)$ x = 48 days

70.

If 16 men or 20 women can do a piece of work in 25 days. In what time will 28 men and 15 women do it ?

(A) $14 \frac{2}{7} days$
(B) $33 \frac{1}{3} days$
(C) $18 \frac{3}{4} days$
(D) $10 days$

Solution: If number of days at work is same $\begin{aligned} So, we can equate directly \\ 16 men = 20 women \\ 4 men = 5 women \\ 5 women = 4 men \\ 1 women = \frac{4}{5} men \\ 15 women = \frac{4}{5} \times 15 \\ 15 women = 12 men \end{aligned}$ ⇒ 16 men complete a work in 25 days 1 men complete a work in 400 days $\begin{aligned} 28 men + 15 women = ? \\ ↓ \\ 28 men + 12 men = 40 men \\ work in = \frac{25 \times 16}{40} \\ = 10 days \\ A l t e r n a t e : \\ 16 m or 20 w \to 25 days \\ 28 m and 15 w \to ? \\ \frac{16 \times 20 \times 25}{(16 \times 15) + (20 \times 28)} \\ = 10 days \end{aligned}$

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Practice MCQ Questions and Answer on Time and Work

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