Practice MCQ Questions and Answer on Time and Work

261.

A and B working together completed a job in 5 days. If A works twice as efficiently as he actually did and B works $\frac{1}{3}$ of actual efficiency, the work would have completed in 3 days. Find the for A to complete the job alone.

(A) $6 \frac{1}{2}$
(B) $6 \frac{1}{4}$
(C) $6 \frac{3}{4}$
(D) $12 \frac{1}{2}$

Solution: One Day's work of A and B together, $\frac{1}{A} + \frac{1}{B} = \frac{1}{5} . . . . . . . . (i)$ When A works with twice efficiency,Then, $\frac{2}{A} + \frac{1}{3 B} = \frac{1}{3} . . . . . . . . (ii)$ on solving equations (i) and (ii), we get $A = \frac{25}{4} = 6 \frac{1}{4}$

262.

25 men with 10 boys can do in 6 days as much work as 21 men with 30 boys can do in days. How many boys must help 40 men to do the same work in 4 days ?

(A) 5
(B) 10
(C) 20
(D) 40

Solution: $\begin{aligned} Let 1 men's 1 day's work = x \\ and 1 boy's 1 day's work = y \\ Then, \\ \Rightarrow 6 (25 x + 10 y) = 5 (21 x + 30 y) \\ \Rightarrow 150 x + 60 y = 105 x + 150 y \\ \Rightarrow 45 x = 90 y \\ \Rightarrow x = 2 y \\ Let, \\ The required number of boys be z \\ Then, \\ \Rightarrow 4 (40 x + z y) = 6 (25 x + 10 y) \\ \Rightarrow 4 (80 y + z y) = 6 (50 y + 10 y) \\ \Rightarrow 80 + z = \frac{6 \times 60}{4} = 90 \\ \Rightarrow z = 10 \end{aligned}$

263.

While working 7 hour a day, A alone can complete a piece of work in 6 days and B alone in 8 days. In what time would they complete it together, 8 hour a day?

(A) 3 days
(B) 4 days
(C) 2.1 days
(D) 3.6 days

Solution: A can complete the work in 7 × 6 = 42 hours 1 hour's work of A = $\frac{1}{42}$ B can complete the work in 7 × 8 = 56 hours 1 hour's work of B = $\frac{1}{56}$ (A + B)'s 1 hour's work $\begin{aligned} = \frac{1}{42} + \frac{1}{56} \\ = \frac{4 + 3}{168} \\ = \frac{7}{168} \end{aligned}$ ∴ Time taken by (A + B) working 8 hours daily $\frac{168}{7 \times 8} = 3 days$

264.

A can build a wall in the same time in which B and C together can do it. If A and B together can do it. If A and B together could do it in 25 days and C alone in 35 days, in what time could B alone do it ?

(A) 91 days
(B) 100 days
(C) 175 days
(D) None of these

Solution: $\begin{aligned} (A + B)'s 1 day's work = \frac{1}{25} \\ C's 1 day's work = \frac{1}{35} \\ (A + B + C)'s 1 day's work \\ = (\frac{1}{25} + \frac{1}{35}) \\ = \frac{12}{175} . . . . . (i) \end{aligned}$ Also, A's 1 day's work = (B + C)'s 1 day's work.....(ii) $\begin{aligned} From (i) and (ii), we get : \\ \Rightarrow 2 \times (A's 1 day's work) = \frac{12}{175} \\ \Rightarrow A's 1 day's work = \frac{6}{175} \\ ∴ B's 1 day's work \\ = (\frac{1}{25} - \frac{6}{175}) \\ = \frac{1}{175} \end{aligned}$

265.

One man or two women or three boys can do a piece of work in 88 days. One man, one woman and one boy will do it in ?

(A) 44 days
(B) 24 days
(C) 48 days
(D) 20 days

Solution: $\begin{aligned} According to the question, \\ 1 man = 2 women = 3 boys \\ 1 man = 2 women \\ 1 man = 3 boys \\ \frac{1}{2} man = 1 women \\ \frac{1}{3} man = 1 boys \\ = 1 man = 1 woman = 1 boys \\ = 1 man = \frac{1}{2} man = \frac{1}{3} man \\ = \frac{11}{6} man \end{aligned}$ 1 man can complete a work in 88 days $\begin{aligned} \frac{11}{6} man can complete a work in \\ = \frac{88}{\frac{11}{6}} = 48 days \end{aligned}$

266.

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 $\begin{aligned} = \frac{96}{13} \\ = 7 \frac{5}{13} days \end{aligned}$

267.

Amit and Sumit can plough a field in 4 days. Sumit alone can plough the field in 6 days. In how many days will Amit alone plough the feild ?

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

Solution: $\begin{aligned} Amit's 1 day's work \\ = (\frac{1}{4} - \frac{1}{6}) \\ = \frac{1}{12} \end{aligned}$ ∴ Amit alone can plough the field in 12 days.

268.

4 men and 10 women were put on a work. They completed $\frac{1}{3}$ of the work in 4 days. After this 2 men and 2 women were increased. They completed $\frac{2}{9}$ more of the work in 2 days. If the remaining work is to be completed in 3 days, then how many more women must be increased ?

(A) 8
(B) 32
(C) 50
(D) 55

Solution: $\begin{aligned} Let 1 man's 1 day's work = x \\ And \\ 1 women's 1 day's work = y \\ Then, \\ \Rightarrow 4 x + 10 y = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} \\ \Rightarrow 4 x + 10 y = \frac{1}{12} \\ \Rightarrow 2 x + 5 y = \frac{1}{24} . . . . . (i) \\ And, \\ \Rightarrow 6 x + 12 y = \frac{1}{9} \\ \Rightarrow 2 x + 4 y = \frac{1}{27} . . . . . (ii) \end{aligned}$ Subtracting (ii) from (i), we get $\begin{aligned} y = \frac{1}{24} - \frac{1}{27} = \frac{1}{216} \\ Now, \end{aligned}$ Now, (6 men + 12 women)'s 3 day's work $\begin{aligned} = (\frac{1}{9} \times 3) \\ = \frac{1}{3} \\ Work completed \\ = (\frac{1}{3} + \frac{2}{9} + \frac{1}{3}) \\ = \frac{8}{9} \\ ∴ Remainig work \\ = (1 - \frac{8}{9}) \\ = \frac{1}{9} \\ 1 women's 3 day's work \\ = (\frac{1}{216} \times 3) \\ = \frac{1}{72} \end{aligned}$ In 3 day's $\frac{1}{72}$ work is done by 1 women. $\begin{aligned} ∴ In 3 day's \frac{1}{9} work is done by \\ = (72 \times \frac{1}{9}) \\ = 8 women . \end{aligned}$

269.

A and B can do a piece of work in 25 days. B alone can do $66 \frac{2}{3} %$ ÃÂÃÂ of the same work in 30 days. In how many days can A alone do $\frac{4}{15}$ part of the same work?

(A) 15
(B) 20
(C) 18
(D) 12

Solution: $\begin{aligned} 66 \frac{2}{3} % \to 30 days \\ \frac{2}{3} \to 30 days \\ 1 \to 45 days \end{aligned}$ $\begin{aligned} A = 4 unit \\ A = \frac{25 \times \frac{4}{15}}{4} \\ A = 15 days \end{aligned}$

270.

A and B together can do a piece of work in 30 days, B and C together can do it in 20 days, A starts the work and works on it for 5 days, B takes up and work for 15 days. Finally C finishes the work in 18 days. The number of days in which C alone can do the work where doing it separately is = ?

(A) 120 days
(B) 40 days
(C) 60 days
(D) 24 days

Solution: L.C.M. of Total Work =60 One day work of A + B = $\frac{60}{30}$ = 2 unit/day One day work of B + C = $\frac{60}{20}$ = 3 unit/day $\begin{aligned} According to the question, \\ \Rightarrow (A + B) . . . . . 30 days \\ \Rightarrow (B + C .) . . . . . 20 days \\ \Rightarrow A . . . . . 5 \\ \Rightarrow B . . . . . 5 + 10 \\ \Rightarrow C . . . . . 10 + 8 days \end{aligned}$ Work done by (A +B) in 5 days = 2 × 5 = 10 units ⇒ Work done by (B + C) in 10 days = 10 × 3 = 30 units ⇒ Total work (finished till now).....40 units ⇒ Remaining work = 60 - 40 = 20 units ⇒ Here we find that C does remaining 20 units in 8 days ⇒ C's efficiency = $\frac{Work}{Day}$ $= \frac{20 Work}{8 Days} = \frac{5}{2}$ ⇒ Therefore, time taken by C alone to complete the work $= \frac{60}{5} \times 2 = 24 days$

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Practice MCQ Questions and Answer on Time and Work

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