Practice MCQ Questions and Answer on Time and Work

201.

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

202.

There was a leakage in the container of the refined oil. If 11 kg oil is leaked out per day then it would have lasted for 50 days, if the leakage was 15 kg per day, then it would have lasted for only 45 days. For how many days would the oil have lasted, if there was no leakage ant it was completely used for eating purpose?

(A) 80 days
(B) 72 days
(C) 100 days
(D) 120 days

Solution: Let x kg of oil be used for the eating purpose, daily, then (x + 11) × 50 = (x +15) × 45 → x = 25 Thus, Total quantity of oil, = (25 +11) × 50 = 1800 Hence, required number of days, = $\frac{1800}{25}$ = 72 days

203.

A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work?

(A) 18 days
(B) 24 days
(C) 30 days
(D) 36 days

Solution: $\begin{aligned} 2(A + B + C)'s 1 day's work \\ = \frac{1}{30} + \frac{1}{24} + \frac{1}{20} \\ = \frac{15}{120} = \frac{1}{8} \\ ∴ (A + B + C)'s 1 day's work \\ = \frac{1}{2 \times 8} = \frac{1}{16} \\ Work done by A, B, C in 10 days \\ = \frac{10}{16} = \frac{5}{8} \\ Remaining work \\ = 1 - \frac{5}{8} = \frac{3}{8} \\ A's 1 day's work \\ = \frac{1}{16} - \frac{1}{24} = \frac{1}{48} \\ Now, \frac{1}{48} work isdone by A in 1 day \\ So, \frac{3}{8} work will be done by A in \\ 48 \times \frac{3}{8} = 18 days \end{aligned}$

204.

A can do $\frac{1}{3}$ rd of a work in 5 days and B can do the do $\frac{2}{5}$ th of this work in 10 days. Both A and B, together can do the work in ?

(A) $7 \frac{3}{8} days$
(B) $8 \frac{4}{5} days$
(C) $9 \frac{3}{8} days$
(D) 10 days

Solution: $\begin{aligned} \frac{1}{3} work in 5 days \\ then a complete work in \\ = 5 \times 3 = 15 days \\ B \to \frac{2}{5} work in 10 days \\ then B completes work in \\ = 10 \times \frac{5}{2} = 25 days \end{aligned}$ L.C.M. of Total Work =75 One day work of A = $\frac{75}{15}$ = 5 unit/day One day work of B = $\frac{75}{25}$ = 3 unit/day $\begin{aligned} (A + B) can do work \\ = \frac{75}{8} = 9 \frac{3}{8} days \end{aligned}$

205.

If there is a reduction in the number of workers in a factory in the ratio 15 : 11 and an increment in their wages in the rate 22 : 25, then the ratio by which the total wages of the workers should be decreased is = ?

(A) 6 : 5
(B) 5 : 6
(C) 3 : 7
(D) 3 : 5

206.

If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:

(A) 4 days
(B) 5 days
(C) 6 days
(D) 7 days

Solution: $\begin{aligned} Let 1 man's 1 day's work = x and \\ 1 boy's 1day's work = y \\ Then, 6 x + 8 y = \frac{1}{10} and \\ 26 x + 48 y = \frac{1}{2} \\ Solving these two equations, we get \\ x = \frac{1}{100} a n d y = \frac{1}{200} \\ (15 men + 20 boy)'s 1 day's work \\ = \frac{15}{100} + \frac{20}{200} = \frac{1}{4} \\ ∴ 15 men and 20 boys can do the work in 4 days \end{aligned}$

207.

A can do a piece of work in 4 hours, B and C together in 3 hours, and A and C together in 2 hours. How long will B alone take to do it ?

(A) 8 hours
(B) 10 hours
(C) 12 hours
(D) 24 hours

Solution: $\begin{aligned} A's 1 hour's work = \frac{1}{4} \\ (B + C)'s 1 hour's work = \frac{1}{3} \\ (A + C)'s 1 hour's work = \frac{1}{2} \\ (A + B + C)'s 1 hour's work \\ = \frac{1}{4} + \frac{1}{3} \\ = \frac{7}{12} \\ ∴ B's 1 hour's work \end{aligned}$ = (A + B + C)'s 1 hour's work - (A + C)'s 1 hour's work $\begin{aligned} = \frac{7}{12} - \frac{1}{2} \\ = \frac{1}{12} \end{aligned}$ So, B alone can complete the work in 12 hours.

208.

A alone can do $\frac{2}{5}$ of a work in 12 days. B is 25 percent more efficient than A. C alone can do the same work in 12 days less than B. D is 25 percent less efficient than C. If they all work together, then the work will be completed in how many days?

(A) $\frac{300}{47}$
(B) $\frac{240}{53}$
(C) $\frac{180}{43}$
(D) $\frac{200}{51}$

209.

Two pipes A and B can fill a tank in 36 min. and 45 min. respectively. Another pipe C can empty the tank in 30 min. First A and B are opened. After 7 minutes, C is also opened. The tank filled up in:

(A) 39 min.
(B) 46 min
(C) 40 min.
(D) 45 min.

Solution: Pipe A can fill empty tank in 36 min. Pipe A can fill the tank = $\frac{100}{36}$ = 2.77% per minute Pipe B can fill empty tank in 45 min. Pipe B can fill the tank = $\frac{100}{45}$ = 2.22% per min. A and B can together fill the tank = (2.77 + 2.22) ≈ 5% per minute So, A and B can fill the tank in 7 min. = 7 × 5 = 35% of the tank Rest tank to be filled = 100 - 35 = 65% C can empty the full tank in 30 min. C can empty the tank = $\frac{100}{30}$ = 3.33% per min. C is doing negative work i.e. emptying the tank A, B and C can together fill the tank, = 2.77% + 2.22% - 3.33% = 1.67% tank per minute So, A, B and C will take time to fill 65% empty tank, = $\frac{65}{1.67}$ = 39 min. (Approx)

210.

A job can be complete by 12 men in 12 days. How many extra days will be needed to complete the job if 6 men leave after working for 6 days ?

(A) 3 days
(B) 6 days
(C) 12 days
(D) 24 days

Solution: According to the question, Total work = 12 M × 12 D = 144 units Work done by 12 men in 6 days = 12 × 6 = 72 units Rest work = 144 - 72 = 72 units Required time for 6 men to complete the work $\begin{aligned} = \frac{72}{6} \\ = 12 days \\ Hence, \\ Total time = 12 + 6 = 18 days \\ Extra time = 18 - 12 = 6 days \end{aligned}$

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Practice MCQ Questions and Answer on Time and Work

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