Practice MCQ Questions and Answer on Time and Work

If two persons, with equal abilities, can do two jobs in two days then 100 persons with equal abilities can do 100 similar jobs in = ?

(A) 100 days
(B) 10 days
(C) 5 days
(D) 2 days

Solution: $\frac{2_{person} \times 2_{days}}{2_{jobs}} = \frac{100 \times D}{10 0_{jobs}}$ (D is number of days required) $\begin{aligned} After solving \\ D = 2 days \end{aligned}$

Two pipes, P and Q can fill a cistern in 12 and 15 minutes respectively. Both are opened together, but at the end of 3 minutes, P is turned off. In how many more minutes will Q fill the cistern?

(A) 7 minutes
(B) $7 \frac{1}{2}$ minutes
(C) 8 minutes
(D) $8 \frac{1}{4}$ minutes

Solution: P can fill cistern in 12 minutes P fills cistern in 1 minute = $\frac{1}{12}$ Q can fill cistern in 15 minutes Q fills cistern in 1 minute = $\frac{1}{15}$ part P and Q together can fill cistern in 1 minute, $= \frac{1}{12} + \frac{1}{15} \Rightarrow \frac{9}{60}$ part So, they can together fill cistern in 3 minute, $= 3 \times \frac{9}{60} \Rightarrow \frac{9}{20}$ part Rest Cistern = $1 - \frac{9}{20}$ = $\frac{11}{20}$ part $\frac{11}{20}$ part cistern must be filled by Q in $\frac{\frac{11}{20}}{\frac{1}{15}} = 8 \frac{1}{4}$ minutes Alternatively, Pipe P can fill cistern in one minute = $\frac{100}{12}$ = 8.33% Pipe Q can fill cistern in one minute = $\frac{100}{15}$ = 6.66% (P + Q) together can fill the cistern in one minute = 15%; In 3 minutes cistern is filled = 45% P become off, then rest of cistern will be filled by Pipe Q in = $\frac{55}{6.66}$ = $8 \frac{1}{4}$ minutes.

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

A can do a piece of work in 10 days, B in 15 days. They work for 5 days. The rest of the work was finished by C in 2 days. If they get Rs. 1500 for the whole work, the daily wages of B and C are ?

(A) Rs. 150
(B) Rs. 225
(C) Rs. 250
(D) Rs. 300

Solution: $\begin{aligned} Part of the done by A \\ = (\frac{1}{10} \times 5) \\ = \frac{1}{2} \\ Part of the done by B \\ = (\frac{1}{15} \times 5) \\ = \frac{1}{3} \\ Part of the done by C \\ = 1 - (\frac{1}{2} + \frac{1}{3}) \\ = \frac{1}{6} \end{aligned}$ So, (A's share) : (B's share) : (C's share) $\begin{aligned} = \frac{1}{2} : \frac{1}{3} : \frac{1}{6} \\ = 3 : 2 : 1 \\ ∴ A's share \\ = Rs . (\frac{3}{6} \times 1500) \\ = Rs .750 \\ B's share \\ = Rs . (\frac{2}{6} \times 1500) \\ = Rs .500 \\ C's share \\ = Rs . (\frac{1}{6} \times 1500) \\ = Rs .250 \\ A's daily wages \\ = Rs . (\frac{750}{5}) \\ = Rs .150 \\ B's daily wages \\ = Rs . (\frac{500}{5}) \\ = Rs .100 \\ C's daily wages \\ = Rs . (\frac{250}{2}) \\ = Rs .125 \\ ∴ Daily wages of B and C \\ = Rs . (100 + 125) \\ = Rs .225 \end{aligned}$

A and B can together finish a piece of work in 30 days. They worked on it 20 days and then B left. The remaining work was done by A alone in 20 days. A alone can finish the work in = ?

(A) 60 days
(B) 54 days
(C) 48 days
(D) 50 days

Solution: (T.W.)60 2 units/day ↓ 30 days(A + B) $\begin{aligned} (A + B) worked for 20 days \\ So, they completed \\ 20 \times 2 = 40 units \\ Work left = 60 - 40 = 20 units \end{aligned}$ (T.W.) 2 0 1 units/day ↓ 20 days (A) A's efficiency 1 units/day A will complete the whole work in $\begin{aligned} = \frac{Total work}{Efficiency} \\ = \frac{60}{1} \\ = 60 days \\ A l t e r n a t e : \\ (A + B) \times 10 = A \times 20 \\ \frac{A + B}{A} = \frac{2}{1} \\ Total work \\ = 2 \times 30 \\ = 60 units \\ Time taken by A alone \\ = \frac{60}{1} \\ = 60 days \end{aligned}$

George takes 8 hours to copy a 50-page manuscript while Sonia can copy the same manuscript in 6 hours. How many hours would it take them to copy a 100-page manuscript, if they work together ?

(A) $6 \frac{6}{7} hours$
(B) $9 hours$
(C) $9 \frac{5}{7} hours$
(D) $14 hours$

Solution: Number of pages typed by Gorge in 1 hour $\begin{aligned} = \frac{50}{8} \\ = \frac{25}{4} \end{aligned}$ Number of pages typed by Sonia in 1 hour $\begin{aligned} = \frac{50}{6} \\ = \frac{25}{3} \end{aligned}$ Number of pages typed by Gorge and Sonia together in 1 hour $\begin{aligned} = (\frac{25}{4} + \frac{25}{3}) \\ = (\frac{75 + 100}{12}) \\ = \frac{175}{12} \\ ∴ Required time \\ = (100 \div \frac{175}{12}) hours \\ = (\frac{100 \times 12}{175}) hours \\ = \frac{48}{7} hours \\ = 6 \frac{6}{7} hours \end{aligned}$

30 men working 8 hours per day can dig a pond in 16 days. By working how many hours per day can 32 men dig the same pond, in 20 days?

(A) 6 hours/day
(B) 5 hours/day
(C) 7 hours/day
(D) 8 hours/day

A, B and C can complete a piece of work in 10, 12 and 15 days respectively. A left the work 5 days before the work was completed and B left 2 days after A had left. Number of days required to complete the whole work was ?

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

Solution: L.C.M. of total work = 60 One day work of A = $\frac{60}{10}$ = 6 unit/day One day work of B = $\frac{60}{12}$ = 5 unit/day One day work of C = $\frac{60}{15}$ = 4 unit/day $\begin{aligned} A leaves before 5 days \\ = 6 \times 5 \\ = 30 \\ B leaves before 3 days \\ = 3 \times 5 \\ = 15 \\ Then total work \\ = 60 + 30 + 15 \\ = 105 \\ Total efficiency \\ = 6 + 5 + 4 \\ = 15 \\ Total days requierd \\ = \frac{105}{15} \\ = 7 days \end{aligned}$

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

10.

If one man or two women or four boys or five girls can finish a work in 39 days, then how many days will one man, one woman, one boy and one girl together take to finish the same work?

(A) 40
(B) 10
(C) 30
(D) 20

Solution: Man × 39 = 2 Women × 39 = 4 Boys × 39 = 5 Girls × 39 $\begin{matrix} \begin{matrix} M \\ ↓ \\ 20 \end{matrix} & \begin{matrix} = \end{matrix} & \begin{matrix} 2W \\ ↓ \\ 10 \end{matrix} & \begin{matrix} = \end{matrix} & \begin{matrix} 4B \\ ↓ \\ 5 \end{matrix} & \begin{matrix} = \end{matrix} & \begin{matrix} 5G \\ ↓ \\ 4 \end{matrix} \\ Efficiency \to & M & W & B & G \end{matrix}$ Total work = 20 × 39 20 × 39 = (M + W + B + G) × d 20 × 39 = 39 × d d = 20 days

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Practice MCQ Questions and Answer on Time and Work

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