Practice MCQ Questions and Answer on Time and Work

A group of workers can complete a piece of work in 50 days, when they are working individually. On the first day one person works, on the second day another person joins him, on the third day one person joins them and this process continues till the work is completed. How many approximate days are needed to complete the work ?

(A) 8 days
(B) 9 days
(C) 10 days
(D) 11 days

Solution: Let a man complete 1 piece of work in a day. Then total work = 50 units Then by statement 1 st day = one man × 1 work/day = 1 Then by statement 2nd day = two men × 1 work/day = 2 Then by statement 3rd day = three men × 1 work/day = 3 Let the whole work will be completed in N days. Then total work 1 + 2 + 3 + ..... + N = 50 $\begin{aligned} \frac{N (N + 1)}{2} = 50 \\ N (N + 1) = 100 \\ Then, N = 10 days (approx) \end{aligned}$

Ramesh and Rahman can do a work in 20 and 25 days respectively. After doing collectively 10 days of work, they leave the work due to illness and Suresh completes rest of the work in 3 days. How many days Suresh alone can take to complete the whole work ?

(A) 32 days
(B) 28 days
(C) 29 days
(D) 30 days

Solution: (Ramesh & Rahman)'s 1 day's work $= \frac{1}{20} + \frac{1}{25} = \frac{5 + 4}{100} = \frac{9}{100}$ ∴ Their 10 day's work $= \frac{90}{100} = \frac{9}{10}$ ∴ Remaining work $= 1 - \frac{9}{10} = \frac{1}{10}$ ∵ Suresh does $\frac{1}{10}$ work in 3 days ∴ Time taken by Suresh in doing 1 work = 3 × 10 = 30 days

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 takes three times as long as B and C together to do a job. B takes four times as long as A and C together to do the work. If all the three, working together can complete the job in 24 days, then the number of days, A alone will take to finish the job is = ?

(A) 100 days
(B) 96 days
(C) 95 days
(D) 90 days

Solution: Let time taken by B and C = x days ∴ Time taken by A = 3x days ∴ Part of work done by A, B and C in 1 day $\begin{aligned} = \frac{1}{x} + \frac{1}{3 x} = \frac{3 + 1}{3 x} = \frac{4}{3 x} \\ ∴ \frac{4}{3 x} = \frac{1}{24} \\ \Rightarrow 3 x = 4 \times 24 \\ \Rightarrow x = \frac{4 \times 24}{3} = 32 days \end{aligned}$ ∴ Time taken by A = 32 × 3 = 96 days. Alternet : Work done by all of them together in $\frac{1}{24}$ Efficiency of A : Efficiency of (B + C) = 1 : 3 Work done by A in 1 day = $\frac{1}{24}$ × $\frac{1}{4}$ = $\frac{1}{96}$ i.e., A alone can finish the job in 96 days. Alternet : Suppose A can complete the job in 3x days and (B + C) can complete the job in x days. $\begin{aligned} \frac{3 x \times x}{3 x + x} = 24 \\ \Rightarrow \frac{3 x}{4} = 24 \\ \Rightarrow x = 32 \\ 3 x = 96 days. \end{aligned}$

A can knit a pair of socks in 3 days. B can knit the same pair of socks in 9 days. If they are knitting together, then in how many days will they knit two pairs of socks ?

(A) 3 days
(B) 4 days
(C) $4 \frac{1}{2}$ days
(D) 5 days

Solution: Number of pairs knit by A and B together in 1 day $\begin{aligned} = (\frac{1}{3} + \frac{1}{9}) \\ = \frac{4}{9} \end{aligned}$ ∴ Required number of days, $\begin{aligned} = (2 \div \frac{4}{9}) \\ = (2 \times \frac{9}{4}) \\ = \frac{9}{2} \\ = 4 \frac{1}{2} days \end{aligned}$

To do a certain work, the ratio of efficiency of A to that of B is 3 : 7. Working together, they can complete the work in $10 \frac{1}{2}$ ÃÂÃÂ days. They work together for 8 days. 60% of the remaining work will be completed by A alone in:

(A) $6 \frac{1}{2} Days$
(B) 4 Days
(C) 5 Days
(D) $5 \frac{1}{2} Days$

Solution: $\begin{matrix} A & B \\ Efficiency \to & 3 & : & 7 \end{matrix}$ $\begin{aligned} Total work = (3 + 7) \times \frac{21}{2} = 105 \\ (A + B) \times 8 = (3 + 7) \times 8 = 80 \\ Remaining work = 105 - 80 = 25 \\ 60 % work done by A = \frac{25 \times 60}{100 \times 3} = 5 days \end{aligned}$

If 72 men can build a wall of 280 m length in 21 days, how many men could take 18 days to build a similar type of wall of length 100 m ?

(A) 30
(B) 10
(C) 18
(D) 28

Solution: $\begin{aligned} Here work is 280 m length of wall \\ And 100 m length of wall \\ Let, M men will finish 100 m wall . \\ \frac{72 \times 21}{280} = \frac{M \times 18}{100} \\ \Leftrightarrow M = 30 \end{aligned}$

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) $\begin{aligned} After solving \\ D = 2 days \end{aligned}$

A group of workers was put on a job. From second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in 55% of the time. How many workers were there in the group?

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

Solution: Let initially X number of workers be there. Now, Using work equivalence method, X + (X - 1) + (X - 2) + . . . . . + 1 = X × 55% of X $\frac{X \times (X + 1)}{2} = \frac{5 X}{100}$ [series is in AP. Sum of AP = {No. of terms (first term + last term)/2}] X = 10 workers.

10.

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

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Practice MCQ Questions and Answer on Time and Work

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