Practice MCQ Questions and Answer on Time and Work

51.

Naman does half the work as Gagan in 4/5 the time. If together they take 16 days to complete a piece of work, then how long will it take Gagan to complete the work?

(A) 23 days
(B) 24 days
(C) 25 days
(D) 26 days

Solution: $\begin{aligned} \frac{Naman \times 4}{\frac{1}{2}} = \frac{Gagan \times 5}{1} \\ \frac{Naman}{Gagan} = \frac{5}{8} \\ Total work = 16 \times (5 + 8) = 16 \times 13 \\ Time taken by Gagan = \frac{16 \times 13}{8} = 26 days \end{aligned}$

52.

A tyre has 3 punctures. The first puncture alone would have made the tyre flat in 9 minutes, the second alone would have done it in 18 minutes, the third alone would have done it in 6 minutes. If the air leaks out at a constant rate, then how long (in minutes) does it take for all the punctures together to make it flat?

(A) 2
(B) 6
(C) 4
(D) 3

Solution: $∴ \frac{18}{2 + 1 + 3} = 3 minutes$

53.

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

54.

A and B can do a job together in 7 days. A is $1 \frac{3}{4}$ ÃÂÃÂ times as efficient as B. The same job can be done by A alone in :

(A) $9 \frac{1}{3}$ days
(B) 11 days
(C) $12 \frac{1}{4}$ days
(D) $16 \frac{1}{3}$ days

Solution: $\begin{aligned} (A's 1 day's work) : (B's 1 day's work) \\ = \frac{7}{4} : 1 = 7 : 4 \\ Let A's and B's 1 day's work be \\ 7 x and 4 x respectively \\ Then, 7 x + 4 x = \frac{1}{7} \\ \Rightarrow 11 x = \frac{1}{7} \\ \Rightarrow x = \frac{1}{77} \\ ∴ A's 1 day's work \\ = \frac{1}{77} \times 7 = \frac{1}{11} \end{aligned}$ So, A will do the work in 11 days

55.

A can do piece of work in 15 days. B is 25% more efficient than A, and C is 40% more efficient than B, A and C work together for 3 days and then C leaves. A and B together will complete the remaining work in:

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

Solution: $\begin{matrix} \begin{matrix} A B C \\ 4 : 5 : 7 \\ 5 : 7 \end{matrix} \\ Efficiency \to & \overset{―}{4 : 5 : 7} \end{matrix}$ Total work = 4 × 15 = 60 units (A + C) × 3 + (A + B)x = 60 11 × 3 + 9 × x = 60 33 + 9x = 60 9x = 27 x = 3 days

56.

Kamal can do a work in 15 days. Bimal is 50% more efficient then Kamal. The number of days, Bimal will take to do the same piece of work, is = ?

(A) $7 \frac{1}{2}$
(B) 10
(C) 12
(D) 14

Solution: Ratio of times taken by Kamal and Bimal $\begin{aligned} = 150 : 100 \\ = 3 : 2 \end{aligned}$ Suppose Bimal takes x days to do the work $\begin{aligned} \Rightarrow 3 : 2 :: 15 : x \\ \Rightarrow x = (\frac{2 \times 15}{3}) \\ \Rightarrow x = 10 days . \end{aligned}$

57.

A alone can do a work in 14 days. B alone can do the same work in 28 days. C alone can do the same work in 56 days. They start the work together and completed the work such that B was not working on last 2 days and A did not work in last 3 days. In how many days (total) was the work completed?

(A) $\frac{72}{7}$
(B) $\frac{79}{7}$
(C) $\frac{65}{7}$
(D) $\frac{82}{7}$

Solution: Work done not done by B = 2 × 2 = 4 Work done not done by A = 3 × 4 = 12 Total work = 56 + 4 + 12 = 72 ∴ Total time $= \frac{72}{4 + 2 + 1} = \frac{72}{7} days$

58.

A can finish a work in 18 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work?

(A) 5
(B) $5 \frac{1}{2}$
(C) 6
(D) 8

Solution: $\begin{aligned} B's 10 day's work \\ = \frac{1}{15} \times 10 = \frac{2}{3} \\ Remaining work \\ = 1 - \frac{2}{3} = \frac{1}{3} \\ Now, \frac{1}{18} work is done by A in 1 day \\ ∴ \frac{1}{3} work is done by A in \\ 18 \times \frac{1}{3} = 6 days \end{aligned}$

59.

The ratio of the efficiencies of A, B and C is 7 : 5 : 8. Working together, they can complete a piece of work in 42 days. B and C worked together for 21 days and the remaining work was completed by A alone. The whole work was completed in:

(A) 96 days
(B) 99 days
(C) 102 days
(D) 93 days

60.

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. $\begin{aligned} ∴ In 3 day's \frac{1}{9} work is done by \\ = (72 \times \frac{1}{9}) \\ = 8 women . \end{aligned}$

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Practice MCQ Questions and Answer on Time and Work

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