Practice MCQ Questions and Answer on Time and Work
351.
A can do three times the work done by B in one day. They together finish $$\frac{2}{5}$$ of the work in 9 days. The number of days by which B can do the work alone are ?
(A) 120 days
(B) 100 days
(C) 30 days
(D) 90 days
Solution:
Efficiency of A = 3 efficiency of B $$\left( {{\text{A}} = 3,{\text{B}} = 1} \right)$$ $$\frac{2}{5}$$ th of the work done by (A + B) in 9 days Total work (A + B) = $$\frac{{45}}{2}$$ days will be completed $$\eqalign{ & {\text{Total work}} \cr & = {\text{days}} \times {\text{efficiency }}\left( {{\text{A}} + {\text{B}}} \right) \cr & = \frac{{45}}{2} \times 4 \cr & = 90 \cr & {\text{Number of days for B }} \cr & {\text{ = }}\frac{{{\text{Total work}}}}{{{\text{Efficiency}}}} \cr & = \frac{{90}}{1} \cr & = 90{\text{ days}} \cr} $$
352.
15 men can finish a piece of work in 20 days, however it takes 24 women to finish it in 20 days. If 10 men and 8 women undertake to complete the work, then they will take ?
(A) 20 days
(B) 30 days
(C) 10 days
(D) 15 days
Solution:
$$\eqalign{ & {\text{According to the question,}} \cr & {\text{15 men}} = {\text{20 days}} \cr & {\text{300 men}} = 1{\text{ days}}.....{\text{(i)}} \cr & {\text{24 women}} = {\text{20 days}} \cr & {\text{480 men}} = 1{\text{ days}}......{\text{(ii)}} \cr & {\text{Compare equation (i) and (ii)}} \cr & {\text{300 men}} = 480{\text{ women}} \cr & {\text{5 men}} = 8{\text{ women}}.....{\text{(iii)}} \cr & {\text{10 men}} + 8{\text{ women}} = ? \cr & {\text{10 men}} + {\text{5 men}} = ? \cr & 15\,{\text{men}} = ? \cr} $$ $${\text{15 men}} \times {\text{20 days}}$$ = $${\text{15 men}}$$ $$ \times $$ $$x{\text{ days}}$$ $$x$$ = 20 days Alternate $$\eqalign{ & {\text{15m}} \times {\text{20 days}} = 24{\text{w}} \times 20{\text{ days}} \cr & \frac{{\text{m}}}{{\text{w}}} = \frac{8}{5} \cr} $$ So, 1 man work 8 units work in one day and 1 woman work 5 units work in one day Total work = 15 × 8 × 20 Hence, (10 men + 8 women) work whole in D days $$\eqalign{ & \left( {{\text{10m}} + {\text{8w}}} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & \left( {{\text{10}} \times {\text{8}} + {\text{8}} \times {\text{5}}} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & \left( {{\text{80}} + 40} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & {\text{D}} = 20{\text{ days}} \cr} $$
353.
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 $$ = \frac{{72}}{{4 + 2 + 1}} = \frac{{72}}{7}{\text{days}}$$
354.
A can complete 25% of a work in 15 days. He works for 15 days and then B alone finishes the remaining work in 30 days. In how many days will A and B working together finish 50% of the same work?
(A) 24
(B) 20
(C) 12
(D) 25
Solution:
$$\frac{{250}}{{100}}A \to 15{\text{ days}}$$ A → 60 days. (Whole work) Let A do 60 work in 60 days. A do 15 day work, remaining work completed by A in 45 days. But B do in 30 days. ∴ 45A = 30B $$\frac{{\text{A}}}{{\text{B}}} = \frac{2}{3}\,\,\,\left( {{\text{efficiency}}} \right)$$ A + B = 5 Total work = A × 60 = 2 × 60 = 120 50% of 120 = 60 A + B = $$\frac{{60}}{5}$$ = 12
355.
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
Solution:
8 × 30 × 16 = 32 × 20 × x x = 6 hours per day