Time And Work - Aptitude MCQ Questions and Solutions with Explanations

11

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 $$10\frac{1}{2}$$ days. They work together for 8 days. 60% of the remaining work will be completed by A alone in:

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

Discuss

Answer & Solution

Answer: Option C

Solution:

\[\begin{array}{*{20}{c}} {}&A&{}&B \\ {{\text{Efficiency}} \to }&3&:&7 \end{array}\]
$$\eqalign{ & {\text{Total work}} = \left( {3 + 7} \right) \times \frac{{21}}{2} = 105 \cr & \left( {A + B} \right) \times 8 = \left( {3 + 7} \right) \times 8 = 80 \cr & {\text{Remaining work}} = 105 - 80 = 25 \cr & 60\% {\text{ work done by }}A = \frac{{25 \times 60}}{{100 \times 3}} = 5{\text{ days}} \cr} $$

12

3 men and 4 women can do a piece of work in 7 days, whereas 2 men and 1 woman can do it in 14 days. 7 women will complete the same work in:

A) 10 days B) 8 days C) 9 days D) 12 days

Discuss

Answer & Solution

Answer: Option A

Solution:

(3m + 4w) × 7 = (2m + 1w) × 14
3m + 4w = 4m + 2w
m = 2w
m : w = 2 : 1
Total work = (3 × 2 + 4 × 1) × 7 = 70 unit
7w = 70 unit = $$\frac{{70}}{7}$$ = 10 days

13

To do a certain work, the ratio of the efficiencies of X and Y is 5 : 4. Working together, they can complete the same work in 10 days. Y alone starts the work and leaves after 5 days. The remaining work will be completed by X alone in:

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

Discuss

Answer & Solution

Answer: Option A

Solution:

\[\begin{array}{*{20}{c}} {}&{\text{X}}&{}&{\text{Y}} \\ {{\text{Efficiency}}}&5&:&4 \end{array}\]
Total work = (5 + 4) × 10 = 90
Y × 5 + X × a = 90
4 × 5 + 5 × a = 90
20 + 5a = 90
5a = 70
a = 14

14

A contractor decided to complete a work in 80 days and employed 60 men at the beginning and 20 men additionally after 20 days and got the work completed as per schedule. If he had not employed, the additional men, how many extra days would he have needed to complete the work (round off to the nearest integer)?

A) 32 days B) 20 days C) 30 days D) 26 days

Discuss

Answer & Solution

Answer: Option B

Solution:

Total work = 80 × 60 + 60 × 20
= 4800 + 1200
= 6000
60 men × d = 6000
d = 100 days
Extra days = 100 - 80 = 20 days

15

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) $$2\frac{1}{2}{\text{ days}}$$ C) 4 days D) $$3\frac{1}{2}{\text{ days}}$$

Discuss

Answer & Solution

Answer: Option A

Solution:

\[\begin{array}{*{20}{c}} {}&\begin{gathered} {\text{A}}\,\,\,\,\,\,{\text{B}}\,\,\,\,\,\,{\text{C}} \hfill \\ {\text{4}}\,\,\,{\text{:}}\,\,\,{\text{5}}\,\,\,{\text{:}}\,\,\,{\text{7}} \hfill \\ {\text{5}}\,\,\,{\text{:}}\,\,\,{\text{7}}\,\,\,\,\,\,\,\,\,\, \hfill \\ \end{gathered} \\ {{\text{Efficiency}} \to }&{\overline {\,4\,:\,5\,:\,7\,} } \end{array}\]
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

16

A and B can do a piece of work in 18 days. B and C together can do it in 30 days. If A is twice as good a workman as C, find the how many days B alone can do the work?

A) 90 days B) 100 days C) 80 days D) 75 days

Discuss

Answer & Solution

Answer: Option A

Solution:

A : C = 2 : 1
A = 4 unit
C = 2 unit
B = 1 unit
B = $$\frac{{90}}{1}$$ = 90 days

17

To do a certain work, the ratio of the efficiencies of A and B is 7 : 5. Working together, they can complete the same work in $$17\frac{1}{2}$$ days. A alone will complete 60% of the same work in:

A) 18 days B) 15 days C) 16 days D) 21 days

Discuss

Answer & Solution

Answer: Option A

Solution:

\[\begin{array}{*{20}{c}} {}&{\text{A}}&:&{\text{B}} \\ {{\text{Efficiency}}}&7&:&5 \end{array}\]
$$\eqalign{ & {\text{A}} + {\text{B}} = 12 \times \frac{{35}}{2} = 35 \times 6{\text{ Total work}} \cr & {\text{A}} = \frac{{35 \times 6 \times \frac{3}{5}}}{7} = 18{\text{ days}} \cr} $$

18

A is as efficient as B and C together. Working together A and B can complete a work in 36 days and C alone can complete it in 60 days. A and C work together for 10 days. B alone will complete the remaining work in:

A) 88 Days B) 84 Days C) 110 Days D) 90 Days

Discuss

Answer & Solution

Answer: Option C

Solution:

\[\begin{array}{*{20}{c}} {}&A&:&{B + C}&{}&{} \\ {{\text{Efficiency}} \to }&{{1_{ \times 4}}}&:&{{1_{ \times 4}}}& \Rightarrow &{{2_{ \times 4}}} \\ {}&{A + B}&:&C&{}&{} \\ {{\text{Days}} \to }&{36}&:&{60}&{}&{} \\ {{\text{Efficiency}} \to }&5&:&3& \Rightarrow &8 \\ {{\text{Efficiency}} \to }&A&:&B&:&C \\ {}&4&:&1&:&3 \end{array}\]
$$\eqalign{ & {\text{Total work}} = 3 \times 60 = 180 \cr & \left( {A + C} \right) \times 10 + B \times x = 180 \cr & 7 \times 10 + 1 \times x = 180 \cr & x = 110{\text{ Days}} \cr} $$

19

A, B and C can complete a piece of work separately in 10, 20 and 40 days, respectively. In how many days will the work be completed if A is assisted by both B and C every third day?

A) $$8\frac{2}{7}$$ B) 9 C) $$10\frac{2}{3}$$ D) 6

Discuss

Answer & Solution

Answer: Option A

Solution:

A's three days work = 4 × 3 = 12
B and C one day work = 3 × 1 = 3
A, B, C three days work = 15 unit
A, B, C six days work = 30 unit
7^th day work = 4 unit
8^th day work = 4 unit
Total = 38 unit
Work left 2 unit
7 units in 1 days
2 units in $$\frac{2}{7}$$ days
Total time = $$8\frac{2}{7}$$ days

20

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

Discuss

Answer & Solution

Answer: Option D

Solution:

$$\therefore \frac{{18}}{{2 + 1 + 3}} = 3{\text{ minutes}}$$