Time and Work

1.If A and B together can complete a piece of work in 15 days and B alone in 20 days, in how many days can A alone complete the work?

**Answer:**Option** 1**

Solution:

A and B complete a work in = 15 days

One day's work of (A + B) =

$$\frac{1}{15}$$

B complete the work in = 20 days;

One day's work of B =

$$\frac{1}{20}$$

Then, A's one day's work

$$\begin{array}{rl}& =\frac{1}{15}-\frac{1}{20}\\ & =\frac{4-3}{6}\\ & =\frac{1}{60}\end{array}$$

Thus, A can complete the work in = 60 days.

(A + B)'s one day's % work =

$$\frac{100}{15}$$

= 6.66%B's one day's % work =

$$\frac{100}{20}$$

= 5%A's one day's % work = 6.66 - 5 = 1.66%

Thus, A need =

$$\frac{100}{1.66}$$

= 60 days to complete the work.
2.If A and B together can complete a work in 18 days, A and C together in 12 days, and B and C together in 9 days, then B alone can do the work in:

**Answer:**Option** 1**

Solution:

$$\begin{array}{rl}& \text{One}\phantom{\rule{thinmathspace}{0ex}}\text{day's}\phantom{\rule{thinmathspace}{0ex}}\text{work}\phantom{\rule{thinmathspace}{0ex}}\text{of}\\ & \left(A+B\right)=\frac{1}{18}\phantom{\rule{thinmathspace}{0ex}}.......(1)\\ & \text{One}\phantom{\rule{thinmathspace}{0ex}}\text{day's}\phantom{\rule{thinmathspace}{0ex}}\text{work}\phantom{\rule{thinmathspace}{0ex}}\text{of}\\ & \left(A+C\right)=\frac{1}{12}\phantom{\rule{thinmathspace}{0ex}}.......\left(2\right)\\ & \text{One}\phantom{\rule{thinmathspace}{0ex}}\text{day's}\phantom{\rule{thinmathspace}{0ex}}\text{work}\phantom{\rule{thinmathspace}{0ex}}\text{of}\\ & \left(B+C\right)=\frac{1}{9}\phantom{\rule{thinmathspace}{0ex}}.......\left(3\right)\\ & \text{Adding}\phantom{\rule{thinmathspace}{0ex}}\left(\text{1}\right)\text{,}\phantom{\rule{thinmathspace}{0ex}}\left(\text{2}\right)\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\left(\text{3}\right)\\ & 2\times \left(A+B+C\right)\\ & =\left\{\left(\frac{1}{18}\right)+\left(\frac{1}{12}\right)+\left(\frac{1}{9}\right)\right\}\\ & =\frac{1}{4}\\ & \text{One day's work of}\\ & \left(A+B+C\right)=\frac{1}{8}\\ & B=\left(\frac{1}{8}\right)-\left(A+C\right)\\ & B=\left(\frac{1}{8}\right)-\left(\frac{1}{12}\right)\\ & \text{One day's work of}\\ & B=\frac{\left(3-2\right)}{24}=\frac{1}{24}\\ & B\phantom{\rule{thinmathspace}{0ex}}\text{need}\phantom{\rule{thinmathspace}{0ex}}24\phantom{\rule{thinmathspace}{0ex}}\text{days}\end{array}$$

3.A and B together can complete a work in 3 days. They start together but after 2 days, B left the work. If the work is completed after two more days, B alone could do the work in

**Answer:**Option** 1**

Solution:

(A+B)'s one day's work =

$$\frac{1}{3}$$

part(A+B) works 2 days together =

$$\frac{2}{3}$$

partRemaining work =

$$1-\frac{2}{3}$$

= $$\frac{1}{3}$$

part$$\frac{1}{3}$$

part of work is completed by A in two daysHence, one day's work of A =

$$\frac{1}{6}$$

Then, one day's work of B =

$$\frac{1}{3}-\frac{1}{6}$$

= $$\frac{1}{6}$$

So, B alone can complete the whole work in 6 days.

(A+B)'s one day's % work =

$$\frac{100}{3}$$

= 33.3%Work completed in 2 days = 66.6%

Remaining work = 33.4%

One day's % work of A =

$$\frac{33.4}{2}$$

= 16.7%One day's work of B = 33.3 - 16.7 = 16.7%

B alone can complete the work in,

=

$$\frac{100}{16.7}$$

= 6 days.
4.Working 5 hours a day, A can Complete a work in 8 days and working 6 hours a day, B can complete the same work in 10 days. Working 8 hours a day, they can jointly complete the work in:

**Answer:**Option** 1**

Solution:

Working 5 hours a day, A can complete the work in 8 days i.e.

= 5 × 8 = 40 hours

Working 6 hours a day, B can complete the work in 10 days i.e.

= 6 × 10 = 60 hours

(A + B)'s 1 hour's work,

$$\begin{array}{rl}& =\frac{1}{40}+\frac{1}{60}\\ & =\frac{3+2}{120}\\ & =\frac{5}{120}\\ & =\frac{1}{24}\end{array}$$

Hence, A and B can complete the work in 24 hours i.e. they require 3 days to complete the work.

% 1 hour's work of A =

$$\frac{100}{40}$$

= 2.5%% 1 hour's work of B =

$$\frac{100}{60}$$

= 1.66(A + B) one hour's % work,

= (2.5 + 1.66) = 4.16%

Time to complete the work,

=

$$\frac{100}{4.16}$$

= 24 hoursThen,

$$\frac{24}{8}$$

= 3 daysThey need 3 days, working 8 hours a day to complete the work.

5.Ganga and Saraswati, working separately can mow field in 8 and 12 hours respectively. If they work in stretches of one hour alternately. Ganga is beginning at 9 a.m., when will the moving be completed?

**Answer:**Option** 1**

Solution:

Whenever, workers are working alternatively on one work, we take 2 as 1 unit

In this case, we take 2 hours as 1 unit.

Part of the field moved by Ganga and Saraswati in 2 hours (1 unit) =

$$\frac{1}{8}$$

+ $$\frac{1}{12}$$

= $$\frac{5}{24}$$

Time taken to complete the work,

=

$$\frac{1}{\frac{5}{24}}$$

= $$\frac{24}{5}$$

unit of timeThen actual time taken by them to complete the work,

=

$$2\times \frac{24}{5}$$

= 9.6 hoursThe work starts at 9 a.m. then it will complete at 6:36 pm

% of the field moved by Ganga and Saraswati in 2 hours (1 unit),

=

$$\left(\frac{100}{8}\right)\mathrm{\%}$$

+ $$\left(\frac{100}{12}\right)\mathrm{\%}$$

= 12.5 + 8.33

= 20.83%; Time taken to move the whole field,

=

$$\frac{100\mathrm{\%}}{20.83\mathrm{\%}}$$

= 4.8 unit of time;

Hence, actual time = 2 × 4.8 = 9.6 hours.

6.If 10 men can do a piece of work in 12 days, the time taken by 12 men to do the same piece of work will be:

**Answer:**Option** 1**

Solution:

Here, we use work equivalence method;10*12 = 12*x;

Or, x = 10 days;

Men Days

10 ↓ 12

12 ↑ x (let)

Here, the two arrows, downward (↓) and upward (↑) show variation between men and days.

Thus,

$$\begin{array}{rl}& \frac{10}{12}=\frac{\text{x}}{12}\\ & \text{or, x}=\frac{10\times 12}{12}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=10\phantom{\rule{thinmathspace}{0ex}}\text{days}\end{array}$$

7.To complete a work, A takes 50% more time than B. If together they take 18 days to complete the work, how much time shall B take to do it?

**Answer:**Option** 1**

Solution:

$$\begin{array}{rl}& \text{We have}\\ & \text{B}=\frac{3}{2}\times \text{A}\\ & \to \text{A}=\frac{2}{3}\times \text{B}\\ & \text{One day's work,}\\ & \text{A}+\text{B}=\frac{1}{18}\\ & \frac{2}{3}\times \text{B}+\text{B}=\frac{1}{18}\\ & \frac{5}{3}\times \text{B}=\frac{1}{18}\\ & \text{One}\phantom{\rule{thinmathspace}{0ex}}\text{day's}\phantom{\rule{thinmathspace}{0ex}}\text{work}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{B}\\ & =\frac{3}{90}\end{array}$$

B alone can complete the work in

$$\begin{array}{rl}& =\frac{90}{3}\\ & =30\phantom{\rule{thinmathspace}{0ex}}\text{days}\end{array}$$

8.A complete

$$\frac{7}{10}$$

of a work in 15 days, then he completed the remaining work with the help of B in 4 days. In how many day A and B can complete entire work together?**Answer:**Option** 1**

Solution:

$$\frac{7}{10}$$

part of work has been completed by A in 15 days. Then,Rest work = 1 -

$$\frac{7}{10}$$

= $$\frac{3}{10}$$

partGiven, That

$$\frac{3}{10}$$

part of the work is completed by A and B together in 4 days. Means, (A + B) completed the

$$\frac{3}{10}$$

of work in 4 daysSo, (A + B)'s 1 day's work =

$$\frac{3}{10\times 4}$$

= $$\frac{3}{40}$$

Hence,

(A + B) can complete the work in

$$\frac{40}{3}$$

= $$13\frac{1}{3}$$

days
9.A can complete a piece of work in 36 days, B in 54 days and C in 72 days. All the three began the work the work together but A left 8 days before the completion of the work and B 12 days before the completion of work. Only C worked up to the end. In how many days was the work completed?

**Answer:**Option** 1**

Solution:

Let the work be completed in x days. C work for x days then A works for (x - 8) days and B works for (x - 12) days.According to the question,

$$\begin{array}{rl}& \left[\frac{\left(x-8\right)}{36}+\frac{\left(x-12\right)}{54}+\frac{x}{72}\right]=1\\ & \left[\frac{\left(6x-48+4x-48+3x\right)}{216}\right]=1\\ & 13x-96=216\\ & 13x=216+96=312\\ & x=\frac{312}{12}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=24\phantom{\rule{thinmathspace}{0ex}}\text{days}\end{array}$$

10.While working 7 hour a day, A alone can complete a piece of work in 6 days and B alone in 8 days. In what time would they complete it together, 8 hour a day?

**Answer:**Option** 1**

Solution:

A can complete the work in 7 × 6 = 42 hours1 hour's work of A =

$$\frac{1}{42}$$

B can complete the work in 7 × 8 = 56 hours

1 hour's work of B =

$$\frac{1}{56}$$

(A + B)'s 1 hour's work

$$\begin{array}{rl}& =\frac{1}{42}+\frac{1}{56}\\ & =\frac{4+3}{168}\\ & =\frac{7}{168}\end{array}$$

∴ Time taken by (A + B) working 8 hours daily

$$\frac{168}{7\times 8}=3\phantom{\rule{thinmathspace}{0ex}}\text{days}$$