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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?

Explanation:

Solution:
1st method:
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.

2nd method:
(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:

Explanation:

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

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

Explanation:

Solution:
1st Method:

(A+B)'s one day's work =
$\frac{1}{3}$
part
(A+B) works 2 days together =
$\frac{2}{3}$
part
Remaining work =
$1-\frac{2}{3}$
=
$\frac{1}{3}$
part
$\frac{1}{3}$
part of work is completed by A in two days
Hence, 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.

2nd Method:
(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:

Explanation:

Solution:
1st Method:
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.

2nd Method:
% 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 hours
Then,
$\frac{24}{8}$
= 3 days
They 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?

Explanation:

Solution:
1st Method:
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 time
Then actual time taken by them to complete the work,
=
$2×\frac{24}{5}$
= 9.6 hours
The work starts at 9 a.m. then it will complete at 6:36 pm

2nd Method:
% 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:

Explanation:

Solution:
Here, we use work equivalence method;
10*12 = 12*x;
Or, x = 10 days;

To understand the work equivalence method, we use a graphic as follows:

Men   Days
10 ↓    12
12     ↑ x (let)
Here, the two arrows, downward (↓) and upward (↑) show variation between men and days.
[If downward arrows show decrements then upward arrows show increments and vice-verse.]
Thus,
$\begin{array}{rl}& \frac{10}{12}=\frac{\text{x}}{12}\\ & \text{or, x}=\frac{10×12}{12}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=10\phantom{\rule{thinmathspace}{0ex}}\text{days}\end{array}$

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?

Explanation:

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

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?

Explanation:

Solution:
$\frac{7}{10}$
part of work has been completed by A in 15 days. Then,
Rest work = 1 -
$\frac{7}{10}$
=
$\frac{3}{10}$
part
Given, 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 days
So, (A + B)'s 1 day's work =
$\frac{3}{10×4}$
=
$\frac{3}{40}$

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?

Explanation:

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?

Explanation:

Solution:
A can complete the work in 7 × 6 = 42 hours
1 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
$\frac{168}{7×8}=3\phantom{\rule{thinmathspace}{0ex}}\text{days}$

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