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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?

A. 60

B. 45

C. 40

D. 30

Discuss

Solution:

1^st 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{aligned} = \frac{1}{15} - \frac{1}{20} \\ = \frac{4 - 3}{6} \\ = \frac{1}{60} \end{aligned}

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

2^nd 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:

A. 18 days

B. 24 days

C. 30 days

D. 40 days

Discuss

Solution:

\begin{aligned} One day's work of \\ (A + B) = \frac{1}{18} . . . . . . . (1) \\ One day's work of \\ (A + C) = \frac{1}{12} . . . . . . . (2) \\ One day's work of \\ (B + C) = \frac{1}{9} . . . . . . . (3) \\ Adding (1), (2) and (3) \\ 2 \times (A + B + C) = \frac{1}{18} + \frac{1}{12} + \frac{1}{9} \\ 2 (A + B + C) = \frac{1}{4} \\ One day's work of \\ A + B + C = \frac{1}{8} \\ B = \frac{1}{8} - (A + C) \\ B = \frac{1}{8} - \frac{1}{12} \\ One day's work of \\ B = \frac{3 - 2}{24} = \frac{1}{24} \\ B need 24 days \end{aligned}

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

A. 5 days

B. 6 days

C. 9 days

D. 10 days

Discuss

Solution:

1^st 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.

2^nd 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 = 33.4 - 16.7 = 16.7%
B alone can complete the work in,
=

\frac{100}{16.7}

= 5.98 days
≈ 6 days.

4. A can complete a piece of work in 18 days, B in 20 days and C in 30 days, B and C together start the work and forced to leave after 2 days. The time taken by A alone to complete the remaining work is:

A. 10 days

B. 12 days

C. 15 days

D. 16 days

Discuss

Solution:

1^st Method:

\begin{aligned} (B + C) 2 days work \\ = 2 \times (\frac{1}{20} + \frac{1}{30}) \\ = 2 \times \frac{3 + 2}{60} \\ = \frac{1}{6} part \\ Remaining work \\ = 1 - \frac{1}{6} \\ = \frac{5}{6} part \\ A's one day's work \\ = \frac{1}{18} part \\ Time taken to complete the work \\ = \frac{\frac{5}{6}}{\frac{1}{18}} days \\ Hence, \\ Time taken to complete the work \\ = \frac{5}{6} \times 18 \\ = 15 days \end{aligned}

2^nd Method:
% of work B completes in one day =

\frac{100}{20}

= 5%;
% of work C completes in one day =

\frac{100}{30}

= 3.33%;
% of work (A + B) completes together in one day = 5 + 3.33 = 8.33%;
% work (A + B) completes together in 2 days = 8.66 × 2 = 16.66%;
Remaining work = 100 - 16.66 = 83.34%;
% of work A completes in 1 day =

\frac{100}{18}

= 5.55%
Time taken to complete the remaining work by A
=

\frac{83.34}{5.55}

= 15 days

5. 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:

A. 3 days

B. 4 days

C. 4.5 days

D. 5.4 days

Discuss

Solution:

1^st 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{aligned} = \frac{1}{40} + \frac{1}{60} \\ = \frac{3 + 2}{120} \\ = \frac{5}{120} \\ = \frac{1}{24} \end{aligned}

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

2^nd 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.

6. 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?

A. 6:20 PM

B. 6:30 PM

C. 6:36 PM

D. 6:42 PM

Discuss

Solution:

According to question,
Ganga begins at 9 am and she does 3 units/hours
Saraswati begins at 10 am and she does 2 units/hours
So by 11 am they complete 5 units
Time

= \frac{T .W .}{3 + 2} = \frac{24}{5}

(4 cycle of 2 hrs each + 4 units left)
And now ganga will complete 3 unit out of 4 units in 1 hr
Now, rest 1 unit work done by =

\frac{1}{2}

hr
Total time = 8 + 1 +

\frac{1}{2}

= 9

\frac{1}{2}

hr
Hence,Work finished at
= 9 am + 9

\frac{1}{2}

hr
= 6:30 PM

Alternate Solution:
Work done by Ganga in 1 hour =

\frac{1}{8}

Work done by Saraswati in 1 hour =

\frac{1}{12}

They are working alternatively with Ganga beginning the job.
Work done in every two hours =

\frac{1}{8}

\frac{1}{12}

\frac{5}{24}

Work done in 4 × 2 = 8 hours =

\frac{5 \times 4}{24}

\frac{5}{6}

Remaining work = 1 -

\frac{5}{6}

\frac{1}{6}

In 9th hour, Ganga starts the work and does

\frac{1}{8}

of the work
Work remaining =

\frac{1}{6}

\frac{1}{8}

\frac{1}{24}

In 10th hour, Saraswati starts the work
Time needed to finish the remaining work

\begin{aligned} = \frac{\frac{1}{24}}{\frac{1}{12}} \\ = \frac{1}{24} \times 12 \end{aligned}

=

0.5 hours

=

30 minutes
i.e., work will be completed in 9 hour 30 minutes, after 9 AM
i.e., at 6:30 PM

7. 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:

A. 12 days

B. 10 days

C. 9 days

D. 8 days

Discuss

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{aligned} \frac{10}{12} = \frac{x}{12} \\ or, x = \frac{10 \times 12}{12} \\ = 10 days \end{aligned}

8. 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?

A. 30 days

B. 35 days

C. 40 days

D. 45 days

Discuss

Solution:

\begin{aligned} We have \\ B = \frac{3}{2} \times A \\ \to A = \frac{2}{3} \times B \\ One day's work, \\ \Rightarrow A + B = \frac{1}{18} \\ \Rightarrow \frac{2}{3} \times B + B = \frac{1}{18} \\ \Rightarrow \frac{5}{3} \times B = \frac{1}{18} \\ One day's work of B \\ = \frac{3}{90} \end{aligned}

B alone can complete the work in

\begin{aligned} = \frac{90}{3} \\ = 30 days \end{aligned}

9. If 10 men or 20 boys can make 260 mats in 20 days, then how many mats will be made by 8 men and 4 boys in 20 days?

A. 260

B. 240

C. 280

D. 520

Discuss

Solution:

10 men = 20 boys
→1 men = 2 boys
8 men = 2 × 8 boys = 16 boys
Then,
(16 boys + 4 boys) = 20 boys can make 260 mats in 20 days
Now,
It can be calculated by work equivalence method:
20 × 260 × 20 = x × 20 × 20
x = 260 mats

10. 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?

A. $10 \frac{1}{2} days$

B. $12 \frac{2}{3} days$

C. $13 \frac{1}{3} days$

D. $8 \frac{1}{4} days$

Discuss

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 \times 4}

\frac{3}{40}

Hence,
(A + B) can complete the work in

\frac{40}{3}

13 \frac{1}{3}

days

1 2 3 4 5 6 7 8 9 10

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