## Time and work-part-7

Problem 1: A can do as much as work in 3 days as c in 5 days. B can do as much work in 3 days as C can do in 2 days. What time would B require to do a work if A takes 24 days to finish it?

Solution:

A  $\rightarrow$ 3                                   5

L.C.M 15

C $\rightarrow$ 5                                     3

$\rightarrow$ 3                                    2

L.C.M 6

C $\rightarrow$2                                      3

A:B:C=5:2:3

$\frac{24\times&space;5}{2}=60$

Problem 2: Ram does a work in 40 days and shaym does same work in 60 days if the efficiency of mohan is 20% less than the combined of ram and shyam together than working alone mohan will complete the whole work in how many days?

Solution:

$\rightarrow$  40                                    3

L.C.M 120

S   $\rightarrow$  60                                    2

M=$\frac{80\times&space;5}{100}=4$

$\frac{120}{4}$  =  30

Problem 3: 2 men and 3 boys can do a piece of work in 10 days while 3m and 2 boys can do the same work in 8 days. In how many days 2 men and 1 boy do the work?

Solution:

2m+3b=10

3m+2b=8

multiply10 with 2 and 3

multi[ply 8 with 3 and 2

20m          30b

– 24m           16b

4m  $\rightarrow$  14b

2m  $\rightarrow$  7b

7b+3b=10

10b=10

2m+1b=?

7b+1b=8b=?

$\frac{10\times&space;10}{8}=12.5$

Problem 4: 24 men or 30 women completes a work in 12 days. 18 men and 40 women together will complete the same work in how many days?

Solution:

24m or 30w=12

18m +40w = ?

d=$\frac{24\times&space;30\times&space;12}{24\times&space;40+18\times&space;30}=\frac{8640}{1500}=5.76$ days

## Time and work-part-6

Problem 1:A and B can do a piece of work in 9 days and 18 days respectively. As they were ill they could do 45% and 90% of their effiiciency. How many days will they take to complete the work together?

Solution:

A   $\overset{100}{\rightarrow}$%   9 days

$\overset{45}{\rightarrow}$%   ? 20 days

$\frac{9&space;\times&space;100}{45}=20$

$B\overset{100}{\rightarrow}$%  18 days

$B\overset{90}{\rightarrow}$%  ? 20 days

$\frac{18\times&space;100}{90}=20$

A+B=$\frac{1}{20}+\frac{1}{20}=\frac{2}{20}=\frac{1}{10}$

Working together they complete the work in 10 days

Problem 2: Due to fever efficiency of ramesh is reduced by 40% hence he takes 12 more days than earlier to complete. If next day he starts working with 50% of his original efficiency. Now in how many days he will complete the total work?

Solution: 40%$\rightarrow&space;\frac{2}{5}$

5-2=3

E       3     :      5

T        5     :      3

5-3=2

2$\rightarrow$12

5$\rightarrow$?        $\frac{12\times&space;5}{2}=30$

3$\rightarrow$?      $\frac{12\times&space;3}{2}=18$

100%$\rightarrow&space;18$

50%$\rightarrow$?

$\frac{18\times&space;50}{100}=36$

He will complete the work in 36 days.

Problem 3: A is 20% less efficient than B. A started the work and work for X days after which B replaced by A and completed the remaining work in (x-7) days. If the ratio of work done by A and B is 3:2 working together they complete the whole work in how many days?

Solution:

A          B

80   :     100

4       :       5

$\frac{4\times&space;X}{5\times&space;(X-7)}=\frac{3}{2}$

8X=15(X-7)

8x=15x-105

7x=105

x=$\frac{105}{7}=15$

4*15=60

5*(15-7)=40

$\frac{60+40}{9}=11\tfrac{1}{9}$ days

Problem 3: A and B can complete the work in half the time of c, While B and C can complete the same work in one-third of time than A. If they together can complete the work in 20 days in how many days A alone can do the same work?

Solution:

A+B=$\frac{1}{2}\times&space;C$  $\Rightarrow$  A+B:C =  1:2 IS T

2:1 IS E

B+C=$\frac{1}{3}\times&space;A$  $\Rightarrow$  B+C:A = 1:3 IS T

3:1 IS E

A+B:C  =  2:1=2+1=3*4

B+C:A  =  3:1=3+1=4*3

To make efficiency equal multiply 3 with 4 and 4 with 3

A+B:C=8:4

B+C:A=9:3

C=4   ,    A=3    ,    B+C=9=9-4=5

4+3+5=12

$\frac{20\times&space;12}{3}=80$

Problem 3: In a garrison there was food from 1000 soldiers for one month, after 10 days, 1000 more soldiers joined the garrison. How long will the soldiers be able to carry on with the remaining food?

Solution:

1000 soldiers *30=30000

10000 soldiers*10=10000

2000  soldiers   $\rightarrow$   20000

$\frac{20000}{2000}=10$ days

## Time and work-part-5

Problems on efficiency

Problem 1: A is 30% more efficient than B. How much time will they work together, take to complete a job which A alone could have done in 23 days?

Solution:

A            B

E      130        100=13:10

T      100         130=10:13

Time and efficiency are inversely proportional to each other

$\frac{13}{23}\times&space;13=13$ days

Problem 2: A is thrice as good a work man as B and therefore is able to finish a job in 60 days less than B, working together they can do it?

Solution:

A              B

E     3              1

T      1             3

2$\rightarrow&space;60$

1$\rightarrow$30

3$\rightarrow&space;30$

A$\rightarrow$30                                        3

L.C.M  90

B$\rightarrow$90                                          1

$\frac{90}{4}=22\tfrac{1}{7}$ dAYS

Problem 3: A does half as much work as B in $\frac{3}{4}$ of the time. If together they take 18 days to complete the work, How much time shall B take to do it alone?

Solution:  let B can do 1 unit of work in 1 day

$B\overset{1}{\rightarrow}1$

$A\overset{\frac{1}{2}}{\rightarrow}\frac{3}{4}$

A can do 1 unit of work in

$A\overset{\frac{1}{2}\times&space;2}{\rightarrow}\frac{3}{4}\times&space;2$

$A\overset{1}{\rightarrow}\frac{3}{2}$

A                  B

1         :          $\frac{3}{2}$

T              2         :           3

E               3         :           2

$18\times&space;\frac{5}{3}=30&space;days$

Problem 3: A and B working together complete a work in 24 days while Band C complete the same work in 30 days. If c is 20% more efficient then B, then how many days working alone A will complete the whole work?

Solution:

A+B$\rightarrow$24

B+C$\rightarrow$30

C          B

120     100

6     :       5   $\rightarrow$ E

B=$30\times&space;\frac{11}{5}=66$

A+B$\rightarrow$24$\frac{1}{24}-\frac{1}{66}=\frac{42}{24\times&space;66}=37.7&space;days$

Problem 4: Anjan is friend of mukul and the ratio of their efficiency is 7:5 working together they complete 40% work in 8 days and remaining work is completed by anjan. If for doing work anjan received RS 24000. Find the amount that mukul will receive?

Solution:

A         M

7      :     5$\rightarrow$ E

8*12=96 is  work done by both

40%$\rightarrow$96

100%$\rightarrow$?

$\frac{96\times&space;100}{40}=240$

Work done by mukul is 5*8=40

240-40=200 is work done by anjan

M      A

40    200

1   :       5

5$\rightarrow$24000

1$\rightarrow$?

$\frac{24000\times&space;1}{5}=4800$

## Time and work-part-4

Formula:$\frac{M_{1}D_{1}H_{1}}{W_{1}}=\frac{M_{2}D_{2}H_{2}}{W_{2}}$

Problem 1: 30 children can do a piece of work in 16 days, how many children would be required to do the same work in 20 days?

Solution: $M_{1}D_{1}=M_{2}D_{2}$

30*16=X*20

X=24

Problem 2:Ketan appointed some workers which can complete total work in 45 days working 10 hours/day. But 30 workers are absent and remaining workers work only for 9 hours/day complete the work in 62.5 days. Find the total number of workers ketan appointed initially?

Solution:$\frac{M_{1}D_{1}H_{1}}{W_{1}}=\frac{M_{2}D_{2}H_{2}}{W_{2}}$

X*45*10=(X-30)*9*62.5

X*50=62.5X-1875

12.5X=1875

X=150

Problem 3:A contractor undertakes to make a road in 80 days and employees 50 men. After 24 days he finds that only $\frac{1}{3}$ of the road is made. How many extra men should be employed so that he is able to complete the work in 16 days?

Solution:

80-(24+16)=40

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

$\frac{50\times&space;24}{\frac{1}{3}}=\frac{x\times&space;40}{\frac{2}{3}}$

x=60

## Time and work-part-3

Problem 1: Two workers A and B engaged to do a work. A alone takes 8 hours more than to complete the job than if both work together, if B worked alone, he would need 4.5 hours more to complete the job than working together. What time would they take to do the work together?

Solution:To solve such kind of problems simply we will use one formula

i.e. $\sqrt{ab}=\sqrt{8\times&space;4.5}=\sqrt{36}=6$

Problem 2:A work is done by three persons A,B and C. A alone takes 10 hours to complete the work but B and C working together takes 4 hours, for the completion of same work. If all of them worked together and completed 14 times of the work, than how many hours have they worked?

Solution:

A$\rightarrow$10                                                     2

L.C.M  20

B+C$\rightarrow$4                                                      5

2+5=7

$\frac{20}{7}\times&space;14=40$

Problem 3: A,B and C can do a piece of work in 20,25,30 days respectively. They started the work together after 2 days of start of work A quits and 3 days after A, B quits. In how many days the total work would be finished ?

Solution:

A$\rightarrow$20                                            15

B$\rightarrow$25             L.C.M  300           12

C$\rightarrow$30                                              10

15+12+10=37

All together worked for 2 days i.e. 37*2=74

2 days after A quits and 3 days after A,B quits i.e 22*3=66

And remaining work will be done by c i.e 300-140=160

$\frac{160}{10}=16$

total time required to complete the work is=16+2+3=21 days

## Time and work-part-2

Problem 1: A can do a piece of work in 20 days and B can complete the same piece of work in 25 days. if they start the work together but after 6 days A left the work. in how many days the total work would be finished ?

Solution:

Both A and B worked together for 6 days so 5+4=9

9*5=54

out of 100. 54 parts of work is finished

100-54=46

A left after 6 days and the remaining work will be finished by B

So

And the total days to complete the work is 11.5+6=17.5 days

Problem 2: A and B can complete a piece of work in 24 days and 36 days respectively. They start the work but 3 days before completion of work, A left. In how many days will the total work be completed?

Solution:

Before 3 days A left so work will be done by B in that 3 days

Work done by B in 3 days is 3*2=6 parts of work is done by B

So remaining work is 72-6=66

66 parts of work is done by both A and B

3+2=5

$\frac{66}{5}=13\frac{1}{5}$

Total days=

$13\frac{1}{5}+3=16\frac{1}{5}$

Problem 3:P and Q can complete a job in 24 days working together. P alone can complete it in 32 days. Both of them worked together for 8 days and then P left. The number of days Q will take to complete the remaining work is?

Solution:

P+Q=4     P=3

3+Q=4

Q=4-3=1

Q=1

P and Q worked together for 8 days so 4*8=32

The remaining work is 96-32=64

p left after 8 days the remaining work will be done by Q

$\frac{64}{1}=64$

days

Problem 4:A can do a piece of work in 9 days and B can do a piece of work in 12 days respectively. If they work for a day alternative, in how many days the work would be finished if A begins the work?

Solution:

(A+B)

$2\text{d→7}$

(4+3)

Multiply with a number that results which is near to 36

$5×2\text{d→5×7}\phantom{\rule{0ex}{0ex}}10\text{d→35}$

35 parts of work is finished by both in 10 days

The remaining work 36-35=1 part of work will be finished by A because the work will begin with A So

$\frac{1}{4}$

Total time required to finish the work is 10+

$\frac{1}{4}=10\frac{1}{4}$

## Time and work-part-1

Problem 1: A can do a piece of work in 10 days while B can do the same piece of work in 15  days in how many days both will complete the same work together?

Solution:A

$\to \frac{1}{10}$

B

$\to \frac{1}{15}$ $=\frac{1}{10}+\frac{1}{15}=\frac{15+10}{10×15}$

=

$\frac{25}{150}=\frac{1}{6}$

Formula=

$\frac{\text{A+B}}{A×B}$

use this formula directly we can get answer.

$\frac{15+10}{15×10}=\frac{25}{150}=\frac{1}{6}$

Method 2: L.C.M method

A

$\to$

10

B

$\to$

15

L.C.M OF 10 and 15 is 30

A

$\to$

10                                                     3

L.C.M 30

B

$\to$

15                                                       2

3+2=5

Problem 2: A can do a piece of work in 7 days of  9 hours each and B can do it in 6 days of  7 hours each. How long will they take to do it working together

$8\frac{2}{5}$

hours a day?

Solution: A

$\stackrel{9}{\to }$

7

$\stackrel{9×7}{\to }$

63 hours

B

$\stackrel{7}{\to }$

6

$\stackrel{7×6}{\to }$

42 hours

L.C.M of 63 and 42 is 126

A

$\stackrel{9}{\to }$

7

$\stackrel{9×7}{\to }$

63 hours                                                     2

L.C.M 126

B

$\stackrel{7}{\to }$

6

$\stackrel{7×6}{\to }$

42 hours                                                      3

2+3=5

$8\frac{2}{5}=\frac{42}{5}$

Problem 3: A and B together can do a work in 18 days. B and C together can do a work in 24 days and A and C together can do a work in 36 days. How long A,B and c can do the work and c alone can do the work?

Solution: A+B=18                                    4

B+C=24         l.c.m is 72         3

A+C=36                                      2

(A+B+B+C+C+A)=4+3+2

2(A+B+C)=9

A+B+C=4.5

C=4.5-(A+B)

C=4.5-4

C=0.5

C=

$\frac{72}{0.5}=144$