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:

$\text{A\u2192205}\phantom{\rule{0ex}{0ex}}\mathrm{LCM}100\phantom{\rule{0ex}{0ex}}\mathrm{B}\to 254$
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

$\frac{46}{4}=11.5\text{days}$
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:

$\text{A\u2192243}\phantom{\rule{0ex}{0ex}}\mathrm{L}.\mathrm{C}.\mathrm{M}72\phantom{\rule{0ex}{0ex}}\mathrm{B}\to 362$
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:

$\text{P+Q\u2192244}\phantom{\rule{0ex}{0ex}}\text{LCM96}\phantom{\rule{0ex}{0ex}}\text{P}\to 323\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{Q1}$
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:

$\text{A\u219294}\phantom{\rule{0ex}{0ex}}\mathrm{L}.\mathrm{C}.\mathrm{M}36\phantom{\rule{0ex}{0ex}}\text{B\u2192123}\phantom{\rule{0ex}{0ex}}$
(A+B)

$2\text{d\u21927}$
(4+3)

Multiply with a number that results which is near to 36

$5\times 2\text{d\u21925\xd77}\phantom{\rule{0ex}{0ex}}10\text{d\u219235}$
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+

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