Micheal and Adam can do together a piece of work in 20 days.

Question

Micheal and Adam can do together a piece of work in 20 days. After they have worked together for 12 days Micheal stops and Adam completes the remaining work in 10 days. In how many days Micheal complete the work separately.

A. 80 days
B. 100 days
C. 120 days
D. 110 days
E. 90 days

A. 80 days
B. 100 days
C. 120 days
D. 110 days
E. 90 days

abhijit_sen · Answer

Suppose rate of work for per day Mike = M & Rate of work per day for Adam = A
Together in a Day they can finish = A + M units
Total Work Done in 20 Days = 20(A+M)
Total Work Done in 12 Days = 12(A+M)
Work Done by Adam in 10 Days = 10A

Since total work is same we can say that 20(A+M) = 12(A+M) + 10A => 8M = 2A => A = 4M
So total work done by them together in 20 Days = 20(4M+M) = 100M
Since Mike does M unit of work per day, it will take him 100 Days to finish up the work.

Answer B.

MHIKER · Answer

Rate of both = 1/20
Together they do = 1/20*12 = 3/5

Left work = 1 - 3/5 = 2/5

Adam completes 2/5 work in 10 day
so he took 10*5/2 = 25 days to complete the left work alone.
Thus the rate of adam is 1/25

Rate of Micheal = 1/20 - 1/25 = 1/100
Thus micheal takes 100 days to complete the whole work.
ans. B.

MHIKER · Answer

I got another way of the same problem
1/20*12+10x=1  
x = 1/25
so, rate of Micheal =1/20-1/25 = 1/100

So Micheal take 100 days
Ans. B

PareshGmat · Answer

Both have completed  \frac{12}{20}  work, means  \frac{8}{20}  work is pending

Adam completed  \frac{8}{20}  work in 10 days

Rate of Adam =  \frac{8}{(20 * 10)} = \frac{8}{200}

So, rate of Michael =  \frac{1}{20} - \frac{8}{200}

=  \frac{2}{200}

=  \frac{1}{100}

So Michael completes the work in 100 Days

Answer = B

mvmarcus7 · Answer

I`ve done in 5 steps:

1 - M+A = X/20 where X is arbitrarily 100, so M+A rate is 5/day of a total of 100.

2 - (M+A) *12 = 60. 100-60 = 40

3 - A completed 40 in 10 days, so 4/day.

4 - M+A = 5, M + 4 = 5, M = 1.

5 - 1 (rate) * 100(total work) = 100 days

Abhishek009 · Answer

Let the total work be 60 ( LCM of 20,12 & 10 )

Efficiency of Micheal and Adam is  \frac{60}{20} = 3

Unit of work completed is 12*3 => 36 units
Units of work left is 60 - 36 => 24 units

So, Adam completes 24 units in 10 days..
Efficiency of Adam = 24/10 =>  \frac{12}{5}  =>  2.4

Efficiency of Micheal and Adam is = 3
Efficiency of Adam is = 2.40
So, Efficiency of Micheal is = 0.60

Thus the time required for Micheal to complete the work is =  \frac{60}{0.60}  => 100
  
Hence correct answer is (B) 100

Wsd65798 · Answer

Since the numbers are conducive, this problem can also be thought through in terms of percentage and can be solved while reading.

Since Micheal and Adam can do a work together in 20 days i.e. 5% in one day. This 5% is the sum of their individual rates of work. Both work for 12 days, meaning they complete 5% * 12 = 60% of work. Then Adam leaves and Michael completes the remaining work, 100% - 60% = 40% in 10 days. Adam completes 40% work in 10 days means he completes 4% in one day; this is Adam's efficiency, which should be assumed to be constant. Now since Michael and Adam together complete 5% work in one day and Adam does 4% of work per day, Micheal does only 1% of work per day. If Michael does 1% in one day, he will need 100 days to complete full work. Therefore, the answer is 100 days. (Option B).

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Micheal and Adam can do together a piece of work in 20 days.

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