Work and Time Math (Emergency Please)

jahid43 wrote:2 men and 3 boys can do a piece of work in 10 days which 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men 1 boy do the work?

This is not reflective of an actual GMAT problem.
That said, we can glean some take-aways from the solution, so here goes:

Let the job = 800 units.

Since 2 men and 3 boys take 10 days to produce 800 units, the rate for 2M + 3B = w/t = 800/10 = 80 units per day.
Thus:
2M + 3B = 80 units per day.

Since 3 men and 2 boys take 8 days to produce 800 units, the rate for 3M + 2B = w/t = 800/8 = 100 units per day.
Thus:
3M + 2B = 100 units per day.

Adding the two equations, we get:
(2M + 3B) + (3M + 2B) = 80 + 100
5M + 5B = 180
M + B = 36 units per day.

Subtracting the first equation from the second, we get:
(3M + 2B - (2M + 3B) = 100 - 80
M - B = 20 units per day.

Adding together M+B=36 and M-B=20, we get:
(M+B) + (M-B) = 36 + 20
2M = 56 units per day.
M = 28 units per day.

Since M=28 and M-B = 20, B = 8 units per day.

Thus:
Rate for 2M + B = 2*28 + 8 = 64 units per day.
Time for 2M + B to produce 800 units = w/r = 800/64 = 100/8 = 12.5 days.

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

(2M + 3B)'s 10 day work = 1

(3M + 2B)'s 8 day work = 1

Therefore,
(2M + 3B) x 10 = (3M + 2B) x 8
20M + 30B = 24M + 16B

4M = 14B
1M = 3.5B

Now,
(Manpower x Time)/ Work = Constant

Therefore, {if a is the number of days taken by 2M and 1B to finish the work}
[(2M + 3B) x 10]/1 = [(2M + 1B) x a]/1
[(7B + 3B) x 10]/1 = [(7B + 1B) x a]/1 {Because 2M = 7B}
[(7B + 3B) x 10]/1 = [(7B + 1B) x a]/1
10B x 10 = 8B x a
a = 100/8 = 12.5 Days

W = R * T
1 = (2M + 3B) * 10 - (1)
1 = (3M + 2B) * 8

To find: (1)/(R)

20M + 30B = 24M + 16B
4M = 14B
M = 3.5B
Using in (1)
B = 1/100

So,
1/ (2M + 1B)
1/(8B)
1/8 * 100
12.5