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Total units of work = 360

1 day work of A+B = 5 units
1 day work of B+C = 3 units
1 day work of A+C = 4 units

So, 1 day work of A+B+C = 6 units (Add the above and divide by 2)

Since, A, B and C together finish 6 units of work in 1 day and B and C together finish 3 units of work in 1 day, A alone will complete 3 units of work in 1 day.

Hence, A will take 120 days (\(\frac{360}{3}\)) to finish the work alone.

The answer will be B.
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Bunuel
A and B can do a piece of work in 72 days; B and C can do it in 120 days; A and C can do it in 90 days. In what time can A alone do it?

(A) 150 days
(B) 120 days
(C) 100 days
(D) 80 days
(E) 70 days
Solution:

If we let a, b, and c be the number of days it takes A, B, and C to each complete the work alone, respectively, we can create the equation:

1/a+ 1/b = 1/72, 1/b + 1/c = 1/120, and 1/a + 1/c = 1/90

If we add the three equations, we have:

2/a + 2/b + 2/c = 1/30

1/a + 1/b + 1/c = 1/60

Now subtract 1/b + 1/c = 1/120 from 1/a + 1/b + 1/c = 1/60, we have:

1/a = 1/120

a = 120

Answer: B
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