aditiniyer wrote:Bill can dig a well in x! hours. Carlos can dig the same well in y! hours. If q is the number of hours it takes for Carlos & Bill to dig the well together, working at their respective rates, is q an integer ?
1) x-y = 1
2) y is a non prime even number
We are given that 1/q = 1/x! + 1/y!
=> 1/q = (x!+y!)/x!y!
=> q = x!y!/(x!+y!)
We have to see whether x!y!/(x!+y!) is an integer.
S1: x-y = 1
=> x = y+1
Thus, q = (y+1)!y! / ((y+1)!+y!)
q = [(y+1)*y!*y!] / [(y+1)*y! +y!]
=> q = [(y+1)*y!] / [(y+1) + 1]; cancelling y!
=> q = (y+1)! / (y+2)
If y = 1, q = (1+1)! / (1+2) = 2/3: not an integer
If y = 4, q = (4+1)! / (4+2) = 120/6: an integer. No unique answer.
S2: y is a non-prime even number.
We do not have any information about x. Insufficient.
S1 and S2:
As per S2, y: {4, 6, 8, 10, ...}. y can be written as y = 2z, where z = a positive integer > 1
Thus, q = (y+1)!/(y+2) = (2z+1)!/(2z+2) = [(2z+1)*
2z*(2z-1)*(2z-2)*....
(z+1)...1] / [
2(z+1)] = Integer; (z+1) and '2' cancell out.
Sufficient.
Answer:
C
Hope this helps!
-Jay
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