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05 Apr 2016, 08:31
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
66% (01:58) correct
34%
(02:04)
wrong
based on 219
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If x, y, and z are positive integers, is (x!)(y!) > z! ?
(1) x^y > z
(2) x + y = z
(1) x^y > z
(2) x + y = z
05 Apr 2016, 10:09
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Bunuel
Hi,
lets see the choices and substitue values to see answer
(1) \(x^y > z\)
x = 3, y=2 and z=6... 3!2!>6!-- NO
x=3, y=2 and z= 3...3!2!>3!-- YES
Different answers -- Insuff
(2) x + y = z
OR x!y!>(x+y)!
It will always be NO, since
x! = 1*2*3...(x-1)*x
y!= 1*2*3...*(y-1)*y
(x+y)!= 1*2*3*...X..y.....(x+y)
x!*y! gives us square of the lower integers, whereas (x+y)! gives us product of higher values..
Examples
x=1 y=2 and z=3..
1!2!>3!----NO
x=3,y=3 and z=6
3!3!>6!---NO
Suff
B
05 Apr 2016, 10:53
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x, y and z are positive integers. Is (x!)(y!) > z! ?
St1: x^y > z
2^1 > 1 --> Is (x!)(y!) > z! ? Yes
2^10 > 1000 --> Is (x!)(y!) > z! ? No
Not sufficient.
St2: x + y = z --> Is (x!)(y!) > (x + y)!
In this case, z! will have all the factors of x!, y! and (x + y)! We will have more multiplication operations on RHS and as a result LHS will always be lesser than RHS.
For ex: (2!)(3!) = (2*1)*(3*2*1) = 12
5! = 5*4*3*2*1 = 120
Hence (x!)(y!) < z!
Sufficient.
Answer: B
St1: x^y > z
2^1 > 1 --> Is (x!)(y!) > z! ? Yes
2^10 > 1000 --> Is (x!)(y!) > z! ? No
Not sufficient.
St2: x + y = z --> Is (x!)(y!) > (x + y)!
In this case, z! will have all the factors of x!, y! and (x + y)! We will have more multiplication operations on RHS and as a result LHS will always be lesser than RHS.
For ex: (2!)(3!) = (2*1)*(3*2*1) = 12
5! = 5*4*3*2*1 = 120
Hence (x!)(y!) < z!
Sufficient.
Answer: B
