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bcd = 770 , given that a<b<c<d> c-b = 3
if d is 77, then c = 5, and b = 2 --> c-b = 3
if d is 22, then c= 7, and b = 5 --> c-b = 2
if d is 14, then c= 11, and b = 5 --> c-b = 6

I see the only choice is A. How did you get c-b = 5 ?

bcd = 770 , given that a<b<c<d> c-b = 3 if d is 77, then c = 5, and b = 2 --> c-b = 3 if d is 22, then c= 7, and b = 5 --> c-b = 2 if d is 14, then c= 11, and b = 5 --> c-b = 6

I see the only choice is A. How did you get c-b = 5 ?

Re: Challenge#23 q25 [#permalink]
22 Jun 2007, 20:23

grad_mba wrote:

a*b*c*d = 770

a,b,c,d are positive integers

a<b<c<d

What is c-b if a=1

a) 3 b) 4 c) 5 d) 7 e) 10

I think more than 1 choice works for the above .. wat do u guys say ?

having worked through a few challenges i am very skeptical of some of the challenge questions; for $80 it is the most expensive lot of practice tests i have paid for and it is obvious they were not written by professional test writers - the explanations to some of the questions are just abysmal. this is a prime example; you can get two answers. i actually went shoonya's route on this one and started plugging in #s from the middle answer choice... and got 5 but i see 3 works too.

gmatclubot

Re: Challenge#23 q25
[#permalink]
22 Jun 2007, 20:23