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08 Sep 2019, 15:43
Kudos
Bookmarks
Hi aviejay,
Based on your approach, you assume that after you divide (re: 120/2 = 60) that exactly half of the 60 remaining options involve situations in which Z is ahead of W. Unfortunately, that is NOT correct (it's actually more than half). Most people would not realize that without actually mapping out all of the options though. To properly do the math, you actually have to do the calculation in a different way (and deal with both 'restrictions' at once, which means that you ultimately divide 120 by 6). I've explained how in an earlier post in this thread.
GMAT assassins aren't born, they're made,
Rich
Based on your approach, you assume that after you divide (re: 120/2 = 60) that exactly half of the 60 remaining options involve situations in which Z is ahead of W. Unfortunately, that is NOT correct (it's actually more than half). Most people would not realize that without actually mapping out all of the options though. To properly do the math, you actually have to do the calculation in a different way (and deal with both 'restrictions' at once, which means that you ultimately divide 120 by 6). I've explained how in an earlier post in this thread.
GMAT assassins aren't born, they're made,
Rich
08 Sep 2019, 18:47
Kudos
Bookmarks
Bunuel
Runners V, W, X, Y, and Z are competing in the Bayville local triathlon. If V finishes before W and W finishes before Z, how many ways can X finish before Y??
A. 5
B. 10
C. 30
D. 60
E. 120
A. 5
B. 10
C. 30
D. 60
E. 120
No of ways = 4C2+4C1= 6+4=10
IMO B
Posted from my mobile device
09 Sep 2019, 07:14
Kudos
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aviejay
VeritasKarishma
Bunuel
Runners V, W, X, Y, and Z are competing in the Bayville local triathlon. If V finishes before W and W finishes before Z, how many ways can X finish before Y??
A. 5
B. 10
C. 30
D. 60
E. 120
A. 5
B. 10
C. 30
D. 60
E. 120
Check this post:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/1 ... s-part-ii/
We extend the same concept to this question. VWZ can be arranged in 3! ways but only 1 way is acceptable. So we divide 120 by 3! to get 20.
X finishes before Y in half of these cases so answer is 10.
Hi VeritasKarishma / Bunuel,
I ttotally understand the solution that you have provided. But I went wrong when I tried to solve the problem myself. Could you please help me understand where did I go wrong? This is what I did.
Total arrangements possible = 5! = 120
In order to find arrangements where v>w, we can simply divide 120 by 2 i.e. 60.
Now these 60 will have only those arrangments where v> w. Also they will have arrangements where w>z and w<z. So if we divide 60 by 2, we get only those arrangements where w> z. similarly, 30/ 2 = 15 gives me only those arrangements where x>y and hence, all conditions are taken care of.
Please explain where am i going wrong in the concept.
The possible sequences are
V W Z
V Z W
Z W V
Z V W
W V Z
W Z V
When you say V W is acceptable, you pick
V W Z
V Z W
Z V W
Now W is before Z in only one of them so dividing by 2 is not correct.
You need to consider all three variables together.
