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Source: — Data Sufficiency |

by Suyog » Sun Feb 03, 2008 8:44 pm
if x = 1 then avg of 1 & 10 = 5.5......I
if x = 3 then avg of 3 & 10 = 6.5.......II

a)
x = 1 & let z = 9 greater than avg ...... from I
x = 1 & let z = 2 smaller than avg ....... from I
Insuff

b)
if x = 1 then z = 5 smaller than avg ....... from I
if x = 3 then z = 15 greater than avg ........ from II
Insuff

Both together Insuff...

Ans E.
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by cris » Mon Feb 04, 2008 10:25 am
The correct answer is A
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by simplyjat » Mon Feb 04, 2008 12:28 pm
Suyog wrote: x = 1 & let z = 9 greater than avg ...... from I
x = 1 & let z = 2 smaller than avg ....... from I
The second option is not possible as Z will be closer to X rather than 10.

Now look at it this way, if Z is equidistant from X and 10, then Z is equal to the average of 10 & X. You need to visualize this....
Now if Z is closer to X, Z is less than average of X & 10, and if Z is closer to 10, Z is greater than the average of X & 10
simplyjat
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by cris » Mon Feb 04, 2008 12:46 pm
simplyjat that was indeed a great great explanation and approach!!

...I am not a fan of picking numbers (even though some times we must)because its both time consuming and you never know if it will work for other numbers or if there ia an exception..so this explanation of why A is SUFF is really the one!.

:D
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by helen123 » Thu Feb 07, 2008 2:17 pm
Was just looking at this one, I think it should be solved like this

1) 10-z < z-x which resolves to 5+x/2 <z> average

SUFF

2) You'd have to sub # in there, cuz you can't use the condition given in 1) for 2)
x =1 , z = 5, A = 5.5, then z <A> A
INSUF

so answer is A
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by xilef » Fri Feb 15, 2008 3:42 pm
helen123 how did you get 10-z < z-x?

did you mean 10-z<10-x?
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by airan » Sun Jul 20, 2008 5:06 am
Z is closer to 10 than it is to x ..so the difference between 10 and z is less than the difference between z and x ..hence
10-z < z-x?
Thanks
Airan
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