Rick can solve x algebra problems per minute. Matt can solve y calculus problems per minute. If y is not equal to zero, is x/y > 3?
x > 3
-(x/y) < 3
x > 3
-(x/y) < 3
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is x/y > 3?
Source: Beat The GMAT — Data Sufficiency |
oki with st 1 i know that x > 3 .think that X is 4 and y is 1 then x/y is > 3 .
however if y is 10 then x/y< 3
with st 2 all we know is X/y > -3
now effectively st 2 does not give me any thing .i know that that both X and Y are positive qty so x/y > 0 . But is X/y > 3 we cannot say
combing together all we have is X/Y >0 and X >3 now think of X as 4 and Y as 1 ---> x/y >3
but think of X as 4 and Y as 15 ---> x/y < 3
hence E
make sense ?
however if y is 10 then x/y< 3
with st 2 all we know is X/y > -3
now effectively st 2 does not give me any thing .i know that that both X and Y are positive qty so x/y > 0 . But is X/y > 3 we cannot say
combing together all we have is X/Y >0 and X >3 now think of X as 4 and Y as 1 ---> x/y >3
but think of X as 4 and Y as 15 ---> x/y < 3
hence E
make sense ?
is x/y > 3? given y <>0
ST1: x > 3
no info about y hence INSUFFICIENT
St2: -(x/y) < 3
-(8/3) < 3 and 8/3 or x/y < 3
-(12/3) = -4 < 3 but 12/3 = 4 > 3
hence insufficient
Together: E ( same reason as given for St2)
ST1: x > 3
no info about y hence INSUFFICIENT
St2: -(x/y) < 3
-(8/3) < 3 and 8/3 or x/y < 3
-(12/3) = -4 < 3 but 12/3 = 4 > 3
hence insufficient
Together: E ( same reason as given for St2)
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Target question: Is x/y > 3?himu wrote:Rick can solve x algebra problems per minute. Matt can solve y calculus problems per minute. If y is not equal to zero, is x/y > 3?
(1) x > 3
(2) -(x/y) < 3
Statement 1: x > 3
There are several pairs of values that meet this condition. Here are two:
Case a: x = 6, and y = 6, in which case x/y is not greater than 3
Case b: x = 6, and y = 1, in which case x/y is greater than 3
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: -(x/y) < 3
There are several pairs of values that meet this condition. Here are two:
Case a: x = 6, and y = 6, in which case x/y is not greater than 3
Case b: x = 6, and y = 1, in which case x/y is greater than 3
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined:
There are still several pairs of values that meet both conditions. Here are two:
Case a: x = 6, and y = 6, in which case x/y is not greater than 3
Case b: x = 6, and y = 1, in which case x/y is greater than 3
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT
Answer = E
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
