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Algebra in DS

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by pareekbharat86 » Mon Nov 04, 2013 8:31 pm
Is xy<6?

(1) x<3 and y<2
(2) 1/2 < x < 3/2 and y*y < 6 .

OA is B.

Source- GMAT Test Code 14
Last edited by pareekbharat86 on Mon Nov 04, 2013 10:15 pm, edited 1 time in total.
Thanks,
Bharat.
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Source: — Data Sufficiency |

by theCodeToGMAT » Mon Nov 04, 2013 9:20 pm
Are you sure of the question???

TO find: xy<6?

Statement 1:
x<3 and y<2
We don't know whether x & y are negative.
For instance:
x = -5 & y = -4, xy = 20 NO
x = 2 & y = 1, xy = 2 YES
INSUFFICIENT


Statement 2:
12 < x < 32 and y*y < 6
So, x > 12

y^2 < 6 ==> y < sqrt(6) => y < 2._
xy = (13)(1) = 13 NO
xy = (13)(-1) = -13 YES
INSUFFICIENT

Combining...
y < 2
x<3 && 12 < x < 32 ---->>>>>> NO OVERLAP.
[spoiler]{E}[/spoiler]
R A H U L
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by Brent@GMATPrepNow » Mon Nov 04, 2013 9:34 pm
pareekbharat86 wrote:Is xy<6?

(1) x<3 and y<2
(2) 12 < x < 32 and y*y < 6 .

OA is B.

Source- GMAT Test Code 14
I think you must have transcribed the question incorrectly. The statements in GMAT DS questions never contradict each other.
Here, statement 1 says that x is less than 3, while statement 2 says that x is greater than 12

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by pareekbharat86 » Mon Nov 04, 2013 10:18 pm
I regret the error.

Statement 2 reads as-

1/2<x<2/3 and y*y=6.

Read the remaining portion as it is.
Thanks,
Bharat.
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by theCodeToGMAT » Mon Nov 04, 2013 11:27 pm
Solution after Question was corrected!

TO find: xy<6?

Statement 1:
x<3 and y<2
We don't know whether x & y are negative.
For instance:
x = -5 & y = -4, xy = 20 NO
x = 2 & y = 1, xy = 2 YES
INSUFFICIENT


Statement 2:
1/2 < x < 3/2 and y*y < 6
x > 0.5
x < 1.5
y*y < 6 ==> y < 2.xx
Considering the max values of x & y
(1.5)(3) = 4.5 == YES
For any negative value of "Y", the result becomes negative... and that would be less than "6"
SUFFICIENT

Answer [spoiler]{B}[/spoiler]
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by mainbhidhruv » Tue Nov 05, 2013 8:34 am
1. x<3 , y<3
if x = y = 1 => xy < 6 but if x=y= -3 => xy > 6 Not Sufficient\

2. 1/2<x<3/2 & y^2<6 => y^2-6<0 => (y-√6)(y+√6)<0 => -√6<y<√6
√6 = √2*√3 = 1.4 * 1.7 = 2.38 ( approx. )

Since x is positive => xy <6 for all the allowed values of x & y hence sufficient

OA = B
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