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2 DS questions !! confused !

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by charmaine » Thu Sep 11, 2008 7:07 pm
14. If x and y are integers, is xy + 1 divisible by 3?
(1) When x is divided by 3, the remainder is 1.
(2) When y is divided by 9, the remainder is 8.

answer. C

i got A ...i thought like one variable can be divided by 3 then it is divisible by 3 ..do not understand


15. If x&#8800;0, is |x| <1?
(1) x2<1
(2) |x| < 1/x

answer: D

can someone be kind enough to explain the absolute part for (2) ...i am still confused with absolute value in DS . thanks a million :)
answer : D
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Source: — Data Sufficiency |
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by aks.anupam » Thu Sep 11, 2008 8:10 pm
15. If x&#8800;0, is |x| <1?
(1) x2<1
(2) |x| < 1/x

In statement 2:

Absolute value of x is smaller than 1/x. Now absolute value of any number is always positive. So if a positive number is smaller than some number, then than number has to be positive. i.e 1/x is positive, which makes x positive in this case. So we have:

x<1/x, which is again x^2<1 which gives x<1. So |x|<1.

Sufficient.
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by cubicle_bound_misfit » Fri Sep 12, 2008 3:47 am
q1.


it is apparent why stmt 1 and 2 separately can not solve it. Taking them together.
LET

x = 3k +1

y = 9z +8

xy = (3k +1)(9z +8) = 3* + 3* + 3* +3* hence xy is divisible by 3 but xy+1 is not.

ANS C. ( NO, xy+1 ain't divisible by 3)

regards,
Cubicle Bound Misfit
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by cubicle_bound_misfit » Fri Sep 12, 2008 3:52 am
q2.

|x|<1,


when it is possible? only when either 0 < x < 1 or -1 < x <0

stmt 1 says x^2 <1

think of 4 different values of x -2, 2, -1/2 1/2. only the last 2 values satisfies the given stmt.
So, if given stmt is true |x| <1

do the same logic in stmt 2 by putting values.

you will see it is sufficient too.
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Re: 2 DS questions !! confused !

by nitin86 » Sat Sep 13, 2008 1:15 am
charmaine wrote: 15. If x&#8800;0, is |x| <1?
(1) x2<1
(2) |x| < 1/x

answer: D

can someone be kind enough to explain the absolute part for (2) ...i am still confused with absolute value in DS . thanks a million :)
answer : D
For statement 2,

|x| < 1/x , we have 2 scenario

1) we know that if x >0 (state X cannot be zero), then |x| = x, so |x| < 1/x => x < 1/x => x^2 < 1 => -1 < x < 1


2) we know that if x<0, then |x| = -x, so |x| < 1/x => -x < 1/x => -x^2 < 1 => x^2 > 1 => either x > 1 or x < -1

Hence, from 1 & 2, we know that x does not satisy -1 < x < 1, so we are clearly able to determine from main statement 2, that |x| is not less than 1.
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by charmaine » Sun Sep 14, 2008 1:40 am
thanks u guys :) got it all ! :)
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by vr4indian » Thu Sep 18, 2008 6:28 am
Answer 15:

Statement 2:

|x| < 1/x => -1/x < x < 1/x

Take X as 2 , -3 , 0.1 (Please take diff. set for + , - for all your example in exam)


-1/2 < 2 < 1/2 (Wrong)

1/3 < -3 < -1/3 (Wrong)

-1/0.1 < 0.1 < 1/0.1 (Write)

So x must be < 1
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by gmat009 » Tue Sep 23, 2008 10:41 am
cubicle_bound_misfit wrote:q1.


it is apparent why stmt 1 and 2 separately can not solve it. Taking them together.
LET

x = 3k +1

y = 9z +8

xy = (3k +1)(9z +8) = 3* + 3* + 3* +3* hence xy is divisible by 3 but xy+1 is not.

ANS C. ( NO, xy+1 ain't divisible by 3)

regards,
I am getting answer as E.
X can take values as 4,7,10,13,16....
Y can take values as 17,26,35......
Therefore XY+1 can take values 69,183,350........
69,183 are divisible by 3 but 350 is not.
Can someone explain where I am doing mistake
Last edited by gmat009 on Tue Sep 23, 2008 11:09 am, edited 1 time in total.
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by flayray » Tue Sep 23, 2008 11:05 am
cubicle_bound_misfit wrote:q1.


it is apparent why stmt 1 and 2 separately can not solve it. Taking them together.
LET

x = 3k +1

y = 9z +8

xy = (3k +1)(9z +8) = 3* + 3* + 3* +3* hence xy is divisible by 3 but xy+1 is not.

ANS C. ( NO, xy+1 ain't divisible by 3)

regards,
xy = 27kz + 24k + 9z +8
so xy + 1 = 27kz + 24k + 9z + 9
which is divisible by 3.
Answer is C
Does that make sense?
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by gmat009 » Tue Sep 23, 2008 11:08 am
flayray wrote:
cubicle_bound_misfit wrote:q1.


it is apparent why stmt 1 and 2 separately can not solve it. Taking them together.
LET

x = 3k +1

y = 9z +8

xy = (3k +1)(9z +8) = 3* + 3* + 3* +3* hence xy is divisible by 3 but xy+1 is not.

ANS C. ( NO, xy+1 ain't divisible by 3)

regards,
xy = 27kz + 24k + 9z +8
so xy + 1 = 27kz + 24k + 9z + 9
which is divisible by 3.
Answer is C
Does that make sense?
Yes I understood this but I am confused what I am doing wrong when using my own method.

X can take values as 4,7,10,13,16....
Y can take values as 17,26,35......
Therefore XY+1 can take values 69,183,350........
69,183 are divisible by 3 but 350 is not.
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by flayray » Tue Sep 23, 2008 11:11 am
gmat009 wrote:
flayray wrote:
cubicle_bound_misfit wrote:q1.


it is apparent why stmt 1 and 2 separately can not solve it. Taking them together.
LET

x = 3k +1

y = 9z +8

xy = (3k +1)(9z +8) = 3* + 3* + 3* +3* hence xy is divisible by 3 but xy+1 is not.

ANS C. ( NO, xy+1 ain't divisible by 3)

regards,
xy = 27kz + 24k + 9z +8
so xy + 1 = 27kz + 24k + 9z + 9
which is divisible by 3.
Answer is C
Does that make sense?
Yes I understood this but I am confused what I am doing wrong when using my own method.

X can take values as 4,7,10,13,16....
Y can take values as 17,26,35......
Therefore XY+1 can take values 69,183,350........
69,183 are divisible by 3 but 350 is not.
I think you want 351, not 350
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by gmat009 » Tue Sep 23, 2008 11:15 am
flayray wrote:
gmat009 wrote:
flayray wrote:
cubicle_bound_misfit wrote:q1.


it is apparent why stmt 1 and 2 separately can not solve it. Taking them together.
LET

x = 3k +1

y = 9z +8

xy = (3k +1)(9z +8) = 3* + 3* + 3* +3* hence xy is divisible by 3 but xy+1 is not.

ANS C. ( NO, xy+1 ain't divisible by 3)

regards,
xy = 27kz + 24k + 9z +8
so xy + 1 = 27kz + 24k + 9z + 9
which is divisible by 3.
Answer is C
Does that make sense?
Yes I understood this but I am confused what I am doing wrong when using my own method.

X can take values as 4,7,10,13,16....
Y can take values as 17,26,35......
Therefore XY+1 can take values 69,183,350........
69,183 are divisible by 3 but 350 is not.
I think you want 351, not 350
Thanks you r right.....
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