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# 1/ Is x > y? (1) (x+y)/2 > 0 (2) (y-x)(y+x) > 0 2/

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1/ Is x > y? (1) (x+y)/2 > 0 (2) (y-x)(y+x) > 0 2/ [#permalink]  22 Jan 2005, 02:53
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1/ Is x > y?
(1) (x+y)/2 > 0
(2) (y-x)(y+x) > 0

2/ One hundred students are taking both law and accounting. how many students are taking only accounting?
(1) 180 students are taking EITHER law OR accounting.
(2) 50 students are taking law but not accounting.

Directions:

A. If statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked;

B. If statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked;

C. If BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient;

D. If EACH statement ALONE is sufficient to answer the question asked;

E. If statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
Intern
Joined: 22 Jan 2005
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Many thanks, HSWKM!

However, for question no. 1, my choice is C.

(2) (y-x)(y+x) > 0
=> (y-x > 0) AND (y+x > 0)
OR (y-x < 0) AND (y+x < 0)

From (1) (x+y)/2 > 0 => x+y > 0

From (1) & (2) => y-x > 0 <=> y > x.

So BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient

Many thanks again.
Intern
Joined: 16 Jan 2005
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C, C
For the second one---

1) gives us 180 = L + A
so we know L (U) A = 180 +100 = 280
not sufficient to know what is A

leaves us with choices B,C and E

2) gives us L= 50
if used alone
we won't be able to find [L (U) A] and [A] in tghe equation-
[L (U) A] = [L] + [A] + [L (I) A]
[L (U) A] = 50 + [A] + 100
not sufficient to know [A]...

but combining 1) and 2) gives us 50 + A = 180
i.e. A = 130
SVP
Joined: 03 Jan 2005
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C,C

Wrong forum though, right?
VP
Joined: 25 Nov 2004
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1. OA should be E.
y>x is not true.
if y=1, and x = -1 satisfies youranswer y>x.

from i, (x+y)/2>0 => x+y>0
but with these values, statement can not be true.
-1+1 is not greater than 0.

from ii, (y-x)(y+x) > 0
with the same values for x and y as in i,
(y-x)(y+x) > 0
(1+1)(1-1)>0

therefore OA should be E.

2/ One hundred students are taking both law and accounting. how many students are taking only accounting?
(1) 180 students are taking EITHER law OR accounting.
(2) 50 students are taking law but not accounting.

It is C.
Current Student
Joined: 28 Dec 2004
Posts: 3392
Location: New York City
Schools: Wharton'11 HBS'12
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Kudos [?]: 180 [0], given: 2

MA
1)
from statement 1, we get y>-x (1>-1) but we dont know if y is greater than X.

from statement 2 we get Y^2-x^2>0, i.e. y^2>x^2, but we still dont know if Y is greatern than +ve X or -ve X...so insufficient

taking the two together we realize that Y>-x but not greater than X, that is still unknown...so I think your answer is correct...

E its

2)
L+A=180, then also we know the overlap is 100, therefore
L(U)A=280= L + A - 100
280= 50+A -100

taking both statements is sufficient...C it is

MA wrote:
1. OA should be E.
y>x is not true.
if y=1, and x = -1 satisfies youranswer y>x.

from i, (x+y)/2>0 => x+y>0
but with these values, statement can not be true.
-1+1 is not greater than 0.

from ii, (y-x)(y+x) > 0
with the same values for x and y as in i,
(y-x)(y+x) > 0
(1+1)(1-1)>0

therefore OA should be E.

2/ One hundred students are taking both law and accounting. how many students are taking only accounting?
(1) 180 students are taking EITHER law OR accounting.
(2) 50 students are taking law but not accounting.

It is C.
SVP
Joined: 03 Jan 2005
Posts: 2251
Followers: 13

Kudos [?]: 213 [0], given: 0

MA wrote:
1. OA should be E.
y>x is not true.
therefore OA should be E.

It doesn't matter if y>x is true or not, as long as if we can determine whether y>x, then it is sufficient.
Current Student
Joined: 28 Dec 2004
Posts: 3392
Location: New York City
Schools: Wharton'11 HBS'12
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Kudos [?]: 180 [0], given: 2

I am confused...Ok tell me what you make of my explination??

HongHu wrote:
MA wrote:
1. OA should be E.
y>x is not true.
therefore OA should be E.

It doesn't matter if y>x is true or not, as long as if we can determine whether y>x, then it is sufficient.
VP
Joined: 18 Nov 2004
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Kudos [?]: 20 [0], given: 0

"C".

(1) (x+y)/2 > 0 .....x+y > 0......x > -y....for x = 1 and y = 1....satisfies....so x = y.....now x = 2 and y = 1.....2 > -1....so x > y....overall insuff

(2) (y-x)(y+x) > 0 ......x = -2 ...y = -3.....satisfies ....x > y
x = 1...y = 3.....satisfoes....y > x......so insuff

combine....

we know x+y > 0 from state 1....for statement 2 to hold true.....y-x > 0...so y > x and x > -y ....so ans is NO.....suff.

Last edited by banerjeea_98 on 23 Jan 2005, 11:04, edited 1 time in total.
VP
Joined: 18 Nov 2004
Posts: 1442
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Kudos [?]: 20 [0], given: 0

MA wrote:
1. OA should be E.
y>x is not true.
if y=1, and x = -1 satisfies youranswer y>x.

from i, (x+y)/2>0 => x+y>0
but with these values, statement can not be true.
-1+1 is not greater than 0.

from ii, (y-x)(y+x) > 0
with the same values for x and y as in i,
(y-x)(y+x) > 0
(1+1)(1-1)>0

therefore OA should be E.

.

MA, ur example X = -1 and Y = 1...doesn't satisfy the statement 1....when u combine the 2 statements.....u actually know that...Y > X AND X > -Y...both of them has to be true and not just Y > X.....X = -1 and Y = 1 is not a valid example.

Last edited by banerjeea_98 on 23 Jan 2005, 11:07, edited 2 times in total.
Current Student
Joined: 28 Dec 2004
Posts: 3392
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 180 [0], given: 2

I am consfused...(inequalities do that to me..LOL), all I can makeout is that Y>-x, but I cannot get how y>+x

statement ii) Y^2>x^2 which means either +,- ve y> +ve x or +,-ve y>-ve X combining the two statements leads us to believe that y>-veX?

banerjeea_98 wrote:
MA wrote:
1. OA should be E.
y>x is not true.
if y=1, and x = -1 satisfies youranswer y>x.

from i, (x+y)/2>0 => x+y>0
but with these values, statement can not be true.
-1+1 is not greater than 0.

from ii, (y-x)(y+x) > 0
with the same values for x and y as in i,
(y-x)(y+x) > 0
(1+1)(1-1)>0

therefore OA should be E.

.

MA, ques is if X is always > Y....ans is X can never be > Y....so No....by combining 1 and 2. Although, Y CAN be > X, X can never be > Y
VP
Joined: 18 Nov 2004
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Kudos [?]: 20 [0], given: 0

fresinha12 wrote:
I am consfused...(inequalities do that to me..LOL), all I can makeout is that Y>-x, but I cannot get how y>+x

statement ii) Y^2>x^2 which means either +,- ve y> +ve x or +,-ve y>-ve X combining the two statements leads us to believe that y>-veX?

banerjeea_98 wrote:
MA wrote:
1. OA should be E.
y>x is not true.
if y=1, and x = -1 satisfies youranswer y>x.

from i, (x+y)/2>0 => x+y>0
but with these values, statement can not be true.
-1+1 is not greater than 0.

from ii, (y-x)(y+x) > 0
with the same values for x and y as in i,
(y-x)(y+x) > 0
(1+1)(1-1)>0

therefore OA should be E.

.

MA, ques is if X is always > Y....ans is X can never be > Y....so No....by combining 1 and 2. Although, Y CAN be > X, X can never be > Y

I edited my comment above...take a look....it will be clear...combined sol is Y > X AND X > -Y (OR Y > -X)...both has to hold true...not just Y > X
Current Student
Joined: 28 Dec 2004
Posts: 3392
Location: New York City
Schools: Wharton'11 HBS'12
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Kudos [?]: 180 [0], given: 2

banerjee....thanks! got it now...

banerjeea_98 wrote:
fresinha12 wrote:
I am consfused...(inequalities do that to me..LOL), all I can makeout is that Y>-x, but I cannot get how y>+x

statement ii) Y^2>x^2 which means either +,- ve y> +ve x or +,-ve y>-ve X combining the two statements leads us to believe that y>-veX?

banerjeea_98 wrote:
MA wrote:
1. OA should be E.
y>x is not true.
if y=1, and x = -1 satisfies youranswer y>x.

from i, (x+y)/2>0 => x+y>0
but with these values, statement can not be true.
-1+1 is not greater than 0.

from ii, (y-x)(y+x) > 0
with the same values for x and y as in i,
(y-x)(y+x) > 0
(1+1)(1-1)>0

therefore OA should be E.

.

MA, ques is if X is always > Y....ans is X can never be > Y....so No....by combining 1 and 2. Although, Y CAN be > X, X can never be > Y

I edited my comment above...take a look....it will be clear...combined sol is Y > X AND X > -Y (OR Y > -X)...both has to hold true...not just Y > X
VP
Joined: 25 Nov 2004
Posts: 1495
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Kudos [?]: 38 [0], given: 0

banerjeea_98 wrote:

I edited my comment above...take a look....it will be clear...combined sol is Y > X AND X > -Y (OR Y > -X)...both has to hold true...not just Y > X

Banerjeea,
i am not clear but i still think that E is the OA.

Last edited by MA on 24 Jan 2005, 07:42, edited 1 time in total.
SVP
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Kudos [?]: 213 [0], given: 0

1/ Is x > y?
(1) (x+y)/2 > 0
(2) (y-x)(y+x) > 0

(1) =>x+y>0
(2) (y-x)(y+x)>0 Combine them, since y+x>0 =>y-x>0 therefore y>x
Sufficient
Manager
Joined: 19 Jun 2004
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C & C

1.

a) x + y > 0 ==> then x>y or x<y satisfied it (insufficient)

b) (y - x)(y + x) >0

then two options: (+)(+) > 0

(y-x) > 0 and (y+x) >0 ..... (1)

or (-)(-) > 0

(y-x) < 0 and (y+x) <0 ...... (2)

from (1) y>x and (y+x) > 0
from (2) y<x and (y+x) < 0

then it could be y>x or y<x (insufficient)

From Both (a) and (b)

from (a) (y+x) > 0
that coincides with (b) (y+x) >0 when y>x

then y>x (sufficient)

C

2. C
Intern
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When we take x = -1 and y = 1 as an example, we certainly cannot satisfy both the data given, i.e. (1) & (2). For a DS question, our job is to decide whether the data given in the statements are sufficient for answering the question.

I surely choose C.
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