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inequalities

by rahul.s » Tue Feb 23, 2010 12:44 am
If x and y are distinct integers, is (x + 7y)^3 > 0?

(1) x^2 > 49y^2
(2) x < 7y

OA: C
Source: Knewton
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by ajith » Tue Feb 23, 2010 1:15 am
rahul.s wrote:If x and y are distinct integers, is (x + 7y)^3 > 0?

(1) x^2 > 49y^2
(2) x < 7y

OA: C
Source: Knewton
(x + 7y)^3 = (x + 7y)^2*(x + 7y) is greater than zero if

x+7y>0

1) (x^2-(7y)^2) >0

(x+7y) (x-7y) >0

either (x+7y), (x-7y) both greater than zero or (x+7y), (x-7y) both less than zero, Insufficient

2) x-7y<0 not sufficient to conclude anything about x+7y

Combining x-7y<0 => x+7y <0 which is sufficient to answer the question

C
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by gauravgundal » Tue Feb 23, 2010 1:30 am
I agree with the answer C

x,y distinct integers to prove : (x+7y)^3>0

Reason
1. Insuff. x^2>49y^2 = x>(+/-)7y = x>7y or x>-7y don't exactly what should be x and y .

2. In suff. x>7y . Now for y= 1 ,x= 3 the (x+7y)^3 =positive ,but for y =-1 and x= -6 the eqN is negative

both

from we get -7y>x>7y
thus we can prove the equation .
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by lunarpower » Tue Feb 23, 2010 6:08 am
rahul.s wrote:If x and y are distinct integers, is (x + 7y)^3 > 0?

(1) x^2 > 49y^2
(2) x < 7y

OA: C
Source: Knewton
this is a pretty cool problem.

REPHRASE THE QUESTION
cubes don't affect positive/negative, so take the cube root:
is x + 7y > 0?
that's your question.

statement (1)
this is equivalent to |x| > |7y|. in other words:
x = +/- BIG NUMBER
7y = +/- SMALL NUMBER
that's all we know.
in this situation, x + 7y could have either sign, so, insufficient.

statement (2)
rearranges to x - 7y < 0.
the sign of x - 7y is not going to help us find the sign of x + 7y.
insufficient.

together
if x < 7y, then 7y must be POSITIVE BIG NUMBER. (see the possibilities outlined in #1)
this means that x + 7y is POSITIVE (since x, even if it is negative, is not big enough to turn this sum negative overall.)
therefore, sufficient.

(c).
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