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joemama142000
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inequality ds 13 Jan 2006, 06:00
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is (x-5)^2 5



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allabout
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inequality ds Updated on: 13 Jan 2006, 10:37
Originally posted by allabout on 13 Jan 2006, 10:13.
Last edited by allabout on 13 Jan 2006, 10:37, edited 1 time in total.
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1) Is clearly insufficient, consider x=-1 and y=0

2)
x+y>5
x>5-y

[(5-y-5)^2 - (y+5)^2<0 ?
.
.
y>-2,5

Since we can't decide whether y is positive or not, we can't know if (x-5)^2<(y+5)^2] this step isn't necessary

It is untrue if |x-5|>|y+5|,
but true if |x-5|<|y+5|

Insufficient

1+2) edit Consider x=-1 y=0, then the statement is untrue
x=1 y=2, then it is true
Well that was stupid it's C
I think it's E

Please critisise my method; what steps can I make more efficiently, where is my method inaccurate...
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inequality ds 13 Jan 2006, 10:32
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i think its C.

statements 1 and 2 on their own are insufficient. Try taking them together, we have the 2 conditions, x<y and x+y>5.

So lets try x=1 and y=6. This satisfies the original equation with a yes answer.

Try x=-7 and y=13. This also satisfies the original equation with a yes answer.
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giddi77
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inequality ds 13 Jan 2006, 10:50
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C?.

Is (x-5)^2<(y+5)^2?
(1) Given x < y
x = 2, y=3
(x-5)^2 = 9
(y+5)^2 = 64 9 < 64 TRUE.
For x=-2, y = -1 FALSE
Clearly INSUFFICIENT.

2) x+y > 5

If we plug x=-3, y=9
(x-5)^2 = 64
(y+5)^2 = 14^2= 196 => TRUE
Also y=-3, x=9 => FALSE.

Combining (1) & (2)
y>x --(I)
x+y>5 --(II)
Subtracting (I) from (II),
x>5-x => x>5/2 [Not sure fi this can be done]
For eg x = 3, y=4 => TRUE.
Even for x = -4, y has to be >9. Hence TRUE

Therefore it is C.

For all values of x> 5/2
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lhotseface
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inequality ds 13 Jan 2006, 14:21
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I got E. I can write down the step if it is the OA.
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inequality ds 14 Jan 2006, 01:55
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I am getting C.

a) If x=1 y=2 the question stem is false
If x=10 and y=12 the question stem is true
so INSUFF

b)If x=1 and y=5 then stmt is False
If x=5 and y=1 then stmt is True
so INSUFF

Combine and we get that x<y and their sum>5
(+ve values)If x=1 and y=5 then stmt is NO
(-ve) values)If x=-3 and y=9 then stmt is NO
(0 values) If x=0 and y=6 then stmt is NO
So all consistent, Hence Ans C
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joemama142000
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inequality ds 14 Jan 2006, 03:12
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oa is C

is the only way to do this is to plug in?
can someone verify giddi's method of subtracting the two ?
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willget800
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inequality ds 14 Jan 2006, 09:30
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It should be B...

Isnt y > 5-x sufficient to show?
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inequality ds 15 Jan 2006, 20:24
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joemama142000
oa is C

is the only way to do this is to plug in?
can someone verify giddi's method of subtracting the two ?
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Unfortunately not. The only valid rule for subtrating is when the two inequalities are in the opp. direction. see HongHu's sticky in this forum.

You can only apply substractions and divisions when their signs are in the opposite directions:
let a>b, d<c
then
a-d>b-c
a/d>b/c
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inequality ds 15 Jan 2006, 23:33
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a) If y = 7, x = 5, then LHS < RHS
If y = 1/2, x = 1/4, then LHS < RHS
If y = -1/2, x = -1/8, then LHS > RHS

Insufficient.

b) x-5 > -y -> say, -0.5y

So LHS = 0.25y^2 RHS = (y+5)^2, then RHS > LHS

Ans B
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Bhai
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inequality ds 15 Jan 2006, 23:44
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ywilfred
a) If y = 7, x = 5, then LHS < RHS
If y = 1/2, x = 1/4, then LHS < RHS
If y = -1/2, x = -1/8, then LHS > RHS

Insufficient.

b) x-5 > -y -> say, -0.5y

So LHS = 0.25y^2 RHS = (y+5)^2, then RHS > LHS

Ans B
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ywili

The OA is C. Can you please check your solution?
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haddy74
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inequality ds 17 Jan 2006, 11:05
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I got it C too……

If we know that X has to be less than Y (first condition) and then we also that X+Y is a positive integer and greater than 5 (second condition), then (X-5)^2 has to be less than (Y+5)^2.

For instance if X+Y is 6 and we also know that X 5, (X-5)^2 will be less than (Y+5)^2. if X is a positive integer then the problem becomes even more simpler.

Please correct me, if I am wrong.



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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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