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xerocoool
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Inequality Based Question

by xerocoool Mon Aug 25, 2014 9:50 pm

If (x-y)^3 > (x-y)^2, then which one of the following must be true ?

A) x^3 < x^5
B) x^5 < y^4
C) x^3 > y^2
D) x^5 > y^4
E) x^3 > y^3

Soln: since (x-y)^3 is greater than (x-y)^2, we can confirm the following -
a) x-y should be positive
b) x-y cannot be equal to 0

So, dividing by (x-2)^2 throughout -
=> x-y > 1
x>1+y

x is greater than y - so the answer could be any one (C,D,E). But the book says "E" (could you please explain how)

Thanks
asishkm
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Re: Inequality Based Question

by asishkm Tue Aug 26, 2014 12:07 am

Hi xero,

x > y + 1
=> x > y (as x > y + 1 > y + 0)
=> x^3 > y^3

Now consider (C) or (D)
x^3 > y^2 or x^5 > y^4
If x is negative, then this inequality will obviously be invalid.
Eg: x = -2, y = -4
tommywallach
Manhattan Prep Staff
 
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Re: Inequality Based Question

by tommywallach Tue Aug 26, 2014 10:41 pm

Asish is absolutely right. This is actually the same error you made on the previous question (i.e. if x > 0, then x > - 1. This is true by LOGIC, not by any algebraic manipulation.)

-t
xerocoool
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Re: Inequality Based Question

by xerocoool Tue Aug 26, 2014 11:57 pm

oh srry. forgot to plug values and check.

so x > 1 + y
x-y > 1

x = 5
y = 3

satisfies condition C,D,E

x = -8
y = -10

x - y = 2

only E is true

Thanks guys ! Got it.
tommywallach
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Posts: 1917
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Re: Inequality Based Question

by tommywallach Fri Aug 29, 2014 1:46 pm

Totally. I actually think my comment isn't totally right, except for the fact that you can use the same logic from that other question to make the jump that asishkm made, from x > y + 1 to x > y.

-t