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roman numeral

Expert replies
by bblast » Mon Aug 22, 2011 11:54 pm
If -1 < c < 0, which of the following inequalities must be true?

I > c^5 < c^7
II> c^7 + c^8 < c^5 + c^4
III>c^8 - c^7 > c^4 - c^5


(A) None

(B) I only

(C) II only

(D) I and II only

(E) I, II and III

D[/img]
Last edited by bblast on Tue Aug 23, 2011 2:52 am, edited 3 times in total.
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Source: — Problem Solving |

by Geva@EconomistGMAT » Mon Aug 22, 2011 11:58 pm
bblast wrote:If -1 < c < 1, which of the following inequalities must be true?

https://my.knewton.com/remote/api/formu ... 921f812499


(A) None

(B) I only

(C) II only

(D) I and II only

(E) I, II and III

D
I see only one inequality - which does not have to be true, as c can equal zero.

EDIT: The question was later changed so that -1<c<0, so the remark about c equal to zero is no longer valid.
Last edited by Geva@EconomistGMAT on Tue Aug 23, 2011 3:13 am, edited 1 time in total.
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by bblast » Tue Aug 23, 2011 12:03 am
Hi Geva,
There was an error posting the image URL so I have typed the question now.

MY query - I solved this by picking numbers and got it in 3 minutes.

But I wanted to know if we can perform algebra on numerals 2 and 3 ?
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by Geva@EconomistGMAT » Tue Aug 23, 2011 12:41 am
You can simplify them by extracting the common factors.

II> c^7 + c^8 < c^5 + c^4

Extract the highest common factor on both sides of the inequality.

on the left, both c^7 and c^8 are divisible by c^7, so extract that.
On the right, both c^5 and c^4 are divisible by c^4, so extract that.

c^7(1+c) < c^4(c+1)

Since you know that -1<c<0, there's no way for c+1 or the equivalent 1+c to be negative - c is a negative fraction less than 1 in absolute value. This means that you can divide by 1+c without changing the direction of the inequality. The result is
c^7 < c^4

EDIT: Since c is a negative fraction, c^7 (off power) will still be negative, while c^4 (even power) will be positive. Thus, II MUST be true for all negative c values of c.

III>c^8 - c^7 > c^4 - c^5
Instead of doing the same here, with the resulting c+1 and 1+c make for a bit of a headache. Bypass the problem by switching the components around until you have only addition on both sides:
c^8 - c^7 > c^4 - c^5 /+c^7 +c^5
c^8 + c^5 > c^7 + c^4

Now extract the highest common factor on both sides

c^5 (c^3+1) > c^4(c^3+1)

Divide by the positive components c^3+1
c^5 > c^4

EDIT: Since c is a negative fraction, c^5 (odd power) will also be negative, while c^4 (even power) will always be positive. Thus, III is never true, since negative cannot be greater than positive.
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by czarczar » Tue Aug 23, 2011 9:06 am
bblast wrote:If -1 < c < 0, which of the following inequalities must be true?

I > c^5 < c^7
II> c^7 + c^8 < c^5 + c^4
III>c^8 - c^7 > c^4 - c^5


(A) None

(B) I only

(C) II only

(D) I and II only

(E) I, II and III

D[/img]
Ans should be D:

You can substitute the value of c as -.2 and try.
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