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Simplify the Expressions

by Cedagmat » Thu Nov 11, 2010 2:07 pm
90. If 0 < n < m < 1, m^2 - n^2 must be less than which of the following expressions?
(A) m - n
(B) (m+n)/2
(C) mn
(D) (m + n)^2
(E) (m - n)^2

OA is D.

Is there any way to simplify these expression to make the solution easier than plugging in numbers?
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by fskilnik@GMATH » Thu Nov 11, 2010 2:23 pm
If 0 < n < m < 1, m^2 - n^2 must be less than which of the following expressions?
(A) m - n
(B) (m+n)/2
(C) mn
(D) (m + n)^2
(E) (m - n)^2
Is there any way to simplify these expression to make the solution easier than plugging in numbers?
Hi, there!

Yep! Please note that the expression given is (m-n)(m+n) , from the difference of squares factorization... now look at the info given: (m+n) is certainly positive and (m-n) too; from the fact that m and n are positive, the former is greater.

Therefore (m+n) > (m-n) and multiplying both sides by (m+n) > 0 we get: (m+n)^2 is greater than (m+n)(m-n) = m^2-n^2 and we are done.

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by pesfunk » Fri Nov 12, 2010 7:40 pm
Sorry, I couldn't understand the answer.

Could someone please explain this in a simple language ?
fskilnik wrote: If 0 < n < m < 1, m^2 - n^2 must be less than which of the following expressions?
(A) m - n
(B) (m+n)/2
(C) mn
(D) (m + n)^2
(E) (m - n)^2
Is there any way to simplify these expression to make the solution easier than plugging in numbers?
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by nubu » Sat Nov 13, 2010 2:32 am
Hi
m^2 - n^2 = (m-n)(m+n) (1)
As m> n>0 so m-n<m+n (2)

From (1) and (2) we have:
m^2 - n^2 = (m-n)(m+n)<(m+n)(m+n)= (m+n)^2

Hope this help!
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by fskilnik@GMATH » Sat Nov 13, 2010 2:47 pm
pesfunk wrote:Sorry, I couldn't understand the answer.

Could someone please explain this in a simple language ?
Let me try to help you myself, pesfunk.

01. We know that n is positive therefore m+n is greater than m-n

:: if you cannot "believe" that, please note that algebraically speaking, m+n > m-n if and only if 2n > 0, and of course it is the case.

02. When we have a> b and we multiply both sides by a positive number, the inequality is preserved, that is:

:: a>b and c>0 implies ac > bc

therefore using a = m+n , b = m-n and c= m+n we have: (m+n)^2 is greater than (m+n)(m-n) = m^2 - n^2.

I hope you got the whole picture.

Regards,
Fabio.
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by pkan51 » Mon Nov 15, 2010 6:43 pm
I just plug in m =.9 and n =.1 since that is the biggest number i can get with m^2 - n^2.

.9x.9 = .81 and .1x.1 = .01
m^2 - n^2 = .8

(A).9-.1=.8 same
(b)(.9+.1)/2=.5 more
(c) .9x.1=.09 more
(d)(.9+.1)^2=1 less
(e)(.9-.1)^2=.64 more

D is the only answer thats more if I plug in .9:.1 not sure if that works or not but thats how I got the answer.
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by goyalsau » Mon Nov 15, 2010 9:20 pm
pkan51 wrote:I just plug in m =.9 and n =.1 since that is the biggest number i can get with m^2 - n^2.

.9x.9 = .81 and .1x.1 = .01
m^2 - n^2 = .8

(A).9-.1=.8 same
(b)(.9+.1)/2=.5 more
(c) .9x.1=.09 more
(d)(.9+.1)^2=1 less
(e)(.9-.1)^2=.64 more

D is the only answer thats more if I plug in .9:.1 not sure if that works or not but thats how I got the answer.
Even i did the same way.
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