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?
Simplify the Expressions
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- fskilnik@GMATH
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Hi, there!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?
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.
Regards,
Fabio.
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- pesfunk
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Sorry, I couldn't understand the answer.
Could someone please explain this in a simple language ?
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?
- fskilnik@GMATH
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Let me try to help you myself, pesfunk.pesfunk wrote:Sorry, I couldn't understand the answer.
Could someone please explain this in a simple language ?
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.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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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.
.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.
- goyalsau
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Even i did the same way.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.
Saurabh Goyal
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EveryBody Wants to Win But Nobody wants to prepare for Win.