If 0

Question

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

chetan2u · Answer

Without too much efforts, one can see that option D is sum of all positive terms, so has to be greater. (m+n)^2=m^2+n^2+2mn . This will surely be greater than m^2-n^2 . But let us see why others are wrong. Let us assume that m^2-n^2 is greater. (A) m^2-n^2>m − n>0…..(m-n)(m+n)>m-n…….(m+n)>1 . We get that by cancelling the positive expression from both sides. Surely m+n>1 is possible. Example m=3/4 and n=1/2 (B) m^2-n^2>\frac{(m + n)}{2}…….m-n>\frac{1}{2} . Possible, and example is m=3/4 and n=1/8 (C) m^2-n^2>mn…….m^2>n^2+mn……m^2>n(m+n) . Possible when n is very small as compared to m, and example is m=3/4 and n=1/16 (D) m^2-n^2>(m + n)^2…….m^2-n^2>m^2+n^2+2mn………..2n^2+2mn<0 …..NOT possible. (E) m^2-n^2>(m - n)^2…….m^2-n^2>m^2+n^2-2mn………..2n^2-2mn<0……..2n(n-m)<0 …..Always true as n-m<0. D

varshas044 · Answer

I did it this way
Basically in such questions only one conflicting answer is enough to eliminate the option

It is given the 0 < n < m < 1 
Let’s take extreme cases 
N = 0.1 and m = 0.9

Case 1: is m^2 – n^2 < m – n ---------- 0.81 - 0.01 < 0.9 – 01 ----- 0.8 < 0.8 ------NO

Case 2: 0.81 – 0.01 < (0.9 + 0.1) /2 -------- 0.80 < 0.5? ---------- NO

Case 3: 0.81 -0.01 < 0.9*0.1 --------- 0.80 < 0.09 --------NO

Case 4: 0.81 – 0.01 < (0.9 +0.1)^2 --------- 0.80 < 1 --------YES

Case 5: 0.81 – 0.01 < (0.9 – 0.1)^2 -------- 0.80 < 0.64 ------ NO

Answer case 4 --- Option D

gmatprepper94111 · Answer

Take it as given that 0 < n < m < 1.

Let's start by factoring m^2 - n^2.
--> m^2 - n^2 = (m + n) * (m - n)

Can we say anything about the sign of this number? We know it's greater than zero. (Unfortunately this doesn't help, but is a good starting place for a question like this IMO)
--> We know that (m + n) > 0 because m > 0, n > 0.
--> We know that (m - n) > 0 because m > n
--> so (m + n) * (m - n) > 0

So let's proceed one by one until we find an expression we can prove.
--> A is easy to invent a counterexample for. I can pick m, n > 0.5 such that (m+n) > 1 ==> (m+n) * (m-n) > (m - n)
--> B doesn't look very promising ... instinctively, I want the proof to use my factored form right away ... skip and come back
--> C doesn't look very promising ... instinctively, I want the proof to use my factored form right away ... skip and come back
--> D looks promising. I'll try this one.
--> Didn't look at E.

Ok let's see if we can't prove D to be true.

We want to show m^2 - n^2 < (m + n)^2; I'll do this one by assuming the opposite and then forcing a contradiction.
--> Suppose it's true that (m+n)^2 <= m^2 - n^2
--> Start by expanding (m+n)^2: (m+n)^2 = (m+n) * (m+n)
--> We can then factor m^2 -n^2 into a similar form: m^2 - n^2 = (m+n) * (m-n)
--> Then we can rewrite our first inequality: (m+n)(m+n) <= (m+n)(m-n)
--> Because m, n > 0, we can just divide both sides by (m+n) safely: (m+n) <= (m-n)
--> Now take out m from both sides: n <= -n
--> And add n to both sides: 2n <= 0
--> And divide by 2: n <= 0
--> Note at the outset that we said n > 0. We've arrived at a contradiction.
--> So It cannot be true that (m+n)^2 <= m^2 - n^2
--> And therefore it must be true that (m+n)^2 > m^2 - n^2

D is the answer.

bumpbot · Answer

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If 0 < n < m < 1, m^2 n^2 must be less than which of the following e

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