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For all nonzero integers l and m, let the operation § be def

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For all nonzero integers l and m, let the operation § be def [#permalink] New post 26 Mar 2018, 15:51
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Question Stats:

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For all nonzero integers l and m, let the operation § be defined by l § m \(= - \mid \frac{l+m}{lm} \mid\).

Quantity A
Quantity B
3§ \(\frac{3}{2}\)
-1


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.

Drill 2
Question: 3
Page: 549
[Reveal] Spoiler: OA

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Re: For all nonzero integers l and m, let the operation § be def [#permalink] New post 04 Apr 2018, 04:18
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In a fake function problem like this we simply need to follow their instructions. In this case, add the terms on either side of the §, then multiply them and divide the former result by the latter result. Then get its absolute value and slap a minus sign on the whole thing. Let's do it.

In quantity A, for the numerator:

3 + 3/2 = 9/2

For the denominator:

3x(3/2) = 9/2

If we divide 3/2 by 3/2 obviously we'll get 1. Since -|1| = -1, the answer must be C.
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1 KUDOS received
GMAT Club Legend
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GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Kudos [?]: 1622 [1] , given: 385

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Re: For all nonzero integers l and m, let the operation § be def [#permalink] New post 04 Apr 2018, 15:59
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Explanation

When a problem gives you a relationship signified by an unfamiliar symbol: Just plug in the given values into the given “function” and solve.

If l § m=\(-|\frac{l+m}{lm}|\) then 3 §\(\frac{3}{2}\)=\(-|\frac{3+ 1.5}{3 \times 1.5}|=-1\).


The quantities are equal, so select choice (C).
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Re: For all nonzero integers l and m, let the operation § be def   [#permalink] 04 Apr 2018, 15:59
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For all nonzero integers l and m, let the operation § be def

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