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If x, y, and z are negative numbers such that z = 40x + 80y and x + y [#permalink]
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07 May 2015, 03:37
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If x, y, and z are negative numbers such that z = 40x + 80y and x + y [#permalink]
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07 May 2015, 09:13
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z = 40x + 80y = 40x+40y +40y = 40 (x+y) +40y = 40 +40y = 40(1  y) = 40 (1 (1x)) = 40(2+x) Statement 1 : x > y Since x + y = 1 x has to be greater than 0.5; When x = 0.5, z = 40(2+ 0.5) = 60 So when x is greater than 0.5 (i.e. 0.49, 0.4 etc) z will always be less than 60. SuffStatement 2 : y < 0.5 When y =0.5 ; z = 40(1 y) = 40(1 0.5) = 60 So when y < 0.5 (i.e. 0.51, 0.6 etc) z will always be less than 60. SuffAnswer : DAmbarish
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Re: If x, y, and z are negative numbers such that z = 40x + 80y and x + y [#permalink]
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07 May 2015, 09:43
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Bunuel wrote: If x, y, and z are negative numbers such that z = 40x + 80y and x + y = 1, is z < 60?
(1) x > y (2) y < 0.5
Kudos for a correct solution. Since x + y = 1, possible values of x and y are {0.1,0.9}; {0.2,0.8}; {0.3,0.7} etc. A) x>y, we can pick 2 extreme values to test this, {x=0.1 and y= 0.9} and {x=0.4 and y= 0.6} => in both these cases, results are 76 and 64 respectively which both are <60  SufficientB) y < 0.5, again we pick two extreme cases, which are same as A, , {x=0.1 and y= 0.9} and {x=0.4 and y= 0.6} => in both these cases, results are 76 and 64 respectively which both are <60  SufficientHence, Answer is D



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Re: If x, y, and z are negative numbers such that z = 40x + 80y and x + y [#permalink]
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11 May 2015, 04:10
Bunuel wrote: If x, y, and z are negative numbers such that z = 40x + 80y and x + y = 1, is z < 60?
(1) x > y (2) y < 0.5
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Many testtakers will see a problem like this and immediately start working with the number properties and testing cases. Could this get you to a correct answer? Sure, but look what happens if we give the warp pipe a try and spend some time with the prompt first. I’d rather work with two variables than three (especially since I only see x and y in the statements), so first I’ll substitute 40x + 80y to replace z in the question inequality. Now the question becomes Is 40x + 80y < 60? I know that x + y = 1, so now I want to manipulate the inequality in such a way that I can make another substitution. If I divide both sides by 40, I get Is x + 2y < 1.5? or, more helpfully, Is x + y + y < 1.5? Now I can substitute 1 for x + y to get Is 1 + y < 1.5? Add 1 to both sides, and we can see that this question really becomes Is y < 0.5? That’s so much simpler than the given prompt, and looking at Statement 2, I can immediately see that the GMAT has given me a room full of bonus coins to reward my work! Is y < 0.5? Why yes, it is – an easy, immediate sufficient. Moving on to Statement 1, I’m still looking for opportunities to substitute with info from the prompt, so I add y to both sides to get x + y > 2y Now I can substitute 1 on the left side 1 > 2y And finally, divide both sides by 2 0.5 > y Done! Another straightforward sufficient, and I can answer D with confidence. This is a question that seems much more difficult than it really is – as we can see, the GMAT rewards testtakers who are patient enough to work with the prompt before moving on to the statements. Take advantage of opportunities to simplify the question, and your room full of coins awaits!
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Re: If x, y, and z are negative numbers such that z = 40x + 80y and x + y [#permalink]
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26 Jan 2016, 10:32
Here is how i worked the problem:
Since x+y=1 and both must be negative, x and y could be {0,5,0,5} or {0,3,0,7}, etc.
z=40x+80y Using x+y=1 we have x=y1 Substituting in the z equation we get: z=40y40
For z to be less than 60 we need "40y" to be less than 20. Thus, we need y to be less than 0.5, so we can get a result less than 20, for example {0.51}.
Statement 1: x>y That means that x must be bigger than 0.5, or 0.49. SUFICIENT
Statement 2: y<0.5 That is exactly what we were looking for. SUFICIENT
ANSWER D



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Re: If x, y, and z are negative numbers such that z = 40x + 80y and x + y [#permalink]
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