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# If x<y, is x(1+x)<y(1+y)?

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 8009
GMAT 1: 760 Q51 V42
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19 Mar 2018, 02:05
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Difficulty:

95% (hard)

Question Stats:

44% (02:27) correct 56% (02:04) wrong based on 62 sessions

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[GMAT math practice question]

If $$x<y$$, is $$x(1+x)<y(1+y)$$?

$$1) x>\frac{1}{2}$$
$$2) x+y>1$$

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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Retired Moderator Joined: 22 Jun 2014 Posts: 1093 Location: India Concentration: General Management, Technology GMAT 1: 540 Q45 V20 GPA: 2.49 WE: Information Technology (Computer Software) Re: If x<y, is x(1+x)<y(1+y)? [#permalink] ### Show Tags 19 Mar 2018, 08:35 MathRevolution wrote: [GMAT math practice question] If $$x<y$$, is $$x(1+x)<y(1+y)$$? $$1) x>\frac{1}{2}$$ $$2) x+y>1$$ Let us re-arrange the equation: $$x (1+x) < y (1+y)$$ $$x + x^2 < y + y^2$$ $$x + x^2 - y - y^2 < 0$$ $$x - y + x^2 - y^2 < 0$$ $$(x-y) + (x-y) (x+y) < 0$$ $$(x-y) (x+y+1) < 0$$ - now this is actually what we need to prove. Already given: $$x<y$$ i.e. $$x-y < 0$$ Statement-1: $$x > 1/2$$ $$x > 1/2$$, it means $$x$$ is +ve. it is given that $$x<y$$, which means y is +ve too. Hence $$(x+y+1)$$ is +ve. with this, We are now left to prove $$(x-y) < 0$$ but it is already given. Sufficient. Statement-2: $$x+y > 1$$ it means $$x+y+1$$ is also positive. we are now left to prove $$(x-y) < 0$$ but it is already given. Sufficient. _________________ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8009 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If x<y, is x(1+x)<y(1+y)? [#permalink] ### Show Tags 21 Mar 2018, 02:49 => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. Now, $$x(1+x)<y(1+y)$$ $$=> x+x^2 - y - y^2 < 0$$ $$=> (x-y) + (x^2 - y^2) < 0$$ $$=> (x-y) + (x-y)(x+y) < 0$$ $$=> (x-y)(1+x+y) < 0$$ $$=> 1+x+y > 0$$, since $$x < y$$. Condition 1) Since $$y > x > \frac{1}{2}$$, we have $$x + y + 1 > 0.$$ Thus, condition 1) is sufficient. Condition 2) Since $$x + y > 1$$, we have $$x + y > 0$$. Thus, condition 2) is sufficient too. Therefore, D is the answer. Answer: D _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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21 Mar 2018, 04:19
MathRevolution wrote:
[GMAT math practice question]

If $$x<y$$, is $$x(1+x)<y(1+y)$$?

$$1) x>\frac{1}{2}$$
$$2) x+y>1$$

I find statement 2. to be insufficient

now, considering statement 2.
mandatory :
x < y (from first line of question)
x+y >1 (part 2 of the question)

Case I : x=2 , y = 3
2 / (2+1) & [align=][/align] 3 / (3+1)
2/3 & 3/4
.66 & .75
solution for is x part is less than y

Case II : x=-3 , y = 5
-3 / (-3+1) & 5 / (5+1)
-3/-2 & 5/6
1.5 & .83
solution for is x part is more y

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pls assess and give your input.
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Re: If x<y, is x(1+x)<y(1+y)?  [#permalink]

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21 Mar 2018, 07:10
GMAT215 wrote:

[/size]2 / (2+1) & [align=][/align] 3 / (3+1)
2/3 & 3/4
.66 & .75[b]

Case II : x=-3 , y = 5
-3 / (-3+1) & 5 / (5+1)

-3/-2 & 5/6
1.5 & .83[b]

.

you are dividing instead of multiplying as mentioned in question.
so it is x * (1+x) NOT x / (1+x)
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Re: If x<y, is x(1+x)<y(1+y)?   [#permalink] 21 Mar 2018, 07:10
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