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

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

by Max@Math Revolution » Mon Mar 19, 2018 1:40 am
[GMAT math practice question]

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

1) x>1/2
2) x+y>1
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Source: — Data Sufficiency |
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Answer

by Vincen » Mon Mar 19, 2018 6:37 am
I think statement (1) is sufficient, but respect to the statement (2), I don't know how to prove if it is or not sufficient.

If x>1/2 then y>x>0. Therefore, if we pick x=1 and y=2 we have that $$1\left(1+1\right) < 2\left(2+1\right)\ \Rightarrow\ \ 1\cdot 2 < 2\cdot3\ \Rightarrow\ 2 < 6\ TRUE.$$ More general, $$x < y\ \Rightarrow\ \ x+1 < y+1\ \Rightarrow\ \ x\left(x+1\right) < x\left(y+1\right)\Rightarrow\ x\left(x+1\right) < y\left(y+1\right).$$ Hence, (1) is sufficient.

Now, I need some help with the statement (2).
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by Max@Math Revolution » Wed Mar 21, 2018 1:43 am
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 > 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
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by Max@Math Revolution » Wed Mar 21, 2018 10:36 am
Vincen wrote:I think statement (1) is sufficient, but respect to the statement (2), I don't know how to prove if it is or not sufficient.

If x>1/2 then y>x>0. Therefore, if we pick x=1 and y=2 we have that $$1\left(1+1\right) < 2\left(2+1\right)\ \Rightarrow\ \ 1\cdot 2 < 2\cdot3\ \Rightarrow\ 2 < 6\ TRUE.$$ More general, $$x < y\ \Rightarrow\ \ x+1 < y+1\ \Rightarrow\ \ x\left(x+1\right) < x\left(y+1\right)\Rightarrow\ x\left(x+1\right) < y\left(y+1\right).$$ Hence, (1) is sufficient.

Now, I need some help with the statement (2).
The modified question is "1+x+y > 0 ?".
From the condition 2), we derived it is true.

Happy Studying !!!
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by Max@Math Revolution » Sat Mar 24, 2018 5:20 pm
Vincen wrote:I think statement (1) is sufficient, but respect to the statement (2), I don't know how to prove if it is or not sufficient.

If x>1/2 then y>x>0. Therefore, if we pick x=1 and y=2 we have that $$1\left(1+1\right) < 2\left(2+1\right)\ \Rightarrow\ \ 1\cdot 2 < 2\cdot3\ \Rightarrow\ 2 < 6\ TRUE.$$ More general, $$x < y\ \Rightarrow\ \ x+1 < y+1\ \Rightarrow\ \ x\left(x+1\right) < x\left(y+1\right)\Rightarrow\ x\left(x+1\right) < y\left(y+1\right).$$ Hence, (1) is sufficient.

Now, I need some help with the statement (2).
Since x + y > 1 from the condition 2), we have x + y + 1 > 2 > 0.
And modified question is x + y + 1 > 0?

Thus condition 2) is sufficient.

Happy Studying !!!
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