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Is √x+x>-√y?

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Is √x+x>-√y?  [#permalink]

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New post 03 Apr 2018, 00:11
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

35% (01:31) correct 65% (01:31) wrong based on 37 sessions

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[GMAT math practice question]

Is \(√x+x>-√y?\)

\(1) √x+√y = 1\)
\(2) x>0\)

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Is √x+x>-√y?  [#permalink]

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New post 03 Apr 2018, 02:59
1
1
Is X^(1/2) +X >- Y^(1/2)?

Since square root function is defined ONLY for NON NEGATIVE NUMBERS, it means X is POSITIVE or ZERO.

Also square root of any number is defined to be NON negative.

Now Question is asking “Is Non negative+ Non negative > - Non negative? “

Is X^(1/2) +X >- Y^(1/2)? Yes, if both are X and Y positive

Is X^(1/2) +X >- Y^(1/2)? Yes , if X = 0, Y positive

Is X^(1/2) +X >- Y^(1/2)? No, if both are ZERO ( in this case both side will be equal to 0)



The answer will always be yes, except when both X and Y are ZERO



St1) X^(1/2) + Y^(1/2)=1

Atleast one of X and Y is positive.

Hence Sufficient.



St2) X>0,

Sufficient.



Answer = D

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Is √x+x>-√y?  [#permalink]

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New post 05 Apr 2018, 04:29
=>

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.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

\(√x+x>-√y
=> √x+√y+ x>0\)
Since \(√x+√y≥0\) is always true, we just need to check if \(x > 0\).
Condition 2) is sufficient and since condition 1) is hard to check and condition 2) is easy to check, the answer is D by CMT (Common Mistake Type) 4B.

Condition 1)

We have \(x ≥ 0\)from \(√x\),
Then \(√x+√y + x ≥ 1 > 0.\)
Thus condition 1) is sufficient.


Condition 2)

Since \(√x+√y ≥ 0\) and \(x > 1\), we have \(√x+√y + x > 1 > 0.\)
Thus condition 2) is sufficient.

Therefore, D is the answer.


Answer: D

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

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