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Is xy + 1 greater than x + y?

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Is xy + 1 greater than x + y?

by Max@Math Revolution » Wed Dec 25, 2019 4:55 am
[GMAT math practice question]

Is xy + 1 greater than x + y?

1) 0 ≤ x < 1 and 0 ≤ y < 1
2) x + y is negative.
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Source: — Data Sufficiency |
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by SampathKp » Thu Dec 26, 2019 3:42 am
From Statement (1)

We have take 4 option for values of X and Y to cover all suitable scenarios.

1. X and Y are lowest possible values
2. X is lowest possible value and Y is highest possible value
3. X is highest possible value and Y is lowest possible value
4. X and Y are highest possible values.

1. X= 0, Y =0 --> XY+1 = 1 , X+Y = 0 so XY+1>X+Y
2. X=0 and Y=.9 ---> XY+1 =1 and X+Y =.9 so XY+1>X+Y
3. X=.9 and Y=0 --> XY+1 = 1 and X+Y =.9 so XY+1 >X+Y
4. X=.9 and Y=.9 ---> XY+1 = 1.81 and X+Y = 1.80 so XY+1>X+Y

Hence we can conclude that Statement (1) is sufficient to answer the question.

From (2) X+ Y is negative

Take 3 options for Values X and Y
1. Both X and Y are negative.
2. X is negative and Y is positive such that X+Y is negative
3. X is positive and Y is negative such that X+Y is negative

1. X=-1 and Y=-1 ---> XY+ 1= 2 , X+Y = -2 so XY+1> X+Y
2. X=-2 and Y=1 ----> XY+1 = -1 , X+Y = -1 so XY+1 = X+Y

So clearly Statement (2) is NOT Sufficient to answer the question.

Answer is Option A i.e Only Statement (1) is sufficient to answer the question and Statement (2) is not sufficient to answer the question.
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by Max@Math Revolution » Thu Dec 26, 2019 11:27 pm
=>

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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

The question xy + 1 > x + y is equivalent to (x - 1)(y - 1)>0 for the following reason.
xy + 1 > x + y
=> xy + 1 - x - y > 0
=> xy - x - y + 1 > 0 (rearranging the equation)
=> x(y - 1) - 1(y - 1) > 0
=> (x - 1)(y - 1) > 0
=> x > 1, y > 1 or x < 1, y < 1

Condition 1)
Condition 1) tells us that 0 ≤ x < 1 and 0 ≤ y < 1. This tells us that x < 1, and y < 1, which fits our modified condition. Therefore, the answer is unique, 'yes,' and the condition is sufficient.

Condition 2)
If x = -1, and y = -1, then xy + 1 = (-1)(-1) + 1 = 1 + 1 = 2, x + y = (-1) + (-1) = -2. In this case, xy + 1 is greater than x + y and the answer is 'yes'.
If x = 2, and y = -3, then xy + 1 = (2)(-3) + 1 = -6 + 1 = -5, x + y = 2 + (-3) = -1. In this case, xy + 1 is less than x + y and the answer is 'no'.

Since condition 2) does not yield a unique solution, it is not sufficient.

Therefore, A is the answer.
Answer: A
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