MathRevolution wrote:
AbdurRakib wrote:
If x and y are positive, is xy > x + y?
(1) x < y
(2) 2 < x
OG Q 2017 New Question(Book Question: 199)
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, and then recheck the question.
xy > x + y
⇔ xy - x - y > 0
⇔ xy - x - y + 1 > 1
⇔ (x-1)(y-1) > 1
⇔ x > 1, y > 1 or 0 < x < 1, 0 < y < 1?Since we have 2 variables (x and y) and 0 equations,C is most likely to be the answer. So, we should consider 1) & 2) first.
Conditions 1) & 2):
Since x > 2, y > x > 2 or y > 2, both conditions together are sufficient.
Since this is an inequality question (one of the key question areas), we should also consider choices A and B by CMT(Common Mistake Type) 4(A).
Condition 1)
x = 2, y = 3: Yes
x = 1/2, y = 1: No
The condition 1) is not sufficient.
Condition 2)
x = 3, y = 4: Yes
x = 3, y = 1/2: No
The condition 2) is not sufficient.
Therefore, the answer is C.
Hi
MathRevolution,
Could you explain me the highlighted step after the below step
(x-1)(y-1)>1
So we have
(x-1)>1
or
(y-1)>1
So we have x> 2 and or we have y>2
What did i miss?
However, this is how I approached
xy>x+y
xy-x-y>0
adding 1 on both sides
xy-x-y+z>1
(x-1)(y-1)>1
so either
x-1>1 more so x>2
or y-1>1 more so y>2
So question is asking IS x>2 and y>2. we can have answer to our question is we know about x and y.
So statement A:
(A) x < y
we know that x>o and y>0, but we dont kow if x>2 and y>2
So insufficient.
Statement B:
(B) 2 < x
We know that x>0 so x>2 but don't have any information about y
so insufficient,
Now combining A and B we have
2<x<y
so we have x>2 and y>2
However just to remember
A * B\(\geq\) A + B.
When A & B \(\leq{0}\), it's always true.
When A & B \(\geq 2\), it's always true.
When A \(\geq{1}\), B \(\leq{1}\) (or B\(\geq{1}\), A\(\leq{1}\)), it's always false.
When 0 < A < 1, B < 0 (or 0 < B < 1, A < 0), it can be either true or false.
When 1 < A < 2, B > 0 (or 1 < B < 2, A > 0), it can be either true or false.
A * B > A + B.
when A>2 and B>2
For this question if we had two statements say
Statement 1: A>2
statement 2: B>2
On combining 1 and 2 we can say that (A*B)> (A+B)
Probus
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Probus
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