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10 Nov 2021, 12:53
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E
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Difficulty:
45%
(medium)
Question Stats:
70% (02:16) correct
30%
(02:18)
wrong
based on 323
sessions
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If 2x + y > 1, and 5 + 4y < x, which of the following must be true?
A) x < 0
B) y < 0
C) y > 0
D) x < y
E) y < x
A) x < 0
B) y < 0
C) y > 0
D) x < y
E) y < x
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11 Nov 2021, 07:41
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BrentGMATPrepNow
When a question provides us with two inequalities, you may be able to apply a property that says the following:
If the inequality symbols of two inequalities are facing the SAME DIRECTION, then we can ADD those inequalities (for more on this, see the video below)
Given:
2x + y > 1
5 + 4y < x
Rewrite the bottom equation as follows:
2x + y > 1
x > 5 + 4y
Subtract 4y from both sides of the bottom inequality:
2x + y > 1
x - 4y > 5
ADD the inequalities to get: 3x - 3y > 6
Divide both sides of the inequality by 3 to get: x - y > 2
At this point, we can conclude that x > y (aka y < x).
Answer: E
Aside: If you're not convinced, just follow these steps:
Take: x - y > 2
Add a second inequality: x - y > 2 > 0
This means: x - y > 0
Add y to both sides to get: x > y
Answer: E
RELATED VIDEO
10 Nov 2021, 23:12
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BrentGMATPrepNow
2x + y > 1 and x > 5 + 4y
In such questions add the similar sides, that is greater to greater and lesser to lesser
\(2x+y+x>1+5+4y\)
\(3x>6+3y\)
\(x>2+y\)
Now \(2+y>y\), so \(x>2+y>y\) or \(x>y\)
E
