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happypuppy
GMAT 1: 720 Q50 V38 (Online)
Posts: 203
05 Jan 2022, 04:56
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
40% (02:27) correct
60%
(02:29)
wrong
based on 144
sessions
History
Date
Time
Result
Not Attempted Yet
If 2x + 3y < 15 and 3x - y ≥ -5, which of the following is true?
(A) x > 0
(B) y < 5
(C) y ≤ 5
(D) A and B
(E) A and C
(A) x > 0
(B) y < 5
(C) y ≤ 5
(D) A and B
(E) A and C
14 Jan 2022, 06:00
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I first thought there is something wrong with the OA, but it's actually not. But feel free to correct me if I am mistaken.
First, it is important to know the rules of manipulating inequalities.
We can add unequal terms only if the signs point into the same direction and subtract them only if they point into opposite directions.
Adding or subtracting a number or variable from both sides does not affect the sign.
Multiplying or dividing both sides with the same positive number keeps the direction, multiplying or dividing by a negative number flips it.
Knowing that, we can do the following
I \(2x + 3y < 15\)
II \(3x - y \geq 5\)
\(I*3-II*2\)
\(11y < 40\)
\(y < \frac{40}{11}\)
We should now be able to rule out answer C) since we have shown that y can not be equal to 5.
Knowing that y can be 3 which sould be close to the upper limit, we can plug this into the formulae.
I \(2x + 9 < 15\)
\(x < 2\)
Now this first made me think that answer A) should also be true, but y can be also a negative number. So let's do that for y = -11
II \(3x + 11 \geq 5\)
\(x \geq -2\)
Therefore we have shown that x does not always has to be a positive number leaving us with the answer B.
Leave some kudos if it helped you
First, it is important to know the rules of manipulating inequalities.
We can add unequal terms only if the signs point into the same direction and subtract them only if they point into opposite directions.
Adding or subtracting a number or variable from both sides does not affect the sign.
Multiplying or dividing both sides with the same positive number keeps the direction, multiplying or dividing by a negative number flips it.
Knowing that, we can do the following
I \(2x + 3y < 15\)
II \(3x - y \geq 5\)
\(I*3-II*2\)
\(11y < 40\)
\(y < \frac{40}{11}\)
We should now be able to rule out answer C) since we have shown that y can not be equal to 5.
Knowing that y can be 3 which sould be close to the upper limit, we can plug this into the formulae.
I \(2x + 9 < 15\)
\(x < 2\)
Now this first made me think that answer A) should also be true, but y can be also a negative number. So let's do that for y = -11
II \(3x + 11 \geq 5\)
\(x \geq -2\)
Therefore we have shown that x does not always has to be a positive number leaving us with the answer B.
Leave some kudos if it helped you
06 Jan 2025, 11:30
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happypuppy
Less than inequality means anything that lies on the left or below that line. Similarly greater than inequality means anything that lies on the right or above that line.
If we plot the graph here, assuming an equals sign in the given linear equations, this becomes quite easy to answer which is nothing but the overlapping area.
Cross verify the intersection point to safe-check the answer choice as option B and C are quite close.
IMO: B
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