How to find absolute value - GRE Quantitative Reasoning
Card 0 of 6
Quantitative Comparison:
Column A
|–3 + 4|
Column B
|–3| + |4|
Quantitative Comparison:
Column A
|–3 + 4|
Column B
|–3| + |4|
The operations in the absolute value are always done first. So in Column A, |–3 + 4| = |1| = 1. In Column B, |–3| + |4| = 3 + 4 = 7.
The operations in the absolute value are always done first. So in Column A, |–3 + 4| = |1| = 1. In Column B, |–3| + |4| = 3 + 4 = 7.
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What is the absolute value of the following equation when 
?
What is the absolute value of the following equation when
(–3)3 = –27. Any time a negative number is cubed, it remains negative. –27 + 5 = –22. The absolute value of any number will ALWAYS be positive so the absolute value of –22 is 22. This is our answer.
(–3)3 = –27. Any time a negative number is cubed, it remains negative. –27 + 5 = –22. The absolute value of any number will ALWAYS be positive so the absolute value of –22 is 22. This is our answer.
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Quantitative Comparison
|x – 3| = 3
Quantity A: x
Quantity B: 2
Quantitative Comparison
|x – 3| = 3
Quantity A: x
Quantity B: 2
It's important to remember that absolute value functions yield two equations, not just one. Here we have x – 3 = 3 AND x – 3 = –3.
Therefore x = 6 or x = 0, so the answer cannot be determined.
If we had just used the quation x – 3 = 3 and forgotten about the second equation, we would have had x = 6 as the only solution, giving us the wrong answer.
It's important to remember that absolute value functions yield two equations, not just one. Here we have x – 3 = 3 AND x – 3 = –3.
Therefore x = 6 or x = 0, so the answer cannot be determined.
If we had just used the quation x – 3 = 3 and forgotten about the second equation, we would have had x = 6 as the only solution, giving us the wrong answer.
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Quantitative Comparison
Quantity A: |10| – |16|
Quantity B: |1 – 5| – |3 – 6|
Quantitative Comparison
Quantity A: |10| – |16|
Quantity B: |1 – 5| – |3 – 6|
Quantity A: |10| = 10, |16| = 16, so |10| – |16| = 10 – 16 = –6.
Quantity B: |1 – 5| = 4, |3 – 6| = 3, so |1 – 5| - |3 – 6| = 4 – 3 = 1.
1 is bigger than –6, so Quantity B is greater.
Quantity A: |10| = 10, |16| = 16, so |10| – |16| = 10 – 16 = –6.
Quantity B: |1 – 5| = 4, |3 – 6| = 3, so |1 – 5| - |3 – 6| = 4 – 3 = 1.
1 is bigger than –6, so Quantity B is greater.
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Quantitative Comparison
Quantity A: (|–4 + 1| + |–10|)2
Quantity B: |(–4 + 1 – 10)2|
Quantitative Comparison
Quantity A: (|–4 + 1| + |–10|)2
Quantity B: |(–4 + 1 – 10)2|
Quantity A: |–4 + 1| = |–3| = 3 and |–10| = 10, so (|–4 + 1| + |–10|)2 = (3 + 10)2 = 132 = 169
Quantity B: |(–4 + 1 – 10)2| = |(–13)2| = 169
The two quantities are equal.
Quantity A: |–4 + 1| = |–3| = 3 and |–10| = 10, so (|–4 + 1| + |–10|)2 = (3 + 10)2 = 132 = 169
Quantity B: |(–4 + 1 – 10)2| = |(–13)2| = 169
The two quantities are equal.
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Quantity A: 
Quantity B: 
Quantity A:
Quantity B:
If 
, then either 
or 
must be negative, but not both. Making them both positive, as in quantity B, and then adding them, would produce a larger number than adding them first and making the result positive.
If
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