How to find absolute value - GRE Quantitative Reasoning

(–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.

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.

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: |–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.

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.

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.

(–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.

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.

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: |–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.

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.

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.

(–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.

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.

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: |–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.

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.

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.

(–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.