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Absolute Value DS Question

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Joined: 06 Oct 2018
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Absolute Value DS Question  [#permalink]

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New post 07 Oct 2018, 08:26
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Question Stats:

50% (01:42) correct 50% (00:10) wrong based on 6 sessions

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Hi All,

I found this question on one of the MGMAT CATs and am very confused about their explanation.

The question is as follows:
What is the distance between x and y on the number line?

(1) |x| – |y| = 5
(2) |x| + |y| = 11

Here is their explanation:

1) This tells us that the difference between the absolute value of x and the absolute value of y is 5. Let’s pick some numbers to prove that this is insufficient. Say, for example, x = 6 and y = 1. Then |x| – |y| = 5 and the distance between x and y is 6 – 1 = 5. However, let’s pick x = 6 and y = -1. Then |x| – |y| = 5 and the distance between x and y is 6 – (-1) = 7. Since we picked two sets of numbers that fit the criteria and got different answers, the statement is insufficient.

If we pick x = 6 and y = -1, then isn't |x| = 6 and |y| = 1?? Absolute value implies taking the difference from that number and 0, hence it'll always be positive. If we pick x = 6 and y = -1, why are we doing the following operation: 6 - (-1) and not 6 - 1??

Obviously, I am aware that if we draw these two points on the number line, we do indeed get a difference of 7 if y = -1 and x = 6. But could someone explain this algebraically or theoretically why if we pick y = -1 we input the minus sign inside the modulus?


Thanks!
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Re: Absolute Value DS Question  [#permalink]

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New post 07 Oct 2018, 08:45
PierTotti17 wrote:
Hi All,

I found this question on one of the MGMAT CATs and am very confused about their explanation.

The question is as follows:
What is the distance between x and y on the number line?

(1) |x| – |y| = 5
(2) |x| + |y| = 11

Here is their explanation:

1) This tells us that the difference between the absolute value of x and the absolute value of y is 5. Let’s pick some numbers to prove that this is insufficient. Say, for example, x = 6 and y = 1. Then |x| – |y| = 5 and the distance between x and y is 6 – 1 = 5. However, let’s pick x = 6 and y = -1. Then |x| – |y| = 5 and the distance between x and y is 6 – (-1) = 7. Since we picked two sets of numbers that fit the criteria and got different answers, the statement is insufficient.

If we pick x = 6 and y = -1, then isn't |x| = 6 and |y| = 1?? Absolute value implies taking the difference from that number and 0, hence it'll always be positive. If we pick x = 6 and y = -1, why are we doing the following operation: 6 - (-1) and not 6 - 1??

Obviously, I am aware that if we draw these two points on the number line, we do indeed get a difference of 7 if y = -1 and x = 6. But could someone explain this algebraically or theoretically why if we pick y = -1 we input the minus sign inside the modulus?


Thanks!


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Re: Absolute Value DS Question &nbs [#permalink] 07 Oct 2018, 08:45
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