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Jmoustafa345
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Guide 6, Chapter 3, Pg. 56

by Jmoustafa345 Tue Mar 17, 2015 7:40 pm

For problem #8, the Answer Key says the answer is B, but I keep getting D. I solved for the absolute value and got that x>5 and x<-1. I tried plugging in several numbers for x (-2 and 6, -10 and 10). With the negative numbers that I plug in for x, quantity A is bigger, and with the positive numbers that I plug in, quantity B is bigger. I can't see how the answer would be B.
n00bpron00bpron00b
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Re: Guide 6, Chapter 3, Pg. 56

by n00bpron00bpron00b Wed Mar 18, 2015 9:44 am

|x-2|>3
Solving this we get, x>5 & x<-1

Let's assume the value of "x" as 5.1 or -1.1 (satisfying the above condition)

Now onto the second part,

|x-3.5| => The distance of "x" is +3.5 units away from origin (or in this case 5.1 or -1.1)

|x-1.5| => The distance of "x" is +1.5 units away from origin (or in this case 5.1 or -1.1)

If x = 5.1
|x-3.5| = 5.1 + 3.5 = 8.6
|x-1.5| = 5.1+1.5 = 6.6

We have to compare minimum possible distance, so (B)

Similarly,
If x = -1.1
|x-3.5| = -1.1+3.5 = 2.4
|x-1.5| = -1.1 + 1.5 = 0.4

Minimum distance (B)
tommywallach
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Re: Guide 6, Chapter 3, Pg. 56

by tommywallach Thu Mar 19, 2015 5:16 pm

Hey J,

It seems like the mistake you made was missing the "minimum" part. So you can't just "take" whatever comes out of the absolute values when you plug in something for x. You have to search out the MINIMUM it can be, and that's your column value. So actually, the question itself makes (D) impossible, because there can only be ONE minimum for A and ONE minimum for B. This means the answer must be A, B, or C. (It can only be (D) if one or both columns could have multiple values.)

-t