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Re: If |y-1/2| < 11/2, which of the following could be a value [#permalink]
06 Aug 2012, 01:50

Expert's post

2

This post was BOOKMARKED

SOLUTION

If \(|y-\frac{1}{2}| < \frac{11}{2}\), which of the following could be a value of y?

(A) -11 (B) -11/2 (C) 11/2 (D) 11 (E) 22

\(|y-\frac{1}{2}| < \frac{11}{2}\), is equivalent to \(-\frac{11}{2}<y-\frac{1}{2}< \frac{11}{2}\).

Add \(\frac{1}{2}\) to each part of the inequality: \(-\frac{11}{2}+\frac{1}{2}<y< \frac{11}{2}+\frac{1}{2}\) --> \(-5<y<6\). Only answer C is from this range.

If \(|y-\frac{1}{2}| < \frac{11}{2}\), which of the following could be a value of y?

(A) -11 (B) -11/2 (C) 11/2 (D) 11 (E) 22

An Easy way for this type of Problem is as follows

When ever you see an in equality with modulus remember these two formulas 1.) |x-b| < c .................This always means that ............ -c+b < x < c +b 2.) |x-b| > c ................. This always means that .......... Either x < -c+b .................... or x > c+b

Use this here and see.

Last edited by Narenn on 25 Mar 2014, 20:57, edited 1 time in total.

Re: If |y-1/2| < 11/2, which of the following could be a value [#permalink]
10 Aug 2012, 03:09

Expert's post

SOLUTION

If \(|y-\frac{1}{2}| < \frac{11}{2}\), which of the following could be a value of y?

(A) -11 (B) -11/2 (C) 11/2 (D) 11 (E) 22

\(|y-\frac{1}{2}| < \frac{11}{2}\), is equivalent to \(-\frac{11}{2}<y-\frac{1}{2}< \frac{11}{2}\).

Add \(\frac{1}{2}\) to each part of the inequality: \(-\frac{11}{2}+\frac{1}{2}<y< \frac{11}{2}+\frac{1}{2}\) --> \(-5<y<6\). Only answer C is from this range.

Re: If |y-1/2| < 11/2, which of the following could be a value [#permalink]
16 Dec 2012, 06:26

1

This post received KUDOS

Bunuel wrote:

SOLUTION

If \(|y-\frac{1}{2}| < \frac{11}{2}\), which of the following could be a value of y?

(A) -11 (B) -11/2 (C) 11/2 (D) 11 (E) 22

\(|y-\frac{1}{2}| < \frac{11}{2}\), is equivalent to \(-\frac{11}{2}<y-\frac{1}{2}< \frac{11}{2}\).

Add \(\frac{1}{2}\) to each part of the inequality: \(-\frac{11}{2}+\frac{1}{2}<y< \frac{11}{2}+\frac{1}{2}\) --> \(5<y<6\). Only answer C is from this range.

Answer: C.

I might sound stupid but I really could not understand why \(-\frac{11}{2}+\frac{1}{2}<y< \frac{11}{2}+\frac{1}{2}\) results in 5<y<6, should not it be -5<y<6

Re: If |y-1/2| < 11/2, which of the following could be a value [#permalink]
16 Dec 2012, 06:33

1

This post received KUDOS

Expert's post

Drik wrote:

Bunuel wrote:

SOLUTION

If \(|y-\frac{1}{2}| < \frac{11}{2}\), which of the following could be a value of y?

(A) -11 (B) -11/2 (C) 11/2 (D) 11 (E) 22

\(|y-\frac{1}{2}| < \frac{11}{2}\), is equivalent to \(-\frac{11}{2}<y-\frac{1}{2}< \frac{11}{2}\).

Add \(\frac{1}{2}\) to each part of the inequality: \(-\frac{11}{2}+\frac{1}{2}<y< \frac{11}{2}+\frac{1}{2}\) --> \(5<y<6\). Only answer C is from this range.

Answer: C.

I might sound stupid but I really could not understand why \(-\frac{11}{2}+\frac{1}{2}<y< \frac{11}{2}+\frac{1}{2}\) results in 5<y<6, should not it be -5<y<6

Yes it should. Typo edited. Thank you. +1. _________________

Re: If |y-1/2| < 11/2, which of the following could be a value [#permalink]
16 Dec 2012, 07:15

Bunuel wrote:

Drik wrote:

Bunuel wrote:

SOLUTION

If \(|y-\frac{1}{2}| < \frac{11}{2}\), which of the following could be a value of y?

(A) -11 (B) -11/2 (C) 11/2 (D) 11 (E) 22

\(|y-\frac{1}{2}| < \frac{11}{2}\), is equivalent to \(-\frac{11}{2}<y-\frac{1}{2}< \frac{11}{2}\).

Add \(\frac{1}{2}\) to each part of the inequality: \(-\frac{11}{2}+\frac{1}{2}<y< \frac{11}{2}+\frac{1}{2}\) --> \(5<y<6\). Only answer C is from this range.

Answer: C.

I might sound stupid but I really could not understand why \(-\frac{11}{2}+\frac{1}{2}<y< \frac{11}{2}+\frac{1}{2}\) results in 5<y<6, should not it be -5<y<6

Yes it should. Typo edited. Thank you. +1.

Thanks Bunnel for such a valuable reply..and +1 for you as well

I was wondering, if the equation stands at -5<y<6, should not it be that both B & C are right.. Thanks

Re: If |y-1/2| < 11/2, which of the following could be a value [#permalink]
15 Apr 2015, 23:09

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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