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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.

WE 1: 7 Yrs in Automobile (Commercial Vehicle industry)

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

Show Tags

06 Aug 2012, 09:33

LalaB wrote:

well, here are 2 methods -

lets just check answers,beginning with ans C

|11/2- (1/2)| <11/2 yes,of course, since u subtract from the number 11/2 some portion.

method 2-

|y-1/2| < 11/2

if y>1/2, then y-1/2 < 11/2 y<6 , so ,we need some number between 1/2 and 6

if y<1/2, then 1/2- y<11/2 , y>6 reject it,since it contradicts with y<1/2

so, now lets check all answer choices and find the answer between 1/2 and 6

btw, do u post 700+ OG13 questions? havent seen them.

If y<1/2, then 1/2-y<11/2 ie 1/2-11/2 <y, ie -5<y. so y lies between -5<y<1/2.

Am I correct???
_________________

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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.

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.

Hi Experts, how much I try I can not refrain from opening the modulus and assigning +ve and -ve values to expressions inside modulus. Is there any short cut approach to this , please see screenshot.

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