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If |y-1/2| < 11/2, which of the following could be a value

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

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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

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

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

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New post 06 Aug 2012, 09:59
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.
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Re: If |y-1/2| < 11/2, which of the following could be a value [#permalink]

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New post 06 Aug 2012, 10:33
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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Re: If |y-1/2| < 11/2, which of the following could be a value [#permalink]

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New post 06 Aug 2012, 10:39
SOURH7WK, yep, u r correct. it is a typo.forgive me all my sins, or only poor computing skills ))
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Re: If |y-1/2| < 11/2, which of the following could be a value [#permalink]

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Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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, 21:57, edited 1 time in total.
Reversed the sign of 2nd inequality

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

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New post 07 Aug 2012, 10:50
We have to see how far the value of y is on the number line
so basically, what we have here is

The distance of y from 1/2 is< 5.5
so it could be from 5 to 6.

hence answer is C

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

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

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New post 14 Dec 2012, 08:53
Why is

|y-1/2| < 11/2 equivalent to -11/2 < |y-1/2| < 11/2

Why am I not seeing this and why didn´t I come up with it?

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

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Donnie wrote:
Why is

|y-1/2| < 11/2 equivalent to -11/2 < |y-1/2| < 11/2

Why am I not seeing this and why didn´t I come up with it?


It seems that you need to brush up fundamentals on absolute value.

Theory: math-absolute-value-modulus-86462.html

PS questions on Absolute Values: search.php?search_id=tag&tag_id=58
DS questions on Absolute Values: search.php?search_id=tag&tag_id=37

Tough inequality and absolute value questions: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.
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If |y-1/2| < 11/2, which of the following could be a value [#permalink]

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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

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

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New post 16 Dec 2012, 08:49
Drik wrote:
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


-11/2=-5.5<-5 and we know that -5<y, thus y cannot be -5.5.

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

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The inequality becomes:
-11/2 < y-1/2 < +11/2

Hence, 1/2 has to be added to each inequality:
-11/2 + 1/2 = -5
11/2 + 1/2 = 6

After simplifying, we get:
-5< y < 6
The only option that lies between -5 and 6 is 11/2, which when simplified is 5.5

Thus, the answer has to be C.

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

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New post 18 Dec 2012, 01:56
Ans:

the first case gives us y<6 and the second case gives –(y-1/2)<11/2 so y >-5 only option c falls in the range of -5<y<6 answer (C).
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Re: If |y-1/2| < 11/2, which of the following could be a value [#permalink]

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New post 16 Dec 2013, 12:53
Bunuel wrote:
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

Practice Questions
Question: 14
Page: 154
Difficulty: 600



\(|y-\frac{1}{2}| < \frac{11}{2}\) <--- basically what this is saying is, "we start at 0.5, and we CANNOT go 5.5 or more steps in either direction"

which means -5 < x < 6

the only one of the options that fit the restriction is 11/2 = 5.5 ... So answer (C)

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

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If |y-1/2| < 11/2

First calculate origin (y-1/2)=0. Origin is +1/2.

This should be read as distance from origin 1/2 is less than 11/2 units. So there is range here.

So, on the right side of 1/2 we add 11/2 units and on the left side we minus 11/2 units

(1/2-11/2= -5)---------------------(1/2)---------------------(1/2+11/2 = 6)

So our answer must be in range of -5 to 6. -5<y<6

only 11/2 lies in this range. Hence answer.

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

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Here's a visual solution for this question:

\(|y-\frac{1}{2}|\) represents the distance of y from the point \(\frac{1}{2}\) = 0.5 on the number line.

The given inequality tells us that the distance of y from the point 0.5 is less than \(\frac{11}{2}\) (that is, 5.5)

The points that are at a distance of 5.5 from 0.5 on the number line are: -5 and 6

Image

This means, -5 < y < 6

Therefore, Option C is the correct answer.
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Re: If |y-1/2| < 11/2, which of the following could be a value [#permalink]

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New post 11 Apr 2016, 16:35
If you are confused by the algebraic solution, then it might be easier to plug in when absolute values are involved.
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Re: If |y-1/2| < 11/2, which of the following could be a value [#permalink]

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Bunuel wrote:
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



Solution:

We must solve the inequality for two cases:

Case 1: (y-1/2) is Positive:

|y – ½| < 11/2

y – ½ < 11/2

Add ½ to both sides of the equation:

y < 6

Case 2: (y-1/2) is Negative:

|y – ½| < 11/2

-(y – ½) < 11/2

-y + ½ < 11/2

Subtract ½ from both sides of the equation:

-y < 5

y > -5

So we know that y is greater than -5 and less than 6. The only answer choice that fits this criterion is answer choice C, 11/2.

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

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New post 30 Aug 2017, 07:48
Bunuel shashankism Engr2012

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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