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Re: Is square root[(y-4)^2] = 4-y? [#permalink]
09 Mar 2012, 05:54

3

This post received KUDOS

Expert's post

Is \(\sqrt{(y-4)^2} = 4-y\)?

Is \(\sqrt{(y-4)^2}=4-y\)? --> is \(|y-4|=4-y\)? is \(y\leq{4}\)?

(1) |y-3| less than or equal to 1 --> \(|y-3|\) is just the distance between 3 and \(y\) on the number line. We are told that this distance is less than or equal to 1: --2----3----4-- so, \(2\leq{y}\leq{4}\). Sufficient.

(2) y*|y|>0 --> just says that \(y>0\). Not sufficient.

Answer: A.

subhashghosh wrote:

sqrt{(y-4)^2} = |y-4|

And |y-4| = 4-y if y-4 < 0

or |y-4| = 4-y if y < 4

Just a little correction, which didn't play a part for this question though may be crucial for others: \(|y-4|=4-y\) if \(y-4\leq{0}\), you missed "=" sign --> if \(y=4\) the equation still holds true. _________________

Just a little correction, which didn't play a part for this question though may be crucial for others: \(|y-4|=4-y\) if \(y-4\leq{0}\), you missed "=" sign --> if \(y=4\) the equation still holds true.

I just cannot understand why do you use the inequality to restate the question? does not is simply states that |y-4| = 4-y -> y-4=4-y -> y=4 ? Otherwise, it is 0=0.

Just a little correction, which didn't play a part for this question though may be crucial for others: \(|y-4|=4-y\) if \(y-4\leq{0}\), you missed "=" sign --> if \(y=4\) the equation still holds true.

I just cannot understand why do you use the inequality to restate the question? does not is simply states that |y-4| = 4-y -> y-4=4-y -> y=4 ? Otherwise, it is 0=0.

Because \(|y-4|=4-y\) holds true for \(y\leq{4}\) and not only for \(y=4\). _________________

Just a little correction, which didn't play a part for this question though may be crucial for others: \(|y-4|=4-y\) if \(y-4\leq{0}\), you missed "=" sign --> if \(y=4\) the equation still holds true.

I just cannot understand why do you use the inequality to restate the question? does not is simply states that |y-4| = 4-y -> y-4=4-y -> y=4 ? Otherwise, it is 0=0.

Because \(|y-4|=4-y\) holds true for \(y\leq{4}\) and not only for \(y=4\).

Bunuel - I did the same thing as JubtaGubar and I'm still confused on how you and subhashghosh came to that conclusion so quickly (that \(y\leq{4}\)).

I think I figured it out --- this \(\sqrt{(y-4)^2}\) must be positive (anything squared and/or taken to its square root) which is essentially like saying that the left side must be positive or absolute value of the left side of the equation and so the right side must also be positive (or > 0). The only way that can happen is when \(y\leq{4}\)

Bunuel - I did the same thing as JubtaGubar and I'm still confused on how you and subhashghosh came to that conclusion so quickly (that \(y\leq{4}\)).

Thank you

aalba005 wrote:

+1, I am confused on the math and how you were able to simplify question step in 2 steps.

destroyerofgmat wrote:

I think I figured it out --- this \(\sqrt{(y-4)^2}\) must be positive (anything squared and/or taken to its square root) which is essentially like saying that the left side must be positive or absolute value of the left side of the equation and so the right side must also be positive (or > 0). The only way that can happen is when \(y\leq{4}\)

Am I thinking about that correctly?

Absolute value properties: When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\);

Next: Must know for the GMAT: \(\sqrt{x^2}=|x|\).

The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples: If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

BACK TO THE ORIGINAL QUESTION:

Question asks: is \(\sqrt{(y-4)^2}=4-y\)?

Now, according to the above \(\sqrt{(y-4)^2}=|y-4|\)? So the question becomes: is \(|y-4|=4-y\)? --> or is \(|y-4|=-(y-4)\)? Which, again according to the properties of absolute value, is true when \(y-4\leq{0}\) or when \(y\leq{4}\).

Re: Is square root[(y-4)^2] = 4-y? [#permalink]
13 May 2013, 11:19

Wait, when you multiply (4-y) by negative one, is that because you assume the absolute value is negative? I mean, I get that when you have absolute value with a variable you have to assume that the product is negative (thus multiplying the result by -1) but why do you do that here?

Bunuel wrote:

destroyerofgmat wrote:

Bunuel - I did the same thing as JubtaGubar and I'm still confused on how you and subhashghosh came to that conclusion so quickly (that \(y\leq{4}\)).

Thank you

aalba005 wrote:

+1, I am confused on the math and how you were able to simplify question step in 2 steps.

destroyerofgmat wrote:

I think I figured it out --- this \(\sqrt{(y-4)^2}\) must be positive (anything squared and/or taken to its square root) which is essentially like saying that the left side must be positive or absolute value of the left side of the equation and so the right side must also be positive (or > 0). The only way that can happen is when \(y\leq{4}\)

Am I thinking about that correctly?

Absolute value properties: When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|\leq{-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|\leq{some \ expression}\). For example: \(|5|=5\);

Next: Must know for the GMAT: \(\sqrt{x^2}=|x|\).

The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples: If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

BACK TO THE ORIGINAL QUESTION:

Question asks: is \(\sqrt{(y-4)^2}=4-y\)?

Now, according to the above \(\sqrt{(y-4)^2}=|y-4|\)? So the question becomes: is \(|y-4|=4-y\)? --> or is \(|y-4|=-(y-4)\)? Which, again according to the properties of absolute value, is true when \(y-4\leq{0}\) or when \(y\leq{4}\).

Re: Is square root[(y-4)^2] = 4-y? [#permalink]
13 May 2013, 15:09

shawndx wrote:

Is \(\sqrt{(y-4)^2} = 4-y\)?

(1) |y-3| less than or equal to 1 (2) y*|y|>0

Please provide a full answer please and thank you

the question basically asks whether or not /y-4/ = 4-y ,i.e. - (y-4) thus the question asks whether y-4<= 0 , is y<=4

from 1

/y-3/<=1 i.e. y is in the range -1<y-3<1 ........... 2<y<4... ............ suff since the whole range is less than 4

from 2

y is +ve and therefore /y/ = y and the inequality becomes y^2>0 .... put another way /y/ >0 ( originally this is equivalent to y<0 or y>0 if we dont know what y is but since we know that y is +ve then y>0 ........insuff ( could be bigger than 4)

Re: Is square root[(y-4)^2] = 4-y? [#permalink]
13 May 2013, 22:36

Expert's post

WholeLottaLove wrote:

Wait, when you multiply (4-y) by negative one, is that because you assume the absolute value is negative? I mean, I get that when you have absolute value with a variable you have to assume that the product is negative (thus multiplying the result by -1) but why do you do that here?

Bunuel wrote:

destroyerofgmat wrote:

Bunuel - I did the same thing as JubtaGubar and I'm still confused on how you and subhashghosh came to that conclusion so quickly (that \(y\leq{4}\)).

Thank you

aalba005 wrote:

+1, I am confused on the math and how you were able to simplify question step in 2 steps.

destroyerofgmat wrote:

I think I figured it out --- this \(\sqrt{(y-4)^2}\) must be positive (anything squared and/or taken to its square root) which is essentially like saying that the left side must be positive or absolute value of the left side of the equation and so the right side must also be positive (or > 0). The only way that can happen is when \(y\leq{4}\)

Am I thinking about that correctly?

Absolute value properties: When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|\leq{-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|\leq{some \ expression}\). For example: \(|5|=5\);

Next: Must know for the GMAT: \(\sqrt{x^2}=|x|\).

The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples: If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

BACK TO THE ORIGINAL QUESTION:

Question asks: is \(\sqrt{(y-4)^2}=4-y\)?

Now, according to the above \(\sqrt{(y-4)^2}=|y-4|\)? So the question becomes: is \(|y-4|=4-y\)? --> or is \(|y-4|=-(y-4)\)? Which, again according to the properties of absolute value, is true when \(y-4\leq{0}\) or when \(y\leq{4}\).

Hope it's clear.

Not sure understand your question. Can you please elaborate?

P.S. Note that the absolute value is ALWAYS non-negative, also I did not multiply anything by -1... _________________

Re: Is square root[(y-4)^2] = 4-y? [#permalink]
16 May 2013, 09:15

Not sure understand your question. Can you please elaborate?

P.S. Note that the absolute value is ALWAYS non-negative, also I did not multiply anything by -1...[/quote]

Sorry for the confusion.

For example, I just did a problem where I was given the absolute value of a variable - an integer. To solve I did the following:

|2x-12| < 10

2x-12 < 10

2x-12 > -10

In other words I solves for cases where 10 is positive and 10 is negative. Why wouldn't I do that with |y-4| = 4-y. Is is because the Sq. Rt. of the square guarantees that y-4 is positive?

Re: Is square root[(y-4)^2] = 4-y? [#permalink]
16 May 2013, 23:02

Expert's post

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

Theory on absolute value is HERE. PS absolute value questions are HERE. DS absolute value questions are HERE. 700+ absolute value questions HERE. _________________

Re: Is square root[(y-4)^2] = 4-y? [#permalink]
11 Jun 2013, 20:38

1

This post received KUDOS

WholeLottaLove wrote:

Is that a valid way of solving for 1) ?

Yes, looks fine!

A lot of people are thinking of this too quantitatively - sometimes it's faster and easier to just think of the idea behind it.

This is essentially what Bunuel did, but I'm just going to throw in some more words:

Looking at the expression |y-3|≤1, if you think about it for a second, it just says the difference between y and 3 is less than or equal to 1, so 2≤y≤4.

Additionally, looking at the question, we're asking "Is the square root of some square difference equal to the negative of that difference?"

Square roots must be non-negative, so the question is essentially asking "Is the right hand side expression non-negative?" I.e, "Is y <= 4?"

The solution comes very easily from there - anyways my point is not to try and do calculations all the time; while I can relate to strict mathematics being the intuitive way to approach problems, it's usually not the most efficient. _________________

Re: Is square root[(y-4)^2] = 4-y? [#permalink]
27 Jun 2013, 16:17

Is √(y-4)^2 = 4-y?

Is |y - 4| = 4 - y? (4 - y) must be positive as it is equal to an absolute value. So, y ≤ 4

(1) |y-3| less than or equal to 1 |y-3| ≤ 1 Two cases:

Positive: y≥3: (y - 3) ≤ 1 y ≤ 4 Invalid: the range of y ≥ 3 but y ≤ 4 meaning y might be greater than three but it might be less than three also.

Negative: y < 3 -(y - 3) ≤ 1 -y + 3 ≤ 1 -y ≤ -2 y ≥ 2 Invalid: as with the positive case, the range is less than 3 and y is greater than or equal to two. It may fall within the range but it may be greater than it as well. Both cases fail. SUFFICIENT

(2) y*|y|>0 for y*|y|>0 y must be positive. Therefore, y > 0 but we don't know if it's 0 < y < 4 or y > 4. INSUFFICIENT.

|Z| = -(Z) , For this to hold Z has to be negative or zero

Rephrase Is Z<=0 = > Is Y <= 4

(1). Tells the same Y <= 4 Sufficient

(2). Tells Y >0 Insufficient

Hence (A). _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Simple substitution of values can also help. Lets consider, LHS: y=5 i.e. y>4 then, sqrt(y-4)^2= sqrt(5-4)^2=1 RHS: (4-y)=> (4-5)=(-1) hence for y>4 this statement will not hold true.---------------(1) Statement1: |y-3| less than or equal to 1=>|y-3|<=1 Be careful when you have modulus. When you remove the modulus you get two equations: eqn1: (y-3)<=1 implies y<=4 eqn1: (y-3)>=(-1) implies y>=2 Remember when the sign in negative on RHS, put the greater than equal to sign. Hence your final result is: 2<=Y<=4, so y is either equal or less than 4 which means as per (1) will make the statement true. Hence this statement is sufficient.

Statement 2: y*|y|>0 which implies y>0 since |y| will always be greater than 0. But this does not ensure that y>4 or y<4, so this statement does not guarantee anything. Hence insufficient.

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