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Absolute Values

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by Bullzi » Fri Oct 02, 2015 10:41 am
What is the value of y?
1. 3|x^2 -4| = y - 2
2. |3 - y| = 11

Answer - Option C

While I was able to work with option 2, I found option 1 difficult to make sense of. Please help!

Thanks
Bullzi
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Source: — Data Sufficiency |

by GMATGuruNY » Fri Oct 02, 2015 11:50 am
What is the value of y?

(1) 3|x2 - 4| = y - 2

(2) |3 - y| = 11
Statement 1: 3|x² - 4| = y - 2.
If x=0, then y=14.
If x=2, then y=2.
Since y can be different values, insufficient.

Statement 2: |3 - y| = 11.
Solving 3-y = 11 and 3-y = -11, we get:
y = -8 or y=14.
Since both y = -8 and y=14 are possible, insufficient.

Statements 1 and 2 combined:
Statement 2 requires that y = -8 or y=14.
Plugging y = -8 into 3|x² - 4| = y - 2, we get:
3|x² - 4| = -8-2
3|x² - 4| = -10
Not possible, since the left side cannot yield a negative result.
Thus, y=14.
Sufficient.

The correct answer is C.
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by Bullzi » Fri Oct 02, 2015 7:34 pm
Thanks Mitch for your response..

I have a quick question, from Statement 1, I see that Y could have multiple values. However, when we combine Statements 1 and 2, we have tested just for the 2 values we have arrived at based on Statement 2. This way, aren't we ignoring probable values we could've have calculated based on Statement 1..?

In short, how do we know for sure that y=14 is the only solution for this problem?

Thanks
Bullzi
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by GMATGuruNY » Sat Oct 03, 2015 2:59 am
Bullzi wrote:Thanks Mitch for your response..

I have a quick question, from Statement 1, I see that Y could have multiple values. However, when we combine Statements 1 and 2, we have tested just for the 2 values we have arrived at based on Statement 2. This way, aren't we ignoring probable values we could've have calculated based on Statement 1..?

In short, how do we know for sure that y=14 is the only solution for this problem?

Thanks
Bullzi
When the statements are combined, AT LEAST ONE value for y must satisfy both statements.
If ONLY ONE value for y satisfies both statements, then the value of y can be determined, and the two statements combined are SUFFICIENT.
If MORE THAN ONE value for y satisfies both statements, then the value of y CANNOT be determined, and the two statements combined are INSUFFICIENT.

In the DS above, only y=-8 and y=14 satisfy statement 2.
Since no other values for y satisfy statement 2, only y=-8 and/or y=14 could satisfy both statements.
y=-8 does not satisfy statement 1.
Implication:
Since AT LEAST ONE value for y must satisfy both statements, it must be true that y=14 satisfies both statements.
Since the value of y can be determined, the two statements combined are SUFFICIENT.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

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by jain2016 » Sun Nov 22, 2015 5:09 am
Hi Experts ,

Can you please check and correct me.

Statement 1 : 3|x^2-4| = y - 2

If I solve this , will get two equation..

a) 3x^2 - y = 10 ....(|)

b) 3x^2 + y = 14 ....(||)

so can we solve these two above equation and get the value of y?

Please advise and correct me..

Thanks,

SJ
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by theCEO » Sun Nov 22, 2015 7:21 am
jain2016 wrote:Hi Experts ,

Can you please check and correct me.

Statement 1 : 3|x^2-4| = y - 2

If I solve this , will get two equation..

a) 3x^2 - y = 10 ....(|)

b) 3x^2 + y = 14 ....(||)

so can we solve these two above equation and get the value of y?

Please advise and correct me..

Thanks,

SJ
Hi SJ,

Inequalities with absolute value can produce 2 equations as you mentioned, however only one of the equations is correct, and they cannot be both correct.

Here is a simple example: 3|x| = 3

Solution 1: 3x = 3; x=1
Solution 2: -3x = 3; x=-1

As you see these two solutions will never be equal.

Therefore only one of the below equations is correct and we cannot use both
a) 3x^2 - y = 10 ....(|)
b) 3x^2 + y = 14 ....(||)
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by jain2016 » Sun Nov 22, 2015 9:43 pm
Hi TheCEO ,

Thanks for your reply.

So how do we come to know , which equation is correct and which one we have to use?

Thanks,

SJ
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by theCEO » Mon Nov 23, 2015 2:11 am
jain2016 wrote:Hi TheCEO ,

Thanks for your reply.

So how do we come to know , which equation is correct and which one we have to use?

Thanks,

SJ
The key to knowing which one is correct is to use the information from the second statement. If the information from the second statement, helps you pick the correct equation then the answer is most likely C; if the information from the second statement doesn't help you pick the correct equation then the answer is most likely E.

Let me know if this help.
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by Matt@VeritasPrep » Fri Nov 27, 2015 1:21 am
jain2016 wrote: Statement 1 : 3|x^2-4| = y - 2

If I solve this , will get two equation..

a) 3x^2 - y = 10 ....(|)

b) 3x^2 + y = 14 ....(||)

so can we solve these two above equation and get the value of y?
Not quite, since x will still be lurking in our equation. We can find that

|x² - 4| = (y - 2)/3

and that either

x² - 4 = (y - 2)/3

or

4 - x² = (y - 2)/3

But there are an infinite number of pairs of solutions to these equations. (Using only integers, we could have x = 1, y = 11 or x = 0, y = 14 or x = -1, y = 11, etc.)
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by Matt@VeritasPrep » Fri Nov 27, 2015 1:23 am
Bullzi wrote:Thanks Mitch for your response..

I have a quick question, from Statement 1, I see that Y could have multiple values. However, when we combine Statements 1 and 2, we have tested just for the 2 values we have arrived at based on Statement 2. This way, aren't we ignoring probable values we could've have calculated based on Statement 1..?

In short, how do we know for sure that y=14 is the only solution for this problem?

Thanks
Bullzi
This is a great question: the short answer is that the correct answer must satisfy BOTH equations. Since S2 only has two solutions, the correct answer must be one of those two. We can then plug those into S1 and see if both work or if only one does, finishing the problem.
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