1 < = x < =0, -1 < = y < = 1, z > = -2, which : Quant Question Archive [LOCKED]
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 08 Dec 2016, 19:32

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# 1 < = x < =0, -1 < = y < = 1, z > = -2, which

Author Message
Senior Manager
Joined: 19 Feb 2004
Posts: 414
Location: Lungi
Followers: 1

Kudos [?]: 30 [0], given: 0

1 < = x < =0, -1 < = y < = 1, z > = -2, which [#permalink]

### Show Tags

20 Feb 2004, 04:51
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

-1 < = x < =0, -1 < = y < = 1, z > = -2, which of the following could have the largest value?

a. x^2 + y ^2
b. x^2 - z^2
c y^2 - z ^2
d. x^2 + y^2 -z^2
e. 2 *( x-z)*(x+z) + 2*y^2
Intern
Joined: 31 May 2003
Posts: 18
Location: Bay Area, CA
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

20 Feb 2004, 11:24
Another vote for A.
Manager
Joined: 26 Dec 2003
Posts: 227
Location: India
Followers: 1

Kudos [?]: 3 [0], given: 0

### Show Tags

20 Feb 2004, 14:43
Senior Manager
Joined: 11 Nov 2003
Posts: 357
Location: Illinois
Followers: 1

Kudos [?]: 5 [0], given: 0

### Show Tags

25 Feb 2004, 09:16
Another vote for E. The greatest possible value for the expression in E is +infinity.

The greatest value for A is just 2.

Batliwala,

I have seen that you are positing a lot of good question these days. Most of them may not need to be confirmed with official naswer. However, there are few questions where official answer is required.

It would be great if you could publish official answer where there is a diagreement among the members.
Director
Joined: 13 Nov 2003
Posts: 790
Location: BULGARIA
Followers: 1

Kudos [?]: 46 [0], given: 0

### Show Tags

25 Feb 2004, 12:50
Please note that 2*(X-Z)*(X+Z)+2Y^2=2X^2-2Z^2+2Y^2 so no matter if Z>= -2 or <=-2 it is always negative, as with the example of gmatblast it should be -inf.Also think it should be A)
Senior Manager
Joined: 19 Feb 2004
Posts: 414
Location: Lungi
Followers: 1

Kudos [?]: 30 [0], given: 0

### Show Tags

26 Feb 2004, 05:53
correct ans is E.

substituting values for x, y, and z and limiting conditions E. has the greatest value =4.
Senior Manager
Joined: 11 Nov 2003
Posts: 357
Location: Illinois
Followers: 1

Kudos [?]: 5 [0], given: 0

### Show Tags

26 Feb 2004, 07:41
batliwala wrote:
correct ans is E.

substituting values for x, y, and z and limiting conditions E. has the greatest value =4.

Well I agree with the answer E but I think the value of the expression in E will be +inf.

Here is how I thought of it:

2 *( x-z)*(x+z) + 2*y^2

To get the maximum value of this expression, we need to select the values as follws:

Maximum possible value of (x-z)
Maximum possible value of (x+z)
Maximum possible value of y

To get the Maximum possible value of (x-z):

Take the maximum possible value of x = 0
Take the minimum possible value of z = -2

Hence, maximum possible value of (x - z) = [0-(-2)] = 2

To get the Maximum possible value of (x+z):

Take the maximum possible value of x = 0
Take the maximum possible value of z = +inf

Hence, the maximum possible value of (x + z) = +inf

To get the Maximum possible value of y:

Take the maximum possible value of y = 1

So the maximum possible value of the given expression = 2 (2) (+inf) + 2(1) = +inf.

What am I missing?

Maximum possible value for x = 0
Minimum possible value for z = +inf
Senior Manager
Joined: 07 Oct 2003
Posts: 353
Location: Manhattan
Followers: 2

Kudos [?]: 20 [0], given: 0

### Show Tags

28 Feb 2004, 15:03
gmatblast wrote:
batliwala wrote:
correct ans is E.

substituting values for x, y, and z and limiting conditions E. has the greatest value =4.

Well I agree with the answer E but I think the value of the expression in E will be +inf.

Here is how I thought of it:

2 *( x-z)*(x+z) + 2*y^2

To get the maximum value of this expression, we need to select the values as follws:

Maximum possible value of (x-z)
Maximum possible value of (x+z)
Maximum possible value of y

To get the Maximum possible value of (x-z):

Take the maximum possible value of x = 0
Take the minimum possible value of z = -2

Hence, maximum possible value of (x - z) = [0-(-2)] = 2

To get the Maximum possible value of (x+z):

Take the maximum possible value of x = 0
Take the maximum possible value of z = +inf

Hence, the maximum possible value of (x + z) = +inf

To get the Maximum possible value of y:

Take the maximum possible value of y = 1

So the maximum possible value of the given expression = 2 (2) (+inf) + 2(1) = +inf.

What am I missing?

Maximum possible value for x = 0
Minimum possible value for z = +inf

You can't have z equal -2 and +inf in the same equation.
Director
Joined: 03 Jul 2003
Posts: 652
Followers: 3

Kudos [?]: 91 [0], given: 0

### Show Tags

28 Feb 2004, 15:59
Ans: E

I'm changing my answer to A.

A simple question and I scrowed up!!
God, Show mercy on my Soul in the GMAT exam !!!

-1 < = x < =0, ==> 0 =< X^2 =< 1
-1 < = y < = 1, ==> 0 =< Y^2 =< 1
z > = -2 ==> 4 =< Z^2 =< +INF

a. x^2 + y ^2 ==> 1 + 1 = 2
b. x^2 - z^2 ==> 1 - 4 = -3
c y^2 - z ^2 ==> 1 - 4 = -3
d. x^2 + y^2 -z^2 ==> 1 + 1 - 4 = -2
e. 2 *( x-z)*(x+z) + 2*y^2
2 ( X^2 + Y^2 - Z^2) ==> 2( 1 + 1 - 4) = -4

Dear Batliwala, the unoffical ETS question banker,

Senior Manager
Joined: 11 Nov 2003
Posts: 357
Location: Illinois
Followers: 1

Kudos [?]: 5 [0], given: 0

### Show Tags

28 Feb 2004, 18:26

Such a simple problem and got screwed up. Thanks for the follow up.
Senior Manager
Joined: 19 Feb 2004
Posts: 414
Location: Lungi
Followers: 1

Kudos [?]: 30 [0], given: 0

### Show Tags

02 Mar 2004, 06:16
The correct choice is E.
x^2 < = 1, Y^2 <=1, Z^2 >=0

Evaluating the maximum value of the answer choices:
a. 1+1 = 2
b. 1-0 =1
c. 1- 0 = 1
d. 1+1 -0 = 2
e. 2(x^2-Z^2) + 2y^2 = 2(1-0)+ 2*1 = 4.

aayi maala wanchwa.
Director
Joined: 03 Jul 2003
Posts: 652
Followers: 3

Kudos [?]: 91 [0], given: 0

### Show Tags

02 Mar 2004, 07:20
batliwala wrote:
The correct choice is E.
x^2 < = 1, Y^2 <=1, Z^2 >=0

Evaluating the maximum value of the answer choices:
a. 1+1 = 2
b. 1-0 =1
c. 1- 0 = 1
d. 1+1 -0 = 2
e. 2(x^2-Z^2) + 2y^2 = 2(1-0)+ 2*1 = 4.

aayi maala wanchwa.

Yes, In deed, Man, Simple problems tend to confuse me!
Intern
Joined: 28 Dec 2005
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

29 Dec 2005, 22:41
***********************************************************
Step 1: 2 *( x-z)*(x+z) + 2*y^2 = 2* (x^2 - z^2) + 2*y^2
Using : (a+b)*(a-b) = a^2-b^2
Step 2: 2* (x^2 - z^2) + 2*y^2 = 2(x^2+y^2-z^2)
Taking 2 as common

Step 3: 2(x^2+y^2-z^2) = 2( (-1^2) + (1^2) - (0^2))
Selecting maximum values for x and y and minimum value for z (Here sign doesnt make any diffrence, since we are squaing the values )
i.e. x = -1; y = 1 or -1; z = 0

Step 4: 2( (-1^2) + (1^2) - (0^2)) = 4
Solving
************************************************************
Note here the maximum value the option A can give is 2 since it is x^2 + y^2 and max values for x and y (ignoring the sign) is 1
Display posts from previous: Sort by