GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 21 Nov 2019, 12:13

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

If x and y are integers such that (x+1)^2 is less than or equal to 36

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59236
If x and y are integers such that (x+1)^2 is less than or equal to 36  [#permalink]

Show Tags

New post 16 Apr 2015, 04:41
5
9
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

64% (02:35) correct 36% (02:54) wrong based on 317 sessions

HideShow timer Statistics

Intern
Intern
avatar
Joined: 18 Feb 2014
Posts: 9
Location: United States
Re: If x and y are integers such that (x+1)^2 is less than or equal to 36  [#permalink]

Show Tags

New post 16 Apr 2015, 08:35
1
Since X and Y are integers and we are trying to min/max XY we need to look for the extremes of the equations (either negative or positive)

(x+1)^2<= 36, therefore (x+1) has to be 6 or -6 and X=5 and X=-7
(y-1)^2 <64, therefore (y+1) has to be less than 8 or greater than -8. As such, either Y=8 or Y=-6

The maximum possible value of xy is X=-7 and Y=-6 (42) and the minimum possible value of xy is X=-7 and Y=8 (-56).

Therefore, 42+ (-56)= -14
Senior Manager
Senior Manager
User avatar
B
Joined: 28 Feb 2014
Posts: 289
Location: United States
Concentration: Strategy, General Management
Reviews Badge
Re: If x and y are integers such that (x+1)^2 is less than or equal to 36  [#permalink]

Show Tags

New post 16 Apr 2015, 09:33
1
(x+1)^2 <= 36
x <= 5
x >= -7

(y-1)^2 < 64
y < 9
y > -7

Max possible value of xy is -7 × -6 = 42

minimum possible value of xy is -7 × 8 = -56

-56 + 42 = -14

Answer : B
Retired Moderator
avatar
B
Status: On a mountain of skulls, in the castle of pain, I sit on a throne of blood.
Joined: 30 Jul 2013
Posts: 299
GMAT ToolKit User Reviews Badge
Re: If x and y are integers such that (x+1)^2 is less than or equal to 36  [#permalink]

Show Tags

New post 16 Apr 2015, 09:43
Bunuel wrote:
If x and y are integers such that (x+1)^2 is less than or equal to 36 and (y-1)^2 is less than 64, what is the sum of the maximum possible value of xy and the minimum possible value of xy?

(A) -16
(B) -14
(C) 0
(D) 14
(E) 16


Kudos for a correct solution.


(x+1)^2<=36
-6<=x+1<=6
-7<=x<=5

(y-1)^2<=64
-8<=y-1<=8
-7<=y<=9

Max xy=-7*-7=49
Min xy=-7*9=-63

49-63=-14

Answer: B
Current Student
User avatar
B
Joined: 25 Nov 2014
Posts: 96
Concentration: Entrepreneurship, Technology
GMAT 1: 680 Q47 V38
GPA: 4
If x and y are integers such that (x+1)^2 is less than or equal to 36  [#permalink]

Show Tags

New post 16 Apr 2015, 11:20
(x+1)^2 <= 36 gives, -6<=x+1<=6 gives, -7<=x<=5
(y-1)^2 <= 64 gives, -7<=y-1<=7 gives, -6<=y<=8

max value of xy = 42 , and min value of xy = -56
Thus sum = -14

Ans B.
_________________
Kudos!!
Manager
Manager
avatar
Joined: 15 May 2014
Posts: 61
GMAT ToolKit User
If x and y are integers such that (x+1)^2 is less than or equal to 36  [#permalink]

Show Tags

New post 17 Apr 2015, 21:32
Given
\((x+1)^2 \leq(36)\)
\(-6 \leq x+1 \leq 6\)
\(-7 \leq x \leq 5\)

\((y-1)^2 \leq(64)\)
\(-8 \leq y-1 \leq 8\)
\(-7 \leq y \leq 9\)

Max value of \(xy = -7*-7\)
Min value of \(xy = -7*9\)
Sum \(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= -7 (-7+9) = -14\)

Answer B
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59236
If x and y are integers such that (x+1)^2 is less than or equal to 36  [#permalink]

Show Tags

New post 20 Apr 2015, 05:56
Bunuel wrote:
If x and y are integers such that (x+1)^2 is less than or equal to 36 and (y-1)^2 is less than 64, what is the sum of the maximum possible value of xy and the minimum possible value of xy?

(A) -16
(B) -14
(C) 0
(D) 14
(E) 16


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION

To get the sum of the maximum and minimum possible values of xy, we need to know the maximum and minimum values of xy. For those, we need to find the values that x and y can take. So first, we should review the information given:

(x + 1)^2 <= 36

(y – 1)^2 < 64

We need to find the values that x and y can take. There are many ways of doing that. We can solve the inequality using the wave method discussed in this post or using the concept of absolute values. Let’s discuss both the methods.

Wave method to solve inequalities:

Solve for x: (x + 1)^2 <= 36

(x + 1)^2 – 6^2 <= 0

(x + 1 + 6)(x + 1 – 6) <= 0

(x + 7)(x – 5) <= 0

-7 <= x <= 5 (Using the wave method)

Solve for y: (y – 1)^2 < 64

(y – 1)^2 – 8^2 < 0

(y – 1 + 8)(y – 1 – 8) < 0

(y + 7)(y – 9) < 0

-7 < y < 9 (Using the wave method)

Or you can solve taking the square root on both sides

Solve for x: (x + 1)^2 <= 36

|x + 1| <= 6

-6 <= x + 1 <= 6 (discussed in your Veritas Algebra book)

-7 <= x <= 5

So x can take values: -7, -6, -5, -4, … 3, 4, 5

Solve for y: (y – 1)^2 < 64

|y – 1| < 8

-8 < y – 1 < 8 (discussed in your Veritas Algebra book)

-7 < y < 9

So y can take values: -6, -5, -4, -3, … 6, 7, 8.

Now that we have the values of x and y, we should try to find the minimum and maximum values of xy.

Note that the values of xy can be positive as well as negative. The minimum value will be the negative value with largest absolute value (largest negative) and the maximum value will be the positive value with the largest absolute value.

Minimum value – For the value to be negative, one and only one of x and y should be negative. Focus on the extreme values: if x is -7 and y is 8, we get xy = -56. This is the negative value with largest absolute value.

Maximum value – For the value to be positive, both x and y should have the same signs. If x = -7 and y = -6, we get xy = 42. This is the largest positive value.

The sum of the maximum value of xy and minimum value of xy is -56 + 42 = -14

Answer (B)

Try to think of it in terms of a number line. x lies in the range -7 to 5 and y lies in the range -6 to 8. The range is linear so the end points give us the maximum/minimum values. Think of what happens when you plot a quadratic – the minimum/maximum could lie anywhere.
_________________
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59236
Re: If x and y are integers such that (x+1)^2 is less than or equal to 36  [#permalink]

Show Tags

New post 20 Apr 2015, 05:59
Bunuel wrote:
If x and y are integers such that (x+1)^2 is less than or equal to 36 and (y-1)^2 is less than 64, what is the sum of the maximum possible value of xy and the minimum possible value of xy?

(A) -16
(B) -14
(C) 0
(D) 14
(E) 16


Kudos for a correct solution.


\((x+1)^2\leq{36}\) --> \({-\sqrt{36}}\leq{x+1}\leq{\sqrt{36}}\) --> \({-6}\leq{x+1}\leq{6}\) --> \({-7}\leq{x}\leq{5}\).

\((y-1)^2<{64}\) --> \({-\sqrt{64}}<{y-1}<{\sqrt{64}}\) --> \({-8}<{y-1}<{8}\) --> \({-7}<{y}<{9}\), as \(y\) is an integer we can rewrite this inequality as \({-6}\leq{y}\leq{8}\).

We should try extreme values of \(x\) and \(y\) to obtain min and max values of \(xy\):

Min possible value of \(xy\) is for \(x=-7\) and \(y=8\) --> \(xy=-56\);
Max possible value of \(xy\) is for \(x=-7\) and \(y=-6\) --> \(xy=42\).

The sum = -56 + 42 = -14.

Answer: B.
_________________
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 13624
Re: If x and y are integers such that (x+1)^2 is less than or equal to 36  [#permalink]

Show Tags

New post 16 Apr 2018, 15:17
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Bot
Re: If x and y are integers such that (x+1)^2 is less than or equal to 36   [#permalink] 16 Apr 2018, 15:17
Display posts from previous: Sort by

If x and y are integers such that (x+1)^2 is less than or equal to 36

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne