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

 It is currently 21 Nov 2019, 12:13 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If x and y are integers such that (x+1)^2 is less than or equal to 36

Author Message
TAGS:

### Hide Tags

Math Expert 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

5
9 00:00

Difficulty:   75% (hard)

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

### HideShow timer Statistics

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.

_________________
Intern  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

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  B
Joined: 28 Feb 2014
Posts: 289
Location: United States
Concentration: Strategy, General Management
Re: If x and y are integers such that (x+1)^2 is less than or equal to 36  [#permalink]

### Show Tags

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

Retired Moderator 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
Re: If x and y are integers such that (x+1)^2 is less than or equal to 36  [#permalink]

### Show Tags

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

Current Student 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

(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  Joined: 15 May 2014
Posts: 61
If x and y are integers such that (x+1)^2 is less than or equal to 36  [#permalink]

### Show Tags

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 $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= -7 (-7+9) = -14$$

Math Expert 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

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

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 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

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.

_________________
Non-Human User 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

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.
_________________ 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  