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

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

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New post 16 Apr 2015, 04:41
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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.

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

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New post 16 Apr 2015, 08:35
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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
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Re: If x and y are integers such that (x+1)^2 is less than or equal to 36  [#permalink]

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New post 16 Apr 2015, 09:33
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(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
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Re: If x and y are integers such that (x+1)^2 is less than or equal to 36  [#permalink]

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

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

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

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

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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.
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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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