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

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25 Jul 2010, 18:51

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If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy.

Re: In equalities how to handle an expression with squares [#permalink]

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25 Jul 2010, 18:58

gmatrant wrote:

If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy.

In equalities how to handle an expression which is squared Does the above equation (x+1)^2 <= 36 mean |x+1| < (+6 or -6)

I then get 4 equations.. and I am am not able to proceed. Can you someone please explain how such questions are to be handled.

You need to brush up quant fundamentals. Buy Mgmat number properties and inequalities book.

\((x+1)^2 <= 36\)

\(x+1 <= 6\) , \(-(x+1) <= 6\) when -ve is multiplied, inequality is reversed. Multiply -1 on both the sides

\(x+1 <= 6\) , \(x+1 >= -6\)

\(x =< 5\) , \(x >= -7\)

|x+1| can never be < -6 as it is always non-negative.
_________________

If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy.

In equalities how to handle an expression which is squared Does the above equation (x+1)^2 <= 36 mean |x+1| < (+6 or -6)

I then get 4 equations.. and I am am not able to proceed. Can you someone please explain how such questions are to be handled.

If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy.

\((y-1)^2<{64}\) --> \({-\sqrt{64}}<{y-1}<{\sqrt{64}}\) --> \({-8}<{y-1}<{8}\) --> \({-7}<{y}<{9}\), 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 \(y=-6\) --> \(xy=42\).

Solving with absolute values gives the same results:

\((x+1)^2\leq{36}\) means \(|x+1|\leq{6}\) --> \({-7}\leq{x}\leq{5}\). \((y-1)^2<{64}\) means \(|y-1|<{8}\) --> \({-7}<{y}<{9}\).

Hi Bunuel, " as y is an integer we can rewrite this inequality as -6<= y<=8 ."

I didn't understand as to how can we change the range of y from -7,9 to -6,8 ?

We are not changing the range here.

We have \({-7}<{y}<{9}\). Now, since \(y\) is an integer, then it can take integer values from -6 to 8, inclusive: -6, -5, ..., 6, 7, 8, which can be written as \({-6}\leq{y}\leq{8}\).

Now solving by the graphical approach, range is between (and inclusive of) -7 and 5.

Similarly solving for the other equation we get the roots in the range of (exclusive of) -7 and 9. Therefore maximum and minimum value of xy can be derived as others before me have explained.

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

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05 Dec 2012, 05:06

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One thing I notice is that you have to be careful with the exclusivity and inclusivity of ranges. In these questions, you will get it wrong if you thought y = 9 is included in the range.

\(|x+1| <= 6\) This means x is within the range of [-1-6,-1+6] = [-7,5]. This is inclusive of -7 and 5.

\(|y-1|<8\) This means y is within the range of (1-8,1+8) = (-7,9). This is exclusive of -7 and 9.

Now to get the extreme values.

Max value or Positive outcomes: -7 * -6 = 42 or 5 * 8 = 40 ==> 42 wins! Min value of Negative outcomes: -7 * 8 = -56 or -6 * 5 = -30 ==> -56 wins!

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

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22 Jul 2013, 09:37

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gmatrant wrote:

If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy.

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

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29 Jul 2013, 02:12

2. If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy. (x+1)^2 < 36 (x+1) < +6 (applying square root on both sides) X < +6-1 X < 5

X >-6-1 X > -7 Range of possible values for x are between -7 and +5

(y-1)^2 < 64 (y-1) <+ 8 (applying square root on both sides) y < +8+1 y < +9

y < -8+1 y > -7 As y is an integer y < 8 or y > -6 Range of possible values for y are between -6 and +8

Range of possible values for xy at their respective highest and lowest levels: -30, 40, 42, -56 Highest value: 42 and lowest value: -56

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

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16 Sep 2014, 20:58

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

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05 Dec 2014, 21:37

Bunuel wrote:

pallavisatsangi wrote:

Hi Bunuel, " as y is an integer we can rewrite this inequality as -6<= y<=8 ."

I didn't understand as to how can we change the range of y from -7,9 to -6,8 ?

We are not changing the range here.

We have \({-7}<{y}<{9}\). Now, since \(y\) is an integer, then it can take integer values from -6 to 8, inclusive: -6, -5, ..., 6, 7, 8, which can be written as \({-6}\leq{y}\leq{8}\).

Hope it's clear.

if it is asking for maximum and minimum value. why did you re-wrote it?

Hi Bunuel, " as y is an integer we can rewrite this inequality as -6<= y<=8 ."

I didn't understand as to how can we change the range of y from -7,9 to -6,8 ?

We are not changing the range here.

We have \({-7}<{y}<{9}\). Now, since \(y\) is an integer, then it can take integer values from -6 to 8, inclusive: -6, -5, ..., 6, 7, 8, which can be written as \({-6}\leq{y}\leq{8}\).

Hope it's clear.

if it is asking for maximum and minimum value. why did you re-wrote it?

Isn't it explained in the very post you are quoting?
_________________

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

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02 Apr 2016, 01:29

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).

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

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04 Apr 2017, 09:33

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).

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