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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.
A. 40 and -30, respectively
B. 40 and -56, respectively
C. 42 and -30, respectively
D. 42 and -56, respectively
E. 49 and -56, respectively
A. 40 and -30, respectively
B. 40 and -56, respectively
C. 42 and -30, respectively
D. 42 and -56, respectively
E. 49 and -56, respectively
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.
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.
gmatrant
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.
\((x+1)^2\leq{36}\) --> \({-\sqrt{36}}\leq{x+1}\leq{\sqrt{36}}\) --> \({-6}\leq{x+1}\leq{6}\) --> \({-7}\leq{x}\leq{5}\).
\((y-1)^2<{64}\) --> \({-\sqrt{64}}<{y-1}<{\sqrt{64}}\) --> \({-8}<{y-1}<{8}\) --> \({-7}<{y}<{9}\), 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 \(y=-6\) --> \(xy=42\).
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 \(|y-1|<{8}\) --> \({-7}<{y}<{9}\).
Hope it's clear.
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Thanks a lot Bunuel.
pallavisatsangi
Joined: 14 Apr 2012
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Concentration: Technology, General Management
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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 ?
" 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 ?
pallavisatsangi
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.
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pallavisatsangi
Joined: 14 Apr 2012
Last visit: 13 Dec 2020
Posts: 16
Location: India
Concentration: Technology, General Management
GMAT Date: 09-12-2012
GPA: 3.5
WE:Information Technology (Retail: E-commerce)
Posts: 16
Ah !! got it now. Thank q so much Bunuel 
GMAT 1: 670 Q48 V35
Posts: 42
Bunuel's attack on the question is obviously by far the best and the least time consuming!
However if you are like me (who tries to avoid getting clawed by mods) here is how I did it (a little more time consuming though)
\((x+1)^2 \leq{36}\)
\((x+1)^2 - 36 \leq{0}\)
\((x+7)(x-5) \leq{0}\)
roots - -7, 5
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.
Hope this helps, too!
However if you are like me (who tries to avoid getting clawed by mods) here is how I did it (a little more time consuming though)
\((x+1)^2 \leq{36}\)
\((x+1)^2 - 36 \leq{0}\)
\((x+7)(x-5) \leq{0}\)
roots - -7, 5
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.
Hope this helps, too!
mbaiseasy
Joined: 13 Aug 2012
Last visit: 29 Dec 2013
Posts: 316
Concentration: Marketing, Finance
Schools: Full Time MBA (A$)
GPA: 3.23
Schools: Full Time MBA (A$)
Posts: 316
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!
adeelahmad
Joined: 13 Sep 2012
Last visit: 05 May 2014
Posts: 6
Schools: Goizueta '15 (D) ISB '14 (D) Darden '15 (WD)
Schools: Goizueta '15 (D) ISB '14 (D) Darden '15 (WD)
Posts: 6
gmatrant
Here's my approach -
(x+1)^2 <= 36 --> Take square root of both sides --> sqrt(x+1)^2 <= sqrt(36) --> |x+1| <= 6 ( Since sqrt(x^2) = |x| )
Similarly, (y-1)^2 < 64 --> |y-1| < 8
Now the two equations formed above can be solved using the three step approach,
|x+1| <= 6
Case 1 (x>-1) --> x+1 <= 6 --> -1<x<=5
Case 2 (x<-1) --> -(x+1) <= 6 --> x+1 >=-6 --> -7<=x<-1
|y-1| < 8
Case 1 (y>1) --> [/b] y-1<8 --> 1<y<9
Case 2 (y<1) --> [/b] -(y-1) < 8 --> y-1>8 --> -7<y<1
The value sets for x and y stack up as below –
x = {-7, -6, -5, -4, -3, -2, -1, 1, 2, 3, 4, 5}
y = {-6, -5, -4, -3, -2, -1, 1, 2, 3, 4, 5, 6, 7, 8}
Therefore,
xymin = -7 * 8 = -56
xymax = -7 * -6 = 42
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.
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