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If x and y are integers such that (x+1)^2 less than equal to
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25 Jul 2010, 18:51
Question Stats:
78% (01:53) correct 22% (03:12) wrong based on 126 sessions
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If x and y are integers such that (x+1)^2 less than equal to 36 and (y1)^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.




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Re: In equalities how to handle an expression with squares
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12 Aug 2010, 13:01
gmatrant wrote: If x and y are integers such that (x+1)^2 less than equal to 36 and (y1)^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 (y1)^2 less than 64. What is the largest possible and minimum possible value of xy.\((x+1)^2\leq{36}\) > \({\sqrt{36}}\leq{x+1}\leq{\sqrt{36}}\) > \({6}\leq{x+1}\leq{6}\) > \({7}\leq{x}\leq{5}\). \((y1)^2<{64}\) > \({\sqrt{64}}<{y1}<{\sqrt{64}}\) > \({8}<{y1}<{8}\) > \({7}<{y}<{9}\), 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 \(x+1\leq{6}\) > \({7}\leq{x}\leq{5}\). \((y1)^2<{64}\) means \(y1<{8}\) > \({7}<{y}<{9}\). Hope it's clear.
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Re: In equalities how to handle an expression with squares
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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 (y1)^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 nonnegative.
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Re: In equalities how to handle an expression with squares
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12 Aug 2010, 11:35
can someone please help with this entire problem?



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Re: In equalities how to handle an expression with squares
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12 Aug 2010, 19:52
Thanks a lot Bunuel.



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Re: In equalities how to handle an expression with squares
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14 Jun 2011, 03:06
x 7 and 5 y 8 and 7
max = 49 min = 56



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Re: In equalities how to handle an expression with squares
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14 Jun 2011, 03:19
amit2k9 wrote: x 7 and 5 y 8 and 7
max = 49 min = 56 Hi Amit, I think it shud be 42 and not 49..( 7<y<9)



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Re: In equalities how to handle an expression with squares
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28 Aug 2012, 00:04
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 ?



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Re: In equalities how to handle an expression with squares
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28 Aug 2012, 04:10
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.
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Re: In equalities how to handle an expression with squares
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28 Aug 2012, 07:14
Ah !! got it now. Thank q so much Bunuel



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Re: If x and y are integers such that (x+1)^2 less than equal to
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30 Aug 2012, 00:53
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) \((x+1)^2 \leq{36}\) \((x+1)^2  36 \leq{0}\) \((x+7)(x5) \leq{0}\) 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!
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Re: If x and y are integers such that (x+1)^2 less than equal to
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05 Dec 2012, 05:06
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 [16,1+6] = [7,5]. This is inclusive of 7 and 5. \(y1<8\) This means y is within the range of (18,1+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!
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Re: If x and y are integers such that (x+1)^2 less than equal to
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22 Jul 2013, 09:37
gmatrant wrote: If x and y are integers such that (x+1)^2 less than equal to 36 and (y1)^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. 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, (y1)^2 < 64 > y1 < 8 Now the two equations formed above can be solved using the three step approach, x+1 <= 6Case 1 (x>1) > x+1 <= 6 > 1<x<=5 Case 2 (x<1) > (x+1) <= 6 > x+1 >=6 > 7<=x<1 y1 < 8Case 1 (y>1) > [/b] y1<8 > 1<y<9 Case 2 (y<1) > [/b] (y1) < 8 > y1>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



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Re: If x and y are integers such that (x+1)^2 less than equal to
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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 (y1)^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 < +61 X < 5
X >61 X > 7 Range of possible values for x are between 7 and +5
(y1)^2 < 64 (y1) <+ 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



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Re: If x and y are integers such that (x+1)^2 less than equal to
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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 rewrote it?



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Re: If x and y are integers such that (x+1)^2 less than equal to
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06 Dec 2014, 05:54
ikishan wrote: 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 rewrote it? Isn't it explained in the very post you are quoting?
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