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If 4<=x<=6 and 2<=y<=3, then the minimum possible value of

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If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

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New post 14 Feb 2019, 12:11
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GMATH practice exercise (Quant Class 11)

If \(\,4 \le x \le 6\,\) and \(\,2 \le y \le 3\) , the minimum possible value of \(\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,\) is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

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Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

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New post 14 Feb 2019, 18:21
fskilnik wrote:
GMATH practice exercise (Quant Class 11)

If \(\,4 \le x \le 6\,\) and \(\,2 \le y \le 3\) , the minimum possible value of \(\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,\) is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

\(?\,\, = \,\,\min \,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,\, = \,\,\min \,\left| {{y^2} - {x^2}} \right|\)


\(2 \le y \le 3\,\,\,\,\, \Rightarrow \,\,\,\,\,4 \le {y^2} \le 9\)

\(4 \le x \le 6\,\,\,\,\, \Rightarrow \,\,\,\,\,16 \le {x^2} \le 36\,\,\,\,\, \Rightarrow \,\,\,\,\, - 36 \le - {x^2} \le - 16\)


\(\left. \matrix{
4 \le {y^2} \le 9 \hfill \cr
- 36 \le - {x^2} \le - 16\,\,\, \hfill \cr} \right\}\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\, - 32 \le {y^2} - {x^2} \le - 7\,\,\,\,\, \Rightarrow \,\,\,\,\,7 \le \left| {{y^2} - {x^2}} \right| \le 32\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 7\)


The correct answer is (D).

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

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New post 17 Feb 2019, 01:56
fskilnik wrote:
GMATH practice exercise (Quant Class 11)

If \(\,4 \le x \le 6\,\) and \(\,2 \le y \le 3\) , the minimum possible value of \(\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,\) is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8



\(4\leq x \leq6\)

\(2\leq y \leq3\)

Given,

\(|y^2 - x^2|\)

but we are asked to find out the minimum value of |y^2 - x^2| .

we must work with corner values / extremes values of x and y.

\(2^2 - 4^2\) = 4 - 16 = - 12 = | -12 | = 12

\(2^2 - 6^2\)= 4 - 36 = -32 = | -32| = 32

\(3^2 - 4^2\)= 9 - 16 = -7 = | - 7| = 7.

\(3^2 - 6^2\)= 9 - 36 = -27 = | -27| = 27.

D is the correct answer.
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If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

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New post 17 Feb 2019, 07:23
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fskilnik wrote:
GMATH practice exercise (Quant Class 11)

If \(\,4 \le x \le 6\,\) and \(\,2 \le y \le 3\) , the minimum possible value of \(\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,\) is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8


For minimum Possible value of ∣(y−x)(y+x)∣ the absolute values of (y-x) and (y+x) must be MINIMUM

Minimum ABSOLUTE Value of y-x = l3-4l = l-1l = 1

Minimum ABSOLUTE Value of y+x = l3+4l = l7l = 7

i.e. Minimum value of l(y-x)(y+x)l = l1*7l = 7

Answer: Option D
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Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

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New post 17 Feb 2019, 08:20
fskilnik wrote:
GMATH practice exercise (Quant Class 11)

If \(\,4 \le x \le 6\,\) and \(\,2 \le y \le 3\) , the minimum possible value of \(\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,\) is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8


So the given expression can be written as

\(|y^2 - x^2|\)

Only case possible is when we maximize y = 4 and minimize x = 3

|9-16|

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Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

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New post 11 Sep 2019, 12:55
GMATinsight wrote:
fskilnik wrote:
GMATH practice exercise (Quant Class 11)

If \(\,4 \le x \le 6\,\) and \(\,2 \le y \le 3\) , the minimum possible value of \(\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,\) is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8


For minimum Possible value of ∣(y−x)(y+x)∣ the absolute values of (y-x) and (y+x) must be MINIMUM

Minimum ABSOLUTE Value of y-x = l3-4l = l-1l = 1

Minimum ABSOLUTE Value of y+x = l3+4l = l7l = 7

i.e. Minimum value of l(y-x)(y+x)l = l1*7l = 7

Answer: Option D



Why can't the Minimum ABSOLUTE Value of y+x = 2+4 = 6? Both y=2 and x=4 are included in the boundaries..
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Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

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New post 17 Sep 2019, 07:04
Why can't the Minimum ABSOLUTE Value of y+x = 2+4 = 6? Both y=2 and x=4 are included in the boundaries.
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Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

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New post 17 Sep 2019, 07:19
fskilnik wrote:
GMATH practice exercise (Quant Class 11)

If \(\,4 \le x \le 6\,\) and \(\,2 \le y \le 3\) , the minimum possible value of \(\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,\) is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8


Values of x can be 4 , 5 & 6 ; Values of y can be 2 , 3

SO, the Possible min value of \(\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,\)

Will be \(\,\left| {\left( {3 - 4} \right)\left( {4 + 3} \right)} \right|\,\) = 1*7 = 7 , Answer must be (D)
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Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

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New post 23 Sep 2019, 05:06
jamalabdullah100 wrote:
GMATinsight wrote:
fskilnik wrote:
GMATH practice exercise (Quant Class 11)

If \(\,4 \le x \le 6\,\) and \(\,2 \le y \le 3\) , the minimum possible value of \(\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,\) is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8


For minimum Possible value of ∣(y−x)(y+x)∣ the absolute values of (y-x) and (y+x) must be MINIMUM

Minimum ABSOLUTE Value of y-x = l3-4l = l-1l = 1

Minimum ABSOLUTE Value of y+x = l3+4l = l7l = 7

i.e. Minimum value of l(y-x)(y+x)l = l1*7l = 7

Answer: Option D



Why can't the Minimum ABSOLUTE Value of y+x = 2+4 = 6? Both y=2 and x=4 are included in the boundaries..


Can someone help with the above please? It doesn't make sense to me..
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Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of   [#permalink] 23 Sep 2019, 05:06
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