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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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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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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

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

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

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

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

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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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.

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

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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

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

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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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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

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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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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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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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

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..
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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