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fskilnik
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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fskilnik
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|

7
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fskilnik
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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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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fskilnik
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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fskilnik
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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fskilnik
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 is Minimum value 4+3 , and why not 4+2 =6, and answer can be 6*1 = 6, and not 7 Bunuel, Kinshook , can you help
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The answer to why min is 4+3 and not 4+2 is.

When we take y=2 and x=4
Min abs |y-x|is = 2 & |y+x|is=6 which gives total value of expression is = 12 hence it is not min. Hope this clarifies.

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Minimum absolute value of |y-x| is 1, and the min abs value of |y+x| is 6 so |(y-x)(y+x)| why the answer is 7 and not 6. Not sure i understand
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