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Author:  fskilnik [ 14 Feb 2019, 12:11 ] 
Post subject:  If 4<=x<=6 and 2<=y<=3, then the minimum possible value of 
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 
Author:  fskilnik [ 14 Feb 2019, 18:21 ] 
Post subject:  Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of 
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. 
Author:  KSBGC [ 17 Feb 2019, 01:56 ] 
Post subject:  If 4<=x<=6 and 2<=y<=3, then the minimum possible value of 
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. 
Author:  GMATinsight [ 17 Feb 2019, 07:23 ] 
Post subject:  If 4<=x<=6 and 2<=y<=3, then the minimum possible value of 
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 (yx) and (y+x) must be MINIMUM Minimum ABSOLUTE Value of yx = l34l = l1l = 1 Minimum ABSOLUTE Value of y+x = l3+4l = l7l = 7 i.e. Minimum value of l(yx)(y+x)l = l1*7l = 7 Answer: Option D 
Author:  KanishkM [ 17 Feb 2019, 08:20 ] 
Post subject:  Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of 
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 = 3 916 7 
Author:  jamalabdullah100 [ 11 Sep 2019, 12:55 ] 
Post subject:  Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of 
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 (yx) and (y+x) must be MINIMUM Minimum ABSOLUTE Value of yx = l34l = l1l = 1 Minimum ABSOLUTE Value of y+x = l3+4l = l7l = 7 i.e. Minimum value of l(yx)(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.. 
Author:  vishakha23 [ 17 Sep 2019, 07:04 ] 
Post subject:  Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of 
Why can't the Minimum ABSOLUTE Value of y+x = 2+4 = 6? Both y=2 and x=4 are included in the boundaries. 
Author:  Abhishek009 [ 17 Sep 2019, 07:19 ] 
Post subject:  Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of 
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) 
Author:  jamalabdullah100 [ 23 Sep 2019, 05:06 ] 
Post subject:  Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of 
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 (yx) and (y+x) must be MINIMUM Minimum ABSOLUTE Value of yx = l34l = l1l = 1 Minimum ABSOLUTE Value of y+x = l3+4l = l7l = 7 i.e. Minimum value of l(yx)(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.. 
Author:  rishab0507 [ 07 Jul 2020, 14:12 ] 
Post subject:  Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of 
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 (yx) and (y+x) must be MINIMUM Minimum ABSOLUTE Value of yx = l34l = l1l = 1 Minimum ABSOLUTE Value of y+x = l3+4l = l7l = 7 i.e. Minimum value of l(yx)(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 
Author:  himtheGMATE [ 07 Jul 2020, 19:38 ] 
Post subject:  Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of 
The answer to why min is 4+3 and not 4+2 is. When we take y=2 and x=4 Min abs yxis = 2 & y+xis=6 which gives total value of expression is = 12 hence it is not min. Hope this clarifies. Posted from my mobile device 
Author:  Subhrajyoti [ 12 Dec 2021, 07:34 ] 
Post subject:  Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of 
Minimum absolute value of yx is 1, and the min abs value of y+x is 6 so (yx)(y+x) why the answer is 7 and not 6. Not sure i understand 
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