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x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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15 Dec 2018, 10:23
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GMATbuster's Weekly Quant Quiz#13 Ques #8 For Questions from earlier quizzes: Click Herex + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy? A.12 B6 C.4 D.2 E.0
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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17 Dec 2018, 07:16
First of all, we need to realize this  Mod is always nonnegative, so maximum possible of Negative of a mod expression will be at the minimum possible value of the mod expression. So we are looking at Min value of xy x < 5 or 5 < x < 5 x is also the distance of x from zero, and in general xa is the distance of x from a. Hence x+3 would be the distance of x from 3: which would translate to x ranging from 5 units to left of3 to 5 units to right of 3. Just imagine that we have shifted the origin to 3 and then applying mod functiontherefore, 8 < x < 2 Similarly, 1 < y < 9 we also have x+3 = y4 so this could happen at x = 1 and y = 2 or at any of the following (x,y) pairs = ( 2.3) (3,4) (4,5) (5,6) ...( 7,8) you can check by putting the values in.Min absolute value of xy would be when x = 1 and y = 2 as y is nonzero integer. Hence , max value of xy will be 2. Option (D) is our choiceRegards, Gladi x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy? A.12 B6 C.4 D.2 E.0
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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16 Dec 2018, 12:07
x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy? x < 5 5<x<5 2<x+3<8 0<x+3<8 y < 5 5<y<5 9<y4<1 0<=y4<1 Now, x+3 = y4 Hence 0<=x+3<1 1<x+3<1 4<x<2 x=3 5<y<5 y = 4,3,2...,3,4 xy = 12 Answer A.12
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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17 Dec 2018, 09:29
gmatbusters wrote: GMATbuster's Weekly Quant Quiz#13 Ques #8 For Questions from earlier quizzes: Click Herex + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy? A.12 B6 C.4 D.2 E.0 The question is best solved by going through the options. Option E can be eliminated as x and y are non zero integers, so xy cannot be zero. Choosing the next maximum value among the options, i.e 2 xy can be 2 or 2. By plugging in values for xy to get 2, we can see x=1 and y=2 solves the given equation. So 2 is the maximum value. No need to even check other options as they are lesser than 2 and question asks for the greatest value. Hence D. Cheers!



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x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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17 Dec 2018, 09:41
Hi Diwakar003, While plugging in values is a very common approach to solve questions quickly, I think it does not benefit the user much here. There would be too many possibilities and stumbling upon the answer without zeroing down the options is a gamble I would rather not take on the exam.For ex. 0 can be eliminated trivially. Apart from that, it so happens that here the next largest option is the correcrt answer. However, if that would not be the case ( in some other scenario) the the number of values to be tested would grow substantially. 2 = 2,2 check x = 1 y = 2 and y = 1 x = 2 4 = 4,4 check x = 1 y =4 ; y = 1 x = 4 ; x = 2 y =2 and three more for negative possibility already 6 cases to test6 = 6,6 check x = 1 y = 6; y = 1 x =6 ; (2,3) (3,2) and four more for negatives. Already 8 cases By the time you have eliminated 3 options, you have already tested 16 cases. This is definitely not recommended and is definitely not the best way to solve the questionHope it makes sense. Best, Gladi Diwakar003 wrote: gmatbusters wrote: GMATbuster's Weekly Quant Quiz#13 Ques #8 For Questions from earlier quizzes: Click Herex + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy? A.12 B6 C.4 D.2 E.0 The question is best solved by going through the options. Option E can be eliminated as x and y are non zero integers, so xy cannot be zero. Choosing the next maximum value among the options, i.e 2 xy can be 2 or 2. By plugging in values for xy to get 2, we can see x=1 and y=2 solves the given equation. So 2 is the maximum value. No need to even check other options as they are lesser than 2 and question asks for the greatest value. Hence D. Cheers!
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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17 Dec 2018, 10:54
The only question I did wrong Great going Gladiator59
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x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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17 Jan 2019, 04:32
x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy? A.12 B6 C.4 D.2 E.0
would anyone please bring some light on this approach?
5<x<5 hence, 2<x+3<8
5<y<5 hence, 9<y4<1
the question says, lx+3l=ly4l; this relation is equal at 0 and 1
when we equate x+3 =0 or 1,we get 3 or 4 respectively when we equate y4 =0 or 1, we get 4 or 3 respectively
multiplying XY(in both cases) produces 12
Hence, It's A.



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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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17 Jan 2019, 04:45
Rupesh1Nonly wrote: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy? A.12 B6 C.4 D.2 E.0
would anyone please bring some light on this approach?
5<x<5 hence, 2<x+3<8
5<y<5 hence, 9<y4<1
the question says, lx+3l=ly4l; this relation is equal at 0 and 1
when we equate x+3 =0 or 1,we get 3 or 4 respectively when we equate y4 =0 or 1, we get 4 or 3 respectively
multiplying XY(in both cases) produces 12
Hence, It's A. You are missing cases there: y  4 is 1 and x + 3 is 1. y  4 is 2 and x + 3 is 2. y  4 is 3 and x + 3 is 3. ... y  4 is 7 and x + 3 is 7. In all those cases x + 3 = y  4.
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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17 Jan 2019, 04:56
gmatbusters wrote: GMATbuster's Weekly Quant Quiz#13 Ques #8 For Questions from earlier quizzes: Click Herex + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy? A.12 B6 C.4 D.2 E.0 If x < 5 and y < 5, from this you get to know the range of x and y 5 < x < 5 & 5 < y < 5 If you can get 2 in some way, that will be the highest value, since you cannot get 0 x =  1 and y = 2 will get you 2 Making D as answer.
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x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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21 Jan 2019, 22:20
Gladiator59 wrote: First of all, we need to realize this  Mod is always nonnegative, so maximum possible of Negative of a mod expression will be at the minimum possible value of the mod expression.
So we are looking at Min value of xy
x < 5 or 5 < x < 5
x is also the distance of x from zero, and in general xa is the distance of x from a.
Hence x+3 would be the distance of x from 3: which would translate to x ranging from 5 units to left of3 to 5 units to right of 3. Just imagine that we have shifted the origin to 3 and then applying mod function
therefore, 8 < x < 2
Similarly, 1 < y < 9
we also have x+3 = y4 so this could happen at x = 1 and y = 2 or at any of the following (x,y) pairs = ( 2.3) (3,4) (4,5) (5,6) ...( 7,8) you can check by putting the values in.
Min absolute value of xy would be when x = 1 and y = 2 as y is nonzero integer. Hence , max value of xy will be 2.
Option (D) is our choice
Regards, Gladi
x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy? A.12 B6 C.4 D.2 E.0 Hi, How did you choose the pairs? ( 2.3) (3,4) (4,5) (5,6) ...( 7,8) Thanks.
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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07 Jul 2019, 18:03
Number line is the best approach.y4is the distance of y from 4 and x+3 is the distance of x from 3.The only value satisfying the equation is x=1 and y=2.so the product xy=2
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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