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If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy? (A) 12 (B) 3 (C) 0 (D) 2 (E) None of the above This is Question 5 of the eGMAT Question Series on Absolute Value.Provide your solution below. Kudos for participation. Happy Solving! Best Regards The eGMAT Team
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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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22 May 2015, 07:55
Official ExplanationCorrect Answer: Cy + 3 ≤ 3 means that the distance of y from the point 3 on the number line is less than, or equal to, 3 This gives us, 6 ≤ y ≤ 0 So, possible values of y (keeping in mind the constraint that y is an integer): 6, 5, 4, 3, 2, 1, 0 We’re given 2y – 3x = 6 So, \(x= \frac{(2y+6)}{3}\) For x to be an integer, y must be a multiple of 3. This means, possible values of y: 6, 3, 0 Corresponding values of x: 2, 0, 2 Corresponding values of xy: 12, 0, 0 Thus, minimum possible value of xy: 0
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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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Updated on: 21 May 2015, 14:20
EgmatQuantExpert wrote: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy? (A) 12 (B) 3 (C) 0 (D) 2 (E) None of the above This is Question 5 of the eGMAT Question Series on Absolute Value.Provide your solution below. Kudos for participation. The Official Answer and Explanation will be posted on 22nd May. Till then, Happy Solving! Best Regards The eGMAT Team How to deal with inequalities involving absolute values? First example shows us the so called "number case"In this case we have y + 3 ≤ 3 which is generalized something ≤ some number. First we solve as if there were no absolute value brackets: y + 3 ≤ 3 y ≤ 0 So y is 0 or negativeSecond scenario  remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign:y + 3 >= 3 y >= 6 Therefore we have a possible range for y: 6=<y<=0Ok, so far so good, we're half way through. What about x? Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x. You can say that 2y + 6 is a multiple of 3 (=3x). So all values which must be integer must also satisfy this constraint. I'm just saying that, so it's easier to evaluate all the possible numbers (6, 3, 0). If you plug in y = 0, x will be 2 and xy = 0 as the lowest possible value. Hence, Answer Choice C is the one to go.
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Originally posted by reto on 20 May 2015, 14:19.
Last edited by reto on 21 May 2015, 14:20, edited 2 times in total.



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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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20 May 2015, 22:43
Eq: 1 y + 3 ≤ 3 and Eq: 2 2y – 3x + 6 = 0, x & y are integersSolve Eq: 1 y + 3 ≤ 3 \(+(y + 3) ≤ 3\) \(y≤ 0\)
\((y + 3) ≤ 3\) \(y\geq{6}\) Solve Eq: 2 2y – 3x + 6 = 0 \(x = \frac{(2y + 6)}{3}\)
as range of y: \(6=<y<=0\)
we get 3 integer value of x , as x & y are integers for y = {6, 3, 0} x = {2, 0, 0} & xy = {12, 0, 0}
lowest xy = 0
Ans : C
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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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21 May 2015, 04:39
reto wrote: How to deal with inequalities involving absolute values? First example shows us the so called "number case" In this case we have y + 3 ≤ 3 which is generalized something ≤ some number. First we solve as if there were no absolute value brackets:
y + 3 ≤ 3 y ≤ 0 So y is 0 or negative
Second scenario  remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign:
y + 3 >= 3 y >= 6
Therefore we have a possible range for y: 6=<y<=0
Ok, so far so good, we're half way through. What about x?
Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x. Now it's all about finding extreme values by combining with 6=<y<=0! If you play with the numbers and plug in y = 6, x will be 2 and the product would be 12. That's not what we want... if you plug in 2 for y, x will be 2/3 and the product xy is 2*2/3 ... since with all trials i don't get any of the numbers in the question steam, I assume the ANSWER is E.
Dear retoVery well explained! Kudos for that. Just want to ask 2 questions: 1. Please refer to the orange part in your solution. Is x = 2/3 a valid value for x? Why/ why not? 2. Is the question asking you to find the value of the product xy? Based on the answers to these 2 questions, you may feel a need to edit the last part of your solution Best Regards Japinder
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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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21 May 2015, 06:23
Answer should be E.
As per the equation y+3 <=3, y can have value from 6 to 0. Now as per the equation x=(2y+6)/3; x will have negative values for all y in {6,5,4,3} hence the xy will be (x)*(y) a positive value. for y in { 2,1,0}, x will have { 2/3,4/3,2}. Hence x*y will have minimum at x=2/3,y=2 or x=4/3,y=1



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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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22 May 2015, 04:28
6<=y<=0 To get the least value of xy, we have to take the least values of both the terms. Taking y=6, x is 2. Xy is 12. Hence, none.
Note : It is clearly mentioned that both x and y are integers.



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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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22 May 2015, 08:03
csirishac wrote: 6<=y<=0 To get the least value of xy, we have to take the least values of both the terms. Taking y=6, x is 2. Xy is 12. Hence, none.
Note : It is clearly mentioned that both x and y are integers. Dear csirishacYou bring up an interesting point by saying the part in red. This part is not true. As you yourself saw, when you minimized the values of both x and y (both negative), you got a positive product. However, if x = +2 and y = 0 (both are actually the maximum values of x and y in this question), the product = 0, far lesser than the product 12 that you got above. So, be careful about what you need to minimize. Best Regards Japinder
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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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10 Nov 2015, 02:50
If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy? (A) 12 (B) 3 (C) 0 (D) 2 (E) None of the above solving y+3<=3 gives 6<=Y<=0 further equation 2y3x+6=0 In order to fiind least value of x we nned to maximize y i,e 0 this gives x =2 And least value of y=6 So product will be 12 Bunuel @VeritasPrepKarsihma Engr2012 Experts please let me know where I am missing something?



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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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03 Dec 2015, 10:38
EgmatQuantExpert wrote: reto wrote: How to deal with inequalities involving absolute values? First example shows us the so called "number case" In this case we have y + 3 ≤ 3 which is generalized something ≤ some number. First we solve as if there were no absolute value brackets:
y + 3 ≤ 3 y ≤ 0 So y is 0 or negative
Second scenario  remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign:
y + 3 >= 3 y >= 6
Therefore we have a possible range for y: 6=<y<=0
Ok, so far so good, we're half way through. What about x?
Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x. Now it's all about finding extreme values by combining with 6=<y<=0! If you play with the numbers and plug in y = 6, x will be 2 and the product would be 12. That's not what we want... if you plug in 2 for y, x will be 2/3 and the product xy is 2*2/3 ... since with all trials i don't get any of the numbers in the question steam, I assume the ANSWER is E.
Dear retoVery well explained! Kudos for that. Just want to ask 2 questions: 1. Please refer to the orange part in your solution. Is x = 2/3 a valid value for x? Why/ why not? 2. Is the question asking you to find the value of the product xy? Based on the answers to these 2 questions, you may feel a need to edit the last part of your solution Best Regards Japinder Japinder, I'm getting back to yours qs: 1) x=2/3 must be rejected as per costraints in the question stem. 2) we are asked to find the least product of xy. This happens when y=0 x=2 or y=3 x=0. Thanks.
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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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07 Dec 2015, 20:38
UJs wrote: Eq: 1 y + 3 ≤ 3 and Eq: 2 2y – 3x + 6 = 0, x & y are integersSolve Eq: 1 y + 3 ≤ 3 \(+(y + 3) ≤ 3\) \(y≤ 0\)
\((y + 3) ≤ 3\) \(y\geq{6}\) Solve Eq: 2 2y – 3x + 6 = 0 \(x = \frac{(2y + 6)}{3}\)
as range of y: \(6=<y<=0\)
we get 3 integer value of x , as x & y are integers for y = {6, 3, 0} x = {2, 0, 0} & xy = {12, 0, 0}
lowest xy = 0
Ans : C is it that we should multiply corresponding pair of XY or can we multiply y= 6 and x= 2 XY =12 pls. clear the doubt. thank in advance.



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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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07 Dec 2015, 21:17
robu wrote: UJs wrote: Eq: 1 y + 3 ≤ 3 and Eq: 2 2y – 3x + 6 = 0, x & y are integersSolve Eq: 1 y + 3 ≤ 3 \(+(y + 3) ≤ 3\) \(y≤ 0\)
\((y + 3) ≤ 3\) \(y\geq{6}\) Solve Eq: 2 2y – 3x + 6 = 0 \(x = \frac{(2y + 6)}{3}\)
as range of y: \(6=<y<=0\)
we get 3 integer value of x , as x & y are integers for y = {6, 3, 0} x = {2, 0, 0} & xy = {12, 0, 0}
lowest xy = 0
Ans : C is it that we should multiply corresponding pair of XY or can we multiply y= 6 and x= 2 XY =12 pls. clear the doubt. thank in advance. Hi, the eq ly+3l<=3 gives following values of y.. y=0 ,1,2,3,4,5,6.. for these values x=2,,,0,,,2...as x has to be an integer.. you have to take the corresponding values only.. i)when y=6, then x =2 ii)and when y=0, x=2.. iii) when y=3, x=0.. ofcourse x as 0 and y as 0 is wrong inputs so you get xy as 12 in (i) case and 0 in (ii) and (iii) case.. so 0 is the answer.. hope it helps
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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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28 Oct 2016, 07:11
EgmatQuantExpert wrote: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy?
(A) 12 (B) 3 (C) 0 (D) 2 (E) None of the above
we are given that 3<= y <= 1. y can take 3 values: 3, 2, 1, and 0. 2y3x +6 = 0 2y+6 = 3x we can test values, but we can clearly see that if y is not 3/0, then x can't be an integer, but we need it to be an integer. therefore, the only cases that actually work are y=3, x=0 or y=0, and x=2 xy = 0 in any case C is the answer.



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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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26 Jul 2017, 05:33
A simple algebraic approach. We have y + 3 less than eq to 3 which gives us 6 <=y<=0. So we have two values which are definitely for y (6 and 0). Now we find x. 2y  3x + 6 = 0 => 2y = 3x  6 => y = 3x  6/2
If we take y = 6; then we have :  6 = 3x  6/ 2 12 = 3x  6 6 = 3x There fore x = 2. X * Y = 2*6 = positive 12 > not the least value
Now lets take y = 0 0 = 3x  6/2 3x = 6 => 6 = 3x so x =2 X*Y = 2*0 = 0
Would this approach be right? Expert comment would be welcome!



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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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26 Jul 2017, 05:46
Madhavi1990 wrote: A simple algebraic approach. We have y + 3 less than eq to 3 which gives us 6 <=y<=0. So we have two values which are definitely for y (6 and 0). Now we find x. 2y  3x + 6 = 0 => 2y = 3x  6 => y = 3x  6/2
If we take y = 6; then we have :  6 = 3x  6/ 2 12 = 3x  6 6 = 3x There fore x = 2. X * Y = 2*6 = positive 12 > not the least value
Now lets take y = 0 0 = 3x  6/2 3x = 6 => 6 = 3x so x =2 X*Y = 2*0 = 0
Would this approach be right? Expert comment would be welcome! Yes Indeed....This approach is correct



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Re: If x and y are integers such that y + 3 ≤ 3 and 2y – 3x + 6 = 0,
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09 Dec 2017, 09:33
If you visualize y+3<=3 on the number line you will get to know that y can take values in the range [6,0]
Next express the give relation in terms of x i.e. y= (2/3)*(y3). and since y is in the range of [6,0] and not in (0,3) .. x will follow y's sign. Thus, the product xy has to be >=0. Minumum being 0 .. check whether one can be 0 .. y=0 falls in the range. min. value of the product =0




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