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x and y are positive integers such that x + 2y > 20 and 3x – 30 < y
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28 Sep 2018, 21:13
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x and y are positive integers such that x + 2y > 20 and 3x – 30 < y. What is the positive difference between the minimum possible value of x and the minimum value of y? A. 6 B. 0 C. 1 D. 4 E. 6
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Re: x and y are positive integers such that x + 2y > 20 and 3x – 30 < y
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29 Sep 2018, 00:50
Given conditions: x + 2y > 20 and 3x  30 < y; x and y are positive integers Objective: To find the difference between the minimum possible value of x and the minimum value of y 3x  30 < y can be written as 3x + y < 30 Please refer to the following 3 graphs. The area that satisfies the first inequality x + 2y > 20 is the region above the red line. The area that satisfies the second inequality 3x + y < 30 is the region below the green line. Because x and y are positive integers, the area of interest is restricted to I quadrant and the area enclosed in the triangle ABC are values of x and y that satisfy both the inequalities. Let us find the coordinates of point C Solve the two equations x + 2y = 20 and 3x + y = 30 3x + 6y = 60 3x + y = 30  5y = 30  Or y = 6. Substituting y = 6 in equation (1), we get x = 8. So, coordinates of point C are (8, 6) It is evident from the graph that lowest point among the three from y coordinates is C, So, the minimum value of y enclosed in the triangle is an integer greater than 6. So, it has to be 7. From the graph, we can also deduce that the minimum value of x enclosed in the triangle is an integer greater than 0. So, it has to be 1. The positive difference between the minimum value of x and minimum value of y is 6.
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Re: x and y are positive integers such that x + 2y > 20 and 3x – 30 < y
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29 Sep 2018, 05:01
seems 1 as 0 is not possible which question limits



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Re: x and y are positive integers such that x + 2y > 20 and 3x – 30 < y
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29 Sep 2018, 05:19
Solution Given:• x and y are positive integers • x + 2y > 20 and 3x – 30 < y To find:• The positive difference between minimum value of x and minimum value of y Approach and Working:• x + 2y > 20 • 3x – 30 < y Or, 3x + y < 30 Or, 3x – y > 30 Or, 6x – 2y > 60 Adding the above two inequalities, we get, • x – 6x + 2y – 2y > 20 – 60 Or, 5x > 40 Or, 5x < 40 Or, x < 8 As x is positive integer, minimum possible value of x is 1 • also, if x < 8, we can say from x + 2y > 20, 2y > 20 – x Or, 2y > 12 Or, y > 6 As y is positive integer, minimum possible value of y is 7 Therefore, the positive difference = 7 – 1 = 6 Hence, the correct answer is option E. Answer: E
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Re: x and y are positive integers such that x + 2y > 20 and 3x – 30 < y
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29 Sep 2018, 05:39
2y > 20 – x Or, 2y > 12
As X<8, how can we consider x=8
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Re: x and y are positive integers such that x + 2y > 20 and 3x – 30 < y
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29 Sep 2018, 05:46
Siddharthachepuri@gmail.com wrote: 2y > 20 – x Or, 2y > 12
As X<8, how can we consider x=8
Posted from my mobile device Hi, we did not consider x = 8. As x is less than 8, we used this to conclude that y is greater than 12. (For example, if x can be maximum 7, then y should be minimum 7 (we wrote it as y > 6).
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Re: x and y are positive integers such that x + 2y > 20 and 3x – 30 < y
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12 Dec 2018, 20:49
workout wrote: x and y are positive integers such that x + 2y > 20 and 3x – 30 < y. What is the positive difference between the minimum possible value of x and the minimum value of y?
A. 6
B. 0
C. 1
D. 4
E. 6 Let us get the range of one variable first and for that we will have to cancel out the other variable. So x+2y>20....(I) 3x30<y..... multiply by 2, so 6x60<2y...(II) Add these two equations.. x+2y2y>20+6x60.......5x<40.....x<8 Now since x is positive integer and x<8, the minimum possible value of x is 1.. Now take the equation for getting the range of the values of y.. x+2y>20...2y>20x.... higher the X, lower the y So take max possible value of x, so 2y>207....y>6.5 3x30<y....y<303x....if you add max value of X, we will get least possible value of y So y<303*7.....y<9 So least possible value is less than 9 but greater than 6.5.. So minimum value of y is 7.. Difference is 71=6
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Re: x and y are positive integers such that x + 2y > 20 and 3x – 30 < y
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12 Dec 2018, 21:53
EgmatQuantExpert wrote: Solution Given:• x and y are positive integers • x + 2y > 20 and 3x – 30 < y To find:• The positive difference between minimum value of x and minimum value of y Approach and Working:• x + 2y > 20 • 3x – 30 < y Or, 3x + y < 30 Or, 3x – y > 30 Or, 6x – 2y > 60 Adding the above two inequalities, we get, • x – 6x + 2y – 2y > 20 – 60 Or, 5x > 40 Or, 5x < 40 Or, x < 8 As x is positive integer, minimum possible value of x is 1 • also, if x < 8, we can say from x + 2y > 20, 2y > 20 – x Or, 2y > 12 Or, y > 6 As y is positive integer, minimum possible value of y is 7 Therefore, the positive difference = 7 – 1 = 6 Hence, the correct answer is option E. Answer: EOnce we calculate value for x, while calculating value for y using first equation we get y>6 and with second equation, we get y<6. How to know which one to take?



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Re: x and y are positive integers such that x + 2y > 20 and 3x – 30 < y
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12 Dec 2018, 22:40
ritu1009 wrote: EgmatQuantExpert wrote: Solution Given:• x and y are positive integers • x + 2y > 20 and 3x – 30 < y To find:• The positive difference between minimum value of x and minimum value of y Approach and Working:• x + 2y > 20 • 3x – 30 < y Or, 3x + y < 30 Or, 3x – y > 30 Or, 6x – 2y > 60 Adding the above two inequalities, we get, • x – 6x + 2y – 2y > 20 – 60 Or, 5x > 40 Or, 5x < 40 Or, x < 8 As x is positive integer, minimum possible value of x is 1 • also, if x < 8, we can say from x + 2y > 20, 2y > 20 – x Or, 2y > 12 Or, y > 6 As y is positive integer, minimum possible value of y is 7 Therefore, the positive difference = 7 – 1 = 6 Hence, the correct answer is option E. Answer: EOnce we calculate value for x, while calculating value for y using first equation we get y>6 and with second equation, we get y<6. How to know which one to take? second equation \(y<303x\)... This will give us the upper limit as we are getting y is less than something. So this should give you the maximum value of y.. and you will get maximum value for y when x is minimum in y<303x, so y<303*1....y<27 so range of values of y is 6<y<27..... Minimum value is 7 and max value is 26 Hope it helps
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Re: x and y are positive integers such that x + 2y > 20 and 3x – 30 < y
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