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x and y are positive integers such that x + 2y > 20 and 3x – 30 < -y

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x and y are positive integers such that x + 2y > 20 and 3x – 30 < -y  [#permalink]

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New post 28 Sep 2018, 21:13
4
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

32% (02:49) correct 68% (02:00) wrong based on 27 sessions

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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  [#permalink]

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New post 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.
Image

The area that satisfies the second inequality 3x + y < 30 is the region below the green line.
Image

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.
Image

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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 29 Sep 2018, 05:46
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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 &nbs [#permalink] 29 Sep 2018, 05:46
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