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

Question

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

WizakoBaskar · Answer

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.

Attachments

Inequalties-3.png [ 28.26 KiB | Viewed 7155 times ]

Inequalties-2.png [ 18.19 KiB | Viewed 7119 times ]

Inequalties-1.png [ 9.25 KiB | Viewed 7280 times ]

faltan · Answer

seems 1 as 0 is not possible which question limits

EgmatQuantExpert · Answer

Solution

Given:

To find:

• The positive difference between minimum value of x and minimum value of y
Approach and Working:

Adding the above two inequalities, we get,

As x is positive integer, minimum possible value of x is 1

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

SiddharthaChepuri · Answer

2y > 20 – x
Or, 2y > 12

As X<8, how can we consider x=8

Posted from my mobile device

EgmatQuantExpert · Answer

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).

chetan2u · Answer

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) 3x-30<-y..... multiply by 2, so 6x-60<-2y...(II) Add these two equations.. x+2y-2y>20+6x-60.......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>20-x.... higher the X, lower the y So take max possible value of x, so 2y>20-7....y>6.5 3x-30<-y....y<30-3x....if you add max value of X, we will get least possible value of y So y<30-3*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 7-1=6

ritu1009 · Answer

Once 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?

chetan2u · Answer

second equation  y<30-3x ... 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<30-3x, so y<30-3*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

hazeljj · Answer

Hi chetan2u , I had the same query as the other user. I was manipulating the equations around and I chanced upon 2 scenarios, 1. When x <8, I found 2 different inequalities for y. Using the expression 3x - 30 < -y -- (1), 3x < 24 -- (2) So (1) - (2): -30 < -y - 24 y < 6 HOWEVER, if (2) - (1): 30 < 24 + y y > 6 I think that was what the other user was getting at disregarding the range of values x can take unless this approach is invalid? I was thinking along the lines of perhaps there are actually 2 possible solutions to y but 1 of them needs to be rejected by taking the other inequality expression in the question into consideration. IF y < 6, 2y <12. And since x < 8, x + 2y < 20 instead of > 20 which is the condition given in the question. Thus y can only be > 6? Your help is greatly appreciated! If egmat or EgmatQuantExpert can leave an input and clarify, that will be great too. Thank you!

bumpbot · Answer

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

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