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 7417 times ]

Inequalties-2.png [ 18.19 KiB | Viewed 7375 times ]

Inequalties-1.png [ 9.25 KiB | Viewed 7541 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

Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions