If x, y, and z are positive numbers such that 3x

Question

If x, y, and z are positive numbers such that 3x < 2y < 4z , which of the following statements could be true?

I. y = z
II. y > z
III. x > z

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

Bunuel · Accepted Answer

Official Solution: If x , y , and z are positive numbers such that 3x < 2y < 4z , which of the following statements could be true? I. y = z II. y > z III. x > z A. I only B. I and II only C. I and III only D. II and III only E. I, II and III Notice that the question asks: " which of the following statements COULD be true? " NOT " which of the following statements MUST be true? " I. y = z . From the stem we know that 2y < 4z . So, y < 2z . It's certainly possible that y = z , for example, y = z=1 . II. y > z From the stem we know that 2y < 4z . So, y < 2z . It's certainly possible that y > z , for example, y = 3 and y=2 . III. x > z . From the stem we know that 3x < 4z . And again, it's possible that x > z , for example, x=10 and y=9 . Answer: E

chetan2u · Answer

Let us take each option.. I. y=z.... Let us compare y and z in the inequality 3x<2y<4z, so 2y<4z or y<2z.. So yes...let y=z=2 and x=1...3*1<2*2<4*2...3<4<8... II. y>z.... Let us compare y and z in the inequality 3x<2y<4z, so 2y<4z or y<2z.. So yes, y can be between z and 2z..let y=3, z=2 and x=1...3*1<2*3<4*2...3<6<8... III. x>z.... Let us compare x and z in the inequality 3x<2y<4z, so 3x<4z or x<4z/3.. So yes, x can be between z and 4z/3...let z=30 and x=31, and y=...3*31<2*50<4*30...91<100<120. All three possible E

Archit3110 · Answer

chetan2u 
thanks for the solution , I had a doubt that since the question has given that x,y,z are +ve numbers so shouldnt we check the given relations with a fraction value as well?

Kinshook · Answer

Asked: If x, y, and z are positive numbers such that 3x < 2y < 4z , which of the following statements could be true? I. y = z 2y<4z If y=z 2y<4y; Take y=1; 2<4 COULD BE TRUE II. y > z 2y<4z Take y = 1.5; z =1 2*1.5=3<4*1=4 COULD BE TRUE III. x > z 3x<4z Take x = 1.1 z = 1 3.3 < 4 COULD BE TRUE IMO E

chetaksatav · Answer

Hello Bunuel, For II condition y>z if I take Y=3, Z=2 and X=1 Then, 3*1 < 2*3 < 4*2 3 < 6 < 8 ..........Correct But if I choose values as below: - X=1 Y=10 & Z=3 Then, 3*1 < 2*10 < 4*3 3 < 20 < 12 .......... Its wrong 20 > 12. So II. Y>Z is may be or may not be condition. Similar for III. X>Z. It is also may be or may not be condition. So don't you think that only I Y=Z i.e. Option A should be the answer. Regards Chetak

Bunuel · Answer

Notice that the question asks: " which of the following statements  COULD  be true? " NOT " which of the following statements  MUST  be true? "

chetaksatav · Answer

Thanks Bunuel for clarifying.

satya2029 · Answer

3x<2y<4z
3/2x<y<2z=1.5x<y<2z
Since all are positive numbers
x=1/2, y =1, z=1 then 0.75<1<2 
x=0.61, y=0.99, z=0.6 then 0.91<0.99<1.2
E:)

GDT · Answer

VeritasKarishma In this ques, since we have to just tell whether a situation involving 2 variables is possible, do we also need to consider what value the 3rd variable can take? (Just want to confirm if doing so would add something) Like for statement 3, I just considered x<4z/3. So take z=3/4=.75 ---> 1>x>.75 take x=.76 (Now do we need to see for what value of y the inequation 3x < 2y < 4z is true?) Thanks in advance!

KarishmaB · Answer

No, you do not need to worry about the third variable here. 
Note that in stmnts I and II, I can just say x = 1/1000 and focus my energies on y and z. 
Again, in statement III, for all values such that 3x < 4z, there will certainly be some values in between these two so it will work for some values of y for sure. 
Note that we might want to be more careful if we were given that x, y and z are positive integers.

ScottTargetTestPrep · Answer

Solution: If we let y = z = 2, then 2y = 4 and 4z = 8; so 2y < 4z is satisfied. This shows that y = z is possible. Notice that y = z = 1 is actually not a valid example since in that case, no positive integer value of x satisfies 3x < 2y. If we let y = 3 and z = 2, then 2y = 6 and 4z = 8; so 2y < 4z is satisfied. This shows that y > z is possible. Finally, if we let x = 10 and z = 9, then 3x = 30 and 4z = 36. Choosing y = 16, we see that 3x < 2y < 4z is satisfied. We see that x > z is also possible. Answer: E

Hoozan · Answer

Hi  ScottTargetTestPrep

You mentioned that "Notice that y = z = 1 is actually not a valid example since in that case, no positive integer value of x satisfies 3x < 2y."

But the question states that x,y and z are positive numbers (not integers) so we very well could have x = 0.5, y = z = 1. In that case 
3x = 1.5
2y = 2
4z = 4

So 3x < 2y

Please could you correct me if I'm wrong

Thanks!

Posted from my mobile device

Crytiocanalyst · Answer

Substituting diferent values should do the trick

I. y = z
when y=z=1 the eqn is satisfied

II. y > z
y=1/2 and z=1/3 eqn is satisfied

III. x > z
x=1/9 and z=1/10 theeqn is satisfied

Therefore IMO E

PolarisTestPrep · Answer

Hey guys,

The key to this problem is to notice the wording

It says that they are numbers and not specifically integers as is often the case on most questions

Keeping this in mind:

I. can be true because y = z = 1 yields 2 < 4, which is true

II. can be true because y = 1.1 and z = 1 yields 2.2 < 4, which is true

III. can be true because x = 1.1 and z = 1 yields 3.3 < 4, which is true

The answer is E

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.

If x, y, and z are positive numbers such that 3x < 2y < 4z , which of

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