Explanation

The best strategy is to inspect each option one by one and substitute values

x = y:

If \(x = 1\), \(y = 1\) \(3x < 2y\) or \((3)(1) < (2)(1)\) or \(3 < 2\) This is incorrect

Will it work with a larger number?

If \(x = 100\), \(y = 100\) \(3x < 2y\) or \((3)(100) < (2)(100)\) or \(300 < 200\) also incorrect.

This one cannot ever be true, no matter how large the number we supply for x an y.

y = z:

If \(y = 1\), \(z = 1\) \(2y < 4z\) or \((2)(1) < (4)(1)\) or \(2 < 4\)

This statement could be true.


y > z:

If y = 1.1, z = 1 2y < 4z or (2)(1.1) < (4)(1) or \(2.2 < 4\)

This statement could be true.

x > z:

If \(x = 1.1\), \(z = 1\) \(3x < 4z\) or \((3)(1.1) < (4)(1)\) or \(3.3 < 4\)

This statement could be true.

Hence option B, C and D are correct.
_________________

A) x = y
If this is true, we can take the given inequality (3x < 2y < 4z) and replace x with y to get: 3y < 2y < 4z
Now let's focus on 3y < 2y
Subtract 2y from both sides to get: y < 0
Hmmm, this says that y is less than 0, HOWEVER, the question tells us that y is POSITIVE
So, it cannot be the case that x = y
ELIMINATE A


B) y = z
If this is true, we can take the given inequality (3x < 2y < 4z) and replace y with z to get: 3x < 2z < 4z
Now let's focus on 2z < 4z
This seems to check out.
So, let's see if we can find some values that satisfy this answer choice.
How about x = 1, y = 3 and z = 3
When we plug those values into the inequality (3x < 2y < 4z) and evaluate, we get: 3 < 6 < 12 (perfect!)
So, it is POSSIBLE that y = z
KEEP B

C) y > z
This is a little trickier.
So, let's just see if we can find some values that satisfy this answer choice.
How about x = 1, y = 3 and z = 2
When we plug those values into the inequality (3x < 2y < 4z) and evaluate, we get: 3 < 6 < 8 (perfect!)
So, it is POSSIBLE that y > z
KEEP C

D) x > z
This is tricky too.
So, let's just see if we can find some values that satisfy this answer choice.
How about x = 10, y = 16 and z = 9
When we plug those values into the inequality (3x < 2y < 4z) and evaluate, we get: 30 < 32 < 36 (perfect!)
So, it is POSSIBLE that x > z
KEEP D

Answers: B, C, and D

RELATED VIDEO

_________________

B, C and D are the answer.
The trick in this question is, the question has asked for Could be true option.
Which means in any range of numbers whether the condition satisfies.
Had it been a Must be true question, then it would be, in every condition whether any number in range satisfies.

Thanks for the explanation guys! Missed out the "could be" part. Something that I need to keep in mind

I didn't get it why is not the First option correct. I mean if we choose x=1/3 and y=1/2. They are positive numbers and it is not mentioned that they can't be fractions so in that case 1st could be true. Because 0.5 is a positive number and so is 0.3333

From the above explanation provided by Brent



Regards

the third & fourth statements could be false right, taking y=10, z=4 we get 20 < 16

The question asks, "which of the following statements could be true?"
So, as long as a statement CAN be true (even just once), then we must select it.


Aside: your values of y=10, z=4 satisfy the given inequality (3x < 2y < 4z), but they don't demonstrate that statements C and D are false.
_________________