If 20x = 49y, which of the following must be true?

A quick glance over says that is correct. Just a different approach. Often the GMAT will do this - offer several ways to solve a problem. No brownie points for doing the more difficult, so if you recognize it, go with whichever you're faster/more comfortable.

I asked here because I was not sure (better i'm sure at 99% ) in particular on statement 2.

But if you said ok ...many Thanks

We have 20x=49y
or x = 49y/20

I need not be true because x is greater than y only when y is positive.

II need not be true because both x and y can be zero. This applies to I also.

III also need not be true because x/7 =49y/140 = 7y/20.
We can see that x/7 need not be an integer as for example in the case of y=1.

This is a flawed question. None of the options must be true.

Notice that 20x = 49y holds true if x=y=0, thus I and II are not always true.

As for III: if x=1 and y=20/49, then x/7 won't be an integer.

If the question were:
If x and y are integers and 20x = 49y, which of the following must be true?

Then in this case III would be always true: RHS is a multiple of 7, thus LHS must also be a multiple of 7 and since 20 is not a multiple of 7, then x must be.

Hope it's clear.

I perfectly understand what you say. Something it was not unclear to me either.

Though, my primary question was is me approach was correct generally speaking.

Thanks,

Regards

If 20x = 49y, which of the following must be true?

I. x > y
II. x^2 > y^2
III. x/7 is an integer

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

OA after some discussion

For x = 1,y = 20/49 and x>y. However, for x=-49,y = -20 and y>x. I is clearly not the right answer.
For y=0, x=0. Thus, even II doesn't hold good.
x/7 = 7y/20. For y=0, x/7 is an integer, however for y=1, it is not.
None of the options are (must be)true

If 20x = 49y, which of the following must be true?

I. x > y
II. x^2 > y^2
III. x/7 is an integer

Just pick \(x=0\) and \(y=0\)

Options I and II are false (0 is not \(> 0\)), only III is true. And holds true for each value of x because \(4*5*x=7*7*y\) so or x=0 or x is a multiple of 7, in both cases \(\frac{x}{7}\) is an integer.
C

PS: given that x, y are integers. Otherwise none must be true

Edit post:
A)I only
B)II only
C) III only
D)I and III
E)I, II, and III

These are Veritas possible answers. The "none" option is not available, as you see. So we must assume that are integers and that the question is not complete (it must specify that x,y are integers). However the question itself , because x and y could be fractions (no integers) has an hypothetical "F)none of the above" as correct answer. No doubt.

The question is incomplete

if x =1/20 and y=1/49 still the equation holds right?

SO at this point of time option 3 also goes wrong...Any opinion on this?

That is not true. It has not been mentioned that x and y are integers.

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