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Re: If 20x = 49y, which of the following must be true? [#permalink]

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01 Jan 2013, 15:10

I'd plug in numbers here. Try the weird ones i.e. 0, -1, 1/2, etc. as it does not note that x or y are integers.

Plugging in 0, We find that statement 1 and 2 are false. Looking at what is left, the solution must be C, Statement III is correct since there is no answer for "None of the above".

Good question, what is the source?
_________________

I attacked the question in that way tell me if is correct

\(20x = 49y\)

\(20\) \(=\) \(2^2\) \(*5\)

\(49\) \(=\) \(7^2\)

so basically we would have the second one in the LHS and viceversa to balance the equation \(BUT\) we should consider also 0 because 0=0 so is equal ( but of course)

1)\(x > y\)true not always \(7^2\)\(>\) \(2^2\) \(*5\) but we have also zero so x > y is not true

2) \(x^2\) \(>\) \(y^2\) we know that x and y are always positive so basically we can reduce the second statement to \(x > y\) and we know already insuff

I attacked the question in that way tell me is correct

\(20x = 49y\)

\(20\) \(=\) \(2^2\) \(*5\)

\(49\) \(=\) \(7^2\)

so basically we would have the second one in the LHS and viceversa to balance the equation \(BUT\) we should consider also 0 because 0=0 so is equal ( but of course)

1)\(x > y\)true not always \(7^2\)\(>\) \(2^2\) \(*5\) but we have also zero so x > y is not true

2) \(x^2\) \(>\) \(y^2\) we know that x and y are always positive so basically we can reduce the second statement to \(x > y\) and we know already insuff

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 attacked the question in that way tell me is correct

\(20x = 49y\)

\(20\) \(=\) \(2^2\) \(*5\)

\(49\) \(=\) \(7^2\)

so basically we would have the second one in the LHS and viceversa to balance the equation \(BUT\) we should consider also 0 because 0=0 so is equal ( but of course)

1)\(x > y\)true not always \(7^2\)\(>\) \(2^2\) \(*5\) but we have also zero so x > y is not true

2) \(x^2\) \(>\) \(y^2\) we know that x and y are always positive so basically we can reduce the second statement to \(x > y\) and we know already insuff

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
_________________

Re: If 20x = 49y, which of the following must be true? [#permalink]

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01 Jan 2013, 16:42

carcass wrote:

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

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.
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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

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.

Re: If 20x = 49y, which of the following must be true? [#permalink]

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02 Jan 2013, 05:31

Bunuel wrote:

carcass wrote:

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

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.

Re: f 20x = 49y, which of the following must be true? [#permalink]

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04 May 2013, 22:49

skamal7 wrote:

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

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

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
_________________

Re: f 20x = 49y, which of the following must be true? [#permalink]

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05 May 2013, 00:05

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
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It is beyond a doubt that all our knowledge that begins with experience.

Re: f 20x = 49y, which of the following must be true? [#permalink]

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05 May 2013, 00:16

1

This post received KUDOS

Zarrolou wrote:

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

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