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If 45x = 121y, which of the following must be true? 16 Sep 2022, 06:37
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If 45x = 121y, which of the following must be true?

I. x > y
II. x^2 > y^2
III. x/11 is an integer

(A) I only
(B) II only
(C) III only
(D) I and III only
(E) None of the above


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If 45x = 121y, which of the following must be true? Updated on: 16 Aug 2025, 02:00
Originally posted by Bunuel on 11 Mar 2023, 02:44.
Last edited by Bunuel on 16 Aug 2025, 02:00, edited 1 time in total.
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If 45x = 121y, which of the following must be true?

I. x > y
II. x^2 > y^2
III. x/11 is an integer

(A) I only
(B) II only
(C) III only
(D) I and III only
(E) None of the above




 


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Official Solution:


If \(45x = 121y\), which of the following must be true?

I. \(x > y\)

II. \(x^2 > y^2\)

III. \(\frac{x}{11}\) is an integer




A. I only
B. II only
C. III only
D. I and III only
E. None of the above


It is important to note that the problem does not specify that \(x\) and \(y\) are necessarily integers, nor does it require \(x\) and \(y\) to be necessarily positive. Therefore, when evaluating each option, it is essential to keep in mind that these variables may take on non-integer or non-positive values!

I. \(x > y\).

This statement is not always true, as \(x\) and \(y\) are not necessarily positive. For instance, consider \(x = -121\) and \(y=-45\), or \(x = 0\) and \(y=0\).

II. \(x^2 > y^2\).

This statement is also not always true. For example, consider \(x = y = 0\).

III. \(\frac{x}{11}\) is an integer.

This statement is also not always true, as \(x\) and \(y\) are not necessarily integers. For instance, consider \(x = \frac{1}{45}\) and \(y=\frac{1}{121}\).


Answer: E
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If 45x = 121y, which of the following must be true? Updated on: 16 Sep 2022, 22:35
Originally posted by gmatophobia on 16 Sep 2022, 10:05.
Last edited by gmatophobia on 16 Sep 2022, 22:35, edited 2 times in total.
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This is a must be true question.

The way the options are presented, we can evaluate I and II first.

Given : 45x = 121y

If both x and y are zero even then the LHS = RHS

Hence Neither I or II can be true.

Let's evaluate III

45x = 121y

Case 1 :

x = 121 & y = 45 ; then yes \(\frac{x}{11}\) is an integer

Case 2 :
However x can well be an fraction and y can adjust accordingly to make the RHS = LHS

Say x = \(\frac{1}{3} \)

45 * x = \(45 * \frac{1}{3}\) = 15

Now y can be \(\frac{15 }{ 121}\) to make LHS = RHS

In this case, \(\frac{x}{11}\) is not an integer.

As it's a must be true question , III can be eliminated.

Option E

Edit : Missed considering zero in the first version. Edited the post to correct the solution.
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If 45x = 121y, which of the following must be true? 16 Sep 2022, 22:17
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gmatophobia
This is a must be true question.

The way the options are presented, we can evaluate II first.

Given : 45x = 121y

\(\frac{x}{y}= \frac{121}{45}\)

The value on the RHS is > 1

This can occur in two scenarios

1) x > y ; x and y are both positive

2) x < y ; x and y are both negative

In either case, \(x^2 > y^2\)

So II is correct and I is incorrect.

Let's evaluate III

45x = 121y

Case 1 :

x = 121 & y = 45 ; then yes \(\frac{x}{11}\) is an integer

Case 2 :
However x can well be an fraction and y can adjust accordingly to make the RHS = LHS

Say x = \(\frac{1}{3} \)

45 * x = \(45 * \frac{1}{3}\) = 15

Now y can be \(\frac{15 }{ 121}\) to make LHS = RHS

In this case, \(\frac{x}{11}\) is not an integer.

As it's a must be true question , III can be eliminated.

Option B
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If x and y are both zero then 45x = 121y, but (I) x>y and (II) x^2 > y^2 are not true.
IMO Answer is E.
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If 45x = 121y, which of the following must be true? 16 Sep 2022, 22:34
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gmatophobia
This is a must be true question.

The way the options are presented, we can evaluate II first.

Given : 45x = 121y

\(\frac{x}{y}= \frac{121}{45}\)

The value on the RHS is > 1

This can occur in two scenarios

1) x > y ; x and y are both positive

2) x < y ; x and y are both negative

In either case, \(x^2 > y^2\)

So II is correct and I is incorrect.

Let's evaluate III

45x = 121y

Case 1 :

x = 121 & y = 45 ; then yes \(\frac{x}{11}\) is an integer

Case 2 :
However x can well be an fraction and y can adjust accordingly to make the RHS = LHS

Say x = \(\frac{1}{3} \)

45 * x = \(45 * \frac{1}{3}\) = 15

Now y can be \(\frac{15 }{ 121}\) to make LHS = RHS

In this case, \(\frac{x}{11}\) is not an integer.

As it's a must be true question , III can be eliminated.

Option B


If x and y are both zero then 45x = 121y, but (I) x>y and (II) x^2 > y^2 are not true.
IMO Answer is E.
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Good catch Aks111! Kudos to you.

P.S - I need to be more careful. :(
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If 45x = 121y, which of the following must be true? 23 Aug 2024, 09:14
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Similar but easier question is here:

https://gmatclub.com/forum/if-x-and-y-a ... 98803.html
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If 45x = 121y, which of the following must be true? 26 Oct 2024, 12:53
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I have a doubt here: Why can't III be true? 45x=11(11y), then x/11 must be a multiple of 11.
Bunuel
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If 45x = 121y, which of the following must be true?

I. x > y
II. x^2 > y^2
III. x/11 is an integer

(A) I only
(B) II only
(C) III only
(D) I and III only
(E) None of the above




 


Enjoy this brand new question we just created for the GMAT Club Tests.

To get 1,600 more questions and to learn more visit: user reviews | learn more

 


Official Solution:


If \(45x = 121y\), which of the following must be true?

I. \(x > y\)

II. \(x^2 > y^2\)

III. \(\frac{x}{11}\) is an integer




A. I only
B. II only
C. III only
D. I and III only
E. None of the above


It is important to note that the problem does not specify that \(x\) and \(y\) are necessarily integers, nor does it require \(x\) and \(y\) to be necessarily positive. Therefore, when evaluating each option, it is essential to keep in mind that these variables may take on non-integer or non-positive values!

I. \(x > y\).

This statement is not always true, as \(x\) and \(y\) are not necessarily positive. For instance, consider \(x = -121\) and \(y=-45\), or \(x = 0\) and \(y=0\).

II. \(x^2 > y^2\).

This statement is not always true, as \(x\) and \(y\) are not necessarily integers. For example, consider \(x = \frac{1}{45}\) and \(y=\frac{1}{121}\). Additionally, since \(x\) and \(y\) can both be 0, this statement is also not true in that case.

III. \(\frac{x}{11}\) is an integer.

This statement is also not always true, as \(x\) and \(y\) are not necessarily integers. For instance, consider \(x = \frac{1}{45}\) and \(y=\frac{1}{121}\).


Answer: E
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If 45x = 121y, which of the following must be true? 26 Oct 2024, 13:01
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raziqsid
I have a doubt here: Why can't III be true? 45x=11(11y), then x/11 must be a multiple of 11.
Bunuel
Bunuel
If 45x = 121y, which of the following must be true?

I. x > y
II. x^2 > y^2
III. x/11 is an integer

(A) I only
(B) II only
(C) III only
(D) I and III only
(E) None of the above




 


Enjoy this brand new question we just created for the GMAT Club Tests.

To get 1,600 more questions and to learn more visit: user reviews | learn more

 


Official Solution:


If \(45x = 121y\), which of the following must be true?

I. \(x > y\)

II. \(x^2 > y^2\)

III. \(\frac{x}{11}\) is an integer




A. I only
B. II only
C. III only
D. I and III only
E. None of the above


It is important to note that the problem does not specify that \(x\) and \(y\) are necessarily integers, nor does it require \(x\) and \(y\) to be necessarily positive. Therefore, when evaluating each option, it is essential to keep in mind that these variables may take on non-integer or non-positive values!

I. \(x > y\).

This statement is not always true, as \(x\) and \(y\) are not necessarily positive. For instance, consider \(x = -121\) and \(y=-45\), or \(x = 0\) and \(y=0\).

II. \(x^2 > y^2\).

This statement is not always true, as \(x\) and \(y\) are not necessarily integers. For example, consider \(x = \frac{1}{45}\) and \(y=\frac{1}{121}\). Additionally, since \(x\) and \(y\) can both be 0, this statement is also not true in that case.

III. \(\frac{x}{11}\) is an integer.

This statement is also not always true, as \(x\) and \(y\) are not necessarily integers. For instance, consider \(x = \frac{1}{45}\) and \(y=\frac{1}{121}\).


Answer: E
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Please check the reasoning below:

III. \(\frac{x}{11}\) is an integer.

This statement is also not always true, as \(x\) and \(y\) are not necessarily integers. For instance, consider \(x = \frac{1}{45}\) and \(y=\frac{1}{121}\).
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If 45x = 121y, which of the following must be true? 22 Apr 2025, 03:20
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Just suppose x=0 and y-0

45 * 0 = 121*0 [The equation is satisfied]

None of the conditions are met
Bunuel
If 45x = 121y, which of the following must be true?

I. x > y
II. x^2 > y^2
III. x/11 is an integer

(A) I only
(B) II only
(C) III only
(D) I and III only
(E) None of the above


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If 45x = 121y, which of the following must be true? 22 Apr 2025, 03:40
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Aabhash777
Just suppose x=0 and y-0

45 * 0 = 121*0 [The equation is satisfied]

None of the conditions are met
Bunuel
If 45x = 121y, which of the following must be true?

I. x > y
II. x^2 > y^2
III. x/11 is an integer

(A) I only
(B) II only
(C) III only
(D) I and III only
(E) None of the above


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In that case, Statement III is actually true, because x/11 = 0, and 0 is an integer.
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If 45x = 121y, which of the following must be true? 15 Aug 2025, 21:03
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Quote:
II. \(x^2 > y^2\).

This statement is not always true, as \(x\) and \(y\) are not necessarily integers. For example, consider \(x = \frac{1}{45}\) and \(y=\frac{1}{121}\). Additionally, since \(x\) and \(y\) can both be 0, this statement is also not true in that case.
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When \(x = \frac{1}{45}\) and \(y=\frac{1}{121}\), we still have \(x^2 > y^2\).
When \(x=0\) and \(y=0\) , we have \(x^2 = y^2\).
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If 45x = 121y, which of the following must be true? 16 Aug 2025, 02:01
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Quote:
II. \(x^2 > y^2\).

This statement is not always true, as \(x\) and \(y\) are not necessarily integers. For example, consider \(x = \frac{1}{45}\) and \(y=\frac{1}{121}\). Additionally, since \(x\) and \(y\) can both be 0, this statement is also not true in that case.
When \(x = \frac{1}{45}\) and \(y=\frac{1}{121}\), we still have \(x^2 > y^2\).
When \(x=0\) and \(y=0\) , we have \(x^2 = y^2\).
Bunuel
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Yes. Consider \(x = y = 0\) to discard.
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