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D
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Difficulty:
65%
(hard)
Question Stats:
52% (01:39) correct
48%
(01:41)
wrong
based on 371
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If x and y are integers 35x = 69y, which of the following must be true?
I. x > y
II. y/7 is an integer
III. x/23 is an integer
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
I. x > y
II. y/7 is an integer
III. x/23 is an integer
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
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23 Aug 2024, 09:11
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Official Solution:
If \(x\) and \(y\) are integers \(35x = 69y\), which of the following must be true?
I. \(x > y\)
II. \(\frac{y}{7}\) is an integer
III. \(\frac{x}{23}\) is an integer
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
It is important to note that the problem does not specify that \(x\) and \(y\) are necessarily positive. Therefore, when evaluating each option, it is essential to keep in mind that these variables may take non-positive values!
I. \(x > y\).
This statement is not always true, as \(x\) and \(y\) are not necessarily positive. For instance, consider \(x = -69\) and \(y=-35\), or \(x = 0\) and \(y=0\).
II. \(\frac{y}{7}\) is an integer
From \(35x = 69y\), we can infer that since the left-hand side, \(35x = 7(5x)\), is a multiple of 7, the right-hand side, \(69y\), must also be a multiple of 7. Since 69 is not a multiple of 7 (prime number), \(y\) must be, making \(\frac{y}{7}\) an integer. Hence, this option is always true.
III. \(\frac{x}{23}\) is an integer
From \(35x = 69y\), we can infer that since the right-hand side, \(69y = 23(3y)\), is a multiple of 23, the left-hand side, \(35x\), must also be a multiple of 23. Since 35 is not a multiple of 23 (prime number), \(x\) must be, making \(\frac{x}{23}\) an integer. Hence, this option is always true.
Answer: D
If \(x\) and \(y\) are integers \(35x = 69y\), which of the following must be true?
I. \(x > y\)
II. \(\frac{y}{7}\) is an integer
III. \(\frac{x}{23}\) is an integer
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
It is important to note that the problem does not specify that \(x\) and \(y\) are necessarily positive. Therefore, when evaluating each option, it is essential to keep in mind that these variables may take non-positive values!
I. \(x > y\).
This statement is not always true, as \(x\) and \(y\) are not necessarily positive. For instance, consider \(x = -69\) and \(y=-35\), or \(x = 0\) and \(y=0\).
II. \(\frac{y}{7}\) is an integer
From \(35x = 69y\), we can infer that since the left-hand side, \(35x = 7(5x)\), is a multiple of 7, the right-hand side, \(69y\), must also be a multiple of 7. Since 69 is not a multiple of 7 (prime number), \(y\) must be, making \(\frac{y}{7}\) an integer. Hence, this option is always true.
III. \(\frac{x}{23}\) is an integer
From \(35x = 69y\), we can infer that since the right-hand side, \(69y = 23(3y)\), is a multiple of 23, the left-hand side, \(35x\), must also be a multiple of 23. Since 35 is not a multiple of 23 (prime number), \(x\) must be, making \(\frac{x}{23}\) an integer. Hence, this option is always true.
Answer: D
General Discussion
Archit3110
Major Poster
Updated on: 17 Sep 2022, 02:02
Originally posted by Archit3110 on 16 Sep 2022, 11:31.
Last edited by Archit3110 on 17 Sep 2022, 02:02, edited 1 time in total.
Last edited by Archit3110 on 17 Sep 2022, 02:02, edited 1 time in total.
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given x & y are integers
7*5 * x= 3*23 * y
x has to be 3*23 and y = 7*5
must be trueif +ve
Else -ve x,y no
I. x > y
II. y/7 is an integer
35*69/7*69 ; true
III. x/23 is an integer
69*35/ 35* 23 ; true
OPTION D
7*5 * x= 3*23 * y
x has to be 3*23 and y = 7*5
must be trueif +ve
Else -ve x,y no
I. x > y
II. y/7 is an integer
35*69/7*69 ; true
III. x/23 is an integer
69*35/ 35* 23 ; true
OPTION D
Bunuel
