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505-555 (Easy)|   Algebra|   Must or Could be True|                           
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Bunuel
If x/y = c/d and d/c = b/a, which of the following must be true?

I. y/x = b/a
II. x/a = y/b
III. y/a = x/b

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

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Here, y, d, c, and a are in denominator, so we can take them to be non zero and then rest would also be non zero

x/y = c/d and d/c = b/a
\(x/y = c/d = a/b\)

I. y/x = b/a True
II. x/a = y/b True
III. y/a = x/b not always true

answer choice C
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Bunuel
If x/y = c/d and d/c = b/a, which of the following must be true?

I. y/x = b/a
II. x/a = y/b
III. y/a = x/b

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

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My Solution:

Given, x/y =c/d and d/c=b/a

If d/c=b/a then c/d =a/b, therefore we can say that x/y=a/b or y/x=b/a. Statement I is true

If x/y=a/b, then x/a=y/b. Statement II is true

As x/y=a/b, then y/a = x/b can't be true. So statement III is not true

Answer is Option C (I & II only)
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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

If x/y = c/d and d/c = b/a, which of the following must be true?

I. y/x = b/a
II. x/a = y/b
III. y/a = x/b

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

d/c = b/a implies c/d=a/b. Together with x/y = c/d --> x/y = c/d = a/b. So y/x= b/a(I is correct). By x/y = a/b --> bx=ay --> x/a = y/b(II is correct and III is not correct). The answer is, therefore, C.
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Substituting numbers for x,y,c,d,b and a is also a viable option.

Considering that \(\frac{x}{y} = \frac{c}{d}\) \(, \frac{c}{d}\) is a multiple of \(\frac{x}{y.}\)
Moreover, considering that \(\frac{d}{c} = \frac{b}{a}\), \(\frac{d}{c}\) is a multiple of \(\frac{b}{a}\) but also the reciprocal of \(\frac{x}{y}\).

So, let's consider the following: \(\frac{x}{y}\) = \(\frac{4}{2}\) = \(\frac{c}{d}\)= \(\frac{8}{4}\) then \(\frac{d}{c}\) = \(\frac{4}{8}\) = \(\frac{b}{a}\)= \(\frac{8}{16}\).

I. y/x = b/a
2/4 = 8/16 --> 2/4 is a multiple of 8/16 therefore TRUE
II. x/a = y/b
4/16 = 2/8 --> 4/16 is a multiple of 2/8 therefore TRUE
III. y/a = x/b
2/16 does not equal 4/8 therefore NOT TRUE.

Only I and II are true. The correct answer is C.
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Bunuel
If x/y = c/d and d/c = b/a, which of the following must be true?

I. y/x = b/a
II. x/a = y/b
III. y/a = x/b

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

Kudos for a correct solution.

We can use some convenient numbers:

x = 2, y = 3

c = 4, d = 6

b = 12, a = 8

Now we can analyze each Roman numeral:

I. 3/2 = 12/8 ?

3/2 = 3/2 → I is true.

II. 2/8 = 3/12 ?

1/4 = 1/4 → II is true.

III. 3/8 = 2/12 ?

3/8 ≠ 1/6 → III is not True.

Answer: C
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Hi All,

While this Roman Numeral question looks a bit complicated, it just involves comparing fractions, so we can use basic Arithmetic rules to answer it (and you can solve it Algebraically or by TESTing VALUES).

To start, you should notice that C/D and D/C appear in the two equations – meaning that we can create one gigantic comparison that uses all of the variables. That would be…

X/Y = C/D = A/B

Since each of the fractions is equal to the two other fractions, we can use cross-multiplication to show that several relationships exist among the variables:

(D)(X) = (C)(Y)
(B)(X) = (A)(Y)
(B)(C) = (A)(D)

These three deductions are important – and we can use them against the three Roman Numerals to quickly prove which are true and which are not…

I. Y/X = B/A

When we cross-multiply this equation, we get… (B)(X) = (A)(Y). This is one of the relationships that we already proved is TRUE (above).

II. X/A = Y/B

When we cross-multiply this equation, we get… (B)(X) = (A)(Y). This is the SAME relationship that we already proved is Roman Numeral 1 – so we know it’s TRUE.

III. Y/A = X/B

When we cross-multiply this equation, we get… (A)(X) = (B)(Y). Based on the original equations, we have no proof that this relationship is actually true (If all of the variables were set to equal 1, then it would be true; if the 6 variables were not the same value though, then the it would be false).

Final Answer:
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Bunuel
If x/y = c/d and d/c = b/a, which of the following must be true?

I. y/x = b/a
II. x/a = y/b
III. y/a = x/b

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


Answer: Option C

Video solution by GMATinsight

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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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EMPOWERgmatRichC
Hi All,

While this Roman Numeral question looks a bit complicated, it just involves comparing fractions, so we can use basic Arithmetic rules to answer it (and you can solve it Algebraically or by TESTing VALUES).

To start, you should notice that C/D and D/C appear in the two equations – meaning that we can create one gigantic comparison that uses all of the variables. That would be…

X/Y = C/D = A/B

Since each of the fractions is equal to the two other fractions, we can use cross-multiplication to show that several relationships exist among the variables:

(D)(X) = (C)(Y)
(B)(X) = (A)(Y)
(B)(C) = (A)(D)

These three deductions are important – and we can use them against the three Roman Numerals to quickly prove which are true and which are not…

I. Y/X = B/A

When we cross-multiply this equation, we get… (B)(X) = (A)(Y). This is one of the relationships that we already proved is TRUE (above).

II. X/A = Y/B

When we cross-multiply this equation, we get… (B)(X) = (A)(Y). This is the SAME relationship that we already proved is Roman Numeral 1 – so we know it’s TRUE.

III. Y/A = X/B

When we cross-multiply this equation, we get… (A)(X) = (B)(Y). Based on the original equations, we have no proof that this relationship is actually true (If all of the variables were set to equal 1, then it would be true; if the 6 variables were not the same value though, then the it would be false).

Final Answer:
GMAT Assassins aren’t born, they’re made,
Rich

EMPOWERgmatRichC

This is very helpful! If you decided to choose numbers to solve this problem, is there a case where you would have to choose additional numbers or will the first set of numbers you choose (assuming all of your math is correct) lead you to rule out choice III.? I am trying to have a better strategy if I should follow what you did for consistency or to choose numbers (if choosing numbers will not always work the first time)
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Hi woohoo921,

You can absolutely TEST VALUES here; there are some specific aspects of the prompt worth noting (that would help you to choose those numbers). First, with so many variables, we would want to keep the work simple - so the first TEST that we'd run would be to set all of the variables EQUAL to one another (for example, make them all 1s or 2s). Second, since we're asked what MUST be true, we're really being asked "what is ALWAYS TRUE no matter how many different examples we can come up with?" Again, with so many variables, we likely wouldn't need to make any extreme choices for the numbers we choose to TEST (and sticking with some mix of small integers would prove that Roman Numeral 3 is NOT always true).

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: [email protected]
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