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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If x and y are integers, is y/x = 17/9? (1) 13/7 < y/x < 19/10 (2) x=9

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Math Expert V
Joined: 02 Sep 2009
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If x and y are integers, is y/x = 17/9? (1) 13/7 < y/x < 19/10 (2) x=9  [#permalink]

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Question Stats: 33% (02:53) correct 67% (02:22) wrong based on 43 sessions

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If x and y are integers, is $$\frac{y}{x} = \frac{17}{9}$$ ?

(1) $$\frac{13}{7} < \frac{y}{x} < \frac{19}{10}$$

(2) $$x = 9$$

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Math Expert V
Joined: 02 Aug 2009
Posts: 8341
Re: If x and y are integers, is y/x = 17/9? (1) 13/7 < y/x < 19/10 (2) x=9  [#permalink]

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If x and y are integers, is $$\frac{y}{x} = \frac{17}{9}$$ ?

(1) $$\frac{13}{7} < \frac{y}{x} < \frac{19}{10}$$
You can see that the numerator is 2(denominator)-1 or of the form $$\frac{2z-1}{z}$$..
As we increase z, the numerator increases by bigger quantity as compared to denominator, so higher the denominator, higher the value of that fraction
so $$\frac{y}{x}$$ can be $$\frac{2*9-1}{9}= \frac{17}{9}$$ as it will lie in the given range or it can be $$\frac{2*8-1}{8}=\frac{15}{8}$$
Insuff

(2) $$x = 9$$

Combined..
If x= 9, y will be 2*9-1=17. Hence $$\frac{y}{x} = \frac{17}{9}$$ .
The moment we take y one less that is 16....$$\frac{y}{x} = \frac{16}{9}<\frac{13}{7}$$ and when we take y one more that is 18....$$\frac{y}{x} = \frac{18}{9}=2>\frac{19}{10}$$

SUff

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Re: If x and y are integers, is y/x = 17/9? (1) 13/7 < y/x < 19/10 (2) x=9  [#permalink]

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Let m be = y/x so stem ---> is m = 17/9 = 1.8888 ?

1. 1.85 < m < 1.9
m could be 1.8888 or m could be 1.86 or 1.89
Not sufficient

Not sufficient

Combined , 16/9 = 1.777 and 18/9 = 2. Both these values contradict statement 1 .
Hence, we have only one unique value of Y i.e. 17.
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Re: If x and y are integers, is y/x = 17/9? (1) 13/7 < y/x < 19/10 (2) x=9  [#permalink]

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Bunuel wrote:
If x and y are integers, is $$\frac{y}{x} = \frac{17}{9}$$ ?

(1) $$\frac{13}{7} < \frac{y}{x} < \frac{19}{10}$$

(2) $$x = 9$$

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Since we have 2 variables (x and y) and 0 equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

Since $$x = 9$$, we should find the value of $$y$$ such that $$13/7 < y/9 < 19/10$$.
When we multiply all sides of the inequality by $$9$$, we have $$16 < 117/7 < y < 171/10 < 18$$ or$$y = 17$$.

Since both conditions together yield a unique solution, they are sufficient.

Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
If$$x = 9, y = 17$$, then the answer is 'yes'.
If $$x = 17 = 7 + 10, y = 32 = 13 + 19$$, then the answer is 'no'.
Note. When we have $$\frac{a}{b} < \frac{c}{d}$$ for positive integers $$a, b, c$$ and $$d$$, we have $$\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$$.

Since condition 1) does not yield a unique solution, it is not sufficient

Condition 2)
Since we don't have any information about y, condition 2) is not sufficient obviously.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
_________________ Re: If x and y are integers, is y/x = 17/9? (1) 13/7 < y/x < 19/10 (2) x=9   [#permalink] 31 Dec 2019, 10:56
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# If x and y are integers, is y/x = 17/9? (1) 13/7 < y/x < 19/10 (2) x=9  