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# If x and y are positive integers, does x/y = 7/9? (1) x = 103 (2) 11/

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If x and y are positive integers, does x/y = 7/9? (1) x = 103 (2) 11/  [#permalink]

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23 Sep 2016, 05:21
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Question Stats:

41% (02:25) correct 59% (02:10) wrong based on 57 sessions

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If x and y are positive integers, does x/y = 7/9?

(1) x = 103
(2) 11/13 < x/y < 8/9

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Re: If x and y are positive integers, does x/y = 7/9? (1) x = 103 (2) 11/  [#permalink]

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23 Sep 2016, 05:37
1
duahsolo wrote:
If x and y are positive integers, does x/y = 7/9?

(1) x = 103
(2) 11/13 < x/y < 8/9

Statement 1 : 103 is a prime number. So, we are 100% sure that whatever the value of y is , we are not going to get 7 at the numerator. Hence, Sufficient.

Statement 2 : 11/13 < x/y < 8/9

We can also write this equation as

99/117 < x/y < 104/117

and 7/9 can be written as 91/117. hence, it will never lie between 99/117 and 104/117. Hence, Sufficient.
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Re: If x and y are positive integers, does x/y = 7/9? (1) x = 103 (2) 11/  [#permalink]

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23 Feb 2017, 05:03
Prompt analysis
x and y are positive integers

Superset
The answer will be either yes or no

Translations
In order to find the value of x/y, we need:
1# exact value of x and y or x/y
2# some relation between x and y

Statement analysis

St 1: x =103, since we don't have the value of y, we cannot say anything. INSUFFICIENT
St 2: 11/13 =0.846, 8/9 = 0.88888, 7/9 = 0.7777. Since x/y lies between 0.846 and 0.888. It definitely is not 7/9.ANSWER

Option B
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If x and y are positive integers, does x/y = 7/9?  [#permalink]

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14 Jan 2018, 06:10
1
Bunuel wrote:
If x and y are positive integers, does x/y = 7/9?

(1) x = 103
(2) 11/13 < x/y < 8/9

Statement 1: $$x=103$$ which is not a multiple of $$7$$. So irrespective of the value of $$y$$, ratio of $$\frac{x}{y}$$ cannot be $$\frac{7}{9}$$. Sufficient.

Statement 2: Here we need to analyse whether $$\frac{7}{9}$$ is greater than $$\frac{11}{13}$$. if Yes then there is a possibility of the ratio to be $$\frac{7}{9}$$ and if no then there is no possibility for $$\frac{x}{y}=\frac{7}{9}$$

so, $$\frac{7}{9}=\frac{7*13}{9*13}$$ & $$\frac{11}{13}=\frac{11*9}{13*9}$$

Clearly $$\frac{11}{13}>\frac{7}{9}$$ so $$\frac{x}{y}$$ cannot be $$\frac{7}{9}$$. Sufficient

Option D
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Re: If x and y are positive integers, does x/y = 7/9?  [#permalink]

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15 Jan 2018, 05:55
Sorry , but i think its not the way wwe evaluate choices in DS questions, as u did in statement 2

If statement leads to multiple solutions then we reject that statement

Correct me if im wrong , thanks in advance

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If x and y are positive integers, does x/y = 7/9?  [#permalink]

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15 Jan 2018, 06:26
AshutoshB wrote:
Sorry , but i think its not the way wwe evaluate choices in DS questions, as u did in statement 2

If statement leads to multiple solutions then we reject that statement

Correct me if im wrong , thanks in advance

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Hi AshutoshB

If you read the question stem carefully then you would realize that it is a "Is/Does" question type that is we need a clear "Yes" or a clear "No" from the statements. If we get sometimes yes and sometimes no, then we reject the statement. we do not need a particular value of x here we only need a confirm possibility

Now as per your understanding is there any possibility for x/y=7/9 from statement 2?
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If x and y are positive integers, does x/y = 7/9?  [#permalink]

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Updated on: 15 Jan 2018, 10:54
Bunuel wrote:
If x and y are positive integers, does x/y = 7/9?

(1) x = 103
(2) 11/13 < x/y < 8/9

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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider 1) & 2) first, since we can save time by first checking whether conditions 1) and 2) are sufficient, when taken together.

Conditions 1) & 2)

We don't have an integer solution of the equation $$\frac{x}{y} = \frac{7}{9} ⇔ 9x = 7y ⇔ 9 \cdot 103 = 7y$$ if $$x = 103$$.
Both conditions together 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)
We can apply the same logic for both conditions together to the condition 1).
We don't have a solution and the answer is "no", The condition 1) is sufficient by CMT (Common Mistake Type) 1, since the answer "no" also means the condition is sufficient.

Condition 2)
Since 7/9 < 11/13, x between 11/13 and 8/9 cannot be equal to 7/9.
The condition 2) is sufficient by CMT (Common Mistake Type) 1, since the answer "no" also means the condition is sufficient.

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.
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Originally posted by MathRevolution on 15 Jan 2018, 10:25.
Last edited by MathRevolution on 15 Jan 2018, 10:54, edited 1 time in total.
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Re: If x and y are positive integers, does x/y = 7/9?  [#permalink]

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15 Jan 2018, 10:32
MathRevolution wrote:
Bunuel wrote:
If x and y are positive integers, does x/y = 7/9?

(1) x = 103
(2) 11/13 < x/y < 8/9

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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider 1) & 2) first, since we can save time by first checking whether conditions 1) and 2) are sufficient, when taken together.

Conditions 1) & 2)

We don't have an integer solution of the equation $$\frac{x}{y} = \frac{7}{9} ⇔ 9x = 7y ⇔ 9 \cdot 103 = 7y$$ if $$x = 103$$.
Both conditions together 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)
We can apply the same logic for both conditions together to the condition 1).
We don't have a solution and the answer is "no", The condition 1) is sufficient by CMT (Common Mistake Type) 1, since the answer "no" also means the condition is sufficient.

Condition 2)
$$\frac{11}{13} < \frac{x}{y} < \frac{8}{9}$$
$$\frac{11 \cdot 9}{13 \cdot 9} < \frac{x}{y} < \frac{8 \cdot 13}{9 \cdot 13}$$
$$\frac{99}{117} < \frac{x}{y} < \frac{104}{117}$$
We have lots of solutions such as $$x = 100, y = 117$$ or $$x = 101, y = 117$$.
Since we don't have a unique solution, the condition 2) is not sufficient.

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.

Hi MathRevolution

from statement 2 it is clear that x/y will never be equal to 7/9. Hence the statement is sufficient to give a clear NO to our question stem.

IMO the answer should be D
Re: If x and y are positive integers, does x/y = 7/9? &nbs [#permalink] 15 Jan 2018, 10:32
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