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What is the value of 2x/3y?

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by VJesus12 » Wed Jul 04, 2018 2:23 am

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Answers

A

B

C

D

E

Stats

Difficulty—

What is the value of 2x/3y?

(1) x^2/y^2 = 36/25
(2) x^5/y^5 > 1

The OA is the option C.

Why is the first statement not sufficient? Could anyone give me some help, please?
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Source: — Data Sufficiency |
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by Vincen » Wed Jul 04, 2018 5:04 am
Hello Vjesus12.

Let's take a look at your question. We want to find the value of $$\frac{2x}{3y}=?$$

First of all, we need to assume that x and y are integers.

First Statement
(1) x^2/y^2 = 36/25
Here we can get at least two different cases:

(i) x=6 and y=5, then $$\frac{2x}{3y}=\frac{2\left(6\right)}{3\left(5\right)}=\frac{4}{5}.$$
(ii) x=-6 and y=5, then $$\frac{2x}{3y}=\frac{2\left(-6\right)}{3\left(5\right)}=-\frac{4}{5}.$$

Since we get two different answers, then this statement is not sufficient.

Second Statement
(2) x^5/y^5 > 1
This statement just tells us that x and y have the same sign. So, x and y can be any value. Therefore, this statement is not sufficient.

First Statement + Second Statement
(1) x^2/y^2 = 36/25
(2) x^5/y^5 > 1
From both statements, we get that x and y have the same sign, therefore we can get just the two following cases:

(i) x=6 and y=5, then $$\frac{2x}{3y}=\frac{2\left(6\right)}{3\left(5\right)}=\frac{4}{5}.$$
(ii) x=-6 and y=-5, then $$\frac{2x}{3y}=\frac{2\left(-6\right)}{3\left(-5\right)}=\frac{4}{5}.$$

So, we get only one answer. Therefore, using both statements together is sufficient.

I hope it helps.
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by Jay@ManhattanReview » Mon Jul 16, 2018 10:53 pm
VJesus12 wrote:What is the value of 2x/3y?

(1) x^2/y^2 = 36/25
(2) x^5/y^5 > 1

The OA is the option C.

Why is the first statement not sufficient? Could anyone give me some help, please?
We have to get the value of 2x/3y. If we get the unique value of x/y, we get the answer.

Question rephrased: What's the value of x/y?

Let's take each statement one by one.

(1) x^2/y^2 = 36/25

=> x/y = ±6/5. No unique value of x/y; it can be 6/5 or -6/5. Insufficient.

(2) x^5/y^5 > 1

=> x/y > 1. No value of x/y. Insufficient.

(1) and (2) together

From (1), if x/y = 6/5 > 1 and if x/y = -6/5 < 1. From (2), we know that x/y > 1; thus, x/y = 6/5. Sufficient.

The correct answer: C

Hope this helps!

-Jay
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by [email protected] » Tue Jul 17, 2018 7:14 pm
Hi All,

We're asked for the value of 2X/3Y. This question can be solved by TESTing VALUES.

1) (X^2)/(Y^2) = 36/25

IF....
X=6, Y=5, then the answer to the question is 12/15 = 4/5
X=6, Y= -5, then the answer to the question is 12/-15 = -4/5
Fact 1 is INSUFFICIENT

2) X^5/Y^5 > 1

With this inequality, we know that X and Y must be the SAME sign (either both positive or both negative).

IF....
X=6, Y=5, then the answer to the question is 12/15 = 4/5
X=6, Y=1, then the answer to the question is 12/3 = 4
Fact 2 is INSUFFICIENT

Combined, we know...
(X^2)/(Y^2) = 36/25
X^5/Y^5 > 1

Since X and Y must be the SAME sign, we have two immediate examples that 'fit' both Facts....
IF....
X=6, Y=5, then the answer to the question is 12/15 = 4/5
X= -6, Y= -5, then the answer to the question is -12/-15 = 4/5

X and Y could be other values though, but the only pairs of values that fit the first equation will be in the ratio of 6:5. For example....
IF....
X=12, Y=10, then the answer to the question is 24/30 = 4/5
X= -12, Y= -10, then the answer to the question is -24/-30 = 4/5
The answer will ALWAYS be 4/5.
Combined, SUFFICIENT

Final Answer: C

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