Data Sufficiency

What is the ratio of x to y?

1) 0.8x + 0.5y = 40

2) 0.625y = 50 - x

My answer is C, but I'm not sure! Could any of you, guys, explain this question to me?

Thanks!

x/y ?

1. 0.8x+0.5y=40

not suff

2.0.625y=50-x

not suff

combine 1 n 2 solve for get x an y ratio hence C

Shiram wrote:What is the ratio of x to y?

1) 0.8x + 0.5y = 40

2) 0.625y = 50 - x

My answer is C, but I'm not sure! Could any of you, guys, explain this question to me?

Thanks!

[1] From a linear equation in two variables of the kind a x + b y + c = 0, the ratio in x and y can be answered only if c = 0, in here c ≠ 0, insufficient

[2] From a linear equation in two variables of the kind p x + q y + r = 0, the ratio in x and y can be answered only if r = 0, in here r ≠ 0, insufficient

Combined, we get a system of linear equations in two variables, which can be solved out for unique x and y values.

[spoiler]C[/spoiler]

Great stuff, Ian - and a really good question. I'd just add that one of my favorite things about Data Sufficiency is that it's a logic game...almost a chess match between you and the author of the question. Based on that, I'm always pretty leery of bait like:

Two variables.

Statement 1 gives one equation.

Statement 2 gives another equation.

With a setup like this, they're trying hard to get you to just take C and move on. But if "both statements together" seems too straightforward, there's an excellent chance that it is, so you should check for:

-Does each statement actually provide NEW value? Or are they, like in this case, just different ways of saying the same thing?

-Does either statement involve a quadratic or any other twist that makes it non-linear, and therefore it would require a third statement to actually solve the question?

-Is there additional information embedded in the question stem or even just the logic of the question that would allow you to solve the question with only one of the statements? For example, if you're dealing with a situation that requires integers or positive numbers ("how many people..."), that may allow you to solve for two variables with just one equation.


Beware the "Easy C",as this problem definitely demonstrates.

Extremely regretful for taking a decision in such a slapdash fashion in spite of already meeting with so many annoying surprises on DS. Thanks Ian