Problem of the Day


Let x and y be real numbers. If 8x+7y=30, what is the value of x?

(1) 3x+5y=16
(2) x and y are positive integers.

Reveal Answer

Answer

D. Each statement alone is sufficient.

See the Solution

Solution

[latexpage]

The given information tells us that $x$ and $y$ are real numbers and that $8x+7y=30$. As it stands, there are infinitely many possible values for $x$. Let’s examine the statements.

Statement 1: SUFFICIENT. Knowing that $3x+5y=16$ allows us to set up a system of equations that we can solve to determine that $x=2$ and $y=2$.

Statement 2: SUFFICIENT. At first it may appear that this doesn’t give us much to work with. However, you should notice that restricting the values of $x$ and $y$ to the set of positive integers severely restricts the possibilities for the values of both variables. $x$ certainly must be less than 4, because otherwise $7y$ would have to be negative. Thus, $x$ can only be 1, 2, or 3. Now, let’s examine each of these possibilities:

$x=1 \rightarrow 8(1)+7y=30 \rightarrow 7y=22 \rightarrow y=\frac{22}{7}$. This contradicts the fact that $y$ must be an integer, so $x$ cannot equal 1.

$x=2 \rightarrow 8(2)+7y=30 \rightarrow 7y=14 \rightarrow y=2$. This works.

$x=3 \rightarrow 8(3)+7y=30 \rightarrow 7y=6 \rightarrow y=\frac{6}{7}$. Again, $y$ fails to be an integer, so $x$ cannot equal 3.

Thus, when $x$ and $y$ are both positive integers, the only possibility is that $x=2$ and $y=2$, making statement #2 sufficient on its own.


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