A container is $1/8$ full of water. After $10$ cups of water are added, the container is $3/4$ full. What is the volume of the container, in cups?

Ok, I wrote out an equation: $$\frac{1}{8}V + 10C = \frac{3}{4}V$$ I realized that it was asking for the volume of the container in CUPS.

Then I changed it to $\frac{1}{8}C + 10 = \frac{3}{4}C$ since I want to find the total volume of the container.

Then the answer came out to be $$10 = \frac{5}{8}C$$ in which $C$ is equal to $16$.

Is this the right way of solving the problem?

  • 3
    $\begingroup$ You can check to see if this result makes sense. One-eighth of 16 cups is 2 cups, the starting volume of water in the container. Adding 10 cups brings that to 12 cups in the container. Twelve cups out of sixteen is $ \ \frac{12}{16} \ = \ \frac{3}{4} \ $ . So the container has been brought to three-quarters full. [It is considered preferable not to write $ \ C \ $ next to the "10" since it could be misunderstood as a variable. You can either leave it out with the understanding that the volume units are cups, or write $ \ \frac{1}{8}V + 10 \ c. \ = \ \frac{3}{4}V \ $ . ] $\endgroup$ Aug 31, 2015 at 5:50
  • $\begingroup$ In each of the two equations, $C$ means a different thing, which is confusing. My middle school teacher made me write something like "Let $C$ be the volume of the container in cups" on the paper whenever I introduced a new variable on a test/homework. I suppose it helps you remember what your variables mean, so that you don't switch them around in the middle. In any case, the container does have $16$ cups of volume. $\endgroup$ Aug 31, 2015 at 18:35
  • $\begingroup$ As a check: It says the container is $\frac18$ full of water at the start, so it has $16\cdot\frac18=2$ cups. You add $10$ cups, so it now has $10+2=12$ cups total. Is the container now $\frac34$ full? Yes, since $16\cdot\frac34=12$. $\endgroup$ Aug 31, 2015 at 18:39

3 Answers 3


The two equations were both basically equivalent, but in the first, you were dealing with volumes $V$ with unspecified unit, and you were explicitly adding $10$ of the unit "cups"

In the second though, you were now dealing with $C$ where $C$ is the volume of the vessel in cups.

Ultimately, it doesn't matter. If you solve the first, you will get $V=16C$, i.e. that the volume is equal to $16$ times the volume of one Cup, which is $16$ cups; the same as for the second, where you got the volume, in cups, as $16$.

In summary, yes, this is correct.


I disagree with your transition from one equation to the next. You go from:

$\frac{1}{8}V + 10C = \frac{3}{4}V$


$\frac{1}{8}C + 10 = \frac{3}{4}C$

So your first equation has units on everything, and now your second equation has a mysterious $10$ that has no units. Moreover, your second equation is saying "$\frac{1}{8}$ cups plus 10 equals $\frac{3}{4}$ cups". This doesn't make any sense in the context of the problem, nor is it the problem you're trying to solve. The transition to this second equation doesn't really help you. Part of the issue is exemplified by your statement:

...in which $C$ is equal to $16$

But $C$ is not a variable(ie not a number); it's the unit you chose for cups.

You could just as easily continue with your first equation:

$10C = \frac{3}{4}V - \frac{1}{8}V$

$10C = \frac{5}{8}V$

From here you can solve for $V$ by multiplying each side by $\frac{8}{5}$ to give you:

$16C = 1V$

Which can be read as "16 cups is equal to [one]the volume of the container", which is exactly what the question was asking for. You ended up getting the right number, but I think you mixed up the idea of units along the way.


$$10\text{ cups }=\frac{3}{4}-\frac{1}{8}=\frac{6-1}{8}=\frac{5}{8}\text{ V} \implies \frac{1}{8}\text{ V} = 2\text{ cups}\implies V=8\cdot2=16\text{ cups} $$


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