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Believe it or not I graduated with a BSc in Computing Science, but apparently that means nothing after being out of school for a year.

The question is:

$\frac{(c+n)}{(t+n)}=\frac{1}{4}$

Solve for $n$.

My attempt:

$c+n=\frac{1}{4}(t+n)$
$c+n= \frac{1}{4}t+\frac{1}{4}n$
$c+\frac{3}{4}n=\frac{1}{4}t$
$\frac{3}{4}n=\frac{1}{4}t-c$
$n=\frac{4}{3}(\frac{1}{4}-c)$

But when I plug in the numbers I have, that doesn't work out, so there must be a mistake in there somewhere, but I can't seem to spot it.

Help?

The numbers I've got are:

$t=3035$
$c=413$

And the answer I expect for $n$ is $461$.

Adola
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mpen
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    In going to the last line, the $t$ disappeared! You want $n=(4/3)(0.25t-c)$. – André Nicolas Jun 21 '11 at 01:30
  • You assumed the worst about your math skills too soon. The more one writes, the more misteaks there will be. – André Nicolas Jun 21 '11 at 01:40
  • @user6312: Haha... but I like me steaks! I think I always always this bad, overlooking such basic mistakes. Once I screwed up the math so bad that the answer I got in the end was actually correct. – mpen Jun 21 '11 at 01:45

2 Answers2

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I think you just forgot the $t$ in your last step. Should be $n=\frac{4}{3}(\frac{1}{4}t-c)$.

Adola
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Steve
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  • Oops. That was just me forgetting to type it though....that's not the only mistake, is it? Edit: Nevermind. That was the mistake. I'm an idiot :D Math works out now. – mpen Jun 21 '11 at 01:31
  • @Mark: you're not an idiot :) We've all been there, done that... – amWhy Jun 21 '11 at 02:27
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In the last line, $t$ has disappeared.

Gerry Myerson
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