# $10=c+d$ and $c$ is one more than $d$.

I was going through a maths paper and found this question:

$10=c+d$

$c$ is one more than $d$

What is the value of $c$?

It looks very simple (a $2$ mark question) but I cannot simply find the answer.

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The second condition gives you $d=c-1$. Can you solve this now? – David Mitra Jun 6 '13 at 17:52
The fact that the two numbers c and d differ by only 1 is a tip-off that they can't be whole numbers, since one of them would have to be even and the other one odd. But the sum of an even and an odd integer can't be an even integer. – RecklessReckoner Jun 6 '13 at 17:54
I think that you may have been trying to find integers (whole numbers) that add to $10$, but differ by one. That's impossible. Two consecutive integers (integers that differ by exactly $1$) will always add to an odd integer: one will have to be even, the other odd. – amWhy Jun 6 '13 at 18:01
Solve $10=2d+1$. – copper.hat Jun 6 '13 at 18:02
This question ended up in the 'hot questions' of StackExchange only because of its title :) Oh... math people. – Youcha Jun 6 '13 at 20:19

$c=1+d$ so

$10=c+d=d+d+1$ so

$9=2d$ and hence $d=4.5,c=5.5$.

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Try $c=d=5$. This clearly doesn't quite work. Now slowly lower the value $d$ while raising the value of $c$ until they are one apart, keeping the total at $10$.

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As above this can be solved immediately. – William Jun 6 '13 at 19:07
@Jason indeed, but this answer seeks to give an intuition about such questions, and I think that is valid. Understanding is worth so much more than calculation in the long run. Without calculation, understanding has no roots ... – Mark Bennet Jun 6 '13 at 19:16
To the down voter - I can guarantee you that I solved OP's math problem quicker mentally this way than by using substitution and a system of equations, mentally or otherwise. – alex.jordan Jun 6 '13 at 20:34
@alex.jordan I wasn't the downvoter, but I have to disagree. There's a lot of assumptions here, that only work because this particular set of equations is relatively simple, but are still very easy to stumble over. And yes, even if I knew to inc/dec by 0.1 beforehand, it would have taken me many times as long with guess-and-check than it would have with simply solving the equations as shown in @James's answer. (I'd've used whole numbers and bumbled around trying to find the sweet spot..) – Izkata Jun 6 '13 at 21:02
I used the same mental process, and +1 for that. – fluffy Jun 6 '13 at 21:03

We have essentially two equations and two unknowns to solve: The fact that "$c$ is one more than $d$" tells you $\;c - d = 1.$

\begin{align} \;\;c + d & = 10 \\ +\;\;c -d & = \;\;1 \\ \hline \\ 2c + 0 & = 11 \end{align}

$$2c = 11 \iff c = 5.5$$ $$c + d = 10\iff d = 10 - c = 10- 5.5 = 4.5$$

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nice answer badge on the way! :-) +1 – Amzoti Jun 7 '13 at 0:30
$\small{Small}$ but $\large{Enough}$ – Babak S. Jun 7 '13 at 7:40
Thanks, @Babak! Nice scripts! – amWhy Jun 9 '13 at 3:00

The answer is fairly easy to see once you notice that $c = d +1$ as indicated by:

$c$ is one more than $d$

So you have:

$10 = d+1+d$

Simplifying we have $10 = 2d + 1$ From this we can see that: $d=4.5$ because

\begin{align} 10-1 = 2d \\ \frac{9}{2} = d \end{align}

or in other words $d= 4.5$.

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+1 that's what I noticed, I even rewrote the whole thing to say this: 10 = c + d where (c = d + 1) it was pretty much an auto conversion in my head. – ryan Jun 6 '13 at 21:02

Here is another way to solve it, a bit different from the answers already discussed. It is slightly overcomplicated for this particular problem, but the canonical solutions have already been given in other answers, so I include this to illustrate a general method in case you run into other things like this (or which may be fun for you to read). The following is how to do it using linear algebra.

You can rewrite your question as $$\begin{eqnarray*}10&=&c+d\\ c&=&d+1\end{eqnarray*}$$ which is equivalent to $$\begin{eqnarray*}c+d&=&10\\ c-d&=&1\end{eqnarray*}$$ This is a system of linear equations, so what we want is to solve the matrix equation $$\left(\begin{array}{cc}1&1\\1&-1\end{array}\right)\left(\begin{array}{c}c\\d\end{array}\right)=\left(\begin{array}{c}10\\1\end{array}\right).$$ To do this, we make an augmented matrix $$\left(\begin{array}{cc|c}1&1&10\\1&-1&1\end{array}\right)$$ and use Gaussian elimination: $$\left(\begin{array}{cc|c}1&1&10\\1&-1&1\end{array}\right)\rightarrow \left(\begin{array}{cc|c}1&1&10\\0&-2&-9\end{array}\right)\rightarrow \left(\begin{array}{cc|c}1&1&10\\0&1&9/2\end{array}\right)\rightarrow\left(\begin{array}{cc|c}1&0&11/2\\0&1&9/2\end{array}\right).$$ This gives us that $c=11/2$, $d=9/2$ is the unique solution for $c$ and $d$, which of course are equal to $5.5$ and $4.5$, respectively.

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This set of answers is getting out of hand! – Sharkos Jun 7 '13 at 2:35
@Sharkos My answer is not a duplicate, if that is what you're implying. I'm casting the problem in a different light to possibly teach the OP something new. – Alexander Gruber Jun 7 '13 at 2:38
Not at all, it's just that it seems unlikely the OP knows about matrices let alone Gaussian elimination! – Sharkos Jun 7 '13 at 2:45
@Sharkos Well, yeah, exactly. Now maybe he'll want to learn about them. – Alexander Gruber Jun 7 '13 at 2:52
Okay, friend! The links help. – Sharkos Jun 7 '13 at 9:53

Another one, why not...

The mean of $c$ and $d$, $m$, is $\frac{c+d}{2}$. So

$$m = 10/2 = 5$$

$c$ being one more than $d$, along with the symmetry of the mean in $c$ and $d$ implies that:

$$c = m+\frac{1}{2}$$ $$d = m-\frac{1}{2}$$

so, $c=5\frac{1}{2}$ and $d=4\frac{1}{2}$.

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## protected by Alexander Gruber♦Jun 7 '13 at 2:28

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