$10=c+d$ 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.
 A: 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 $d= 4.5$.
A: 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.
A: $c=1+d$ so 
$10=c+d=d+d+1$ so
$9=2d$ and hence $d=4.5,c=5.5$.
A: 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 $10$.
A: 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} 
$$2c = 11 \iff c = 5.5$$ $$c + d = 10\iff d = 10 - c = 10- 5.5 = 4.5$$
A: 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 $d=4\frac{1}{2}$.
