concerning a cheque A man went into a bank to cash a check. In handling over the money the cashier, by mistake, gave him dollars for cents and cents for dollars. He pocketed the money without examining it and on the way home he spent a nickel. Later on examining it, he found that he had twice the amount of money written on the check.He had no money in his pocket before going to the bank. 
How can we solve this problem? I tried by solving it by making equations but it does not work. The answer is given as $31.63$ on which he received $63.31$ and spent a nickel. But other than guessing, is there any way to find the answer of this problem?

Let $D$ be the check's actual number of dollars and $C$ be its actual number of cents.  Then the actual amount of the check, expressed in pennies, is 
$$A=100D+C$$  
The amount the man is given is 
$$G=100C+D$$  
The pertinent equation is 
$$G-5=2A$$  
Can you take it from there?
Added later:  A couple of people correctly admonished me for leaving the hard part of the solution for the OP to do.  That wasn't really my intention; I had miscounted the three equations as having three unknowns.  Let me try to atone for that by suggesting a fairly slick way to get to the final answer.
As others have found, the problem boils down to finding a solution in non-negative integers to the equation
$$98C-199D=5$$
with $C\lt100$.  Since $98$ and $199$ have no common factor, the equation has a unique solution with $C\lt199$.  Moreover, if you can find any integer solution, then you get to the solution with $C\lt199$ by subtracting an appropriate multiple of $199$.  (We might note at this point that there's no guarantee that the solution with $C\lt199$ will actually satisfy $C\lt100$.  If the problem had specified some amount other than a nickel, there might not be a solution.)
The standard way to solve the equation $98C-199D=5$ is to run the Euclidean Algorithm on it.  But let's see if we can eyeball our way more quickly.  The fact that $199$ is close to $2\times98=196$ suggests a clever multiplication of the equation by $2$:
$$196C-199(2D)=10$$
which can be rewritten as
$$199(C-2D)-3C=10$$
If we now note that
$$1990-1980=10$$
we see that
$$C={1980\over3}=660$$
is a solution (with $660-2D=10$ giving an integer value to $D$).  To get it below $199$, we need to subtract the appropriate multiple of $199$:
$$660-3\times199=63$$
The corresponding value of $D$ is now
$${98\times63-5\over199}=31$$
