concerning a cheque

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

$${98\times63-5\over199}=31$$

Well let the number of Dollars in the amount he should have got be $D$ and the number of Cents $C$.

He should have got $100D+C$. He actually got $100C+D$. After his spending spree he had $100C+D-5$ and this was equal to $200D+2C$

Let's assume for the moment that $D\ge 5$ and $2C\lt 100$ so there are no carries.

Then we have $100C=200D$ or $C=2D$ equating the dollars, and $D-5=2C$ equating the cents, and that makes $3D=-5$ which is no solution.

So there has to be a carry. Let's assume now that $2C\ge 100, D\ge 5$

That gives us $C=2D+1$ for the dollars, and $D-5=2C-100=2(2D+1)-100$ for the cents, from which we obtain $D=31, C=63$ and check that $2C\ge 100$ as required.

The cases with $D\lt 5$ can be. checked similarly.

Let $D$ be the number of dollars and $C$ be the number of cents that man actually got. The amount he got is then $100D+C$ cents, and the amount he wanted was $100C+D$ cents. After spending the nickel, he had $100D+C-5$ cents, and we know $$100D+C-5 = 2(100C+D)$$ which we can rearrange to $$98D-5 = 199C.$$

The left-hand side is odd, so the right-hand side must be also, so $C$ is odd.

The left-hand side is at most $98\cdot 99 -5 = 9697$, so $C \le \frac{9697}{199} < 49$.

Now consider both sides mod 49. We get $$-5 \equiv 3C\pmod{49}$$

(Because $98D\equiv 0\pmod{49}$ and $199\equiv 3\pmod{49}$.)

We should find out what value will work on the left-hand side; it must be equivalent to $-5\pmod{49}$. Since the right-hand side is a multiple of 3, the left-hand side must be also. $-5$ is not a multiple of 3, and neither is $-5+49 = 44$, but $-5+49+49 =93$ works; we get equality if we take $C=31$. Similarly $-5+5\cdot 49 = 240$ on the left gives $C=31+49=80$. But we already know that $C=80$ is too big and not odd, so $C=31$ is the only solution. Then we have $$98D-5 = 199\cdot 31$$ which gives us $D=63$.

So the answer is: The man asked for $\$31.63$, received $\$63.31$, and after spending a nickel had $\$62.26 = 2\cdot\$31.63$.

(If you would like me to explain any part of this in more details, please leave a comment.)