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?
 A: 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
$${98\times63-5\over199}=31$$
A: 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.
A: 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.)
A: The way to solve any problem of this type is to write some equations. Words are slippery and hard to work with; equations are much better. In this case, using $d$ for the number of dollars on the check and $c$ for the number of cents, you can write expressions for the total amount of money the check was worth, and the later process described: $$ 100c + d - 5 = 2(100d + c) $$
Simplify this equation a bit to get:
$$ 98c - 199 d = 5 $$
Now, one equation in two unknowns isn't enough for us. However we also have the constraints that $c, d < 100$ and $c, d \in \mathbb{N}$, which limit the possible solutions considerably: we need to find a (small positive) multiple of 98 which is 5 more than a multiple of 199. Look up "Diophantine equations" for some hints on how to do this. In particular, linear Diophantine equations (which this is) are well understood. You can use the extended Euclidean algorithm to work this out: it gives you $98(132) - 199(65) = 1$, so $98(660) - 199(325) = 5$. 660 and 325 aren't legal solutions, but we can change them by remembering that $98(199) - 199(98) = 0$, so $$5 = 98(660) - 199(325) = 98(660-3(199)) - 199(325-3(98)) = 98(63) - 199(31).$$ Therefore $c = 63, d = 31$ is a legal solution (in fact, the only one).
A: Nothing new here other than (marginally) formalizing the other approaches:
As in the other questions, the equation is $98c+199d = 5$, with $c \in \{0,...,99\}$.
We have $\gcd(98,199) = 1$ so Bézout's identity says there are $x,y$ such that
$xc+yd = 1$, hence a solution to the above equation is $(5x,5y)$. If there are two solutions, they must differ by some multiple of $(199,98)$, hence the general
solution is $(5x+199k, 5y+98k)$.
The Euclidean algorithm can be used to compute a solution $x=-67,y=-33$, then we need to choose $k$ so that $c$ satisfies the above constraint. It is straightforward to check that $k=2$ is the only $k$ that works.
A: Original cheque amount d dollars and c cents
Bank gave  c dollars and d cents
Remaining  amount c dollars and d-5 cents
This is double of original amount. 
If we take double amount as 2d and 2c and equate dollars and cents we get negative answers. Then we take that there is carry in doubling the amount. Then we take the double amount as 2d+1 and 2c-100.
Now the equations to solve are 2d+1=c and 2c-100=d-5
You get the solution 
