My question:

A cashier has a total of $114$ bills, made up of tens and hundreds. The total value of the money is $\$6450.00$. How much of each kind does he have?

He has [BLANK] tens and [BLANK] hundreds.

I'm pretty sure there is an algebra equation for this. There is the hard way of doing it on paper (that takes long) or the short way of putting it equation.



If all 114 were 10's then their sum is 1140, which is 5310 too low. So we need to swap some 10's for 100's. Each swap adds 90, so we need 5310/90 = 59 swaps, yielding 59 100's, $\:$ hence 114 - 59 = 55 10's.


Let $T$ be the number of tens, and let $H$ be the number of hundreds. You want to find integer solutions to the equation $$ 10T + 100H = 6450.$$ This is what is called a linear Diophantine equation.

Added: And in fact, you have another piece of information which I missed on my first read-through, as noted by jericson: $$H+T = 114.$$ Once you have the two equations, this is no longer a Diophantine equation problem, but a simple system of two linear equations in two unknowns.

  • 2
    $\begingroup$ And $H +T =114$ ;) $\endgroup$ – jericson Oct 3 '10 at 23:33
  • $\begingroup$ The machinery for solving linear Diophantine equations isn't even needed here, thankfully. :) Simply solving the two equations in two unknowns works. $\endgroup$ – J. M. is a poor mathematician Oct 3 '10 at 23:40
  • $\begingroup$ Yup; I missed that condition on the first read through. $\endgroup$ – Arturo Magidin Oct 3 '10 at 23:41
  • $\begingroup$ So.. what do I do exactly? :\ $\endgroup$ – Dan Oct 4 '10 at 0:15
  • $\begingroup$ You don't know how to solve a system of two linear equations with two unknowns? Use one equation to "solve for one variable" in terms of the other; substitute into the other equation and solve, then solve for the other variable. $\endgroup$ – Arturo Magidin Oct 4 '10 at 1:40


Once you have $$H+T=114$$ and $$10T+100H=6450$$ as above, write $$H=114-T$$

Then substitute this into the second equation above


Expand, and simplify.

  • $\begingroup$ You mean distribute... right? $\endgroup$ – Dan Oct 4 '10 at 1:10
  • $\begingroup$ I am getting lost. .. $\endgroup$ – Dan Oct 4 '10 at 1:41
  • $\begingroup$ Dan - $a(b+c)=ab+ac$, so... $\endgroup$ – Juan S Oct 4 '10 at 2:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy