- #1
bergausstein
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A wallet has $\$460$, in $\$5$, $\$10$, and $\$20$ bills. the number of $\$5$ bills exceeds twice the number of $\$10$ bills by $4$, while the number of $\$20$ bills is 6 fewer than the number of $\$10$ bills. how many bills of each type are there?
i solved this problem in two ways by 1st choosing the unknown represent the number of $\$10$ bills and 2nd by choosing the unknown represent the number of $\$20$ bills.
and i got these answers from my two methods, $14$ $\$10$ bills, $8$ $\$20$ bills and $32$ $\$5$ bills.
on my third method i chose the unknown represent the number of $\$5$ bills.
and here how it goes,
let $x=$ number of $\$5$ bills,
$\frac{x}{2}-4=$ number of $\$10$ bills
$\frac{x}{2}-4-6=$ number of $\$20$ bills
$5x+10\left(\frac{x}{2}-4\right)+20\left(\frac{x}{2}-10\right)=460$
$5x+5x-40+10x-200=460$
$20x-240=460$
$20x=700$
then, $x=35$ ---> from here the number of $\$5$ bill is bigger than my previous result. can you pinpoint where is the mistake here? thanks!
p.s use only one variable.
i solved this problem in two ways by 1st choosing the unknown represent the number of $\$10$ bills and 2nd by choosing the unknown represent the number of $\$20$ bills.
and i got these answers from my two methods, $14$ $\$10$ bills, $8$ $\$20$ bills and $32$ $\$5$ bills.
on my third method i chose the unknown represent the number of $\$5$ bills.
and here how it goes,
let $x=$ number of $\$5$ bills,
$\frac{x}{2}-4=$ number of $\$10$ bills
$\frac{x}{2}-4-6=$ number of $\$20$ bills
$5x+10\left(\frac{x}{2}-4\right)+20\left(\frac{x}{2}-10\right)=460$
$5x+5x-40+10x-200=460$
$20x-240=460$
$20x=700$
then, $x=35$ ---> from here the number of $\$5$ bill is bigger than my previous result. can you pinpoint where is the mistake here? thanks!
p.s use only one variable.