Question

A parking meter contains nickels and dimes worth ​$3.35. There are 47 coins in all. Find how many of each there are?

Answer 1

Method 1 of 2

20 dimes and 27 nickels

Explanation:

10 dimes = $1
20 nickels = $1

If all coins are dimes then the total value is: `$47/10=$4.70`

If all coins are nickels then the total dollar is `$47/20 =$2.35`

Target value is `$3.35`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

If you consider just the count of dimes then the count of nickels is directly implied as `"count of nickels "=47 - "count of dimes"`

Using rates of change in total value relating to count of dimes like the gradient in a straight line graph.

`("change in value")/("count of dimes") = (4.70-2.35)/(47-0) = 2.35/47`

Rate of change of part is the same as rate of change for all

`("change in value")/("count of dimes") = 2.35/47 = ("Target-start")/x = (3.35-2.35)/x`

`("change in value")/("count of dimes") =2.35/47=1/x`

`"count in dimes "=x=47/2.35 = 20`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check:

dimes `->20/10=$2.00`
nickels`->27/20=ul($1.35 larr" Add")`
`color(white)("ddddddddddddd")$3.35" "`as required

Answer 2

Method 2 - This is how most people would solve it.

dimes count `=20`
nickel count `=27`

Explanation:

10 dimes = $1
20 nickels = $1

Let the count of dimes be `x`

Then the count if nickels is `47-x`

convert dimes to dollars `->x xx1/10`
convert nickels to dollars `->(47-x)xx1/20`

Given that the combined value (target) is `$3.35` we have:

`x/10+1/20(47-x)=3.35`

`x/10+2.35-x/20=3.35`

`(2x-x)/20=3.35-2.35`

`x=20(1)`

dimes count `=x=20`
nickel count `=47-20=27`

All as in method 1