Taylor has 6.70$ in quarters and dimes. If the total number of coins in her wallet is 17 how many quarters does she have
This problem has a typo. I think the number of coins should be 37,
certainly not 17. I'll first do it with 17 to show you that there can be
no way to have $6.70 in quarters and dimes with only 17 coins. If
you think about it, you know that's not possible because if they were all
quarters, 17 quarters is only $4.50
Then below that I'll do it changing the 17 to 37.
Let the number of quarters be x
Then the number of dimes, using
ONE PART = TOTAL MINUS OTHER PART,
is 17-x.
Value Value
Type Number of of
of of EACH ALL
coin coins coin coins
-------------------------------------------
quarters x $0.25 $0.25x
dimes 17-x $0.10 $0.25(17-x)
-------------------------------------------
TOTALS 17 ----- $6.70
The equation comes from the column on the right
0.25x + 0.10(17-x) = 6.70
Get rid of decimals by multiplying every term by 100
25x + 10(17-x) = 670
25x + 170 - 10x = 670
15x + 170 = 670
15x = 500
x = 33.3333333 = the number of quarters
The number of dimes is 17-x or 17-33.3333333 or -16.3333333 dimes
You cannot have a fraction number of quarters and a negative number
of dimes. There is no solution.
-------------------------------------------------------
Now here's the same problem using 37 coins instead of 17 coins.
Let the number of quarter be x
Then the number of dime, using
ONE PART = TOTAL MINUS OTHER PART,
is 37-x.
Value Value
Type Number of of
of of EACH ALL
coin coins coin coins
-------------------------------------------
quarter x $0.25 $0.25x
dime 37-x $0.1 $0.1(37-x)
-------------------------------------------
TOTALS 37 ----- $6.70
The equation comes from the column on the right
0.25x + 0.10(37-x) = 6.70
Get rid of decimals by multiplying every term by 100
25x + 10(37-x) = 670
25x + 370 - 10x = 670
15x + 370 = 670
15x = 300
x = 20 = the number of quarters
The number of dimes is 37-x or 37-20 or 17 dimes
Edwin
