SOLUTION: Wanda has 40 coins, all nickels, dimes, and quarters. She has 1 less than twice the number of quarters in nickels. If she has $4.45, how many of each type of coin does she have?

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: Wanda has 40 coins, all nickels, dimes, and quarters. She has 1 less than twice the number of quarters in nickels. If she has $4.45, how many of each type of coin does she have?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 755530: Wanda has 40 coins, all nickels, dimes, and quarters. She has 1 less than twice the number of quarters in nickels. If she has $4.45, how many of each type of coin does she have?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +n+ = number of nickels
Let +d+ = number of dimes
Let +q+ = number of quarters
-------------
given:
(1) +n+%2B+d+%2B+q+=+40+
(2) +n+=+2q+-+1+
(3) +5n+%2B+10d+%2B+25q+=+445+ ( in cents )
---------------------------
You have 3 equations and 3 unknowns
so it's solvable
Substitute (2) into (1)
(1) +2q+-+1+%2B+d+%2B+q+=+40+
(1) +d+%2B+3q+=+41+
(1) +d+=+41+-+3q+
Now substitute (1) and (2) into (3)
(3) +5%2A%28+2q+-+1+%29+%2B+10%2A%28+41+-+3q+%29+%2B+25q+=+445+
(3) +10q+-+5+%2B+410+-+30q+%2B+25q+=+445+
(3) +-20q+%2B+25q+=+445+-+405+
(3) +5q+=+40+
(3) +q+=+8+
and, since
(1) +d+=+41+-+3q+
(1) +d+=+41+-+3%2A8+
(1) +d+=+41+-+24+
(1) +d+=+17+
and
(2) +n+=+2q+-+1+
(2) +n+=+2%2A8+-+1+
(2) +n+=+15+
She has 15 nickels, 17 dimes, and 8 quarters
check:
(3) +5n+%2B+10d+%2B+25q+=+445+
(3) +5%2A15+%2B+10%2A17+%2B+25%2A8+=+445+
(3) +75+%2B+170+%2B+200+=+445+
(3) +445+=+445+
OK