SOLUTION: A jar of coins has 48 coins worth $4.50 in it. All the coins are nickels and dimes. Determine how many nickels and dimes are in the jar.

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: A jar of coins has 48 coins worth $4.50 in it. All the coins are nickels and dimes. Determine how many nickels and dimes are in the jar.      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 487951: A jar of coins has 48 coins worth $4.50 in it. All the coins are nickels and dimes. Determine how many nickels and dimes are in the jar.
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
----------
given:
(1) +n+%2B+d+=+48+
(2) +5n+%2B+10d+=+450+
------------------
Multiply both sides of (1) by 5
and subtract (1) from (2)
(2) +5n+%2B+10d+=+450+
(1) +-5n+-+5d+=+-240+
+5d+=+210+
+d+=+42+
and, since
(1) +n+%2B+d+=+48+
(1) +n+%2B+42+=+48+
(1) +n+=+48+-+42+
(1) +n+=+6++
There are 6 nickels and 42 dimes in the jar
check:
(2) +5n+%2B+10d+=+450+
(2) +5%2A6+%2B+10%2A42+=+450+
(2) +30+%2B+420+=+450+
(2) +450+=+450+
OK