SOLUTION: A bank teller notices that he has 50 coins all of which are 5c and 10c pieces. He finds that the value of the coins is $4.20. How many of each must he have?
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Question 751866: A bank teller notices that he has 50 coins all of which are 5c and 10c pieces. He finds that the value of the coins is $4.20. How many of each must he have?
You can put this solution on YOUR website! x = 5c: y = 10c
x + y = 50.......1
5x + 10y = 420......2
Multiply (1) by 10
10x + 10y = 500.....3
5x + 10y = 420.....2
Subtract 2 from 3
5x = 80
x = 16
Substitute x = 16 into Eqn (1)
16 + y = 50
y = 50 - 16
y = 34
Therefore there are 16 x 5c and 34 x 10c
Hope this helps.
:-)
You can put this solution on YOUR website! there are 50 coins altogether
let n = the amount of 5c pieces and d the 10c pieces
so n+d =50
rearranging
d=50-n
now the value of n = 5 (cents), and d =10 (cents)
so 5n +10d = 420 (cents)
substituting d=50-n instead of d in the equation
5n + 10 (50-n) = 420
multiplying out and rearranging you get
80/5 =n
so n = 16
therefore d = 34
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