SOLUTION: A box contains $6.75 in nickels, dimes,and quarters. There are 40 coins in all, and the sum of the numbers of nickels and dimes is 2 less than the number of quarters. How many coin
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Question 1082619: A box contains $6.75 in nickels, dimes,and quarters. There are 40 coins in all, and the sum of the numbers of nickels and dimes is 2 less than the number of quarters. How many coins of each kind are there?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
let n = no. of nickels
let d = no. of dimes
let q = no. of quarters
:
Write an equation for each statement
:
A box contains $6.75 in nickels, dimes,and quarters.
.05n + .10d + .25q = 6.75
;
There are 40 coins in all,
n + d + q = 40
:
and the sum of the numbers of nickels and dimes is 2 less than the number of quarters.
n + d = q - 2
n + d - q = -2
subtract from the 1st equation
n + d + q = 40
n + d - q = -2
----------------Subtraction eliminates n and d, find q
0 + 0 + 2q = 42
q = 42/2
q = 21 quarters
:
Create two 2 unknown equations by replacing q with 21
n + d + 21 = 40
n + d = 19
and
.05n + .10d + .25 (21) = 6.75
.05n + .10d = 6.75 - 5.25
.05n + .10d = 1.50
multiply by 10
.5n + 1d = 15.00
Subtract from the 1st 2 unknown equation
n + d = 19
.5n + d = 15
------------------eliminates d
.5n = 4
n = 8 nickels
:
Find d
8 + d = 19
d = 11 dimes
:
How many coins of each kind are there? 8 nickels, 11 dimes, 21 quarters
:
:
Check this for yourself: .05(8) + .10(11) + .25(21) =
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