SOLUTION: A cashier goes to the bank to get $10 worth of change. He requests twice as many quarters as half dollars, twice as many dimes as quarters, three times as many nickels as dimes,

Question 231061: A cashier goes to the bank to get $10 worth of change. He requests twice as many
quarters as half dollars, twice as many dimes as quarters, three times as many
nickels as dimes, and no pennies or dollars. How many of each coin did the cashier
get?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
f = number of dollars
h = number of half dollars
q = number of quarters
d = number of dimes
n = number of nickels
p = number of pennies.

total money equals $10.00

f * 1 = amount of money in dollars
h * .5 = amount of money in half dollars
q * .25 = amount of money in quarters
d * .1 = amount of money in dimes
n * .05 = amount of money in nickels
p * .01 = amount of money in pennies

Total money collected equals $10.00

1*f + .5*h + .25*q + .10*d + .05*n + .01*p = 10

f and p are equal to 0, so this equation becomes:

1*0 + .5*h + .25*q + .10*d + .05*n + .01*0 = 10

Simplify this equation to get:

.5*h + .25*q + .10*d + .05*n = 10

Since:

q = 2*h (number of quarters equals 2 times the number of half dollars)
d = 2*q (number of dimes equals 2 times the number of quarters)
n = 3*d (number of nickels equals 3 times the number of dimes)

Replace n with 3*d to get:

.5*h + .25*q + .10*d + .05*3*d = 10

Replace d with 2*q to get:

.5*h + .25*q + .10*2*q + .05*3*2*q = 10

Replace q with 2*d to get:

.5*h + .25*2*h + .10*2*2*h + .05*3*2*2*h = 10

Simplify to get:

.5*h + .5*h + .4*h + .6*h = 10

Combine like terms to get:

2*h = 10

Divide both sides by 2 to get:

h = 5

Since q = 2*h, then q = 10
Since d = 2*q, then d = 20
Since n = 3*d, then n = 60

You have:

h = 5 * .5 = $2.50
q = 10 * .25 = $2.50
d = 20 * .1 = $1.00
n = 60 * .05 = $3.00
Total = $10.00 which is correct.

Cashier gets:

5 half dollars
10 quarters
20 dimes
60 nickels

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