SOLUTION: Kimberly bought 90 stamps in denominations of 41 cents, 58 cents and 75 cents. To test her daughter, who is taking an algebra class, she said that she bought three times as many 4
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Question 1103997: Kimberly bought 90 stamps in denominations of 41 cents, 58 cents and 75 cents. To test her daughter, who is taking an algebra class, she said that she bought three times as many 41 cent stamps as 58 cent stamps and that the total cost of the stamps was $43.70. How many stamps of each denomination did she buy?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
let a = 41 cent stamps
let b = 58 cent stamps
let c = 75 cent stamps
:
write an equation for each statement
;
"Kimberly bought 90 stamps"
a + b + c = 90
three times as many 41 cent stamps as 58 cent stamps
a = 3b
and that the total cost of the stamps was $43.70."
.41a + .58b + .75c = 43.70
How many stamps of each denomination did she buy?
:
Replace a with 3b in the first equation
3b + b + c = 90
4b + c = 90
c = (90-4b) use this for substitution in the 3rd equation
.41a + .58b + .75c = 43.70
.41a + .58b + .75(90-4b) = 43.70
.41a + .58b + 67.5 - 3b = 43.70
combine like terms
.41a - .58b - 3b = 43.70 - 67.50
.41a - 2.42b = -23.80
replace a with 3b
.41(3b) - 2.42b = -23.80
1.23b -2.42b = -23.80
-1.19b = -23.80
b = -23.80/-1.19
b = +20 ea .58 cent stamps
then
3*20 = 60 ea .41 cent stamps
and
90 - 20 - 60 = 10 ea .75 cent stamps
:
:
Check this in the 3rd equation
.41(60) + .58(20) + .75(10) =
24.60 + 11.69 + 7.50 = 43.70
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