Tuesday, June 30, 2015
100 coin (or ticket) problems
a) 100 coins, all quarters and pennies, total = $13.72
b) 100 tickets, all children ($6) and adult ($12), total =$912
c) 100 coins, all quarters and nickels, total = $8.20
d) 100 coins, all dimes and pennies, total = $3.79
e) 100 coins, all dimes and nickels, total = $7.85
f) 100 coins, all quarters and dimes, total = $14.65
Answers in the comments
Subscribe to:
Post Comments (Atom)
1 comment:
a) 100 coins, all quarters and pennies, total = $13.72
p + q = 100
p + 25q = 1372
subtracting the first equation from the second we get
24q = 1272
q = 1272/24 = 53
53 quarters, 47 pennies
b) 100 tickets, all children ($6) and adult ($12), total =$912
c + a = 100
6c + 12a = 912
Mulitply the first equation by 6 to get
6c + 6a = 600
subtract this from the second equation to get
6a = 312
a = 312/6 = 52
52 adults, 48 children
c) 100 coins, all quarters and nickels, total = $8.20
n + q = 100
5n + 25q = 820
5 times the first equation
5n + 5q = 500
subtract from second equation to get
20q = 320
q = 320/20 = 16
16 quarters, 84 nickels
d) 100 coins, all dimes and pennies, total = $3.79
p + d = 100
p + 10d = 379
subtract first from second to get
9d = 279
d = 279/9 = 31
31 dimes, 69 pennies
e) 100 coins, all dimes and nickels, total = $7.85
d + n = 100
10d + 5n = 785
multiply first equation by 5 to get
5d + 5n = 500
subtract from 2nd equation
5d = 285
d = 285/5 = 57
57 dimes, 43 nickels
f) 100 coins, all quarters and dimes, total = $14.65
q + d = 100
25q + 10d = 1465
10q + 10d = 1000
subtract.
15q = 465
q = 465/15 = 31
31 quarters, 69 dimes
Post a Comment