## 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

#### 1 comment:

Prof. Hubbard said...

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