Kimberly spent
311 of her money on 4 cookies and 6 wafers. If each cookie costs thrice as much as a wafer, how many wafers could Kimberly buy with the rest of her money?
| |
Cookies |
Wafers |
| Number |
4 |
6 |
| Value |
3 u |
1 u |
| Total value |
12 u |
6 u |
Cost of 1 wafer = 1 u
Cost of 1 cookie = 3 u
Total cost of 4 cookies and 6 wafers
= 4 x 3 u + 6 x 1 u
= 18 u
Fraction of Kimberly's money left
= 1 -
311 =
811311 of Kimberly's money = 18 u
111 of Kimberly's money = 18 u ÷ 3 = 6 u
811 of Kimberly's money = 6 u x 8 = 48 u
Number of wafers that Kimberly could buy
= 48 u ÷ 1 u
= 48
Answer(s): 48
