Let the number of bike be x and the number of skates be y, then
21x + 20y ≥ 362 . . . (1)
2y = x . . . (2)
Putting (2) into (1), then
21(2y) + 20y ≥ 362
42y + 20y ≥ 362
62y ≥ 362
y ≥ 5.84
The least number of pairs of skates they need to rent each day to make their minimum is 6.
Answer:
its C
Step-by-step explanation:
Both of the problems equal 600
A is the answer, because 3/5=.60 and 37.5/62.5=60
