Morris the cat finds a 100$ bill in an alley way and decides to go shopping at Rodent Warehouse. He wants to purchase white mice, and brown hamsters. The mice are 3$ each, and the hamsters are 8$ each. If he wants to buy twice as many mice as hamsters, what is the most of each that he can buy whiteout spending more than 100$
3m + 8h ≤ 100
but h = 2m
3m + 8(2m) ≤ 100
19m ≤ 100
m ≤ 5.25
so he can buy at most 5 mice and 10 hamsters
testing:
5 mice and 10 hamsters would cost 15 + 80 = $95
6 mice and 12 hamsters would cost 18 + 96 = 114$ , he does not have that much
3 m + 8 h ≤ 100
m = 2 h
substituting ... 3 (2 h) + 8 h ≤ 100 ... 14 h ≤ 100 ... h = 7
substitute back to find m
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To find out the maximum number of white mice and brown hamsters Morris can buy without spending more than $100, we need to set up some equations.
Let's assume Morris buys 'x' white mice and 'y' brown hamsters.
Given that the mice cost $3 each and the hamsters cost $8 each, we can set up the following equations:
Cost of white mice = $3 * x
Cost of brown hamsters = $8 * y
We know that he wants to buy twice as many mice as hamsters, so we can also set up the equation:
x = 2y
Now, let's calculate the maximum quantity of each that Morris can buy without spending more than $100.
We can substitute the value of x from the third equation into the first equation:
Cost of white mice = $3 * (2y)
Cost of white mice = $6y
The total cost should not exceed $100, so we can write the equation:
$6y + $8y ≤ $100
Combine like terms:
$14y ≤ $100
Divide both sides by $14 to solve for y:
y ≤ $100 / $14
y ≤ 7.14
Since y represents the number of brown hamsters, it cannot be a decimal or fraction. Therefore, y can be at most 7.
Substitute this value of y into the equation x = 2y:
x = 2 * 7
x = 14
So, the maximum number of white mice Morris can buy is 14, and the maximum number of brown hamsters he can buy is 7, without spending more than $100.