Mr. Carnegie will buy pencils and pens for his students for the school year. Use the following information to determine the number of pencils and the number of pens he can buy.
He can spend no more than
$
30
on pencils and pens.
Pencils cost
$
0
.
15
each, and pens cost
$
0
.
75
each.
He needs to buy at least
3
times as many pencils as pens.
Let's represent the number of pencils as P and the number of pens as N.
The cost of the pencils is $0.15 each, so the cost of P pencils will be 0.15P.
The cost of the pens is $0.75 each, so the cost of N pens will be 0.75N.
He can spend no more than $30 on pencils and pens, so the total cost should be less than or equal to $30.
0.15P + 0.75N ≤ 30
He needs to buy at least 3 times as many pencils as pens, so P ≥ 3N.
Now we can solve the system of inequalities:
0.15P + 0.75N ≤ 30
P ≥ 3N
Let's find the maximum value of P and N that satisfies these inequalities.
First, we can simplify the first inequality by dividing both sides by 0.15:
P + 5N ≤ 200
Next, we can simplify the second inequality by dividing both sides by 3:
P/3 ≥ N
Now we have the following system of inequalities:
P + 5N ≤ 200
P/3 ≥ N
To find the maximum value of P and N, we need to find the intersection of these two inequalities.
Let's assume P/3 = N and substitute it into the first inequality:
P + 5(P/3) ≤ 200
Multiply both sides by 3 to remove the fraction:
3P + 5P ≤ 600
8P ≤ 600
P ≤ 75
Since P should be an integer, the maximum value for P is 75.
Substituting P = 75 into P/3 ≥ N:
75/3 ≥ N
25 ≥ N
So, the maximum value for N is 25.
Therefore, Mr. Carnegie can buy a maximum of 75 pencils and 25 pens for his students for the school year.