A school store must spend no more than $225 for an order of pens and pencils. Each pen costs $0.75, and each pencil costs $0.25. This relationship is shown in the graph. The store needs at least 100 more pencils than pens. Which choice is a reasonable solution?
300 pens and 900 pencils
300 pens and 200 pencils
200 pens and 300 pencils
100 pens and 200 pencils
A school store must spend no more than $225 for an order of pens and pencils. Each pen costs $0.75, and each pencil costs $0.25. This relationship is shown in the graph. The store needs at least 100 more pencils than pens. Which choice is a reasonable solution
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To determine which choice is reasonable, we can set up the constraints based on the information provided:
1. Cost Constraint:
– Let ( p ) represent the number of pens and ( c ) represent the number of pencils.
– The cost equation is:
[
0.75p + 0.25c leq 225
]
2. Pencil Constraint:
– The store needs at least 100 more pencils than pens, which can be expressed as:
[
c geq p + 100
]
Now, let’s analyze the provided options:
1. 300 pens and 900 pencils:
– Cost: ( 0.75(300) + 0.25(900) = 225 + 225 = 450 ) (exceeds budget)
2. 300 pens and 200 pencils:
– Cost: ( 0.75(300) + 0.25(200) = 225 + 50 = 275 ) (exceeds budget)
3. 200 pens and 300 pencils:
– Cost: ( 0.75(200) + 0.25(300) = 150 + 75 = 225 ) (fits budget)
– Pencil constraint: ( 300 geq 200 + 100 ) (