Penelope went to the store to buy x pounds of broccoli for $1.10 per pound and y cans of soup for $2.50 each. In total, she spent less than $10. The inequality relating the purchases she made and the total purchase price is 1.10x + 2.50y < 10. Which are reasonable solutions for this situation? Check all that apply. (–1, 4) (0, 2) (3, 2.5) (2, 4) (0.5, 3.78) (1.5, 3)
Penelope went to the store to buy x pounds of broccoli for
2.50 each. In total, she spent less than $10. The inequality relating the purchases she made and the total purchase price is 1.10x + 2.50y < 10. Which are reasonable solutions for this situation? Check all that apply. (–1, 4) (0, 2) (3, 2.5) (2, 4) (0.5, 3.78) (1.5, 3)
The Correct Answer and Explanation is:
To solve this problem, we are given the inequality: 1.10x+2.50y<101.10x + 2.50y < 10
Where:
- xx = pounds of broccoli bought at $1.10 per pound
- yy = cans of soup bought at $2.50 each
- The total spent must be less than $10.
We are given several coordinate pairs (x, y) to test as possible solutions. For a solution to be reasonable, it must:
- Satisfy the inequality 1.10x+2.50y<101.10x + 2.50y < 10
- Make sense in a real-world context, where xx and yy must be non-negative (you can’t buy negative quantities)
Test Each Option:
(–1, 4)
1.10(−1)+2.50(4)=−1.10+10=8.90<10⇒Satisfies inequality but x=−1 is not valid1.10(-1) + 2.50(4) = -1.10 + 10 = 8.90 < 10 \Rightarrow \text{Satisfies inequality but } x = -1 \text{ is not valid}
Reject (Negative broccoli is not possible)
(0, 2)
1.10(0)+2.50(2)=0+5=5<101.10(0) + 2.50(2) = 0 + 5 = 5 < 10
Accept (Reasonable and satisfies the inequality)
(3, 2.5)
1.10(3)+2.50(2.5)=3.30+6.25=9.55<101.10(3) + 2.50(2.5) = 3.30 + 6.25 = 9.55 < 10
Accept (Reasonable even with fractional soup cans, though rare in real life, still mathematically valid)
(2, 4)
1.10(2)+2.50(4)=2.20+10=12.20>101.10(2) + 2.50(4) = 2.20 + 10 = 12.20 > 10
Reject (Exceeds $10 budget)
(0.5, 3.78)
1.10(0.5)+2.50(3.78)=0.55+9.45=10.00≮101.10(0.5) + 2.50(3.78) = 0.55 + 9.45 = 10.00 \not< 10
Reject (Equals $10, but not less than $10)
(1.5, 3)
1.10(1.5)+2.50(3)=1.65+7.50=9.15<101.10(1.5) + 2.50(3) = 1.65 + 7.50 = 9.15 < 10
Accept
✅ Final Answers:
- (0, 2)
- (3, 2.5)
- (1.5, 3)
Explanation :
This problem involves identifying valid real-world solutions to a linear inequality modeling a shopping budget. Penelope is buying xx pounds of broccoli at $1.10 per pound and yy cans of soup at $2.50 each. The constraint is that her total purchase must be less than $10, which gives the inequality: 1.10x+2.50y<101.10x + 2.50y < 10
To determine which points are valid, each given pair (x, y) is substituted into the inequality. Only those combinations that result in a value strictly less than 10 qualify.
Additionally, because this is a real-life situation involving buying physical goods, values of xx and yy must be non-negative. Negative quantities like –1 pounds of broccoli aren’t realistic, so such points are discarded even if they mathematically satisfy the inequality.
Evaluating each point:
- (–1, 4) satisfies the inequality mathematically but is not realistic.
- (0, 2), (3, 2.5), and (1.5, 3) are all reasonable and satisfy the budget constraint.
- (2, 4) exceeds the $10 limit.
- (0.5, 3.78) hits exactly $10, which doesn’t meet the “less than $10” requirement.
Thus, the reasonable solutions that satisfy both the inequality and the context of the situation are:
(0, 2), (3, 2.5), and (1.5, 3).
