Let h = hamburgers and d = hot dogs in the original plan. We have h + d = 200 and 5h + 3d = 800. From the first equation: d = 200 - h. Substituting: 5h + 3(200 - h) = 800, so 5h + 600 - 3h = 800, giving 2h = 200 and h = 100. Originally planned: 100 hamburgers, 100 hot dogs. Actually sold: 80 hamburgers (20 fewer), 120 hot dogs (20 more to reach 200 total). Actual revenue: 5(80) + 3(120) = 400 + 360 = $760. Difference: $800 - $760 = $40. Choice A miscalculates the price difference. Choice C uses wrong quantities. Choice D doubles the correct answer.

Let w = width and l = length. We have 2w + 2l = 84 and l = 2w + 6. Substituting: 2w + 2(2w + 6) = 84, so 2w + 4w + 12 = 84, giving 6w = 72 and w = 12 feet. Then l = 2(12) + 6 = 30 feet. The perimeter is indeed 84 feet. Fence cost = 84 × $3.50 = $294. Money left = $300 - $294 = $6. Choice B doubles the correct remainder. Choice C uses wrong cost calculation. Choice D results from perimeter calculation errors.

This is a linear programming problem where you need to find the optimal combination of two products given limited resources. When you see problems involving maximizing profit with multiple constraints, set up a system of equations representing your resource limitations.

Let $$w$$ = widgets produced and $$g$$ = gadgets produced. Your constraints are:

• Labor: $$3w + 2g = 240$$
• Material: $$2w + 4g = 280$$
• Profit function: $$P = 15w + 12g$$

Since you must use exactly all resources, solve this system of equations. From the material constraint: $$w + 2g = 140$$, so $$w = 140 - 2g$$. Substituting into the labor constraint: $$3(140 - 2g) + 2g = 240$$, which gives $$420 - 6g + 2g = 240$$, so $$4g = 180$$ and $$g = 45$$. Therefore $$w = 140 - 2(45) = 50$$.

Maximum profit = $$15(50) + 12(45) = 750 + 330 = 1080$$.

Answer choice A ($$960$$) likely comes from using suboptimal production quantities. Answer choice B ($$1020$$) might result from calculation errors in solving the constraint equations. Answer choice C ($$1140$$) could come from ignoring one of the constraints entirely, perhaps assuming you can produce more than the resources allow.

The correct answer is D: $$1080$$.

Remember that linear programming problems require you to work within all given constraints simultaneously. Always verify your solution satisfies every constraint before calculating the final objective value.

For pencils: Need 480 pencils, boxes contain 24 each. Number of boxes needed = 480 ÷ 24 = 20 boxes exactly. Cost for pencils = 20 × $6 = $120. For erasers: Need 360 erasers, boxes contain 18 each. Number of boxes needed = 360 ÷ 18 = 20 boxes exactly. Cost for erasers = 20 × $4 = $80. Total cost = $120 + $80 = $200. Since both divide evenly, there are no leftover supplies and this is the minimum cost. Choice B adds unnecessary markup. Choice C assumes extra boxes needed. Choice D includes incorrect calculations.

Let b = Car B's speed and a = Car A's speed = b + 15. After 2.5 hours: 2.5b + 2.5(b + 15) = 275. This gives 2.5b + 2.5b + 37.5 = 275, so 5b = 237.5 and b = 47.5 mph. Car A's speed = 62.5 mph. In the next hour, Car A travels at 57.5 mph and Car B at 47.5 mph. Additional distance = 57.5 + 47.5 = 105 miles. Total distance apart = 275 + 105 = 380 miles. Choice A uses original speeds incorrectly. Choice B forgets Car A's speed reduction. Choice D adds the speed reduction instead of subtracting.

Pipe A fills 1/8 of the pool per hour, Pipe B fills 1/12 per hour. Together they fill 1/8 + 1/12 = 3/24 + 2/24 = 5/24 of the pool per hour. In 3 hours working together, they fill 3 × 5/24 = 15/24 = 5/8 of the pool. Remaining to fill = 1 - 5/8 = 3/8 of the pool. Pipe B alone fills 1/12 per hour, so time needed = (3/8) ÷ (1/12) = 3/8 × 12 = 36/8 = 4.5 hours. Choice A uses incorrect combined rate. Choice B forgets to account for work already done. Choice D uses Pipe A's rate instead of Pipe B's.

This problem tests setting up and solving systems from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables with units (let x = older sibling's age in years, y = younger sibling's age in years), (2) writing an equation from the sum constraint (x + y = 45), (3) writing an equation from the age difference (x = y + 5 or x - y = 5), (4) solving the system (substitute x = y + 5 into first: y + 5 + y = 45, 2y = 40, y = 20, x = 25), (5) interpreting (older is 25 years, younger is 20 years), and (6) verifying (25 + 20 = 45, 25 - 20 = 5). The correct solution is x=25, y=20, which matches choice C. The equations are correct as they capture the total age and the difference stated. Common errors include reversing the difference (y = x + 5) or arithmetic slips leading to other pairs like in A or D. To set up: identify constraints carefully, define variables with context, form equations accurately, solve and check for sensibility (ages positive and logical).

This question tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns with context interpretation, then verifying a proposed solution. The process involves: (1) defining variables such as x for large posters and y for small, (2) writing x + y = 11 for total posters, (3) writing 5x + 2y = 34 for revenue, (4) checking if (4,7) satisfies both, (5) interpreting if it's the solution, and (6) verifying calculations. Plugging in: 4 + 7 = 11 (true) and 5(4) + 2(7) = 20 + 14 = 34 (true), so yes. Common errors include miscalculation like 20 + 14 = 33 or ignoring one equation. To avoid errors: set up system first, plug values into both equations carefully, and confirm both are satisfied before concluding.

This question tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables such as $n$ for notebooks and $p$ for pens, (2) writing $n + p = 18$ for total items, (3) writing $3n + p = 38$ for total cost, (4) solving, (5) interpreting (number of notebooks), and (6) verifying. Solving correctly: subtract first from second: $(3n + p) - (n + p) = 38 - 18$, $2n = 20$, $n = 10$, then $p = 8$. These satisfy: $10 + 8 = 18$ and $3(10) + 8 = 30 + 8 = 38$. Common errors include incorrect cost equation or not solving the system fully. To avoid errors: define clearly, translate costs and counts, use elimination or substitution, interpret the asked quantity, and verify.

This problem tests setting up and solving systems from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables with units (let a = liters of apple juice, g = liters of grape juice), (2) writing total volume equation (a + g = 10), (3) writing total cost equation (3a + 5g = 38), (4) solving (a = 10 - g, 3(10 - g) + 5g = 30 - 3g + 5g = 30 + 2g = 38, 2g = 8, g = 4, a = 6), (5) interpreting (6 liters apple, 4 liters grape), (6) verifying (6 + 4 = 10, 18 + 20 = 38). This matches choice A. Equations correctly represent totals. Errors: swapping costs or wrong solving like B. Setup: identify quantities, define, translate, solve, interpret, verify. Ensure non-negative sensible values.