First, find the actual distance: $$\frac{8.5 \text{ inches}}{1.5 \text{ inches}} \times 12 \text{ miles} = \frac{8.5 \times 12}{1.5} = \frac{102}{1.5} = 68$$ miles. Then find the time: $$\frac{68 \text{ miles}}{51 \text{ mph}} = \frac{68}{51} = \frac{4}{3} = 1\frac{1}{3}$$ hours $$= 1$$ hour $$20$$ minutes. Choice B uses incorrect scale calculation ($$8.5 \times 12 = 102$$ miles directly). Choice C incorrectly calculates $$\frac{68}{52}$$ instead of $$\frac{68}{51}$$. Choice D uses $$8.5 \times 1.5 \times 12 = 153$$ miles.

When you encounter recipe scaling problems, you need to work with ratios and proportions. The key insight is finding how much flour is needed per person, then determining how many people your available flour can serve.

First, let's find the flour needed per person. The recipe serves 6 people with $$2\frac{1}{4}$$ cups of flour. Converting to an improper fraction: $$2\frac{1}{4} = \frac{9}{4}$$ cups. So each person needs $$\frac{9/4}{6} = \frac{9}{4} \times \frac{1}{6} = \frac{9}{24} = \frac{3}{8}$$ cups of flour.

Now, with $$4\frac{3}{4}$$ cups available, let's see how many people we can serve. Converting: $$4\frac{3}{4} = \frac{19}{4}$$ cups. Dividing by the per-person amount: $$\frac{19/4}{3/8} = \frac{19}{4} \times \frac{8}{3} = \frac{152}{12} = 12\frac{2}{3}$$ people.

Since you can't serve a fraction of a person, you can serve a maximum of 12 people.

Choice (A) 15 people would require $$15 \times \frac{3}{8} = \frac{45}{8} = 5\frac{5}{8}$$ cups—more than you have. Choice (B) 13 people would need $$13 \times \frac{3}{8} = \frac{39}{8} = 4\frac{7}{8}$$ cups—still too much. Choice (C) 14 people would require $$14 \times \frac{3}{8} = \frac{42}{8} = 5\frac{1}{4}$$ cups—way more than available. Choice (D) 12 people needs exactly $$12 \times \frac{3}{8} = 4\frac{1}{2}$$ cups, which is less than your $$4\frac{3}{4}$$ cups.

Remember: in "maximum" problems, always round down to the nearest whole number when dealing with people or discrete items.

The scale information and model length are red herrings - they don't affect the speed calculation. The real locomotive travels $$280$$ miles in $$4$$ hours $$40$$ minutes. Convert time: $$4$$ hours $$40$$ minutes $$= 4\frac{40}{60} = 4\frac{2}{3} = \frac{14}{3}$$ hours. Speed $$= \frac{280}{\frac{14}{3}} = 280 \times \frac{3}{14} = \frac{840}{14} = 60$$ mph. Choice A incorrectly uses $$5$$ hours instead of $$4\frac{2}{3}$$ hours. Choice C uses $$4.5$$ hours instead of $$4\frac{2}{3}$$ hours. Choice D uses $$4$$ hours, ignoring the extra $$40$$ minutes.

When you encounter work rate problems, think in terms of total work units and how they're distributed over time. The key insight is that work rate remains constant per worker.

First, calculate the total work needed. If 8 workers complete the job in 15 days, the total work equals $$8 \times 15 = 120$$ worker-days.

Next, determine how much work gets completed in the first 6 days. With 8 workers working for 6 days, that's $$8 \times 6 = 48$$ worker-days of completed work.

This leaves $$120 - 48 = 72$$ worker-days of remaining work.

After 6 days, 2 workers are reassigned, leaving $$8 - 2 = 6$$ workers. To find how long these 6 workers need to complete the remaining 72 worker-days of work: $$\frac{72}{6} = 12$$ days.

Looking at the wrong answers: Choice (A) 18 days likely comes from incorrectly calculating $$\frac{72}{4}$$ workers instead of 6. Choice (B) 15 days represents the original timeline, ignoring both the work already completed and the reduced workforce. Choice (D) 21 days might result from adding the original 15 days to the 6 days already worked, which double-counts the time.

The correct answer is (C) 12 days.

Strategy tip: In work rate problems, always convert to total work units first (worker-days, worker-hours, etc.), then track what's completed versus what remains. This systematic approach prevents confusion when workforce size changes mid-project.

Initial rate: $$\frac{450 \text{ widgets}}{3 \text{ hours}} = 150$$ widgets per hour. The rate increases by $$20%$$, so new rate $$= 150 \times 1.20 = 180$$ widgets per hour. In $$2$$ hours at this rate: $$180 \times 2 = 360$$ widgets. Choice A incorrectly uses the original rate for $$2$$ hours: $$150 \times 2 = 300$$. Choice C incorrectly calculates $$450 \times 1.20 = 540$$. Choice D uses an incorrect rate of $$300$$ widgets per hour.

When you encounter problems involving proportional enlargement or reduction, you're working with similar figures and scale factors. The key insight is that when dimensions change by a scale factor, the area changes by the square of that scale factor.

Start by finding the original area: $$4 \times 6 = 24$$ square inches. The enlarged photograph has an area of $$150$$ square inches, so the area scale factor is $$\frac{150}{24} = 6.25$$. Since area scales by the square of the linear scale factor, the linear scale factor is $$\sqrt{6.25} = 2.5$$. This means each dimension is multiplied by $$2.5$$.

The new dimensions are: $$4 \times 2.5 = 10$$ inches and $$6 \times 2.5 = 15$$ inches. Let's verify: $$10 \times 15 = 150$$ square inches. ✓

Now examine why the other choices fail. Choice A gives $$8 \times 12 = 96$$ square inches, which is too small and represents a scale factor of only $$2$$. Choice B yields $$12 \times 18 = 216$$ square inches, which exceeds our target area and represents a scale factor of $$3$$. Choice D lists the same dimensions as C but in reverse order ($$15 \times 10$$), which gives the correct area but doesn't follow the conventional format of listing the smaller dimension first when maintaining the original orientation.

Remember: when dealing with similar figures, always work backwards from the area ratio to find the linear scale factor by taking the square root. This prevents calculation errors and helps you systematically approach enlargement problems.