Setting up the equation: 12h = 42, where h is hours. Dividing both sides by 12: h = 42 ÷ 12 = 3.5 hours. Choice A (2.5) would cost $30. Choice C (4.5) would cost $54. Choice D (5.5) would cost $66.

Setting up the equation: x + 2/3 = 5/6. Subtracting 2/3 from both sides: x = 5/6 - 2/3. Converting to common denominator: x = 5/6 - 4/6 = 1/6. Choice B (1/3 = 2/6) plus 2/3 = 2/6 + 4/6 = 6/6 = 1. Choice C (1/2 = 3/6) plus 2/3 = 3/6 + 4/6 = 7/6. Choice D (2/3 = 4/6) plus 2/3 = 4/6 + 4/6 = 8/6.

When you see a problem where a collected amount represents "times" a goal, you're working with multiplication relationships that need to be reversed using division.

Here, the school collected $$240$$ canned goods, and this amount was $$1.5$$ times their original goal. To find the goal, you need to work backwards. If the goal multiplied by $$1.5$$ equals $$240$$, then the goal equals $$240 \div 1.5$$.

To calculate $$240 \div 1.5$$, you can convert the decimal to a fraction: $$1.5 = \frac{3}{2}$$. Dividing by $$\frac{3}{2}$$ is the same as multiplying by $$\frac{2}{3}$$. So: $$240 \times \frac{2}{3} = \frac{480}{3} = 160$$. The original goal was $$160$$ canned goods, making D correct.

Let's check why the other answers don't work. If you multiply A) $$150$$ by $$1.5$$, you get $$225$$, not $$240$$. Choice B) $$200$$ times $$1.5$$ equals $$300$$, which is too high. Option C) $$180$$ times $$1.5$$ gives you $$270$$, also too high.

The most common mistake here is confusing which operation to use. Some students might add $$1.5$$ to something or subtract it, but "times" always indicates multiplication. When you know a result and the multiplier, always divide the result by the multiplier to find the original amount.

Remember: "collected amount ÷ times factor = original goal." This division approach works for any similar "times the goal" problem.

Setting up the equation: 6d = 18, where d is the drop per hour. Solving: d = 18 ÷ 6 = 3°F per hour. Choice A would give a total drop of 6 × 2 = 12°F. Choice C would give a total drop of 6 × 4 = 24°F. Choice D would give a total drop of 6 × 6 = 36°F.

This is a rate problem where you need to find how far Roberto walks per minute. When you see questions about constant speed or rate, think about the relationship between distance, time, and speed.

To find Roberto's walking rate per minute, you need to divide the total distance by the total time. He walks $$\frac{3}{4}$$ mile in 15 minutes, so his rate is $$\frac{3/4}{15}$$ miles per minute. To divide by 15, you multiply by $$\frac{1}{15}$$: $$\frac{3}{4} \times \frac{1}{15} = \frac{3}{60} = \frac{1}{20}$$ mile per minute.

Let's examine why the other answers are incorrect. Answer A) $$\frac{1}{15}$$ mile comes from incorrectly thinking that if it takes 15 minutes total, he walks $$\frac{1}{15}$$ mile per minute - this ignores the actual distance of $$\frac{3}{4}$$ mile. Answer B) $$\frac{1}{10}$$ mile might result from miscalculating $$\frac{3}{4} \div 15$$ or confusing the relationship between the given numbers. Answer C) $$\frac{1}{12}$$ mile could come from incorrectly using 12 instead of 15 in your calculation, perhaps by misreading the problem.

The key strategy for rate problems is to always set up the division correctly: rate = distance ÷ time. Make sure you're dividing the total distance by the total time to find the unit rate. Double-check your fraction arithmetic, especially when dividing fractions or multiplying by reciprocals.

Since Maria has 3/4 as many stickers as her brother, we multiply the brother's stickers (s) by 3/4 to get Maria's stickers: (3/4)s = 18. Choice B and C incorrectly add 3/4 to s instead of multiplying. Choice D subtracts 3/4 from s, which doesn't represent the relationship described.

Jake used 0.8 cups more than required, so he used 2.5 + 0.8 cups total. To find f, we need f - 0.8 = 2.5. Choice B would mean Jake used less than required. Choice C incorrectly subtracts the recipe amount. Choice D treats 0.8 as a multiplier rather than an added amount.

When you see a word problem asking you to write an equation, your job is to translate the English words into mathematical symbols. The key is identifying what operation is being described and which quantity is unknown.

The problem states "When $$\frac{1}{4}$$ is added to a number, the result is $$\frac{7}{8}$$." Let's break this down: you have some unknown number (let's call it $$n$$), and when you add $$\frac{1}{4}$$ to it, you get $$\frac{7}{8}$$. The phrase "added to" tells you this is addition, so you're looking for: unknown number + $$\frac{1}{4}$$ = $$\frac{7}{8}$$. This translates directly to $$n + \frac{1}{4} = \frac{7}{8}$$, which is choice C.

Let's examine why the other options don't work. Choice A shows $$n \times \frac{1}{4} = \frac{7}{8}$$, which represents multiplication instead of addition. Choice B gives us $$n - \frac{1}{4} = \frac{7}{8}$$, which would mean "when $$\frac{1}{4}$$ is subtracted from a number" rather than added. Choice D shows $$\frac{1}{4} - n = \frac{7}{8}$$, which represents "$$\frac{1}{4}$$ minus some number," completely reversing the relationship described in the problem.

When translating word problems into equations, pay careful attention to the order of operations mentioned. "Added to a number" means the number comes first, then you add something to it. Practice identifying key phrases like "added to," "subtracted from," "multiplied by," and "divided by" to build your equation-writing skills.

Setting up the equation: 8w = 96, where w is the weekly savings amount. Solving: w = 96 ÷ 8 = $12 per week. Choice A would result in 8 × $10 = $80 total. Choice B would result in 8 × $11 = $88 total. Choice D would result in 8 × $14 = $112 total.

This question tests writing and solving one-step equations of the form x + p = q or p x = q from real-world problems involving nonnegative rational numbers, using inverse operations to solve. One-step equations include the addition form x + p = q solved by subtracting p from both sides to get x = q - p, for example, if x + 15 = 42, then x = 42 - 15 = 27; or the multiplication form p x = q solved by dividing both sides by p to get x = q / p, for example, if 6x = 18, then x = 18 ÷ 6 = 3; inverse operations undo the original operation, so addition is undone by subtraction and multiplication by division, with nonnegative rational numbers like whole numbers (x=27, p=15, q=42), fractions (x=1/2 from x + 1/4 = 3/4), or decimals (x=4 from 2.5x = 10), all greater than or equal to zero. For example, Maria has x dollars, earns $15, and now has $42, so the equation is x + 15 = 42, solved by subtracting 15 to get x = 27, meaning she started with $27; or for 6 items at $x each totaling $18, the equation is 6x = 18, solved by dividing by 6 to get x = 3, so each costs $3; or with fractions, x + 1/4 = 3/4, solved as x = 3/4 - 1/4 = 1/2 cup. In this problem, the correct equation is 2.5x = 15, solved by dividing by 2.5 to get x = 6, meaning the runner ran for 6 days. A common error is setting up addition instead like choice B, or dividing wrong like choice C (15 × 2.5 = 37.5) or choice D (perhaps 15 - 10 or miscalculation). To solve these problems: (1) write the equation from the context, identifying the operation such as 'each day' meaning multiplication, (2) identify the form as x + p = q or p x = q, (3) apply the inverse operation like subtracting p or dividing by p, (4) calculate accurately, such as 15 ÷ 2.5 = 6, (5) verify by substituting back, like 2.5 × 6 = 15, and (6) interpret the solution in context, such as x = 6 means 6 days. Remember, all values are nonnegative, including whole numbers, fractions, or decimals greater than or equal to zero, and avoid mistakes like using addition instead of multiplication, decimal division errors, or failing to verify.