When you need to compare ratios, think of the ratios as fractions that need common denominators.The value of c in each ratio will need to become 28, and the other values will need to change accordingly.That will make b:c = 105:28 after multiplying the original values by 7, and a:c = 12:28 after multiplying the original values by 4.This means a:b = 12:105. Simplify, and you get 4:35.

C and F

C and F Since red is common to the given ratios, you’ll want to multiply the red:green ratio by 3 so that red is 6 in both. Now you can put it all together in one ratio—red:green:yellow = 6:15:5. More importantly, you can put them in one Ratio Box:

No need to finish the rest of the Ratio Box—you have all you need. Look for answer choices that are multiples of 26. Only choices (C) and (F) work.

To simplify this problem, ignore the chart column about the greatest single employee donation and plug in easy values for the number of employees for each company.Try 15 employees for Company A and 1 employee for Company K: Now you have a total donation of 15 × 34.6 = 519 per year from Company A, and a total donation of 104.4 per year from the single employee of Company K.The ratio is thus 519:104.4, which reduces almost exactly to 5:1, so choice (E) is correct. If you selected choice (A), you may have solved for the averages rather than the actual amount of the donations; if you selected choice (D), you may have solved for the period 1980–1990.

According to the given definition,∆M is 455.2 – 422.4 = 32.8, and ∆B is 40.2 – 18.2 = 22.Round ∆M to 33, and the ratio of∆M to ∆B is 33 to 22, which reduces to 3 to 2.Choice (B) is correct.

¹⁄₃

¹⁄₃ Set up a ratio box. Put the recipe’s original ratio of ingredients in the top row and start with 2 eggs in the “actual” row.This makes your multiplier ²⁄₃. Fill out the rest of the ratio box to determine the number of cups of butter.

A,B, D, and E

Set up your ratio box.The number given is the actual total number of players, so put 24 there.Then start Plugging In the Answers into your ratio row to see which could work.The ratio of 1:2 in choice (A) would yield a ratio total of 3; this works with a multiplier of 8, so you know choice (A) works.Choice (B) gives a ratio total of 4, which would work with a multiplier of 6; choice (B) works. Choice (C), however, gives a ratio total of 5; since 24 isn’t a multiple of 5, it would yield a fractional multiplier, and thus fractional juniors and seniors.Eliminate choice (C).Choices (D) and (E) would work with multipliers of 4 and 3, respectively.Choice (F) yields a ratio total of 11, which will again yield a fractional multiplier, so eliminate choice (F).

27

Make a ratio box and fill in what you know: the ratio and the total number of salmon.You then total number of fish in the ratio row, and find your multiplier in the salmon column. Multiply down to figure out the actual total number of fish, which is 27.

Plug in values. If a = 5 and c = 1, then b = 20. In that situation Quantity A is larger; eliminate choices (B) and (C). Plug In again to see if this is always the case. If a = 100 and c = 2, then b = 800. Quantity A is still larger.There isn’t anything else you can try that would change the values, so choice (A) is the best answer.

Because 7 and 17 are prime and have no common factor greater than 1, their least common multiple will be their product: 7 × 17 = 119.The least possible value for (a – 1), then, is 17, so a = 18; likewise, the least possible value for (b – 1) is 7, so b = 8.The least possible sum for a and b, therefore, is 18 + 8 = 26.Be careful if you selected choice (A): Although Plugging In a value of 1 for both a and b would yield 0 on both sides of the equation, the problem specifies that the product of a and b be greater than 1.

Try Plugging In. If there are 10 cats at the beginning of the day, then there are 11 dogs; at the end of the day, there would be 15 cats and 15 dogs. In this case, Quantity A and Quantity B are equal.Eliminate choices (A) and (B). However, there could be 20 cats and 22 dogs at the beginning of the day; then there would be 25 cats and 26 dogs at the end of the day. In that situation, Quantity B is greater; eliminate choice (C). Only choice (D) remains.