Leah wants to shrink her photos to 3t a computer screen. Her photos currently have a width of 1,024 pixels and a height of 768 pixels. If she reduces the width to 800 pixels, then to what height, in pixels, must she reduce the photos to preserve the same ratio of width to height?
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Solution
Set up a proportion so that the original ratio equals the final ratio:\(\frac{1024}{800}=\frac{768}{x}\).Cross-multiply and then divide both sides by 1024 to find x = 600.
| Quantity A | Quantity B |
| 16 percent of 83 | 83 percent of 16 |
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Solution
To find 16 percent of 83, multiply 83 by 0.16.To find 83 percent of 16, multiply 16 by 0.83.Both expressions yield 13.28, so choice (C) is correct.
One cup of nuts that contains exactly half peanuts and half cashews is added to a bowl of nuts that is exactly one third peanuts, one third cashews, and one third almonds.This results in a three-cup mixture of nuts. What fraction of the new nut mixture is peanuts?
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Solution
\(\frac{7}{18}\)
The total mixture contains three cups, so the second bowl must contain two cups.This 2-cup bowl of nuts divided into even thirds consists of 2⁄3 cups peanuts,2⁄3 cups cashews, and 2⁄3 cups almonds.Combining this with the 1-cup mixture of 1⁄2 cup peanuts and 1⁄2 cup cashews results 7⁄6 in cups peanuts in a 3-cup mixture. So,\(\frac{7}{\frac{6}{3}}=\frac{7}{18}\) of the new nut mixture is peanuts.
Which expression is equivalent to \(\frac{1}{y-\frac{1}{y}}y\)?
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Solution
This is a good problem for Plugging In. If y = 2, then the expression becomes equal to − 4⁄3.Choice (B) is the only choice that gives you −4⁄3 when you replace y with 2.
27 percent of p is 100.
p is q percent of 100.
| Quantity A | Quantity B |
| q | 400 |
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Solution
To find exact values for p and q, apply percent translation:\(\frac{27}{100}\times p=100\), so \(\frac{27p}{100}=100\), 27p = 10,000, and p = 370.37; p is q percent of 100, so q = 370.37 as well. Quantity B is greater. Alternatively, you could avoid the calculation altogether and Ballpark this one all the way through: 100 is more than 25% (or 1⁄4) of p, so p must be less than 400—and so must q.
| Quantity A | Quantity B |
| The total value of 100 dollars after it is invested for m months at 8 percent simple annual interest | \(100\left ( 1+\frac{0.08}{m} \right )\) dollars |
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Solution
Plug in values for m. When m = 1, Quantity B is larger. When m = 12, Quantity A is larger.
Which of the following is equal to 1⁄5 of the reciprocal of 0.004 percent?
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Solution
Solve this problem in chunks.To find the numerical value for 0.004 percent, divide by 100: 0.004 ÷ 100 = 0.00004.The reciprocal of that is \(\frac{1}{0.00004}=25,000\). So 1⁄5 of the result is 1⁄5 × 25,000 = 5,000.
The annual interest rate on a certain savings account increases from 1.25% to 1.5%. What percent increase in the annual interest rate does this change represent?
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Solution
The percent change formula is \(\frac{difference}{original}\)×100, so plugging the numbers from the problem into the formula yields \(\frac{0.25}{1.25}\times 100=\frac{25}{125}\times 100=\frac{1}{5}\times 100=20\). If you selected choice (E), you may have used the wrong value as the original:Remember, in a percent increase, the original number is the smaller value.
Halfway through the season, Antonio’s scoring average per game was 20% higher than David’s. The two scored the same number of
points in the second half of the season.
| Quantity A | Quantity B |
| 90% of Antonio’s scoring average for the whole season | David’s scoring average for the whole season |
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Solution
Plug in some real numbers to compare quantities. For example, plug in 10 for the number of games in the season. For the first 5 games of the season, try an average of 10 points for David (for a total of 50 points), which makes an average of 12 points for Antonio (for a total of 60 points). Next, try a total of 0 points for each player for the second half of the season; now Antonio’s average for the season is 6 points, and David’s is 5.Because 90% of 6 is 5.4, Quantity A is greater, so eliminate choices (B) and (C). Finally, try a total of 100 points for each player for the second half of the season; now Antonio has scored a total of 160 points in 10 games, for an average of 16 points, and David has scored a total of 150 points in 10 games, for an average of 15 points.Because 90% of 16 is 14.4, Quantity B is now greater, so eliminate choice (A), and you’re left with choice (D).
n > 0
\(\frac{6n}{15},0.3n,\frac{19n}{50},\frac{n}{4}\)
| Quantity A | Quantity B |
| The positive difference between the greatest and least values above | Three times the positive difference between the two least values above |
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Solution
The first thing you need to do is to clean up these expressions.You have 15th, 50th, and decimals, so it is very difficult to compare values.\(\frac{6n}{15}\) can be reduced to \(\frac{2n}{5}\). 0.3n is the same as .Change your first expression from \(\frac{2n}{5}\) to \(\frac{4n}{10}\),\(\frac{19n}{50}\). is pretty close to \(\frac{20n}{50}\) or \(\frac{2n}{5}\), the first expression, but a bit smaller.Because n⁄4 is clearly the smallest expression and you need only concern yourself with the smallest, the second smallest, and the biggest, you can ignore \(\frac{19n}{50}\).Convert n⁄4 to \(\frac{5n}{20}\), and convert your other expressions to 20ths as well.You now have \(\frac{8n}{20}\),\(\frac{6n}{20}\), and \(\frac{5n}{20}\).The difference between the smallest and largest is \(\frac{3n}{20}\).Three times the difference between the two smallest is also 3.The answer is choice (C).