The price of Mabel’s car, not including interest, is 12 percent more than the price of Rose’s car. Combined, Mabel and Rose’s cars cost $53,000. If Mabel’s car loan interest rate was 5.20 percent, what was the total cost of Mabel’s car, including interest?
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Solution
29,456
First set up the first part of the question to calculate Mabel’s car cost without interest. Let Mabel = m and Rose = r. So, 1.12r = m and r + m = 53,000. Substitute: Since r + 1.12r = 53,000, then 2.12r = 53,000. Solve: r = 25,000 and m = 28,000. Now, you can multiply m by the interest rate. (28,000)(.052) = 1,456 of interest. Add it to the price of the car to get 28,000 + 1,456 = 29,456.
If each of three grocery stores receives 1⁄4 of a farmer’s potato crop, a farmer’s market receives 1⁄3 of the remaining, and a local fast food restaurant receives the remaining 200 pounds, how many pounds of potatoes were in the farmer’s crop?
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Solution
Since you know there are 200 lbs remaining after the grocery stores and farmer’s market get their shares, you cannot plug in your own number. So plug in the answers, and start with middle choice (C):
Carmen wants to open a special savings account through her work. If Carmen invests $7,000 at 6 percent simple annual interest in January, and no other money is added to or removed from the account, which of the following is true?
Indicate \(\underline{all}\) possible values.
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Solution
B and D
To calculate the interest earned, multiply the principal by the annual interest rate: in one year,Carmen will earn $7000 × 0.06 = $420 in interest, so choice (B) works.To calculate her interest for any part of the year, divide $420 by the appropriate fraction of a year. At the end of April,1⁄3 of the year has passed, so Carmen will have earned 1⁄3 × $420 = $140; eliminate choice A. At the end of six months,Carmen will have earned 1⁄2 × $420 = $210; eliminate choice (C). At the end of three months,1⁄4 of the year has passed, so Carmen has earned 1⁄4 × $420 = $105; there will be a total of $7000 + $105 = $7105 in the account, so choice (D) works.
Ben’s music album sold 5⁄6 the number of copies as Regina’s album. If Regina’s album sold at least 1,500 copies more than Ben’s album, how many copies of Ben’s album were sold?
Indicate \(\underline{all}\) possible values.
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Solution
(A) and (C)
A and C To solve this question, plug in the answers. In choice (D), if Ben,B, sold 6500 copies and B = 5⁄6R, then 6500 = 5⁄6R, so Regina, R, sold 7800 copies, thus giving a difference of 7800 − 6500 = 1300. Since this is too small, eliminate choices (D), (E), and (F) and try larger numbers. In choice (C), if B = 9000, then 9000 = 5⁄6R, so R = 10,800, thus giving 10,800 − 9000 = 1800; since this is bigger than 1500, keep choice (C). In choice (B), if B = 11,244, then 11,244 = 5⁄6R, so R = 13,492.8 copies; eliminate choice (B), as it is impossible to sell a fraction of an album. In choice (A), if B = 12,000, then 12,000 = 5⁄6R, so R = 14,400 and 14,400 − 12,000 = 2400. Since this is bigger than 1500, keep choice (A).The correct answers are choices (A) and (C).
If \(\frac{\left ( x^{\frac{3}{2}} \right )^{2}}{x^{6}}=8^{-1}\), then what is the value of x?
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Solution
2
To solve this question, remember the rules of exponents. When an exponent is outside a parentheses, it gets multiplied to any exponents inside the parentheses; thus \(\left ( x^{\frac{3}{2}} \right )^{2}\) = x3. Next, since when two numbers of the same base are divided, the exponents are subtracted, x3 divided by x5 = x3 − 6 = x−3. Since a negative exponent is equal to its reciprocal, then x−3 = = 8–1 = and thus x3 = 8.Thus, x = 2, the correct answer
An investment club has had an average rate of return of 15% per year for the past 6 years. If Teresa invests $1000 today and neither adds nor subtracts money from the club, how much will Teresa have invested after 5 years assuming that the rate of return does not change?
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Solution
To solve this question, remember that the formula for finding the result of periodic increases at a certain rate is (Original Amount)(1 + rate)number of periods. In this case, the final amount would be 1000(1.15)5.The correct answer is choice (B). If you forgot the formula, you could calculate the final amount after 5 years, and then calculate all the answers for a match.
The sum \(\frac{3}{10}+\frac{43}{100}+\frac{17}{1000}\) is equivalent to which of the following sums?
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Solution
To solve this question, convert the fractions into decimals and carefully add them.\(\frac{3}{10}+\frac{43}{100}+\frac{17}{1000}\) = 0.3 + 0.43 + 0.017 = 0.747; this is your target answer.Choice (A) equals 0.7 + 0.4 + 0.07 = 1.17.Choice (B) equals 0.6 + 0.12 + 0.037 = 0.757.Choice (C) equals 0.07 + 0.4 + 0.007 = 0.477.Choice (D) equals 0.07 + 0.004 + 0.7 = 0.774.Choice (E) equals 0.32 + 0.4 + 0.027 = 0.747.The correct answer is choice (E).
67.345×1015 is equivalent to which of the following?
Indicate \(\underline{all}\) possible values
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Solution
C, D, and E
To solve this question, remember that for every space the decimal in the base number moves to right, the power of ten should decrease by 1, and vice versa. For choice (A), 100013 = 1015, so the decimal should not have moved. In choice (B), moving the decimal 1 space to the right means that 1015 should decrease by 1 to become 1014.Eliminate choices (A) and (B). For choice (C), moving the decimal 1 space to the left means that the power should increase by 1; this is correct. For choice (D), 10014 is equivalent to 1015, so this is correct. For choice (E), moving the decimal 3 spaces to the left means that the power should increase by 3; this is correct.The correct answers are (C), (D), and (E).
If during a one-day period the Q train arrives at the station 30% less frequently than the B train, and the B train arrives 10% less frequently than the F train, then the Q train’s frequency is what percent of the F train’s frequency?
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Solution
To solve this question Plug In. Since the question deals with percents, try 100. If the F train arrives 100 times per day, then the B will arrive 10% fewer times than 100:B = (1 – \(\frac{10}{100}\)) F = (\(\frac{90}{100}\))100 = 90 times, and the Q will arrive 30% fewer times than 90: Q = (1– \(\frac{30}{100}\)) B = (\(\frac{70}{100}\))90 = 63 times.Translating the question “the Q train’s frequency is what percentage of the F train’s” gives Q = (\(\frac{x}{100}\))F and thus 63 = (\(\frac{x}{100}\))100, which means that the Q’s frequency is 63% of the F’s.The correct answer is choice (D).
Which of the following are greater than 1?
Indicate \(\underline{all}\) possible values.
| A.\(\frac{4(3+0.07)}{11.092}\) |
| B.\(\frac{\sqrt{82}-17^{2}}{\sqrt{34}}\) |
| C.\(\frac{9978.4-0.0083}{101^{2}}\) |
| D.\(\frac{\sqrt{143}\times \sqrt[3]{7}}{24.034}\) |
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Solution
A and B
A and B You can use your on-screen calculator, but it will probably be faster to ballpark on at least some of these—for example, think of \(\sqrt{82}\) as slightly more than 9, or \(\sqrt{82}\) as slightly less than 6. In order to figure out which fractions are greater than one, just figure out if the top part of the fraction is bigger than the bottom part:
\(\frac{4(3+0.7)}{11.0.92}\): top bigger than 12, bottom smaller, so choice (A) works;
\(\frac{\sqrt{82}-1.7^{2}}{\sqrt{34}}\): top bigger than 6, bottom smaller, so choice (B) works;
\(\frac{9978.4-.0083}{101^{2}}\): top smaller than 10,000, bottom bigger, so eliminate choice (C);
\(\frac{\sqrt{143}\times \sqrt[3]{7}}{24.0.34}\): top smaller than 24—you might think of it as less than 12 times less than 2—bottom bigger, so eliminate choice (D).