\(\frac{xz}{y}=420\)
In the equation above, x is an integer with 3 distinct prime factors, and y is a positive integer with no prime factors. If z is a positive, non-prime number, what is one possible value of z?
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Solution
4, 6, 10, and 14
Don’t forget that you can use your on-screen calculator.There’s only one positive integer with no prime factors, the number 1. Therefore, y = 1, and xz = 420.Create a prime factor tree to get the prime factors of 420: 2, 2, 3, 5, and 7. Pick 3 distinct values from that list, such as 2, 3, and 5, and multiply them to find one possible value of x. One example is 2 × 3 × 5 = 30, or one possible value for x. If 30z = 420, then z = 14.Confirm that 14 is not prime, then enter it in the field. If you chose 2, 3, 7, then x is 42 and z is 10 and also correct, and so on.
If j is a multiple of 12, and k is a multiple of 21, then jk must be a multiple of which of the following?
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Solution
Plug in values for j and k. Since every number is a multiple of itself, go ahead and start with j = 12 and k = 21; jk is now 252.You can use your on-screen calculator to determine that, of the answer choices, only 28 divides evenly into 252.Choice (D) is correct.
When x is divided by 3, the remainder is 1. When x is divided by 7, the remainder is 2. How many positive integers less than 100 could be values for x?
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Solution
4
To solve this question, write it out. Since there are fewer numbers that yield a remainder of 2 when divided by 7, start there.The first such number is 2, and thereafter they increase by 7; the rest of the list is thus 9, 16, 23, 30, 37, 44, 51, 58, 65, 72, 79, 86, and 93. Rather than list out all the numbers that yield a remainder of 1 when divided by 3, just select the numbers that meet the requirement from the list you already have: Only 16, 37, 58, and 79 do, so there are 4 values for x.
What is the smallest common multiple of 160 and 240?
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Solution
480 Make lists of the multiples for each number. Work on 240 first, then list the multiples of 160 until you find one on the list for 240.
If |–3x + 1| < 7, then which of the following represents all possible values of x?
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Solution
You can Plug In or solve on this problem.To Plug In, choose a value that fits one of the answer choices, such as x = 2, which would fit in the range for choice (C). If x = 2, then |–3x + 1| = 5, which is true, so we can eliminate any answer choice that doesn’t include x = 2: choices (A), (D), and (E). Logically, it doesn’t make sense that an inequality with a < sign would have a ≤ sign when it’s been solved, but to be sure, check x = −2. In that case, |–3x + 1| = 7, and is not < 7, so the answer must be choice (B). If you solve this problem, remember that you have to solve both −3x + 1 < 7, and −3x + 1 > 7. Also remember that you must flip the sign any time you multiply or divide both sides of an inequality by a negative number.
If xaybzc equals the product of 154 and 56, z > y > x, and a > b > c, then what is the value of axbycz?
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Solution
To solve this question, find the prime factors:The prime factors of 154 are 2, 7, and 11, and the prime factors of 56 are 2, 2, 2, and 7. Thus, the product of 154 and 56 will have the prime factors 2, 2, 2, 2, 7, 7, and 11, or (24)(72)(111). Line up your bases and exponents with the inequalities, and you get a = 4, b = 2, and c = 1 for the bases, and x = 2, y = 7, z = 11 for the exponents. Now axbycz = (42)(27)(1), which equals 16 × 128 × 1, or 2,048.The correct answer is choice (B).
If when x is divided by z, the result is y remainder q, then which of the following must be true?
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Solution
As soon as you see variables in the answer choices, set up your scratch paper to Plug In. If x = 16 and z = 5, then 16 ÷ 5 = 3 remainder 1, so y = 3 and q = 1. Plug your values into the answer choices, and only choice (B) works:\(\frac{16}{5}-3=\frac{1}{5}\).
The integer m is a multiple of 154, 250, and 264. Which of the following do NOT have to be factors of m?
Indicate all possible values.
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Solution
A,B, and E
To solve this question, turn large numbers into small numbers by working with factors.The prime factors of 154 are 2, 7, and 11; the prime factors of 264 are 2, 2, 2, 3, and 11; and the prime factors of 250 are 2, 5, 5, and 5.The only numbers that must be a factor of m are those made up of factors contained in the other three numbers.You can’t recount factors that overlap in the different numbers, so you know that m is made up of, at least, three 2’s, one 3, three 5’s, one 7, and one 11. Now check the answers.The prime factors of 176 are 2, 2, 2, 2, and 11, which is one 2 too many, so choice (A) is not a factor; since the question asks you to identify which choices are not factors, choice (A) is part of the credited response.The prime factors of 242 are 2, 11, and 11, which is one 11 too many, so (B) is also not a factor.The prime factors of 275 are 5, 5, and 11, so choice (C) is a factor.The prime factors of 924 are 2, 2, 3, 7, and 11, so choice (D) is a factor.The prime factors of 2,500 are 2, 2, 5, 5, 5, and 5, which is one 5 too many, so choice (E) is not a factor. And, finally, the prime factors of 7,000 are 2, 2, 2, 5, 5, 5, and 7, so choice (F) is a factor.The correct answers are choices (A), (B), and (E).
a is the product of 3 and the square root of 2, b is the product of 2 and the square root of 3, and c is the product of 2 and the square root of 6. If x is the square of the sum of a and b, y is the product of 6 and the difference of 5 and c, and z is the product of 2 squared and 3 squared, what is \(\frac{xy}{z}\)?
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Solution
Let’s translate this question, one step at a time.
a=3√2 b=2√3 c=2√6
x=(3√2 + 2√3)2 = (3√2 + 2√3)(3√2 + 2√3) =
(9)(2) + 6√6 + 6√6 + (4)(3) = 18 + 12√6 + 12 = 30 + 12√6
y = 6(5 - 2√6) = 30 - 12√6
z = 22 × 32 = 62 = 36
\(\frac{xy}{z}=\frac{(30+12\sqrt{6})(30-12\sqrt{6})}{36}=\frac{900+360\sqrt{6}-144(6)}{36}=\frac{900-864}{36}=\frac{36}{36}=1\)
Last year, Melania had a total of $20,000 invested in two mutual funds, Capital Growth Fund and Venture Index Fund. At the end of the year, she analyzed her investments and found that her earnings on her shares of Capital Growth Fund were three times half of her earnings on her investment in Venture Index Fund. If she earned a total of $1,250 on her investments in the two funds, and had three times as much money invested in Capital Growth Fund as in Venture Index Fund, what percent interest did Melania earn on her investment in Venture Index Fund? (percent interest = \(\left ( \frac{earning}{investment} \right )\) × 100
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Solution
Start by translating her earnings on her shares of Capital Growth Fund were three times half of her earnings on her investment in Venture Index Fund from English to math.Represent her earnings in Capital Growth Fund as C. Were translates to “ = .” Three times half is “3 × 1⁄2,” and of is “×” (multiplication).Represent Melania’s earnings from Venture Index Fund as V, and the resulting equation is C = 3⁄2V.Total earnings on the two funds were $1,250, so C + V = $1,250, and since C = 3⁄2V, that equation can be rewritten as 3⁄2V + V = $1,250, or 5⁄2V = $1,250. Solve this to find that V = $500 earned on Venture Index Fund. Notice that this is a partial answer that is included among the answer choices.Continuing to solve, represent the amount of money invested in Venture Index Fund as x. Melania had three times as much money invested in Capital Growth Fund as in Venture Index Fund; therefore she had 3x dollars invested in Capital Growth Fund. She had a total of $20,000 invested in the two funds; therefore 3x + x = $20,000. Solve to find that x = $5,000 invested in Venture Index Fund. Now, plug in those numbers to the percent interest formula given in the problem,\(\left ( \frac{500}{5000} \right )\) ×100, which equals the credited answer, choice (D), 10.You will arrive at the remaining, wrong answer choices if you mistakenly solve for the percent interest earned on Capital Growth Fund, and/or you represent your answer as a multiplier, rather than a percent (i.e., 0.01 versus 10%).