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MTEL General Curriculum Mathematics Practice
Question 1 |
There are six gumballs in a bag — two red and four green. Six children take turns picking a gumball out of the bag without looking. They do not return any gumballs to the bag. What is the probability that the first two children to pick from the bag pick the red gumballs?
\( \large \dfrac{1}{3}\) Hint: This is the probability that the first child picks a red gumball, but not that the first two children pick red gumballs. | |
\( \large \dfrac{1}{8}\) Hint: Are you adding things that you should be multiplying? | |
\( \large \dfrac{1}{9}\) Hint: This would be the probability if the gumballs were returned to the bag. | |
\( \large \dfrac{1}{15}\) Hint: The probability that the first child picks red is 2/6 = 1/3. Then there are 5 gumballs in the bag, one red, so the probability that the second child picks red is 1/5. Thus 1/5 of the time, after the first child picks red, the second does too, so the probability is 1/5 x 1/3 = 1/15. |
Question 1 Explanation:
Topic: Calculate the probabilities of simple and compound events and of
independent and dependent events (Objective 0026).
Question 2 |
The histogram below shows the frequency of a class's scores on a 4 question quiz.
What was the mean score on the quiz?
\( \large 2.75\) Hint: There were 20 students who took the quiz. Total points earned: \(2 \times 1+6 \times 2+ 7\times 3+5 \times 4=55\), and 55/20 = 2.75. | |
\( \large 2\) Hint: How many students are there total? Did you count them all? | |
\( \large 3\) Hint: How many students are there total? Did you count them all? Be sure you're finding the mean, not the median or the mode. | |
\( \large 2.5\) Hint: How many students are there total? Did you count them all? Don't just take the mean of 1, 2, 3, 4 -- you have to weight them properly. |
Question 2 Explanation:
Topics: Analyze and interpret various graphic
representations, and use measures of central tendency (e.g., mean, median, mode) and
spread to describe and interpret real-world data (Objective 0025).
Question 3 |
Use the table below to answer the question that follows:
Each number in the table above represents a value W that is determined by the values of x and y. For example, when x=3 and y=1, W=5. What is the value of W when x=9 and y=14? Assume that the patterns in the table continue as shown.
\( \large W=-5\) Hint: When y is even, W is even. | |
\( \large W=4\) Hint: Note that when x increases by 1, W increases by 2, and when y increases by 1, W decreases by 1. At x=y=0, W=0, so at x=9, y=14, W has increased by \(9 \times 2\) and decreased by 14, or W=18-14=4. | |
\( \large W=6\) Hint: Try fixing x or y at 0, and start by finding W for x=0 y=14 or x=9, y=0. | |
\( \large W=32\) Hint: Try fixing x or y at 0, and start by finding W for x=0 y=14 or x=9, y=0. |
Question 3 Explanation:
Topic: Recognize and extend patterns using a variety of representations (e.g., verbal, numeric, pictorial, algebraic) (Objective 0021)
Question 4 |
What is the greatest common factor of 540 and 216?
\( \large{{2}^{2}}\cdot {{3}^{3}}\) Hint: One way to solve this is to factor both numbers: \(540=2^2 \cdot 3^3 \cdot 5\) and \(216=2^3 \cdot 3^3\). Then take the smaller power for each prime that is a factor of both numbers. | |
\( \large2\cdot 3\) Hint: This is a common factor of both numbers, but it's not the greatest common factor. | |
\( \large{{2}^{3}}\cdot {{3}^{3}}\) Hint: \(2^3 = 8\) is not a factor of 540. | |
\( \large{{2}^{2}}\cdot {{3}^{2}}\) Hint: This is a common factor of both numbers, but it's not the greatest common factor. |
Question 4 Explanation:
Topic: Find the greatest common factor of a set of numbers (Objective 0018).
Question 5 |
Taxicab fares in Boston (Spring 2012) are $2.60 for the first \(\dfrac{1}{7}\) of a mile or less and $0.40 for each \(\dfrac{1}{7}\) of a mile after that.
Let d represent the distance a passenger travels in miles (with \(d>\dfrac{1}{7}\)). Which of the following expressions represents the total fare?
\( \large \$2.60+\$0.40d\) Hint: It's 40 cents for 1/7 of a mile, not per mile. | |
\( \large \$2.60+\$0.40\dfrac{d}{7}\) Hint: According to this equation, going 7 miles would cost $3; does that make sense? | |
\( \large \$2.20+\$2.80d\) Hint: You can think of the fare as $2.20 to enter the cab, and then $0.40 for each 1/7 of a mile, including the first 1/7 of a mile (or $2.80 per mile).
Alternatively, you pay $2.60 for the first 1/7 of a mile, and then $2.80 per mile for d-1/7 miles. The total is 2.60+2.80(d-1/7) = 2.60+ 2.80d -.40 = 2.20+2.80d. | |
\( \large \$2.60+\$2.80d\) Hint: Don't count the first 1/7 of a mile twice. |
Question 5 Explanation:
Topic: Identify variables and derive algebraic expressions that represent real-world situations (Objective 0020), and select the linear equation that best models a real-world situation (Objective 0022).
There are 5 questions to complete.
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