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MTEL General Curriculum Mathematics Practice
Question 1 |
The letters A, and B represent digits (possibly equal) in the ten digit number x=1,438,152,A3B. For which values of A and B is x divisible by 12, but not by 9?
\( \large A = 0, B = 4\) Hint: Digits add to 31, so not divisible by 3, so not divisible by 12. | |
\( \large A = 7, B = 2\) Hint: Digits add to 36, so divisible by 9. | |
\( \large A = 0, B = 6\) Hint: Digits add to 33, divisible by 3, not 9. Last digits are 36, so divisible by 4, and hence by 12. | |
\( \large A = 4, B = 8\) Hint: Digits add to 39, divisible by 3, not 9. Last digits are 38, so not divisible by 4, so not divisible by 12. |
Question 2 |
Exactly one of the numbers below is a prime number. Which one is it?
\( \large511 \) Hint: Divisible by 7. | |
\( \large517\) Hint: Divisible by 11. | |
\( \large519\) Hint: Divisible by 3. | |
\( \large521\) |
Question 3 |
Kendra is trying to decide which fraction is greater, \( \dfrac{4}{7}\) or \( \dfrac{5}{8}\). Which of the following answers shows the best reasoning?
\( \dfrac{4}{7}\) is \( \dfrac{3}{7}\)away from 1, and \( \dfrac{5}{8}\) is \( \dfrac{3}{8}\)away from 1. Since eighth‘s are smaller than seventh‘s, \( \dfrac{5}{8}\) is closer to 1, and is the greater of the two fractions. | |
\( 7-4=3\) and \( 8-5=3\), so the fractions are equal.Hint: Not how to compare fractions. By this logic, 1/2 and 3/4 are equal, but 1/2 and 2/4 are not. | |
\( 4\times 8=32\) and \( 7\times 5=35\). Since \( 32<35\) , \( \dfrac{5}{8}<\dfrac{4}{7}\)Hint: Starts out as something that works, but the conclusion is wrong. 4/7 = 32/56 and 5/8 = 35/56. The cross multiplication gives the numerators, and 35/56 is bigger. | |
\( 4<5\) and \( 7<8\), so \( \dfrac{4}{7}<\dfrac{5}{8}\)Hint: Conclusion is correct, logic is wrong. With this reasoning, 1/2 would be less than 2/100,000. |
Question 4 |
Use the samples of a student's work below to answer the question that follows:
This student divides fractions by first finding a common denominator, then dividing the numerators.
\( \large \dfrac{2}{3} \div \dfrac{3}{4} \longrightarrow \dfrac{8}{12} \div \dfrac{9}{12} \longrightarrow 8 \div 9 = \dfrac {8}{9}\) \( \large \dfrac{2}{5} \div \dfrac{7}{20} \longrightarrow \dfrac{8}{20} \div \dfrac{7}{20} \longrightarrow 8 \div 7 = \dfrac {8}{7}\) \( \large \dfrac{7}{6} \div \dfrac{3}{4} \longrightarrow \dfrac{14}{12} \div \dfrac{9}{12} \longrightarrow 14 \div 9 = \dfrac {14}{9}\)Which of the following best describes the mathematical validity of the algorithm the student is using?
It is not valid. Common denominators are for adding and subtracting fractions, not for dividing them.Hint: Don't be so rigid! Usually there's more than one way to do something in math. | |
It got the right answer in these three cases, but it isn‘t valid for all rational numbers.Hint: Did you try some other examples? What makes you say it's not valid? | |
It is valid if the rational numbers in the division problem are in lowest terms and the divisor is not zero.Hint: Lowest terms doesn't affect this problem at all. | |
It is valid for all rational numbers, as long as the divisor is not zero.Hint: When we have common denominators, the problem is in the form a/b divided by c/b, and the answer is a/c, as the student's algorithm predicts. |
Question 5 |
Below is a portion of a number line:
Point B is halfway between two tick marks. What number is represented by Point B?
\( \large 0.645\) Hint: That point is marked on the line, to the right. | |
\( \large 0.6421\) Hint: That point is to the left of point B. | |
\( \large 0.6422\) Hint: That point is to the left of point B. | |
\( \large 0.6425\) |
Question 6 |
Use the expression below to answer the question that follows.
\( \large 3\times {{10}^{4}}+2.2\times {{10}^{2}}\)
Which of the following is closest to the expression above?
Five millionHint: Pay attention to the exponents. Adding 3 and 2 doesn't work because they have different place values. | |
Fifty thousandHint: Pay attention to the exponents. Adding 3 and 2 doesn't work because they have different place values. | |
Three millionHint: Don't add the exponents. | |
Thirty thousandHint: \( 3\times {{10}^{4}} = 30,000;\) the other term is much smaller and doesn't change the estimate. |
Question 7 |
Below is a portion of a number line.
Point A is one-quarter of the distance from 0.26 to 0.28. What number is represented by point A?
\( \large0.26\) Hint: Please reread the question. | |
\( \large0.2625\) Hint: This is one-quarter of the distance between 0.26 and 0.27, which is not what the question asked. | |
\( \large0.265\) | |
\( \large0.27\) Hint: Please read the question more carefully. This answer would be correct if Point A were halfway between the tick marks, but it's not. |
Question 8 |
The histogram below shows the number of pairs of footware owned by a group of college students.
Which of the following statements can be inferred from the graph above?
The median number of pairs of footware owned is between 50 and 60 pairs.Hint: The same number of data points are less than the median as are greater than the median -- but on this histogram, clearly more than half the students own less than 50 pairs of shoes, so the median is less than 50. | |
The mode of the number of pairs of footware owned is 20.Hint: The mode is the most common number of pairs of footwear owned. We can't tell it from this histogram because each bar represents 10 different numbers-- perhaps 8 students each own each number from 10 to 19, but 40 students own exactly 6 pairs of shoes.... or perhaps not.... | |
The mean number of pairs of footware owned is less than the median number of pairs of footware owned.Hint: This is a right skewed distribution, and so the mean is bigger than the median -- the few large values on the right pull up the mean, but have little effect on the median. | |
The median number of pairs of footware owned is between 10 and 20.Hint: There are approximately 230 students represented in this survey, and the 41st through 120th lowest values are between 10 and 20 -- thus the middle value is in that range. |
Question 9 |
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 10 |
Which of the following is equivalent to \( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{2}{8}\times \dfrac{1}{2}?\)
\( \large \dfrac{7}{16}\) Hint: Multiplication comes before addition and subtraction in the order of operations. | |
\( \large \dfrac{1}{2}\) Hint: Addition and subtraction are of equal priority in the order of operations -- do them left to right. | |
\( \large \dfrac{3}{4}\) Hint: \( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{2}{8}\times \dfrac{1}{2}\)=\( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{1}{8}\)=\( \dfrac{3}{4}+-\dfrac{1}{8}+\dfrac{1}{8}\)=\( \dfrac{3}{4}\) | |
\( \large \dfrac{3}{16}\) Hint: Multiplication comes before addition and subtraction in the order of operations. |
Question 11 |
| I. \(\large \dfrac{1}{2}+\dfrac{1}{3}\) | II. \( \large .400000\) | III. \(\large\dfrac{1}{5}+\dfrac{1}{5}\) |
| IV. \( \large 40\% \) | V. \( \large 0.25 \) | VI. \(\large\dfrac{14}{35}\) |
Which of the lists below includes all of the above expressions that are equivalent to \( \dfrac{2}{5}\)?
I, III, V, VIHint: I and V are not at all how fractions and decimals work. | |
III, VIHint: These are right, but there are more. | |
II, III, VIHint: These are right, but there are more. | |
II, III, IV, VI |
Question 12 |
Which of the following inequalities describes all values of x with \(\large \dfrac{x}{2}\le \dfrac{x}{3}\)?
\( \large x < 0\) Hint: If x =0, then x/2 = x/3, so this answer can't be correct. | |
\( \large x \le 0\) | |
\( \large x > 0\) Hint: If x =0, then x/2 = x/3, so this answer can't be correct. | |
\( \large x \ge 0\) Hint: Try plugging in x = 6. |
Question 13 |
Which of the following is equal to one million three hundred thousand?
\(\large1.3\times {{10}^{6}}\)
| |
\(\large1.3\times {{10}^{9}}\)
Hint: That's one billion three hundred million. | |
\(\large1.03\times {{10}^{6}}\)
Hint: That's one million thirty thousand. | |
\(\large1.03\times {{10}^{9}}\) Hint: That's one billion thirty million |
Question 14 |
A family went on a long car trip. Below is a graph of how far they had driven at each hour.
Which of the following is closest to their average speed driving on the trip?
\( \large d=20t\) Hint: Try plugging t=7 into the equation, and see how it matches the graph. | |
\( \large d=30t\) Hint: Try plugging t=7 into the equation, and see how it matches the graph. | |
\( \large d=40t\) | |
\( \large d=50t\) Hint: Try plugging t=7 into the equation, and see how it matches the graph. |
Question 15 |
What is the least common multiple of 540 and 216?
\( \large{{2}^{5}}\cdot {{3}^{6}}\cdot 5\) Hint: This is the product of the numbers, not the LCM. | |
\( \large{{2}^{3}}\cdot {{3}^{3}}\cdot 5\) 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 for each prime that's a factor of either number, use the largest exponent that appears in one of the factorizations. You can also take the product of the two numbers divided by their GCD. | |
\( \large{{2}^{2}}\cdot {{3}^{3}}\cdot 5\) Hint: 216 is a multiple of 8. | |
\( \large{{2}^{2}}\cdot {{3}^{2}}\cdot {{5}^{2}}\) Hint: Not a multiple of 216 and not a multiple of 540. |
Question 16 |
The column below consists of two cubes and a cylinder. The cylinder has diameter y, which is also the length of the sides of each cube. The total height of the column is 5y. Which of the formulas below gives the volume of the column?
\( \large 2{{y}^{3}}+\dfrac{3\pi {{y}^{3}}}{4}\) Hint: The cubes each have volume \(y^3\). The cylinder has radius \(\dfrac{y}{2}\) and height \(3y\). The volume of a cylinder is \(\pi r^2 h=\pi ({\dfrac{y}{2}})^2(3y)=\dfrac{3\pi {{y}^{3}}}{4}\). Note that the volume of a cylinder is analogous to that of a prism -- area of the base times height. | |
\( \large 2{{y}^{3}}+3\pi {{y}^{3}}\) Hint: y is the diameter of the circle, not the radius. | |
\( \large {{y}^{3}}+5\pi {{y}^{3}}\) Hint: Don't forget to count both cubes. | |
\( \large 2{{y}^{3}}+\dfrac{3\pi {{y}^{3}}}{8}\) Hint: Make sure you know how to find the volume of a cylinder. |
Question 17 |
Below are four inputs and outputs for a function machine representing the function A:
Which of the following equations could also represent A for the values shown?
\( \large A(n)=n+4\) Hint: For a question like this, you don't have to find the equation yourself, you can just try plugging the function machine inputs into the equation, and see if any values come out wrong. With this equation n= -1 would output 3, not 0 as the machine does. | |
\( \large A(n)=n+2\) Hint: For a question like this, you don't have to find the equation yourself, you can just try plugging the function machine inputs into the equation, and see if any values come out wrong. With this equation n= 2 would output 4, not 6 as the machine does. | |
\( \large A(n)=2n+2\) Hint: Simply plug in each of the four function machine input values, and see that the equation produces the correct output, e.g. A(2)=6, A(-1)=0, etc. | |
\( \large A(n)=2\left( n+2 \right)\) Hint: For a question like this, you don't have to find the equation yourself, you can just try plugging the function machine inputs into the equation, and see if any values come out wrong. With this equation n= 2 would output 8, not 6 as the machine does. |
Question 18 |
A biology class requires a lab fee, which is a whole number of dollars, and the same amount for all students. On Monday the instructor collected $70 in fees, on Tuesday she collected $126, and on Wednesday she collected $266. What is the largest possible amount the fee could be?
$2Hint: A possible fee, but not the largest possible fee. Check the other choices to see which are factors of all three numbers. | |
$7Hint: A possible fee, but not the largest possible fee. Check the other choices to see which are factors of all three numbers. | |
$14Hint: This is the greatest common factor of 70, 126, and 266. | |
$70Hint: Not a factor of 126 or 266, so couldn't be correct. |
Question 19 |
In March of 2012, 1 dollar was worth the same as 0.761 Euros, and 1 dollar was also worth the same as 83.03 Japanese Yen. Which of the expressions below gives the number of Yen that are worth 1 Euro?
\( \large {83}.0{3}\cdot 0.{761}\) Hint: This equation gives less than the number of yen per dollar, but 1 Euro is worth more than 1 dollar. | |
\( \large \dfrac{0.{761}}{{83}.0{3}}\) Hint: Number is way too small. | |
\( \large \dfrac{{83}.0{3}}{0.{761}}\) Hint: One strategy here is to use easier numbers, say 1 dollar = .5 Euros and 100 yen, then 1 Euro would be 200 Yen (change the numbers in the equations and see what works). Another is to use dimensional analysis: we want # yen per Euro, or yen/Euro = yen/dollar \(\times\) dollar/Euro = \(83.03 \times \dfrac {1}{0.761}\) | |
\( \large \dfrac{1}{0.{761}}\cdot \dfrac{1}{{83}.0{3}}\) Hint: Number is way too small. |
Question 20 |
There are 15 students for every teacher. Let t represent the number of teachers and let s represent the number of students. Which of the following equations is correct?
\( \large t=s+15\) Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation? | |
\( \large s=t+15\) Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation? | |
\( \large t=15s\) Hint: This is a really easy mistake to make, which comes from transcribing directly from English, "1 teachers equals 15 students." To see that it's wrong, plug in s=2; do you really need 30 teachers for 2 students? To avoid this mistake, insert the word "number," "Number of teachers equals 15 times number of students" is more clearly problematic. | |
\( \large s=15t\) |
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