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I used the official objectives and sample test to construct these questions, but cannot promise that they accurately reflect what’s on the real test. Some of the sample questions were more convoluted than I could bear to write. See terms of use. See the MTEL Practice Test main page to view questions on a particular topic or to download paper practice tests.
MTEL General Curriculum Mathematics Practice
Question 1 |
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 2 |
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 3 |
At a school fundraising event, people can buy a ticket to spin a spinner like the one below. The region that the spinner lands in tells which, if any, prize the person wins.
If 240 people buy tickets to spin the spinner, what is the best estimate of the number of keychains that will be given away?
40Hint: "Keychain" appears on the spinner twice. | |
80Hint: The probability of getting a keychain is 1/3, and so about 1/3 of the time the spinner will win. | |
100Hint: What is the probability of winning a keychain? | |
120Hint: That would be the answer for getting any prize, not a keychain specifically. |
Question 4 |
The student used a method that worked for this problem and can be generalized to any subtraction problem.Hint: Note that this algorithm is taught as the "standard" algorithm in much of Europe (it's where the term "borrowing" came from -- you borrow on top and "pay back" on the bottom). | |
The student used a method that worked for this problem and that will work for any subtraction problem that only requires one regrouping; it will not work if more regrouping is required.Hint: Try some more examples. | |
The student used a method that worked for this problem and will work for all three-digit subtraction problems, but will not work for larger problems.Hint: Try some more examples. | |
The student used a method that does not work. The student made two mistakes that cancelled each other out and was lucky to get the right answer for this problem.Hint: Remember, there are many ways to do subtraction; there is no one "right" algorithm. |
Question 5 |
Which of the numbers below is not equivalent to 4%?
\( \large \dfrac{1}{25}\) Hint: 1/25=4/100, so this is equal to 4% (be sure you read the question correctly). | |
\( \large \dfrac{4}{100}\) Hint: 4/100=4% (be sure you read the question correctly). | |
\( \large 0.4\) Hint: 0.4=40% so this is not equal to 4% | |
\( \large 0.04\) Hint: 0.04=4/100, so this is equal to 4% (be sure you read the question correctly). |
Question 6 |
Which of the lists below contains only irrational numbers?
\( \large\pi , \quad \sqrt{6},\quad \sqrt{\dfrac{1}{2}}\) | |
\( \large\pi , \quad \sqrt{9}, \quad \pi +1\) Hint: \( \sqrt{9}=3\) | |
\( \large\dfrac{1}{3},\quad \dfrac{5}{4},\quad \dfrac{2}{9}\) Hint: These are all rational. | |
\( \large-3,\quad 14,\quad 0\) Hint: These are all rational. |
Question 7 |
Cell phone plan A charges $3 per month plus $0.10 per minute. Cell phone plan B charges $29.99 per month, with no fee for the first 400 minutes and then $0.20 for each additional minute.
Which equation can be used to solve for the number of minutes, m (with m>400) that a person would have to spend on the phone each month in order for the bills for plan A and plan B to be equal?
\( \large 3.10m=400+0.2m\) Hint: These are the numbers in the problem, but this equation doesn't make sense. If you don't know how to make an equation, try plugging in an easy number like m=500 minutes to see if each side equals what it should. | |
\( \large 3+0.1m=29.99+.20m\) Hint: Doesn't account for the 400 free minutes. | |
\( \large 3+0.1m=400+29.99+.20(m-400)\) Hint: Why would you add 400 minutes and $29.99? If you don't know how to make an equation, try plugging in an easy number like m=500 minutes to see if each side equals what it should. | |
\( \large 3+0.1m=29.99+.20(m-400)\) Hint: The left side is $3 plus $0.10 times the number of minutes. The right is $29.99 plus $0.20 times the number of minutes over 400. |
Question 8 |
Given that 10 cm is approximately equal to 4 inches, which of the following expressions models a way to find out approximately how many inches are equivalent to 350 cm?
\( \large 350\times \left( \dfrac{10}{4} \right)\) Hint: The final result should be smaller than 350, and this answer is bigger. | |
\( \large 350\times \left( \dfrac{4}{10} \right)\) Hint: Dimensional analysis can help here: \(350 \text{cm} \times \dfrac{4 \text{in}}{10 \text{cm}}\). The cm's cancel and the answer is in inches. | |
\( \large (10-4) \times 350
\) Hint: This answer doesn't make much sense. Try with a simpler example (e.g. 20 cm not 350 cm) to make sure that your logic makes sense. | |
\( \large (350-10) \times 4\) Hint: This answer doesn't make much sense. Try with a simpler example (e.g. 20 cm not 350 cm) to make sure that your logic makes sense. |
Question 9 |
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 10 |
Use the expression below to answer the question that follows:
\( \large \dfrac{\left( 7,154 \right)\times \left( 896 \right)}{216}\)
Which of the following is the best estimate of the expression above?
2,000Hint: The answer is bigger than 7,000. | |
20,000Hint: Estimate 896/216 first. | |
3,000Hint: The answer is bigger than 7,000. | |
30,000Hint: \( \dfrac{896}{216} \approx 4\) and \(7154 \times 4\) is over 28,000, so this answer is closest. |
Question 11 |
A solution requires 4 ml of saline for every 7 ml of medicine. How much saline would be required for 50 ml of medicine?
\( \large 28 \dfrac{4}{7}\) ml Hint: 49 ml of medicine requires 28 ml of saline. The extra ml of saline requires 4 ml saline/ 7 ml medicine = 4/7 ml saline per 1 ml medicine. | |
\( \large 28 \dfrac{1}{4}\) ml Hint: 49 ml of medicine requires 28 ml of saline. How much saline does the extra ml require? | |
\( \large 28 \dfrac{1}{7}\) ml Hint: 49 ml of medicine requires 28 ml of saline. How much saline does the extra ml require? | |
\( \large 87.5\) ml Hint: 49 ml of medicine requires 28 ml of saline. How much saline does the extra ml require? |
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 |
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 14 |
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 15 |
Here is a student's work solving an equation:
\( x-4=-2x+6\)
\( x-4+4=-2x+6+4\)
\( x=-2x+10\)
\( x-2x=10\)
\( x=10\)
Which of the following statements is true?
The student‘s solution is correct.Hint: Try plugging into the original solution. | |
The student did not correctly use properties of equality.Hint: After \( x=-2x+10\), the student subtracted 2x on the left and added 2x on the right. | |
The student did not correctly use the distributive property.Hint: Distributive property is \(a(b+c)=ab+ac\). | |
The student did not correctly use the commutative property.Hint: Commutative property is \(a+b=b+a\) or \(ab=ba\). |
Question 16 |
| 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 17 |
Use the table below to answer the question that follows:
Gordon wants to buy three pounds of nuts. Each of the stores above ordinarily sells the nuts for $4.99 a pound, but is offering a discount this week. At which store can he buy the nuts for the least amount of money?
Store AHint: This would save about $2.50. You can quickly see that D saves more. | |
Store BHint: This saves 15% and C saves 25%. | |
Store C | |
Store DHint: This is about 20% off, which is less of a discount than C. |
Question 18 |
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. | |
\( \large s=15t\) |
Question 19 |
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 20 |
M is a multiple of 26. Which of the following cannot be true?
M is odd.Hint: All multiples of 26 are also multiples of 2, so they must be even. | |
M is a multiple of 3.Hint: 3 x 26 is a multiple of both 3 and 26. | |
M is 26.Hint: 1 x 26 is a multiple of 26. | |
M is 0.Hint: 0 x 26 is a multiple of 26. |
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