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MTEL General Curriculum Mathematics Practice
Question 1 |
On a map the distance from Boston to Detroit is 6 cm, and these two cities are 702 miles away from each other. Assuming the scale of the map is the same throughout, which answer below is closest to the distance between Boston and San Francisco on the map, given that they are 2,708 miles away from each other?
21 cmHint: How many miles would correspond to 24 cm on the map? Try adjusting from there. | |
22 cmHint: How many miles would correspond to 24 cm on the map? Try adjusting from there. | |
23 cmHint: One way to solve this without a calculator is to note that 4 groups of 6 cm is 2808 miles, which is 100 miles too much. Then 100 miles would be about 1/7 th of 6 cm, or about 1 cm less than 24 cm. | |
24 cmHint: 4 groups of 6 cm is over 2800 miles on the map, which is too much. |
Question 2 |
What is the length of side \(\overline{BD}\) in the triangle below, where \(\angle DBA\) is a right angle?

\( \large 1\) Hint: Use the Pythagorean Theorem. | |
\( \large \sqrt{5}\) Hint: \(2^2+e^2=3^2\) or \(4+e^2=9;e^2=5; e=\sqrt{5}\). | |
\( \large \sqrt{13}\) Hint: e is not the hypotenuse. | |
\( \large 5\) Hint: Use the Pythagorean Theorem. |
Question 3 |
The "houses" below are made of toothpicks and gum drops.
Which of the following does not represent the number of gumdrops in a row of h houses?
\( \large 2+3h\) Hint: Think of this as start with 2 gumdrops on the left wall, and then add 3 gumdrops for each house. | |
\( \large 5+3(h-1)\) Hint: Think of this as start with one house, and then add 3 gumdrops for each of the other h-1 houses. | |
\( \large h+(h+1)+(h+1)\) Hint: Look at the gumdrops in 3 rows: h gumdrops for the "rooftops," h+1 for the tops of the vertical walls, and h+1 for the floors. | |
\( \large 5+3h\) Hint: This one is not a correct equation (which makes it the correct answer!). Compare to choice A. One of them has to be wrong, as they differ by 3. |
Question 4 |
How many lines of reflective symmetry and how many centers of rotational symmetry does the parallelogram depicted below have?
4 lines of reflective symmetry, 1 center of rotational symmetry.Hint: Try cutting out a shape like this one from paper, and fold where you think the lines of reflective symmetry are (or put a mirror there). Do things line up as you thought they would? | |
2 lines of reflective symmetry, 1 center of rotational symmetry.Hint: Try cutting out a shape like this one from paper, and fold where you think the lines of reflective symmetry are (or put a mirror there). Do things line up as you thought they would? | |
0 lines of reflective symmetry, 1 center of rotational symmetry.Hint: The intersection of the diagonals is a center of rotational symmetry. There are no lines of reflective symmetry, although many people get confused about this fact (best to play with hands on examples to get a feel). Just fyi, the letter S also has rotational, but not reflective symmetry, and it's one that kids often write backwards. | |
2 lines of reflective symmetry, 0 centers of rotational symmetry.Hint: Try cutting out a shape like this one from paper. Trace onto another sheet of paper. See if there's a way to rotate the cut out shape (less than a complete turn) so that it fits within the outlines again. |
Question 5 |
Use the expression below to answer the question that follows.
\(\large \dfrac{\left( 155 \right)\times \left( 6,124 \right)}{977}\)
Which of the following is the best estimate of the expression above?
100Hint: 6124/977 is approximately 6. | |
200Hint: 6124/977 is approximately 6. | |
1,000Hint: 6124/977 is approximately 6. 155 is approximately 150, and \( 6 \times 150 = 3 \times 300 = 900\), so this answer is closest. | |
2,000Hint: 6124/977 is approximately 6. |
Question 6 |
Aya and Kendra want to estimate the height of a tree. On a sunny day, Aya measures Kendra's shadow as 3 meters long, and Kendra measures the tree's shadow as 15 meters long. Kendra is 1.5 meters tall. How tall is the tree?
7.5 metersHint: Here is a picture, note that the large and small right triangles are similar: ![]() One way to do the problem is to note that there is a dilation (scale) factor of 5 on the shadows, so there must be that factor on the heights too. Another way is to note that the shadows are twice as long as the heights. | |
22.5 metersHint: Draw a picture. | |
30 metersHint: Draw a picture. | |
45 metersHint: Draw a picture. |
Question 7 |
What is the probability that two randomly selected people were born on the same day of the week? Assume that all days are equally probable.
\( \large \dfrac{1}{7}\) Hint: It doesn't matter what day the first person was born on. The probability that the second person will match is 1/7 (just designate one person the first and the other the second). Another way to look at it is that if you list the sample space of all possible pairs, e.g. (Wed, Sun), there are 49 such pairs, and 7 of them are repeats of the same day, and 7/49=1/7. | |
\( \large \dfrac{1}{14}\) Hint: What would be the sample space here? Ie, how would you list 14 things that you pick one from? | |
\( \large \dfrac{1}{42}\) Hint: If you wrote the seven days of the week on pieces of paper and put the papers in a jar, this would be the probability that the first person picked Sunday and the second picked Monday from the jar -- not the same situation. | |
\( \large \dfrac{1}{49}\) Hint: This is the probability that they are both born on a particular day, e.g. Sunday. |
Question 8 |
Which of the following nets will not fold into a cube?
![]() Hint: If you have trouble visualizing, cut them out and fold (during the test, you can tear paper to approximate). | |
![]() | |
![]() Hint: If you have trouble visualizing, cut them out and fold (during the test, you can tear paper to approximate). | |
![]() Hint: If you have trouble visualizing, cut them out and fold (during the test, you can tear paper to approximate). |
Question 9 |
Which of the graphs below represent functions?
I.
II.
III.
IV. 
I and IV only.Hint: There are vertical lines that go through 2 points in IV . | |
I and III only.Hint: Even though III is not continuous, it's still a function (assuming that vertical lines between the "steps" do not go through 2 points). | |
II and III only.Hint: Learn about the vertical line test. | |
I, II, and IV only.Hint: There are vertical lines that go through 2 points in II. |
Question 10 |
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 11 |
Use the graph below to answer the question that follows.
Which of the following is a correct equation for the graph of the line depicted above?
\( \large y=-\dfrac{1}{2}x+2\) Hint: The slope is -1/2 and the y-intercept is 2. You can also try just plugging in points. For example, this is the only choice that gives y=1 when x=2. | |
\( \large 4x=2y\) Hint: This line goes through (0,0); the graph above does not. | |
\( \large y=x+2\) Hint: The line pictured has negative slope. | |
\( \large y=-x+2\) Hint: Try plugging x=4 into this equation and see if that point is on the graph above. |
Question 12 |
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 13 |
Here are some statements:
I) 5 is an integer II)\( -5 \) is an integer III) \(0\) is an integer
Which of the statements are true?
I only | |
I and II only | |
I and III only | |
I, II, and IIIHint: The integers are ...-3, -2, -1, 0, 1, 2, 3, .... |
Question 14 |
In each expression below N represents a negative integer. Which expression could have a negative value?
\( \large {{N}^{2}}\) Hint: Squaring always gives a non-negative value. | |
\( \large 6-N\) Hint: A story problem for this expression is, if it was 6 degrees out at noon and N degrees out at sunrise, by how many degrees did the temperature rise by noon? Since N is negative, the answer to this question has to be positive, and more than 6. | |
\( \large -N\) Hint: If N is negative, then -N is positive | |
\( \large 6+N\) Hint: For example, if \(N=-10\), then \(6+N = -4\) |
Question 15 |
A family has four children. What is the probability that two children are girls and two are boys? Assume the the probability of having a boy (or a girl) is 50%.
\( \large \dfrac{1}{2}\) Hint: How many different configurations are there from oldest to youngest, e.g. BGGG? How many of them have 2 boys and 2 girls? | |
\( \large \dfrac{1}{4}\) Hint: How many different configurations are there from oldest to youngest, e.g. BGGG? How many of them have 2 boys and 2 girls? | |
\( \large \dfrac{1}{5}\) Hint: Some configurations are more probable than others -- i.e. it's more likely to have two boys and two girls than all boys. Be sure you are weighting properly. | |
\( \large \dfrac{3}{8}\) Hint: There are two possibilities for each child, so there are \(2 \times 2 \times 2 \times 2 =16\) different configurations, e.g. from oldest to youngest BBBG, BGGB, GBBB, etc. Of these configurations, there are 6 with two boys and two girls (this is the combination \(_{4}C_{2}\) or "4 choose 2"): BBGG, BGBG, BGGB, GGBB, GBGB, and GBBG. Thus the probability is 6/16=3/8. |
Question 16 |
A map has a scale of 3 inches = 100 miles. Cities A and B are 753 miles apart. Let d be the distance between the two cities on the map. Which of the following is not correct?
\( \large \dfrac{3}{100}=\dfrac{d}{753}\) Hint: Units on both side are inches/mile, and both numerators and denominators correspond -- this one is correct. | |
\( \large \dfrac{3}{100}=\dfrac{753}{d}\) Hint: Unit on the left is inches per mile, and on the right is miles per inch. The proportion is set up incorrectly (which is what we wanted). Another strategy is to notice that one of A or B has to be the answer because they cannot both be correct proportions. Then check that cross multiplying on A gives part D, so B is the one that is different from the other 3. | |
\( \large \dfrac{3}{d}=\dfrac{100}{753}\) Hint: Unitless on each side, as inches cancel on the left and miles on the right. Numerators correspond to the map, and denominators to the real life distances -- this one is correct. | |
\( \large 100d=3\cdot 753\) Hint: This is equivalent to part A. |
Question 17 |
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 18 |
Below are front, side, and top views of a three-dimensional solid.
Which of the following could be the solid shown above?
A sphereHint: All views would be circles. | |
A cylinder | |
A coneHint: Two views would be triangles, not rectangles. | |
A pyramidHint: How would one view be a circle? |
Question 19 |
A family on vacation drove the first 200 miles in 4 hours and the second 200 miles in 5 hours. Which expression below gives their average speed for the entire trip?
\( \large \dfrac{200+200}{4+5}\) Hint: Average speed is total distance divided by total time. | |
\( \large \left( \dfrac{200}{4}+\dfrac{200}{5} \right)\div 2\) Hint: This seems logical, but the problem is that it weights the first 4 hours and the second 5 hours equally, when each hour should get the same weight in computing the average speed. | |
\( \large \dfrac{200}{4}+\dfrac{200}{5} \) Hint: This would be an average of 90 miles per hour! | |
\( \large \dfrac{400}{4}+\dfrac{400}{5} \) Hint: This would be an average of 180 miles per hour! Even a family of race car drivers probably doesn't have that average speed on a vacation! |
Question 20 |
Elena is going to use a calculator to check whether or not 267 is prime. She will pick certain divisors, and then find 267 divided by each, and see if she gets a whole number. If she never gets a whole number, then she's found a prime. Which numbers does Elena NEED to check before she can stop checking and be sure she has a prime?
All natural numbers from 2 to 266.Hint: She only needs to check primes -- checking the prime factors of any composite is enough to look for divisors. As a test taking strategy, the other three choices involve primes, so worth thinking about. | |
All primes from 2 to 266 .Hint: Remember, factors come in pairs (except for square root factors), so she would first find the smaller of the pair and wouldn't need to check the larger. | |
All primes from 2 to 133 .Hint: She doesn't need to check this high. Factors come in pairs, and something over 100 is going to be paired with something less than 3, so she will find that earlier. | |
All primes from \( \large 2\) to \( \large \sqrt{267}\).Hint: \(\sqrt{267} \times \sqrt{267}=267\). Any other pair of factors will have one factor less than \( \sqrt{267}\) and one greater, so she only needs to check up to \( \sqrt{267}\). |
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