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MTEL General Curriculum Mathematics Practice
Question 1 |
The "houses" below are made of toothpicks and gum drops.
How many toothpicks are there in a row of 53 houses?
212Hint: Can the number of toothpicks be even? | |
213Hint: One way to see this is that every new "house" adds 4 toothpicks to the leftmost vertical toothpick -- so the total number is 1 plus 4 times the number of "houses." There are many other ways to look at the problem too. | |
217Hint: Try your strategy with a smaller number of "houses" so you can count and find your mistake. | |
265Hint: Remember that the "houses" overlap some walls. |
Question 2 |
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 3 |
A car is traveling at 60 miles per hour. Which of the expressions below could be used to compute how many feet the car travels in 1 second? Note that 1 mile = 5,280 feet.
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot 5280\dfrac{\text{feet}}{\text{mile}}\cdot 60\dfrac{\text{minutes}}{\text{hour}}\cdot 60\dfrac{\text{seconds}}{\text{minute}}
\) Hint: This answer is not in feet/second. | |
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot 5280\dfrac{\text{feet}}{\text{mile}}\cdot \dfrac{1}{60}\dfrac{\text{hour}}{\text{minutes}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}}
\) Hint: This is the only choice where the answer is in feet per second and the unit conversions are correct. | |
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot \dfrac{1}{5280}\dfrac{\text{foot}}{\text{miles}}\cdot 60\dfrac{\text{hours}}{\text{minute}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}}\) Hint: Are there really 60 hours in a minute? | |
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot \dfrac{1}{5280}\dfrac{\text{mile}}{\text{feet}}\cdot 60\dfrac{\text{minutes}}{\text{hour}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}}\) Hint: This answer is not in feet/second. |
Question 4 |
The expression \( \large {{7}^{-4}}\cdot {{8}^{-6}}\) is equal to which of the following?
\( \large \dfrac{8}{{{\left( 56 \right)}^{4}}}\) Hint: The bases are whole numbers, and the exponents are negative. How can the numerator be 8? | |
\( \large \dfrac{64}{{{\left( 56 \right)}^{4}}}\) Hint: The bases are whole numbers, and the exponents are negative. How can the numerator be 64? | |
\( \large \dfrac{1}{8\cdot {{\left( 56 \right)}^{4}}}\) Hint: \(8^{-6}=8^{-4} \times 8^{-2}\) | |
\( \large \dfrac{1}{64\cdot {{\left( 56 \right)}^{4}}}\) |
Question 5 |
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 6 |
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 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 |
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 9 |
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 10 |
Which of the numbers below is a fraction equivalent to \( 0.\bar{6}\)?
\( \large \dfrac{4}{6}\) Hint: \( 0.\bar{6}=\dfrac{2}{3}=\dfrac{4}{6}\) | |
\( \large \dfrac{3}{5}\) Hint: This is equal to 0.6, without the repeating decimal. Answer is equivalent to choice c, which is another way to tell that it's wrong. | |
\( \large \dfrac{6}{10}\) Hint: This is equal to 0.6, without the repeating decimal. Answer is equivalent to choice b, which is another way to tell that it's wrong. | |
\( \large \dfrac{1}{6}\) Hint: This is less than a half, and \( 0.\bar{6}\) is greater than a half. |
Question 11 |
The letters A, B, and C represent digits (possibly equal) in the twelve digit number x=111,111,111,ABC. For which values of A, B, and C is x divisible by 40?
\( \large A = 3, B = 2, C=0\) Hint: Note that it doesn't matter what the first 9 digits are, since 1000 is divisible by 40, so DEF,GHI,JKL,000 is divisible by 40 - we need to check the last 3. | |
\( \large A = 0, B = 0, C=4\) Hint: Not divisible by 10, since it doesn't end in 0. | |
\( \large A = 4, B = 2, C=0\) Hint: Divisible by 10 and by 4, but not by 40, as it's not divisible by 8. Look at 40 as the product of powers of primes -- 8 x 5, and check each. To check 8, either check whether 420 is divisible by 8, or take ones place + twice tens place + 4 * hundreds place = 18, which is not divisible by 8. | |
\( \large A =1, B=0, C=0\) Hint: Divisible by 10 and by 4, but not by 40, as it's not divisible by 8. Look at 40 as the product of powers of primes -- 8 x 5, and check each. To check 8, either check whether 100 is divisible by 8, or take ones place + twice tens place + 4 * hundreds place = 4, which is not divisible by 8. |
Question 12 |
In the triangle below, \(\overline{AC}\cong \overline{AD}\cong \overline{DE}\) and \(m\angle CAD=100{}^\circ \). What is \(m\angle DAE\)?

\( \large 20{}^\circ \) Hint: Angles ACD and ADC are congruent since they are base angles of an isosceles triangle. Since the angles of a triangle sum to 180, they sum to 80, and they are 40 deg each. Thus angle ADE is 140 deg, since it makes a straight line with angle ADC. Angles DAE and DEA are base angles of an isosceles triangle and thus congruent-- they sum to 40 deg, so are 20 deg each. | |
\( \large 25{}^\circ \) Hint: If two sides of a triangle are congruent, then it's isosceles, and the base angles of an isosceles triangle are equal. | |
\( \large 30{}^\circ \) Hint: If two sides of a triangle are congruent, then it's isosceles, and the base angles of an isosceles triangle are equal. | |
\( \large 40{}^\circ \) Hint: Make sure you're calculating the correct angle. |
Question 13 |
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 14 |
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 15 |
The polygon depicted below is drawn on dot paper, with the dots spaced 1 unit apart. What is the perimeter of the polygon?

\( \large 18+\sqrt{2} \text{ units}\) Hint: Be careful with the Pythagorean Theorem. | |
\( \large 18+2\sqrt{2}\text{ units}\) Hint: There are 13 horizontal or vertical 1 unit segments. The longer diagonal is the hypotenuse of a 3-4-5 right triangle, so its length is 5 units. The shorter diagonal is the hypotenuse of a 45-45-90 right triangle with side 2, so its hypotenuse has length \(2 \sqrt{2}\). | |
\( \large 18 \text{ units}
\) Hint: Use the Pythagorean Theorem to find the lengths of the diagonal segments. | |
\( \large 20 \text{ units}\) Hint: Use the Pythagorean Theorem to find the lengths of the diagonal segments. |
Question 16 |
Which of the following sets of polygons can be assembled to form a pentagonal pyramid?
2 pentagons and 5 rectangles.Hint: These can be assembled to form a pentagonal prism, not a pentagonal pyramid. | |
1 square and 5 equilateral triangles.Hint: You need a pentagon for a pentagonal pyramid. | |
1 pentagon and 5 isosceles triangles. | |
1 pentagon and 10 isosceles triangles. |
Question 17 |
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 18 |
The least common multiple of 60 and N is 1260. Which of the following could be the prime factorization of N?
\( \large2\cdot 5\cdot 7\) Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM. | |
\( \large{{2}^{3}}\cdot {{3}^{2}}\cdot 5 \cdot 7\) Hint: 1260 is not divisible by 8, so it isn't a multiple of this N. | |
\( \large3 \cdot 5 \cdot 7\) Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM. | |
\( \large{{3}^{2}}\cdot 5\cdot 7\) Hint: \(1260=2^2 \cdot 3^2 \cdot 5 \cdot 7\) and \(60=2^2 \cdot 3 \cdot 5\). In order for 1260 to be the LCM, N has to be a multiple of \(3^2\) and of 7 (because 60 is not a multiple of either of these). N also cannot introduce a factor that would require the LCM to be larger (as in choice b). |
Question 19 |
The chart below gives percentiles for the number of sit-ups that boys of various ages can do in 60 seconds (source , June 24, 2011)
Which of the following statements can be inferred from the above chart?
95% of 12 year old boys can do 56 sit-ups in 60 seconds.Hint: The 95th percentile means that 95% of scores are less than or equal to 56, and 5% are greater than or equal to 56. | |
At most 25% of 7 year old boys can do 19 or more sit-ups in 60 seconds.Hint: The 25th percentile means that 25% of scores are less than or equal to 19, and 75% are greater than or equal to 19. | |
Half of all 13 year old boys can do less than 41 sit-ups in 60 seconds and half can do more than 41 sit-ups in 60 seconds.Hint: Close, but not quite. There's no accounting for boys who can do exactly 41 sit ups. Look at these data: 10, 20, 41, 41, 41, 41, 50, 60, 90. The median is 41, but more than half can do 41 or more. | |
At least 75% of 16 year old boys can only do 51 or fewer sit-ups in 60 seconds.Hint: The "at least" is necessary due to duplicates. Suppose the data were 10, 20, 51, 51. The 75th percentile is 51, but 100% of the boys can only do 51 or fewer situps. |
Question 20 |
What set of transformations will transform the leftmost image into the rightmost image?
A 90 degree clockwise rotation about (2,1) followed by a translation of two units to the right.Hint: Part of the figure would move below the x-axis with these transformations. | |
A translation 3 units up, followed by a reflection about the line y=x.Hint: See what happens to the point (5,1) under this set of transformations. | |
A 90 degree clockwise rotation about (5,1), followed by a translation of 2 units up. | |
A 90 degree clockwise rotation about (2,1) followed by a translation of 2 units to the right.Hint: See what happens to the point (3,3) under this set of transformations. |
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