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MTEL General Curriculum Mathematics Practice
Question 1 |
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 2 |
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 3 |
Use the graph below to answer the question that follows.
If the polygon shown above is reflected about the y axis and then rotated 90 degrees clockwise about the origin, which of the following graphs is the result?
![]() Hint: Try following the point (1,4) to see where it goes after each transformation. | |
![]() | |
Hint: Make sure you're reflecting in the correct axis. | |
![]() Hint: Make sure you're rotating the correct direction. |
Question 4 |
How many factors does 80 have?
\( \large8\) Hint: Don't forget 1 and 80. | |
\( \large9\) Hint: Only perfect squares have an odd number of factors -- otherwise factors come in pairs. | |
\( \large10\) Hint: 1,2,4,5,8,10,16,20,40,80 | |
\( \large12\) Hint: Did you count a number twice? Include a number that isn't a factor? |
Question 5 |
Here is a number trick:
1) Pick a whole number
2) Double your number.
3) Add 20 to the above result.
4) Multiply the above by 5
5) Subtract 100
6) Divide by 10
The result is always the number that you started with! Suppose you start by picking N. Which of the equations below best demonstrates that the result after Step 6 is also N?
\( \large N*2+20*5-100\div 10=N\) Hint: Use parentheses or else order of operations is off. | |
\( \large \left( \left( 2*N+20 \right)*5-100 \right)\div 10=N\) | |
\( \large \left( N+N+20 \right)*5-100\div 10=N\) Hint: With this answer you would subtract 10, instead of subtracting 100 and then dividing by 10. | |
\( \large \left( \left( \left( N\div 10 \right)-100 \right)*5+20 \right)*2=N\) Hint: This answer is quite backwards. |
Question 6 |
The picture below represents a board with pegs on it, where the closest distance between two pegs is 1 cm. What is the area of the pentagon shown?

Question 7 |
The window glass below has the shape of a semi-circle on top of a square, where the side of the square has length x. It was cut from one piece of glass.

What is the perimeter of the window glass?
\( \large 3x+\dfrac{\pi x}{2}\) Hint: By definition, \(\pi\) is the ratio of the circumference of a circle to its diameter; thus the circumference is \(\pi d\). Since we have a semi-circle, its perimeter is \( \dfrac{1}{2} \pi x\). Only 3 sides of the square contribute to the perimeter. | |
\( \large 3x+2\pi x\) Hint: Make sure you know how to find the circumference of a circle. | |
\( \large 3x+\pi x\) Hint: Remember it's a semi-circle, not a circle. | |
\( \large 4x+2\pi x\) Hint: Only 3 sides of the square contribute to the perimeter. |
Question 8 |
Which of the lists below is in order from least to greatest value?
\( \large \dfrac{1}{2},\quad \dfrac{1}{3},\quad \dfrac{1}{4},\quad \dfrac{1}{5}\) Hint: This is ordered from greatest to least. | |
\( \large \dfrac{1}{3},\quad \dfrac{2}{7},\quad \dfrac{3}{8},\quad \dfrac{4}{11}\) Hint: 1/3 = 2/6 is bigger than 2/7. | |
\( \large \dfrac{1}{4},\quad \dfrac{2}{5},\quad \dfrac{2}{3},\quad \dfrac{4}{5}\) Hint: One way to look at this: 1/4 and 2/5 are both less than 1/2, and 2/3 and 4/5 are both greater than 1/2. 1/4 is 25% and 2/5 is 40%, so 2/5 is greater. The distance from 2/3 to 1 is 1/3 and from 4/5 to 1 is 1/5, and 1/5 is less than 1/3, so 4/5 is bigger. | |
\( \large \dfrac{7}{8},\quad \dfrac{6}{7},\quad \dfrac{5}{6},\quad \dfrac{4}{5}\) Hint: This is in order from greatest to least. |
Question 9 |
An above-ground swimming pool is in the shape of a regular hexagonal prism, is one meter high, and holds 65 cubic meters of water. A second pool has a base that is also a regular hexagon, but with sides twice as long as the sides in the first pool. This second pool is also one meter high. How much water will the second pool hold?
\( \large 65\text{ }{{\text{m}}^{3}}\) Hint: A bigger pool would hold more water. | |
\( \large 65\cdot 2\text{ }{{\text{m}}^{3}}\) Hint: Try a simpler example, say doubling the sides of the base of a 1 x 1 x 1 cube. | |
\( \large 65\cdot 4\text{ }{{\text{m}}^{3}}\) Hint: If we think of the pool as filled with 1 x 1 x 1 cubes (and some fractions of cubes), then scaling to the larger pool changes each 1 x 1 x 1 cube to a 2 x 2 x 1 prism, or multiplies volume by 4. | |
\( \large 65\cdot 8\text{ }{{\text{m}}^{3}}\) Hint: Try a simpler example, say doubling the sides of the base of a 1 x 1 x 1 cube. |
Question 10 |
P is a prime number that divides 240. Which of the following must be true?
P divides 30Hint: 2, 3, and 5 are the prime factors of 240, and all divide 30. | |
P divides 48Hint: P=5 doesn't work. | |
P divides 75Hint: P=2 doesn't work. | |
P divides 80Hint: P=3 doesn't work. |
Question 11 |
Each individual cube that makes up the rectangular solid depicted below has 6 inch sides. What is the surface area of the solid in square feet?
\( \large 11\text{ f}{{\text{t}}^{2}}\) Hint: Check your units and make sure you're using feet and inches consistently. | |
\( \large 16.5\text{ f}{{\text{t}}^{2}}\) Hint: Each square has surface area \(\dfrac{1}{2} \times \dfrac {1}{2}=\dfrac {1}{4}\) sq feet. There are 9 squares on the top and bottom, and 12 on each of 4 sides, for a total of 66 squares. 66 squares \(\times \dfrac {1}{4}\) sq feet/square =16.5 sq feet. | |
\( \large 66\text{ f}{{\text{t}}^{2}}\) Hint: The area of each square is not 1. | |
\( \large 2376\text{ f}{{\text{t}}^{2}}\) Hint: Read the question more carefully -- the answer is supposed to be in sq feet, not sq inches.
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Question 12 |
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 13 |
Four children randomly line up, single file. What is the probability that they are in height order, with the shortest child in front? All of the children are different heights.
\( \large \dfrac{1}{4}\) Hint: Try a simpler question with 3 children -- call them big, medium, and small -- and list all the ways they could line up. Then see how to extend your logic to the problem with 4 children. | |
\( \large \dfrac{1}{256}
\) Hint: Try a simpler question with 3 children -- call them big, medium, and small -- and list all the ways they could line up. Then see how to extend your logic to the problem with 4 children. | |
\( \large \dfrac{1}{16}\) Hint: Try a simpler question with 3 children -- call them big, medium, and small -- and list all the ways they could line up. Then see how to extend your logic to the problem with 4 children. | |
\( \large \dfrac{1}{24}\) Hint: The number of ways for the children to line up is \(4!=4 \times 3 \times 2 \times 1 =24\) -- there are 4 choices for who is first in line, then 3 for who is second, etc. Only one of these lines has the children in the order specified. |
Question 14 |
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 15 |
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 16 |
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 17 |
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 18 |
\( \large \dfrac{17}{24}\) Hint: You might try adding segments so each quadrant is divided into 6 pieces with equal area -- there will be 24 regions, not all the same shape, but all the same area, with 17 of them shaded (for the top left quarter, you could also first change the diagonal line to a horizontal or vertical line that divides the square in two equal pieces and shade one) . | |
\( \large \dfrac{3}{4}\) Hint: Be sure you're taking into account the different sizes of the pieces. | |
\( \large \dfrac{2}{3}\) Hint: The bottom half of the picture is 2/3 shaded, and the top half is more than 2/3 shaded, so this answer is too small. | |
\( \large \dfrac{17}{6} \) Hint: This answer is bigger than 1, so doesn't make any sense. Be sure you are using the whole picture, not one quadrant, as the unit. |
Question 19 |
A class is using base-ten block to represent numbers. A large cube represents 1000, a flat represents 100, a rod represents 10, and a little cube represents 1. Which of these is not a correct representation for 2,347?
23 flats, 4 rods, 7 little cubesHint: Be sure you read the question carefully: 2300+40+7=2347 | |
2 large cubes, 3 flats, 47 rodsHint: 2000+300+470 \( \neq\) 2347 | |
2 large cubes, 34 rods, 7 little cubesHint: Be sure you read the question carefully: 2000+340+7=2347 | |
2 large cubes, 3 flats, 4 rods, 7 little cubesHint: Be sure you read the question carefully: 2000+300+40+7=2347 |
Question 20 |
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. |
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