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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 speed of sound in dry air at 68 degrees F is 343.2 meters per second. Which of the expressions below could be used to compute the number of kilometers that a sound wave travels in 10 minutes (in dry air at 68 degrees F)?
\( \large 343.2\times 60\times 10\) Hint: In kilometers, not meters. | |
\( \large 343.2\times 60\times 10\times \dfrac{1}{1000}\) Hint: Units are meters/sec \(\times\) seconds/minute \(\times\) minutes \(\times\) kilometers/meter, and the answer is in kilometers. | |
\( \large 343.2\times \dfrac{1}{60}\times 10\) Hint: Include units and make sure answer is in kilometers. | |
\( \large 343.2\times \dfrac{1}{60}\times 10\times \dfrac{1}{1000}\) Hint: Include units and make sure answer is in kilometers. |
Question 3 |
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 4 |
The picture below shows identical circles drawn on a piece of paper. The rectangle represents an index card that is blocking your view of \( \dfrac{3}{5}\) of the circles on the paper. How many circles are covered by the rectangle?

4Hint: The card blocks more than half of the circles, so this number is too small. | |
5Hint: The card blocks more than half of the circles, so this number is too small. | |
8Hint: The card blocks more than half of the circles, so this number is too small. | |
12Hint: 2/5 of the circles or 8 circles are showing. Thus 4 circles represent 1/5 of the circles, and \(4 \times 5=20\) circles represent 5/5 or all the circles. Thus 12 circles are hidden. |
Question 5 |
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 6 |
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 7 |
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 8 |
Which of the lists below is in order from least to greatest value?
\( \large -0.044,\quad -0.04,\quad 0.04,\quad 0.044\) Hint: These are easier to compare if you add trailing zeroes (this is finding a common denominator) -- all in thousandths, -0.044, -0.040,0 .040, 0.044. The middle two numbers, -0.040 and 0.040 can be modeled as owing 4 cents and having 4 cents. The outer two numbers are owing or having a bit more. | |
\( \large -0.04,\quad -0.044,\quad 0.044,\quad 0.04\) Hint: 0.04=0.040, which is less than 0.044. | |
\( \large -0.04,\quad -0.044,\quad 0.04,\quad 0.044\) Hint: -0.04=-0.040, which is greater than \(-0.044\). | |
\( \large -0.044,\quad -0.04,\quad 0.044,\quad 0.04\) Hint: 0.04=0.040, which is less than 0.044. |
Question 9 |
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 10 |
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 11 |
Which of the following values of x satisfies the inequality \( \large \left| {{(x+2)}^{3}} \right|<3?\)
\( \large x=-3\) Hint: \( \left| {{(-3+2)}^{3}} \right|\)=\( \left | {(-1)}^3 \right | \)=\( \left | -1 \right |=1 \) . | |
\( \large x=0\) Hint: \( \left| {{(0+2)}^{3}} \right|\)=\( \left | {2}^3 \right | \)=\( \left | 8 \right | \) =\( 8\) | |
\( \large x=-4\) Hint: \( \left| {{(-4+2)}^{3}} \right|\)=\( \left | {(-2)}^3 \right | \)=\( \left | -8 \right | \) =\( 8\) | |
\( \large x=1\) Hint: \( \left| {{(1+2)}^{3}} \right|\)=\( \left | {3}^3 \right | \)=\( \left | 27 \right | \) = \(27\) |
Question 12 |
A homeowner is planning to tile the kitchen floor with tiles that measure 6 inches by 8 inches. The kitchen floor is a rectangle that measures 10 ft by 12 ft, and there are no gaps between the tiles. How many tiles does the homeowner need?
30Hint: The floor is 120 sq feet, and the tiles are smaller than 1 sq foot. Also, remember that 1 sq foot is 12 \(\times\) 12=144 sq inches. | |
120Hint: The floor is 120 sq feet, and the tiles are smaller than 1 sq foot. | |
300Hint: Recheck your calculations. | |
360Hint: One way to do this is to note that 6 inches = 1/2 foot and 8 inches = 2/3 foot, so the area of each tile is 1/2 \(\times\) 2/3=1/3 sq foot, or each square foot of floor requires 3 tiles. The area of the floor is 120 square feet. Note that the tiles would fit evenly oriented in either direction, parallel to the walls. |
Question 13 |
The table below gives data from various years on how many young girls drank milk.
Based on the data given above, what was the probability that a randomly chosen girl in 1990 drank milk?
\( \large \dfrac{502}{1222}\) Hint: This is the probability that a randomly chosen girl who drinks milk was in the 1989-1991 food survey. | |
\( \large \dfrac{502}{2149}\) Hint: This is the probability that a randomly chosen girl from the whole survey drank milk and was also surveyed in 1989-1991. | |
\( \large \dfrac{502}{837}\) | |
\( \large \dfrac{1222}{2149}\) Hint: This is the probability that a randomly chosen girl from any year of the survey drank milk. |
Question 14 |
Which of the following is equal to eleven billion four hundred thousand?
\( \large 11,400,000\) Hint: That's eleven million four hundred thousand. | |
\(\large11,000,400,000\) | |
\( \large11,000,000,400,000\) Hint: That's eleven trillion four hundred thousand (although with British conventions; this answer is correct, but in the US, it isn't). | |
\( \large 11,400,000,000\) Hint: That's eleven billion four hundred million |
Question 15 |
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 16 |
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 17 |
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 18 |
The first histogram shows the average life expectancies for women in different countries in Africa in 1998; the second histogram gives similar data for Europe:
How much bigger is the range of the data for Africa than the range of the data for Europe?
0 yearsHint: Range is the maximum life expectancy minus the minimum life expectancy. | |
12 yearsHint: Are you subtracting frequencies? Range is about values of the data, not frequency. | |
18 yearsHint: It's a little hard to read the graph, but it doesn't matter if you're consistent. It looks like the range for Africa is 80-38= 42 years and for Europe is 88-64 = 24; 42-24=18. | |
42 yearsHint: Read the question more carefully. |
Question 19 |
Solve for x: \(\large 4-\dfrac{2}{3}x=2x\)
\( \large x=3\) Hint: Try plugging x=3 into the equation. | |
\( \large x=-3\) Hint: Left side is positive, right side is negative when you plug this in for x. | |
\( \large x=\dfrac{3}{2}\) Hint: One way to solve: \(4=\dfrac{2}{3}x+2x\) \(=\dfrac{8}{3}x\).\(x=\dfrac{3 \times 4}{8}=\dfrac{3}{2}\). Another way is to just plug x=3/2 into the equation and see that each side equals 3 -- on a multiple choice test, you almost never have to actually solve for x. | |
\( \large x=-\dfrac{3}{2}\) Hint: Left side is positive, right side is negative when you plug this in for x. |
Question 20 |
If two fair coins are flipped, what is the probability that one will come up heads and the other tails?
\( \large \dfrac{1}{4}\) Hint: Think of the coins as a penny and a dime, and list all possibilities. | |
\( \large \dfrac{1}{3} \) Hint: This is a very common misconception. There are three possible outcomes -- both heads, both tails, and one of each -- but they are not equally likely. Think of the coins as a penny and a dime, and list all possibilities. | |
\( \large \dfrac{1}{2}\) Hint: The possibilities are HH, HT, TH, TT, and all are equally likely. Two of the four have one of each coin, so the probability is 2/4=1/2. | |
\( \large \dfrac{3}{4}\) Hint: Think of the coins as a penny and a dime, and list all possibilities. |
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