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MTEL General Curriculum Mathematics Practice
Question 1 |
If x is an integer, which of the following must also be an integer?
\( \large \dfrac{x}{2}\) Hint: If x is odd, then \( \dfrac{x}{2} \) is not an integer, e.g. 3/2 = 1.5. | |
\( \large \dfrac{2}{x}\) Hint: Only an integer if x = -2, -1, 1, or 2. | |
\( \large-x\) Hint: -1 times any integer is still an integer. | |
\(\large\sqrt{x}\) Hint: Usually not an integer, e.g. \( \sqrt{2} \approx 1.414 \). |
Question 2 |
Kendra is trying to decide which fraction is greater, \( \dfrac{4}{7}\) or \( \dfrac{5}{8}\). Which of the following answers shows the best reasoning?
\( \dfrac{4}{7}\) is \( \dfrac{3}{7}\)away from 1, and \( \dfrac{5}{8}\) is \( \dfrac{3}{8}\)away from 1. Since eighth‘s are smaller than seventh‘s, \( \dfrac{5}{8}\) is closer to 1, and is the greater of the two fractions. | |
\( 7-4=3\) and \( 8-5=3\), so the fractions are equal.Hint: Not how to compare fractions. By this logic, 1/2 and 3/4 are equal, but 1/2 and 2/4 are not. | |
\( 4\times 8=32\) and \( 7\times 5=35\). Since \( 32<35\) , \( \dfrac{5}{8}<\dfrac{4}{7}\)Hint: Starts out as something that works, but the conclusion is wrong. 4/7 = 32/56 and 5/8 = 35/56. The cross multiplication gives the numerators, and 35/56 is bigger. | |
\( 4<5\) and \( 7<8\), so \( \dfrac{4}{7}<\dfrac{5}{8}\)Hint: Conclusion is correct, logic is wrong. With this reasoning, 1/2 would be less than 2/100,000. |
Question 3 |
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 4 |
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. |
Question 5 |
The Americans with Disabilties Act (ADA) regulations state that the maximum slope for a wheelchair ramp in new construction is 1:12, although slopes between 1:16 and 1:20 are preferred. The maximum rise for any run is 30 inches. The graph below shows the rise and runs of four different wheelchair ramps. Which ramp is in compliance with the ADA regulations for new construction?

AHint: Rise is more than 30 inches. | |
BHint: Run is almost 24 feet, so rise can be almost 2 feet. | |
CHint: Run is 12 feet, so rise can be at most 1 foot. | |
DHint: Slope is 1:10 -- too steep. |
Question 6 |
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 7 |
Here is a student's work on several multiplication problems:
For which of the following problems is this student most likely to get the correct solution, even though he is using an incorrect algorithm?
58 x 22Hint: This problem involves regrouping, which the student does not do correctly. | |
16 x 24Hint: This problem involves regrouping, which the student does not do correctly. | |
31 x 23Hint: There is no regrouping with this problem. | |
141 x 32Hint: This problem involves regrouping, which the student does not do correctly. |
Question 8 |
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. |
Question 9 |
\( \large \dfrac{17}{24}\) 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 10 |
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 11 |
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 12 |
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 13 |
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 14 |
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 15 |
A teacher has a list of all the countries in the world and their populations in March 2012. She is going to have her students use technology to compute the mean and median of the numbers on the list. Which of the following statements is true?
The teacher can be sure that the mean and median will be the same without doing any computation.Hint: Does this make sense? How likely is it that the mean and median of any large data set will be the same? | |
The teacher can be sure that the mean is bigger than the median without doing any computation.Hint: This is a skewed distribution, and very large countries like China and India contribute huge numbers to the mean, but are counted the same as small countries like Luxembourg in the median (the same thing happens w/data on salaries, where a few very high income people tilt the mean -- that's why such data is usually reported as medians). | |
The teacher can be sure that the median is bigger than the mean without doing any computation.Hint: Think about a set of numbers like 1, 2, 3, 4, 10,000 -- how do the mean/median compare? How might that relate to countries of the world? | |
There is no way for the teacher to know the relative size of the mean and median without computing them.Hint: Knowing the shape of the distribution of populations does give us enough info to know the relative size of the mean and median, even without computing them. |
Question 16 |
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 17 |
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 18 |
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 19 |
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 20 |
What is the perimeter of a right triangle with legs of lengths x and 2x?
\( \large 6x\) Hint: Use the Pythagorean Theorem. | |
\( \large 3x+5{{x}^{2}}\) Hint: Don't forget to take square roots when you use the Pythagorean Theorem. | |
\( \large 3x+\sqrt{5}{{x}^{2}}\) Hint: \(\sqrt {5 x^2}\) is not \(\sqrt {5}x^2\). | |
\( \large 3x+\sqrt{5}{{x}^{{}}}\) Hint: To find the hypotenuse, h, use the Pythagorean Theorem: \(x^2+(2x)^2=h^2.\) \(5x^2=h^2,h=\sqrt{5}x\). The perimeter is this plus x plus 2x. |
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