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MTEL General Curriculum Mathematics Practice
Question 1 |
Use the solution procedure below to answer the question that follows:
\( \large {\left( x+3 \right)}^{2}=10\)
\( \large \left( x+3 \right)\left( x+3 \right)=10\)
\( \large {x}^{2}+9=10\)
\( \large {x}^{2}+9-9=10-9\)
\( \large {x}^{2}=1\)
\( \large x=1\text{ or }x=-1\)
Which of the following is incorrect in the procedure shown above?
The commutative property is used incorrectly.Hint: The commutative property is \(a+b=b+a\) or \(ab=ba\). | |
The associative property is used incorrectly.Hint: The associative property is \(a+(b+c)=(a+b)+c\) or
\(a \times (b \times c)=(a \times b) \times c\). | |
Order of operations is done incorrectly. | |
The distributive property is used incorrectly.Hint: \((x+3)(x+3)=x(x+3)+3(x+3)\)=\(x^2+3x+3x+9.\) |
Question 2 |
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 3 |
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 4 |
Below is a portion of a number line:
Point B is halfway between two tick marks. What number is represented by Point B?
\( \large 0.645\) Hint: That point is marked on the line, to the right. | |
\( \large 0.6421\) Hint: That point is to the left of point B. | |
\( \large 0.6422\) Hint: That point is to the left of point B. | |
\( \large 0.6425\) |
Question 5 |
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 6 |
AHint: \(\frac{34}{135} \approx \frac{1}{4}\) and \( \frac{53}{86} \approx \frac {2}{3}\). \(\frac {1}{4}\) of \(\frac {2}{3}\) is small and closest to A. | |
BHint: Estimate with simpler fractions. | |
CHint: Estimate with simpler fractions. | |
DHint: Estimate with simpler fractions. |
Question 7 |
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. |
Question 8 |
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 9 |
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 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 |
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 12 |
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 13 |
\( \large \dfrac{17}{24}\) 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 14 |
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 15 |
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 16 |
Which of the numbers below is not equivalent to 4%?
\( \large \dfrac{1}{25}\) Hint: 1/25=4/100, so this is equal to 4% (be sure you read the question correctly). | |
\( \large \dfrac{4}{100}\) Hint: 4/100=4% (be sure you read the question correctly). | |
\( \large 0.4\) Hint: 0.4=40% so this is not equal to 4% | |
\( \large 0.04\) Hint: 0.04=4/100, so this is equal to 4% (be sure you read the question correctly). |
Question 17 |
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 18 |
Here is a mental math strategy for computing 26 x 16:
Step 1: 100 x 16 = 1600
Step 2: 25 x 16 = 1600 ÷· 4 = 400
Step 3: 26 x 16 = 400 + 16 = 416
Which property best justifies Step 3 in this strategy?
Commutative Property.Hint: For addition, the commutative property is \(a+b=b+a\) and for multiplication it's \( a \times b = b \times a\). | |
Associative Property.Hint: For addition, the associative property is \((a+b)+c=a+(b+c)\) and for multiplication it's \((a \times b) \times c=a \times (b \times c)\) | |
Identity Property.Hint: 0 is the additive identity, because \( a+0=a\) and 1 is the multiplicative identity because \(a \times 1=a\). The phrase "identity property" is not standard. | |
Distributive Property.Hint: \( (25+1) \times 16 = 25 \times 16 + 1 \times 16 \). This is an example of the distributive property of multiplication over addition. |
Question 19 |
Use the table below to answer the question that follows:
Each number in the table above represents a value W that is determined by the values of x and y. For example, when x=3 and y=1, W=5. What is the value of W when x=9 and y=14? Assume that the patterns in the table continue as shown.
\( \large W=-5\) Hint: When y is even, W is even. | |
\( \large W=4\) Hint: Note that when x increases by 1, W increases by 2, and when y increases by 1, W decreases by 1. At x=y=0, W=0, so at x=9, y=14, W has increased by \(9 \times 2\) and decreased by 14, or W=18-14=4. | |
\( \large W=6\) Hint: Try fixing x or y at 0, and start by finding W for x=0 y=14 or x=9, y=0. | |
\( \large W=32\) Hint: Try fixing x or y at 0, and start by finding W for x=0 y=14 or x=9, y=0. |
Question 20 |
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? |
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