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MTEL General Curriculum Mathematics Practice
Your answers are highlighted below.
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 1 Explanation:
Topic: Recognize and apply connections between algebra and geometry
(e.g., the use of coordinate systems, the Pythagorean theorem) (Objective 0024).
Question 2 |
In March of 2012, 1 dollar was worth the same as 0.761 Euros, and 1 dollar was also worth the same as 83.03 Japanese Yen. Which of the expressions below gives the number of Yen that are worth 1 Euro?
\( \large {83}.0{3}\cdot 0.{761}\) Hint: This equation gives less than the number of yen per dollar, but 1 Euro is worth more than 1 dollar. | |
\( \large \dfrac{0.{761}}{{83}.0{3}}\) Hint: Number is way too small. | |
\( \large \dfrac{{83}.0{3}}{0.{761}}\) Hint: One strategy here is to use easier numbers, say 1 dollar = .5 Euros and 100 yen, then 1 Euro would be 200 Yen (change the numbers in the equations and see what works). Another is to use dimensional analysis: we want # yen per Euro, or yen/Euro = yen/dollar \(\times\) dollar/Euro = \(83.03 \times \dfrac {1}{0.761}\) | |
\( \large \dfrac{1}{0.{761}}\cdot \dfrac{1}{{83}.0{3}}\) Hint: Number is way too small. |
Question 2 Explanation:
Topic: Analyze the relationships among proportions, constant rates, and linear
functions (Objective 0022).
Question 3 |
Use the problem below to answer the question that follows:
T shirts are on sale for 20% off. Tasha paid $8.73 for a shirt. What is the regular price of the shirt? There is no tax on clothing purchases under $175.
Let p represent the regular price of these t-shirt. Which of the following equations is correct?
\( \large 0.8p=\$8.73\) Hint: 80% of the regular price = $8.73. | |
\( \large \$8.73+0.2*\$8.73=p\) Hint: The 20% off was off of the ORIGINAL price, not off the $8.73 (a lot of people make this mistake). Plus this is the same equation as in choice c. | |
\( \large 1.2*\$8.73=p\) Hint: The 20% off was off of the ORIGINAL price, not off the $8.73 (a lot of people make this mistake). Plus this is the same equation as in choice b. | |
\( \large p-0.2*\$8.73=p\) Hint: Subtract p from both sides of this equation, and you have -.2 x 8.73 =0. |
Question 3 Explanation:
Topics: Use algebra to solve word problems involving percents and identify variables, and derive algebraic expressions that represent real-world situations (Objective 0020).
Question 4 |
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 4 Explanation:
Converting between fractions, decimals, and percents (Objective 0017).
Question 5 |
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 5 Explanation:
Topic: Recognize and apply connections between algebra and geometry
(e.g., the use of coordinate systems, the Pythagorean theorem) (Objective 0024).
Question 6 |
Which of the following is equivalent to \( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{2}{8}\times \dfrac{1}{2}?\)
\( \large \dfrac{7}{16}\) Hint: Multiplication comes before addition and subtraction in the order of operations. | |
\( \large \dfrac{1}{2}\) Hint: Addition and subtraction are of equal priority in the order of operations -- do them left to right. | |
\( \large \dfrac{3}{4}\) Hint: \( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{2}{8}\times \dfrac{1}{2}\)=\( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{1}{8}\)=\( \dfrac{3}{4}+-\dfrac{1}{8}+\dfrac{1}{8}\)=\( \dfrac{3}{4}\) | |
\( \large \dfrac{3}{16}\) Hint: Multiplication comes before addition and subtraction in the order of operations. |
Question 6 Explanation:
Topic: Operations on Fractions, Order of Operations (Objective 0019).
Question 7 |
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 7 Explanation:
Topics: Estimation, simplifying fractions (Objective 0016, overlaps with other objectives).
Question 8 |
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? |
Question 8 Explanation:
Topic: Match three-dimensional figures and their two-dimensional
representations (e.g., nets, projections, perspective drawings) (Objective 0024).
Question 9 |
Use the samples of a student's work below to answer the question that follows:
\( \large \dfrac{2}{3}\times \dfrac{3}{4}=\dfrac{4\times 2}{3\times 3}=\dfrac{8}{9}\) \( \large \dfrac{2}{5}\times \dfrac{7}{7}=\dfrac{7\times 2}{5\times 7}=\dfrac{2}{5}\) \( \large \dfrac{7}{6}\times \dfrac{3}{4}=\dfrac{4\times 7}{6\times 3}=\dfrac{28}{18}=\dfrac{14}{9}\)Which of the following best describes the mathematical validity of the algorithm the student is using?
It is not valid. It never produces the correct answer.Hint: In the middle example,the answer is correct. | |
It is not valid. It produces the correct answer in a few special cases, but it‘s still not a valid algorithm.Hint: Note that this algorithm gives a/b divided by c/d, not a/b x c/d, but some students confuse multiplication and cross-multiplication. If a=0 or if c/d =1, division and multiplication give the same answer. | |
It is valid if the rational numbers in the multiplication problem are in lowest terms.Hint: Lowest terms is irrelevant. | |
It is valid for all rational numbers.Hint: Can't be correct as the first and last examples have the wrong answers. |
Question 9 Explanation:
Topic: Analyze Non-Standard Computational Algorithms (Objective 0019).
Question 10 |
A cylindrical soup can has diameter 7 cm and height 11 cm. The can holds g grams of soup. How many grams of the same soup could a cylindrical can with diameter 14 cm and height 33 cm hold?
\( \large 6g\) Hint: You must scale in all three dimensions. | |
\( \large 12g\) Hint: Height is multiplied by 3, and diameter and radius are multiplied by 2. Since the radius is squared, final result is multiplied by \(2^2\times 3=12\). | |
\( \large 18g\) Hint: Don't square the height scale factor. | |
\( \large 36g\) Hint: Don't square the height scale factor. |
Question 10 Explanation:
Topic: Determine how the characteristics (e.g., area, volume) of geometric
figures and shapes are affected by changes in their dimensions (Objective 0023).
Question 11 |
Taxicab fares in Boston (Spring 2012) are $2.60 for the first \(\dfrac{1}{7}\) of a mile or less and $0.40 for each \(\dfrac{1}{7}\) of a mile after that.
Let d represent the distance a passenger travels in miles (with \(d>\dfrac{1}{7}\)). Which of the following expressions represents the total fare?
\( \large \$2.60+\$0.40d\) Hint: It's 40 cents for 1/7 of a mile, not per mile. | |
\( \large \$2.60+\$0.40\dfrac{d}{7}\) Hint: According to this equation, going 7 miles would cost $3; does that make sense? | |
\( \large \$2.20+\$2.80d\) Hint: You can think of the fare as $2.20 to enter the cab, and then $0.40 for each 1/7 of a mile, including the first 1/7 of a mile (or $2.80 per mile).
Alternatively, you pay $2.60 for the first 1/7 of a mile, and then $2.80 per mile for d-1/7 miles. The total is 2.60+2.80(d-1/7) = 2.60+ 2.80d -.40 = 2.20+2.80d. | |
\( \large \$2.60+\$2.80d\) Hint: Don't count the first 1/7 of a mile twice. |
Question 11 Explanation:
Topic: Identify variables and derive algebraic expressions that represent real-world situations (Objective 0020), and select the linear equation that best models a real-world situation (Objective 0022).
Question 12 |
Use the graph below to answer the question that follows:
The graph above best matches which of the following scenarios:
George left home at 10:00 and drove to work on a crooked path. He was stopped in traffic at 10:30 and 10:45. He drove 30 miles total.Hint: Just because he ended up 30 miles from home doesn't mean he drove 30 miles total. | |
George drove to work. On the way to work there is a little hill and a big hill. He slowed down for them. He made it to work at 11:15.Hint: The graph is not a picture of the roads. | |
George left home at 10:15. He drove 10 miles, then realized he‘d forgotten something at home. He turned back and got what he‘d forgotten. Then he drove in a straight line, at many different speeds, until he got to work around 11:15.Hint: A straight line on a distance versus time graph means constant speed. | |
George left home at 10:15. He drove 10 miles, then realized he‘d forgotten something at home. He turned back and got what he‘d forgotten. Then he drove at a constant speed until he got to work around 11:15. |
Question 12 Explanation:
Topic: Use qualitative graphs to represent functional relationships in the real world (Objective 0021).
Question 13 |
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 13 Explanation:
Topic: Determine how the characteristics (e.g., area, volume) of geometric
figures and shapes are affected by changes in their dimensions (Objective 0023).
Question 14 |
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 14 Explanation:
Topic: Use unit conversions and dimensional analysis to solve measurement
problems (Objective 0023).
Question 15 |
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 15 Explanation:
Topic: Analyze the relationships among proportions, constant rates, and linear
functions (Objective 0022).
Question 16 |
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 16 Explanation:
Topic: Apply knowledge of combinations and permutations to
the computation of probabilities (Objective 0026).
Question 17 |
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 17 Explanation:
Topic: Applying fractions to word problems (Objective 0017) This problem is similar to one on the official sample test for that objective, but it might fit better into unit conversion and dimensional analysis (Objective 0023: Measurement)
Question 18 |
The prime factorization of n can be written as n=pqr, where p, q, and r are distinct prime numbers. How many factors does n have, including 1 and itself?
\( \large3\) Hint: 1, p, q, r, and pqr are already 5, so this isn't enough. You might try plugging in p=2, q=3, and r=5 to help with this problem. | |
\( \large5\) Hint: Don't forget pq, etc. You might try plugging in p=2, q=3, and r=5 to help with this problem. | |
\( \large6\) Hint: You might try plugging in p=2, q=3, and r=5 to help with this problem. | |
\( \large8\) Hint: 1, p, q, r, pq, pr, qr, pqr. |
Question 18 Explanation:
Topic: Recognize uses of prime factorization of a number (Objective 0018).
Question 19 |
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 19 Explanation:
Topic: Apply knowledge of combinations and permutations to
the computation of probabilities (Objective 0026).
Question 20 |
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 Explanation:
Topic: Analyze and interpret various graphic and nongraphic data
representations (e.g., frequency distributions, percentiles) (Objective 0025).
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