Hints will display for most wrong answers; explanations for most right answers. You can attempt a question multiple times; it will only be scored correct if you get it right the first time.
I used the official objectives and sample test to construct these questions, but cannot promise that they accurately reflect what’s on the real test. Some of the sample questions were more convoluted than I could bear to write. See terms of use. See the MTEL Practice Test main page to view questions on a particular topic or to download paper practice tests.
MTEL General Curriculum Mathematics Practice
Question 1 |
Which of the following inequalities describes all values of x with \(\large \dfrac{x}{2}\le \dfrac{x}{3}\)?
\( \large x < 0\) Hint: If x =0, then x/2 = x/3, so this answer can't be correct. | |
\( \large x \le 0\) | |
\( \large x > 0\) Hint: If x =0, then x/2 = x/3, so this answer can't be correct. | |
\( \large x \ge 0\) Hint: Try plugging in x = 6. |
Question 2 |
The chairs in a large room can be arranged in rows of 18, 25, or 60 with no chairs left over. If C is the smallest possible number of chairs in the room, which of the following inequalities does C satisfy?
\( \large C\le 300\) Hint: Find the LCM. | |
\( \large 300 < C \le 500 \) Hint: Find the LCM. | |
\( \large 500 < C \le 700 \) Hint: Find the LCM. | |
\( \large C>700\) Hint: The LCM is 900, which is the smallest number of chairs. |
Question 3 |
The table below gives the result of a survey at a college, asking students whether they were residents or commuters:
Based on the above data, what is the probability that a randomly chosen commuter student is a junior or a senior?
\( \large \dfrac{34}{43}\) | |
\( \large \dfrac{34}{71}\) Hint: This is the probability that a randomly chosen junior or senior is a commuter student. | |
\( \large \dfrac{34}{147}\) Hint: This is the probability that a randomly chosen student is a junior or senior who is a commuter. | |
\( \large \dfrac{71}{147}\) Hint: This is the probability that a randomly chosen student is a junior or a senior. |
Question 4 |
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 5 |
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 6 |
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 7 |
The expression \( \large{{8}^{3}}\cdot {{2}^{-10}}\) is equal to which of the following?
\( \large 2\) Hint: Write \(8^3\) as a power of 2. | |
\( \large \dfrac{1}{2}\) Hint: \(8^3 \cdot {2}^{-10}={(2^3)}^3 \cdot {2}^{-10}\) =\(2^9 \cdot {2}^{-10} =2^{-1}\) | |
\( \large 16\) Hint: Write \(8^3\) as a power of 2. | |
\( \large \dfrac{1}{16}\) Hint: Write \(8^3\) as a power of 2. |
Question 8 |
The histogram below shows the frequency of a class's scores on a 4 question quiz.
What was the mean score on the quiz?
\( \large 2.75\) Hint: There were 20 students who took the quiz. Total points earned: \(2 \times 1+6 \times 2+ 7\times 3+5 \times 4=55\), and 55/20 = 2.75. | |
\( \large 2\) Hint: How many students are there total? Did you count them all? | |
\( \large 3\) Hint: How many students are there total? Did you count them all? Be sure you're finding the mean, not the median or the mode. | |
\( \large 2.5\) Hint: How many students are there total? Did you count them all? Don't just take the mean of 1, 2, 3, 4 -- you have to weight them properly. |
Question 9 |
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 10 |
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 11 |
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 12 |
| I. \(\large \dfrac{1}{2}+\dfrac{1}{3}\) | II. \( \large .400000\) | III. \(\large\dfrac{1}{5}+\dfrac{1}{5}\) |
| IV. \( \large 40\% \) | V. \( \large 0.25 \) | VI. \(\large\dfrac{14}{35}\) |
Which of the lists below includes all of the above expressions that are equivalent to \( \dfrac{2}{5}\)?
I, III, V, VIHint: I and V are not at all how fractions and decimals work. | |
III, VIHint: These are right, but there are more. | |
II, III, VIHint: These are right, but there are more. | |
II, III, IV, VI |
Question 13 |
Which of the lines depicted below is a graph of \( \large y=2x-5\)?

aHint: The slope of line a is negative. | |
bHint: Wrong slope and wrong intercept. | |
cHint: The intercept of line c is positive. | |
dHint: Slope is 2 -- for every increase of 1 in x, y increases by 2. Intercept is -5 -- the point (0,-5) is on the line. |
Question 14 |
A publisher prints a series of books with covers made of identical material and using the same thickness of paper for each page. The covers of the book together are 0.4 cm thick, and 125 pieces of the paper used together are 1 cm thick.
The publisher uses a linear function to determine the total thickness, T(n) of a book made with n sheets of paper. What are the slope and intercept of T(n)?
Intercept = 0.4 cm, Slope = 125 cm/pageHint: This would mean that each page of the book was 125 cm thick. | |
Intercept =0.4 cm, Slope = \(\dfrac{1}{125}\)cm/pageHint: The intercept is how thick the book would be with no pages in it. The slope is how much 1 extra page adds to the thickness of the book. | |
Intercept = 125 cm, Slope = 0.4 cmHint: This would mean that with no pages in the book, it would be 125 cm thick. | |
Intercept = \(\dfrac{1}{125}\)cm, Slope = 0.4 pages/cmHint: This would mean that each new page of the book made it 0.4 cm thicker. |
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 16 |
The column below consists of two cubes and a cylinder. The cylinder has diameter y, which is also the length of the sides of each cube. The total height of the column is 5y. Which of the formulas below gives the volume of the column?
\( \large 2{{y}^{3}}+\dfrac{3\pi {{y}^{3}}}{4}\) Hint: The cubes each have volume \(y^3\). The cylinder has radius \(\dfrac{y}{2}\) and height \(3y\). The volume of a cylinder is \(\pi r^2 h=\pi ({\dfrac{y}{2}})^2(3y)=\dfrac{3\pi {{y}^{3}}}{4}\). Note that the volume of a cylinder is analogous to that of a prism -- area of the base times height. | |
\( \large 2{{y}^{3}}+3\pi {{y}^{3}}\) Hint: y is the diameter of the circle, not the radius. | |
\( \large {{y}^{3}}+5\pi {{y}^{3}}\) Hint: Don't forget to count both cubes. | |
\( \large 2{{y}^{3}}+\dfrac{3\pi {{y}^{3}}}{8}\) Hint: Make sure you know how to find the volume of a cylinder. |
Question 17 |
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 18 |
The equation \( \large F=\frac{9}{5}C+32\) is used to convert a temperature measured in Celsius to the equivalent Farentheit temperature.
A patient's temperature increased by 1.5° Celcius. By how many degrees Fahrenheit did her temperature increase?
1.5°Hint: Celsius and Fahrenheit don't increase at the same rate. | |
1.8°Hint: That's how much the Fahrenheit temp increases when the Celsius temp goes up by 1 degree. | |
2.7°Hint: Each degree increase in Celsius corresponds to a \(\dfrac{9}{5}=1.8\) degree increase in Fahrenheit. Thus the increase is 1.8+0.9=2.7. | |
Not enough information.Hint: A linear equation has constant slope, which means that every increase of the same amount in one variable, gives a constant increase in the other variable. It doesn't matter what temperature the patient started out at. |
Question 19 |
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 20 |
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 21 |
Which of the numbers below is the decimal equivalent of \( \dfrac{3}{8}?\)
0.38Hint: If you are just writing the numerator next to the denominator then your technique is way off, but by coincidence your answer is close; try with 2/3 and 0.23 is nowhere near correct. | |
0.125Hint: This is 1/8, not 3/8. | |
0.375 | |
0.83Hint: 3/8 is less than a half, and 0.83 is more than a half, so they can't be equal. |
Question 22 |
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 23 |
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 24 |
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 25 |
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 26 |
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 27 |
The letters A, and B represent digits (possibly equal) in the ten digit number x=1,438,152,A3B. For which values of A and B is x divisible by 12, but not by 9?
\( \large A = 0, B = 4\) Hint: Digits add to 31, so not divisible by 3, so not divisible by 12. | |
\( \large A = 7, B = 2\) Hint: Digits add to 36, so divisible by 9. | |
\( \large A = 0, B = 6\) Hint: Digits add to 33, divisible by 3, not 9. Last digits are 36, so divisible by 4, and hence by 12. | |
\( \large A = 4, B = 8\) Hint: Digits add to 39, divisible by 3, not 9. Last digits are 38, so not divisible by 4, so not divisible by 12. |
Question 28 |
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 29 |
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 30 |
The pattern below consists of a row of black squares surrounded by white squares.
How many white squares would surround a row of 157 black squares?
314Hint: Try your procedure on a smaller number that you can count to see where you made a mistake. | |
317Hint: Are there ever an odd number of white squares? | |
320Hint: One way to see this is that there are 6 tiles on the left and right ends, and the rest of the white tiles are twice the number of black tiles (there are many other ways to look at it too). | |
322Hint: Try your procedure on a smaller number that you can count to see where you made a mistake. |
Question 31 |
In January 2011, the national debt was about 14 trillion dollars and the US population was about 300 million people. Someone reading these figures estimated that the national debt was about $5,000 per person. Which of these statements best describes the reasonableness of this estimate?
It is too low by a factor of 10Hint: 14 trillion \( \approx 15 \times {{10}^{12}} \) and 300 million \( \approx 3 \times {{10}^{8}}\), so the true answer is about \( 5 \times {{10}^{4}} \) or $50,000. | |
It is too low by a factor of 100 | |
It is too high by a factor of 10 | |
It is too high by a factor of 100 |
Question 32 |
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 33 |
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 34 |
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 35 |
The letters A, B, and C represent digits (possibly equal) in the twelve digit number x=111,111,111,ABC. For which values of A, B, and C is x divisible by 40?
\( \large A = 3, B = 2, C=0\) Hint: Note that it doesn't matter what the first 9 digits are, since 1000 is divisible by 40, so DEF,GHI,JKL,000 is divisible by 40 - we need to check the last 3. | |
\( \large A = 0, B = 0, C=4\) Hint: Not divisible by 10, since it doesn't end in 0. | |
\( \large A = 4, B = 2, C=0\) Hint: Divisible by 10 and by 4, but not by 40, as it's not divisible by 8. Look at 40 as the product of powers of primes -- 8 x 5, and check each. To check 8, either check whether 420 is divisible by 8, or take ones place + twice tens place + 4 * hundreds place = 18, which is not divisible by 8. | |
\( \large A =1, B=0, C=0\) Hint: Divisible by 10 and by 4, but not by 40, as it's not divisible by 8. Look at 40 as the product of powers of primes -- 8 x 5, and check each. To check 8, either check whether 100 is divisible by 8, or take ones place + twice tens place + 4 * hundreds place = 4, which is not divisible by 8. |
Question 36 |
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 37 |
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 38 |
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 39 |
Use the graph below to answer the question that follows:
The graph above represents the equation \( \large 3x+Ay=B\), where A and B are integers. What are the values of A and B?
\( \large A = -2, B= 6\) Hint: Plug in (2,0) to get B=6, then plug in (0,-3) to get A=-2. | |
\( \large A = 2, B = 6\) Hint: Try plugging (0,-3) into this equation. | |
\( \large A = -1.5, B=-3\) Hint: The problem said that A and B were integers and -1.5 is not an integer. Don't try to use slope-intercept form. | |
\( \large A = 2, B = -3\) Hint: Try plugging (2,0) into this equation. |
Question 40 |
The least common multiple of 60 and N is 1260. Which of the following could be the prime factorization of N?
\( \large2\cdot 5\cdot 7\) Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM. | |
\( \large{{2}^{3}}\cdot {{3}^{2}}\cdot 5 \cdot 7\) Hint: 1260 is not divisible by 8, so it isn't a multiple of this N. | |
\( \large3 \cdot 5 \cdot 7\) Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM. | |
\( \large{{3}^{2}}\cdot 5\cdot 7\) Hint: \(1260=2^2 \cdot 3^2 \cdot 5 \cdot 7\) and \(60=2^2 \cdot 3 \cdot 5\). In order for 1260 to be the LCM, N has to be a multiple of \(3^2\) and of 7 (because 60 is not a multiple of either of these). N also cannot introduce a factor that would require the LCM to be larger (as in choice b). |
Question 41 |
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. |
Question 42 |
Use the samples of a student's work below to answer the question that follows:
This student divides fractions by first finding a common denominator, then dividing the numerators.
\( \large \dfrac{2}{3} \div \dfrac{3}{4} \longrightarrow \dfrac{8}{12} \div \dfrac{9}{12} \longrightarrow 8 \div 9 = \dfrac {8}{9}\) \( \large \dfrac{2}{5} \div \dfrac{7}{20} \longrightarrow \dfrac{8}{20} \div \dfrac{7}{20} \longrightarrow 8 \div 7 = \dfrac {8}{7}\) \( \large \dfrac{7}{6} \div \dfrac{3}{4} \longrightarrow \dfrac{14}{12} \div \dfrac{9}{12} \longrightarrow 14 \div 9 = \dfrac {14}{9}\)Which of the following best describes the mathematical validity of the algorithm the student is using?
It is not valid. Common denominators are for adding and subtracting fractions, not for dividing them.Hint: Don't be so rigid! Usually there's more than one way to do something in math. | |
It got the right answer in these three cases, but it isn‘t valid for all rational numbers.Hint: Did you try some other examples? What makes you say it's not valid? | |
It is valid if the rational numbers in the division problem are in lowest terms and the divisor is not zero.Hint: Lowest terms doesn't affect this problem at all. | |
It is valid for all rational numbers, as long as the divisor is not zero.Hint: When we have common denominators, the problem is in the form a/b divided by c/b, and the answer is a/c, as the student's algorithm predicts. |
Question 43 |
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 44 |
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 45 |
Elena is going to use a calculator to check whether or not 267 is prime. She will pick certain divisors, and then find 267 divided by each, and see if she gets a whole number. If she never gets a whole number, then she's found a prime. Which numbers does Elena NEED to check before she can stop checking and be sure she has a prime?
All natural numbers from 2 to 266.Hint: She only needs to check primes -- checking the prime factors of any composite is enough to look for divisors. As a test taking strategy, the other three choices involve primes, so worth thinking about. | |
All primes from 2 to 266 .Hint: Remember, factors come in pairs (except for square root factors), so she would first find the smaller of the pair and wouldn't need to check the larger. | |
All primes from 2 to 133 .Hint: She doesn't need to check this high. Factors come in pairs, and something over 100 is going to be paired with something less than 3, so she will find that earlier. | |
All primes from \( \large 2\) to \( \large \sqrt{267}\).Hint: \(\sqrt{267} \times \sqrt{267}=267\). Any other pair of factors will have one factor less than \( \sqrt{267}\) and one greater, so she only needs to check up to \( \sqrt{267}\). |
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