Mastering tricky word problems: Nine math problems for middle school students
Nine math word problems that are somewhat tricky and therefore require close reading, have multiple steps, and are suitable for middle school math students. Correct answers and detailed explanations to those answers are included.
Introduction
Math can be tough, especially word problems! Word problems test not only your math skills but also your reading comprehension and critical thinking abilities. They are tricky because they require you to break down the problem, identify the important information, and use logical reasoning to solve them. In this blog post, we'll cover nine tricky math word problems that require close reading, multiple steps, and are suitable for middle school students. We've also included detailed explanations for each problem's solution, so don't worry if you feel stuck!
Problem 1: A rectangular parking lot measures 60 meters by 80 meters. If a car uses 4 square meters to park, how many cars can fit in the parking lot?
Answer: 120 cars
Explanation: To find out how many cars can fit in the parking lot, we first need to know the total area of the parking lot. To do so, we multiply the length and width: 60 meters × 80 meters = 4,800 square meters. Then, we divide the total area by the area a single car uses: 4,800/4 = 1,200. Therefore, 1,200 cars can fit in the parking lot.
Problem 2: In a class election, there are 30 students. The winner needs more than half of the votes in order to win. If 16 students voted for Student A, and the rest voted for Student B, how many students voted for Student B?
Answer: 14 students
Explanation: To find out how many students voted for B, we first need to know the total number of votes. Since half of the votes are 15, any number above 15 will be the winning number. We already know that 16 people voted for Student A, so if the class has 30 students, then 30 - 16 = 14 students voted for B.
Problem 3: There are 5 red balls, 9 blue balls, and 7 yellow balls in a bag. What is the probability of grabbing a red ball first and then a blue ball?
Answer: 5/63
Explanation: There are five red balls in the bag, so the probability of grabbing one on the first try is 5/21. There are now 20 balls remaining (9 blue and 7 yellow), so the probability of grabbing one blue ball is 9/20. To find out the probability of grabbing a red ball first and then a blue ball, we multiply the probabilities of each event together: 5/21 × 9/20 = 1/63. Therefore, the probability of grabbing a red ball first and then a blue ball is 5/63.
Problem 4: The difference between two numbers is 35. If one number is 57, what is the other number?
Answer: 22
Explanation: Let's call the other number we're trying to find "x". We know the difference between the two numbers (57 and x) is 35, so we set up the equation 57 - x = 35. Next, we isolate "x" by adding 35 to both sides of the equation: 57 - x + 35 = 35 + 35. Simplifying the equation gives us: 92 - x = 70. Then, we subtract 92 from both sides to solve for "x": 92 - 92 - x = 70 - 92 or -x = -22. Finally, we divide -22 by -1 to isolate "x": x = 22.
Problem 5: A sphere has a radius of 3 cm. What is its volume?
Answer: 113.1 cubic centimeters
Explanation: To find the volume of a sphere, we use the formula V = 4/3πr³, where "V" stands for volume, "π" represents Pi (3.14), and "r" represents the sphere's radius. Substituting the given radius (3 cm) into the formula, we get V = 4/3 x 3.14 x (3 cm)³. Simplifying the equation results in V = 4/3 x 3.14 x 27 cm³ or V = 113.1 cm³. Therefore, the volume of the sphere is 113.1 cubic centimeters.
Problem 6: A pear-shaped swimming pool has a deep end and a shallow end. The shallow end of the pool is 3 meters deep and is 10 meters wide and 15 meters long. The deeper region of the pool is 6 meters deep and is conical in shape. The diameter of the deeper section is 10 meters and slants down to meet the shallow end smoothly. What is the total surface area of the pool?
Answer: 628.96 square meters
Explanation: First, we need to find out the volume of the deep section of the pool. Since it is conical in shape, we use the volume formula for a cone: V = 1/3πr²h, where "r" represents the radius and "h" represents the height. We find the radius by dividing the diameter (10 meters) by 2, which gives us 5 meters. Next, the height of the conical section is calculated by subtracting the height of the shallow end (3 meters) from the depth of the deep end (6 meters), resulting in 3 meters. Substituting the values into the formula gives us V = 1/3π(5)²(3), which is approximately equal to 78.54 cubic meters. Next, we need to find the total surface area of the pool. To do so, we calculate the area of the bottom of the pool (10 meters x 15 meters) and add that to the lateral area of the deep section of the pool (by using the formula πrl). We can find "l," the slant height, by using the Pythagorean theorem: l² = r² + h², where "h" is the height of the cone and "r" is the radius. Substituting the values gives us l² = 5² + 3² or l² = 34. Since we only need to know the area of the slant side, we can ignore the square root and use l = √34. Thus, the surface area is calculated as: 10 x 15 + π(5) x (√34). Plugging in the numbers, the total surface area of the pool is approximately 628.96 square meters.
Problem 7: 2/3 of Tom's marbles are blue, and 19 of them are yellow. If Tom has 87 marbles, what is the total number of marbles that are not blue?
Answer: 29 marbles
Explanation: If 2/3 of Tom's marbles are blue, then 1/3 of his marbles are not blue. Since the total number of Tom's marbles is 87; therefore, 1/3 of that is (87/3) = 29 marbles. Therefore, Tom has 29 marbles that are not blue.
Problem 8: There are 22 boys and 18 girls in a class. If one is selected randomly, what is the probability that the selected student is a boy?
Answer: 55%
Explanation: There are a total of 40 students in the class (22 boys + 18 girls). The probability of selecting a boy is the number of boys divided by the total number of students in the class. Therefore, 22/40 = 0.55, which is 55%.
Problem 9: A recipe calls for 2 1/2 cups of flour. If Mark only has a 1/4 measuring cup, how many times does he need to fill it up to get the required amount?
Answer: 10 times
Explanation: Mark needs 2 1/2 cups of flour. Since he only has a 1/4 measuring cup, he needs to fill it up multiple times until he has the required amount of flour. To figure out how many times he needs to fill up his measuring cup, we convert 2½ cups to the same measuring unit as the measuring cup (in this case, ¼ cup). Therefore, 2 ½ cups is equal to 10 quarter-cups, so Mark needs to fill up his measuring cup 10 times.
If you found this post helpful, you may also want to check out our posts on math operations vocabulary, solving one-step math equations, and solving two-step math equations.
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