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When Riya’s alarm went off on the first day of her senior year in high school, she hit the snooze button and slept for another 10 minutes. When the alarm sounded the next time, Riya got up and immediately brushed and flossed her teeth, which took 5 minutes. A half hour later, Riya was showered and dressed. She then spent 45 minutes getting her hair and make-up just right. Next, Riya had a glass of milk and a bowl of microwave oatmeal for breakfast. Riya spent 30 minutes preparing and eating her breakfast. Twenty-four minutes later, at 8:39 a.m., Riya arrived at school ready to begin the new school year. At what time did Riya’s alarm go off the first time?
After Riya’s alarm went off the first time, and before she arrived at school, Riya did the following:
- Snoozed 10 minutes
- Brushed & flossed 5 minutes
- Showered & dressed 30 minutes
- Fixed hair & make-up 45 minutes
- Prepared & ate breakfast 30 minutes
- Traveled to school 24 minutes
TOTAL ELAPSED TIME: 2 hours 24 minutes
If Riya arrived at school at 8:39, her alarm went off the first time at 2 hours and 24 minutes before 8:39, which would have been 8:39 − 2:24 = 6:15 a.m.
There are two different routes that Riya can take to get to school. The first route, which is 12 miles, gets her to school in 20 minutes. The second route is only 8 miles, but has more traffic lights, so it takes 24 minutes. Based on this information, in miles per hour, what is the absolute difference in Riya’s average speeds for both routes?
For the first route, Riya can travel 12 miles in 20 minutes. Since 20 minutes is 20/60 = 1/3 hour, Riya’s rate would be 12 ÷ (1/3) = 12 × 3 = 36 mi/h. For the second route, Riya can travel 8 miles in 24 minutes. Since 24 minutes is 24/60 = 2/5 hour, Riya’s rate would be 8 ÷ (2/5) = 8 × (5/2) = 20 mi/h. The absolute difference in these rates is |36 − 20| = 16 mi/h.
At Riya’s school, there are three parking lots with a total of 175 parking spaces. Three-fifths of these spaces are reserved for juniors and seniors who drive to school. There are twice as many parking spaces assigned to seniors as there are assigned to juniors. How many of the parking spaces at Riya’s school are not assigned to seniors?
We are told that 3/5 of the spaces are reserved for juniors and seniors. We are also told that the number of spaces for seniors is twice the number of spaces for juniors, meaning that 2/3 of the spaces reserved for juniors and seniors are for seniors. It follows, then, that 2/3 of 3/5 of the 175 spaces, or (2/3) × (3/5) = 2/5 of the spaces are reserved for seniors. That means that 3/5 of the spaces, or (3/5) × 175 = 105 spaces are not reserved for seniors.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Ms. Cross leads The National Math Club at the middle school where she teaches. At the first club meeting of the school year, 60% of the students in attendance were 7th graders. If there was one fewer 8th grader than 7th grader in attendance, how many students attended the first club meeting?
Let a and b represent the number of 7th graders and 8th graders who attended the first club meeting, respectively. We know that a = b + 1 and a = 0.6(a + b). Solving for a in the second equation yields a = 0.6a + 0.6b → 0.4a = 0.6b → a = 1.5b. Substituting this value for a in the first equation, we get 1.5b = b + 1 → 0.5b = 1 → b = 2. So, there were 2 8th graders and 2 + 1 = 3 7th graders in attendance at the first club meeting. Therefore, a total of 2 + 3 = 5 students attended the first club meeting.
At the second club meeting of the school year, Ms. Cross noticed that among the students in attendance, there were equal numbers of 7th graders and 8th graders. Ms. Cross also noticed that all the students from the first club meeting were in attendance, along with some new students who weren’t at the first club meeting. If twice as many new 8th graders attended the second club meeting as new 7th graders, how many new students attended the second club meeting?
From the previous problem, we know that there were 3 7th graders and 2 8th graders in attendance at the first club meeting. It follows, then, from the information given, that if x new 7th graders attended the second club meeting, 2x new 8th graders attended that meeting. Since the same number of 8th graders and 7th graders attended the second meeting, we can write 3 + x = 2 + 2x and solve to get x = 1. That means 1 new 7th grader and 2 × 1 = 2 new 8th graders attended the second meeting. Therefore, a total of 1 + 2 = 3 new students attended the second meeting.
At the third meeting of The National Math Club, Ms. Cross noticed that 60% of the students in attendance were 8th graders and that the total number of students in attendance was double that of the first club meeting. Ms. Cross was pleased to see that all the students who attended the second club meeting also attended the third club meeting. Of the students in attendance at the third club meeting who did not attend the first two club meetings, what is the absolute difference between the number of 8th graders and the number of 7th graders?
From the first problem, we know that a total of 5 students attended the first club meeting. Based on the information provided, twice that number, or 5 × 2 = 10 students attended the third club meeting. Since 8th graders accounted for 60% of these 10 students, it follows that 0.6 × 10 = 6 8th graders and 10 – 6 = 4 7th graders attended the third club meeting. Since 4 7th graders and 4 8th graders attended the second club meeting, there were 4 – 4 = 0 new 7th graders and 6 – 4 = 2 new 8th graders at the third club meeting. The absolute difference between these two quantities is 2 – 0 = 2.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Each week, Devin buys school lunch on Friday because the cafeteria serves pizza every Friday. Monday through Thursday, Devin brings his lunch to school. The lunch he brings always includes a peanut butter and jelly sandwich, a fruit and a drink. Devin has enough fruit and drink options to make a different lunch every day that he brings lunch for the next four weeks. If Devin has the same number of fruit options as he has drink options and he takes no days off from school, how many different fruit options does he have?
In four weeks, there are 20 school days. Devin will be eating school lunch on 4 of those days. So, he has enough fruit and drink options to make 20 − 4 = 16 different lunches. If Devin only had one option for fruit and one option for his drink, that would be enough to make 1 unique meal. If he had two options for each, Devin could make 2 × 2 = 4 unique meals since each fruit option could be paired with each of the drink options (F1D1, F1D2, F2D1, F2D2). This is the Fundamental Principle of Counting, which says that if you have m distinct varieties of one item and n distinct varieties of another item, there are m × n distinct pairs consisting of one of each of the items. So, to get 16 different fruit and drink pairs, Devin must have √16 = 4 different options for fruit (and 4 different options for drink).
Devin’s twin brother, Kevin, doesn’t eat pizza, so he brings a tuna salad sandwich, a fruit and a drink for lunch every Friday. Monday through Thursday, however, Kevin buys lunch in the school cafeteria. Kevin has twice as many fruit options as he does drink options. If Kevin has four fruit options, how many different lunches can Kevin bring to school for his Friday lunch?
Kevin has 4 fruit options. That means Kevin has 4 ÷ 2 = 2 drink options. With 4 fruit options and 2 drink options, there are 2 × 4 = 8 different lunches that Kevin could bring to school for his Friday lunch.
Each four-week period, Devin spends a total of $10.56 to buy pizza from the school cafeteria. Kevin pays the same amount for each meal he buys from the school cafeteria and spends a total of $33.60 each four-week period. How much more money is spent per meal by the twin who buys the more expensive school lunch?
Each four-week period, Devin buys 4 school lunches and pays a total of $10.56. That’s 10.56 ÷ 4 = $2.64 per meal. Kevin pays a total of $33.60 each four-week period to buy 16 school lunches. That’s 33.60 ÷ 16 = $2.10 per meal. So, Devin spends 2.64 − 2.10 = $0.54 more per meal than Kevin.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
The Washington Monument in Washington, D.C. is an obelisk. An obelisk is a tall monument with four gently sloping sides and a pyramid shape on top. The obelisk structure first appeared in Ancient Egyptian architecture.
The Great Pyramid of Giza, another famous architectural design of the Ancient Egyptians, has a square base of side length 756 feet. The square base of the pyramid on top of the Washington Monument has side length 34 feet. The perimeter of the square base of the Great Pyramid is how many times bigger than the square base of the pyramid at the top of the Washington Monument? Express your answer to the nearest whole number.
The perimeter of the square base of the Great Pyramid of Giza is 4 × 756 feet = 3024 feet. The perimeter of the base of the pyramid on top of the Washington Monument is 4 × 34 feet = 136 feet. Compared to the perimeter of the base of the pyramid on top of the Washington Monument, the perimeter of the base of the Great Pyramid is 3024 feet ÷ 136 feet ≈ 22 times larger.
The Washington Monument is 100 feet taller than the Great Pyramid of Giza. If the Washington Monument and the Great Pyramid each were 355 feet shorter, the Washington Monument would be twice as tall as the Great Pyramid. How many feet tall is the Washington Monument?
Using the information in the problem, we can write two equations. Let w represent the height of the Washington Monument, in feet and p represent the height of the Great Pyramid of Giza, in feet. From the first part of the problem, which states that the Washington Monument is 100 feet taller than the Great Pyramid, we can write w = p + 100. From the second part of the problem, we can write 2 × (p – 355) = w − 355. If we isolate p in the first equation by subtracting 100 from both sides, we get p = w – 100. We can then plug this into the second equation and solve for w:
2 × (w – 100 – 355) = w – 355
2 × (w – 455) = w – 355
2w – 910 = w – 355
w = 555
Since we defined w as the height of the Washington Monument, in feet, the Washington Monument is 555 feet tall.
The entire Washington Monument is ten times as high as the pyramid on top. Based on the previous problem, if this pyramid was removed, how tall would the Washington Monument be? Express your answer to the nearest tenth of a foot.
If the entire Washington Monument is ten times the height of the pyramid on top, the height of the pyramid must be 555 feet ÷ 10 = 55.5 feet. If the pyramid was removed from the top of the monument, the new height would be 555 feet – 55.5 feet = 499.5 feet.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Nancy and Tim have an apple cider stand at their school every morning during the fall to raise money for their club. Since September 23 is the Autumnal Equinox, the first day of fall and their first day of business will be September 23.
Several weeks ago, in preparation for their 2023 opening, Nancy and Tim went to pick their apples from a local orchard (all of their cider is homemade). Nancy picked apples at a rate of 30 apples per hour and Tim picked apples at a rate of 25 apples per hour. Tim and Nancy picked the same number of apples. Nancy picked apples for 5 hours, so how many hours must Tim have spent picking apples?
If Nancy picked 30 apples per hour for 5 hours, Nancy picked 5 × 30 = 150 apples. If Tim picks apples at a rate of 25 apples per hour, we can divide 150 apples by his rate of 25 and get 150 ÷ 25 = 6 hours.
Each gallon of cider produced requires one bushel of apples (40 apples), which cost Nancy and Tim $30 at the orchard. Anxious to calculate how much they will make, Nancy decides to calculate their profits, based on serving 6 oz cups of cider (in cups that were donated by a friend) and charging $2.00 per cup. Assuming they sell all of the cider they can make, how much profit will they make from the apples they picked? Note: There are 128 oz in 1 gallon.
First, take Nancy and Tim’s total of 300 apples and divide that by 40 apples per bushel to get 7.5 bushels of apples that they picked. Since one gallon of cider requires one bushel of apples, 7.5 bushels would produce 7.5 gallons of cider. Knowing there are 128 ounces per gallon, Nancy and Tim can make 7.5(128) = 960 ounces of cider. Since each cup is 6 ounces, Nancy and Tim have 960/6 = 160 cups of cider to sell. At $2.00 per cup, Nancy and Tim would make 160($2.00) = $320.00. They spent $30 per bushel, so in total, $30 × 7.5 bushels = $225.00. Therefore, in profits, Nancy and Tim would make $320 – $225 = $95.
Hoping to make more than what the calculations show, Nancy ponders what will happen if they add some water to the cider. She decides to calculate how much they would profit if they added 1.5 gallons of water to the 7.5 gallons of cider. If they sell the diluted cider for the same price as they had planned to sell for ($2.00 per 6-ounce cup), how much additional profit will be made?
Since they are adding 1.5 gallons of water, they will have an additional 1.5 gallons of liquid to sell without any additional expenses. Thus, those (1.5 gallons)(128 oz/gal)/(6 oz per cup) = 32 extra cups are pure profit. This will result in an additional profit of 32($2.00) = $64.00.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
It’s Autumn, and that means leaves are falling! Amber’s house occupies 2650 ft2 in the middle of a 50-foot by 80-foot parcel of land, and the leaves that fall cover the ground all around her house. If it takes Amber’s family 4.5 hours to rake all the leaves, how many square feet of leaves did they rake each hour?
The entire parcel of land on which Amber’s house sits measures 50 × 80 = 4000 ft2. Since leaves cover the ground all around the house, it follows that the family raked 4000 − 2650 = 1350 ft2 of leaves in 4.5 hours. That means each hour they raked 1350/4.5 = 300 ft2 of leaves.
Last year, when Amber’s family raked the leaves, they used one bag per 100 square feet raked. This year, since there were more fallen leaves, they ended up using 50% more bags. How many more bags did they use this year?
Let’s assume that they raked 1350 ft2 of fallen leaves last year, and used one bag per 100 square feet raked. Since 1350/100 = 13.5, they would have used 14 bags. Since they used 50% more bags than they did last year, this year they must have used an additional 0.5 × 14 = 7 bags, or 21 bags total.
Amber especially loves this time of year, when the leaves on the trees in her yard begin changing color to display an array of vibrant colors. There are yellow hickory tree leaves, orange maple tree leaves, purple cherry tree leaves, yellow ash tree leaves and scarlet dogwood tree leaves. Before raking, Amber collected some of the fallen leaves for a craft project. If the leaves Amber collected include at least one leaf from each type of tree, at least two purple leaves and at least three yellow leaves, how many such collections of 10 leaves are possible?
From the information given, we know that five of the leaves Amber collected were a yellow hickory, a orange maple, a purple cherry, a yellow ash and a scarlet dogwood, and these five leaves include one purple and two yellow leaves. We also know that her collection must include at least one more purple cherry leaf. We now know that any such collection of 10 leaves must include the six leaves we’ve already identified. So, we need only focus on the four leaves needed to complete a collection of 10 leaves, noting that at least one of the four leaves must be yellow.
While we could count how many groups of four leaves that contain at least one yellow leaf, it might be easier to subtract the number of groups of four that do not include any yellow leaves from the total number of groups of four leaves. We can do this quickly using the “stars and bars” technique. In this case, we have 4 stars, representing the four leaves, and 4 bars used to separate them into five bins (one for each type of leaf). For example, * | * | * | * | shows a star in each of the first four bins, depicting a group made up of one yellow hickory leaf, one orange maple, one purple cherry and one yellow ash. Similarly, * | * | | | * * shows a star in the first two bins and two stars in the fifth bin, depicting a group made up of one yellow hickory leaf, one orange maple and two scarlet dogwoods.
Now, to find the total number of groups of four leaves, we need to find the number of permutations of * * * * | | | | (4 stars and 4 bars). Recall that the number of permutations of m objects with n objects of one kind and m − n objects of another kind is m!/(n!(m − n)!). So, the total number of groups of four leaves is 8!/(4! × 4!) = (8 × 7 × 6 × 5)/(4 × 3 × 2 × 1) = 70 groups. Next, if we want to count only groups of four that exclude yellow leaves, there are three types of leaves to consider, or three bins with only 2 bars needed to separate them. So, to determine the number of groups of four that do not include any yellow leaves, we need to find the number of permutations of * * * * | | (4 stars and 2 bars). Doing so, we see that there are 6!/(4! × 2!) = (6 × 5)/(2 × 1) = 15 groups of four that do not include any yellow leaves. Subtracting, we get 70 − 15 = 55 of the collections of 10 leaves described.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
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When Riya’s alarm went off on the first day of her senior year in high school, she hit the snooze button and slept for another 10 minutes. When the alarm sounded the next time, Riya got up and immediately brushed and flossed her teeth, which took 5 minutes. A half hour later, Riya was showered and dressed. She then spent 45 minutes getting her hair and make-up just right. Next, Riya had a glass of milk and a bowl of microwave oatmeal for breakfast. Riya spent 30 minutes preparing and eating her breakfast. Twenty-four minutes later, at 8:39 a.m., Riya arrived at school ready to begin the new school year. At what time did Riya’s alarm go off the first time?
After Riya’s alarm went off the first time, and before she arrived at school, Riya did the following:
- Snoozed 10 minutes
- Brushed & flossed 5 minutes
- Showered & dressed 30 minutes
- Fixed hair & make-up 45 minutes
- Prepared & ate breakfast 30 minutes
- Traveled to school 24 minutes
TOTAL ELAPSED TIME: 2 hours 24 minutes
If Riya arrived at school at 8:39, her alarm went off the first time at 2 hours and 24 minutes before 8:39, which would have been 8:39 − 2:24 = 6:15 a.m.
There are two different routes that Riya can take to get to school. The first route, which is 12 miles, gets her to school in 20 minutes. The second route is only 8 miles, but has more traffic lights, so it takes 24 minutes. Based on this information, in miles per hour, what is the absolute difference in Riya’s average speeds for both routes?
For the first route, Riya can travel 12 miles in 20 minutes. Since 20 minutes is 20/60 = 1/3 hour, Riya’s rate would be 12 ÷ (1/3) = 12 × 3 = 36 mi/h. For the second route, Riya can travel 8 miles in 24 minutes. Since 24 minutes is 24/60 = 2/5 hour, Riya’s rate would be 8 ÷ (2/5) = 8 × (5/2) = 20 mi/h. The absolute difference in these rates is |36 − 20| = 16 mi/h.
At Riya’s school, there are three parking lots with a total of 175 parking spaces. Three-fifths of these spaces are reserved for juniors and seniors who drive to school. There are twice as many parking spaces assigned to seniors as there are assigned to juniors. How many of the parking spaces at Riya’s school are not assigned to seniors?
We are told that 3/5 of the spaces are reserved for juniors and seniors. We are also told that the number of spaces for seniors is twice the number of spaces for juniors, meaning that 2/3 of the spaces reserved for juniors and seniors are for seniors. It follows, then, that 2/3 of 3/5 of the 175 spaces, or (2/3) × (3/5) = 2/5 of the spaces are reserved for seniors. That means that 3/5 of the spaces, or (3/5) × 175 = 105 spaces are not reserved for seniors.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Ms. Cross leads The National Math Club at the middle school where she teaches. At the first club meeting of the school year, 60% of the students in attendance were 7th graders. If there was one fewer 8th grader than 7th grader in attendance, how many students attended the first club meeting?
Let a and b represent the number of 7th graders and 8th graders who attended the first club meeting, respectively. We know that a = b + 1 and a = 0.6(a + b). Solving for a in the second equation yields a = 0.6a + 0.6b → 0.4a = 0.6b → a = 1.5b. Substituting this value for a in the first equation, we get 1.5b = b + 1 → 0.5b = 1 → b = 2. So, there were 2 8th graders and 2 + 1 = 3 7th graders in attendance at the first club meeting. Therefore, a total of 2 + 3 = 5 students attended the first club meeting.
At the second club meeting of the school year, Ms. Cross noticed that among the students in attendance, there were equal numbers of 7th graders and 8th graders. Ms. Cross also noticed that all the students from the first club meeting were in attendance, along with some new students who weren’t at the first club meeting. If twice as many new 8th graders attended the second club meeting as new 7th graders, how many new students attended the second club meeting?
From the previous problem, we know that there were 3 7th graders and 2 8th graders in attendance at the first club meeting. It follows, then, from the information given, that if x new 7th graders attended the second club meeting, 2x new 8th graders attended that meeting. Since the same number of 8th graders and 7th graders attended the second meeting, we can write 3 + x = 2 + 2x and solve to get x = 1. That means 1 new 7th grader and 2 × 1 = 2 new 8th graders attended the second meeting. Therefore, a total of 1 + 2 = 3 new students attended the second meeting.
At the third meeting of The National Math Club, Ms. Cross noticed that 60% of the students in attendance were 8th graders and that the total number of students in attendance was double that of the first club meeting. Ms. Cross was pleased to see that all the students who attended the second club meeting also attended the third club meeting. Of the students in attendance at the third club meeting who did not attend the first two club meetings, what is the absolute difference between the number of 8th graders and the number of 7th graders?
From the first problem, we know that a total of 5 students attended the first club meeting. Based on the information provided, twice that number, or 5 × 2 = 10 students attended the third club meeting. Since 8th graders accounted for 60% of these 10 students, it follows that 0.6 × 10 = 6 8th graders and 10 – 6 = 4 7th graders attended the third club meeting. Since 4 7th graders and 4 8th graders attended the second club meeting, there were 4 – 4 = 0 new 7th graders and 6 – 4 = 2 new 8th graders at the third club meeting. The absolute difference between these two quantities is 2 – 0 = 2.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Each week, Devin buys school lunch on Friday because the cafeteria serves pizza every Friday. Monday through Thursday, Devin brings his lunch to school. The lunch he brings always includes a peanut butter and jelly sandwich, a fruit and a drink. Devin has enough fruit and drink options to make a different lunch every day that he brings lunch for the next four weeks. If Devin has the same number of fruit options as he has drink options and he takes no days off from school, how many different fruit options does he have?
In four weeks, there are 20 school days. Devin will be eating school lunch on 4 of those days. So, he has enough fruit and drink options to make 20 − 4 = 16 different lunches. If Devin only had one option for fruit and one option for his drink, that would be enough to make 1 unique meal. If he had two options for each, Devin could make 2 × 2 = 4 unique meals since each fruit option could be paired with each of the drink options (F1D1, F1D2, F2D1, F2D2). This is the Fundamental Principle of Counting, which says that if you have m distinct varieties of one item and n distinct varieties of another item, there are m × n distinct pairs consisting of one of each of the items. So, to get 16 different fruit and drink pairs, Devin must have √16 = 4 different options for fruit (and 4 different options for drink).
Devin’s twin brother, Kevin, doesn’t eat pizza, so he brings a tuna salad sandwich, a fruit and a drink for lunch every Friday. Monday through Thursday, however, Kevin buys lunch in the school cafeteria. Kevin has twice as many fruit options as he does drink options. If Kevin has four fruit options, how many different lunches can Kevin bring to school for his Friday lunch?
Kevin has 4 fruit options. That means Kevin has 4 ÷ 2 = 2 drink options. With 4 fruit options and 2 drink options, there are 2 × 4 = 8 different lunches that Kevin could bring to school for his Friday lunch.
Each four-week period, Devin spends a total of $10.56 to buy pizza from the school cafeteria. Kevin pays the same amount for each meal he buys from the school cafeteria and spends a total of $33.60 each four-week period. How much more money is spent per meal by the twin who buys the more expensive school lunch?
Each four-week period, Devin buys 4 school lunches and pays a total of $10.56. That’s 10.56 ÷ 4 = $2.64 per meal. Kevin pays a total of $33.60 each four-week period to buy 16 school lunches. That’s 33.60 ÷ 16 = $2.10 per meal. So, Devin spends 2.64 − 2.10 = $0.54 more per meal than Kevin.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
The Washington Monument in Washington, D.C. is an obelisk. An obelisk is a tall monument with four gently sloping sides and a pyramid shape on top. The obelisk structure first appeared in Ancient Egyptian architecture.
The Great Pyramid of Giza, another famous architectural design of the Ancient Egyptians, has a square base of side length 756 feet. The square base of the pyramid on top of the Washington Monument has side length 34 feet. The perimeter of the square base of the Great Pyramid is how many times bigger than the square base of the pyramid at the top of the Washington Monument? Express your answer to the nearest whole number.
The perimeter of the square base of the Great Pyramid of Giza is 4 × 756 feet = 3024 feet. The perimeter of the base of the pyramid on top of the Washington Monument is 4 × 34 feet = 136 feet. Compared to the perimeter of the base of the pyramid on top of the Washington Monument, the perimeter of the base of the Great Pyramid is 3024 feet ÷ 136 feet ≈ 22 times larger.
The Washington Monument is 100 feet taller than the Great Pyramid of Giza. If the Washington Monument and the Great Pyramid each were 355 feet shorter, the Washington Monument would be twice as tall as the Great Pyramid. How many feet tall is the Washington Monument?
Using the information in the problem, we can write two equations. Let w represent the height of the Washington Monument, in feet and p represent the height of the Great Pyramid of Giza, in feet. From the first part of the problem, which states that the Washington Monument is 100 feet taller than the Great Pyramid, we can write w = p + 100. From the second part of the problem, we can write 2 × (p – 355) = w − 355. If we isolate p in the first equation by subtracting 100 from both sides, we get p = w – 100. We can then plug this into the second equation and solve for w:
2 × (w – 100 – 355) = w – 355
2 × (w – 455) = w – 355
2w – 910 = w – 355
w = 555
Since we defined w as the height of the Washington Monument, in feet, the Washington Monument is 555 feet tall.
The entire Washington Monument is ten times as high as the pyramid on top. Based on the previous problem, if this pyramid was removed, how tall would the Washington Monument be? Express your answer to the nearest tenth of a foot.
If the entire Washington Monument is ten times the height of the pyramid on top, the height of the pyramid must be 555 feet ÷ 10 = 55.5 feet. If the pyramid was removed from the top of the monument, the new height would be 555 feet – 55.5 feet = 499.5 feet.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Nancy and Tim have an apple cider stand at their school every morning during the fall to raise money for their club. Since September 23 is the Autumnal Equinox, the first day of fall and their first day of business will be September 23.
Several weeks ago, in preparation for their 2023 opening, Nancy and Tim went to pick their apples from a local orchard (all of their cider is homemade). Nancy picked apples at a rate of 30 apples per hour and Tim picked apples at a rate of 25 apples per hour. Tim and Nancy picked the same number of apples. Nancy picked apples for 5 hours, so how many hours must Tim have spent picking apples?
If Nancy picked 30 apples per hour for 5 hours, Nancy picked 5 × 30 = 150 apples. If Tim picks apples at a rate of 25 apples per hour, we can divide 150 apples by his rate of 25 and get 150 ÷ 25 = 6 hours.
Each gallon of cider produced requires one bushel of apples (40 apples), which cost Nancy and Tim $30 at the orchard. Anxious to calculate how much they will make, Nancy decides to calculate their profits, based on serving 6 oz cups of cider (in cups that were donated by a friend) and charging $2.00 per cup. Assuming they sell all of the cider they can make, how much profit will they make from the apples they picked? Note: There are 128 oz in 1 gallon.
First, take Nancy and Tim’s total of 300 apples and divide that by 40 apples per bushel to get 7.5 bushels of apples that they picked. Since one gallon of cider requires one bushel of apples, 7.5 bushels would produce 7.5 gallons of cider. Knowing there are 128 ounces per gallon, Nancy and Tim can make 7.5(128) = 960 ounces of cider. Since each cup is 6 ounces, Nancy and Tim have 960/6 = 160 cups of cider to sell. At $2.00 per cup, Nancy and Tim would make 160($2.00) = $320.00. They spent $30 per bushel, so in total, $30 × 7.5 bushels = $225.00. Therefore, in profits, Nancy and Tim would make $320 – $225 = $95.
Hoping to make more than what the calculations show, Nancy ponders what will happen if they add some water to the cider. She decides to calculate how much they would profit if they added 1.5 gallons of water to the 7.5 gallons of cider. If they sell the diluted cider for the same price as they had planned to sell for ($2.00 per 6-ounce cup), how much additional profit will be made?
Since they are adding 1.5 gallons of water, they will have an additional 1.5 gallons of liquid to sell without any additional expenses. Thus, those (1.5 gallons)(128 oz/gal)/(6 oz per cup) = 32 extra cups are pure profit. This will result in an additional profit of 32($2.00) = $64.00.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
It’s Autumn, and that means leaves are falling! Amber’s house occupies 2650 ft2 in the middle of a 50-foot by 80-foot parcel of land, and the leaves that fall cover the ground all around her house. If it takes Amber’s family 4.5 hours to rake all the leaves, how many square feet of leaves did they rake each hour?
The entire parcel of land on which Amber’s house sits measures 50 × 80 = 4000 ft2. Since leaves cover the ground all around the house, it follows that the family raked 4000 − 2650 = 1350 ft2 of leaves in 4.5 hours. That means each hour they raked 1350/4.5 = 300 ft2 of leaves.
Last year, when Amber’s family raked the leaves, they used one bag per 100 square feet raked. This year, since there were more fallen leaves, they ended up using 50% more bags. How many more bags did they use this year?
Let’s assume that they raked 1350 ft2 of fallen leaves last year, and used one bag per 100 square feet raked. Since 1350/100 = 13.5, they would have used 14 bags. Since they used 50% more bags than they did last year, this year they must have used an additional 0.5 × 14 = 7 bags, or 21 bags total.
Amber especially loves this time of year, when the leaves on the trees in her yard begin changing color to display an array of vibrant colors. There are yellow hickory tree leaves, orange maple tree leaves, purple cherry tree leaves, yellow ash tree leaves and scarlet dogwood tree leaves. Before raking, Amber collected some of the fallen leaves for a craft project. If the leaves Amber collected include at least one leaf from each type of tree, at least two purple leaves and at least three yellow leaves, how many such collections of 10 leaves are possible?
From the information given, we know that five of the leaves Amber collected were a yellow hickory, a orange maple, a purple cherry, a yellow ash and a scarlet dogwood, and these five leaves include one purple and two yellow leaves. We also know that her collection must include at least one more purple cherry leaf. We now know that any such collection of 10 leaves must include the six leaves we’ve already identified. So, we need only focus on the four leaves needed to complete a collection of 10 leaves, noting that at least one of the four leaves must be yellow.
While we could count how many groups of four leaves that contain at least one yellow leaf, it might be easier to subtract the number of groups of four that do not include any yellow leaves from the total number of groups of four leaves. We can do this quickly using the “stars and bars” technique. In this case, we have 4 stars, representing the four leaves, and 4 bars used to separate them into five bins (one for each type of leaf). For example, * | * | * | * | shows a star in each of the first four bins, depicting a group made up of one yellow hickory leaf, one orange maple, one purple cherry and one yellow ash. Similarly, * | * | | | * * shows a star in the first two bins and two stars in the fifth bin, depicting a group made up of one yellow hickory leaf, one orange maple and two scarlet dogwoods.
Now, to find the total number of groups of four leaves, we need to find the number of permutations of * * * * | | | | (4 stars and 4 bars). Recall that the number of permutations of m objects with n objects of one kind and m − n objects of another kind is m!/(n!(m − n)!). So, the total number of groups of four leaves is 8!/(4! × 4!) = (8 × 7 × 6 × 5)/(4 × 3 × 2 × 1) = 70 groups. Next, if we want to count only groups of four that exclude yellow leaves, there are three types of leaves to consider, or three bins with only 2 bars needed to separate them. So, to determine the number of groups of four that do not include any yellow leaves, we need to find the number of permutations of * * * * | | (4 stars and 2 bars). Doing so, we see that there are 6!/(4! × 2!) = (6 × 5)/(2 × 1) = 15 groups of four that do not include any yellow leaves. Subtracting, we get 70 − 15 = 55 of the collections of 10 leaves described.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
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