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As the weather is turning cooler and daylight is arriving later in the morning and trees are starting to lose their leaves, many of us are noticing that summer is over and we’re well into fall. Or is it autumn? Mr. Kravis surveyed his class, and out of 28 students (with every student picking exactly one of the two options), the ratio of the number of students who called the season “fall” to the number of students who called the season “autumn” was 3:1. How many students called the season “fall?”
The ratio tells us that for every three students who call the season “fall”, there is one student who calls the season “autumn”. We know this is true for every group of four students. In the class of 28 students, there are 28 ÷ 4 = 7 of these groups of four students. This means that there are 7 × 3 = 21 students who call the season “fall”. An equation that we could have written and used is 3x + 1x = 28. Then we could solve for x and find the value of 3x.
If we were now to create a pie chart showing the two groups of students determined during this survey, what would be the degree measure of the central angle of the sector of the circle representing the students who call the season “autumn?”
Still using the information we gained from the initial ratio, we see that there were 7 students of 28 that call the season “autumn”. This is 25% of the students. (We could also get this by knowing that one of every four students used “autumn”.) If 25% of the pie chart is then devoted to this group of students, the sector of the pie chart would be 25% of 360 degrees, which is 0.25 × 360 = 90 degrees.
Mr. Kravis then extended his survey to include the entire school. Every student responded with the term they use the most (fall or autumn). He found that in this larger survey of the entire school, the number of students who refer to the season as “fall” is five times the number of students who refer to the season as “autumn.” What fraction of the total number of students in the school refers to the season as “autumn?” Express your answer as a common fraction.
If we allow x to be the number of “autumn students”, then 5x would represent the number of “fall students”. This is a total of x + 5x = 6x students. Then x of the 6x students use the term “autumn” or x/6x = 1/6 of the students use the term “autumn”.
If there are 360 students in the school who answered Mr. Kravis’ survey, how many of them use the term “fall?”
As we saw with the solution to the previous problem, our total number of students is represented by 6x, which we now know to be equal to 360. If 6x = 360, then dividing both sides by 6 shows x = 60. The number of students using the term “fall” was represented by 5x, which we find is 5(60) = 300 students.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
The letters of the words GHOST, GHOUL, TRICK, TREAT and CANDY were written on 25 index cards, one letter per card. If the cards are shuffled and one card is selected at random, what is the probability that the card contains one of the letters in the word GHOST?
The 25 index cards contain the letters A, A, C, C, D, E, G, G, H, H, I, K, L, N, O, O, R, R, S, T, T, T, T, U and Y. Since two of the 25 cards contain the letter G, two contain the letter H, two contain the letter O, one contains the letter S and four contain the letter T, it follows that 2 + 2 + 2 + 1 + 4 = 11 of the 25 cards contain a letter in the word GHOST. Therefore, the probability of randomly selecting a card that contains a letter in the word GHOST is 11/25.
If all 25 cards are shuffled again, what is the probability of selecting, in any order and without replacement, five cards that contain the letters G, H, O, S and T, in the first five selections? Express your answer as a common fraction.
Since two of the cards contain the letter G, the probability of randomly selecting one of those cards is 2/25. If a card with the letter G is selected, the probability of randomly selecting next a card with the letter H is 2/24. The probability that the next randomly selected card contains the letter O is 2/23, since two of the cards contain the letter O. Since only one card contains the letter S, the probability that the next randomly selected card contains the letter S is 1/22. There are four cards with the letter T, so the probability that the fifth randomly selected card contains the letter T is 4/21. Therefore, the probability of selecting the letters G, H, O, S and T (in that order) is (2/25) × (2/24) × (2/23) × (1/22) × (4/21) = 32/6,375,600 = 2/398,475. But we are asked to determine the probability of randomly selecting, in any order, cards containing these letters. Since any five cards can be selected in 5! = 120 different orders, the probability is 2/398,475 × 120 = 240/398,475 = 16/26,565.
Another approach would be to consider all the ways that cards with the letters G, H, O, S and T can be selected and divide that by the total number of ways any five cards can randomly be selected. There are two ways to randomly select cards with each of the letters G, H and O. There is one way to randomly select a card with the letter S, and there are four ways to randomly select a card with the letter T. Therefore, the number of ways to select cards with the letters G, H, O, S and T is 2 × 2 × 2 × 1 × 4 = 32 ways. But cards with these five letters can be selected in 5! = 120 different orders, for a total of 32 × 120 = 3840 ways. The number of ways to randomly select five lettered cards is 25P5 = 25!/20! = 25 × 24 × 23 × 22 × 21 = 6,375,600. Thus, the probability of randomly selecting cards with the letters G, H, O, S and T is 3840/6,375,600 = 16/26,565.
If all 25 cards are shuffled once more, what is the probability of selecting, in any order and without replacement, five cards that contain the letters T, R, I, C and K or five cards that contain the letters T, R, E, A and T, in the first five selections? Express your answer as a common fraction.
First, let’s determine the probability of randomly selecting the letters T, R, I, C and K. Since four cards contain the letter T, the probability of randomly selecting one of those cards is 4/25. The probability of next randomly selecting one of the two cards containing the letter R is 2/24. The probability that the next card selected contains the letter I is 1/23. The probability that the next randomly selected card contains the letter C is 2/22, and the probability that the fifth card randomly selected contains the letter K is 1/21. Therefore, the probability of randomly selecting five cards containing the letter T, R, I, C and K (in that order) is (4/25) × (2/24) × (1/23) × (2/22) × (1/21) = 16/6,375,600 = 1/398,475. Since any five cards can be selected in 5! = 120 different orders, the probability of randomly selecting five cards with the letters T, R, I, C and K is 1/398,475 × 120 = 120/398,475 = 8/26,565.
Now, we can determine the probability of randomly selecting cards containing the letters T, R, E, A and T. Again, the chance of randomly selecting one of the four cards containing the letter T is 4/25, and the probability of next randomly selecting one of the two cards containing the letter R is 2/24. The probability that the next randomly selected card contains the letter E is 1/23, and the probability that the card randomly selected after that one contains the letter A is 2/22. Finally, the probability that the fifth randomly selected card also contains the letter T is 3/21, since one T has already been removed. Therefore, the probability of randomly selecting five cards containing the letters T, R, E, A and T (in that order) is (4/25) × (2/24) × (1/23) × (2/22) × (3/21) = 48/6,375,600 = 1/132,825. Five different cards can be selected in 5! = 120 different orders, but since the two cards containing the letter T are indistinguishable, we must divide this by 2! = 2. Therefore, these five cards can actually be selected in 120/2 = 60 different orders. So, the probability of randomly selecting five cards with the letters T, R, E, A and T is 1/132,825 × 60 = 60/132,825 = 12/26,565.
So, the probability of randomly selecting five cards that contain the letters T, R, I, C and K or five cards that contain the letters T, R, E, A and T is 8/26,565 + 12/26,565 = 20/26,565 = 4/5313.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
In honor of Veteran’s Day, people around the country are flying American flags. For many people, finding flags to buy in stores has been next to impossible due to the great demand. Therefore, many of us are getting creative, and have started making our own handmade flags out of paper, pins, construction paper, fabric, etc. When flying horizontally, the overall width (top to bottom) and length (left to right) of the flag is in the ratio of 1:1.9. There are 13 stripes of equal width. The width (top to bottom) of the blue field containing the stars is 7/13 of the overall width of the flag; that is, from the top of the flag to the bottom of the 7th stripe. The length (left to right) of blue field is .76 of the overall width of the flag.
Ana wants to draw a US flag on a standard 8½ by 11-inch sheet of construction paper. What are the dimensions, in inches, of the largest flag she can draw if she approximates the ratio of the width:length as 1:2?
The length is limited to 11 inches, so the width must be half of that, or 5.5 inches. Therefore, the largest dimensions for the flag will be 5.5 by 11 inches.
Antonio is making a US flag with 2-inch-wide stripes. If his flag has the correct ratios, what is the area of the entire rectangular blue field (including the stars)?
If each stripe is 2 inches wide, the total width of the flag is 2 × 13 = 26 inches. The length of the blue field is 0.76 × 26 = 19.76 inches. The width of the blue field is 7 × 2 = 14 inches. Therefore, the area of the blue field is 19.76 × 14 = 276.64 in2.
Our current flag has 50 stars, arranged in nine alternating rows of six and five stars. However, the original flag had 13 stars arranged in a circle, such that the outermost tip of each of the 13 stars was on the circumference of the circle. If the distance between the outermost tips of consecutive stars is 4 inches on the circumference of the circle, what is the radius of the circle, in inches? Express your answer in terms of pi.
If the points of the 13 stars are 4 inches apart, the circumference of the circle measures 13 × 4 = 52 inches. Since C = 2πr, we get 52 = 2πr. Solving for r, by dividing both sides of the equation by 2π, gives us a radius of 26/π inches.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Abby, Brenda, Celia, Denise and Elizabeth are going to their town’s annual hayride. Tickets cost $8.00 for people 12 years old and older but only $5.00 for people 11 years old and younger. The total price for their five tickets is $31.00. How many people in the group are 12 years old or older?
First let’s set up equations based on the information in the question. Let’s use a to represent the number of tickets for people 12 years old and older sold and b to represent the number of tickets for people 11 years old and younger sold.
a + b = 5
8a + 5b = $31.00
Now that we have two variables and two equations, we can use substitution to solve.
a + b = 5 → a = 5 – b
Substitute: 8(5 – b) + 5b = $31.00
40 – 8b + 5b = 31 → –3b = –9 → b = 3 people 11 years old or younger
Thus, 5 – 3 = 2 people are 12 years old or older.
The group of five girls managed to secure five seats in a row on the crowded hayride. In how many orders can the five girls sit if they all sit in a row?
Any of the five girls could sit in the first seat; any of the remaining four girls could sit in the second seat, and so on. Thus, there are (5)(4)(3)(2)(1) = 120 orders the five girls could sit in.
In an attempt to minimize the lines this year, the hayride operators decided to run 3 flatbeds at a time. Initially, they use 20 bales of hay on each flatbed. They find that at the end of each day they need to replace half of the hay on each of the flatbeds to be ready for the next day’s business. How many total bales of hay will the hayride operators have used on the 3 flatbeds after they replenish the hay at the end of the fifth day of business?
On the first day, they start with 20(3) = 60 bales of hay. At the end of each business day, they replenish half of the hay (which would be 10 bales per flatbed, or 30 bales). After replenishing at the end of the first day, they have used a total of 60 + 30 = 90 bales. After replenishing at the end of the second day, they have used a total of 90 + 30 = 120 bales. Continuing this process, we see that at the end of the fifth day, they have used 60 + 30 + 30 + 30 + 30 + 30 = 60 + 150 = 210 bales of hay.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
In preparation for Thanksgiving dinner, Mrs. Cobble decided to get a new tablecloth for the dining room table. Her rectangular table measures 3.5 feet by 6 feet, and she would like the tablecloth to hang exactly 9 inches over each side. What is the area, in square feet, of the ideal tablecloth for Mrs. Cobble?
If Mrs. Cobble would like the tablecloth to hang exactly 9 inches over each side of the rectangular table, then she will need to add 18 inches to both the width and the length of the table’s dimensions. The table is 3.5 feet by 6 feet, and we are asked to give the answer in square feet. Therefore, let’s change 18 inches into its equivalent in feet. We can calculate that 18 inches is 18 ÷ 12 = 1.5 feet. The dimensions of the tablecloth will then be 3.5 + 1.5 = 5 feet by 6 + 1.5 = 7.5 feet. This rectangle has an area of 5 × 7.5 = 37.5 square feet.
Now Mrs. Cobble will turn her attention to the actual meal. According to an article she read, she should purchase 1¼ pounds of turkey per person. She figures this is a good amount for an adult, but a child would need only 2/3 of this amount. There will be 8 adults and 6 children at Thanksgiving dinner. According to Mrs. Cobble’s logic, how many pounds of turkey should she purchase?
We know that an adult will require 1¼ pounds of turkey. (We can use 1¼ or 5/4 or 1.25 as we proceed.) We are told that each child is only going to get 2/3 of this amount of turkey. So, we need to calculate 2/3 of 5/4. To do this, we find the product of the two fractions and see that each child will require 10/12 = 5/6 of a pound of turkey. Now we have eight adults each requiring 5/4 pounds and six children each requiring 5/6 of a pound. Mrs. Cobble will then need to buy a turkey that is 8(5/4) + 6(5/6) = 10 + 5 = 15 pounds.
When Thanksgiving has finally arrived, the whole family sits down to a Thanksgiving dinner that includes turkey, mashed potatoes, dressing, cranberry sauce and gravy. Uncle Bob’s favorite part of the meal, though, is the olive tray. Mrs. Cobble put the same number of green olives and black olives on the tray, and Uncle Bob was the first person to select from the tray. After Uncle Bob took eight green olives, the ratio of green olives to black olives was 3:5. How many total olives were on the tray before Uncle Bob took his olives?
We know that there used to be the same number of green olives and black olives, but after Uncle bob took 8 green olives, the ratio is then 3:5. This means that in every group of 8 olives, 3 of them are green and 5 of them are black. To bring this group of 8 olives back to “even,” Uncle Bob would need to put back 2 of his green olives, which would give us a total of 10 olives (5 green; 5 black). If it takes 2 olives for Uncle Bob to complete every uneven group of 8 olives, he could fix 4 groups of uneven olives with his 8 olives. Each of these 4 groups would then be 10 olives big, for a total of 40 olives.
Another way of looking at this...
The ratio of 3 green olives to 5 black olives leads us to believe that the number of olives now is a multiple of 8. So, there are now 8x olives. The ratio before Uncle Bob took his olives was 5 green olives to 5 black olives, or a total of 10x olives. There is a difference of 10x – 8x = 2x olives. So, the 8 olives he took is our 2x. We see 2x = 8 results in x = 4. So, there are now 8(4) = 32 olives, and there were originally 10(4) = 40 olives.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Melrose’s hot air balloon is a sphere of diameter 100 feet. How many cubic feet of air will Melrose’s hot air balloon hold when completely filled? Express your answer as a decimal to the nearest tenth.
The formula for the volume of a sphere is (4/3) × π × r3. Since we know the diameter of the balloon is 100 feet and the radius r is half the diameter, it follows that the radius is 50 feet. Substituting, we see that, completely filled, Melrose’s balloon will hold (4/3) × π × (50)3 = 523,598.8 ft3 of air.
Coco takes a hot air balloon trip over part of the Alleghany Mountains with her dad. They travel at an average speed of 40 mi/h for 120 miles before they land to take a 45-minute lunch break. After lunch, they fly 175 miles at an average speed of 35 mi/h. Including the lunch break, how many hours was their entire trip? Express your answer as a decimal to the nearest hundredth.
To solve this problem, we will be using the equation distance = rate × time. We’ll let T1 be the number of hours it took them to travel 120 miles at an average speed of 40 mi/h. So, we have 120 = 40T1 → T1 = 120/40 = 3 hours. We’ll let T2 be the number of hours it took them to travel 175 miles at an average speed of 35 mi/h. So, we have 175 = 35T2 → T2 = 5 hours. The lunch break was 45 minutes = 45/60 = 0.75 hours. Therefore, the entire trip, including the lunch break, took 3 + 0.75 + 5 = 8.75 hours.
Rhonda found that 10 super-sized helium balloons provide just enough lift to carry her 3-pound toy. At that same rate, how many super-sized balloons would be required to carry Rhonda’s 81-pound Alaskan malamute?
If 10 balloons lift exactly 3 pounds, then each balloon lifts the equivalent of 3/10 pound. Dividing 81 pounds by 3/10 pound, we see that the number of balloons required to lift her dog is 81/(3/10) = 81 × 10/3 = 810/3 = 270 balloons.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
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As the weather is turning cooler and daylight is arriving later in the morning and trees are starting to lose their leaves, many of us are noticing that summer is over and we’re well into fall. Or is it autumn? Mr. Kravis surveyed his class, and out of 28 students (with every student picking exactly one of the two options), the ratio of the number of students who called the season “fall” to the number of students who called the season “autumn” was 3:1. How many students called the season “fall?”
The ratio tells us that for every three students who call the season “fall”, there is one student who calls the season “autumn”. We know this is true for every group of four students. In the class of 28 students, there are 28 ÷ 4 = 7 of these groups of four students. This means that there are 7 × 3 = 21 students who call the season “fall”. An equation that we could have written and used is 3x + 1x = 28. Then we could solve for x and find the value of 3x.
If we were now to create a pie chart showing the two groups of students determined during this survey, what would be the degree measure of the central angle of the sector of the circle representing the students who call the season “autumn?”
Still using the information we gained from the initial ratio, we see that there were 7 students of 28 that call the season “autumn”. This is 25% of the students. (We could also get this by knowing that one of every four students used “autumn”.) If 25% of the pie chart is then devoted to this group of students, the sector of the pie chart would be 25% of 360 degrees, which is 0.25 × 360 = 90 degrees.
Mr. Kravis then extended his survey to include the entire school. Every student responded with the term they use the most (fall or autumn). He found that in this larger survey of the entire school, the number of students who refer to the season as “fall” is five times the number of students who refer to the season as “autumn.” What fraction of the total number of students in the school refers to the season as “autumn?” Express your answer as a common fraction.
If we allow x to be the number of “autumn students”, then 5x would represent the number of “fall students”. This is a total of x + 5x = 6x students. Then x of the 6x students use the term “autumn” or x/6x = 1/6 of the students use the term “autumn”.
If there are 360 students in the school who answered Mr. Kravis’ survey, how many of them use the term “fall?”
As we saw with the solution to the previous problem, our total number of students is represented by 6x, which we now know to be equal to 360. If 6x = 360, then dividing both sides by 6 shows x = 60. The number of students using the term “fall” was represented by 5x, which we find is 5(60) = 300 students.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
The letters of the words GHOST, GHOUL, TRICK, TREAT and CANDY were written on 25 index cards, one letter per card. If the cards are shuffled and one card is selected at random, what is the probability that the card contains one of the letters in the word GHOST?
The 25 index cards contain the letters A, A, C, C, D, E, G, G, H, H, I, K, L, N, O, O, R, R, S, T, T, T, T, U and Y. Since two of the 25 cards contain the letter G, two contain the letter H, two contain the letter O, one contains the letter S and four contain the letter T, it follows that 2 + 2 + 2 + 1 + 4 = 11 of the 25 cards contain a letter in the word GHOST. Therefore, the probability of randomly selecting a card that contains a letter in the word GHOST is 11/25.
If all 25 cards are shuffled again, what is the probability of selecting, in any order and without replacement, five cards that contain the letters G, H, O, S and T, in the first five selections? Express your answer as a common fraction.
Since two of the cards contain the letter G, the probability of randomly selecting one of those cards is 2/25. If a card with the letter G is selected, the probability of randomly selecting next a card with the letter H is 2/24. The probability that the next randomly selected card contains the letter O is 2/23, since two of the cards contain the letter O. Since only one card contains the letter S, the probability that the next randomly selected card contains the letter S is 1/22. There are four cards with the letter T, so the probability that the fifth randomly selected card contains the letter T is 4/21. Therefore, the probability of selecting the letters G, H, O, S and T (in that order) is (2/25) × (2/24) × (2/23) × (1/22) × (4/21) = 32/6,375,600 = 2/398,475. But we are asked to determine the probability of randomly selecting, in any order, cards containing these letters. Since any five cards can be selected in 5! = 120 different orders, the probability is 2/398,475 × 120 = 240/398,475 = 16/26,565.
Another approach would be to consider all the ways that cards with the letters G, H, O, S and T can be selected and divide that by the total number of ways any five cards can randomly be selected. There are two ways to randomly select cards with each of the letters G, H and O. There is one way to randomly select a card with the letter S, and there are four ways to randomly select a card with the letter T. Therefore, the number of ways to select cards with the letters G, H, O, S and T is 2 × 2 × 2 × 1 × 4 = 32 ways. But cards with these five letters can be selected in 5! = 120 different orders, for a total of 32 × 120 = 3840 ways. The number of ways to randomly select five lettered cards is 25P5 = 25!/20! = 25 × 24 × 23 × 22 × 21 = 6,375,600. Thus, the probability of randomly selecting cards with the letters G, H, O, S and T is 3840/6,375,600 = 16/26,565.
If all 25 cards are shuffled once more, what is the probability of selecting, in any order and without replacement, five cards that contain the letters T, R, I, C and K or five cards that contain the letters T, R, E, A and T, in the first five selections? Express your answer as a common fraction.
First, let’s determine the probability of randomly selecting the letters T, R, I, C and K. Since four cards contain the letter T, the probability of randomly selecting one of those cards is 4/25. The probability of next randomly selecting one of the two cards containing the letter R is 2/24. The probability that the next card selected contains the letter I is 1/23. The probability that the next randomly selected card contains the letter C is 2/22, and the probability that the fifth card randomly selected contains the letter K is 1/21. Therefore, the probability of randomly selecting five cards containing the letter T, R, I, C and K (in that order) is (4/25) × (2/24) × (1/23) × (2/22) × (1/21) = 16/6,375,600 = 1/398,475. Since any five cards can be selected in 5! = 120 different orders, the probability of randomly selecting five cards with the letters T, R, I, C and K is 1/398,475 × 120 = 120/398,475 = 8/26,565.
Now, we can determine the probability of randomly selecting cards containing the letters T, R, E, A and T. Again, the chance of randomly selecting one of the four cards containing the letter T is 4/25, and the probability of next randomly selecting one of the two cards containing the letter R is 2/24. The probability that the next randomly selected card contains the letter E is 1/23, and the probability that the card randomly selected after that one contains the letter A is 2/22. Finally, the probability that the fifth randomly selected card also contains the letter T is 3/21, since one T has already been removed. Therefore, the probability of randomly selecting five cards containing the letters T, R, E, A and T (in that order) is (4/25) × (2/24) × (1/23) × (2/22) × (3/21) = 48/6,375,600 = 1/132,825. Five different cards can be selected in 5! = 120 different orders, but since the two cards containing the letter T are indistinguishable, we must divide this by 2! = 2. Therefore, these five cards can actually be selected in 120/2 = 60 different orders. So, the probability of randomly selecting five cards with the letters T, R, E, A and T is 1/132,825 × 60 = 60/132,825 = 12/26,565.
So, the probability of randomly selecting five cards that contain the letters T, R, I, C and K or five cards that contain the letters T, R, E, A and T is 8/26,565 + 12/26,565 = 20/26,565 = 4/5313.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
In honor of Veteran’s Day, people around the country are flying American flags. For many people, finding flags to buy in stores has been next to impossible due to the great demand. Therefore, many of us are getting creative, and have started making our own handmade flags out of paper, pins, construction paper, fabric, etc. When flying horizontally, the overall width (top to bottom) and length (left to right) of the flag is in the ratio of 1:1.9. There are 13 stripes of equal width. The width (top to bottom) of the blue field containing the stars is 7/13 of the overall width of the flag; that is, from the top of the flag to the bottom of the 7th stripe. The length (left to right) of blue field is .76 of the overall width of the flag.
Ana wants to draw a US flag on a standard 8½ by 11-inch sheet of construction paper. What are the dimensions, in inches, of the largest flag she can draw if she approximates the ratio of the width:length as 1:2?
The length is limited to 11 inches, so the width must be half of that, or 5.5 inches. Therefore, the largest dimensions for the flag will be 5.5 by 11 inches.
Antonio is making a US flag with 2-inch-wide stripes. If his flag has the correct ratios, what is the area of the entire rectangular blue field (including the stars)?
If each stripe is 2 inches wide, the total width of the flag is 2 × 13 = 26 inches. The length of the blue field is 0.76 × 26 = 19.76 inches. The width of the blue field is 7 × 2 = 14 inches. Therefore, the area of the blue field is 19.76 × 14 = 276.64 in2.
Our current flag has 50 stars, arranged in nine alternating rows of six and five stars. However, the original flag had 13 stars arranged in a circle, such that the outermost tip of each of the 13 stars was on the circumference of the circle. If the distance between the outermost tips of consecutive stars is 4 inches on the circumference of the circle, what is the radius of the circle, in inches? Express your answer in terms of pi.
If the points of the 13 stars are 4 inches apart, the circumference of the circle measures 13 × 4 = 52 inches. Since C = 2πr, we get 52 = 2πr. Solving for r, by dividing both sides of the equation by 2π, gives us a radius of 26/π inches.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Abby, Brenda, Celia, Denise and Elizabeth are going to their town’s annual hayride. Tickets cost $8.00 for people 12 years old and older but only $5.00 for people 11 years old and younger. The total price for their five tickets is $31.00. How many people in the group are 12 years old or older?
First let’s set up equations based on the information in the question. Let’s use a to represent the number of tickets for people 12 years old and older sold and b to represent the number of tickets for people 11 years old and younger sold.
a + b = 5
8a + 5b = $31.00
Now that we have two variables and two equations, we can use substitution to solve.
a + b = 5 → a = 5 – b
Substitute: 8(5 – b) + 5b = $31.00
40 – 8b + 5b = 31 → –3b = –9 → b = 3 people 11 years old or younger
Thus, 5 – 3 = 2 people are 12 years old or older.
The group of five girls managed to secure five seats in a row on the crowded hayride. In how many orders can the five girls sit if they all sit in a row?
Any of the five girls could sit in the first seat; any of the remaining four girls could sit in the second seat, and so on. Thus, there are (5)(4)(3)(2)(1) = 120 orders the five girls could sit in.
In an attempt to minimize the lines this year, the hayride operators decided to run 3 flatbeds at a time. Initially, they use 20 bales of hay on each flatbed. They find that at the end of each day they need to replace half of the hay on each of the flatbeds to be ready for the next day’s business. How many total bales of hay will the hayride operators have used on the 3 flatbeds after they replenish the hay at the end of the fifth day of business?
On the first day, they start with 20(3) = 60 bales of hay. At the end of each business day, they replenish half of the hay (which would be 10 bales per flatbed, or 30 bales). After replenishing at the end of the first day, they have used a total of 60 + 30 = 90 bales. After replenishing at the end of the second day, they have used a total of 90 + 30 = 120 bales. Continuing this process, we see that at the end of the fifth day, they have used 60 + 30 + 30 + 30 + 30 + 30 = 60 + 150 = 210 bales of hay.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
In preparation for Thanksgiving dinner, Mrs. Cobble decided to get a new tablecloth for the dining room table. Her rectangular table measures 3.5 feet by 6 feet, and she would like the tablecloth to hang exactly 9 inches over each side. What is the area, in square feet, of the ideal tablecloth for Mrs. Cobble?
If Mrs. Cobble would like the tablecloth to hang exactly 9 inches over each side of the rectangular table, then she will need to add 18 inches to both the width and the length of the table’s dimensions. The table is 3.5 feet by 6 feet, and we are asked to give the answer in square feet. Therefore, let’s change 18 inches into its equivalent in feet. We can calculate that 18 inches is 18 ÷ 12 = 1.5 feet. The dimensions of the tablecloth will then be 3.5 + 1.5 = 5 feet by 6 + 1.5 = 7.5 feet. This rectangle has an area of 5 × 7.5 = 37.5 square feet.
Now Mrs. Cobble will turn her attention to the actual meal. According to an article she read, she should purchase 1¼ pounds of turkey per person. She figures this is a good amount for an adult, but a child would need only 2/3 of this amount. There will be 8 adults and 6 children at Thanksgiving dinner. According to Mrs. Cobble’s logic, how many pounds of turkey should she purchase?
We know that an adult will require 1¼ pounds of turkey. (We can use 1¼ or 5/4 or 1.25 as we proceed.) We are told that each child is only going to get 2/3 of this amount of turkey. So, we need to calculate 2/3 of 5/4. To do this, we find the product of the two fractions and see that each child will require 10/12 = 5/6 of a pound of turkey. Now we have eight adults each requiring 5/4 pounds and six children each requiring 5/6 of a pound. Mrs. Cobble will then need to buy a turkey that is 8(5/4) + 6(5/6) = 10 + 5 = 15 pounds.
When Thanksgiving has finally arrived, the whole family sits down to a Thanksgiving dinner that includes turkey, mashed potatoes, dressing, cranberry sauce and gravy. Uncle Bob’s favorite part of the meal, though, is the olive tray. Mrs. Cobble put the same number of green olives and black olives on the tray, and Uncle Bob was the first person to select from the tray. After Uncle Bob took eight green olives, the ratio of green olives to black olives was 3:5. How many total olives were on the tray before Uncle Bob took his olives?
We know that there used to be the same number of green olives and black olives, but after Uncle bob took 8 green olives, the ratio is then 3:5. This means that in every group of 8 olives, 3 of them are green and 5 of them are black. To bring this group of 8 olives back to “even,” Uncle Bob would need to put back 2 of his green olives, which would give us a total of 10 olives (5 green; 5 black). If it takes 2 olives for Uncle Bob to complete every uneven group of 8 olives, he could fix 4 groups of uneven olives with his 8 olives. Each of these 4 groups would then be 10 olives big, for a total of 40 olives.
Another way of looking at this...
The ratio of 3 green olives to 5 black olives leads us to believe that the number of olives now is a multiple of 8. So, there are now 8x olives. The ratio before Uncle Bob took his olives was 5 green olives to 5 black olives, or a total of 10x olives. There is a difference of 10x – 8x = 2x olives. So, the 8 olives he took is our 2x. We see 2x = 8 results in x = 4. So, there are now 8(4) = 32 olives, and there were originally 10(4) = 40 olives.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Melrose’s hot air balloon is a sphere of diameter 100 feet. How many cubic feet of air will Melrose’s hot air balloon hold when completely filled? Express your answer as a decimal to the nearest tenth.
The formula for the volume of a sphere is (4/3) × π × r3. Since we know the diameter of the balloon is 100 feet and the radius r is half the diameter, it follows that the radius is 50 feet. Substituting, we see that, completely filled, Melrose’s balloon will hold (4/3) × π × (50)3 = 523,598.8 ft3 of air.
Coco takes a hot air balloon trip over part of the Alleghany Mountains with her dad. They travel at an average speed of 40 mi/h for 120 miles before they land to take a 45-minute lunch break. After lunch, they fly 175 miles at an average speed of 35 mi/h. Including the lunch break, how many hours was their entire trip? Express your answer as a decimal to the nearest hundredth.
To solve this problem, we will be using the equation distance = rate × time. We’ll let T1 be the number of hours it took them to travel 120 miles at an average speed of 40 mi/h. So, we have 120 = 40T1 → T1 = 120/40 = 3 hours. We’ll let T2 be the number of hours it took them to travel 175 miles at an average speed of 35 mi/h. So, we have 175 = 35T2 → T2 = 5 hours. The lunch break was 45 minutes = 45/60 = 0.75 hours. Therefore, the entire trip, including the lunch break, took 3 + 0.75 + 5 = 8.75 hours.
Rhonda found that 10 super-sized helium balloons provide just enough lift to carry her 3-pound toy. At that same rate, how many super-sized balloons would be required to carry Rhonda’s 81-pound Alaskan malamute?
If 10 balloons lift exactly 3 pounds, then each balloon lifts the equivalent of 3/10 pound. Dividing 81 pounds by 3/10 pound, we see that the number of balloons required to lift her dog is 81/(3/10) = 81 × 10/3 = 810/3 = 270 balloons.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
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