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Here is a table showing the number of tiles on which each letter appears in a standard English Scrabble game. Note that the blank tiles can be used as any letter a player wants.
Jennifer and Pete are going to play Scrabble with their brand new Scrabble game. They have decided that in order to determine who will go first, they will each randomly select one letter, without replacement, from the bag containing all of the letter tiles except the 2 blank tiles. The person who selects a letter closest to A goes first (if they both draw the same letter they will draw again). Jennifer draws first and selects an E. What is the probability that Pete’s selection will result in Pete going first (without having to redraw)?
There are 98 tiles that have letters printed on them, so after Jennifer selects her letter, Pete is left with 97 tiles to draw from. In order for Pete to go first, he must select an A, B, C or D. There are 9 + 2 + 2 + 4 = 17 such tiles. Thus, the probability that Pete will go first is 17/97.
After determining who goes first, Jennifer and Pete each put their letter back in the bag, added the 2 blank tiles, and then shook the bag to mix the letters. Now each person will draw seven letters (drawing one letter at a time) to be used on their first turn. Pete selects his 7 letters first and ends up with 2 Es, 1 R, 1 H, 1 S, 1 K and 1 O. Now it is Jennifer’s turn to select 7 letters. What is the probability that the first letter Jennifer selects will be a vowel (A, E, I, O or U)? Express your answer as a common fraction.
There are 100 tiles in the bag before anyone makes a selection, thus after Pete selects his 7 tiles, there are 93 tiles from which Jennifer gets to draw. Also, notice that there are 9 + 12 + 9 + 8 + 4 = 42 vowels when all of the tiles are in the bag, but since Pete drew 2 + 1 = 3 vowels, there are 42 – 3 = 39 vowels left in the bag when Jennifer goes to make her first draw. Thus, the probability of Jennifer drawing a vowel on her first draw is 39/93 = 13/31.
Jennifer ended up drawing 1 A, 2 Is, 1 S, 1 D, 1 U and 1 N. If the 2 I tiles are indistinguishable, in how many distinct orders can the 7 tiles be placed on Jennifer’s letter tray?
If each of the letter tiles were distinct, the number of orders would be 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040, since there would be seven distinct options for the first spot, leaving 6 distinct options for the second spot, and so on. However, in this case, there are 2 tiles that are the same. We can handle this issue by dividing 7! by 2, because half of the orders counted in 7! are due to the 2 identical Is swapping places. Thus, there are 7! ÷ 2 = 2520 distinct orders.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
According to “A History of Valentine’s Day Cards in America” by T.M. Wilson, in 1847 Esther Howland was the first to mass-produce Valentine’s Day cards. She made them out of lace, paint and expensive paper, and each one was individually written by a skilled calligrapher. The average card sold for $7.50 while others cost as much as $50. If 10 cents in 1847 would be equivalent to $3.74 today, how much would the average card and most expensive card have cost today?
Since we are given a ratio of 10 cents to $3.74, we can set up 2 more ratios to find what $7.50 and $50 would convert to. Just remember to convert 10 cents to $0.10 before beginning. An extended proportion would say that 0.10/3.74 = 7.50/x = 50/y. By cross-multiplying and solving for x and y, we would see people were spending what would be equivalent to $280.50 and $1,870 for us today!
Kelly decided to celebrate Valentine’s Day for an entire month. She started giving her Valentine 1 candy heart on Jan. 14th, 2 candy hearts on Jan. 15th, 4 candy hearts on Jan. 16th, and continued doubling the number of hearts each day until Feb. 14th. If 200 candy hearts come in a bag, how many bags of candy hearts would Kelly need just for Feb. 14th?
This is an exponential growth problem that shows how quickly an amount can grow when repeatedly doubled. The first day, she gave 1 candy. The second day, she gave 1 × 2 candies. The third day, she gave 1 × 2 × 2 candies. She will keep multiplying by 2 until she gets to the 32nd day. Therefore, the amount of candy she’ll need just for Feb. 14th is 1 × 231. This is 2,147,483,648 pieces of candy. Dividing this by 200 for each bag of candy means she’ll need 10,737,419 bags just to cover Valentine’s Day!
For Valentine’s Day, Kevin wanted to send Mary Beth 11 balloons, since that was her favorite number. In the store, plain-colored balloons cost $0.75, multi-colored balloons cost $1.30, and extra-large balloons cost $1.50. How many different combinations of 11 balloons can Kevin buy if he only has $12.00?
Making an orderly chart may be the best way to approach this problem. Start with buying as many of the extra-large balloons as possible, then methodically subtract an extra-large balloon, and so on. Though he can afford 8 extra-large balloons, he then could not afford 3 more to make the 11 balloons needed, So, the most extra-large balloons he can afford is 5 ($7.50), leaving him just enough to buy 6 plain-colored balloons ($4.50). Then, find possibilities with 4 extra-large balloons. Notice exchanging a multi-colored balloon for a plain-colored balloon raises the cost $0.55. This may help when determining possibilities and finding patterns. Eventually, you will find 24 possible combinations!
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
The Iditarod sled-dog race is run on a trail that was originally a mail-supply route. In 1925, part of the trail became a lifesaving highway for the children who lived in Nome.
The Iditarod is sometimes called “The Last Great Race on Earth.” Every year, it begins in Anchorage, Alaska during the first weekend in March. Each team of 12 to 16 dogs and a musher covers the distance to Nome in approximately 9 to 20 days.
There are two different routes used for the Iditarod. There is a northern route, which is run on even-numbered years, and a southern route, which is run on odd-numbered years. The exact measured distance of the race varies, but according to the official website the northern route is 975 miles long, and the southern route is 998 miles long.
Each of the eight letters in the word “IDITAROD” is written on a card. The cards are put into a bowl. The cards are drawn at random one at a time without replacement and placed from left to right in the order in which they are drawn. What is the probability the letters on the cards correctly spell “IDITAROD”? Express your answer as a common fraction.
The probability of drawing each letter in the correct order to spell IDITAROD is shown in the table. The probability of correctly spelling IDITAROD is the product of the probability of each letter being in the correct spot.
(2/8) × (2/7) × (1/6) × (1/5) × (1/4) × (1/3) × (1/2) × (1/1) = 1/10,080.
In 2002, Martin Buser, from Big Lake, Alaska, won the Iditarod in a time of 8 days, 22 hours, 46 minutes, and 2 seconds, setting a new record. In 2006, Jeff King, from Denali, Alaska, won the Iditarod in a time of 9 days, 11 hours, 11 minutes, and 36 seconds. Assume the length of the race is 975 miles. What is the positive difference between their mean speeds in miles per hour? Express your answer as a decimal to the nearest hundredth.
Convert each time to hours and divide the distance by the time to find the mean speed of each musher.
Buser: (8 × 24) + 22 + (46 ÷ 60) + (2 ÷ 3600) ≈ 214.7672 hours
(975 ÷ 214.7672) ≈ 4.5398 miles per hour
King: (9 × 24) + 11 + (11 ÷ 60) + (36 ÷ 3600) ≈ 227.1933 hours
(975 ÷ 227.1933) ≈ 4.2915 miles per hour
4.5398 – 4.2915 = 0.2483
The positive difference in their mean speeds is 0.25 mph, to the nearest hundredth.
The 2023 Iditarod followed the southern route. Assume the length of the race was 998 miles. If a musher and sled dog team had a mean speed of 5.3 miles per hour, what was their projected finish time? Express your answer in the form w days: x hours: y minutes: z seconds.
Divide the distance by the mean speed to find the number of hours. Convert the hours to days, hours, minutes and seconds.
(998 ÷ 5.3) ≈ 188.3019 hours
(188.3019 ÷ 24) ≈ 7.8459 days
(0.8459 × 24) = 20.3016 hours
(0.3016 × 60) = 18.096 minutes
(0.096 × 60) = 5.76 seconds
The projected finish time is 7 days: 20 hours: 18 minutes: 5.76 seconds.
To learn more about the Iditarod go to the following website: https://iditarod.com/about/
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Happy Surveyors Week! Surveyors play an essential role in construction, land development, the creation of maps, and other land-based projects. They measure property boundaries, natural land features, and precise locations. This week, we celebrate all the surveyors out there who are dedicated to accuracy and ensure that building projects move forward safely. Sponsors like the National Council of Examiners for Engineering and Surveying (NCEES), help us provide free math resources (like this Problem of the Week) and incredible program experiences for students!
In their day-to-day roles, surveyors use all kinds of math, including proportions and ratios, coordinate geometry, and measurement. To work your surveyor’s brain, below are some practice problems on these math topics!
For St. Patrick’s Day, Martin’s family is having a barbecue. Martin is making the lemonade. If the directions call for 2 tablespoons of mix per quart of water, how many tablespoons of mix are needed to make 1.5 gallons of lemonade?
Since there are 4 quarts in a gallon, it follows that 4 × 2 = 8 tablespoons of mix are required per gallon of water, and half that amount, or 8/2 = 4 tablespoons of mix are required per 0.5 gallon of water. Therefore, to make 1.5 gallons of lemonade, Martin needs to use 8 + 4 = 12 tablespoons of mix.
On some graph paper, graph the following segments:
y = x, for 0 ≤ x ≤ 2
y = 2x − 2, for 2 ≤ x ≤ 3
x = 3, for 4 ≤ y ≤ 6
y = −x + 9, for 2 ≤ x ≤ 3
y = 7, for 1 ≤ x ≤ 2
y = x + 6, for 0 ≤ x ≤ 1
Now reflect each of the segments over the y-axis. What popular shape have you drawn?
The first segment connects the points (0, 0) and (2, 2). The second segment connects the points (2, 2) and (3, 4). The third segment connects the points (3, 4) and (3, 6). The fourth segment connects the points (3, 6) and (2, 7). The fifth segment connects the points (2, 7) and (1, 7). Finally, the sixth segment connects the points (1, 7) and (0, 6). This should appear as half a heart. Once the reflection is done, you should have the shape of a heart with the y-axis running down the center.
A particular construction crew places orange barrels on both sides of a road that is under construction such that the centers of adjacent barrels on the same side of the road are 15 feet apart. If the crew does this for a 1.5 mile stretch of roadway, how many barrels will be placed on the two sides of the road in total?
There are 1.5 × 5280 = 7920 feet on each side of the road. We start with one barrel at the "zero" mark and then add 7920 ÷ 15 = 528 barrels on each side of the road, for a total of 528 + 1 = 529 barrels per side. So for both sides there are a total of 2 × 529 = 1058 barrels.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Thursday is Pi Day! For all of the following calculations, use 3.14 for pi.
In honor of Pi Day, Wilma’s class is having a pie party. Wilma brought in her favorite kind of pie – apple. When she cuts her slice of pie, it has a central angle of 60 degrees. If the diameter of the pie was 10 inches, what is the area of Wilma’s slice of pie? Express your answer as a decimal to the nearest tenth.
First, find the area of the whole pie.
A = πr2
A = (3.14)(10/2)2 = 78.5 square inches
Now, multiply by the fraction of the pie that Wilma took.
78.5(60/360) ≈ 13.1 square inches
A buggy has front wheels with a diameter of 12 inches and back wheels with a diameter of 18 inches. If Molly pushes the buggy for 300 yards, how many more revolutions does each of the front wheels make than each of the back wheels? Express your answer as a decimal to the nearest tenth.
First, find the circumference of each of the wheels.
C = πd
C = (3.14)(12) = 37.68 inches
C = (3.14)(18) = 56.52 inches
Now, convert the number of yards traveled by the buggy to inches.
300 yds (3 ft/yd) (12 in/ft) = 10,800 inches
Divide the distance traveled (in inches) by the distance traveled in one revolution of each wheel.
10,800/37.68 ≈ 286.624
10,800/56.52 ≈ 191.083
Finally, find the difference.
286.624 – 191.083 ≈ 95.5 revolutions, to the nearest tenth
Semi-circle A’s radius is twice as long as semi-circle B’s radius. The length of semi-circle B’s radius is 30 percent of the length of semi-circle C’s radius. None of the semi-circles overlap any of the others. If A’s radius is 10 cm, what is the sum of the areas of semi-circles A, B and C?
If semi-circle A’s radius is 10 cm, we know that B’s radius is 5 cm and that C’s radius is 16.66667 cm. Now, we can find the area of each semi-circle.
A = (πr2)/2
A = (3.14)(10)2/2 = 157
A = (3.14)(5)2/2 = 39.25
A = (3.14)(16.66667)2/2 = 436.111...
Now, find the sum.
(436.111…) + (157) + (39.25) = 632.36111...
NOTE: Your final answer may differ SLIGHTLY because of differences in rounding on this problem.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
If you took the DoD STEM Alumni Survey for MATHCOUNTS with your Competition Series coach, and you are 13 years old or older, please complete the webform below if you are interested in being interviewed! Our evaluators are looking for additional insight on your experience in MATHCOUNTS. Each interview will take approximately 15 minutes, and while it will take place via Zoom, you are not required to have your camera on. In order to participate, a signed parent/guardian permission form is required (provided once you have completed this webform and it is confirmed that additional interviewees are still needed).
Any student who participates in an interview will receive a $15 gift card!
The following information and problems were submitted by a MATHCOUNTS volunteer, Emily Koithan. Thank you, Emily!
Neuroscientists use a special type of brain scanning called functional magnetic resonance imaging, or fMRI, to answer questions about how our brains work. It can tell us what areas of the brain are active when we move, listen and think.
A researcher asks a study participant to tap their finger during an fMRI scan and observes activity in 10 different brain regions. If the activity in each region is compared to the activity in every other region, how many comparisons are made?
Let’s say the 10 brain regions are labeled with the numbers 1 through 10. We can methodically count the comparisons made by starting with region 1. Region 1 is separately compared to regions 2, 3, 4, 5, 6, 7, 8, 9 and 10, which is a total of 9 comparisons. Moving on to region 2, we see that the comparison between region 1 and region 2 has already taken place. Thus, region 2 still needs to be compared to regions 3, 4, 5, 6, 7, 8, 9 and 10, which is a total of 8 comparisons. At this point, region 3 has already been compared to regions 1 and 2, so region 3 still needs to be compared to regions 4, 5, 6, 7, 8, 9 and 10. This is a total of 7 comparisons. We can continue with this pattern to find that if the activity in each brain region is compared to the activity in every other brain region, there will be 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45 comparisons.
Alternatively, each of the 10 brain regions will be compared with 9 other brain regions (every region but itself). However, if we simply multiply 10 × 9, we will have double-counted each of the comparisons. For example, we will have counted when region 1 is compared to region 2 and when region 2 is compared to region 1. This is actually the same comparison, though, since each brain region is only sharing one comparison with each other region. Thus, we’ll need to account for this by dividing by 2, so there will be (10 × 9)/2 = 45 comparisons.
Alternatively, we can use the combinations formula, which says that the total number of combinations (in this case, comparisons) is n!/[r!(n – r)!], where n = the total number of brain regions and r = the number of brain regions to select for each comparison. Since there are 10 brain regions and 2 regions in a single comparison, using the combinations formula, we find that there will be 10!/[2!(10 – 2)!] = 10!/[2!8!] = (10 × 9)/(2 × 1) = 90/2 = 45 comparisons.
The researcher then plays music and looks at activity in the participant’s auditory cortex, one part of the brain active when you hear sounds. She plays the music louder, and the activity in the auditory cortex increases by 60%. Then, she plays the music softer, and the activity in the auditory cortex decreases by 14%. If x represents the original activity level, by what percent does the activity in the participant’s auditory cortex increase while the music is playing? Express your answer as a decimal to the nearest tenth.
Let the original activity in response to the music be given by x. Then, when the music is played louder, the increase in activity of 60% can be represented by 1.6x. When the music is lowered and the activity decreases by 14%, this can be represented by 0.14(1.6x). Thus, the activity at the end is 1.6x – 0.14(1.6x) = 1.6x – 0.224x = 1.376x. So, the activity is 1.376 times higher at the end than it was when the music originally started to play, which is an increase of 37.6%.
Researchers often study the brain using voxels, which are tiny three-dimensional cubes. The researcher finds that in one participant, the inferior parietal lobule (a region of the brain active when performing arithmetic) has a volume of 22 cubic centimeters. Suppose the inferior parietal lobule in that participant contains 2,750 voxels. How long is the side of one voxel in millimeters?
The volume of one voxel is given by 22 cm3 / 2750 = 22,000 mm3 / 2750 = 8 mm3. Since the volume of a cube is given by length × width × height, the side of each voxel is 2 millimeters long.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
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Here is a table showing the number of tiles on which each letter appears in a standard English Scrabble game. Note that the blank tiles can be used as any letter a player wants.
Jennifer and Pete are going to play Scrabble with their brand new Scrabble game. They have decided that in order to determine who will go first, they will each randomly select one letter, without replacement, from the bag containing all of the letter tiles except the 2 blank tiles. The person who selects a letter closest to A goes first (if they both draw the same letter they will draw again). Jennifer draws first and selects an E. What is the probability that Pete’s selection will result in Pete going first (without having to redraw)?
There are 98 tiles that have letters printed on them, so after Jennifer selects her letter, Pete is left with 97 tiles to draw from. In order for Pete to go first, he must select an A, B, C or D. There are 9 + 2 + 2 + 4 = 17 such tiles. Thus, the probability that Pete will go first is 17/97.
After determining who goes first, Jennifer and Pete each put their letter back in the bag, added the 2 blank tiles, and then shook the bag to mix the letters. Now each person will draw seven letters (drawing one letter at a time) to be used on their first turn. Pete selects his 7 letters first and ends up with 2 Es, 1 R, 1 H, 1 S, 1 K and 1 O. Now it is Jennifer’s turn to select 7 letters. What is the probability that the first letter Jennifer selects will be a vowel (A, E, I, O or U)? Express your answer as a common fraction.
There are 100 tiles in the bag before anyone makes a selection, thus after Pete selects his 7 tiles, there are 93 tiles from which Jennifer gets to draw. Also, notice that there are 9 + 12 + 9 + 8 + 4 = 42 vowels when all of the tiles are in the bag, but since Pete drew 2 + 1 = 3 vowels, there are 42 – 3 = 39 vowels left in the bag when Jennifer goes to make her first draw. Thus, the probability of Jennifer drawing a vowel on her first draw is 39/93 = 13/31.
Jennifer ended up drawing 1 A, 2 Is, 1 S, 1 D, 1 U and 1 N. If the 2 I tiles are indistinguishable, in how many distinct orders can the 7 tiles be placed on Jennifer’s letter tray?
If each of the letter tiles were distinct, the number of orders would be 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040, since there would be seven distinct options for the first spot, leaving 6 distinct options for the second spot, and so on. However, in this case, there are 2 tiles that are the same. We can handle this issue by dividing 7! by 2, because half of the orders counted in 7! are due to the 2 identical Is swapping places. Thus, there are 7! ÷ 2 = 2520 distinct orders.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
According to “A History of Valentine’s Day Cards in America” by T.M. Wilson, in 1847 Esther Howland was the first to mass-produce Valentine’s Day cards. She made them out of lace, paint and expensive paper, and each one was individually written by a skilled calligrapher. The average card sold for $7.50 while others cost as much as $50. If 10 cents in 1847 would be equivalent to $3.74 today, how much would the average card and most expensive card have cost today?
Since we are given a ratio of 10 cents to $3.74, we can set up 2 more ratios to find what $7.50 and $50 would convert to. Just remember to convert 10 cents to $0.10 before beginning. An extended proportion would say that 0.10/3.74 = 7.50/x = 50/y. By cross-multiplying and solving for x and y, we would see people were spending what would be equivalent to $280.50 and $1,870 for us today!
Kelly decided to celebrate Valentine’s Day for an entire month. She started giving her Valentine 1 candy heart on Jan. 14th, 2 candy hearts on Jan. 15th, 4 candy hearts on Jan. 16th, and continued doubling the number of hearts each day until Feb. 14th. If 200 candy hearts come in a bag, how many bags of candy hearts would Kelly need just for Feb. 14th?
This is an exponential growth problem that shows how quickly an amount can grow when repeatedly doubled. The first day, she gave 1 candy. The second day, she gave 1 × 2 candies. The third day, she gave 1 × 2 × 2 candies. She will keep multiplying by 2 until she gets to the 32nd day. Therefore, the amount of candy she’ll need just for Feb. 14th is 1 × 231. This is 2,147,483,648 pieces of candy. Dividing this by 200 for each bag of candy means she’ll need 10,737,419 bags just to cover Valentine’s Day!
For Valentine’s Day, Kevin wanted to send Mary Beth 11 balloons, since that was her favorite number. In the store, plain-colored balloons cost $0.75, multi-colored balloons cost $1.30, and extra-large balloons cost $1.50. How many different combinations of 11 balloons can Kevin buy if he only has $12.00?
Making an orderly chart may be the best way to approach this problem. Start with buying as many of the extra-large balloons as possible, then methodically subtract an extra-large balloon, and so on. Though he can afford 8 extra-large balloons, he then could not afford 3 more to make the 11 balloons needed, So, the most extra-large balloons he can afford is 5 ($7.50), leaving him just enough to buy 6 plain-colored balloons ($4.50). Then, find possibilities with 4 extra-large balloons. Notice exchanging a multi-colored balloon for a plain-colored balloon raises the cost $0.55. This may help when determining possibilities and finding patterns. Eventually, you will find 24 possible combinations!
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
The Iditarod sled-dog race is run on a trail that was originally a mail-supply route. In 1925, part of the trail became a lifesaving highway for the children who lived in Nome.
The Iditarod is sometimes called “The Last Great Race on Earth.” Every year, it begins in Anchorage, Alaska during the first weekend in March. Each team of 12 to 16 dogs and a musher covers the distance to Nome in approximately 9 to 20 days.
There are two different routes used for the Iditarod. There is a northern route, which is run on even-numbered years, and a southern route, which is run on odd-numbered years. The exact measured distance of the race varies, but according to the official website the northern route is 975 miles long, and the southern route is 998 miles long.
Each of the eight letters in the word “IDITAROD” is written on a card. The cards are put into a bowl. The cards are drawn at random one at a time without replacement and placed from left to right in the order in which they are drawn. What is the probability the letters on the cards correctly spell “IDITAROD”? Express your answer as a common fraction.
The probability of drawing each letter in the correct order to spell IDITAROD is shown in the table. The probability of correctly spelling IDITAROD is the product of the probability of each letter being in the correct spot.
(2/8) × (2/7) × (1/6) × (1/5) × (1/4) × (1/3) × (1/2) × (1/1) = 1/10,080.
In 2002, Martin Buser, from Big Lake, Alaska, won the Iditarod in a time of 8 days, 22 hours, 46 minutes, and 2 seconds, setting a new record. In 2006, Jeff King, from Denali, Alaska, won the Iditarod in a time of 9 days, 11 hours, 11 minutes, and 36 seconds. Assume the length of the race is 975 miles. What is the positive difference between their mean speeds in miles per hour? Express your answer as a decimal to the nearest hundredth.
Convert each time to hours and divide the distance by the time to find the mean speed of each musher.
Buser: (8 × 24) + 22 + (46 ÷ 60) + (2 ÷ 3600) ≈ 214.7672 hours
(975 ÷ 214.7672) ≈ 4.5398 miles per hour
King: (9 × 24) + 11 + (11 ÷ 60) + (36 ÷ 3600) ≈ 227.1933 hours
(975 ÷ 227.1933) ≈ 4.2915 miles per hour
4.5398 – 4.2915 = 0.2483
The positive difference in their mean speeds is 0.25 mph, to the nearest hundredth.
The 2023 Iditarod followed the southern route. Assume the length of the race was 998 miles. If a musher and sled dog team had a mean speed of 5.3 miles per hour, what was their projected finish time? Express your answer in the form w days: x hours: y minutes: z seconds.
Divide the distance by the mean speed to find the number of hours. Convert the hours to days, hours, minutes and seconds.
(998 ÷ 5.3) ≈ 188.3019 hours
(188.3019 ÷ 24) ≈ 7.8459 days
(0.8459 × 24) = 20.3016 hours
(0.3016 × 60) = 18.096 minutes
(0.096 × 60) = 5.76 seconds
The projected finish time is 7 days: 20 hours: 18 minutes: 5.76 seconds.
To learn more about the Iditarod go to the following website: https://iditarod.com/about/
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Happy Surveyors Week! Surveyors play an essential role in construction, land development, the creation of maps, and other land-based projects. They measure property boundaries, natural land features, and precise locations. This week, we celebrate all the surveyors out there who are dedicated to accuracy and ensure that building projects move forward safely. Sponsors like the National Council of Examiners for Engineering and Surveying (NCEES), help us provide free math resources (like this Problem of the Week) and incredible program experiences for students!
In their day-to-day roles, surveyors use all kinds of math, including proportions and ratios, coordinate geometry, and measurement. To work your surveyor’s brain, below are some practice problems on these math topics!
For St. Patrick’s Day, Martin’s family is having a barbecue. Martin is making the lemonade. If the directions call for 2 tablespoons of mix per quart of water, how many tablespoons of mix are needed to make 1.5 gallons of lemonade?
Since there are 4 quarts in a gallon, it follows that 4 × 2 = 8 tablespoons of mix are required per gallon of water, and half that amount, or 8/2 = 4 tablespoons of mix are required per 0.5 gallon of water. Therefore, to make 1.5 gallons of lemonade, Martin needs to use 8 + 4 = 12 tablespoons of mix.
On some graph paper, graph the following segments:
y = x, for 0 ≤ x ≤ 2
y = 2x − 2, for 2 ≤ x ≤ 3
x = 3, for 4 ≤ y ≤ 6
y = −x + 9, for 2 ≤ x ≤ 3
y = 7, for 1 ≤ x ≤ 2
y = x + 6, for 0 ≤ x ≤ 1
Now reflect each of the segments over the y-axis. What popular shape have you drawn?
The first segment connects the points (0, 0) and (2, 2). The second segment connects the points (2, 2) and (3, 4). The third segment connects the points (3, 4) and (3, 6). The fourth segment connects the points (3, 6) and (2, 7). The fifth segment connects the points (2, 7) and (1, 7). Finally, the sixth segment connects the points (1, 7) and (0, 6). This should appear as half a heart. Once the reflection is done, you should have the shape of a heart with the y-axis running down the center.
A particular construction crew places orange barrels on both sides of a road that is under construction such that the centers of adjacent barrels on the same side of the road are 15 feet apart. If the crew does this for a 1.5 mile stretch of roadway, how many barrels will be placed on the two sides of the road in total?
There are 1.5 × 5280 = 7920 feet on each side of the road. We start with one barrel at the "zero" mark and then add 7920 ÷ 15 = 528 barrels on each side of the road, for a total of 528 + 1 = 529 barrels per side. So for both sides there are a total of 2 × 529 = 1058 barrels.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Thursday is Pi Day! For all of the following calculations, use 3.14 for pi.
In honor of Pi Day, Wilma’s class is having a pie party. Wilma brought in her favorite kind of pie – apple. When she cuts her slice of pie, it has a central angle of 60 degrees. If the diameter of the pie was 10 inches, what is the area of Wilma’s slice of pie? Express your answer as a decimal to the nearest tenth.
First, find the area of the whole pie.
A = πr2
A = (3.14)(10/2)2 = 78.5 square inches
Now, multiply by the fraction of the pie that Wilma took.
78.5(60/360) ≈ 13.1 square inches
A buggy has front wheels with a diameter of 12 inches and back wheels with a diameter of 18 inches. If Molly pushes the buggy for 300 yards, how many more revolutions does each of the front wheels make than each of the back wheels? Express your answer as a decimal to the nearest tenth.
First, find the circumference of each of the wheels.
C = πd
C = (3.14)(12) = 37.68 inches
C = (3.14)(18) = 56.52 inches
Now, convert the number of yards traveled by the buggy to inches.
300 yds (3 ft/yd) (12 in/ft) = 10,800 inches
Divide the distance traveled (in inches) by the distance traveled in one revolution of each wheel.
10,800/37.68 ≈ 286.624
10,800/56.52 ≈ 191.083
Finally, find the difference.
286.624 – 191.083 ≈ 95.5 revolutions, to the nearest tenth
Semi-circle A’s radius is twice as long as semi-circle B’s radius. The length of semi-circle B’s radius is 30 percent of the length of semi-circle C’s radius. None of the semi-circles overlap any of the others. If A’s radius is 10 cm, what is the sum of the areas of semi-circles A, B and C?
If semi-circle A’s radius is 10 cm, we know that B’s radius is 5 cm and that C’s radius is 16.66667 cm. Now, we can find the area of each semi-circle.
A = (πr2)/2
A = (3.14)(10)2/2 = 157
A = (3.14)(5)2/2 = 39.25
A = (3.14)(16.66667)2/2 = 436.111...
Now, find the sum.
(436.111…) + (157) + (39.25) = 632.36111...
NOTE: Your final answer may differ SLIGHTLY because of differences in rounding on this problem.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
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Neuroscientists use a special type of brain scanning called functional magnetic resonance imaging, or fMRI, to answer questions about how our brains work. It can tell us what areas of the brain are active when we move, listen and think.
A researcher asks a study participant to tap their finger during an fMRI scan and observes activity in 10 different brain regions. If the activity in each region is compared to the activity in every other region, how many comparisons are made?
Let’s say the 10 brain regions are labeled with the numbers 1 through 10. We can methodically count the comparisons made by starting with region 1. Region 1 is separately compared to regions 2, 3, 4, 5, 6, 7, 8, 9 and 10, which is a total of 9 comparisons. Moving on to region 2, we see that the comparison between region 1 and region 2 has already taken place. Thus, region 2 still needs to be compared to regions 3, 4, 5, 6, 7, 8, 9 and 10, which is a total of 8 comparisons. At this point, region 3 has already been compared to regions 1 and 2, so region 3 still needs to be compared to regions 4, 5, 6, 7, 8, 9 and 10. This is a total of 7 comparisons. We can continue with this pattern to find that if the activity in each brain region is compared to the activity in every other brain region, there will be 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45 comparisons.
Alternatively, each of the 10 brain regions will be compared with 9 other brain regions (every region but itself). However, if we simply multiply 10 × 9, we will have double-counted each of the comparisons. For example, we will have counted when region 1 is compared to region 2 and when region 2 is compared to region 1. This is actually the same comparison, though, since each brain region is only sharing one comparison with each other region. Thus, we’ll need to account for this by dividing by 2, so there will be (10 × 9)/2 = 45 comparisons.
Alternatively, we can use the combinations formula, which says that the total number of combinations (in this case, comparisons) is n!/[r!(n – r)!], where n = the total number of brain regions and r = the number of brain regions to select for each comparison. Since there are 10 brain regions and 2 regions in a single comparison, using the combinations formula, we find that there will be 10!/[2!(10 – 2)!] = 10!/[2!8!] = (10 × 9)/(2 × 1) = 90/2 = 45 comparisons.
The researcher then plays music and looks at activity in the participant’s auditory cortex, one part of the brain active when you hear sounds. She plays the music louder, and the activity in the auditory cortex increases by 60%. Then, she plays the music softer, and the activity in the auditory cortex decreases by 14%. If x represents the original activity level, by what percent does the activity in the participant’s auditory cortex increase while the music is playing? Express your answer as a decimal to the nearest tenth.
Let the original activity in response to the music be given by x. Then, when the music is played louder, the increase in activity of 60% can be represented by 1.6x. When the music is lowered and the activity decreases by 14%, this can be represented by 0.14(1.6x). Thus, the activity at the end is 1.6x – 0.14(1.6x) = 1.6x – 0.224x = 1.376x. So, the activity is 1.376 times higher at the end than it was when the music originally started to play, which is an increase of 37.6%.
Researchers often study the brain using voxels, which are tiny three-dimensional cubes. The researcher finds that in one participant, the inferior parietal lobule (a region of the brain active when performing arithmetic) has a volume of 22 cubic centimeters. Suppose the inferior parietal lobule in that participant contains 2,750 voxels. How long is the side of one voxel in millimeters?
The volume of one voxel is given by 22 cm3 / 2750 = 22,000 mm3 / 2750 = 8 mm3. Since the volume of a cube is given by length × width × height, the side of each voxel is 2 millimeters long.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
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