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In 1998, Nkem Chukwu gave birth to the first set of surviving octuplets. The eight babies weighed from 11.3 ounces to 28.6 ounces at birth. Leonard Weisman, the Texas Children’s Hospital’s chief of neonatal services, said it was imperative they gain lots of weight before they go home. “For instance, the 500-gram babies are going to have to quadruple their weight [to reach a healthy level],” he said. How many ounces will the heaviest baby have to gain to reach a healthy level? Express your answer as a decimal to the nearest tenth.
There are 454 grams in a pound. Likewise, there are 16 ounces in one pound. Hence, one ounce is 454/16 = 28.375 grams. A healthy level is 4 × 500 = 2000 grams, or 2000/28.375 ≈ 70.5 ounces. The heaviest of the babies weighed 28.6 ounces at birth, and will have to gain 70.5 – 28.6 = 41.9 ounces, an increase of almost 150% its weight at birth.
The birth of octuplets is no small matter, but Mrs. Fyodor Vassilyev of Shuya, Russia (1707-1782) gave birth to 16 sets of twins, in addition to 7 sets of triplets, and 4 sets of quadruplets. She gave birth to a total of 69 children. What is the probability that one of her 69 children, selected at random, was not a twin, triplet or quadruplet?
The number of children who were twins, triplets or quadruplets is (16 × 2) + (7 × 3) + (4 × 4) = 69. She had no children born by themselves! So, the probability is zero (0).
The average length of pregnancy is 39 weeks for a single gestation; 35 weeks for twins; 33 weeks for triplets; and 29 weeks for quadruplets. If this trend in data continues, how long might you expect the length of pregnancy to be for octuplets to the nearest week?
It’s difficult to estimate these things accurately, so this is mostly an academic exercise! From a single gestation to the length of pregnancy for quadruplets, the decrease is 10 weeks, which is an average decrease of 10/3 = 3.3 weeks per additional child. So, it might be reasonable to assume that the length of pregnancy for octuplets is 39 – 7(3.3) ≈ 16 weeks.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
The third Monday of January is the holiday commemorating the birthday of Martin Luther King, Jr. The first time Martin Luther King Jr. Day was celebrated was in 1986. Try to solve these problems without using a calendar!
How many times has Martin Luther King Jr. Day been celebrated?
Martin Luther King Jr. Day has been celebrated 2024 – 1986 = 38 times, plus 1 (because we are including the Martin Luther King Jr. Day in 1986). So, Martin Luther King Jr. Day has been celebrated 39 times.
How many times since 1986 has Martin Luther King Jr. Day actually been celebrated on his birthday, January 15th?
Martin Luther King Jr. Day has been celebrated on his birthday, January 15th, 6 times (in 1990, 1996, 2001, 2007, 2018, and 2024). There is a range of dates in which the third Monday of January can fall – any day on or between the 15th and the 21st of January (see solution to next problem). If Martin Luther King Jr. Day is on January 15th this year (2024), then we can find that it was on the 16th last year, the 17th the year before that, the 18th the year before, and so on. You can continue in this way to solve this problem, but remember that you must account for leap years!
What is the probability Martin Luther King Jr.’s actual birth date falls on the third Monday in January?
Each day of the week must occur twice before a third Monday can occur. Therefore, 15 days must occur to arrive at a third Monday. The earliest date in January the nation can celebrate Martin Luther King Jr.’s birthday is January 15th. If January 1st occurs on a Tuesday, the third Monday will occur two weeks and six days later, or 20 days later. The latest date in January the nation can celebrate Martin Luther King Jr’s birthday is January 21st. The third Monday can fall on any of the seven dates January 15th through 21st. His actual birth date is on one of these days, January 15th. The probability that Martin Luther King Jr.’s actual birth date falls on the third Monday in January is 1/7.
Does a decade exist in which Martin Luther King Jr. Day will not be celebrated at least once on his birthday, January 15th? On which days could Martin Luther King Jr. Day be celebrated throughout this decade?
There is a pattern for the date of Martin Luther King Jr. Day. If we didn’t need to worry about leap years, the pattern would look like this: 21, 20, 19, 18, 17, 16, 15, 21, 20, 19, 18, 17, 16, 15, 21, 20, and so on. Unfortunately, leap years do play a role, though, so every fourth year actually moves back two numbers instead of one to account for the leap day. So, for example, the pattern of dates for successive Martin Luther King Jr. Days might be: 21, 20, 19, 17, 16, 15, 21, 19, 18, 17, 16, 21, 20, 19, 18, 16, 15, 21, 20, and so on. Hence, the following string of 10 successive holidays could occur: 21, 19, 18, 17, 16, 21, 20, 19, 18, 16, which is a string of 10 days without a 15 among them. Hence, it could happen that there will be a decade without a Martin Luther King Jr. Day falling on King’s actual birthday.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Cara was born on January 1, 2010, and her mother, Sydney, was born on January 1, 1982. In what year will Sydney’s age be twice Cara’s age?
If we let Cara’s age be C, then Sydney’s age is S = C + 28, since Cara was born 28 years after Sydney. We are interested in determining when Sydney’s age is twice Cara’s age, in other words, when S = 2C. Substituting, we have C + 28 = 2C. Solving for C we get C = 28. Therefore, Sydney’s age will be twice Cara’s age in the year 2010 + 28 = 2038.
In 2018, Cara’s brother, Nile, celebrated a birthday on January 4th. Cara’s age at that time was 4/5 Nile’s age at that time. How old was Nile’s mom, Sydney, when he was born?
Cara’s age in 2018 was 2018 – 2010 = 8 years old. We are told that Cara’s age in 2018 was 4/5 Nile’s age in 2018, N. That means (4/5)N = 8. Solving for N, we see that Nile’s age in 2018 was N = (5/4)8 = 10 years old. Sydney’s age in 2018, S, was 2018 – 1982 = 36 years. Therefore, ten years prior, when Nile was born, Sydney’s age was 36 – 10 = 26 years old.
The sum of the ages of Cara, Nile and Sydney each year forms an arithmetic progression. The sum of their ages in 2018 was 54. In what year will the sum of their ages be 78?
From the previous problem, we know that Cara’s, Nile’s and Sydney’s ages in 2018 were 8, 10 and 36, respectively. The sum of these ages is 8 + 10 + 36 = 54. The following year, the sum of their ages was 9 + 11 + 37 = 57. The following year, the sum of their ages was 10 + 12 + 38 = 60. Notice that the common difference in the arithmetic progression is +3, since each year the ages of Cara, Nile and Sydney each increase by 1. To determine in how many years the sum of the ages will be 78, we can find how many times 3 is added to 54 to get to 78. In other words, we can solve the equation 54 + 3x = 78 → 3x = 24 → x = 8. So, the sum of the ages of Cara, Nile and Sydney will be 78 in the year 2018 + 8 = 2026.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
The Sprint Round of the MATHCOUNTS Chapter Competition has 30 questions and students are given 40 minutes to complete the round. Though it isn’t expected that most students will finish all 30 questions, what is the average time a student can spend on each of the 30 questions, in minutes:seconds per question?
The 40-minute time limit is equivalent to 40 × 60 = 2400 seconds. Dividing this time equally among the 30 questions yields 2400 ÷ 30 = 80 seconds per question. This is one minute (60 seconds) and 20 seconds, or 1:20.
Some chapters will hold a Countdown Round for the highest-scoring 25% of the students at the competition or the top 10 students at the competition, whichever is fewer students. What is the greatest number of students at a competition for which “25% of the students (to the nearest whole number)” is fewer students than “the top 10 students?”
We can first figure out when 25% of the students is equal to 10 students. Let x be the number of students in the competition. Then we are solving 0.25x = 10 or (1/4)x = 10. Multiplying both sides by 4 (or dividing both sides by 0.25) tells us x = 40. Now we know that if there are 40 students at the competition, taking the top 25% or the top 10 will both result in 10 students participating in the Countdown Round. It’s true that 0.25(39) ˂ 10, but 0.25(39) = 9.75, which is still 10 students when rounded to the nearest whole number. We can also see that 0.25(38) = 9.5, which is again 10 when rounded to the nearest whole number. It’s not until our total number of students reaches 37 that the number of students (9, rounded to the nearest whole number) when taking the top 25% is fewer students than if we take the top 10 students.
MATHCOUNTS competitions are very different from tests students take in class. For a MATHCOUNTS competition, a score of 23 out of 46 (or 50%) is absolutely fantastic! The Target Round of a MATHCOUNTS competition has four pairs of problems. If we’re told that a student answered exactly half of the Target Round questions correctly, and answered one question in each of the pairs of questions correctly, how many different combinations of questions could she have answered correctly? (One combination is questions #1, 3, 5 and 7.)
We could try to list them all out or use the Counting Principle. We know there are four pairs of questions, she answered four questions correctly, and she answered one question correctly in each pair of questions. That means that in the first pair of questions, there are two options. She either got #1 or #2 correct. When considering the second pair of questions, she either got #3 or #4 correct, and what she did on the first pair has no bearing on the second pair. So, there are 2 options for the first pair and 2 options for the second pair, which makes a total of 2 × 2 = 4 combinations so far (#1 and #3 correct, #1 and #4 correct, #2 and #3 correct, #2 and #4 correct). Now the third pair of questions has 2 options, again neither of which is dependent on what happened with the first two pairs of questions. So now we’re up to 2 × 2 × 2 = 8 combinations. You can see that we can take the four combinations listed previously, and add #5 to each of them and then add #6 to each of them, thus doubling the number of combinations. Finally, we have our fourth pair of questions, which gives us another two options. We now have a total of 2 × 2 × 2 × 2 = 16 possible combinations of questions our competitor could have answered correctly if she answered exactly four questions correctly with one correct answer in each pair of questions.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Hydraulic engineering is a subfield of civil engineering that centers around the transport and management of water resources. Hydraulic engineers design things like channels, canals, dams and levees. They must consider many factors before beginning a project, including the collection, storage, flow, measurement and use of the water, in order to ensure its control, safety and cleanliness.
In a canal, its flow rate can be defined as the amount of water that is passing through a specific cross-section of the canal every second. The flow rate is important to monitor and manage. A flow rate that is too high can cause flooding and damage to the canal itself, while a flow rate that is too low can result in unsanitary conditions.
2.1 Marcos is working on a canal that has been designed to transport water across a field. He locates and measures the below cross-section of the canal in order to determine its flow rate. What is the cross-sectional area of the canal? Express your answer as a decimal to the nearest hundredth.
2.2 A velocimeter is a tool that uses laser beams to estimate the speed of water. Marcos stands on the edge of the canal with the velocimeter and points it at the water just in front of his feet. He finds that the water is moving at 1.5 ft/s. If flow rate is equal to the cross-sectional area of the canal multiplied by the speed of the water, what is the flow rate of the canal in cubic feet per second (cfs)? Express your answer to the nearest whole number.
2.3 Nadine, Marcos’s colleague, uses the velocimeter to take a second measurement, this time pointing it towards the center of the canal. Nadine’s reading indicates that the speed of the water is 2 ft/s. Because she and Marcos got different measurements, Nadine suggests using a better method for estimating flow rate called Manning’s Equation. Manning’s Equation for this canal can be simplified as:
Flow rate = 1.571 × A × R 2/3
where A is the cross-sectional area of the canal and R is the hydraulic radius, a ratio that indicates how efficiently the water is flowing. If the hydraulic radius for this canal is 2.29 ft, what is the estimated flow rate of the canal using Manning’s Equation? Express your answer as a decimal to the nearest tenth.
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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In 1998, Nkem Chukwu gave birth to the first set of surviving octuplets. The eight babies weighed from 11.3 ounces to 28.6 ounces at birth. Leonard Weisman, the Texas Children’s Hospital’s chief of neonatal services, said it was imperative they gain lots of weight before they go home. “For instance, the 500-gram babies are going to have to quadruple their weight [to reach a healthy level],” he said. How many ounces will the heaviest baby have to gain to reach a healthy level? Express your answer as a decimal to the nearest tenth.
There are 454 grams in a pound. Likewise, there are 16 ounces in one pound. Hence, one ounce is 454/16 = 28.375 grams. A healthy level is 4 × 500 = 2000 grams, or 2000/28.375 ≈ 70.5 ounces. The heaviest of the babies weighed 28.6 ounces at birth, and will have to gain 70.5 – 28.6 = 41.9 ounces, an increase of almost 150% its weight at birth.
The birth of octuplets is no small matter, but Mrs. Fyodor Vassilyev of Shuya, Russia (1707-1782) gave birth to 16 sets of twins, in addition to 7 sets of triplets, and 4 sets of quadruplets. She gave birth to a total of 69 children. What is the probability that one of her 69 children, selected at random, was not a twin, triplet or quadruplet?
The number of children who were twins, triplets or quadruplets is (16 × 2) + (7 × 3) + (4 × 4) = 69. She had no children born by themselves! So, the probability is zero (0).
The average length of pregnancy is 39 weeks for a single gestation; 35 weeks for twins; 33 weeks for triplets; and 29 weeks for quadruplets. If this trend in data continues, how long might you expect the length of pregnancy to be for octuplets to the nearest week?
It’s difficult to estimate these things accurately, so this is mostly an academic exercise! From a single gestation to the length of pregnancy for quadruplets, the decrease is 10 weeks, which is an average decrease of 10/3 = 3.3 weeks per additional child. So, it might be reasonable to assume that the length of pregnancy for octuplets is 39 – 7(3.3) ≈ 16 weeks.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
The third Monday of January is the holiday commemorating the birthday of Martin Luther King, Jr. The first time Martin Luther King Jr. Day was celebrated was in 1986. Try to solve these problems without using a calendar!
How many times has Martin Luther King Jr. Day been celebrated?
Martin Luther King Jr. Day has been celebrated 2024 – 1986 = 38 times, plus 1 (because we are including the Martin Luther King Jr. Day in 1986). So, Martin Luther King Jr. Day has been celebrated 39 times.
How many times since 1986 has Martin Luther King Jr. Day actually been celebrated on his birthday, January 15th?
Martin Luther King Jr. Day has been celebrated on his birthday, January 15th, 6 times (in 1990, 1996, 2001, 2007, 2018, and 2024). There is a range of dates in which the third Monday of January can fall – any day on or between the 15th and the 21st of January (see solution to next problem). If Martin Luther King Jr. Day is on January 15th this year (2024), then we can find that it was on the 16th last year, the 17th the year before that, the 18th the year before, and so on. You can continue in this way to solve this problem, but remember that you must account for leap years!
What is the probability Martin Luther King Jr.’s actual birth date falls on the third Monday in January?
Each day of the week must occur twice before a third Monday can occur. Therefore, 15 days must occur to arrive at a third Monday. The earliest date in January the nation can celebrate Martin Luther King Jr.’s birthday is January 15th. If January 1st occurs on a Tuesday, the third Monday will occur two weeks and six days later, or 20 days later. The latest date in January the nation can celebrate Martin Luther King Jr’s birthday is January 21st. The third Monday can fall on any of the seven dates January 15th through 21st. His actual birth date is on one of these days, January 15th. The probability that Martin Luther King Jr.’s actual birth date falls on the third Monday in January is 1/7.
Does a decade exist in which Martin Luther King Jr. Day will not be celebrated at least once on his birthday, January 15th? On which days could Martin Luther King Jr. Day be celebrated throughout this decade?
There is a pattern for the date of Martin Luther King Jr. Day. If we didn’t need to worry about leap years, the pattern would look like this: 21, 20, 19, 18, 17, 16, 15, 21, 20, 19, 18, 17, 16, 15, 21, 20, and so on. Unfortunately, leap years do play a role, though, so every fourth year actually moves back two numbers instead of one to account for the leap day. So, for example, the pattern of dates for successive Martin Luther King Jr. Days might be: 21, 20, 19, 17, 16, 15, 21, 19, 18, 17, 16, 21, 20, 19, 18, 16, 15, 21, 20, and so on. Hence, the following string of 10 successive holidays could occur: 21, 19, 18, 17, 16, 21, 20, 19, 18, 16, which is a string of 10 days without a 15 among them. Hence, it could happen that there will be a decade without a Martin Luther King Jr. Day falling on King’s actual birthday.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
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Cara was born on January 1, 2010, and her mother, Sydney, was born on January 1, 1982. In what year will Sydney’s age be twice Cara’s age?
If we let Cara’s age be C, then Sydney’s age is S = C + 28, since Cara was born 28 years after Sydney. We are interested in determining when Sydney’s age is twice Cara’s age, in other words, when S = 2C. Substituting, we have C + 28 = 2C. Solving for C we get C = 28. Therefore, Sydney’s age will be twice Cara’s age in the year 2010 + 28 = 2038.
In 2018, Cara’s brother, Nile, celebrated a birthday on January 4th. Cara’s age at that time was 4/5 Nile’s age at that time. How old was Nile’s mom, Sydney, when he was born?
Cara’s age in 2018 was 2018 – 2010 = 8 years old. We are told that Cara’s age in 2018 was 4/5 Nile’s age in 2018, N. That means (4/5)N = 8. Solving for N, we see that Nile’s age in 2018 was N = (5/4)8 = 10 years old. Sydney’s age in 2018, S, was 2018 – 1982 = 36 years. Therefore, ten years prior, when Nile was born, Sydney’s age was 36 – 10 = 26 years old.
The sum of the ages of Cara, Nile and Sydney each year forms an arithmetic progression. The sum of their ages in 2018 was 54. In what year will the sum of their ages be 78?
From the previous problem, we know that Cara’s, Nile’s and Sydney’s ages in 2018 were 8, 10 and 36, respectively. The sum of these ages is 8 + 10 + 36 = 54. The following year, the sum of their ages was 9 + 11 + 37 = 57. The following year, the sum of their ages was 10 + 12 + 38 = 60. Notice that the common difference in the arithmetic progression is +3, since each year the ages of Cara, Nile and Sydney each increase by 1. To determine in how many years the sum of the ages will be 78, we can find how many times 3 is added to 54 to get to 78. In other words, we can solve the equation 54 + 3x = 78 → 3x = 24 → x = 8. So, the sum of the ages of Cara, Nile and Sydney will be 78 in the year 2018 + 8 = 2026.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
The Sprint Round of the MATHCOUNTS Chapter Competition has 30 questions and students are given 40 minutes to complete the round. Though it isn’t expected that most students will finish all 30 questions, what is the average time a student can spend on each of the 30 questions, in minutes:seconds per question?
The 40-minute time limit is equivalent to 40 × 60 = 2400 seconds. Dividing this time equally among the 30 questions yields 2400 ÷ 30 = 80 seconds per question. This is one minute (60 seconds) and 20 seconds, or 1:20.
Some chapters will hold a Countdown Round for the highest-scoring 25% of the students at the competition or the top 10 students at the competition, whichever is fewer students. What is the greatest number of students at a competition for which “25% of the students (to the nearest whole number)” is fewer students than “the top 10 students?”
We can first figure out when 25% of the students is equal to 10 students. Let x be the number of students in the competition. Then we are solving 0.25x = 10 or (1/4)x = 10. Multiplying both sides by 4 (or dividing both sides by 0.25) tells us x = 40. Now we know that if there are 40 students at the competition, taking the top 25% or the top 10 will both result in 10 students participating in the Countdown Round. It’s true that 0.25(39) ˂ 10, but 0.25(39) = 9.75, which is still 10 students when rounded to the nearest whole number. We can also see that 0.25(38) = 9.5, which is again 10 when rounded to the nearest whole number. It’s not until our total number of students reaches 37 that the number of students (9, rounded to the nearest whole number) when taking the top 25% is fewer students than if we take the top 10 students.
MATHCOUNTS competitions are very different from tests students take in class. For a MATHCOUNTS competition, a score of 23 out of 46 (or 50%) is absolutely fantastic! The Target Round of a MATHCOUNTS competition has four pairs of problems. If we’re told that a student answered exactly half of the Target Round questions correctly, and answered one question in each of the pairs of questions correctly, how many different combinations of questions could she have answered correctly? (One combination is questions #1, 3, 5 and 7.)
We could try to list them all out or use the Counting Principle. We know there are four pairs of questions, she answered four questions correctly, and she answered one question correctly in each pair of questions. That means that in the first pair of questions, there are two options. She either got #1 or #2 correct. When considering the second pair of questions, she either got #3 or #4 correct, and what she did on the first pair has no bearing on the second pair. So, there are 2 options for the first pair and 2 options for the second pair, which makes a total of 2 × 2 = 4 combinations so far (#1 and #3 correct, #1 and #4 correct, #2 and #3 correct, #2 and #4 correct). Now the third pair of questions has 2 options, again neither of which is dependent on what happened with the first two pairs of questions. So now we’re up to 2 × 2 × 2 = 8 combinations. You can see that we can take the four combinations listed previously, and add #5 to each of them and then add #6 to each of them, thus doubling the number of combinations. Finally, we have our fourth pair of questions, which gives us another two options. We now have a total of 2 × 2 × 2 × 2 = 16 possible combinations of questions our competitor could have answered correctly if she answered exactly four questions correctly with one correct answer in each pair of questions.
♦ Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Hydraulic engineering is a subfield of civil engineering that centers around the transport and management of water resources. Hydraulic engineers design things like channels, canals, dams and levees. They must consider many factors before beginning a project, including the collection, storage, flow, measurement and use of the water, in order to ensure its control, safety and cleanliness.
In a canal, its flow rate can be defined as the amount of water that is passing through a specific cross-section of the canal every second. The flow rate is important to monitor and manage. A flow rate that is too high can cause flooding and damage to the canal itself, while a flow rate that is too low can result in unsanitary conditions.
2.1 Marcos is working on a canal that has been designed to transport water across a field. He locates and measures the below cross-section of the canal in order to determine its flow rate. What is the cross-sectional area of the canal? Express your answer as a decimal to the nearest hundredth.
2.2 A velocimeter is a tool that uses laser beams to estimate the speed of water. Marcos stands on the edge of the canal with the velocimeter and points it at the water just in front of his feet. He finds that the water is moving at 1.5 ft/s. If flow rate is equal to the cross-sectional area of the canal multiplied by the speed of the water, what is the flow rate of the canal in cubic feet per second (cfs)? Express your answer to the nearest whole number.
2.3 Nadine, Marcos’s colleague, uses the velocimeter to take a second measurement, this time pointing it towards the center of the canal. Nadine’s reading indicates that the speed of the water is 2 ft/s. Because she and Marcos got different measurements, Nadine suggests using a better method for estimating flow rate called Manning’s Equation. Manning’s Equation for this canal can be simplified as:
Flow rate = 1.571 × A × R 2/3
where A is the cross-sectional area of the canal and R is the hydraulic radius, a ratio that indicates how efficiently the water is flowing. If the hydraulic radius for this canal is 2.29 ft, what is the estimated flow rate of the canal using Manning’s Equation? Express your answer as a decimal to the nearest tenth.
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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