Math Problem of the Day--ANSWERS
Math Problem of the Day 3/24/2011
On the first day of school, Mr. Vilani gave his third
grade students 5 new words to spell. On
each day of school after that, he gave the students 3 new words to spell. In the first 20 days of school, how many new
words had he given the students to spell?
From an old ACT exam.
ANSWER:
62 words. On the first day the students learned to spell five words, and then they learned 3 words a day for 19 days. 5 + 3*19 = 5+57 = 62 words
From an old ACT exam.
ANSWER:
62 words. On the first day the students learned to spell five words, and then they learned 3 words a day for 19 days. 5 + 3*19 = 5+57 = 62 words
Math Problem of the Day 3/23/2011
Sorry there have not been Math Problems of the Day lately, but I just
got caught up from spring break!! Without further ado....
Lisa checks the statistics for her favorite puzzle game on her iPod. She sees that she has won exactly 72% of the 2400 puzzles that she has played. What is the minimum number of puzzles that Lisa must win before she has won at least 73% of the total number of puzzles played?
Coffey, M. (Ed.). (2011). April "Calendar" Problems. MATHEMATICS TEACHER , 104 (8), 600-601.
ANSWER:
Lisa would have to win at least 89 consecutive games in order to win 73% of the total number of puzzles played.
89. Achieving a winning percentage of at least 73% after playing as few additional puzzle games as possible implies that Lisa wins every additional game she plays. She has already won 0.72 •2400 = 1728 games. Let w = additional number of games won. We have the following:
Lisa checks the statistics for her favorite puzzle game on her iPod. She sees that she has won exactly 72% of the 2400 puzzles that she has played. What is the minimum number of puzzles that Lisa must win before she has won at least 73% of the total number of puzzles played?
Coffey, M. (Ed.). (2011). April "Calendar" Problems. MATHEMATICS TEACHER , 104 (8), 600-601.
ANSWER:
Lisa would have to win at least 89 consecutive games in order to win 73% of the total number of puzzles played.
89. Achieving a winning percentage of at least 73% after playing as few additional puzzle games as possible implies that Lisa wins every additional game she plays. She has already won 0.72 •2400 = 1728 games. Let w = additional number of games won. We have the following:
[1728+x] / [2400+x] ≥ 0.73
1728 + w ≥ 0.73(2400 + w)
1728 + w ≥ 1752 + 0.73w
0.27w ≥ 24
w ≥ 88.89
Lisa must win at least 89 consecutive additional puzzle games.
1728 + w ≥ 0.73(2400 + w)
1728 + w ≥ 1752 + 0.73w
0.27w ≥ 24
w ≥ 88.89
Lisa must win at least 89 consecutive additional puzzle games.
Math Problem of the Day 3/11/2011
H. Constant is the hero of a particularly successful film -rates of the
Pythagorean, whose climactic scene involved Constant “walking the
planck.” On its premier weekend, the film grossed $31.4 million. If its
sequel is expected to surpass this amount by 15.9%, how much will the
sequel gross?
Coffey, M. (Ed.). (2011). March "Calendar" Problems. MATHEMATICS TEACHER , 104 (7), 520-521.
ANSWER
Approximately $36.4 million
If the sequel made 15.9% more than the first film, then the gross of the sequel is 115.9% of the gross from the first film.
31.4 (115.9%)=(31.4)(1.159)= 36.3926
Therefore, the sequel will make approximately $36.4 million.
Planck's constant, which relates a photon's energy and the frequency of its electromagnetic wave is used in quantum mechanics and is represented by h.
Coffey, M. (Ed.). (2011). March "Calendar" Problems. MATHEMATICS TEACHER , 104 (7), 520-521.
ANSWER
Approximately $36.4 million
If the sequel made 15.9% more than the first film, then the gross of the sequel is 115.9% of the gross from the first film.
31.4 (115.9%)=(31.4)(1.159)= 36.3926
Therefore, the sequel will make approximately $36.4 million.
Planck's constant, which relates a photon's energy and the frequency of its electromagnetic wave is used in quantum mechanics and is represented by h.
Math Problem of the Day 3/10/2011
Kaytlyn the Pi-rate
was famed for her collection of
gold and silver jewelry. Which would
hold more of her loot—a right cylindrical barrel with diameter d and height 2d or a chest in the shape of a rectangular prism with height d and
width and length of 2d?
Coffey, M. (Ed.). (2011). March "Calendar" Problems. MATHEMATICS TEACHER , 104 (7), 520-521.
ANSWER:
The CHEST will hold more than the BARREL.
Coffey, M. (Ed.). (2011). March "Calendar" Problems. MATHEMATICS TEACHER , 104 (7), 520-521.
ANSWER:
The CHEST will hold more than the BARREL.
Math Problem of the Day 3/9/2011
Captain Hook’s parrot is 4 times as tall as Peter Pan’s parrot. A
parrot owned by another Pi-rate on the Jolly Roger is 5/8 the height of
Hook’s bird. The total height of all these birds is 225cm. If d
represents the height of Peter’s parrot, find d.
From the March Issue of the Mathematics Teacher
Coffey, M. (Ed.). (2011). March "Calendar" Problems. MATHEMATICS TEACHER , 104 (7), 520-521.
ANSWER:
Peter's Parrot is 30 cm
From the March Issue of the Mathematics Teacher
Coffey, M. (Ed.). (2011). March "Calendar" Problems. MATHEMATICS TEACHER , 104 (7), 520-521.
ANSWER:
Peter's Parrot is 30 cm
Math Problem of the Day 3/8/2011
A data set of integers has 5 members. The median of the data set is 6.
The mode of the data set is 2. The mean of the data set is 5. Which
of the following statements must be true about the data set?
a) The sum of the five integers is 30
b) The sum of the five integers is 10
c) The median of the three largest integers is 6
d) The largest integer is 9
e) The largest integer is 8
This problem came from an old ACT Exam
ANSWER
Choice E is the Correct Response
a) The sum of the five integers is 30
b) The sum of the five integers is 10
c) The median of the three largest integers is 6
d) The largest integer is 9
e) The largest integer is 8
This problem came from an old ACT Exam
ANSWER
Choice E is the Correct Response
Math Problem of the Day 3/4/2011
A rectangular box is to be made from a rectangular piece of sheet metal
20 inches wide and 30 inches long by cutting a square piece of equal
size from each of the corners and bending up the sides so that they make
a 90 degree angle with the base. The base of the finished box will be
200 square inches. About how many inches high will the box be?
a) 5 inches
b) 10 inches
c) 15 inches
d) 20 inches
e) 30 inches
ANSWER
CHOICE A -- 5 inches
a) 5 inches
b) 10 inches
c) 15 inches
d) 20 inches
e) 30 inches
ANSWER
CHOICE A -- 5 inches
Math Problem of the Day 3/3/2011
In a large high school, some teachers teach only one subject and some
teachers teach more than one subject. Using the information given below
about math, science, and gym teachers in the school, how many teach
math only?
ANSWER:
There are only 6 teachers that only teach math.
- 10 teachers teach at least one class of gym
- 16 teachers teach at least one class of math
- 12 teachers teach at least one class of science
- 5 teachers teach both math and science but not gym
- 2 teachers teach both gym and science but not math
- 3 teachers teach gym only
- 1 teacher teaches math, gym, and science
ANSWER:
There are only 6 teachers that only teach math.
Math Problem of the Day 3/2/2011
Use the formula for continuously compounded interest (below) for the
questions that follow
P = initial amount A = amount at time t r =
interest rate t=time in years
e is approximately equal to 2.7183 ln(2) is approximately 0.69315
(2010). Section 8-7: Base e and Natural Logarithms. In J. A. Carter, D. Roger, C. Malloy, & G. J. Cuevas, Algebra 2 (pp. 525-532). Columbus: Glencoe McGraw-Hill.
ANSWERS
1. $1001.86
800*e^(0.045*5)=1001.86
2. about 15.4 yrs
2=e^(0.045*t)
ln(2)=ln[e^(0.045*t)]
ln(2)=(0.045*t)*ln(e)
ln(2)=(0.045*t)
t = ln(2)/(0.045)
t=15.4
3. about 7.7%
2=e^(r * 9)
ln(2)=ln[e^(r * 9)]
ln(2)=(r * 9)*ln(e)
ln(2)=(r * 9)
r = ln(2)/(9)
r =0.077 , which is 7.7%
4. about $5655.25
10,000 = P* e^(0.045*12)
P = 10,000 / (e^(0.045*12))
P= 5827.48
e is approximately equal to 2.7183 ln(2) is approximately 0.69315
- If you deposited $800 in an account paying 4.5% interest compounded continuously, how much money would be in the account in 5 years?
- How long would it take you to double your money?
- If you want to double your money in 9 years, what rate would you need?
- If you want to open an account that pays 4.75% interest compounded continuously and have $10,000 in the account 12 years after your deposit, how much would you need to deposit?
(2010). Section 8-7: Base e and Natural Logarithms. In J. A. Carter, D. Roger, C. Malloy, & G. J. Cuevas, Algebra 2 (pp. 525-532). Columbus: Glencoe McGraw-Hill.
ANSWERS
1. $1001.86
800*e^(0.045*5)=1001.86
2. about 15.4 yrs
2=e^(0.045*t)
ln(2)=ln[e^(0.045*t)]
ln(2)=(0.045*t)*ln(e)
ln(2)=(0.045*t)
t = ln(2)/(0.045)
t=15.4
3. about 7.7%
2=e^(r * 9)
ln(2)=ln[e^(r * 9)]
ln(2)=(r * 9)*ln(e)
ln(2)=(r * 9)
r = ln(2)/(9)
r =0.077 , which is 7.7%
4. about $5655.25
10,000 = P* e^(0.045*12)
P = 10,000 / (e^(0.045*12))
P= 5827.48
Math Problem of the Day 2/28/2011
Most people think that Chicago is the windiest city in the U.S. It's
not. Boston is the windiest city. The numbers below indicate the miles
per hour of the winds for Boston for a 2-week period. Use the
information to make a stem and leaf plot. Then find the range, median
and mode. 22, 35, 27, 15, 38, 32, 22, 35, 25, 29,
22, 18, 24, 36
[For a stem and leaf plot the number 35 would have a stem of 3 and a leaf of 5]
Malen, M. L., & Holtz, G. (n.d.). Blow Wind, Blow! Retrieved Feburary 28, 2011, from PUZZLING AND PERPLEXING PROBLEMS: http://www.fi.edu/school/math/wind.html
ANSWERS
I first reorganize the information in numerical order: 15, 18, 22, 22, 22, 24, 25, 27, 29, 32, 35, 35, 36, 38
Then I constructed a Stem and Leaf Plot:
[For a stem and leaf plot the number 35 would have a stem of 3 and a leaf of 5]
Malen, M. L., & Holtz, G. (n.d.). Blow Wind, Blow! Retrieved Feburary 28, 2011, from PUZZLING AND PERPLEXING PROBLEMS: http://www.fi.edu/school/math/wind.html
ANSWERS
I first reorganize the information in numerical order: 15, 18, 22, 22, 22, 24, 25, 27, 29, 32, 35, 35, 36, 38
Then I constructed a Stem and Leaf Plot:
range (Biggest Value - Smallest Value)=38 - 15=23;
median (the number in the middle of the set)=26;
mode (the number that occurs the most)=22
median (the number in the middle of the set)=26;
mode (the number that occurs the most)=22
Math Problem of the Day 2/24/2011 and 2/25/2011
1) Penelope bought 17 large packs and 5 small packs of identical pens.
When she got home, her little sister opened all the packages onto the
floor. If a total of 234 pens were strewn across the floor, how many
pens did a large pack contain?
2) A mile-long train is moving at 60 mph when it reaches a mile-long tunnel. How long does it take for the entire train to pass through the tunnel, assuming that it maintains its current speed?
From Februray 2011 Edition of the MATHEMATICS TEACHER
Coffey, M. (Ed.). (2011). Feburary "Calendar" Problems. MATHEMATICS TEACHER , 104 (6), 440-441.
ANSWER:
1)12 pens.
2) A mile-long train is moving at 60 mph when it reaches a mile-long tunnel. How long does it take for the entire train to pass through the tunnel, assuming that it maintains its current speed?
From Februray 2011 Edition of the MATHEMATICS TEACHER
Coffey, M. (Ed.). (2011). Feburary "Calendar" Problems. MATHEMATICS TEACHER , 104 (6), 440-441.
ANSWER:
1)12 pens.
2) 2
minutes. The train must travel 2
miles for the engine to enter and the caboose to exit the tunnel.
Since 60mph is 1 mile per minute, it will take the train 2 minutes to
pass through the tunnel.
Math Problem of the Day 2/23/2011
Chelsea had made six of seventeen free-throws attempts. How many
consecutive free throws must she make to raise her percentage of free
throws made to exactly 50 percent?
Problems from Mathematics Teacher. (2011). Retrieved Februrary 17, 2011, from National Council of Teachers of Mathematics Website: http://www.nctm.org/publications/calendar/index.aspx?journal_id=2
Answer:
Five Free Throws. Suppose that Chelsea makes her next x free throws. Then her percent is (6+x)/(17+x)=1/2 or 50%.
We can cross multiply to obtain:
2(6+x)=1(17+x)
12+2x=17+x
x=5
Problems from Mathematics Teacher. (2011). Retrieved Februrary 17, 2011, from National Council of Teachers of Mathematics Website: http://www.nctm.org/publications/calendar/index.aspx?journal_id=2
Answer:
Five Free Throws. Suppose that Chelsea makes her next x free throws. Then her percent is (6+x)/(17+x)=1/2 or 50%.
We can cross multiply to obtain:
2(6+x)=1(17+x)
12+2x=17+x
x=5
Math Problem of the Day 2/22/2011
A typical high school student consumes 67.5 pounds of sugar per year.
As part of a new nutrition plan, each member of a track team plans to
lower the sugar he or she consumes by at least 20% for the coming year.
Assuming each track member had consumed sugar at the level of a typical
high school student and will adhere to this plan for the coming year,
what is the maximum number of pounds of sugar to be consumed by each
track team member in the coming year?
A.14
B.44
C.48
D.54
E.66
Answer:
54 (Choice D) is the correct answer. For each member of the track team to consume 20% less sugar, the track member will consume 100% – 20% = 80% of the level of a typical high school student. 80% of 67.5 = 0.80(67.5) = 54.
A.14
B.44
C.48
D.54
E.66
Answer:
54 (Choice D) is the correct answer. For each member of the track team to consume 20% less sugar, the track member will consume 100% – 20% = 80% of the level of a typical high school student. 80% of 67.5 = 0.80(67.5) = 54.
Math Problem of the Day 2/18/2011
A census taker came to the house with no one home. He asked a neighbor
the ages of the three inhabitants and was told, "The product of their
ages is 252 and the sum of their ages is their house number." Later the
census taker returned, saying, "I need more information." The reply
was, "The twins are named Derek and Doreen." The Census taker wrote down
the ages and left. What were the ages?
Problems from Mathematics Teacher. (2011). Retrieved Februrary 17, 2011, from National Council of Teachers of Mathematics Website: http://www.nctm.org/publications/calendar/index.aspx?journal_id=2
Answer:
3,3, 28. The number 252 has prime factorization 2*2*3*3*7. We let the three ages be x, y, and z. Let us next consider all possible triples of positive integers whose product is 252. Twenty possible triples occur, all of which have unique sums except the triples (1, 12, 21) and (3,3,28), both of whose sums are 34. Since the mathematician neighbor needed to give more information to express their ages, the house number must be 34, and since the inhabitants include a set of twins, the triple (3,3,28) must represent the ages.
**SIDE NOTE: This problem is a bit tricky because of the way the word problem is set up. As the audience, we were not given all the information (i.e. the house number), however the census taker did have this knowledge (otherwise he is a bad census taker). That is why the key to the problem for us, the audience, is that all the possible triples have unique sums except for two of them. From there we can deduce that the house number had to have been 34 in order for the census taker to ask for more information.
Problems from Mathematics Teacher. (2011). Retrieved Februrary 17, 2011, from National Council of Teachers of Mathematics Website: http://www.nctm.org/publications/calendar/index.aspx?journal_id=2
Answer:
3,3, 28. The number 252 has prime factorization 2*2*3*3*7. We let the three ages be x, y, and z. Let us next consider all possible triples of positive integers whose product is 252. Twenty possible triples occur, all of which have unique sums except the triples (1, 12, 21) and (3,3,28), both of whose sums are 34. Since the mathematician neighbor needed to give more information to express their ages, the house number must be 34, and since the inhabitants include a set of twins, the triple (3,3,28) must represent the ages.
**SIDE NOTE: This problem is a bit tricky because of the way the word problem is set up. As the audience, we were not given all the information (i.e. the house number), however the census taker did have this knowledge (otherwise he is a bad census taker). That is why the key to the problem for us, the audience, is that all the possible triples have unique sums except for two of them. From there we can deduce that the house number had to have been 34 in order for the census taker to ask for more information.
Math Problem of the Day 2/17/2011
- The average age of three students is twenty years old. If all the students are at least eighteen years old, what are the possible ages for the oldest of the three?
- A man invests some money at 8 percent interest and twice as much at 6 percent. If the total income from the two investments is $250, how much is invested at the lower rate?
- At a resort, eight people rented a motorboat. If four more people had shared the cost, each person would have paid $1 less. How much did each person have to pay?
Problems from Mathematics Teacher. (2011). Retrieved Februrary 17, 2011, from National Council of Teachers of Mathematics Website: http://www.nctm.org/publications/calendar/index.aspx?journal_id=2
Answer:
- Between twenty and twenty-four years old. The largest of three numbers must be at least as big as their average, A A , since otherwise all three numbers would be less than A A , so their sum would be less than 3A A and their average would be less than A A . Therefore, the oldest student is at least twenty years old. We know that the sum of the three students’ ages is 3 × 20 = 60 and that each of the ages is at least eighteen. To find the largest possible age, we need to set the other two ages equal to the minimum possible value, 18. If the two younger students were both 18, then the third student would be 60 - 2(18), or twenty-four, years old. The oldest of the three students is then between twenty and twenty-four years old.
- $2,500. Let I1 = interest earned at 8 percent and I2 = interest earned at 6 percent. If x = amount of money invested at 8 percent, then 2x represents the amount of money invested at 6 percent. So, I1 + I2 = x(0.08) + 2x(0.06) = 250. Solving, 0.20x = 250 and x = 1250. So 2x = 2500.
- $3. The eight people paid x dollars each, for a total of 8x
dollars. If the number of people had been twelve, then each would have
paid x – 1 dollars, for a total of 12x – 12 dollars.
Setting these two expressions equal gives 8x = 12x – 12,
or x = 3. Each person paid three dollars. (See Philip Haber, Mathematical
Puzzles and Pastimes [Mount Vernon, N.Y.: Peter Pauper Press,
1957].)
Math Problem of the Day 2/16/2011
Use the Diagram below to answer the following
problem:
Answer:
The correct response is choice D.
Math Problem of the Day 2/15/2011
Points A, B, C, and D are on a line such that B is between A and C, and C is between B and D. The distance from A to B is 6 units. The distance from B to C is twice the distance from A to B, and the distance from C to D is twice the distance from B to C. What is the distance, in units, from the midpoint of BC to the midpoint of CD ?
F.18
G.14
H.12
J. 9
K. 6
Answer:
The correct response is Choice F. BC = 2AB = 2(6) = 12 and CD = 2BC = 2(12) = 24. The distance between the midpoints of BC and CD is BC + CD = (12) + (24) = 18.
For those of you who are visual learners here is another way to get the answer.
Math Problem of the Day 2/14/2011 (Happy Valentine's Day!)
Val picked a chrysanthemum with a Fibonacci number
[click for definition] of petals. He plucked one petal: "She loves
me." He plucked a second petal: "She loves me not." He plucked a third
petal: "She loves me." He continued in this way until he reached the
last petal. What is the probability that the last petal was a
"she-loves-me" petal?
From Februray 2011 Edition of the MATHEMATICS TEACHER
Coffey, M. (Ed.). (2011). Feburary "Calendar" Problems. MATHEMATICS TEACHER , 104 (6), 440-441.
ANSWER:
Val's first petal was a "she-loves-me" petal. Hence, all odd petals will be "she-loves-me"petals. The Fibonacci sequence begins 1,1,2,3,5,8,.... We observe that every third number is even and that the other numbers are odd. Therefore, the probability that the chrysanthemum has an odd number of petals is 2/3.
From Februray 2011 Edition of the MATHEMATICS TEACHER
Coffey, M. (Ed.). (2011). Feburary "Calendar" Problems. MATHEMATICS TEACHER , 104 (6), 440-441.
ANSWER:
Val's first petal was a "she-loves-me" petal. Hence, all odd petals will be "she-loves-me"petals. The Fibonacci sequence begins 1,1,2,3,5,8,.... We observe that every third number is even and that the other numbers are odd. Therefore, the probability that the chrysanthemum has an odd number of petals is 2/3.
Math Problem of the Day 2/11/2011
Jim has
$9.60 in his pocket. He has an equal number of quarters, dimes, and
nickels but no other coins or bills. How many coins does Jim have?
From Februray 2011 Edition of the MATHEMATICS TEACHER
Coffey, M. (Ed.). (2011). Feburary "Calendar" Problems. MATHEMATICS TEACHER , 104 (6), 440-441.
Answer:
Let "x" represent the number of each type of coin.
Then $0.25x+$0.10x+$0.05x =$9.60
$0.40x = $9.60
Then x=24
Therefore, the number of coins Jim had in his pockets were 3*x= 3(24)= 72 coins
From Februray 2011 Edition of the MATHEMATICS TEACHER
Coffey, M. (Ed.). (2011). Feburary "Calendar" Problems. MATHEMATICS TEACHER , 104 (6), 440-441.
Answer:
Let "x" represent the number of each type of coin.
Then $0.25x+$0.10x+$0.05x =$9.60
$0.40x = $9.60
Then x=24
Therefore, the number of coins Jim had in his pockets were 3*x= 3(24)= 72 coins
Math Problem of the Day 2/7/2011
Today's Math Problem will be simple math riddles. These riddles deal
more with logic than mathematical computation
CHALLENGE QUESTION
Two Riddles in one!To solve this riddle you don't want to manually do all of the math but rather try to figure out a pattern.
A. What digit is the most frequent between the numbers 1 and 1,000?
B. What digit is the least frequent between the numbers 1 and 1,000?
Answer:
CHALLENGE QUESTION
Two Riddles in one!To solve this riddle you don't want to manually do all of the math but rather try to figure out a pattern.
A. What digit is the most frequent between the numbers 1 and 1,000? One is the most frequent number
B. What digit is the least frequent between the numbers 1 and 1,000? Zero is the least frequent number
Reasoning for Challenge:
The digits 0 through 9 all follow the same pattern there is exactly 1 occurance of each digit for every ten numbers.
- Which weighs more? A pound of iron or a pound of feathers?
- What is the difference between a new penny and an old quarter?
- If you take three apples from five apples, how many do you have?
- It happens once in a minute, twice in a week, and once in a year. What is it?
CHALLENGE QUESTION
Two Riddles in one!To solve this riddle you don't want to manually do all of the math but rather try to figure out a pattern.
A. What digit is the most frequent between the numbers 1 and 1,000?
B. What digit is the least frequent between the numbers 1 and 1,000?
Answer:
- Which weighs more?
A pound of iron or a pound of feathers? They both weigh the same (one pound).
- What is the
difference
between a new penny and an old quarter? 24 cents
- If you take
three apples
from five apples, how many do you have? You have three apples.
- It happens
once in a
minute, twice in a week, and once in a year. What
is it? The letter "e"
CHALLENGE QUESTION
Two Riddles in one!To solve this riddle you don't want to manually do all of the math but rather try to figure out a pattern.
A. What digit is the most frequent between the numbers 1 and 1,000? One is the most frequent number
B. What digit is the least frequent between the numbers 1 and 1,000? Zero is the least frequent number
Reasoning for Challenge:
The digits 0 through 9 all follow the same pattern there is exactly 1 occurance of each digit for every ten numbers.
- For instance the digit 2 appears once between 10 and 19, at 12. And 2 appears once between, 30 and 39 at 32.
- However, each of the digits 1 through 9 also appear in other numbers in the tens and hundres place Again, let's look at 2 which appears in 20,21,22, 23, etc.. as well as 200,201, 202,203.. So to figure out how to answer the first riddle you had to see what distinguishes the number 1? Only that we are including 1,000 which would be the first '1' in a new series of ten! In other words, the digit 1 only has a single extra occurance (301 occurences) compared to 2 or 3 or 9 which each have exactly 300 occurences.
Math Problem of the Day 2/1/2011
Tribbles are cute, furry creatures that exist in the world of Star Trek. Tribbles can produce a litter of 10 babies every 12 hours. One tribble was trapped in the grain hold of Space Station K-7 for 3 days. How many tribbles were in the grain hold at the end of the third day (assuming none of them died)?
Bonus Question:
Use a graphing calculator (or Microsoft Excel), to prepare a scatter plot for the data ( # of tribbles vs. # hours) Find the exponential regression equation for this data.
Answer:
Below is a data table that marks # of tribbles that were existing on the space station every 12 hours.
Bonus Question:
Use a graphing calculator (or Microsoft Excel), to prepare a scatter plot for the data ( # of tribbles vs. # hours) Find the exponential regression equation for this data.
Answer:
Below is a data table that marks # of tribbles that were existing on the space station every 12 hours.
After three days there would be 1,771,561 tribbles on the space station after 3 days.
Bonus Question:
Bonus Question:
Problem of the Day 1/31/2011
In the movie, Legally Blonde, Elle uses reasoning to prove to her fellow lawyers that their client did not kill her husband. Elle states, "Exercising gives you endorphins, endorphins make you happy, happy people just don't kill their husbands, they just don't." Elle demonstrates the law of syllogism, which allows her to state a conclusion from two true conditional statements when the conclusion of one statement is the hypothesis of the other. Assuming that Elle's logic is sound, which of the following statements is NOT true?
a) People who exercise don't kill their husbands.
b) People who exercise are happy.
c) People who do not kill their husbands are happy.
d) People who kill their husbands are unhappy.
Answer:
a) People who exercise don't kill their husbands.
b) People who exercise are happy.
c) People who do not kill their husbands are happy.
d) People who kill their husbands are unhappy.
Answer:
Math Problem of the day 1/27/2011
Three brothers divide 17 candy bars. The oldest child gets 1/2, the middle child 1/3, and the youngest child 1/9. So that they do not have to cut any bars, Mom provides an extra one. The oldest child then takes 9 candy bars, the middle child 6, and the youngest 2, and they return the extra one. Which child received more than his fair share?
Answer:
All three children receive over their fair share.
The oldest child should have received 17/2 = 8.5 candy bars,
the middle child 17/3 = 5.67, and
the youngest child 17/9 = 1.89.
The oldest child received the most over his expected amount.
Coffey, M. (Ed.). (2011). January "Calendar" Problems. MATHEMATICS TEACHER , 104 (5), 368-369.
Answer:
All three children receive over their fair share.
The oldest child should have received 17/2 = 8.5 candy bars,
the middle child 17/3 = 5.67, and
the youngest child 17/9 = 1.89.
The oldest child received the most over his expected amount.
Coffey, M. (Ed.). (2011). January "Calendar" Problems. MATHEMATICS TEACHER , 104 (5), 368-369.
Math Problem of the Day 1/25/2011
a) How many ways can I rearrange the letters in the word "MATH"?
b) How many ways can I rearrange the letters in the word "JAZZ"?
c) Why do I get two different answers for these two different four letter words?
Bonus Question
How many ways can I rearrange the letters in the word "MISSISSIPPI"?
ANSWERS
DISCLAIMER-- There are several ways to reach the solution to these problems. I attempt to show both the most direct and quickest solution.
a) 4!=4*3*2*1= 24 different ways I can rearrange the letters in the word "MATH"
b) (4!)/(2!) =(4*3*2*1)/(2*1) =12 different ways I can rearrange the letters in the word "JAZZ"
REASONING: There are 4letters. So there are 4! ways to arrange those letters. You divide by 2! because there are 2 Z's in JAZZ. Z's in themselves can be arranged in 2! different ways (which are included in the 4! by default). But they are not unique, which is why we divide them out.
c) You get two different answers because "MATH" has four distinct letters, while "JAZZ" has three distinct letter.
Bonus Question: (11!)/(4!*4!*2!)=34,650 ways to rearrange the letters in "MISSISSIPPI".
REASONING: There are 11 letters. So there are 11! ways to arrange those letters. You divide by 4! because there are 4 S's in Mississippi. S's in themselves can be arranged in 4! different ways (which are included in the 11! by default). But they are not unique, which is why we divide them out. The same is true of the 4 I's and the 2 P's
b) How many ways can I rearrange the letters in the word "JAZZ"?
c) Why do I get two different answers for these two different four letter words?
Bonus Question
How many ways can I rearrange the letters in the word "MISSISSIPPI"?
ANSWERS
DISCLAIMER-- There are several ways to reach the solution to these problems. I attempt to show both the most direct and quickest solution.
a) 4!=4*3*2*1= 24 different ways I can rearrange the letters in the word "MATH"
b) (4!)/(2!) =(4*3*2*1)/(2*1) =12 different ways I can rearrange the letters in the word "JAZZ"
REASONING: There are 4letters. So there are 4! ways to arrange those letters. You divide by 2! because there are 2 Z's in JAZZ. Z's in themselves can be arranged in 2! different ways (which are included in the 4! by default). But they are not unique, which is why we divide them out.
c) You get two different answers because "MATH" has four distinct letters, while "JAZZ" has three distinct letter.
Bonus Question: (11!)/(4!*4!*2!)=34,650 ways to rearrange the letters in "MISSISSIPPI".
REASONING: There are 11 letters. So there are 11! ways to arrange those letters. You divide by 4! because there are 4 S's in Mississippi. S's in themselves can be arranged in 4! different ways (which are included in the 11! by default). But they are not unique, which is why we divide them out. The same is true of the 4 I's and the 2 P's
Math Problem of the Day 1/24/2011
During the Superbowl XXXII, the Packers scored 3 touchdowns, 3 extra points, and 1 field goals. The Denver Broncos scored 4 touchdowns, 4 extra points, and 1 field goal. What's the score? By how many points did the Denver Broncos beat the Packers?
(Touchdown=6pts and Field goal = 3pts)
Answer:
Packers: 3(6pts)+3(1pt)+1(3pts) = 18pts + 3pts +3pts = 24pts
Broncos: 4(6pts)+4(1pt)+1(3pts) =24pts + 4pts +3pts = 31pts
The Score was 24 Packers to 31 Broncos.
The Broncos beat the Packers by 7 points (or one touch down and an extra point)
(Touchdown=6pts and Field goal = 3pts)
Answer:
Packers: 3(6pts)+3(1pt)+1(3pts) = 18pts + 3pts +3pts = 24pts
Broncos: 4(6pts)+4(1pt)+1(3pts) =24pts + 4pts +3pts = 31pts
The Score was 24 Packers to 31 Broncos.
The Broncos beat the Packers by 7 points (or one touch down and an extra point)
Math Problem of the Day 1/21/2011
A dartboard has the numbers 3,6,9,12,15,19,21,25,27, and 30 on it. What is the fewest number of darts that could be thrown if two darts cannot land on the same number and if the sum of the scores must be 50?
Answer:
A quick glance should indicate that neither one nor two darts will suffice, so we consider three darts. Since most of the target numbers are odd and three odd numbers will not sum to 50, we try two odd and one even targets. When we consider 12, we find that no two of the remaining targets will add to 50, since two darts cannot land on 19. So we try 6. Subsequent searching produces 19 and 25 as the other two targets.
Therefore it would take three darts, which would be 6, 19, and 25.
Coffey, M. (Ed.). (2011). January "Calendar" Problems. MATHEMATICS TEACHER , 104 (5), 368-369.
Answer:
A quick glance should indicate that neither one nor two darts will suffice, so we consider three darts. Since most of the target numbers are odd and three odd numbers will not sum to 50, we try two odd and one even targets. When we consider 12, we find that no two of the remaining targets will add to 50, since two darts cannot land on 19. So we try 6. Subsequent searching produces 19 and 25 as the other two targets.
Therefore it would take three darts, which would be 6, 19, and 25.
Coffey, M. (Ed.). (2011). January "Calendar" Problems. MATHEMATICS TEACHER , 104 (5), 368-369.
Math Problem of the Day 1/20/2011
In honor of the premiere of American Idol, Mr. Kruczinski proudly presents a music themed math problem.
A number of girls from a singing group. All but 3 of the girls are blondes, all but 4 are brunettes, and all but 5 are redheads. How many girls are in the group?
Answer:
If we use x, y, and z to represent the number of blonds, brunettes, and redheads, respectively, we obtain three equations:
y+z=3 (because all except 3 girls were blond, so the rest had to be brunettes [y] or redheads [z])
x+z=4 (because all except 4 girls were brunette, so the rest had to be blondes [x] or redheads [z])
x+y=5 (because all except 5 girls were redhead, so the rest had to be blondes [x] or brunettes [y])
Add the three equations together and get : 2x+2y+2z=12
Divide both sides of the equation by 2 and get: x+y+z =6
Therefore, there are 6 girls in the singing group.
Coffey, M. (Ed.). (2011). January "Calendar" Problems. MATHEMATICS TEACHER , 104 (5), 368-369.
A number of girls from a singing group. All but 3 of the girls are blondes, all but 4 are brunettes, and all but 5 are redheads. How many girls are in the group?
Answer:
If we use x, y, and z to represent the number of blonds, brunettes, and redheads, respectively, we obtain three equations:
y+z=3 (because all except 3 girls were blond, so the rest had to be brunettes [y] or redheads [z])
x+z=4 (because all except 4 girls were brunette, so the rest had to be blondes [x] or redheads [z])
x+y=5 (because all except 5 girls were redhead, so the rest had to be blondes [x] or brunettes [y])
Add the three equations together and get : 2x+2y+2z=12
Divide both sides of the equation by 2 and get: x+y+z =6
Therefore, there are 6 girls in the singing group.
Coffey, M. (Ed.). (2011). January "Calendar" Problems. MATHEMATICS TEACHER , 104 (5), 368-369.
Math Problem of the Day 1/19/2011
Mr. Kruczinski's Website is proud to introduce our first multiple part math problem of the day!!
Two regular six sided dice are rolled, find the probability that the sum is ....
a) equal to or less than 8
b) less than 5
c) equal to 7
Answer:
Before we can do much we have to write all the possible pairs that we can get if we roll two regular dice...
Two regular six sided dice are rolled, find the probability that the sum is ....
a) equal to or less than 8
b) less than 5
c) equal to 7
Answer:
Before we can do much we have to write all the possible pairs that we can get if we roll two regular dice...
a) Now we have to locate all the pairs that are equal to and less than 8
Therefore the probability of rolling a pair of dice that the sum is equal to or less than 8 is 26/36 = 13/18 or approximately 72.22%
b) Now we have to locate all the pairs that are less than 5
Therefore the probability of rolling a pair that is less than 5 is 6/36 = 1/6 or approximately 16.67%
c) Now we have to locate all the pairs that are equal to 7
Therefore the probabiltiy of rolling a pair that is equal to 7 is 6/36 = 1/6 or approximately 16.67%
Math Problem of the Day 1/18/2010
There are about 20 potatoes in a 5 pound bag. A restaurant uses about 2 potatoes per order of French Fries. They charge $.95 for an order of Fries. How much money does the restaurant take in on a day that they use 400 pounds of potatoes?
ANSWER: 400 pounds *(20 potatoes/5 pounds)*(1 order/2 potatoes) * ($.95/1 order) = $760.00
ANSWER: 400 pounds *(20 potatoes/5 pounds)*(1 order/2 potatoes) * ($.95/1 order) = $760.00
Math Problem of the Day 12/15/2010
Santa Claus averages 27 miles per gallon of magic dust for his sleigh. If magic dust costs $4.04 per gallon, which of the following is closest to how much the magic dust would cost for this sleigh to travel 2,727 miles?
A.$ 44.44
B.$109.08
C.$118.80
D.$408.04
E.$444.40
ANSWER:
If you divide 2,727 miles by 27 miles per gallon you will get the number of gallons that Santa Needs to have: = 101. Then, multiply the number of gallons by the cost per gallon: 101(4.04) = 408.04. CHOICE D
Question provided by ACT website with a minor edit from Mr. Kruczinski.
A.$ 44.44
B.$109.08
C.$118.80
D.$408.04
E.$444.40
ANSWER:
If you divide 2,727 miles by 27 miles per gallon you will get the number of gallons that Santa Needs to have: = 101. Then, multiply the number of gallons by the cost per gallon: 101(4.04) = 408.04. CHOICE D
Question provided by ACT website with a minor edit from Mr. Kruczinski.
Math Problem of the Day 12/14/2010
The word game DOUBLET was invented by Lewis Carroll, author of Alice in Wonderland. The object of the game is to change a word into another word by successively changing only one letter at the time to form an intermediate valid word. For example, the following process changes the word TEN to JOY:
T E N
T O N
T O Y
J O Y
T O N
T O Y
J O Y
Using a similar process, change the word
Gift to Love
Gift to Love
ANSWER:
G I F T
L I F T
L I F E
L I V E
L O V E
L I F T
L I F E
L I V E
L O V E
This problem came from LogicVille.
Math Problem of the Day 12/13/2010
How many gifts are given in the song: The Twelve Days of Christmas?
HINT:
Remember how the song goes...
First Verse
On the first day of Christmas, my true love gave to me...A Partridge in a Pear Tree.
Second Verse:
On the second day of Christmas, my true love gave to me...2 Turtle Doves AND a Partridge in a Pear Tree.
So by the second day...there were 4 gifts given (1 from day one and 3 from day two)
ANSWER:
Work:
Day1: 1+
Day2: 1+2+
Day3: 1+2+3+
Day4: 1+2+3+4+
Day5: 1+2+3+4+5+
Day6: 1+2+3+4+5+6+
Day7: 1+2+3+4+5+6+7+
Day8: 1+2+3+4+5+6+7+8+
Day9: 1+2+3+4+5+6+7+8+9+
Day10: 1+2+3+4+5+6+7+8+9+10+
Day11: 1+2+3+4+5+6+7+8+9+10+11+
Day12: 1+2+3+4+5+6+7+8+9+10+11+12
Look there are 12 ones, 11 twos, 10 threes, etc...
1(12)+2(11)+3(10)+4(9)+5(8)+6(7)+7(6)+8(5)+9(4)+10(3)+11(2)+12(1)=364 Gifts in total
You can solve this problem a number of ways
Wild about Math!
HINT:
Remember how the song goes...
First Verse
On the first day of Christmas, my true love gave to me...A Partridge in a Pear Tree.
Second Verse:
On the second day of Christmas, my true love gave to me...2 Turtle Doves AND a Partridge in a Pear Tree.
So by the second day...there were 4 gifts given (1 from day one and 3 from day two)
ANSWER:
Work:
Day1: 1+
Day2: 1+2+
Day3: 1+2+3+
Day4: 1+2+3+4+
Day5: 1+2+3+4+5+
Day6: 1+2+3+4+5+6+
Day7: 1+2+3+4+5+6+7+
Day8: 1+2+3+4+5+6+7+8+
Day9: 1+2+3+4+5+6+7+8+9+
Day10: 1+2+3+4+5+6+7+8+9+10+
Day11: 1+2+3+4+5+6+7+8+9+10+11+
Day12: 1+2+3+4+5+6+7+8+9+10+11+12
Look there are 12 ones, 11 twos, 10 threes, etc...
1(12)+2(11)+3(10)+4(9)+5(8)+6(7)+7(6)+8(5)+9(4)+10(3)+11(2)+12(1)=364 Gifts in total
You can solve this problem a number of ways
- adding up each row then adding them all together
- using a formula... n(n+1)(2n+1)/6
Wild about Math!
Math Problem of the Day 12/10/2010
Mr. Green is putting lights around 8 windows; each window is 3 and 1/2 feet wide and 5 feet long. How many feet of lights does he need?
Answer: (2*3.5+2*5)*8= 136 ft (17ft per window)
Answer: (2*3.5+2*5)*8= 136 ft (17ft per window)
Math Problem of the Day 12/9/2010
Macy's has hired 400 store Santas. If each Santa sees 125 children a day for 30 days, how many children are seen by Macy's Santas?
Answer: 400*125*30=1,500,000 children
Answer: 400*125*30=1,500,000 children