getbullish shop

Monday Math Problem: The Answer to “Mischievous Multiples”

May 21, 2010

Every Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Here is Monday’s problem again — click “Read More” for the explanation, and then come back Monday for a new quantitative challenge!

This Week’s Math Problem:

The Abingdon twins, the Bernstein twins, the Cullen triplets, and the Dalton twins are all lining up, in a single row, for a photo. All of the multiples really are identical — that is, the photographer cannot tell any multiple from his/her siblings. As such, the kids think it’s funny to switch places with their own siblings every time the photographer turns his back. In how many distinct combinations can the kids rearrange themselves from their original lineup without the cameraman noticing anything funny?

Read more

Monday Math Problem: Mischievous Multiples

May 17, 2010

Every Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Give this one a shot and post your solution in the comments!

This Week’s Math Problem:

The Abingdon twins, the Bernstein twins, the Cullen triplets, and the Dalton twins are all lining up, in a single row, for a photo. All of the multiples really are identical — that is, the photographer cannot tell any multiple from his/her siblings. As such, the kids think it’s funny to switch places with their own siblings every time the photographer turns his back. In how many distinct combinations can the kids rearrange themselves from their original lineup without the cameraman noticing anything funny?

Monday Math Problem: The Answer to “Evil Barbershop”

May 14, 2010

Every Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Here is Monday’s problem again — click “Read More” for the explanation, and then come back Monday for a new quantitative challenge!

This Week’s Math Problem:

At the evil barbershop, when you request “just a trim,” you get 1.5 inches off. When you request that a specific amount be taken off, you get 1.75 times that amount taken off — that is, unless you are the barber’s son, in which case you get exactly what you asked for. If a dozen customers come in, one of which may or may not be the barber’s son, none have more than 6 inches of hair to start, and everyone requests that at least 1/2 inch be taken off but no one requests that more than half his hair length be taken off, what is the minimum and maximum total length of hair that could have been cut off all twelve customers, in inches?

Read more

Monday Math Problems: Evil Barbershop

May 10, 2010

Monday math is back! I had a deadline last week and didn’t get to it, but this week, enjoy the evil barbershop! Give this one a shot and post your solution in the comments. I’ll post a solution on Friday.

This Week’s Math Problem:

At the evil barbershop, when you request “just a trim,” you get 1.5 inches off. When you request that a specific amount be taken off, you get 1.75 times that amount taken off — that is, unless you are the barber’s son, in which case you get exactly what you asked for. If a dozen customers come in, one of which may or may not be the barber’s son, none have more than 6 inches of hair to start, and everyone requests that at least 1/2 inch be taken off but no one requests that more than half his hair length be taken off, what is the minimum and maximum total length of hair that could have been cut off all twelve customers, in inches?

Monday Math Problems: The Answer to “A Hop, Skip, and a Jump”

April 30, 2010

Every Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Here is Monday’s problem again — click “Read More” for the explanation, and then come back Monday for a new quantitative challenge!

This Week’s Math Problem:

Just how long is a hop, skip, and a jump? If 1 hop is equal to 0.4 skips and 1 skip is equal to .7 jumps, how far away, in hops, is something that is 7 jumps and one-half a skip away?

Read more

Monday Math Problems: A Hop, Skip, and a Jump

April 27, 2010

images.jpegEvery Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Give this one a shot and post your solution in the comments!

This Week’s Math Problem: Just how long is a hop, skip, and a jump? If 1 hop is equal to 0.4 skips and 1 skip is equal to .7 jumps, how far away, in hops, is something that is 7 jumps and one-half a skip away?

Monday Math Problems: The Answer to “Abdul’s Zucchini”

April 23, 2010

Every Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Here is Monday’s problem again — click “Read More” for the explanation, and then come back Monday for a new quantitative challenge!

This Week’s Math Problem:

Abdul’s garden is in the shape of a regular hexagon with integer side lengths (as measured in feet), and its area is greater than 200 square feet. If Abdul joins three vertices of the garden to rope off an area (CORRECTION: I really should have said “triangle” instead of “area”) in which to grow zucchini, and the cost of fertilizing garden space is 20¢ per square foot rounded up to the nearest square foot, what is the minimum amount of money Abdul would save by choosing to fertilize only the zucchini area and not the entire garden?

Before I get into this, though, I’d like to say that next week’s math problem will be MUCH easier, for several reasons. One is that my purpose in posting math problems in the first place was more to interest those of general nerdy inclinations in humorous situations that could be made more sense of through math. However, it also sort of seemed natural for me to escalate the problems as the weeks went by, just because, since the Enlightenment, we pretty much just like to push things forward for their own sake (thanks, Voltaire!) So, I think I shall resist that impulse in the future. For instance, I was pretty happy with The One About Being a Player.

Also, having to post lots of square-root signs and ^ symbols in the comments of a blog makes geometry problems a bit impractical for this platform. Also, when people write things like, “I’d explain how to derive this formula from first principles, but neither the readership nor the editorship of this blog seems to be interested in proof methodology,” I know I’ve strayed from my original mission.

For instance, once my mom called me, asking how to find the circumference of a circle. She had a real-life reason! She wanted to buy lace to trim a round tablecloth. Working off some vague high school memories, she had already measured the diameter of the tablecloth when she called. How cool was that? Um, I thought it was awesome.

And in answer to a question posed, my idea of a good solution is one that can be understood as readily and quickly as possible by as regular a person (math-wise) as possible, and the knowledge of which would allow such a person to confidently solve a similar problem in the future. I doubt that that goal will be achieved on this humdinger of a problem, but that’s the math ideal on jenniferdziura.com. That’s what you get from a GMAT instructor.

So, I’ll run through this one, but next week, look forward to being able to do the problem on your fingers.

Read more

Monday Math Problems: Abdul’s Zucchini

April 19, 2010

Every Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Give this one a shot and post your solution in the comments!

This Week’s Math Problem:

Abdul’s garden is in the shape of a regular hexagon with integer side lengths (as measured in feet), and its area is greater than 200 square feet. If Abdul joins three vertices of the garden to rope off an area in which to grow zucchini, and the cost of fertilizing garden space is 20¢ per square foot rounded up to the nearest square foot, what is the minimum amount of money Abdul would save by choosing to fertilize only the zucchini area and not the entire garden?

Monday Math Problem: The Answer to “International Handshakes”

April 9, 2010

Every Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Here is Monday’s problem again — click “Read More” for the explanation, and then come back Monday for a new quantitative challenge!

This Week’s Math Problem:

There are n people in a room. Each person is a French, Danish, Zambian, Norwegian, Swedish, Latvian, Japanese, Chinese, Singaporean, Spanish, Welsh, Greek, Italian, Hungarian, Austrian, or Comorian citizen. There exists in the room at least one citizen of each of the listed nationalities, and no one in the room is a citizen of more than one nation. If, as part of an international friendship exercise, every citizen of a European nation is to shake hands once with every citizen of a non-European nation, and if n is less than 60 and has exactly four prime factors (not necessarily distinct), what is the range of the numbers of possible handshakes that could take place?

Read more

Monday Math Problem: International Handshakes

April 5, 2010

Every Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Give this one a shot and post your solution in the comments! (“International Handshakes” might sound like an innuendo about something, but as you are soon to discover, it is totally literal!)

This Week’s Math Problem:

There are n people in a room. Each person is a French, Danish, Zambian, Norwegian, Swedish, Latvian, Japanese, Chinese, Singaporean, Spanish, Welsh, Greek, Italian, Hungarian, Austrian, or Comorian citizen. There exists in the room at least one citizen of each of the listed nationalities, and no one in the room is a citizen of more than one nation. If, as part of an international friendship exercise, every citizen of a European nation is to shake hands once with every citizen of a non-European nation, and if n is less than 60 and has exactly four prime factors (not necessarily distinct), what is the range of the numbers of possible handshakes that could take place?

Monday Math Problem: The Answer to “A Very Personal Question”

April 2, 2010

Every Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Here is Monday’s problem again — click “Read More” for the explanation, and then come back Monday for a new quantitative challenge!

This Week’s Math Problem:

If Jennifer Dziura was five years and two months old when the Macintosh computer was first introduced, and k3 is divisible by the year of Jennifer Dziura’s birth, what is the minimum integer value k2 could be?


Read more

Monday Math Problem: A Very Personal Question

March 29, 2010

Every Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Give this one a shot and post your solution in the comments!

This Week’s Math Problem:

If Jennifer Dziura was five years and two months old when the Macintosh computer was first introduced, and k3 is divisible by the year of Jennifer Dziura’s birth, what is the minimum integer value k2 could be?

Monday Math Problem: The Answer to “Joey’s Syllabic Obsession”

March 26, 2010

Every Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Here is Monday’s problem again — click “Read More” for the explanation, and then come back Monday for a new quantitative challenge!

This Week’s Math Problem:

Joey’s obsessive-compulsive disorder requires him to put a nickel in a jar every time he says a syllable, and an extra penny every time he completes a sentence. For instance, if he said, “I have to go to the bathroom,” he would put 41¢ in the jar. Fortunately for the sake of this math problem, Joey speaks very little. Every morning, he says “Good morning.” (This counts as a sentence). Every night, he says “Buenas noches.” (This also counts as a sentence). Mid-day on every Monday, he says “The best type of lava is a’a.” Mid-day on every Tuesday, he says “I have a simile about the muse Terpsichore.” Mid-day every Wednesday, he says, “I have not only sangfroid, but gravitas and avoirdupois — and a little je ne sais quoi.” Mid-day every Thursday, he says — in perfect French — “Merde!” Mid-day every Friday, he says, “My schadenfreude is downright Kierkegaardian.” Mid-day every Saturday, he sings the first verse of the song “Yellow Submarine” (which he considers to be all one sentence), just as he learned it from listening to the “Revolver” album. Mid-day every Sunday, he says, “Please leave me alone to listen to Sade.” If April begins on a Thursday and Joey begins with an empty change jar, how much money will he have put in the jar by the end of the month? Read more

Monday Math Problem: Joey’s Syllabic Obsession

March 22, 2010

Every Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Give this one a shot and post your solution in the comments!

This Week’s Math Problem:

Joey’s obsessive-compulsive disorder requires him to put a nickel in a jar every time he says a syllable, and an extra penny every time he completes a sentence. For instance, if he said, “I have to go to the bathroom,” he would put 41¢ in the jar. Fortunately for the sake of this math problem, Joey speaks very little. Every morning, he says “Good morning.” (This counts as a sentence). Every night, he says “Buenas noches.” (This also counts as a sentence). Mid-day on every Monday, he says “The best type of lava is a’a.” Mid-day on every Tuesday, he says “I have a simile about the muse Terpsichore.” Mid-day every Wednesday, he says, “I have not only sangfroid, but gravitas and avoirdupois — and a little je ne sais quoi.” Mid-day every Thursday, he says — in perfect French — “Merde!” Mid-day every Friday, he says, “My schadenfreude is downright Kierkegaardian.” Mid-day every Saturday, he sings the first verse of the song “Yellow Submarine” (which he considers to be all one sentence), just as he learned it from listening to the “Revolver” album. Mid-day every Sunday, he says, “Please leave me alone to listen to Sade.” If April begins on a Thursday and Joey begins with an empty change jar, how much money will he have put in the jar by the end of the month?

Monday Math Problem: The Answer to “Danica’s Endorsement Deals”

March 19, 2010

Every Monday, I’ll be posting a math problem on this site; every Friday, I’ll post the answer to that week’s problem. Here is Monday’s problem again — click “Read More” for the explanation, and then come back Monday for a new quantitative challenge!

This Week’s Math Problem:

Danica leaves her house and drives exactly 30 miles due west at 45 miles per hour before realizing that, as she is on her way to the racetrack to drop off her racecar, she really ought to be driving the racecar rather than the Prius. So she turns around and drives back home over the same route at the same speed.

When she arrives back home, she gets in the racecar and drives due west for 120 miles at 360 miles per hour.

After dropping off the racecar, she gets on a bus going due north and goes 202.5 miles at 15 mph, then gets off the bus and transfers to a Segway, on which she travels (also due north) 67.5 miles at 22.5 miles per hour.

Read more

Next Page »