Kiss those Math Headaches GOODBYE!

Posts tagged ‘Challenge Problem’

How to Simplify Fractions — FAST


So c’mon … everything that can be said about simplifying fractions has been said … right?

Not quite! Here’s something that might just be original … a hack to smack those fractions down to size.

Suppose you’re staring at an annoying-looking fraction:  96/104, and it’s annoying the heck out of you, particularly because it’s smirking at you!

But it won’t smirk for long. For you open up your bag of hacks (obtained @ mathchat.me) and …

1st)  Subtract to get the difference between numerator and denominator. I also like to call this the gap between the numbers. Difference (aka, gap) = 104 – 96 = 8.

NOTE: Turns out that this gap, 8, is the upper limit for any numbers that can possibly go into BOTH 96 and 104. No number larger than 8 can go into both. And this is a … HACK FACT:  The gap represents the largest number that could possibly go into BOTH numerator and denominator. In other words, the gap is the largest possible greatest common factor (GCF).

2nd)  Try 8. Does 8 go into both 96 and 104? Turns out it does, so smack the numerator and denominator down to size:  96 ÷ 8 = 12, and 104 ÷ 8 = 13.

3rd)  State the answer:  96/104 = 12/13.

Is it still smirking? I think … NOT!

Try another. Say you’re now puzzling over:  74/80.

1st)  Subtract to get the gap. 80 – 74 = 6. So 6 is the largest number that can possibly go into BOTH 74 and 80.

2nd)  So try 6. Does it go into both 74 and 80? No, in fact it goes into neither number.

NOTE:  Turns out that even though 6 does NOT go into 74 OR 80, the fact that the gap is 6 still says something. It tells us that the only numbers that can possibly go into both 74 and 80 are the factors of 6:  6, 3 and 2. This, it turns out, is another … HACK FACT:  Once you know the gap, the only numbers that can possibly go into the two numbers that make the gap are either the factors of the gap, or the gap number itself.

3rd)  So now, try the next largest factor of 6, which just happens to be 3. Does 3 go into both 74 and 80? No. Like 6, 3 goes into neither 74 nor 80. But that’s actually a good thing because now there’s only one last factor to test, 2. Does 2 go into both 74 and 80? Yes! At last you’ve found a number that goes into both numerator and denominator.

4th)  Hack the numbers down to size:  74 ÷ 2 = 37, and 80 ÷ 2 = 40.

5th)  State the answer. 74/80 gets hacked down to 37/40, and that fraction, my dear friends, is the answer. 37/40 the final, simplified form of 74/80. 

O.K., are you ready to smack some of those fractions down to size? I believe you are. So here are some problems that will let you test out your new hack.

As you slash these numbers down, remember this rule. In some of these problems the gap number itself is the number that divides into numerator and denominator. But in other problems, it’s not the gap number itself, but rather a factor of the gap number that slashes both numbers down to size. So if the gap number itself doesn’t work, don’t forget to check out its factors.

Ready then? Here you go … For each problem, state the gap and find the largest number that goes into both numerator and denominator. Then write the simplified version of the fraction.

a)   46/54
b)   42/51
c)   48/60
d)   45/51
e)   63/77

Answers:

a)   46/54:  gap = 8. Largest common factor (GCF) = 2. Simplified form = 23/27
b)   42/51:  gap = 9. Largest common factor (GCF) = 3. Simplified form = 14/17
c)   48/60:  gap = 12. Largest common factor (GCF) = 12. Simplified form = 4/5
d)   45/51:  gap = 6. Largest common factor (GCF) = 3. Simplified form = 15/17
e)   63/77:  gap = 14. Largest common factor (GCF) = 7. Simplified form = 9/11

Josh Rappaport is the author of five math books, including the wildly popular Algebra Survival Guide and its trusty sidekick, the Algebra Survival Workbook. Josh has been tutoring math for more years than he can count — even though he’s pretty good at counting after all that tutoring — and he now tutors students in math, nationwide, by Skype. Josh and his remarkably helpful wife, Kathy, use Skype to tutor students in the U.S. and Canada, preparing them for the “semi-evil” ACT and SAT college entrance tests. If you’d be interested in seeing your ACT or SAT scores rise dramatically, shoot an email to Josh, addressing it to: josh@SingingTurtle.com  We’ll keep an eye out for your email, and our tutoring light will always be ON.

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Category:

College Entrance Exams, Help with the ACT, Help with the SAT, SAT, Tutoring

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Fun Math Problem #2


Here is the second in my series of “Fun Math Problems.”

Feel free to try these problems. Share them with friends and colleagues. Use them however you see fit! I will post the answer to the problems two days later, after people have had time to respond.

To provide your response, simply send an email to me @ info@SingingTurtle.com
and make your Subject: Fun Problem.
Please show how you worked the problem. Thanks. I will post the names of the first three people who get this right.

The Problem:  Before you go out to lunch, you glance at the clock above your desk. When you come back from lunch, you glance at the clock again, and you notice something strange. The minute and the hour hand have exchanged places from the positions they had just before you went to lunch.

The question is:  how long were you away?

Rusting clock face

Image by The Hidaway (Simon) via Flickr

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Category:

Algebra, Challenge Problem, Formulas, FUN PROBLEM, Integers, Linear Equations, Making Math Fun, Practice Problems, Problem of the Month, Problem of the Week, Puzzles, Turtle Talk, Uncategorized

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FUN MATH PROBLEM — Circling the Square & Vice-Versa


From time to time I will post interesting math problems.

Feel free to try these problems. Share them with friends and colleagues. Use them however you see fit!

I will post the answer to the problems two days later, after people have had time to respond.

To post your response, simply send an email to me @ info@SingingTurtle.com
and make your Subject: Fun Problem.

The problem: Which provides the fuller fit? Putting a circular peg in a square hole, or putting a square peg in a circular hole? To get credit, show all work, and justify your answer by expressing each “fit” as a percent.

A few term-clarifications, to help you do this correctly:

a) By “fit,” I mean the ratio of the smaller shape to the larger shape, expressed as a percent. For
example, if a ratio is 4 to 5, that would represent a “fit” of 80 percent.

b) For the circular peg in the square hole, assume that the diameter of the circle equals the side of the
square. For the square peg in a circular hole, assume that the diameter of the circle equals the diagonal of the square.

c) By “fuller fit,” I mean the larger of the two ratios.

Have fun!

Category:

Area, Formulas, Formulas, FUN PROBLEM, Geometry, Making Math Fun, Math in General, Math Instruction Techniques, Practice Problems, Probability, Proofs, Proportions, Puzzles, Ratios, Uncategorized

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Problem of the Week-10/25/2010


Here’s one of those:  “Can you make it?” problems.

Using exactly six toothpicks of equal length, how can you put them together to create four congruent equlateral triangles?

Send your answers as comments to this blog post. You need not worry about incorrect answers being posted. I will post only those answers that are correct, and I will post the first five correct answers on Monday.

Feel free to share this problem with anyone who might like to try it.

Category:

Geometry, Uncategorized

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Problem of the Week — 9/24/2010


Problem of the Week —  9/24/2010

At Gamesville High, students love their clubs. While 20% of the children in the Hex Club are also members of the Backgammon Club, 80% of children in the Backgammon Club are also members of the Hex Club. The Backgammon Club has 35 children. The question:  how many children are in the Hex Club?

To get the honor of being put into the Winner’s Circle, you need to get the correct answer and show how you arrived at it.

Please write your answers as comments on the blog post. Or alternatively, you can send it as an email to me:  josh@SingingTurtle.com

I will post the answer and name the five who make it to the Winner’s Circle on Monday.

P.S.:   Hex is a great board game, invented by two mathematicians. If you’d like to read a post about it, go here:

https://mathchat.wordpress.com/2010/01/27/play-a-game-meet-john-nash/

Screenshot from the program GNU Backgammon (Fr...

Backgammon, Image via Wikipedia

Category:

Probability, Problem of the Week, Uncategorized

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ANSWER TO FRIDAY’S PROBLEM:


ANSWER TO FRIDAY’S PROBLEM:

The problem, once again:

For any polygon, a “diagonal” is defined as a line segment that runs from one vertex  to another, running  through the polygon’s interior. Find a formula that determines the number of diagonals in any convex polygon with n sides. Once you have the formula, use it to figure out the number of diagonals in a convex polygon with 1,000 sides (don’t try this by hand! — that’s why algebra was invented).

The winning answer was provided by Chris Mark. The formula, for a convex polygon with n sides, is this:  Number of Diagonals =  [n(n– 3)]/2. For n = 1000, the number of diagonals = 498,500.

The reasoning behind the formula. A polygon has as many vertices as sides. So a polygon with n sides also has n vertices. Now, consider any vertex of the n-gon. From that vertex the number of diagonals that can be drawn is (n – 3). That is because we cannot draw a diagonal to 3 vertices:  the vertex chosen, and the two adjacent vertices. So the expression (n – 3) = the number of diagonals that can be drawn from any vertex. We multiply (n – 3) by n to obtain the total number of diagonals that can be drawn from all n vertices. But if we simply multiply n by (n – 3), we’d be counting each diagonal twice. To eliminate that problem, we divide the product, [ n(n – 3)], by 2, and that provides the correct formula:

Diagonals  =  [n(n – 3)]/2

Applying this formula to a convex polygon with 1,000 sides, we see that the number of diagonals =  498,500.

In addition to providing the answer, Chris pointed out that the problem need not be restricted to regular polygons, as it was when posted. This formula works for all convex polygons, regular or not.

Thanks, Chris. And thanks to everyone who submitted answers.

Category:

Algebra, Formulas, Formulas, Geometry, Logic of Geometry, Problem of the Month, Proofs

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Challenge Problem – the ANSWER


Hi everyone,

Here is the answer to yesterday’s challenge problem, the probability problem about catching fish.

First, there are 6 ways to catch the three fish in three casts, getting exactly one trout, one carp and one bass.

You could get the fish in any of these six orders:

TCB / TBC / CTB / CBT / BTC / BCT

Next you find the probability of getting one of these possibilities. Let’s take the first one, TCB.

Keep in mind that after you catch the first fish, the number of fish left goes down by one, to 11; after catching the second fish, the number of fish goes down to 10.

The probability for the TCB possibility is calculated by multiplying:  6/12 x 4/11 x 2/10 = 2/55

When you think about the five other ways to catch the fish, you’ll see that the order of the numerators changes, but the denominators remain 12, 11, 10. So the probability for catching the three fish in any of the six ways is always the same:  2/55.

To get the probability for all six catches, just multiply the probability of one catch by 6:

6/1 x 2/55  =  12/55

And that is the answer: the probability of catching exactly one trout, one carp and one bass is 12/55, which works out to about 21.8%, meaning that this should happen a little more than 1/5 of the time.

I didn’t see anyone submit any answers, but feel free to send them in. Remember that I will post only the correct answers, so no one has to worry about seeing an incorrect answer posted.

Have a great day!

—  Josh

Category:

Algebra, Challenge Problem, Probability, Puzzles, Solving Equations, Uncategorized, Word Problems

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Interactive Challenge Problem — Send in responses


O.K., time to wake up the ol’ brain cells.

Here’s a little challenge problem. I’ll post the problem today, and then I’ll post the answer the next day. My thought is that this would be relevant for many kinds of people.

Teachers can use this as a fun class-opener. Homeschoolers can use it to start their math studies. And anyone who enjoys math can use it to sharpen math skills. So enjoy.

It would also be fun to see how many people get it right, so please send in your answers as comments. I will post only the correct answers.

The problem:

A lake contains exactly 12 fish: 6 trout, 4 carp, and 2 bass.
On any given cast, you catch exactly one fish, and no kind of fish is biting any more than any other kind. (i.e.:  Your odds of catching fish are governed by mathematical probability alone.)
What is the probability that in three casts you will catch exactly one trout, one carp and one bass?
To get credit, you must provide the correct answer and show how you solved the problem.

Category:

Algebra, Challenge Problem, Probability, Problem of the Month, Word Problems

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