Watermelons and New Years' Parties (Puzzle)

Here’s a fun math puzzle involving percents.  The solution is not terribly difficult – but it is surprising how many people end up getting the answer wrong.  Why not give it a try:

The owner of a farm stand puts a bunch of watermelons on display at the beginning of the day.  The watermelons have a mass of 200 kg.  99% of the watermelons’ mass is water.  It is hot, and during the day water evaporates from the melons.  At the end of the day, 98% of the watermelons’ mass is water.  Assuming that no watermelons were sold during the day, what is the mass of the watermelons at the end of the day?

If you’d like to check your answer, the correct solution is below.

 

If you’re having a hard time, consider this puzzle.  The math involved the same as the watermelon problem, but the way in which the puzzle is presented tends to make the solution easier to find.

Some guys planned a New Years’ party and invited only women.  At the party, there were a total of 200 people, 99% of whom were women.  A lot of the women thought that the party was lame, so they left.  After one hour, 98% of the people left at the party were women.  How many total people were at the party after one hour?

This puzzle is slightly easier to grasp because it explicitly states that there are men and women.  (And, it states that only women leave the party.)

 

See below for the solutions.

 

 

 

Solution: New Years’ party puzzle

It’s not very hard to determine how many men and women there were at the beginning of the party:

Women:  99% of 200 = 198           Men:  1% of 200 = 2

Since only women left the party, there are still two men after one hour.  However, these two men now comprise 2% of the total party crowd.  2 is 2% of 100… therefore, there are 100 people at the party after one hour.

 

Solution: Watermelon puzzle

The watermelon puzzle is a little trickier simply because the problem doesn’t explicitly state that the watermelon is made up of water and solids.  Therefore, many people get hung up focusing on the water.

If 99% of the watermelon mass is water, then 1% is made up of solids.  1% of 200kg is 2 kg – so there are 2 kg of solids.  Solids do not evaporate, so at the end of the day, there are still 2 kg of solids.  This now comprises 2% of the total watermelon mass.  2 kg is 2% of 100 kg.  Therefore, there are 100 kg of watermelons left at the end of the day.

Happy New Years’, Everyone!

 

------------------------------

 

For Math Tutorials and Silly Math Songs, visit   www.onlinemathpro.com

 

 

 

Affiliate Disclosure

Ready, Go, SET!

Riddles and puzzles are a good way to challenge yourself and increase your logic skills.  They also give valuable practice in solving the story problems that seem to haunt so many math students.

Here’s a brain teaser that can be solved with simple math.  (As is the case with many math puzzles, there is more than one way to solve the problem.)  If you think you’re up to it, try to answer the question before reading the solutions and explanations at the bottom of this page.

There are 340 players in a SET* tournament.  Each game in the tournament is played with 4 people.  For each game, there is one winner who moves on to the next round.  Rounds will continue until there is just one winner left – the champion.  In any round, if the number of players is not a multiple of 4, then some players will be chosen randomly to advance automatically so that the number of people playing in that round is divisible by 4.  (This will ensure that each game that is played will have 4 players.)  How many total games will be played in the tournament?

{Scroll down for solutions}

--------------------------------------------------

*I love card games.  I also love activities that help people learn and/or increase their brain power.  SET is a card game that is simple to learn, fun for a wide range of players (recommended for players 6 years old and above), and helps build brain power.  As a teacher and card game enthusiast, I give SET my full recommendation.  (And now SET-Junior is available for younger players.)

--------------------------------------------------

Solutions/Explanations

Method 1 (slow)

To solve the problem, many people get out pencil and paper and begin calculating the results, round by round.  For each round, they keep track of the number of games, and then add these up at the end to find the answer to the question.  This method can be quite time consuming.  One such solution is shown:

Method 2 (fast)

Of course, there is another way to solve this problem – a way that takes much less time.  In the previous solution, the focus was on the WINNERS from each round.  (The number of players in rounds 2 – 5 was determined by figuring out how many winners there were in the previous round and adding this amount to the number of people who were lucky enough to advance automatically.)  A much quicker solution involves focusing on LOSERS:

We know each game that is played will have 4 players.  This means that each game is guaranteed to have 3 losers.  The tournament begins with 340 players, and 339 of those players will be a loser at some point.  All we need to do is divide 339 by 3 to find our answer: There will be 113 games.

When looking at a tough problem, we often hurt ourselves by focusing on the wrong part of the problem.  Sometimes it is helpful to take a step back, reexamine the problem, and figure out if our focus is in the right place.


For math tutorials and silly math songs, visit www.onlinemathpro.com




Affiliate Disclosure

Percentiles?!

We take tests throughout our school years and most of us understand how the scoring works.  Take the number of points earned, divide by the number of points possible, move the decimal a couple of spaces right and, voila, we have a percent that represents our score.  (If you want to learn more about percents, there's a song for that.)

But then, once a year, we take a standardized test and when the score is reported, we are given a percentile.  What in the world is a percentile?

Unfortunately, we don’t learn much about percentiles in school.  However, we do learn about the median, which I discussed in my last blog entry, and also about quartiles.  Let’s review briefly:

If all of the values in a data set are arranged in order, then the middle number (or, if there are two middle numbers, their average) is the median.  This number divides the data set into two equal groups:  Half of the values in the set are above the median, and half of the values are below the median.

If all of the values in a data set are arranged in order, then the quartiles (there are three of them) are the values that split the data set into four equal groups:

One-fourth (25%) of the values in the set are less than or equal to the 1^st quartile, and three-fourths (75%) of the values are greater than or equal to the 1^st quartile.

Half (50%) of the values in the set are less than or equal to the 2^nd quartile, and half (50%) of the values are greater than or equal to the 2^nd quartile.  (The 2^nd quartile is the same as the median.)

Three-fourths (75%) of the values in the set are less than or equal to the 3^rd quartile, and one-fourth (25%) of the values are greater than or equal to the 3^rd quartile.

 

If you understand how quartiles work, then it’s not much of a leap to understand percentiles.

If all of the values in a data set are arranged in order, then the percentiles (there are 99 of them) are the values that split the data into 100 equal groups.  (Note: some people call these centiles.)

1% of the values in the set are less than or equal to the 1^st percentile, and 99% of the values are greater than or equal to the 1^st percentile.

2% of the values in the set are less than or equal to the 2^nd percentile, and 98% of the values are greater than or equal to the 2^nd percentile.

And so on…

Because they split the data set into so many groups, percentiles are only useful in analyzing data sets that are very large – like the number of students who take standardized tests.

If we understand how percentiles work, then interpreting standardized test scores becomes much easier.

If a score is reported as 20^th percentile, then 20% of test-takers scored at or below that level.

If a score is reported as 90^th percentile, then 90% of test-takers scored at or below that level.

That wasn’t so hard was it?

So, why are there so many people who struggle with percentiles?

I blame politicians.

 
In the recent years, politicians arguing over tax rates have consistently discussed “Americans in the top two percent of earners”.  They would do us all a great service if they would instead argue about “Americans at or above the 98^th percentile of earnings”.  It wouldn’t make politicians any more likely to work well together, but at least people would understand standardized test scores better!



Affiliate Disclosure

NFL Salaries (part 2) - Is Average really Average?

In my last blog entry, I discussed average NFL salaries.  I had initially hoped to talk about median salaries but, with my limited internet searching skills, was not able to find enough information about the history of median NFL salaries.

When analyzing data, why do we often focus on the median instead of the average (mean)?

The fact is that the average (mean) is not always so average (typical).  If we want to get a good picture of what a typical NFL player makes, we should analyze median salaries. 

Indianapolis Colts – 2012 Salaries
Rk	Player	Salary
1	Dwight Freeney	$19,035,000
2	Antoine Bethea	$6,225,000
3	Robert Mathis	$5,750,000
4	Andrew Luck	$4,015,000
5	Reggie Wayne	$3,500,000
6	Adam Vinatieri	$3,200,000
7	Antonio Johnson	$2,000,000
8	Samson Satele	$1,866,667
9	Anthony Castonzo	$1,818,431
10	Cory Redding	$1,716,667
11	Jerry Hughes	$1,691,250
12	Winston Justice	$1,500,000
13	Austin Collie	$1,417,388
14	Donald Brown	$1,385,000
15	Pat McAfee	$1,322,500
16	Drew Stanton	$1,225,000
17	Pat Angerer	$1,132,500
18	Tom Zbikowski	$1,033,333
19	Coby Fleener	$971,000
20	Vontae Davis	$957,500
21	Fili Moala	$873,000
22	Trai Essex	$825,000
23	Mewelde Moore	$825,000
24	Jerraud Powers	$784,625
25	Darius Butler	$770,000
26	Mike McGlynn	$700,000
27	Moise Fokou	$627,438
28	Drake Nevis	$622,278
29	Donnie Avery	$615,000
30	Justin King	$615,000
31	Delone Carter	$571,301
32	Dwayne Allen	$565,826
33	Ricardo Mathews	$551,628
34	Kavell Conner	$551,309
35	Josh Gordy	$540,000
36	Jeff Linkenbach	$540,000
37	Seth Olsen	$540,000
38	Matt Overton	$540,000
39	T.Y. Hilton	$521,750
40	Sergio Brown	$490,000
41	Cassius Vaughn	$490,000
42	Joe Lefeged	$467,667
43	Kris Adams	$465,000
44	Mario Addison	$465,000
45	Martin Tevaseu	$465,000
46	Joe Reitz	$455,000
47	Vick Ballard	$426,140
48	LaVon Brazill	$409,670
49	Chandler Harnish	$401,474
50	Mario Harvey	$390,000
51	Justin Hickman	$390,000
52	Dominique Jones	$390,000
53	Nathan Palmer	$390,000
54	Bradley Sowell	$390,000
55	Jerrell Freeman	$375,000
	Source: Fox Sports*

As an example, let’s look at the salaries for the 2012 Indianapolis Colts.

This team is somewhat of an anomaly because they have the highest-paid player in the league (Dwight Freeney) and the lowest total team payroll in the league.  This makes the Colts an exaggerated microcosm of the NFL – a league in which most players earn a high salary, but a few players earn an insanely high salary.

So, how much money does the typical player on the Colts earn?

If we use average salary (rounded to the nearest dollar) the answer is $1,469,115 – but is this really typical?  43 players, representing78 percent of the team, make less than this amount.

The average salary is artificially inflated by the top-earning players on the list.  If we take the top three salaries (Freeney, Bathea, and Mathis) off of the list and recalculate the average, it becomes $957,526 (again, rounded to the nearest dollar).  Removing three players caused the average salary to decrease by about 35%.

It seems more reasonable to use median salary and state that the typical player on this team earns $622,278.  This amount is not artificially inflated by the top earners on the list.  If we remove the top three salaries, the median becomes $615,000.  This number is about 1% lower than the original median.

If the example of the Colts does not convince you that median is the better way to go, consider this example:

A group of one hundred people decide to look in their wallets to see how much money each person is carrying.  Ninety-nine of the people are carrying $1 and one person is carrying $1,000,000.  The average amount of money per person is $10,000.99 ($1,000,099 ÷ 100).  Does this mean that the typical person in this group is carrying over ten thousand dollars in his/her wallet?  Of course not.

In situations like these, median gives the best approximation of the typical case – the average is not really very average.

----------------------------------------------------------------------------

For Math Tutorials and Silly Math Songs, visit

www.onlinemathpro.com

(*Source: http://msn.foxsports.com/nfl/team/indianapolis-colts/salary/67048?q=indianapolis-colts)

Affiliate Disclosure