Saturday, August 4, 2012

Find Math Word Problems To Be Difficult?



It is always inspiring to notice some difficulties that math sometimes sets forth. To put it in a better way, I would rather call it an occasional math challenge. Have you ever stumbled upon a challenging math problem? A word problem perhaps?

Some individuals suffer from word problems due to infrequent translations from words to math symbols. Believe it or not, mathematics is most importantly a language of our everyday life, just as music is a language of our soul. No matter what we calculate or estimate, words describing real-life problems give us a good perspective of mathematical symbols that we can use. Very often, translation to mathematical symbols simplifies a given problem that might seem wordy or blurry at first. By understanding and mastering the uses of algebra, geometry and trigonometry we can implement our vocabulary of mathematical symbols and successfully apply them to real-life word problems. For example, we can consider a sample word problem (assuming one has elementary knowledge of math symbols):

A baseball team won 50 of the first 92 games played in a season. If the season consists of 152 games, how many more games must the team win to finish the season winning 62.5% of games played?

First, we have to understand what solution method this problem requires. If one reads it very carefully, it is easy to see that this is a direct proportion problem. Direct proportion involves two quantities, each increasing with respect to another increasing, or each decreasing with respect to another decreasing. Each direct proportion includes a constant (a quantity or number that does not change) such as the total number of games played (152 games in a season), or 100% of total games played. In the context of this problem, we need to know how many more games must be won (when playing additional 60 games) to match the 62.5% of all season games won. We must set up two fractions in direct proportion. The number of additional games that need to be won will be our x - the variable that we need to find.

We set it up as follows:

We know how many games we won out of 92 games already played = 50. We add the variable x to 50. So, all games won in a season of 152 games will become 50 + x. We must divide this by the total number of games played (152) in order to express how many games are won out of the total games played.

The second fraction has everything that we need: 62.5% out of 100%, which is a translation of numbers to percent value. So, we divide 62.5% by 100% just as we did with the first fraction since 62.5% will represent the percentage of games won out of the total 152 games played (same as 50 + x divided by 152).

So (50 + x) equals 62.5%, and 152 equals 100%. The percentage is just another way of expressing the number of games won divided by the total number of games played.

Now, we write what we described symbolically:

[ (50 + x) / 152 ] = [ 62.5 / 100 ] 

and then we cross-multiply (to get the product of means to equal the product of extremes), and solve for x to get
x = 45. So, in order to finish with a 62.5% winning percentage of the entire game season consisting of 152 games - having already won 50 games out of 92 games played - we must win 45 more games in the remaining 60 games of the season.

Of course, explaining word problems in person is much simpler. Words and verbal communication often simplify and shorten written description and explanation. There are many word problems that involve fractions and other skills, which frequently require some thinking.

This is why our company has excellent tutors in Mathematics. We are here to help you achieve your goals.

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