Fraction Word Problem

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phylispatt

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In an adult concominium complex, 2/3 of the men are married to 3/5 of the women. What part of the residents are married?

(Note from instructor: There are many ways to solve this problem. At least one is algebraic, at least two are arithmetic, at least one solves the problem pictorially. There is no set way to do this problem. Just see what you think. See what you can do with it. Have fun!) Answer is worth 25 big fat extra credit points!!

Rebecca
 
phylispatt said:
In an adult concominium complex, 2/3 of the men are married to 3/5 of the women. What part of the residents are married?

(Note from instructor: There are many ways to solve this problem. At least one is algebraic, at least two are arithmetic, at least one solves the problem pictorially. There is no set way to do this problem. Just see what you think. See what you can do with it. Have fun!) Answer is worth 25 big fat extra credit points!!

Rebecca
Click to expand...

For 25 big fat extra credit points, I assume you've done SOMETHING!!

Please show us your thoughts and your work, and we'll be happy to add our comments.
 
So far, I have a big fat ZERO! I'm a 61 year old algebra student relying on passing my math equivelancy to earn a BA in social work, sitting in a room of 30 18-10 year olds, some of whom are fresh from high school algebra. My last math class was 1962, long before pre-algebra was thought of.

I've tried to think of this problem from a proportion standpoint of 2/3 X 3/5, or a division problem 2/3 divided ty 3/5, but I don't seem to get anywhere. From a practical stand point, I have worked with Senior Citizens for 6 years and know that there are more women in senior complexes than men but that isn't included in the problem.

Also, the instructor "advised" us today to keep in mind whatever answer we came up with, since the married "residents" would be coupled, that would make the answer reduced by half. So if we came up with 60% as the total married people, they would represent 1/2 of that number of residents.

Sorry, but I'm lost from any other angle.

Rebecca
 
phylispatt said:
Also, the instructor "advised" us today to keep in mind whatever answer we came up with, since the married "residents" would be coupled, that would make the answer reduced by half. So if we came up with 60% as the total married people, they would represent 1/2 of that number of residents.
Click to expand...
Wonder why the instructor told you that...

Take a simple case:
9 men only; so 2/3 * 9 = 6 are married
So there must be also 6 women married, right? (unless some old guy cheated and married 2 old women!)
And 6 is 3/5 of 10: so there are 10 women in total.

So 12 of 19 residents are married, or 12/19 = .63157... or 63% rounded.
No need to cut nothing!!
 
Right, I get that... and I didn't see her reasoning in saying it. But, what I don't get is how we are expected to come up with a number of residents, married or single, since we don't have a starting number. Say, you had 100 Residents, if 2/3 of the men were married to 3/5 of the women, you would at least have a starting point, right? But, since we don't know how many men or women or total residents to start with, where can we take this question/problem? Or am I making it more difficult than it needs to be?

All comments gratefully accepted. I haven't lived 61 years without earning a thick skin...but, unfortunately, a thick head as well. What am I not seeing?

Rebecca
 
No, Rebecca: you are not expected to come up with actual number of residents; here's the problem:
"What part of the residents are married?". Don't argue: YOU typed that!!

So you only need to set up a test case to find the percentage: which I did for you: 63%;
that will not change, regardless of the number of residents...kapish?
 
Hello, Rebecca!

I have an algebraic solution . . . Hope you can follow it.

In an adult concominium complex, 2/3 of the men are married to 3/5 of the women.
What part of the residents are married?
Click to expand...

Let \(\displaystyle M\) = number of men.
Let \(\displaystyle W\) = number of women.
. . The total number of people is: .\(\displaystyle M+W\)

\(\displaystyle \text{Two-thirds of the men are married: }\:\tfrac{2}{3}M \text{ married men.}\)
\(\displaystyle \text{Three-fifths of the women are married: }\:\tfrac{3}{5}W \text{ married women.}\)
. . \(\displaystyle \text{The total number of married people is: }\:\tfrac{2}{3}M + \tfrac{3}{5}W\)

\(\displaystyle \text{The portion (fraction) of married people is: }\:\frac{\frac{2}{3}M + \frac{3}{5}W}{M+W}\) .[1]


The married men and married women are married to each other, so their numbers are equal.
. . \(\displaystyle \text{So we have: }\:\tfrac{2}{3}M \:=\:\tfrac{3}{5}W\)

\(\displaystyle \text{Multiply }\tfrac{3}{2}\!:\;\; \tfrac{3}{2}\left(\tfrac{2}{3}M\right) \:=\:\tfrac{3}{2}\left(\tfrac{3}{5}W\right) \quad\Rightarrow\quad M \:=\:\tfrac{9}{10}W\)


Substitute this into [1]:

. . \(\displaystyle \frac{\frac{2}{3}\left(\frac{9}{10}W\right) + \frac{3}{5}W} {\frac{9}{10}W + W} \;=\;\frac{\frac{3}{5}W + \frac{3}{5}W}{\frac{19}{10}W} \;=\;\frac{\frac{6}{5}W}{\frac{19}{10}W} \;=\;\frac{\frac{6}{5}}{\frac{19}{10}} \;=\;\frac{6}{5}\cdot\frac{10}{19} \;=\;\boxed{\frac{12}{19}}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Let's test this on a specific group of people.

Suppose there are 18 men and 20 women.

\(\displaystyle \text{Then: }\:\tfrac{2}{3}\cdot18 \:=\:12\text{ men are husbands.}\)

. \(\displaystyle \text{and: }\:\tfrac{3}{5}\cdot20 \:=\:12\text{ women are their wives.}\)


There are 24 married people among 38 residents.

. . \(\displaystyle \text{The fraction of married people is: }\:\frac{24}{38} \:=\:\frac{12}{19}\)

 
Thank you both very much! It appears that in my inexperience, I was trying to overthink it by assuming I had to come up with a "headcount" instead of an expression to apply to any given number.

I very much appreciate the help and you will probably be hearing from me again. In the meantime, I took the opportunity to check out a few of the other listings and found that I wasn't the only one overthinking things. I guess the KISS theory wins again.

Again, thanks much and have a great weekend.

Rebecca
 
In an adult concominium complex, 2/3 of the men are married to 3/5 of the women. What part of the residents are married?

Two ways:

--2/3rds of the men in a village are married and 3/5ths of the women are married. What percentage of the total population are married?

Lets take it step by step.
Blow your screen up to full size.

2/3 of the men are married
3/5 of the women are married
1--Let M = men, W = women, MM = married men, and MW = married women
2--Let M + W = a unit population of 1 or M + W = 1 or
............................MEN....................................................WOMEN
........................M = (1-W)...................and.......................W = (1-M).
3--...............2/3M are married..............and..................3/5W are married
4--Therefore, since there is 1 married man for every married woman, the number of married men must be equal to the number of married women and
.....................2M/3 = 3W/5.................and.....................2M/3 = 3W/5
5--Then........2M/3 = 3(1-M)/5.......................................2(1-W)/3 = 3W/5
6--.............2M/3 + 3M/5 = 3/5...................................2W/3 + 3W/5 = 2/3
7--............10M + 9M)/15 = 3/5.................................(10W + 9W)/15 = 2/3
8--....................19M = 9.................................................19W = 10
9--.....M = 9/19ths of the population...................W = 10/19ths of the population
10--.......MM = 2/3(9/19) = 6/19ths........................Mw = 3/5(10/19) = 6/19ths
11--Therefore, MM + MW = 12/19ths of the population.

Simpler
1--Assume M + W = 1
2--Given 2M/3 = 3W/5
3--Then, 10M/15 - 9W/15 = 0;.
4--Then, 150M - 135W = 0
5--Multiply (1) by 135 and add to (4) giving
....285M = 135 or M = 135/285 of the total.
6--Thus, W = 150/285 of the total.
7--(2/3)135/285 = 90/285 are married
8--(3/5)150/285 = 90/285 are married
9--Thus, (90 + 90)/285 = 180/285 = 12/19ths are married.
 
Thank you very much for your help. It appears my inexperience got in my way, assuming I didn't have enough information to solve the problem, when, in fact, there was enough information in the instructor's statement of the problem. I just needed to read and apply the facts to come up with an expression instead of a headcount.

Live and learn.

Rebecca
 
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