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11-04-28 10:09 PM
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Fraction Action! [Problem #17060]
If you add the denominator to both the numerator and denominator of a fraction, you get a new fraction that sometimes is larger than the
original fraction.
Find a fraction where the new fraction is four times as large as the original fraction after you add the denominator to both the numerator
and denominator.
Extra: In order for the new fraction to be n times greater than the original fraction, how many times larger than the original numerator
must the original denominator be? Express your answer in terms of n.
Comments and Sample Solutions
Fraction Action! proved to be a pretty challenging problem, and the number of solutions submitted was well below normal. Many of the solutions that we
did get used trial and error rather than algebra to find the fraction.
Part of what made the algebra hard was that you had to use two variables in your work - one to represent the numerator of the original fraction and one to
represent the denominator. That meant that all the algebra and equation work that followed involved keeping track of two variables. Additionally, the
equations you produced involved fractions with variable expressions in the denominators, which is something you probably have not had a lot of experience
with.
But we did have some folks who solved the problem successfully! Two of them are shown below. Take a look at the algebra that Ryan and Jaehoon each
did. Ryan got the equation into the form of two equal fractions, which means it will behave like a proportion. Since in any proportion the products of the
means and extremes are equal, Ryan used "cross-multiplication" to get rid of the fractions. Jaehoon approached it slightly differently, choosing instead to
multiply both sides of the equation by the common denominator of 2d, which cleared all the fractions. Both students had started with an original fraction of
n/d and discovered at the end of their equation work that d = 7n.
At that point Ryan chose to let n = 1 and solve for d, resulting in the fraction 1/7. Notice that had Ryan chosen to let n = 2 instead, the resulting fraction
would be 2/14, which would then reduce to 1/7. An n value of 3 would lead to 3/21, which would again reduce to 1/7. In fact, the result of d = 7n means
that as long as the original denominator is 7 times as large as the original numerator, the problem will work. And any fraction where the denominator is 7
times as large as the numerator will always reduce to 1/7.
Instead of picking a value for n, Jaehoon chose to substitute 7n for d in the original fraction and then reduce the result, which turns out to be 1/7. That's a
very convincing way to show that for all possible values of n, the original fraction will always reduce to 1/7.
Congratulations to Ryan, Jaehoon, and anyone else who used algebra to solve this challenging problem.
- Riz
From: Ryan V , age 13 , Long Valley Middle School
The starting fraction is 1/7.
You can set up an equation for this problem. The original fraction could be n/d.
The n stands for numerator and the d stands for denominator. To get a fraction four times larger than the original, the question says I have to add the
denominator to the numerator and denominator. So then the equation would look like (d+n) / (d+d) = 4n / d. After using cross multiplicaton(because it puts
the fractions into simpler but equal from), the new equation would look like
8nd = d(d+n). After cancelling the d, the equation would look like 8n = d+n. After subtracting the n, the equation turns into 7n =d. If you turn the n into 1 :
7(1)=d so d = 7. So then you get 1/7 and have to add the denominator of 7 to the numerator and denominator : (1+7) /( 7+7) = 8/14 = 4/7. 1/7 is the
original fraction.
1/7 x 4/1 = 4/7.
From: Jaehoon Y , age 15 , Rancho San Joaquin Middle School
I think that the fraction is 1/7.
To find the answer to the problem, I made an equation. When you add the denominator to both the numerator(n) and the denominator(d), then you have to
get four times the amount of the original fraction. It should look something like this:
(n+d)/2d=4(n/d).
http://mathforum.org/pows/solution.htm?publication=3783
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The Math Forum @ Drexel University
11-04-28 10:09 PM
Next, I change the equation into n/2d +1/2=4n/d because d/2d=1/2. Subtracting 1/2 from both sides, I get n/2d=4n/d-1/2. Multiplying 2d on both sides, I get
n=2d(4n/d-1/2) which adds up to n=8n-d.
Then, I subtract 8n from both sides, resulting in -7n=-d which then is 7n=d.
To prove that this is true, I replace all d with 7n in the unkown fraction. I get n/7n which is 1/7. Adding the denominator to both numerator and denominator,
I get 4/7, which is four times 1/7.
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