Fractions Word Problem Help: Solving 800/512

[Duckfan]

I'm kinda glad I found this place. Hope it will help. Prepping for state exam and math is on test.

Getting stuck on fraction word problem, and probably many more to come. But struggling to figure how to set up this prob:

Of 800 adults surveyed, 512 replied that they have used drugs. What fractional part of the adults, blah, blah...expressed in lowest terms.

Believing this is division problem. Saw what answer is...a mixed number but can't figure how to get to mixed number. I know I don't multiply. I did 800 over 1 div. by 512 over 1 then flip to multiply?

 

[MarkFL]

Hello and welcome to MHB, Duckfan! (Wave)

To find the fractional part $F$ of the adults in the survey who stated they had used drugs, we would take the number who stated drug use, and divide by the total:

\(\displaystyle F=\frac{512}{800}\)

Now, to reduce the fraction, we can employ a prime factorization of the numerator and denominator:

\(\displaystyle F=\frac{2^9}{2^5\cdot5^2}\)

Now we can reduce:

\(\displaystyle F=\frac{\cancel{2^5}\cdot2^4}{\cancel{2^5}\cdot5^2}=\frac{2^4}{5^2}=\frac{16}{25}\)

Does this make sense?

 

[Duckfan]

Hello and welcome to MHB, Duckfan! (Wave)

To find the fractional part $F$ of the adults in the survey who stated they had used drugs, we would take the number who stated drug use, and divide by the total:

\(\displaystyle F=\frac{512}{800}\)

Now, to reduce the fraction, we can employ a prime factorization of the numerator and denominator:

\(\displaystyle F=\frac{2^9}{2^5\cdot5^2}\)

Now we can reduce:

\(\displaystyle F=\frac{\cancel{2^5}\cdot2^4}{\cancel{2^5}\cdot5^2}=\frac{2^4}{5^2}=\frac{16}{25}\)

Does this make sense?

Well part of it does. First I did it backwards. I tried to divide 800/512. And I used 8 as LCD for N/D since I thought it was "lowest LCD". I goofed there. But what do I know. I was in a galaxy long time ago, far, far away since I really had to do this. I also understand it has to be factored with LCD for numerator and denominator(these quote marks are making it hard for me to read your response). That I understand. I noticed you squared. I don't understand that part unless you are multiplying 2x9 in numerator? Top LCD is different than bottom. The answer is correct as I have it in back of book. But seeing you do what looks like squaring is throwing me off.

 

[MarkFL]

I rewrote 512 as 2 to the 9th power:

\(\displaystyle 512=2^9\)

If the exponents are throwing you off, then think of it as:

\(\displaystyle F=\frac{512}{800}=\frac{32\cdot16}{32\cdot25}=\frac{\cancel{32}\cdot16}{\cancel{32}\cdot25}=\frac{16}{25}\)

I like to use prime factorizations as it works in all cases without having to know the GCD. :)

 

[Duckfan]

I rewrote 512 as 2 to the 9th power:

\(\displaystyle 512=2^9\)

If the exponents are throwing you off, then think of it as:

\(\displaystyle F=\frac{512}{800}=\frac{32\cdot16}{32\cdot25}=\frac{\cancel{32}\cdot16}{\cancel{32}\cdot25}=\frac{16}{25}\)

I like to use prime factorizations as it works in all cases without having to know the GCD. :)

Okay, now I see it in clearer terms. So I guess I didn't use proper factorization? I've been programmed to use "lowest" LCD which is why I used 8. So 32 is correct LCD? I want to say I get the gist of what I'm supposed to do as far as finding factor for both numerator and denominator, but it's when I get wrong answer or when something looks wrong, I get lost then get frustrated.
Thank you for your advice. I'll work the other problems and try to report back. Thank you. (Cool)

 

[MarkFL]

When reducing fractions, you look for the GCD, which is the largest quantity that divides both the numerator and the denominator...in the case of 512 and 800, this is 32...no larger integer will go into both. But, if you use a smaller common divisor, you can still reduce the fraction, you will just have to reduce again. For example, if you had used 8, then you would have:

\(\displaystyle F=\frac{512}{800}=\frac{8\cdot64}{8\cdot100}=\frac{64}{100}\)

Now we can see that 4 goes into both, so we could write:

\(\displaystyle F=\frac{64}{100}=\frac{4\cdot16}{4\cdot25}=\frac{16}{25}\)

Using 8 as a common divisor and then 4 is equivalent to using $8\cdot4=32$, but using 32 is quicker, as it only requires 1 step...however, there's nothing wrong with using multiple steps.

If you use prime factorizations, then you will always get the GCD. :D

 

[Duckfan]

Hope this is not a stupid question. What is the simplest way to find a factored number that goes into a numerator and denominator if it's 3 digits?

Like 240 over 600 or 192 over 268, etc.

 

[MarkFL]

Hope this is not a stupid question. What is the simplest way to find a factored number that goes into a numerator and denominator if it's 3 digits?

Like 240 over 600 or 192 over 268, etc.

Well, let's start with $\dfrac{240}{600}$

We see that:

\(\displaystyle 240=(24)(10)=(8)(3)(2)(5)=(2)(2)(2)(3)(2)(5)=2^4\cdot3\cdot5\)

\(\displaystyle 600=(60)(10)=(12)(5)(2)(5)=(3)(4)(5)(2)(5)=(3)(2)(2)(5)(2)(5)=2^3\cdot3\cdot5^2\)

It doesn't matter which factors you initially find, you will always end with the same prime factorization. Now, we want to grab the prime factors present in both numbers, and use the smaller exponent, hence:

\(\displaystyle \gcd(240,600)=2^3\cdot3\cdot5=120\)

And so we find:

\(\displaystyle \frac{240}{600}=\frac{120\cdot2}{120\cdot5}=\frac{2}{5}\)

See if you can use a similar method to reduce $\dfrac{192}{268}$. :)

 

[Duckfan]

Thanks Mark. I have been reviewing basic fractions past 2 weeks and have not worked with larger numbers yet. So that's reason for asking. Examples have been smaller and easier to work with and factor. Still working on exercises from last night. Not done yet. Thank you. I will try to provide samples of stuff when I do finish.