Please help math question!

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<p>no worries, this doesn't have viruses i promise LLOL</p>

<p>I was wondering, how do we know if this is a direct variation or indirect variation/ proportion question?
and what is the correct answer? I don't have the key</p>

<p>[url=<a href="http://www.mediafire.com/?4ua4j9x2cqta6m3%5Dkjdf.jpeg.pdf%5B/url"&gt;http://www.mediafire.com/?4ua4j9x2cqta6m3]kjdf.jpeg.pdf[/url&lt;/a&gt;]&lt;/p>

<p>Thank you!</p>

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<p>It would probably be better to upload as a picture or just write it out or something, so that people do not have to download that…</p>

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<p>i can’t convert it though :(</p>

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<p>A certain job take p persons h hours to complete. How long would it take n persons to complete the same job.</p>

<p>Job = ph;
If p = 2 and h = 12 then Job = 24;
Now let’s set n = 4;
24 = 4(h2);
h2 = 6; (this is what we want the answer to be)</p>

<p>Let’s review our variables: p = 2; n = 4; h = 12;</p>

<p>(C) hp/n = 6;</p>

<p>The answer, I believe, is C.</p>

<p>800 Math, 80 PSAT Math, 800 SAT II Math II…and I still don’t know how to solve it algebraically. XD</p>

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<p>^ How do you arbitrarily choose n to be 4? Why is it 2x p?</p>

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<p>Ideally you want the “job” to be divisible by the new n so that you don’t have to work with decimals. Other than that, you can set n to anything you want…since the equation should hold true for all values n. If you’re feeling extra paranoid on test day, think up a few other values and test them just in case a certain answer choice doesn’t “accidentally” work for your arbitrary value.</p>

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<p>If you plug the number you made up into ALL of the answer choices, and only ONE of them matches, then you are done. The only time you have to test other values is if your first value produces more than one match.</p>

<p>"800 Math, 80 PSAT Math, 800 SAT II Math II…and I still don’t know how to solve it algebraically. XD " – that’s pretty funny.</p>

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<p>Okay, thanks. Does anyone know how to do this algebraically? My mind tends to work better that way.</p>

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<p>Here goes. I think I didnt make a mistake but this is how I solved it with algebra.</p>

<p>So I call the certain job or whatever j. If you Think of j as output then the output or job = p (persons)*h (time) *r (rate). </p>

<p>Now, for n number of people j remains the same as does r, but h changes so let me call h now h2.</p>

<p>j = n<em>h2</em>r and from the first equation j = p<em>h</em>r</p>

<p>so since j=j: n<em>h2</em>r= p<em>h</em>r</p>

<p>therefore h2 = hp/n</p>

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<p>Here’s a quick algebraic solution: </p>

<p>Solve for x in the following equation: hp = xn</p>

<p>Simply divide both sides by n to get hp/n = x, choice (C).</p>

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<p>Remark: It is easy to tell that this is an inverse proportion because as one variable gets larger, the other gets smaller.</p>

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<p>Thank you! That made much more sense.</p>