Need help with distance problems!

Test Preparation SAT Preparation
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<p>Having difficulty with these problems and I would really appreciate some help.</p>

<p>So here are two practice problems from Gruber's SAT Math Workbook (Second Edition).
They are #3 and #5 from section 2-31 "To do problems dealing with distance"</p>

<h1>3) A boat can reach its destination in 3 hours. If it goes twice as fast, it can reach 10 miles more than its destination, in 2 hours. How fast does it originally go?</h1>

<h1>5) A man can drive a certain distance in 5 hours. If he increased his speed by 10 miles per hour, he could travel the same distance in 4 and 1/6 hours. What is the distance he travels?</h1>

2

<h1>3) The answer, as you probably know, is 10 miles per hour. But do you get to 10 MPH? Here’s the lowdown :):</h1>

<p>Distance = rate * time, or simply d=rt</p>

<p>**Scenario #1: **</p>

<p>distance = x <br>
rate = y (this is what you are solving for - “how fast does it originally go”?) <br>
time = 3 (“3 hours”)</p>

<p>thus,
x = 3y</p>

<p>**Scenario #2 **</p>

<p>distance = x + 10 (“it can reach 10 miles more”)
rate = 2y (“if it goes twice as fast”)
time = 2 (“2 hours”)</p>

<p>thus,
x + 10 = 4y</p>

<p>Now, solve for y, since y = rate (see scenario #1). </p>

<p>Here are your two equations:</p>

<p>x + 10 = 4y
x = 3y </p>

<p>Substitute 3y for x: </p>

<p>3y + 10 = 4y
y = 10 = rate of boat in scenario #1. </p>

<p>Ultimately, this is a classic distance/rate problem a trick mixed in - you have to set up a system of equations and do some substitution.</p>

<p>I’ll let you try #5 on your own. It’s the same problem, only with different numbers! Good luck, and I hope my explanation helped :)!</p>

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<p>

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<p>Rephrase the problem as A boat can reach its destination in 3 hours. It can reach 10 miles farther in 4 hours. How fast does it go?</p>

<p>The SAT is a reasoning test!</p>

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<p>^ nice…</p>

<p>But can you come up with as slick a way to do the other one? I don’t see one…which is why the first one is MUCH more SAT-like than the second one…Read, think, play…</p>

<p>Here’s a challenge: find me a college-board problem (actual SAT, PSAT, blue book, online course, etc) that REQUIRES you to solve a system of two equations and two unknowns.</p>

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<p>Thank you so much for your help IceQube. I am now able to solve both of the problems with ease.</p>

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<p>^No problem :)! I’m always ready to help :).</p>

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<p>

</p>

<p>Haha, I fully expected someone to ask me that. The second problem does not permit a very simple rewrite to find the solution directly. Every problem is different. However, this is how I would approach it:</p>

<p>

</p>

<p>First, let’s get rid of the fractions as the only difficulty comes from 4 1/6. I multiply everything by 6 and set x as the original speed. </p>

<p>30x = 25(x+10)
30x - 25x = 250
x = 50 </p>

<p>In this case, the traditional method is quick and logical, especially after eliminating the fraction. As far as a reasoning method, here is a different approach, namely an analysis of the ratios of rate and time.</p>

<p>Since we know that d is constant, we know that an increase in speed causes a proportional decrease in time. In this case, the decrease in time was 5/6, representing a 20 percent decrease. The equation should be “balanced” by an equal 20 percent increase in speed . The solution then becomes:</p>

<p>10 = .20x
x = 50. </p>

<p>PS For what is worth, the way I solved is “weirder” as all I did was looking at the ratios in this manner:</p>

<p>I looked for a fraction proportional to 5/6 where the two terms would yield a difference of 10. The answer was immediate: 50/60. Unfortunately explaining why this works would take longer than solving it the traditional way.</p>

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<p>I didn’t read this thread… but… for all distance questions, the setup goes like this…</p>

<ol>
<li><p>Plug everything in distance = time x speed formula. (you will probably need subscripts for two equations)</p></li>
<li><p>Cancel and set like terms equal to each other.</p></li>
<li><p>Solve.</p></li>
</ol>