<p>I'm going through Dr. Chung's SAT math book and I came across this question which I completely don't understand. Math nerds please help!</p>
<p>If you have the book: pg 130 practice test 2 section 3 question 20</p>
<p>Question: maxim telephone company charges k cents for the first t minutes of a call and charges for any additional time at the rate of r cents per minute. If a certain customer pays $10 which of the following could be the length of that phone call in minutes?</p>
<p>Answer:</p>
<p>D) (1000-k)/r + t</p>
<p>explanation:
k cents --> up to t minutes
for the (1000-k) cents ---> r cents per minute
it follows that the minute is (1000-k)/r
therefore total length of phone call is (1000-k)/r + t</p>
<p>Just read the explanation thoroughly and you should understand it. You pay k cents for the first t minutes, so you pay 1000-k cents for the remaining minutes. The 1000-k cents is at a rate of r cents/min, so the number of “extra” minutes is (1000-k)/r. Add this to t to get (1000-k)/r + t.</p>
<p>The other possible value for the # of minutes occurs when t < 10, in which the answer is some other number.</p>
<p>Edit: If you want a purely algebraic solution, refer to fogcity’s solution below.</p>
<p>Let L be the length of the phone call. Assume that L > t.</p>
<p>The cost of this phone call (in cents) is k + (L - t)r.</p>
<p>We know that this is 1000 (converting $10 to cents). So 1000 = k + (L - t) r. Solve for L:</p>
<p>L = (1000 - k)/r + t</p>
<p>Try this (and with any problem with variables in both the question and answer choices that has you stumped):</p>
<p>Q. Maxim telephone company charges k cents for the first t minutes of a call and charges for any additional time at the rate of r cents per minute. If a certain customer pays $10 which of the following could be the length of that phone call in minutes?</p>
<p>Suppose: The company charges 500 cents (k = 500) for the first 10 minutes (t = 10) and charges for any additional time 100 cents (r = 100) per minute. If a certain customer pays $10, what is the length of the phone call.</p>
<ol>
<li>$10 = 1000 cents available to spend (get our units the same)</li>
<li>first 10 minutes, spent 500 cents</li>
<li>next minute, spent 100 cents (11 minutes total, 600 cents spent)</li>
<li>next minute, spent 100 cents (12 minutes total, 700 cents spent)</li>
<li>next minute, spent 100 cents (13 minutes total, 800 cents spent)</li>
<li>next minute, spent 100 cents (14 minutes total, 900 cents spent)</li>
<li>next minute, spent 100 cents (15 minutes total, 1000 cents spent)</li>
</ol>
<p>15 minutes!</p>
<p>Now to the equations (check all of them and D works):</p>
<p>(1000 - 500) / 100 + 10
500 /100 + 10
5 + 10 = 15 (Equation is valid for our numbers, so D is correct!)</p>
<p>@flashandcrash, that solution works, but in my opinion it involves too much “bashing.” It becomes a much longer and time-consuming solution.</p>
<p>I suggest simply working with the variables (you have to get used to it in college anyway).</p>
<p>I picked really simple-to-demonstrate numbers that would be clear, even though long-winded. Its very fast to move through picking larger numbers closer to 1000 in the first place. For students having trouble with the algebra, it’s a really simple approach that saves time. The SAT isn’t the time to learn Math, better to just focus on a foolproof way of getting it right. Making up your own numbers for variable problems or plugging in answer choices when there is confusion in algebra is always a smart move.</p>
<p>Edit: Yes, it is ideal to become comfortable with variables. But it’s good to see how this can be worked out in case that escapes a student on the test. Plus working it out this way can help show how the equation is derived for students who then study it.</p>
<p>Yeah, there are certain (certain) problems that I would attempt plugging in random numbers. But what if there is a problem like:</p>
<p>Q: n is a positive integer such that when n is divided by 5, a remainder of 3 results. When n is divided by 11, a remainder of 9 results. What is the smallest possible value of n?</p>
<p>Note: this isn’t an SAT question, just some random problem I made up…</p>
<p>I honestly would again suggest plugging in. First number after 11 with a remainder of 9? 20. Remainder of 0 with 5. That won’t work. Next, 31. Nope. 42. Nope. 53, yup.</p>
<p>Or just note that n+2 must be divisible by 5 and 11 (55), therefore n=53 is the smallest positive integer solution.</p>
<p>I don’t think that’s entirely intuitive. It requires an attention to detail and an understanding of why remainders work the way that they do which most students will probably not remember under the stressful SAT situation. Which is why I suggest (for problems when the algebra isn’t intuitive) make up numbers that follow the rules given and just churn through them.</p>
<p>If the algebra is intuitive, there’s nothing wrong with using it, however. :)</p>
<p>Yeah, it really depends when to plug in numbers. Very rarely have I ever done that though. I’d only do that if the algebra is too long/can’t see a solution right away.</p>