  # 2 Quick Math Questions

<p>So I have two math questions from a practice PSAT from McGraw Hill's prep book. They are:</p>

<ol>
<li><p>If the maximum value in the range of the function y=f(x) is 6, what is the maximum value in the range of the function y=3f(x-1)? I thought the answer was 15 (I tried plugging in 6 for x into both equations and got that the answer is 15, but it's actually 18).</p></li>
<li><p>A jar contains only red, white, and blue marbles. If the number of red marbles is 5/6 the number of white marbles and the number of red marbles is 6/7 the number of blue marbles, what is the least possible number of marbles in the jar? The answer is 101, but I don't understand the given explanation.</p></li>
</ol>

<p>To be honest, I don't really understand the answers for either questions, and I got 800 on the actual SAT (so don't stress-- the second one at least is more difficult than the vast majority of SAT questions). I believe the first question to be a typo, unless they had an explanation that verifies 18 as the correct answer. </p>

<p>For the second question, the answer I got was 108. I'll explain what I did, but perhaps some CC math whizzes can prove 101 as the right answer. This is how I did it:</p>

<p>There is a 5:6 ratio of red to white marbles and a 6:7 ratio of red to blue marbles. When I see the words "least possible number", I immediately look for the lowest common multiple between 5 and 6. This number happens to be 30, so I mark this as the lowest possible amount of red marbles that satisfies the boundaries of the problem. </p>

<p>To keep the ratios constant for the blue and white marbles, I multiply the white marble number (the 6 in 5/6) by 6, and I multiply the blue marble number (the 7 in 6:7) by 6. I get 36 and 42, respectively. When I add 30+36+42, I get 108. I don't know how they got 101, but if you have the explanation, I really would like to see it.</p>

<p>these two questions are like double as hard as the actual sat math questions lol</p>

<p>red = 5/6 x White
red = 6/7 x Blue</p>

<p>R,W,B are all integers</p>

<p>5/6 x W = 6/7 x B
35/6 x W = 6 x B
35 x W = 36 x B</p>

<p>For W and B to be integers, B = 35 and W = 36; thus R = 30</p>

<p>35 + 36 + 30 = </p>

<p>Hopefully I was clear.</p>

<p>maximum of f(x) = 6
Find max of 3f(x-1)</p>

<p>Think of fuction with max value of 6... simply -x^2 + 6</p>

<p>therefore 3f(x-1) = 3(-x^2 + 5) = -3x^2 + 15</p>

<p>Now find max of -3x^2 + 15 = f(x)</p>

<p>f'(x) = -6x = 0 => x = 0
f(0) = </p>

<p>I used a bit of calculus and I also got 15, so the answer to number 1.) is definitely 15. Your book is wrong.</p>

<p>Ok, thanks everyone! I was worried that there would be questions this hard on the PSAT and kinda freaked out....hopefully, that won't be the case!</p>

<p>For #1, the answer is indeed 18. I have a couple of methods, pick whichever one makes the most sense to you.</p>

<p>First, think of what 3*f(x-1) actually means. The (x-1) is within the function, so it is just a shift one unit to the right, which does not affect the maximum value. Therefore, the maximum value of f(x-1) is equal to the max value of f(x), which is 6. 3(6)=18, which is the answer.</p>

<p>GreedIsGood's method of making a function also works, but a mistake was made. (x-1)^2 does not equal x^2 - 1, it equals x^2 - 2x + 1. Forgetting to factoris a common mistake, but a mistake nonetheless. I will use the same original function as GIG, -x^2 + 6 in my example. Correctly worked with calculus, the problem becomes:
y= 3(f(x-1)) = 3(-(x-1)^2 + 6) = 3(-(x^2 - 2x + 1)+6) = -3x^2 + 6x + 15
Now find max of y:
y'= -6x + 6 = 0 => x=1.
The value of 3(f(x-1)) is 18.</p>

<p>The correct answer is 18, but nonetheless, the others are correct in saying that both of these problems are much harder than one would expect to find on the SAT/PSAT.</p>

<p>Great job everyone who could help the OP.</p>

<p>

</p>

<p>Generally, problems in prep books are purposely made more difficult to "adequately" prepare you, but I don't think this is a good way to go about it. If you are practicing for the PSAT, I recommend buying previous years' PSAT adminstrations from the College Board (from their website) which cost only \$3, and the Answers also cost only \$3.</p>

<p>^Really? That's good to know! I'm only a rising sophomore, but I'll probably have exhausted the PR 11 tests and BB by the time I'm a junior and I'll be in need of those before taking my second PSAT.</p>

<p>I think those questions are fair game difficulty-wise for the PSAT. There were some pretty tricky ones in past years. </p>

<p>The first one is quite simple: if y=3f(x-1), think how y differs from f(x). the (x-1) means the graph is just shifted right one unit, so range (vertical values) is not affected by that. the 3 makes the graph three times as tall, so if the highest value of f(x) is 6, three times that is 18.</p>

<p>The second is also relatively simple. They give both the proportions in terms of red marbles. If the number of red marbles is 5/6ths of one number and 6/7ths of another number, it must be a common multiple of both 5 and 6. The LCM of 5 and 6 is 30, so that is the number of reds. Then just multiple 30 by 6/5 to get the number of whites and by 7/6 to get the number of blues, 36 and 35 respectively, and add them all up to 101.</p>

<p>Thank you MathBeast! The first explanation was very clear and concise. Also, thanks for the advice Schoolisfun; I was actually thinking about doing that, but I forgot about it until you posted! :)</p>

<p>

</p>

<p>Oh that's right! I forgot about that. But either way, prep company tests cannot be fully dependable upon. </p>

<p>

</p>

<p>No problem :)</p>