<p>New Question:
Of 200 families surveyed, 75% have at least one car and 20% of those with cars have more than two cars. If 50 families each have exactly two cars, how many families have exactly one car?</p>
<p>Post the level of difficulty you also think it is</p>
<ol>
<li>If a and b are positive integers, what is the SMALLEST value of a + b for which 2a + 3b is divisible by 24</li>
</ol>
<p>75% of 200 is 150 families that have at least one car. 20% of this figure is 30 families that have more than two cars. So if 150 families have at least one car and this includes 30 families with >2 cars and 50 families with exactly two cars, there are 70 families left with exactly one car. I’d give it a 4.</p>
<p>The smallest value will occur when 2a+3b=24. Note that 2a must be divisible by 3 and that 3b must be divisible by 2. This means that a=3m for some integer m and that b=2n for some integer n. This gives m+n=4. The quantity we are trying to maximize is 3m+2n, which will be smallest when n is as large as possible. m=1 and n=3 (both need to be positive as given in the problem) gives the minimum of 1<em>3+2</em>3=9. This corresponds to 6+18=24. I’d give this a 5.</p>
<p>" If a and b are positive integers, what is the SMALLEST value of a + b for which 2a + 3b is divisible by 24 ?"</p>
<p>I agree that this is a 5. And it is very SAT-like in the sense that there is a non-algebra way to solve it:</p>
<p>It’s like a game where blue coins are worth 3 and red coins are worth 2. How do you get to 24 exactly with the fewest coins? You use as many of the larger value coins as you can. 8 blue coins would do it but then the rules say you can’t have zero red coins. 7 blue coins leaves you at 21 and you can’t get to exactly 24. 6 blue coins gets you to 18 and then you need 3 red ones to get you up to 24. So 9 is the smallest total.</p>
<br>
<br>
<p>Not sure why students ask about the level of a question when that level is given in the answer sheet. No need to waste time on questions that do not come with the ETS scale. </p>
<p>Fwiw, that question should be no higher than a three and probably a two. It requires one simple conversion. </p>
<p>If Cameron was born on a Friday, and Martha was born 300 days later, what day was Martha born? </p>
<p>Please see if my method for these type of questions is valid. Thank you!</p>
<p>300/7= 42…
7x42= 294</p>
<p>The 294th day is a Thursday, since it’s at the end of the cycle so the answer is Wednesday. </p>
<p>Just a problem type I’ve seen – made my own type</p>
<p>Looks good. What you’re doing here is modulo 7 arithmetic (since there are 7 days in a week), or more simply, you’re simply finding the remainder when dividing by 7. </p>
<p>Yeah, 300 is congruent to 6 (or -1) mod 7, so there you go.</p>
<p>Here’s another number theory problem for you guys (not too difficult):</p>
<p>What is the remainder when 2^2014 is divided by 11?</p>
<p>2^2014=2^2000<em>2^10</em>2^4=(2^10)^200<em>2^10</em>2^4. By Fermat’s little theorem, 2^10=2^(11-1) leaves a remainder of 1 when divided by 11, so the only term that contributes to the remainder is 2^4=16, which leaves a remainder of 5 when divided by 11. Of course FLT isn’t ever needed for an SAT-level question but it’s faster than writing out remainders of powers of two divided by 11 and looking for periodicity.</p>
<br>
<br>
<p>Please see if my method for these type of questions is valid. Thank you!</p>
<p>300/7= 42…
7x42= 294</p>
<p>The 294th day is a Thursday, since it’s at the end of the cycle so the answer is Wednesday.<<<</p>
<p>What is that end of a cycle being Thursday? </p>
<p>It has been a long day and I might miss something but if today is Friday, 300 days from today would not be a Wednesday. 300 days from today would be Thursday May 28th. </p>
<p>Fwiw, you should use 301 as 280 + 21 is easy. </p>
<p>What I meant by end of the cycle is: </p>
<p>Friday, Saturday, Sunday…,Thursday which is 7 days. So I treated Friday like 1 and Thursday like 7. That’s what I meant by “cycle”. </p>
<p>How did you get May 28th specifically?</p>
<p>I’m sorry, I’m a little confused now.</p>
<p>^ You are really not as confused as you think. Your method was fine. You saw that 300 is 6 more than a multiple of 7 so you knew to count 6 steps into the pattern. But you started at the wrong place. Cameron was born on Friday. Martha was born 300 days AFTER that. The first of those days is a Saturday. That’s where you should have begun your count. </p>
<p>As for how Xiggi got the actual date…well, I think he noticed that he was answering your question on a Friday – Friday, August 1 to be specific. 300 days later…</p>
<p>
</p>
<p>I am also sorry to confuse you with the example of looking at 300 days after August 1st. I probably should have been clearer in addressing your answer. </p>
<p>Unless I read your problem statement incorrectly, I really do not think the answer to “If Cameron was born on a Friday, and Martha was born 300 days later, what day was Martha born?” could be … Wednesday. Look at my calculations and it is Thursday. If you follow PCK’s logic (which one might) and stress the AFTER by adding a day, the answer is Friday. </p>
<p>I asked about the "start of the cycle because I am afraid you had your series wrong. Try the same problem with an easier pattern. Use 13 or 20 instead of 300 as it is the same thing. And you might simply respond to this: </p>
<br>
<br>
<p>HTH</p>
<p>Ok, I see. In response to the 6 day problem, the answer is Thursday. I understand pckeller’s logic–I didn’t consider the AFTER very closely.</p>
<p>A link to a one day crash course for basic calc?</p>
<p>thnaks in advance!</p>
<p>@2200andbeyondXD <a href=“Calculus I in 20 Minutes (The Original) by Thinkwell - YouTube”>Calculus I in 20 Minutes (The Original) by Thinkwell - YouTube;
<p>@MITer94 isn’t this NOT the course itself?!</p>
<p>ok I don’t understand how the answer was reached in this question(it is a grid-on circle question from Barron’s)</p>
<p>The circumference of a circle is a(pi) unites, and the area of the circle is b(pi),what is the radius of the circle?
The answer is 2, but I don’t understand why.</p>
<p>There isn’t enough information to answer this question. For example, if a=2 and b=1, then r=1. But also, if a=4 and b=4, then r=2. </p>
<p>In general, a=r/2, and b=r^2. So b=(2a)^2=4a^2. o you can choose any positive value for a, and you would get a unique value for b. Since r=2a, r can be any positive number.</p>
<p>In other words, this is a bad question.</p>
<p>My mistake I missed on small tidbit a=b. I just reviewed the problem and get it now. I guess I was just tired. Thanks though!</p>
<p>can u guys resolve and explain this question the,‘what is the remainder when 2^2014 is divided by 11’</p>