<p>@DrSteve Sorry, my post had more of a confrontational tone than I intended. </p>
<p>I agree that it takes practice, but so does most of the math on the SAT. For some reason remainders aren’t really taught as well as most math topics on the test, so students don’t have as much practice. I don’t think remainders are intrinsically harder.</p>
<p>I didn’t take it as confrontational at all. I agree that remainders are generally not dealt with too much in high school. But this is true of many types of SAT math questions. Also, I think it’s great for students to have a better understanding of remainders (in fact, I’ve written several blog posts on this), and I don’t see any problem with SAT tutors covering this topic in detail with all their students. I personally choose to focus on more basic strategies whenever possible with weaker students, and only get into details like this for more advanced students.</p>
<p>I’ve dominated this thread enough for now, so I’ll contribute a problem myself. I’m not the best at assigning difficulties (especially to problems I write), but maybe level 3/4 geometry? </p>
<p>A square is inscribed in a circle. The circle is inscribed in another square. What is the ratio of the area of the larger square to the area of the smaller square?</p>
<p>That’s a nice problem. I would say it’s at least Level 4.</p>
<p>Nice problem indeed. I would say that is both a level 5 question for your typical SAT student and a … 1 for a student who has prepared a small arsenal of easy to remember rules and typical ratio. Why can this be a 1? Because anyone can rotate the smaller square by 90 degrees and have the corners of the smaller square at the middle of the larger sides. That or even better intuitively know the answer is 2 to 1 because of a small series of ratios. </p>
<p>Fwiw, remember the circle is entirely trivial in this problem. </p>
<p>And for your pleasure, here is one= A square is inscribed in a circle. Within this square is inscribed a smaller circle. What is the ratio of the area of the larger circle to that of the smaller circle?</p>
<p>If you had 5 seconds, what would guess and would you? </p>
<p>To make it clearer, here is a good page </p>
<p><a href=“http://calculus-geometry.hubpages.com/hub/Circle-Inscribed-in-Square-Geometry-Problems-Solutions”>http://calculus-geometry.hubpages.com/hub/Circle-Inscribed-in-Square-Geometry-Problems-Solutions</a></p>
<p>“Because anyone can rotate the smaller square by 90 degrees and have the corners of the smaller square at the middle of the larger sides.” I didn’t explicitly state how the squares were oriented with respect to one another, but I understand. This is definitely the more elegant way to to do it. </p>
<p>I don’t think I’d give it a difficulty of 1, no matter how knowledgeable the student. To my mind, a level 1 is something like “If 5 boxes each have the same weight and 2 of them weigh 6 pounds together, how much do all 5 weight together?”</p>
<p>“A square is inscribed in a circle. Within this square is inscribed a smaller circle. What is the ratio of the area of the larger circle to that of the smaller circle?” 2:1 and I’d be pretty confident about it, though I never like writing an answer without checking it (of course if I had 5 seconds I choose it).</p>
<p>OK, I suppose I need to come up with a new problem. Level 4/5 counting:</p>
<p>The diagonals of a square divide its interior into 4 congruent, non-overlapping, triangles. Given the three colors red, yellow, and green, how many ways can a color be assigned to each of these triangles in such a way that any two triangles sharing a side do not have the same color?</p>
<p>@DrSteve Hi, can I request help on this question?</p>
<p>The figure above shows a portion of the graph of the function f. If f(x+5) = f(x) for all values of x, then f(x) = 0 for how many different values of x between 0 and 12?</p>
<p>A) 8
B) 9
C) 10
D) 11
E) 12</p>
<p>[GRAPH FOUND ON PG 369 OF SAT BLUE BOOK] #8</p>
<p>@medicsz I’m not DrSteve, but…</p>
<p>Looking at the graph, there are four values that make f zero between x=0 and x=5. f(x) will be zero if and only if x differs from one of these zeroes by a multiple of 5. We can add 5 to each of these zeroes to get 4 new zeroes (all are still less than 12). But we can also add 5 again to the smallest zero because if 1<x<2, then 11<x+10<12. This is the largest zero which is still smaller than 12, so we get that the total number of zeroes is 4+4+1=9.</p>
<p>@RandomHSer Thanks so much! I understand the first part, up until the explanation where you add 5 again to the smallest zero.</p>
<p>Could you re-explain that?</p>
<p>Sure. After we add 5 to the all the zeroes, we get that there is a zero: between 6 and 7, between 7 and 8, approximately at 8, and between 9 and 10. But if we have a zero between x=6 and x=7, then we also have a zero between x=6+5 and x=7+5, or between x=11 and x=12. This is smaller than 12, so it counts towards the problem.</p>
<p>OHH, I see what you mean between 6 and 7 and so on… Because on the graph, the zeroes before the change clearly show that the values are in-between…</p>
<p>Thank you so much for your time!</p>
<p>if 3^a+3^a+3^a=3^b , what is b in terms of a ?</p>
<p>answer is a+1
I don’t get why?
Really bad with tricky exponent rules?
I know you can simplify the right to
3 X 3^a =
but idk how to solve for b in terms of a?</p>
<p>Please also post what level of difficulty do you think this is from 1-5 like the SAT does!
thanks
(please explain fully!!)</p>
<p>When you add something 3 times you get 3 of those things. For example an apple plus an apple plus an apple is 3 apples. Here 3^a + 3^a + 3^a = 3(3^a)=(3^1)(3^a)=3^(1+a).</p>
<p>Do you see why the answer is a+1 now?</p>
<p>Now I’m guessing this is a multiple choice question, so you can actually just pick a number here. For example, if you let a=4, then the left hand side becomes 3^4+3^4+3^4= 81 + 81 + 81 = 243. So we have 3^b=243=3^5. So b=5. It looks like b is 1 more than a (but of course you should check all 5 answer choices and eliminate any that do not come out to 5).</p>
<p>This is a Level 4 or 5 question, leaning toward level 5.</p>
<p>yes thank you makes sense!</p>
<p>The equation of a certain parabola is f(x)=ax2+bx+c, where a, b, and c are constants. If the minimum of the parabola is at x=5and its y-intercept is 12, then all of the following must be true EXCEPT</p>
<p>(A)
f(6)<12</p>
<p>(B)
f(10)=12</p>
<p>©
c=12</p>
<p>(D)
a>1</p>
<p>(E)
b≠0</p>
<p>???</p>
<p>Can someone explain to me what does the A and B stand for in a parabola equation?? thanks</p>
<p>@Busybee123</p>
<p>A has to be true because the parabola is opening upwards (since it has a minimum). This means that as you get x gets farther from 5, f(x) gets larger. Combining this with symmetry about the vertex f(6)=f(4)<f(0)=12.</p>
<p>B has to be true by symmetry about the vertex. Since the vertex occurs at x=5 f(5-5)=f(5+5) or f(10)=f(0)=12.</p>
<p>C has to be true because f(0)=12 as shown by the fact that the y-intercept is at (0,12). f(0)=c=12.</p>
<p>D There’s no reason why D should be true. The simplest counterexample would be something like f(x)=(x-5)^2-13=x^2-10x+25-13=x^2-10x+12 and we have a counterexample where a=1 (not larger than 1).</p>
<p>D has to be true because the conditions tell us that f(x)=a(x-5)^2+h for some a and h. Expanding this gives us a “b-term” of -10ax, which is never zero unless a=0, which we know isn’t the case.</p>
<p>So D is the answer.</p>
<p>x = -b/(2a) is the x-coordinate of the vertex of the parabola</p>
<p>thanks you guys LOVE YOUR explanations</p>
<p>One More please?</p>
<p>11 is both the median and the mode of a set of 5 positive integers. What is the least possible value of the average (arithmetic mean) of the set?</p>
<p>Since it is the mode, we know that there must be at least two 11s. If the median is 11, then we know that the third number (when they’re arranged in numerical order) must be 11. The fourth and fifth terms are minimized if they’re both 11. This gives us 3 11s, so we can have the lowest two numbers both be 1s. Then our numbers are: 1,1,11,11,11 (two 1s and 3 11s). This makes the average (2+33)/5=7.</p>
<p>Sam’s job is to deliver pizza, and he keeps careful track of the tips he earns. One week, he noted that his average (arithmetic mean) tip total for Monday, Tuesday, and Wednesday was $75. How much would Sam have to earn in tips on Thursday and Friday to bring his 5-day average tip total to $100???</p>