"More vibrato!  Now, you try..."

When I posted that question writing contest a few weeks back, I thought I’d get some pretty good stuff. Because you guys are smart. And then, for a little while, not much happened. I was all </3. But then some great stuff started rolling in and I was all \(*o*)//. I bet, reading this paragraph, you’re all o_O. Right?

To the submissions!

From Debbie Stier:

  • Candy bars were passed out among 10 students. If the average (arithmetic mean) number of candy bars that the students received was 11, what is the greatest possible number of candy bars that one student could have received?

A) 11
B) 20

C) 101

D) 109

E) 110

Commenter Katie (fresh off a weekend challenge win) posted a few great grid-ins, too:

  • Dan wrote a 7 digit phone number on a piece of paper. He tore the paper accidentally and the last two digits were lost. What is the max number of arrangements of two digits, using digits 0 through 9, he could use to find the correct number?
  • If x + (1/x) = 4, what is the value of x2 + 1/(x2)?
  • The lengths of the side of an isosceles triangle are 30, n, n. What is the smallest possible perimeter of the triangle?
  • If (a/4) + (b/8) + (c/24) = 1, what is one possible value for abc?

And of course, this great question was submitted to me by email.

Two problems
  1. Some of these require some teensy clarifications before they’re ready for prime time.
  2. I have no solutions!
Commenters, this is where you come in.

Comments (3)

Let’s try these:

Debbie’s candy bar question –
I’m assuming “passed out” means that everybody gets at least one. If the average is 11, then 11 bars*10 people = 110 is the total number of candy bars handed out. The MOST a person could have would be if everyone else only got one. So 9 people got 1 bar, leaving 101 left for the last person. I pick ‘C’. (Did I get it? Or  is it a trick?)

Grid-ins –
Phone #: Am I missing something, or is this just asking how many numbers there are between 00 – 99? Because the answer to that is 100.

x+(1/x)=4 – conveniently, when you square both sides you end up with the middle term dropping out since x(1/x) = 1. So you get your target expression + 2 on the left and 16 on the right, so the answer is 14.

Isosceles – this problem made me realize I forget how to do geometry. To minimize the perimeter, n should be as small as possible, but not so small that the two equal sides of the triangle can’t touch. So I’m imagining the point of the triangle getting closer and closer to the base, so that (n+n) gets closer and closer to the length of the base, 30. In this case the perimeter would then be getting closer and closer to 60. But if the point of the triangle ever actually touches the base, it’s no longer a triangle. So the lower LIMIT of the perimeter should be 60, but I would think that’s not actually the lowest POSSIBLE value for the perimeter because it’s not actually possible. Unless I missed something and a line can be a triangle now. I guess I would put 60.

(a/4) + (b/8) + (c/24) = 1 – if I’m not mistaken, this is a great candidate for working backwards. You just need ANY values so that the 3 fractions add up to 1. At first I thought to just make them all equal to 1/3, but that’s not convenient for 2 of them, so if we make a=1, b=2, and c=12, then you have 1/4 + 1/4 + 1/2 which is also 1. In that case abc = 24.

Nice work. I think you touched on a few things about these questions that are important with regards to clarity. You’re right that the candy bar question should explicitly mention that each kid gets at least one. 

The triangle inequality theorem question requires a bit of revising, too. The perimeter can’t really be 60 or you really just have line segments lying on top of each other. It should be stipulated that n is an integer, to avoid anything that resembles limits (which don’t appear on the SAT). The answer should be 62. 

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