# Daily Quiz #2: March 23, 2014

As of writing this post, it looks like many of you have not seen the contest yet. I’m hoping those of you that do know about it will tell the others, and I’ll give you a day to get points for the questions you answer on Quiz #1. Remember, I’m preparing you for your competition this weekend!

6. Willy Wonka’s famous ice cream shop has three types of ice cream cones: the Rocky Road Regular, the Tiger Stripe Tall, and the White Chocolate Wide. Yum. They all have the same dimensions, except the tall’s height is three times the regular’s, and the wide’s diameter is twice the regular’s. Does the tall cone or the wide cone hold more ice cream, and how much more? Express your answer as a common fraction (ratio of the bigger cone to the smaller one). (10 pts.)

7. Given a standard, fair coin, what is the probability of flipping the coin three times and getting exactly two heads? (10 pts.)

8. If the sum of 10 consecutive numbers is 395, what is the sum of the odd numbers among them? (10 pts.)

9. Bonus. Joh can mow half an acre of lawn in 1 hour. His good friend Num can mow a full acre in one and a half hours. If they start mowing a 10.5-acre lawn at 10:00 AM together, at what time will they finish? (20 pts.)

10. Bonus. The height of a falling object in feet as time passes is given by h = -16t2vts, where v is the initial velocity of the object and s is the initial height. If a projectile is launched off a 24-foot building at an initial velocity of 40 feet per second, how many seconds later will the projectile hit the ground? (20 pts.)

# Daily Quiz #1: March 22, 2014

The state competition is coming up! We’re already making great progress towards doing well at state – now we just need to keep up the skills until next weekend. So, here’s what I have in mind: every day from now until Thursday, I’ll post five problems on this blog. You, the Mathlete, should solve everything you can and post your answers (along with your name) in the comments. The next day, the answers will be posted so you can check your work.

Here’s the game, though: you will receive points for each question you answer correctly. Every day, I’ll post each person’s rankings along with the answers. Who will be the best Mathlete of all? We shall see….

1. I bought a one-year membership to the Abscissa Nature Park the other day for $99. The membership lets me visit the park for only$5, while non-members must pay \$18. How many visits do I have to make this year to make buying the membership worth my money? (10 pts.)

2. Srikar is looking for books at the Broadmoor Library, and there are 70 fiction books and 80 nonfiction books on the nearest shelf. If no two books are the same, what is the probability that he will randomly choose one book of each kind without replacement? (10 pts.)

3. Sunjay is thinking of three positive integers. He tells his friend Robert that the sum of the first and second numbers is 8 less than the sum of the first and third, and 14 less than the sum of the second and third. If the second number is three times the first number, what is the sum of all three numbers? (10 pts.)

4. Bonus. What is the sum of 17268 and 3246? Express your answer in base 10. (20 pts.)

5. Bonus. The plot of land for Caroline’s new house is a trapezoid with two right angles, two sides of length 300 m, and a diagonal of length 500 m. What is the length of the other diagonal? Express your answer as a decimal to the nearest tenth of a meter. (20 pts.)?

# Problem of the Day: 3/15/13

What is the probability that a term selected at random from the binomial expansion of $(a+b)^{21}$ has a coefficient that is a multiple of 2? (edited)

Solution to yesterday’s problem:

The trick to sum of sequence problems is to find the pattern of the sums. In this problem, notice that the sum of the first and third terms, 71, is the same as the sum of the second and fourth terms. It’s also the sum of the third-last and last terms. So we can assume that the sum of two terms that have one term between them is 71.

The next step is to find the number of these pairs in the sequence. Well, we notice that the first two numbers starts counting down from 35, and each set of two numbers separated by two numbers continues this pattern down to 1 at the end. The numbers between the decreasing numbers start at 36 and increase to 70. It takes 35 numbers to count from 35 to 1, and 35 numbers to count from 36 to 70. Therefore there are 70 numbers and 35 pairs. Multiplying the number of pairs by the sum of each pair, 35 x 71 = 2485.

Another way this problem can be solved is to notice that the series essentially just contains the integers from 1 to 70. The sum of these integers is half the product of the biggest number and the next number after that.

$frac{(70)(71)}{2}$

$frac{70}{2}times71$

$35times71$

We wind up with the same answer, which is 2485.

# Problem of the Day: 3/14/13

What is the sum of the terms in the sequence 35+34+36+37+33+32+38+39+ … +2+1+69+70?

Solution to yesterday’s problem:

We can assume that triangle PQR has integer sides because more information was not specified. This would mean that PQR is a Pythagorean triple with a short side of 9: a 9-12-15. So the hypotenuse of PQR is 15 cm.

Now we’re going to do some more estimating! You can sort of see that the small triangle QUT takes about a third of the hypotenuse. This goes along with our theory about integer lengths because it would make the hypotenuse 5, which leads us to believe that QUT is a 3-4-5 triangle. Also, the larger triangle RST takes up two-thirds of the hypotenuse, or 10 cm. This suggests that RST is a 6-8-10 triangle. So let’s show the triangle again with the lengths labeled:

The length we actually want to find is US, conveniently the only segment we don’t know. We can use the Pythagorean theorem for triangle STU.

$4^2+6^2=x^2$

$16+36=x^2$

$52=x^2$

$\sqrt{52}=x$

$\sqrt{4\times13}=x$

$\sqrt{4}\times\sqrt{13}=x$

$2\sqrt{13}=x$

# Problem of the Day: 3/13/13

In the figure below showing three right triangles, segment PR measures 9 cm. What is the measure of segment US? Express your answer in simplest radical form.

Solution to yesterday’s problem:

Let’s use the blanks method to solve this problem. We have 10 people to choose from and 10 seats.

__10__ __9__ __8__ __7__ __6__ __5__ __4__ __3__ __2__ __1__

However, we have to divide by 10 because each arrangement could start at any of the 10 seats around the table. Multiplying the blanks and dividing by 10, we find a total of 362,880 ways.

# Problem of the Day: 3/12/13

In how many different ways can 10 people be seated at a round table? Note that arranging the people the same way at different positions around the table does not count.

Solution to yesterday’s problem:

Since the common factor of the two numbers is 4, both numbers are divisible by 4. We can therefore say that the first number is 4 times a number, or 4m. The second number is 4 times another number, or 4n. The problem tells us that

$4m+4n=52$

Factoring out a 4 on the left side,

$4(m+n)=52$

We can now divide by 4:

$m+n=13$

So the range of possible values for m and n is such that their sum is 13: 1 and 12, 2 and 11. However, logic tells us that since we want the smallest possible difference between the numbers, we should choose numbers close to 13/2; namely, 6 and 7. Using 6 and 7, our original numbers would be 24 and 28. These do have a sum of 52 and a GCF of 4, and their difference is 4.

# Problem of the Day: 3/11/13

Two positive integers have a sum of 52. If their greatest common factor is 4, what is the least possible positive difference between the numbers?

Solution to yesterday’s problem:

If Carlo needs two cans of paint to coat three walls, then his rate of “paint usage” is $\frac{2}{3}$ cans per wall. If he needs to paint eight 4-walled rooms, he has to paint 32 walls total. He also needs to paint two coats per wall, which is equivalent to painting 32 x 2 = 64 walls total. Using the rate mentioned above, we can write

$\frac{2}{3}\times64$

to represent the number of cans of paint he needs. This evaluates to $\frac{128}{3}$, or $42\frac{2}{3}$. Remember that Carlo couldn’t very well buy two-thirds of a can of paint, so we round up to 43 cans.