Solution for probability, Problem for rate

I'm sorry for not posting in awhile. I went on a family reunion to Yosemite for a week. It was nice :) Also, I'm not sure I'll be able to post as often as I have before, since school has started and I may be busy with homework.


Thanks everyone for trying out the problem and posting in the comments!


Problem:
At a competition with N players, the number of players given elite status is equal to 2^{1+⌊log₂(N-1)⌋} - N. Suppose that 19 players are given elite status. What is the sum of the two smallest possible values of N?


Solution:
We are given the equation  2^{1+⌊log₂(N-1)⌋} - N for the number of players given elite status and we are also told that 19 players have been given elite status. Thus we can make  2^{1+⌊log₂(N-1)⌋} - N equal to 19.


 2^{1+⌊log₂(N-1)⌋} - N = 19


Now we can simply it a little bit.


2^{1+⌊log₂(N-1)⌋} - N
2¹*2^{⌊log₂(N-1)⌋} - N = 19
2^{⌊log₂(N-1)⌋} - N/2 = 19/2
2^{⌊log₂(N-1)⌋} = (19 + N)/2


Now we can put this equation into log form.


2^{⌊log₂(N-1)⌋} = (19 + N)/2
log₂((19+N)/2) = ⌊log₂(N-1)⌋


Because the ⌊log₂(N-1)⌋ has to be an integer (because ⌊x⌋ is the nearest integer ≤ x) we can say that log₂((19 + N)/2) also has to be an integer, because they are equal to each other. Now we can start plugging in values of N that are greater than 19 (there are at least 19 players because 19 of them were given elite status). From the log equation we have (19 + N)/2 = 2^x. Because the N has to be greater than 19 the lowest value 2^x can be is 32. (Try out 16, you can see that it is too small)


(19 + N)/2 = 32
19 + N = 64
N = 45


We have to find the two lowest values of N so now we make 2^x equal to 64


(19 + N)/2 = 64
19 + N = 128
N = 109


We can see that the 2 lowest values of N are 45 and 109. The sum of those two numbers is 154, thus the answer is 154.


Here's the next problem
Problem:
A canoeist paddled upstream for 2 hours, then downstream for 3. The rate of the current was 2 mph. When she stopped, the canoeist realized she was 20 miles downstream form her starting point. How many hours will it take her to paddle back to her starting point?


Good luck! and remember, no calculators are allowed. :)

Solution for probability, Problem for sum

Thanks everyone for trying out the problem, and congrats to those who got the problem right!

Problem:

A pair of standard 6-sided dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference.

Solution:

First of all, we need to find all the possible diameters of the circle. The question  states the numerical value of the area of the circle is less than the numerical value of the circle's circumference. If that is the case, then the radii will be less than 2 because

πr² < 2πr

πr < 2π

r < 2 

If a pair of standard 6-sided die is rolled once, then the diameter will have to be either 2 or 3, because the lowest value you can roll is a 2, and a 4 won't work since the radii has to be less than 2.

There is only 1 possible way to roll a 2 (1,1), and 2 possible ways to roll a 3 (1,2) (2,1).

There are 3 possible rolls out of 36 (6*6) different rolls. Thus, the answer is 1/12.

Here's the next problem:

Problem:

At a competition with N players, the number of players given elite status is equal to 2^{1+⌊log₂(N-1)⌋} - N. Suppose that 19 players are given elite status. What is the sum of the two smallest possible values of N?

⌊x⌋ is the nearest integer ≤ x

Good luck! and remember, no calculators are allowed. :)

Solution for free throws, Problem for probability

Thanks everyone for trying out the problem, and congrats to those who got the problem right!

Problem:

The players on a basketball team made some three-point shots, some two-points shots, and some one-point free throws. They scored as may points with two-points shots as with three-points shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?

Solution:

First of all, we will need to set up a system of equations to solve this problem. Let's make x= # of free throws, y= # of two-point shots, and z= # of three-point shots. Now, we have x + 2y + 3z = 61. If they scored as many points with two-point shots as three-point shots, then 2y=3z. Their number of successful three throws was one more than their number of successful two-point shots, which means x = y + 1. Now we can substitute these into the original equation.

x + 2y + 3z = 61

(y + 1) + 2y + (2y) = 61

y + 1 + 2y + 2y = 61

5y + 1 = 61

5y = 60

y = 12

We can now plug in 12 in for y into x = y + 1 to solve for the number of free throws.

x = y + 1

x = 12 + 1

x = 13

The answer is 13 free throws.

Here's the next problem

Problem:

A pair of standard 6-sided dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference.

Good luck! and remember, calculators are not allowed. :) 

Solution for the maximum value, Problem for free throws

Thanks everyone for trying out the problem and posting in the comments! I'll try to put up a little easier problem this time.

Problem:

Suppose that |x + y| + |x - y| = 2. What is the maximum possible value of x^2 - 6x + y^2?

Solution:

When you first look at x^2 - 6x + y^2 you'll realize that if you want the maximum value, x would have to be a negative number to cancel out the -6x. Also, for the equation to be the maximum value, x and y would have to be the greatest absolute value possible. 

From the equation |x + y| + |x - y| = 2, you can see that |x + y| and |x - y| has to be less than 2. Basically, from looking at the problem for a bit, you can see that if x=-1 and y=1 the equation works.

|x + y| + |x - y| = 2

|-1 + 1| + |-1 - 1| = 2

|0| + |-2| = 2

|-2| = 2

2 = 2

If you plug in -1 for x and 1 for y into  x^2 - 6x + y^2 you get 8 which is the answer. 

Problem:

The players on a basketball team made some three-point shots, some two-points shots, and some one-point free throws. They scored as may points with two-points shots as with three-points shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?

Good luck! and remember calculators are not allowed. :)

Solution for the cost of pencils, Problem for maximum value

Thanks everyone for trying out the problem and posting in the comments. Even if you didn't get the problem right, it's great that you tried and I encourage you to do the other problems! If you keep doing them you'll eventually get a lot better with these types of problems. :)

Problem:

A majority of the 30 students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the 

pencils was $17.71. What was the cost of a pencil in cents?

Solution:

This problem may seem hard when you first look at it because you're hardly given any information. All you basically know is

cost of pencil in cents > # of pencils > 1

and

# of students who bought the pencil > 15

(because it stated a majority of the 30 students)

Also, you can figure out that the products of all the unknown variables will equal 1771

(# of pencils)*(cost in cents)*(# of students) = 1771

All you can do now is start trying to find the prime factors of 1771. If you know your divisibility rules for numbers 1-10 you'll realize pretty quickly that 7 is the lowest prime factor of 1771.

1771/7

253

Now we have to find the factors of 253. Once again, we can use the divisibility rules and go through the numbers rather quickly. You find out that 11 is the next prime factor of 253.

253/11

23

Now, we know (7)(11)(23)=1771 From our inequality equations that we made earlier we can see that 23 is the number of students in the class, the cost of pencils in cents is 11, and the number of pencils each student bought is 7 pencils.

Therefore, each pencils costs 11 cents which makes it the answer. 

If you didn't know the divisibility rules then the only way you could do it is just divide each number. It would only just take a little bit longer to do the problem. Here are the divisibility rules for numbers 1-10

1: Any number (15)

2: If the number is even (64)

3: Add up the digits in the number and if the sum is divisible by 3 (3453)

4: If the last 2 digits is divisible by 4 (236)

5: If the last digit is 0 or 5 (35)

6: If the number is divisible by 2 and 3 (1542)

7: Multiply the digits from the right hand side with numbers 1, 3, 2, 6, 4, 5, and add the products. If the sum is divisible by 7 then the number is divisible by 7 (2016)

8: If the last 3 digits is divisible by 8 (5248) 

9: Add up the digits in the number and if the sum is divisible by 9 (855)

10: If the number ends in 0

Hopefully, this can help you do simple division problems much faster and may help you on tests in school or other things.

Here's the next problem!

Problem:

Suppose that |x + y| + |x - y| = 2. What is the maximum possible value of x^2 - 6x + y^2?

Good luck! and remember, calculators are not allowed.

Solution for the area, Problem for the cost of pencils

I'm sorry for not posting in awhile. I kind of got a little lazy. Thanks everyone for trying out the problem and posting in the comments!


Problem:A square of side length 1 and a circle of radius (√3)/3 share the same center. What is the area inside the circle, but outside the square?


Solution:
π is the pi symbol
You can easily find the distance from the center of the square to the corner of the square to be 1/√(2). 1/√(2) is greater than 1/√(3) therefore the circle does not extend out of the corner of the square. However the length from the center of the square to the side of the square is 1/2. Thus, we get something like this:

(We're solving for the shaded regions)

The method I'm going to use is find the area of one of the shaded regions (top) and multiply that by 4 to get the total area of all 4 regions. We can plot this onto a graph with the center of the circle and square as (0,0). For the circle we have an equation of x^2 + y^2 = 1/3. For the line of the square at the top we have y = 1/2. We can now find the points of intersection of the square and the circle by substituting 1/2 for y.  

x^2 + y^2 = 1/3

x^2 + (1/2)^2 = 1/3

x^2 + 1/4 = 1/3

x^2 = 1/4 - 1/3

x^2 = 1/12

x = ±1/√(12)

We can draw a triangle from the center of the square to the intersection points like this:

The height of the triangle is 1/2 (because it is half the length of the square) and the length of the base of the triangle is 1/√3. Also, the other two sides of the triangle are 1/√3 because it is the radius of the circle. Using simple trigonometry, you can find the angles of the intersection points to be 60°. That means the angle at the center is also 60° because 180-60-60=60. Now, we can take the area of the part of the circle and subtract the area of the triangle to find the shaded region.

πr^2

π/3

π/3 is the area of the entire circle. However, we only want the area of the 60° slice of the circle. Therefore we multiply the total area by 1/6 because 60° is 1/6 of 360°.

π/3 * 1/6 = π/18

Now we have to subtract the area of the triangle from π/18.

b*h/2

1/√3 * (1/2)/2

1/4√3

Thus, the area of one of the shaded regions is π/18 - 1/4√3. We must now multiply it by 4 to find the total area of all the shaded regions.

4(π/18) - 4(1/4√3)

2π/9 - 1/√3

Therefore the answer is 2π/9 - 1/√3

Problem:

A majority of the 30 students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the 

pencils was $17.71. What was the cost of a pencil in cents?

Good luck! and remember, calculators are not allowed.

Solution for probability Problem for area

Thanks everyone for trying the problem and posting in the comments, and congrats for those of you who got the right answer!


Problem:
Suppose a and b are single-digit positive integers chosen independently and at random. What is the probability that the point (a,b) lies above the parabola y=ax^2-bx.


Solution:
Because the point (a,b) needs to be above the parabola ax^2-bx, the value of y has to be greater than the function itself. Therefore we get y>ax^2-bx. Next, we're finding the probability that point (a,b) is above the parabola, so we just plug in a for x and b for y. Thus, we get b>a*a^2-b*a = a^3-ab. Now we can solve for b in terms of a.


b>a^3-ab 
b+ab>a^3
b(1+a)>a^3
b>(a^3)/(1+a) (We can divide by (1+a) because in the problem it states a is positive)


Now we have the inequality b>(a^3)/(1+a). From this you can see that (a^3)/(1+a) is an increasing function, since a^3 grows faster than 1+a. From this, you see that the largest integer you can plug in for a is 3 because plugging in a 4 for a requires b to be a double digit integer. We have restricted a to 1≤a≤3. Now, we plug in values of a and see what positive single digit integers of b can make the inequality true.


If we plug in 1 for a, we get b>1/4. All numbers 1-9 can be plugged in for b for the inequality to hold. Which gives 9 possible ordered pairs. 
If we plug in 2 for a, we get b>8/3. Numbers 3-9 can be plugged in for b which gives 7 ordered pairs.
If we plug in 3 for a, we get b>27/4. Number 7-9 can be plugged in for b which gives 3 ordered pairs.


In total, the number of possible ordered pairs is 9+7+3=19. This is out of the total number of 9*9=81 ordered pairs. Hence the probability is 19/81.


Here's the next problem:
Problem:
A square of side length 1 and a circle of radius (√3)/3 share the same center. What is the area inside the circle, but outside the square?


Good luck! and remember calculators are not allowed. 

AMC12A 2011 Problem #14

I've gotten several comments stating that the problems I've been doing have been too simple, and that they want more of a challenge. For this reason, I have decided to skip some of the simpler problems on the test, and move on to a more challenging section.
Another change I will be making, is I will now be posting the next day's question at the bottom of each post, for those of you who want to try and do the problems by yourself.


Problem:
Suppose a and b are single-digit positive integers chosen independently and at random. What is the probability that the point (a,b) lies above the parabola y=ax^2-bx.


Good luck! and remember, calculators are not allowed.

AMC12A 2011 Problem #2

I don't think anyone will have trouble doing this problem, but I wrote a solution anyway.


Problem:
There are 5 coins placed flat on a table according to the figure. What is the order of the coins from top to bottom.

Solution:
From looking at this picture, you can easily tell that coin C is the coin at the very top because it overlaps coins D, E, and A, and coins A and D are overlapping coin B. Coin D is overlapping coin A and B and overlapped by coin E. That makes coin E the second in the order of top to bottom and coin D the third. Because coin A overlaps coin B, coin A is going to be the 4th and coin B is the 5th. Therefore the order is (C,E,D,A,B).

AMC12A 2011 Problem #1

I'm going to do math problems from various tests such as the AMC or the SAT on this blog, and hopefully it can help people out with tests. If there are any requests to do specific problems just leave them in the comments.
I'm starting off with problems from AMC12A that we had earlier this year. 


Problem:
A cell phone plans costs $20 each month, plus 5 cents per text message sent, plus 10 cents for each minute used over 30 hours. In January Michelle sent 100 text messages and talked for 30.5 hours. How much did she have to pay?


Solution:
$20.00+($0.05*100)+($0.10*30)=$28.00


Because the cell phone plan costs $20 each month, we can start out the cost at $20. Michelle sent 100 text messages in that month, and the cost of text messaging is 5 cents per text message. Thus, the extra cost from text messaging is $0.05*100 which is equal to $5.00. Now the total cost is $25. The problem states that there is an additional cost of 10 cents for each minute used over 30 hours. Michelle talked on the phone for 30.5 hours which means she spent an additional .5 hours or 30 minutes on the phone. Because the cost of an extra minute on the phone is 10 cents, we multiply $0.10*30 which equals $3.00. Now we add that to the total cost and we get $28.00.