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.