# Math Help - 3 math problems

1. ## 3 math problems

I have three hard math problems that I can not figure out no matter how hard I try. I would be nice if some one would help me and get an answer for each (or maybe just give me one) and how to do the problem.

1. Three people play a game in which one person loses and two people win each game. The one who loses must double the amount of money that each of the other two players has the time. The three players agree to play three games. At the end of the three games, each player has lost one game each person has $8. What was the original stake of each player? 2. In a ball game where only 10 and 13 points can be scored at any one time, what is the largest final score that cannot be obtained? 3. Suppose five bales of hay are weighed two at a time in all possible ways. The weights in pounds are 102, 104, 105, 106, 107, 108, 109, 110, 112, 113. How much does each bale weigh? (Bale weight must be whole numbers.) 2. Wouldn't the stake be$2?

2 2 2
2 4 4
4 4 8
8 8 8

3. Originally Posted by JamesNannen
I have three hard math problems that I can not figure out no matter how hard I try. I would be nice if some one would help me and get an answer for each (or maybe just give me one) and how to do the problem.

1. Three people play a game in which one person loses and two people win each game. The one who loses must double the amount of money that each of the other two players has the time. The three players agree to play three games. At the end of the three games, each player has lost one game each person has $8. What was the original stake of each player? 2. In a ball game where only 10 and 13 points can be scored at any one time, what is the largest final score that cannot be obtained? 3. Suppose five bales of hay are weighed two at a time in all possible ways. The weights in pounds are 102, 104, 105, 106, 107, 108, 109, 110, 112, 113. How much does each bale weigh? (Bale weight must be whole numbers.) I do not give partial answers, so I won't give an answer if you did not say "(or maybe just give me one)"-----because I cannot solve the #2 and #3 problems just yet. I would have asked you to post the 3 problems separately in 3 postings, but, what the heck, I will just give you one as you mentioned. For the question #1, here is one way of doing it. Let x = original stake of one person or person A. y = original stake of second person or person B. z = original stake of third person or person C. Person A lost. So, Person B will get y amount from A. New money of B is y+y = (2y). Person C will get z amount from A. New money of C is z+z = (2z). New money of A is (x -y -z). Then person B lost. Person A will get (x -y -z) from B. New money of A is 2(x -y -z) = [2x -2y -2z]. Person C will get 2z from B. New money of C is 2(2z) = [4z]. New money of B is (2y) -(x -y -z) -(2z) = 2y -x +y +z -2z = [3y -x -z]. Then person C lost. Person A will get [2x -2y -2z] from C. New money of A is 2[2x -2y -2z] = {4x -4y -4z}. Person B will get [3y -x -z] from C. New money of B is 2[3y -x -z] = {6y -2x -2z}. New money of C is [4z] -[2x -2y -2z] -[3y -x -z] = 4z -2x +2y +2z -3y +x +z = {7z -x -y}. The problem says that each of these 3 new moneys is$8, so,
4x -4y -4z = 8 ------(1)
6y -2x -2z = 8 ------(2)
7z -x -y = 8 --------(3)

I know that you can take over from here, but let me finish the solution anyway.

Rearranging and simplifying those 3 equations,
x -y -z = 2 -----------(1a)
-x +3y -z = 4 ----------(2a)
-x -y +7z = 8 ----------(3a)

We eliminate x first,

(1a) plus (2a),
2y -2z = 6
y -z = 3 -------------(4)

(1a) plus (3a),
-2y +6z = 10
-y +3z = 5 ------------(5)

Then, (4) plus (5),
2z = 8
z = 8/2 = 4 dollars -----------------------------------***
And, from (4), y = 3 +z = 3+4 = 7 dollars -------------***
And, from (1a), x = 2 +y +z = 2+7+4 = 13 dollars -------***

Check,
Person A has initially $13; person B has initially$7; person C has initially $4. Person A lost. New money of B = 2(7) =$14.
New money of C = 2(4) = $8. New money of A = 13 -7 -4 =$2.

Person B lost.
New money of C = 2(8) = $16. New money of A = 2(2) =$4.
New money of B = 14 -8 -2 = $4. Person C lost. New money of A = 2(4) =$8.
New money of B = 2(4) = $8. New money of C = 16 -4 -4 =$8.

Therefore, depending on who lose first, who lose second, the first person to lose has $13 initially, the second person to lose has$7 initially, and the third person has $4 initial stake. ----------answer. 4. #1, ticbolb, you got it right (i think) the answer is$13, $7,$4.

#2. I think the answer is 107 but i dont really understand it.

#3 I gave up on that one after awhile, I made a giant chart matching each bale but it got all messy and i got frustrated and gave up.

So i still need help on #3 and it would be nice if some would could check out #2 because it do not really understand it, even if thats the right answer