Statistics Problem Solving

[desparate]

I have 2 problems that I can not figure out how to work the problem to find the answer. Any help is GREATLY appreciated.

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 at the time. The three players agree to play three games. At the end of the three games each player has lost one game and each person has $8. What is the original stake of each player?

2. In a ball game where only 7 and 10 points can be scored at any one time, what is the largest final score that CANNOT be obtained?

Thanks

[robertR0931]

I never was any good at solving story problems and haven't the slightest clue about statistics. Quite honestly I was surprised to find a statistical question here because you really don't have any conversion going on. That being said, with all the spam appearing on the first page I am going to bump this one up to try and help you get an answer.

[convert-me.com]

I have 2 problems that I can not figure out how to work the problem to find the answer. Any help is GREATLY appreciated.

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 at the time. The three players agree to play three games. At the end of the three games each player has lost one game and each person has $8. What is the original stake of each player?

2. In a ball game where only 7 and 10 points can be scored at any one time, what is the largest final score that CANNOT be obtained?

Thanks

1. Since each player lost one game of the three, it means each one won 2 games. Hence the stake of each player was doubled twice. That means the initial stake of each player was $2 ($2 = cubic root of $8).

2. We need to find the largest possible number that cannot be constructed as a sum of 10s and 7s. Multiplies of 10 give us only round numbers ending with 0. Multiplies of 7 can produce numbers ending in all digits from 1 to 9 (7*1 = 7, 7*2 = 14, 7*3 = 21, 7*4 = 28, 7*5 = 35, 7*6 = 42, 7*7 = 49, 7*8 = 56, 7*9 = 63). So to construct any big enough score (e.g. 154) we pick as many 7s to get the ending 4 (7*2=14) and add the rest with 10s (154 = 7*2 + 10*14). The biggest number of 7s is needed to construct final score ending with 3 - it's 7*9. While we can construct 63 itself by means of 7s only (no 10s needed at all), the previous number ending with 3 (53) cannot be constructed. So the answer to this question is 53.

Having said all the above I still have to warn this subject being offtopic here. It is not related to units conversion :wink:

[convert-me.com]

... with all the spam appearing on the first page...

You're right. :oops: We're cleaning spam several times a week and are using some sophisticated filters. Unfortunately it seems impossible to keep the first page tidy all the time.