Another Math Question

[Zetta_x]

On a certain game, your average score out of 427 games is 13,000.

1) After scoring below average on the next game, your new average (approximated to two decimal places) is 12,980.14. If you can only score integer numbers, what was your score on the 428th game?

2) What must you score on the 429th game in order to raise your average back up to 13,000?

For courtesy of people who do not want to see the answers, I ask that you put your answers in spoilers unless you are the jerkass who clearly acknowledges that you are spoiling the answers.

 

[Snailface]

Spoiler

1. 4500
2. 21500

 

[lufere7]

Spoiler

4500 and 21500, just used my calculator, dunno if there was another method

 

[Nathan Drake]

Spoiler

13,000 per game for 427 games yields an average of 13,000.
The average after game 428 is 12,980.14.
Average back to 13,000 after game 429.

1)
12,980.14 x 428 = 5,555,499.92
13,000 x 427 = 5,551,000
Difference = 4499.92 which, rounded up, is 4500

Total after 428 games: 5,555,500 w/ avg 12,980.14

2)
13,000 x 429 = 5,577,000
5,577,000 - 5,555,500 = 21,500

((13000 x 427) + 4500 + 21500)/429 = 13,000

1) 4,500
2) 21,500

Huh, should have considered that the total for 428 and 429 would have to equal 26,000 to maintain an average of 13,000 at the end of problem two. Could have saved myself a bit of subtraction and the like if I had recognized that before the fact. Oh well.

 

[jimmyemunoz]

Spoiler

5551000 + x/428 = 12980.14 solution x= 4500
5551000+4500 (5,555,500)
5,555,500 + x/429 = 13000 solution x= 21,500

 

[mechadylan]

Spoiler

(427*13000)/427=13000
5551000=5551000

((427*13000)+x)/428=12980.14
5551000+x=5555499.92
x=4499.92 x~4500

((427*13000)+x+y)/429=13000
5551000+4500+y=5577000
5555500+y=5577000
y=21500

 

[Zetta_x]

Spoiler

Actually Nathan Drake hit it right on the head what I was looking for. There are two ways of getting number 2, one way is much easier and the other way is a tad bit more complicated. If you used similar mathematics to get number 1 to also get number 2, you took a more complicated root. You use the facts that if something had an average of X with k1 sample size and another thing had an average of X with k2 sample size, then the average of these would also be X regardless of what k1 and k2 are equal to.

I had set this question up carefully, to see which path people would take.

Notice if I had said, game 428 was answer Y, what would game 429 have to be? Then people would take the easier path. This exact logical chain structure is exactly what I was looking for.