Can you solve this?

srican wrote: ↑

Mon Jan 15, 2007 7:43 pm

A certain basketball team that has played 2/3 of its games has a record of 17 wins and 3 losses. What is the greatest number of the remaining games that the team can lose and still win at least 3/4 of all of its games?

(A) 7
(B) 6
(C) 5
(D) 4
(E) 3

Total Game = 30 (based on 20 games played so far). I'm struggling with how to split it into 3/4 ratio.

Thanks

Since 20 games (17 wins and 3 losses) represent 2/3 of the total games played, we can say:

(2/3)(total) = 20

Total = 30

Thus, there are 10 games left.

We need to determine how many games can be lost so that at least 3/4 of the games are won, or, in other words, so that at most 1/4 of the games are lost. We can let x = the number of games that can be lost, and we have:

(x + 3)/30 ≤ 1/4

4x + 12 ≤ 30

4x ≤ 18

x ≤ 4.5

Since x must be an integer, the largest integer less than or equal to 4.5 is 4.

Answer: D

srican wrote: ↑

Mon Jan 15, 2007 7:43 pm

A certain basketball team that has played 2/3 of its games has a record of 17 wins and 3 losses. What is the greatest number of the remaining games that the team can lose and still win at least 3/4 of all of its games?

(A) 7
(B) 6
(C) 5
(D) 4
(E) 3

Total Game = 30 (based on 20 games played so far). I'm struggling with how to split it into 3/4 ratio.

Thanks

Given:
The team has played 2/3 of its games. The team has played 20 games.
If 2/3 of the total number of games = 20, then the total number of games = 30.
This means that there are 10 games remaining.


We want the team to win at least 3/4 of its games.
3/4 of 30 = 22.5
So, in order to win at least 3/4 of the games, the team must win a total of 23 games or more.

The team has already won 17 games, so it needs to win at least 6 of the remaining 10 games.
Or we can say that the team can lose 4 games at most.

Answer = D

Cheers,
Brent