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# A college basketball team has won 60% of its games, with 15

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A college basketball team has won 60% of its games, with 15 [#permalink]

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14 May 2007, 06:52
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A college basketball team has won 60% of its games, with 15 games remaining on the season's schedule. If the team is to win at least 60% of its scheduled games for the entire season, at most how many of the remaining games can the team lose?
A. 6
B. 7
C. 8
D. 9
E. 10

A school supply store sells only one kind of desk and one kind of chair, at a uniform cost per desk or per chair. If the total cost of 3 desks and 1 chair is twice that of 1 desk and 3 chairs, then the total cost of 4 desks and 1 chair is how many times that of 1 desk and 4 chairs?
A. 5
B. 3
C. 8/3
D. 5/2
E. 7/3
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14 May 2007, 07:07
A college basketball team has won 60% of its games, with 15 games remaining on the season's schedule. If the team is to win at least 60% of its scheduled games for the entire season, at most how many of the remaining games can the team lose?
A. 6
B. 7
C. 8
D. 9
E. 10

15 games is 40% of all games then 100% 15/0.4= 38

38*. 60= 23 games they have to win to get 60%

they won 60% of 60% so 23*0.6= 14 games they won already so the answer is 23-14=9

P.S. probably you have a faster way to solve the problem but this one makes more sense to me.
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14 May 2007, 07:10
A school supply store sells only one kind of desk and one kind of chair, at a uniform cost per desk or per chair. If the total cost of 3 desks and 1 chair is twice that of 1 desk and 3 chairs, then the total cost of 4 desks and 1 chair is how many times that of 1 desk and 4 chairs?
A. 5
B. 3
C. 8/3
D. 5/2
E. 7/3

3d+1c=2(1d+3c); 3d+1c=2d+6c; d=5c

4d+1c=x(1d+4c) substitute d with 5c and get 21c=x*9c x=21c/9c=7/3

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14 May 2007, 07:13
Jamesk486 wrote:
A school supply store sells only one kind of desk and one kind of chair, at a uniform cost per desk or per chair. If the total cost of 3 desks and 1 chair is twice that of 1 desk and 3 chairs, then the total cost of 4 desks and 1 chair is how many times that of 1 desk and 4 chairs?
A. 5
B. 3
C. 8/3
D. 5/2
E. 7/3

(E) it is

3d + c = 2(d + 3c)
=> d = 5c

=> 4d + c = 21c and d + 4c = 9c

=> 4d + c = 21/9 (d + 4c)
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14 May 2007, 22:25
Manager
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14 May 2007, 22:36
A college basketball team has won 60% of its games, with 15 games remaining on the season's schedule. If the team is to win at least 60% of its scheduled games for the entire season, at most how many of the remaining games can the team lose?
A. 6
B. 7
C. 8
D. 9
E. 10

simple again from 15 games, team has to win 60% to maintain overall 60% figure.
so it can loose 40% at max. 15*0.4=6
A
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14 May 2007, 23:32
A college basketball team has won 60% of its games, with 15 games remaining on the season's schedule. If the team is to win at least 60% of its scheduled games for the entire season, at most how many of the remaining games can the team lose?
A. 6
B. 7
C. 8
D. 9
E. 10

The team needs to win this amount of games at seasons end: .6(.6x + 15). .6(.6x+15) = .36x + 9. In order for the team to maintain their 6/10 win ratio they'll have to win 9 of the 15 games. They can afford to lose 6. Does anyone object to this method? Does anyone have a better one?
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14 May 2007, 23:36
good approach by Apache

The ans is A i.e 6

Sergey when you say

Quote:
15 games is 40% of all games then 100% 15/0.4= 38

it is not correct as we don't know if 15 games are 40% of the games.

The question says that the team had won 60% for the games played

Let the games played till now be x

Games won = 0.6x
Games lost = 0.4x

Total no of games = x + 15
60% of entire season games = 0.6(x+15) = 0.6x + 9

No of games it needs to win = 0.6x +9 - 0.6x = 9.

Games it can lose = 15 - 9 = 6

A 6 it is.
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15 May 2007, 02:22
guys this is a very simple que.

just check this fact.
60%of A+60% of B =k
then K = 60% of (A+B)
15 May 2007, 02:22
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