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eqn :-

=> 8 + 1/4*(x-8)=1/2 * x

=> x = 24

x total number of games
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Bunuel
A basketball team won its first 8 games of the season, and then won 25% of its remaining games. If it won 50% of its games for the entire season, how many games did it play that season?

A. 12
B. 16
C. 18
D. 20
E. 24

let t=total games
8+(t-8)/4=t/2
t=24
E
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8+x/4/8+x=1/2
32+x/4/8+x=1/2
32+x/4(8+x)=1/2
32=2x
x=16
Answer=16+8=24=E
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Bunuel
A basketball team won its first 8 games of the season, and then won 25% of its remaining games. If it won 50% of its games for the entire season, how many games did it play that season?

A. 12
B. 16
C. 18
D. 20
E. 24

Let X = total games

8 + 1/4 (X - 8) = 1/2 X

8 + (1/4) X - 2= 1/2 X

1/4 X = 6...........X =24

Answer: E
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8+x/4-2 = x/2
So X= 24
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We can use answer choices as well.
Let us start from the middle.
Total games played is=18
50% won=9
First 8 games won then game left is 1 now this cannot be 25% of the rest of the 25% games won.
Let us take a higher no.
Let total no of games Played=24
50% of games won=12
first 8 games won left is 4.(Total games won is 12)
Rest of the games left is 16(24-8) and 25% of 16 is 4. Total games won -=8+4=12 Answer.
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I think this question can be solved by weighted averages.

Let the # of the remaining games be R then

8 + 1/4 (R) = 1/2 (R+8)

Solve for R.
After solving, R = 16
Total = R+8 = 16+8 = 24 = E
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A basketball team won its first 8 games of the season, and then won 25% of its remaining games. If it won 50% of its games for the entire season, how many games did it play that season?

A. 12
B. 16
C. 18
D. 20
E. 24

We could use alligation
A1 = 100 (Win % = 100 as the basketball team won all of its first 8 games)
A2 = 25(Win % = 25 for remaining games - stem)
Aavg = 50%(stem)
\(A2-Aavg/Aavg-A1\) =\( 25-50/50-100\) = \(1/2\)
\(1/3\)(Total) = 8
Total = 24

E
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