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During a certain season, a team won 80 percent of its first 100 games

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During a certain season, a team won 80 percent of its first 100 games

by BTGModeratorVI » Fri Jul 03, 2020 7:12 am

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During a certain season, a team won 80 percent of its first 100 games and 50 percent of its remaining games. If the team won 70 percent of its games for the entire season, what was the total number of games that the team played?

(A) 180
(B) 170
(C) 156
(D) 150
(E) 105

Answer: D
Source: Official guide
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Re: During a certain season, a team won 80 percent of its first 100 games

by Brent@GMATPrepNow » Fri Jul 03, 2020 8:12 am
BTGModeratorVI wrote: ↑
Fri Jul 03, 2020 7:12 am
 During a certain season, a team won 80 percent of its first 100 games and 50 percent of its remaining games. If the team won 70 percent of its games for the entire season, what was the total number of games that the team played?

(A) 180
(B) 170
(C) 156
(D) 150
(E) 105

Answer: D
Source: Official guide
If we let G = the TOTAL number of games played in the ENTIRE SEASON, then ...
G - 100 = the number of games REMAINING after the first 100 have been played

We can now start with a "word equation":
(# of wins in 1st 100 games) + (# of wins in remaining games) = (# of wins in ENTIRE season)
We get: (80% of 100) + (50% of G-100) = 70% of G
Rewrite as 80 + 0.5(G - 100) = 0.7G
Expand: 80 + 0.5G - 50 = 0.7G
Simplify: 30 = 0.2G
Solve: G = 150
Answer: D
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Re: During a certain season, a team won 80 percent of its first 100 games

by swerve » Sat Jul 04, 2020 7:13 am
BTGModeratorVI wrote: ↑
Fri Jul 03, 2020 7:12 am
 During a certain season, a team won 80 percent of its first 100 games and 50 percent of its remaining games. If the team won 70 percent of its games for the entire season, what was the total number of games that the team played?

(A) 180
(B) 170
(C) 156
(D) 150
(E) 105

Answer: D
Source: Official guide
First \(100\) games \(\Longrightarrow\) win \(80\% = 80\)
Remaining games \(\Longrightarrow X\)
Remaining games won \(\Longrightarrow (50/100)X\)
Total games \(100\) first \(+ X \Longrightarrow\) won \(\dfrac{70}{100} (100+X)\)
So.

\(80+\dfrac{50}{100}x= \dfrac{70}{100} (100+x)\)
\(80-70 ={70}{100}x-\dfrac{50}{100}x\)
\(10 = \dfrac{20}{100}x\)

\(x=50\) (remaining games) \(\Longrightarrow\) Total \(= 100 +50=150 \Longrightarrow\) D
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