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

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During a certain season, a team won 80 percent of its first [#permalink]

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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
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Re: During a certain season, a team won 80 percent of its first [#permalink]

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Walkabout wrote:
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


This is simple weighted average question.

Let the # of the remaining games be \(x\) then \(0.8*100+0.5*x=0.7*(100+x)\) --> \(x=50\) --> total # of games thus equal to \(100+x=100+50=150\).

Answer: D.
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Re: During a certain season, a team won 80 percent of its first [#permalink]

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New post 31 Jul 2014, 19:53
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First...................... Remain ................. Total
100 ........................ x ............................ 100+x

Won
80 ............................. \(\frac{50x}{100}\) ...................... \(\frac{70}{100} * (100+x)\)

Equation would be

\(80 + \frac{50x}{100} = 70 + \frac{70x}{100}\)

x = 50

Total games = 100 + 50 = 150

Answer = D
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Re: During a certain season, a team won 80 percent of its first [#permalink]

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Walkabout wrote:
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


Total # of games = x
# of games won in first 100 = .8 * 100
# of games won in remaining games = .5 * (x-100)
# of games won in entire season = .7x

Now, .7x = .8 * 100 + .5 * (x-100)
.7x - .5x = 80 - 50
.2x = 30
x = 300/20 = 150
Ans: D

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Re: During a certain season, a team won 80 percent of its first [#permalink]

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Quote:
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


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 [#permalink]

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New post 15 Nov 2015, 23:43
Walkabout wrote:
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


Given: A team won 80 percent of its first 100 games and 50 percent of its remaining games.
Team won 70 percent of its total games
Required: Total number of games played?

Assume that the remaining games = 100x
Total games won = 80 + 50x
This is 70% of the total games played.

0.7*(100x + 100) = 80 + 50x
70x + 70 = 80 +50x
x = 0.5

Hence 100x = 50

Total games played = 100 + 100x = 150 Option D

Note: We used 100x to avoid the usage of unitary method caused by assuming 100

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Re: During a certain season, a team won 80 percent of its first [#permalink]

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New post 28 Jun 2016, 04:51
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Walkabout wrote:
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


We are first given that a team won 80 percent of its first 100 games. This means the team won 0.8 x 100 = 80 games out of its first 100 games.

We are next given that the team won 50 percent of its remaining games. If we use variable T to represent the total number of games in the season, then we can say T – 100 equals the number of remaining games in the season. Thus we can say:

0.5(T – 100) = number of wins for remaining games

0.5T – 50 = number of wins for remaining games

Lastly, we are given that team won 70 percent of all games played in the season. That is, they won 0.7T games in the entire season. With this we can set up the equation:

Number of first 100 games won + Number of games won for remaining games = Total Number of games won in the entire season

80 + 0.5T – 50 = 0.7T

30 = 0.2T

300 = 2T

150 = T

Answer is D.
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Re: During a certain season, a team won 80 percent of its first [#permalink]

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New post 27 Nov 2016, 13:44
The extremes are 50 % and 80 %. 70% is the overall percent:

50 -------------70-------80
...........20.......10......

The difference between these numbers are in blue, above.

The ratio of the initial set (100 games) and the second set (x games) will be:
100/x = 20/10.
Thus x = 50. Total number of games played = 100 + x = 150
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During a certain season, a team won 80 percent of its first [#permalink]

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New post 05 Jan 2017, 17:10
First 100 games -> win 80% = 80
Remaining games -> X
Remaining games won-> (50/100)X
Total games 100first + X -> won 70/100 (100+X)
So..

80+(50/100)x= 70/100 (100+x)
80-70 =(70/100)x-(50/100)x
10 = (20/100)x
x=50 (remaining games) --> Total = 100 +50=150 D

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Re: During a certain season, a team won 80 percent of its first [#permalink]

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New post 01 Apr 2017, 08:41
Walkabout wrote:
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


Total games= 100+x
Won= 0.80*100= 80
and also won= 0.50*x

80+0.5x=0.70(100+x)
160+x=(7/5)* (100+x)
800+5x=700+7x
x=50

Therefore total number of games played by the team are (100+50=150).

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Re: During a certain season, a team won 80 percent of its first [#permalink]

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New post 16 Dec 2017, 02:58
Let the games above 100 be : Y

What we know. 80% won from 100 & 50 % won from Y games.
Final winning average is 70 %

Equation becomes

.8 * 100 + .5 * Y= .7 * ( 100 + Y)

Solve for y , answer is D = 150

What does the equation mean
LHS: 80 % of 100 games Plus 50 % of remaning Y games will tell us how many total games the team won.
RHS: Tells us if we multiply final winning % with Total games, we will find out how many games the team won.

Since LHS & RHS are telling us the same thing, they become equatable.

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Re: During a certain season, a team won 80 percent of its first   [#permalink] 16 Dec 2017, 02:58
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