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# In the first half of last year, a team won 60 percent of the games it

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In the first half of last year, a team won 60 percent of the games it  [#permalink]

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09 May 2018, 00:55
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Difficulty:

75% (hard)

Question Stats:

64% (01:25) correct 36% (23:34) wrong based on 74 sessions

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In the first half of last year, a team won 60 percent of the games it played. In the second half of last year, the team played 20 games, winning 3 of them. If the team won 50 percent of the games it played last year, what was the total number of games the team played last year?

A) 60

B) 70

C) 80

D) 90

E) 100

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In the first half of last year, a team won 60 percent of the games it  [#permalink]

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09 May 2018, 12:12
1
x = number of games in the first half of last year
putting the questions into words.
0.6x+3=0.5(x+20)
0.6x+3=0.5x+10
0.1x=7
x=70

The total game played is number of games in the first half of last year + number of games in the second half of last year = 70+20 =90

Hence, option D = 90 is the answer.
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Re: In the first half of last year, a team won 60 percent of the games it  [#permalink]

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05 Jul 2018, 16:47
Bunuel wrote:
In the first half of last year, a team won 60 percent of the games it played. In the second half of last year, the team played 20 games, winning 3 of them. If the team won 50 percent of the games it played last year, what was the total number of games the team played last year?

A) 60

B) 70

C) 80

D) 90

E) 100

We can n = the number of games played in the first half. Thus, we see that 0.6n is the number of games won in the first half of the year and (0.6n + 3) is the total number of games won for the entire year. We also see that (n + 20) is the total number of games played during the entire year Thus, we can create the equation:

(0.6n + 3)/(n + 20) = 1/2

2(0.6n + 3) = n + 20

1.2n + 6 = n + 20

0.2n = 14

n = 70

Therefore, they played a total of 70 + 20 = 90 games last year.

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In the first half of last year, a team won 60 percent of the games it  [#permalink]

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05 Jul 2018, 19:09
1
Bunuel wrote:
In the first half of last year, a team won 60 percent of the games it played. In the second half of last year, the team played 20 games, winning 3 of them. If the team won 50 percent of the games it played last year, what was the total number of games the team played last year?

A) 60

B) 70

C) 80

D) 90

E) 100

Alternate ways:-

1) logical - short and crisp

Had the team won 12 out of team they would have remained at 60%.
BUT they win 12-3=9 matches less
Loss of 9 matches - winning % drops from 60% to 50%
so 10% of total matches including 20 played in second half = 9
total = $$9*\frac{100}{10}=90$$

2) substitute-
check from centre value
c) 80
so 60% of 60 + 3 = 36+3=39
It is NOT 50% of 80 or rather < 50%, so look for higher TOTAL
d) 90
so 60% of 70 + 3 = 42+3=45
It is 50% of 90

Algebraic method
let the initial games be x, so 60% of x = 0.6x
so Total games = x+20 and games won = 0.6x+3
as per info given - $$0.5(x+20)=0.6x+3..........0.5x+10=0.6x+3..............0.1x=7........x=70$$so total = 70+20=90

Best way for you to do - SUBSTITUTE
logical method is best if you have mastered it

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Re: In the first half of last year, a team won 60 percent of the games it  [#permalink]

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05 Jul 2018, 23:07
Bunuel wrote:
In the first half of last year, a team won 60 percent of the games it played. In the second half of last year, the team played 20 games, winning 3 of them. If the team won 50 percent of the games it played last year, what was the total number of games the team played last year?

A) 60

B) 70

C) 80

D) 90

E) 100

Given, First half winnings = 60%

Second half winnings = (3/20)*100 = 15%

Average winnings for the year = 50%

Let the total games played in the first half of the year = X

Using the Weighted Average / Alligation approach,

60%............50%...................15%

(50-15).........50%................(60-50)

We get, 35/10 = X/20

Hence, X = 70

Therefore Total # of games played for the year = 70 + 20 = 90

Thanks,
GyM
Re: In the first half of last year, a team won 60 percent of the games it &nbs [#permalink] 05 Jul 2018, 23:07
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