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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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New post 09 May 2018, 00:55
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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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New post 09 May 2018, 12:12
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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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New post 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.

Answer: D
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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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New post 05 Jul 2018, 19:09
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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
our ANSWER

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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New post 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


Answer D.



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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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New post 21 Sep 2018, 06:46
Top Contributor
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


Let T = TOTAL number of games played last year

The team won 50 percent of the games it played last year
So, TOTAL NUMBER OF WINS = 0.5T


In the SECOND HALF of last year, the team played 20 games, winning 3 of them
Number of WINS IN SECOND HALF = 3

If the team plays 20 games in the SECOND HALF, then T - 20 = the number of games played in the FIRST HALF

In the first half of last year, a team won 60 percent of the games it played.
So, the team won 60% of the T - 20 games it played in the FIRST HALF
Number of WINS IN FIRST HALF = (0.6)(T - 20)

We know that: WINS IN FIRST HALF + WINS IN SECOND HALF = TOTAL NUMBER OF WINS
So: (0.6)(T - 20) + 3 = 0.5T
Expand: 0.6T - 12 + 3 = 0.5T
Simplify left side: 0.6T - 9 = 0.5T
Subtract 0.6T from both sides to get: -9 = -0.1T
Solve: T = 90

Answer: D

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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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New post 21 Sep 2018, 07:12
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

From the fact that 60% = 3/5 and 50% = 1/2, let´s call 10M the number of games played in the first half.
Besides that, do not forget (in the table below) that half (10M+20) equals 5M+10. Therefore:

\(\begin{array}{*{20}{c}}
{} \\
{{1^{{\text{st}}}}\,{\text{half}}} \\
{{2^{{\text{nd}}}}\,{\text{half}}} \\
{{\text{Total}}}
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
{{\text{won}}} \\
{\frac{3}{5}\left( {10M} \right) = \boxed{6M}} \\
{\boxed3} \\
{\boxed{5M + 10}}
\end{array}\,\,\,\,\,\begin{array}{*{20}{c}}
{{\text{lost}}} \\
{\frac{2}{5}\left( {10M} \right) = 4M} \\
{17} \\
{5M + 10}
\end{array}\,\,\,\,\,\begin{array}{*{20}{c}}
{{\text{Total}}} \\
{10M} \\
{20} \\
{10M + 20}
\end{array}\)


\(? = 10M + 20\)


\(6M + 3 = 5M + 10\,\,\,\,\, \Rightarrow \,\,\,\,\,M = 7\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 90\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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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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New post 21 Sep 2018, 09:11
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



let x be total number of games played

let y be total number of games won


1st half / games played =x / games won 0.6x

2nd half / games played = 20 / games won 3


total games won = 0.5x
---------------------------------------------------

total games played 20+x = ? (plug in the equation below)

total games won 0.6x +3 = 0.5x

0.6x +3 = 0.5(20+x)
x = 70

hence 20+70 = 90 :-) i made it :)
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In the first half of last year, a team won 60 percent of the games it   [#permalink] 21 Sep 2018, 09:11
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