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

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In the first half of last year, a team won 60 percent of the [#permalink]  05 Mar 2017, 09:11
Expert's post
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Question Stats:

52% (01:46) correct 47% (01:34) wrong based on 87 sessions

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
[Reveal] Spoiler: OA

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Re: In the first half of last year, a team won 60 percent of the [#permalink]  11 Mar 2017, 20:13
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Expert's post
Explanation

Total games last year = $$x$$

Games in 2nd half of year = $$20$$

Games in 1st half of year = $$x - 20$$

60% of games in first half + 3 games in second half = 50% of the total games

$$0.60 \times (x - 20) + 3 = 0.50 \times x$$

$$0.60(x - 20) + 3 = 0.50x$$

$$0.60x - 12 + 3 = 0.50x$$

$$0.60x - 9 = 0.50x$$

$$-9 = -0.10x$$

$$x =90$$

Hence option D is correct.
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Re: In the first half of last year, a team won 60 percent of the [#permalink]  17 Feb 2018, 04:17
Please could you correct me since there is possibility that I can make this mistake again …

I did the following
X = number of total games in the first half of the year
0.6x+3=0.5(x+20)
0.6x+3=0.5x+10
0.1x=7
X=70, which is not correct
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Joined: 07 Jun 2014
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GRE 1: Q167 V156
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Re: In the first half of last year, a team won 60 percent of the [#permalink]  17 Feb 2018, 12:35
Expert's post
yell2012prime wrote:
Please could you correct me since there is possibility that I can make this mistake again …

I did the following
X = number of total games in the first half of the year
0.6x+3=0.5(x+20)
0.6x+3=0.5x+10
0.1x=7
X=70, which is not correct

X=70 is correct!. Now add the 20 to get toatal games.
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Sandy
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Re: In the first half of last year, a team won 60 percent of the [#permalink]  17 Feb 2018, 16:15
sandy wrote:
yell2012prime wrote:
Please could you correct me since there is possibility that I can make this mistake again …

I did the following
X = number of total games in the first half of the year
0.6x+3=0.5(x+20)
0.6x+3=0.5x+10
0.1x=7
X=70, which is not correct

X=70 is correct!. Now add the 20 to get toatal games.

trial and error made me take something like 2.5 minutes to answer this question... guess that's too long, yikes
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Re: In the first half of last year, a team won 60 percent of the [#permalink]  17 Feb 2018, 17:03
1
KUDOS
sandy wrote:
yell2012prime wrote:
Please could you correct me since there is possibility that I can make this mistake again …

I did the following
X = number of total games in the first half of the year
0.6x+3=0.5(x+20)
0.6x+3=0.5x+10
0.1x=7
X=70, which is not correct

X=70 is correct!. Now add the 20 to get toatal games.

Thank you, this is why I have to REread prompt again and don't rush
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Kudos [?]: 17 [0], given: 9

Re: In the first half of last year, a team won 60 percent of the [#permalink]  17 Feb 2018, 17:10
matayo wrote:
sandy wrote:
yell2012prime wrote:
Please could you correct me since there is possibility that I can make this mistake again …

I did the following
X = number of total games in the first half of the year
0.6x+3=0.5(x+20)
0.6x+3=0.5x+10
0.1x=7
X=70, which is not correct

X=70 is correct!. Now add the 20 to get toatal games.

trial and error made me take something like 2.5 minutes to answer this question... guess that's too long, yikes

If you solve 20 such quest you can do it in 30 sec (after 3rd one), but we should be more careful with final answer bc there is possibility to make careless mistake (like I did when I took full cat)
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Re: In the first half of last year, a team won 60 percent of the [#permalink]  23 Feb 2018, 00:11
Brilliant technique!

sandy wrote:
Explanation

Total games last year = $$x$$

Games in 2nd half of year = $$20$$

Games in 1st half of year = $$x - 20$$

60% of games in first half + 3 games in second half = 50% of the total games

$$0.60 \times (x - 20) + 3 = 0.50 \times x$$

$$0.60(x - 20) + 3 = 0.50x$$

$$0.60x - 12 + 3 = 0.50x$$

$$0.60x - 9 = 0.50x$$

$$-9 = -0.10x$$

$$x =90$$

Hence option D is correct.
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Re: In the first half of last year, a team won 60 percent of the [#permalink]  05 Jul 2018, 15:56
1
KUDOS
Expert's post
Carcass 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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Re: In the first half of last year, a team won 60 percent of the [#permalink]  21 Sep 2018, 05:46
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Expert's post
Carcass 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

Cheers,
Brent
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Re: In the first half of last year, a team won 60 percent of the [#permalink]  01 Oct 2018, 03:22
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

every time when I need help with multiple choice questions or other assignments I use [url="https://assignmentmasters.org/"]"https://assignmentmasters.org/"[/url]
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Re: In the first half of last year, a team won 60 percent of the [#permalink]  21 May 2019, 06:24
1
KUDOS
yell2012prime wrote:
Please could you correct me since there is possibility that I can make this mistake again …

I did the following
X = number of total games in the first half of the year
0.6x+3=0.5(x+20)
0.6x+3=0.5x+10
0.1x=7
X=70, which is not correct

this is your total game, NOT X only.
So, you are correct if you calculate X+20=> 70+20=> 90
I guess, here lies the tricks of the math. If we take Sandy's way, then we will not fall into the GRE trap!
Re: In the first half of last year, a team won 60 percent of the   [#permalink] 21 May 2019, 06:24
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