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Bunuel
During a certain season, a team won 80 percent of its first 80 games and 56 percent of its remaining games. If the team won of its games in the entire season, what was the total number of games that the team played?

A. 180
B. 170
C. 156
D. 120
E. 105

Is the question complete. I see If the team won of its games in the entire season, what was the total number of games that the team played? I think the total no. of games the team won is missing..

Pls check.
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Fixed the typo. Thank you.
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Bunuel
During a certain season, a team won 80 percent of its first 80 games and 56 percent of its remaining games. If the team won 2/3 of its games in the entire season, what was the total number of games that the team played?

A. 180
B. 170
C. 156
D. 120
E. 105

Total no. of wins by the team will be (80/100)*80+(56/100)(x-80)
which is equal to (2/3)x
So 64+(56/100)x-224/5 = (2/3)x
=> 96/5 = (32/300)x
=> x = (96/5)*(300/32) = 180

Ans. is A.
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the team won 80% of its first 80 games and 56% of the remaining games... let the unknown be "X"..

as per the question.. 80*80/100 + X*56/100 = [2/3]*[80+X]
[6400+56X]/100 = [2/3]*[80+X]
19200 + 168X= 16000 + 200X
X=100

total games played = 80+X => 80+100 = 180 ..

A
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Let the total games played be \(x\)

Total wins = \(\frac{2}{3}x\)

Initial games played = 80

80% wins of 80 games is 64 wins

Remaining games = \(x-80\)

Wins from remaining games = \(\frac{56}{100}(x-80)\)

So we are given that

\(64+\frac{56}{100}(x-80)=\frac{2}{3}x\)

\(64+\frac{14}{25}x-\frac{224}{5}=\frac{2}{3}x\)

\(\frac{2}{3}x-\frac{14}{25}x=64-\frac{224}{5}\)

\(\frac{50-42}{75}x=\frac{320-224}{5}\)

\(\frac{8}{75}x=\frac{96}{5}\)

\(x=180\)

Answer is (A)
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Hi, my answer is A and my solution is similar to those above posts; however, it is quite time-consuming. So I'm waiting for the fastest solution for this question.
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Hi, my answer is A and my solution is similar to those above posts; however, it is quite time-consuming. So I'm waiting for the fastest solution for this question.

Hi, DarleneTran

Here's an alternate approach.

From the question, we can see that the total number of games must be a multiple of 3. So we can eliminate B right away

And then, (Total games minus 80) must be a multiple of 25 since 56/100 = 14/25

We are left with A and E now

At this point, we can plug in 180 and 105 to see which one satisfies the conditions given in the question and we'll end up with A

Hope this helps :)
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Hi firas92

Thanks for your explanation. "From the question, we can see that the total number of games must be a multiple of 3" I am wondering whether you mean that checking from each answer choice? Cause I couldn't get this point from the question stem, just only from given answer choices.
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DarleneTran
Hi firas92

Thanks for your explanation. "From the question, we can see that the total number of games must be a multiple of 3" I am wondering whether you mean that checking from each answer choice? Cause I couldn't get this point from the question stem, just only from given answer choices.

We are given that the team won 2/3 of its total games, and we're dealing with integer values here because a team cannot win 1.3 or 2.5 games or so.. So 2/3*(Total number of games) must be an integer. Obviously 2 is not a multiple of 3 and so, the total number of games must be a multiple of 3 so as to make 2/3*(Total number of games) an integer
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80-------------------56
66
(66-56) (80-56)
10 14


the number of 80 percent games to 56 % games should be in the ratio 10:14 ie 5:7

therefore no. of 56% games = (80/5)*7= 112

total around 180.

( if we take exact 66.66 for two third. total will be exact 180
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