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# In an intercollegiate tournament, Rosebud college won 70% of the first

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Math Expert
Joined: 02 Sep 2009
Posts: 60605
In an intercollegiate tournament, Rosebud college won 70% of the first  [#permalink]

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10 Dec 2019, 05:47
00:00

Difficulty:

45% (medium)

Question Stats:

65% (02:37) correct 35% (02:16) wrong based on 22 sessions

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In an intercollegiate tournament, Rosebud college won 70% of the first 80 games played during the first session of the year. In order for the college to win at least 80% of the total games played during the entire year, how many more it will have to play in the remaining sessions of the year? (Note: The college wins all the games in the remaining sessions)

I. 35
II. 40
III. 50

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III

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Math Expert
Joined: 02 Aug 2009
Posts: 8341
Re: In an intercollegiate tournament, Rosebud college won 70% of the first  [#permalink]

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10 Dec 2019, 23:10
1
Bunuel wrote:
In an intercollegiate tournament, Rosebud college won 70% of the first 80 games played during the first session of the year. In order for the college to win at least 80% of the total games played during the entire year, how many more it will have to play in the remaining sessions of the year? (Note: The college wins all the games in the remaining sessions)

I. 35
II. 40
III. 50

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III

Let the minimum games required be x, so total games played = 80+x..
Thus, 80% of (80+x)=70% of 80 +x......$$\frac{80}{100}*(80+x)=\frac{70}{100}*80+x................6400+80x=5600+100x........800=20x...x=40$$

So 40 and 50 will ensure at least 80% wins

D
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Joined: 12 Apr 2019
Posts: 359
Re: In an intercollegiate tournament, Rosebud college won 70% of the first  [#permalink]

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10 Dec 2019, 23:53
1
This is a question where good old Algebra works very well. The most important key word in this question is ‘at least 80%’. Keep in mind that the overall percentage of wins need not be an integer value, as long as it is more than or equal to 80%.

Rosebud college won 70% of the first 80 games. This means that they won 56 of the first 80 games, whilst losing the rest.
Let Rosebud play ‘x’ games for the remaining sessions of the year. In this case,

Total number of games played during the entire year = 80 + x.

Total number of wins during the entire year = 56 + x ( since the question says that the college wins all the games in the remaining sessions).

Since the college has to win at least 80% of the total games played during the year,

56 + x ≥ $$\frac{4}{5}$$ (80 +x) {Note that the fraction $$\frac{4}{5}$$ is the equivalent of 80% and we are using the ≥ since this symbol represents 'at least'}. Solving the above inequality, we get x≥40.
Therefore, the possible values for x are 40 or 50. Statement I is not a possible value for x.

The correct answer option is D.

Note that answer option B is a trap answer for all of you who missed the ‘at least 80%’ part of the question stem and went ahead and found out 80% of the total games. This is a very common mistake that can happen in this question, so make sure that you re-read the question just to make sure that your answer matches the constraints.

Hope that helps!
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Re: In an intercollegiate tournament, Rosebud college won 70% of the first   [#permalink] 10 Dec 2019, 23:53
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