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total games won till now ; .8*15 = 12
and total games target 45 games
we are left with ; 50-15 ; 35 games
and 45-12 ; 33 to be won to achieve target
so 35-33 ; 2 games he can lose
IMO C

Bunuel
Anand has won 80 % of the games he has played so far in the tournament. His goal is to win 90% of all the games he has to play in the tournament. If he has already played 15 out of the total 50 games that he has to play, what is the maximum number of games he can afford to loose in the remaining games and yet meet his goal?

A. 0
B. 1
C. 2
D. 3
E. 4
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Solution



Given
In this question, we are given that
    • Anand has won 80% of the games he has played so far in the tournament
    • His goal is to win 90% of all the games he has to play in the tournament
    • He has already played 15 out of the total 50 games that he has to play

To find
We need to determine
    • The maximum number of games Anand can afford to lose in the remaining games and yet meet his goal

Approach and Working out
In the 15 games played, Anand won 80% of the games
    • Thus, the number of games Anand lost so far = 20% of 15 = 3

If he plays total 50 games with 90% win, the maximum number of loss possible = 10% of 50 = 5
    • Hence, the maximum number of losses Anand can afford in the remaining matches = 5 – 3 = 2

Thus, option C is the correct answer.

Correct Answer: Option C
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Deconstructing the Question

Total games = 50.
Goal: win 90% of all games.
He has already played 15 games and won 80% of them.

Step-by-step

Total wins required:

\(0.9 \times 50 = 45\)

Wins already achieved:

\(0.8 \times 15 = 12\)

Wins still needed:

\(45 - 12 = 33\)

Remaining games:

\(50 - 15 = 35\)

Maximum losses allowed:

\(35 - 33 = 2\)

Answer: 2
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