Question

In the recently concluded Olympics, Germany played a total of m football matches. It did not win p
games and it did not lose q games. What fraction of the games it played were a draw?
(p+q)/m-1
(p-q)/m +1
(m-p-9)/m
Insufficient Data​

Answer 1

Correct Option:

(c) $\frac{(m-p-q)}{m}$

Answer:

$\frac{(m-p-q)}{m}$ is the fraction of the games Germany played was a draw. So, the correct option is (c)  $\frac{(m-p-q)}{m}$.

Given:

The total number of matches Germany played is m.

The total number of matches not won by Germany is p games.

The total number of matches not lost by Germany is q games.

To find:

The fraction needs to be determined in which Germany played was a draw.

Solution:

Let's assume that the games played were as a draw is d.

Given the total number of games played by Germany is m.

Germany did not win the games meaning the games lost by Germany are p.

Germany did not lose the games meaning the games won by Germany are q.

Therefore, the total number of games played will be equal to the sum of the games lost by Germany, the games won by Germany, and the games in which Germany played were a draw.

⇒ The total number of games played = Games lost + Games won + Games which was draw

⇒ m = p + q + d ......... (i)

⇒ d = m - p - q ........... (ii)

Therefore, the number of games drawn which were played by Germany is d = m - p - q.

So, the fraction of the games it played was a draw = $\frac{The number of games drawn}{The total number of games played}$ = $\frac{d}{m}$.

From equations (i) and (ii) put the value of m and d in the above equation.

⇒ Fraction = $\frac{d}{m}$ = $\frac{m-p-q}{p+q+d}$ or $\frac{m-p-q}{m}$.

Hence, the fraction of the games that were drawn which were played by Germany was in option (c) $\frac{m-p-q}{m}$.

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