A certain game was played by 18 players as a team, in which ${\frac{2}{3}}^{r d}$ of the team actually played the game, while the rest were reserved players - ready to replace the main members who got injured. How many reserved players did this team have?

$12$

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$\frac{2}{3} \times 18$

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$\frac{1}{3} \times 18$

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$6$

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Solution

The correct option is D
$6$

Total number of players $= 18$
As per the given statement, the number of main players $= \frac{2}{3} \times 18 = 12$

So, the number of reserved players $=$ Total number of players $-$ Number of main players
$= 18 - 12 = 6$

Alternatively,
Fraction of main players $= \frac{2}{3}$
So, the fraction of reserved players $= 1 - \frac{2}{3} = \frac{1}{3}$

Hence, number of reserved players $= \frac{1}{3} \times 18 = 6$

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