A certain game was played by 18 players as a team, in which only ${\frac{2}{3}}^{r d}$ of the team actually played the game. The rest were reserve players , ready to replace the main members if they get injured. How many reserve players did this team have?

$12$

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$6$

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

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

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Solution

The correct option is D
$\frac{1}{3} \times 18$

The team had $18 \times \frac{2}{3}$ = 12 main players. So, (18 - 12) = 6 players were reserve players.

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A certain game was played by 18 players as a team, in which only ${\frac{2}{3}}^{r d}$ of the team actually played the game. The rest were reserve players , ready to replace the main members if they get injured. How many reserve players did this team have?

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Q. A game was played by a team of 12 players, in which only

{\frac{2}{3}}^{r d}

of the team actually played the game. The rest were reserve players. How many reserve players did this team have?

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A certain game was played by 18 players as a team, in which only 23rd of the team actually played the game. The rest were reserve players , ready to replace the main members if they get injured. How many reserve players did this team have?

The correct option is D 13×18 The team had 18×23 = 12 main players. So, (18 - 12) = 6 players were reserve players.

A certain game was played by 18 players as a team, in which only ${\frac{2}{3}}^{r d}$ of the team actually played the game. The rest were reserve players , ready to replace the main members if they get injured. How many reserve players did this team have?

The correct option is D
$\frac{1}{3} \times 18$

The team had $18 \times \frac{2}{3}$ = 12 main players. So, (18 - 12) = 6 players were reserve players.