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 reserve players – ready to replace the main members who got injured. How many reserve players did this team have?

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Solution

The correct options are
B
6

D

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 ${\frac{2}{3}}^{r d}$ of the team actually played the game, while the rest were reserve players – ready to replace the main members who got injured. 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 (23)rd of the team actually played the game, while the rest were reserve players – ready to replace the main members who got injured. How many reserve players did this team have?

The correct options are B 6 D 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 $(\frac{2}{3})^{r d}$ of the team actually played the game, while the rest were reserve players – ready to replace the main members who got injured. How many reserve players did this team have?

The correct options are
B
6

D

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