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

Given,
The team had players $= 18$
Number of players who actually played
$= \frac{2}{3} \times Total number of player in team$
$= 18 \times \frac{2}{3}$
$= 12$ main players.
Therefore, number of reserve players are
$(18 - 12) = 6$
The correct options are (B) and (D).

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A certain game was played by 18 players as a team, in which 23rd 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?

Given,The team had players =18Number of players who actually played=23×Total number of player in team =18×23 =12main players. Therefore, number of reserve players are(18−12)=6The correct options are (B) and (D).

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?

Given,
The team had players $= 18$
Number of players who actually played
$= \frac{2}{3} \times Total number of player in team$
$= 18 \times \frac{2}{3}$
$= 12$ main players.
Therefore, number of reserve players are
$(18 - 12) = 6$
The correct options are (B) and (D).