In a game, A can give 36 points to B, A can give 42 point to C, B can give 10 points to C. How many points make the game?

90 points

The correct answer is Option (3) → 90 points

Let the game be of $N$ points.

A can give 36 points to B implies $A : B = N : (N-36)$

A can give 42 points to C implies $A : C = N : (N-42)$

B can give 10 points to C implies $B : C = N : (N-10)$

From first ratio: $\frac{A}{B}=\frac{N}{N-36}$

From second ratio: $\frac{A}{C}=\frac{N}{N-42}$

From third ratio: $\frac{B}{C}=\frac{N}{N-10}$

Now, $\frac{A}{B}\times\frac{B}{C}=\frac{A}{C}$

$\frac{N}{N-36}\times\frac{N}{N-10}=\frac{N}{N-42}$

$\frac{N^2}{(N-36)(N-10)}=\frac{N}{N-42}$

Cross-multiply:

$N^2(N-42)=N(N-36)(N-10)$

Cancel $N$ (since $N\neq 0$):

$N(N-42)=(N-36)(N-10)$

$N^2-42N=N^2-46N+360$

Bring terms together:

$-42N=-46N+360$

$4N=360$

$N=90$

Answer: The game is of 90 points.