Question

To divide a line segment $AB$ internally in the ratio $3:5$, first a ray $AX$ is drawn so that $BAX$ is an acute angle and then at equal distances points are marked on the ray $AX$ such that the minimum number of these points is :

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

Answer: (D)

$8$

To divide a line segment $AB$ in the ratio $m:n$, a ray $AX$ making an acute $\angle BAX$ is drawn and then $(m+n)$ points are marked at equal distances on ray $AX$

$m=3$ and $n=5$

$\therefore$Minimum number of points to be marked on $AX$

$\Rightarrow m+n=3+5=8$