Given that: Distance from X-ville to Y-ville = 70km
Speeds are constant throughout the journey
Ann's speed > Bea's speed

Target question => When they meet, how far has Bea traveled?

Statement 1 => Ann's speed is 30km/hr greater than Bea's speed

Let distance traveled by Bea = x
Let Bea's speed = y
Let Ann's speed = (y + 30)km/hr
Let distance traveled by Ann on the journey back to X-ville = a

Total distance traveled by Ann = (70+a)km
Total distance traveled by Bea = (70-a)km

Ann's travel time =- Bea's travel time
$$where\ Ann's\ travel\ time=\frac{Ann's\ total\ dis\tan ce}{Ann's\ speed}=\frac{70+a}{y+30}$$
$$and\ Bea's\ time=\frac{Bea's\ total\ dis\tan ce}{Bea's\ speed}=\frac{70-a}{y}$$
$$\frac{70+a}{y+30}=\frac{70-a}{y}$$
$$y\left(70+a\right)=\left(y+30\right)\left(70-a\right)$$
$$70y+ay=70y-ay+2100-30a$$
$$ay+ay+30a=2100$$
$$2ay+30a=2100$$
$$\frac{2a\left(y+15\right)}{2}=\frac{2100}{2}$$
$$a\left(y+15\right)=1050$$
$$Value\ of\ a\ and\ y\ are\ unknown\ ,statement\ 1\ is\ NOT\ SUFFICIENT$$

Statement 2 => Ann's speed is twice Bea's speed
Let distance traveled by Ann on the journey back = a
Total distance traveled by Ann = (70+a)km
Total distance traveled by Bea = (70-a)km
Let Bea's speed = y
Ann's speed = 2y
Ann's travel time = Bea's travel time

$$\frac{70+a}{2y}=\frac{70-a}{y}$$
$$y\left(70+a\right)=2y\left(70-a\right)$$
$$70y+ay=140y-2ay$$
$$ay+2ay=140y-70y$$
$$\frac{3ay}{3y}=\frac{70y}{3y}$$
$$a=\frac{70}{3}$$
$$total\ dis\tan ce\ traveled\ by\ Bea\ =\ 70-a$$
$$where\ a\ =\frac{70}{3}$$
$$=70\ -\frac{70}{3}$$
$$=\frac{210-70}{3}=\frac{140}{3}=46.67km$$
$$Statement\ 2\ alone\ IS\ SUFFICIENT$$
$$Answer\ =\ B$$