From two places, 60 km apart, A and B start towards each other at the same time and meet each other after 6 hour. If A traveled with $$\frac{2}{3}$$ of his speed and B traveled with double of his speed, they would have met after 5 hours. The speed of A is:

A →_______60Km_________← B
Let the speed of A = x kmph and that of B = y kmph

According to the question;
x × 6 + y × 6 = 60
Or, x + y = 10 --------- (i)
And,
$$\left( {\frac{{2{\text{x}}}}{3} \times 5} \right) + \left( {2{\text{y}} \times 5} \right) = 60$$
Or, 10x + 30y = 180
Or, x + 3y = 18 ---------- (ii)
From equation (i) × 3 - (ii)
3x + 3y - x - 3y = 30 - 18
Or, 2x = 12
Hence, x = 6 kmph

Alternate

$$\because $$ They meet after 6 hours if they walk towards each other i.e., their speed will be added.
So, their relative speed in opposite direction
$$ = \frac{{{\text{Distance }}}}{{{\text{Time }}}} = \frac{{60}}{6}$$
Relative speed in opposite direction :
$$\left( \rightleftharpoons \right) = 10{\text{ km/h}}.....{\text{(i)}}$$
According to the question,
$$\eqalign{ & \Rightarrow \frac{2}{3}A + 2B = \frac{{60}}{5} \cr & \Rightarrow \frac{2}{3}A + 2B = 12 \cr & \Rightarrow A + 3B = 18 \cr & \Rightarrow B's{\text{ Speed = }}\frac{{18 - A}}{3} \cr & \Rightarrow A + B = 10 \cr & \Rightarrow A + \frac{{18 - A}}{3} = 10 \cr & \Rightarrow 3A + 18 - A = 30 \cr & \Rightarrow 2A = 12 \cr & \Rightarrow A{\text{'s speed = 6 km/h}} \cr} $$