Question

A particle moving in a straight line covers one-third of
total distance with 50 km/h and the remaining two third
with 60 km/h. The average speed of the particle is
​

Answer 1

Let the Total Distance to be traveled by the Particle be : 3D

$\mathtt{\bigstar\;\;One-third\;of\;the\;total\;distance = \dfrac{3D}{3} = D}$

Given : Particle covers One - third of total distance with 50 kmph

$\mathtt{\bigstar\;\;Time\;taken = \dfrac{Distance\;traveled}{Speed}}$

$\mathtt{\implies Time\;taken = \dfrac{D}{50}}$

$\mathtt{\bigstar\;\;Two-third\;of\;the\;total\;distance = \dfrac{3D \times 2}{3} = 2D}$

Given : Particle covers Two - third of total distance with 60 kmph

$\mathtt{\implies Time\;taken = \dfrac{2D}{60} = \dfrac{D}{30}}$

$\mathtt{\bigstar\;\;Average\;Speed = \dfrac{Total\;Distance\;traveled}{Total\;Time\;taken}}$

$\mathtt{\implies Average\;Speed = \dfrac{3D}{\dfrac{D}{50} + \dfrac{D}{30}}}$

$\mathtt{\implies Average\;Speed = \dfrac{3D}{\dfrac{3D + 5D}{150}}}$

$\mathtt{\implies Average\;Speed = \dfrac{3D}{\dfrac{8D}{150}}}$

$\mathtt{\implies Average\;Speed = \dfrac{3D \times 150}{8D}}$

$\mathtt{\implies Average\;Speed = \dfrac{3 \times 150}{8}}$

$\mathtt{\implies Average\;Speed = ({3 \times 18.75})}$

$\mathtt{\implies Average\;Speed = 56.25\;kmph}$

Answer 2

Let Total Distance =x

According to Question

1/3 of X covered with speed of 50 km/h

Distance Covered at 50 km/h=1/3×x=x/3

Time=Distance÷Speed

$\frac{x}{3} \div 50 \\ \\ \frac{x}{3} \times \frac{1}{50} \\ \\ \frac{x}{150} \: hours$

Time=X/150 hours

Now 2/3 of X covered at Speed of 60 km/h

Distance covered=2/3×x=2x/3

Speed=60 km/h

Time=Distance÷Speed

$\frac{2x}{3} \div 60 \\ \\ \frac{2x}{3} \times \frac{1}{60} \\ \\ \frac{2x}{18 0} \\ \\ = \frac{x}{90} \: hours$

Now We know

$\mathtt{average \: speed = \frac{total \: distance}{total \: time} }$

Total Distance=x km

Total Time=x/150+x/90

LCM of 150 and 90=450

$\mathtt{ \frac{x}{150} + \frac{x}{90} } \\ \\ \frac{3x + 5x}{450} \\ \\ \frac{8x}{450}$

Total Time=8x/450 hours

$\mathtt{average \: speed = x \div \frac{8x}{450} } \\ \\ = \mathtt{ x \times \frac{450}{8x} } \\ \\ = \frac{450}{8} \\ \\ = \frac{225}{4} \\ \\ = 56.25$

$\underline{\mathtt{Average\:Speed=56.25\:km/h}}$