Question

A car moving on a straight road covers one third of a certain distance with 20 km/h and the rest with 60 km/h. The average speed is : 1 point (A) 40 km/h (B) 80 km/h (C) 46 km/h (D) 36 km/h​

Answer 1

AnswEr :

Let us assume the total distance covered by the car is x km.

― According to the question, it has been given to use that, it covers ⅓ of the certain distance with 20 km/h. This means, in the first case,

$\longmapsto \rm { S_1 = \dfrac{1}{3}(Certain \; Distance) } \\ \\ \longmapsto \underline{\bf { S_1 = \dfrac{1}{3}x \; km }}$

And,

• $\rm { V_1 = 20 \; kmh^{-1} }$

As we know that, Time = Distance ÷ Speed

Thus, time taken to cover ⅓ of the certain distance with 20 km/h :

$\longmapsto \rm { T_1 = S_1 \div V_1} \\ \\ \longmapsto \rm { T_1 = \Bigg ( \dfrac{1}{3}x \div 20 \Bigg ) \; h} \\ \\ \longmapsto \rm { T_1 = \Bigg ( \dfrac{1}{3}x \times \dfrac{1}{20} \Bigg ) \; h} \\ \\ \longmapsto \underline{\bf { T_1 = \dfrac{x}{60} \; h}}$

Also, it covers rest distance with the speed of 60 km/h. So, in the second case

$\longmapsto \rm { S_2 = S_{(Total)} - S_1 } \\ \\ \longmapsto \rm { S_2 = \Bigg ( x - \dfrac{1}{3}x \Bigg ) \; km } \\ \\ \longmapsto \rm { S_2 = \Bigg ( \dfrac{3x - x}{3}\Bigg ) \; km } \\ \\ \longmapsto \underline{\bf { S_2 = \dfrac{2x}{3} \; km}}$

And,

• $\rm { V_2 = 60 \; kmh^{-1} }$

As we know that, Time = Distance ÷ Speed

Thus, time taken to cover ⅔ of the certain distance with 60 km/h :

$\longmapsto \rm { T_2 = S_2 \div V_2} \\ \\ \longmapsto \rm { T_2 = \Bigg ( \dfrac{2}{3}x \div 60 \Bigg ) \; h} \\ \\ \longmapsto \rm { T_2 = \Bigg ( \dfrac{2}{3}x \times \dfrac{1}{60} \Bigg ) \; h} \\ \\ \longmapsto \underline{\bf { T_2 = \dfrac{x}{90} \; h}}$

Now, by using the formula of average speed,

$\longmapsto \underline{\boxed {\rm{Speed_{(Avg)} = \dfrac{Total \; distance}{Total \; time} }}} \\ \\ \longmapsto\rm{Speed_{(Avg)} = \Bigg \{ S_{(Total)} \div \Bigg (T_1 + T_2 \Bigg ) \Bigg \} \; kmh^{-1} } \\ \\ \longmapsto\rm{Speed_{(Avg)} = \Bigg \{x \div \Bigg (\dfrac{x}{60}+ \dfrac{x}{90} \Bigg ) \Bigg \} \; kmh^{-1} } \\ \\ \longmapsto\rm{Speed_{(Avg)} = \Bigg \{x \div \Bigg ( \dfrac{3x+2x}{180} \Bigg ) \Bigg \} \; kmh^{-1} } \\ \\ \longmapsto\rm{Speed_{(Avg)} = \Bigg \{x \div \Bigg ( \dfrac{5x}{180} \Bigg ) \Bigg \} \; kmh^{-1} } \\ \\ \longmapsto\rm{Speed_{(Avg)} = \Bigg \{x \div \Bigg ( \dfrac{1x}{36} \Bigg ) \Bigg \} \; kmh^{-1} } \\ \\ \longmapsto\rm{Speed_{(Avg)} = \Bigg \{\cancel{x} \times \dfrac{36}{\cancel{1x}} \Bigg \} \; kmh^{-1} } \\ \\ \longmapsto\underline{\boxed{\bf{Speed_{(Avg)} = 36 \; kmh^{-1}} } } \; \red{\bigstar}$

Therefore, the average speed is 36 km/h.

(Option D)