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1. 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. 4 km/h

B. 6 km/h

C. 10 km/h

D. 12 km/h

Solution:

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,

(\frac{2 x}{3} \times 5) + (2 y \times 5) = 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
Speed Time and Distance mcq solution image

∵

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{Distance}{Time} = \frac{60}{6}

Relative speed in opposite direction :

(⇌) = 10 km/h . . . . . (i)

According to the question,

\begin{aligned} \Rightarrow \frac{2}{3} A + 2 B = \frac{60}{5} \\ \Rightarrow \frac{2}{3} A + 2 B = 12 \\ \Rightarrow A + 3 B = 18 \\ \Rightarrow B^{'} s Speed  = \frac{18 - A}{3} \\ \Rightarrow A + B = 10 \\ \Rightarrow A + \frac{18 - A}{3} = 10 \\ \Rightarrow 3 A + 18 - A = 30 \\ \Rightarrow 2 A = 12 \\ \Rightarrow A's speed  =  6 km/h \end{aligned}

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