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  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

Show Answer Answer: Option B
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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, $\begin{array}{l} \left(\frac{2 x}{3} \times 5\right)+(2 y \times 5)=60 \\ \text { Or, } 10 x+30 y=180 \\ \text { Or, } x+3 y=18-\ldots \ldots \text { (ii) } \\ \text { From equation (i) } \times 3-\text { (ii) } \\ 3 x+3 y-x-3 y=30-18 \\ \text { Or, } 2 x=12 \\ \text { Hence, } x=6 \text { kmph } \end{array}$
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 $=\frac{\text{ Distance }}{\text{ Time }}=6$ Relative speed in opposite direction : $(\rightleftharpoons)=10 \mathrm{~km} / \mathrm{h} \ldots \ldots$ According to the question, $\begin{array}{l}\Rightarrow\frac{2}{3}A+2B=\frac{60}{5}\\ \Rightarrow\frac{2}{3}A+2B=12\\ \Rightarrow A+3B=18\\ \Rightarrow B^{\prime}s\text{ Speed }=\frac{18-}{3}\\ \Rightarrow A+B=10\\ \Rightarrow A+\frac{18-A}{3}=10\\ \Rightarrow3A+18-A=30\\ \Rightarrow2A=12\\ \Rightarrow A^{\prime}\text{ s speed }=6\mathrm{~km}/\mathrm{h}\end{array}$

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