(C) Let the average speed of walking be $x \, km/hr$.Total distance covered $= \text{Average speed} \times \text{Total time} = 32 \times 3 = 96 \, km$

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(C) Let the average speed of walking be $x \, km/hr$.Total distance covered $= \text{Average speed} \times \text{Total time} = 32 \times 3 = 96 \, km$.Distance walked $= 6 \, km$.Distance covered by train $= 96 - 6 = 90 \, km$.Time taken for walking $= \frac{6}{x} \, hours$.Time taken by train $= \frac{90}{60} = 1.5 \, hours$.Total time $= \frac{6}{x} + 1.5 = 3 \, hours$.$\frac{6}{x} = 3 - 1.5 = 1.5$.$\frac{6}{x} = \frac{3}{2}$.$3x = 12$,so $x = 4 \, km/hr$.

Competitive Quantitative Aptitude Average Average

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Mr. Patel walked $6 \, km$ to reach the station from his house,then he boarded a train whose average speed was $60 \, km/hr$ and thus he reached his destination. In this way,he took a total of $3 \, hours$. If the average speed of the entire journey was $32 \, km/hr$,then the average speed of walking is (in $km/hr$):

A
$3$
B
$4.5$
C
$4$
D
$5$

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Of the three numbers,the first is twice the second and the second is twice the third. The average of the reciprocals of the numbers is $\frac{7}{72}$. The numbers are:

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