Question

a person covers the first one fourth of the distance at 8 km per hour the next three fifth at 6 km per hour and the remaining distance 15 km/h. find the average speed during the entire journey

Answer 1

let total distance be x
8=0.25x/t¹
15=0.75x/t²
total time=0.25x/8+0.75x/15
total time=x/32+x/20
total time=20x+32x/640
total time=52x/640
total time=13x/160

Average speed= dis./time=x÷13x/160=160/13
Average speed = 12.30 km/h

Answer 2

Answer:

The average speed of entire journey is 7.92 km per hour.

CONCEPT:

1. Distance = speed × time
2. Time taken = distance / speed
3. Average speed = total distance / total time taken

GIVEN:

• covers first one fourth of distance at 8 km per hour
• covers next three fifth of distance at 6 km per hour
• covers rest of the distance at 15 km per hour

TO FIND:

    Average speed of the entire journey
SOLUTION:

Let the total distance of the journey be x.

T1 = Time taken to cover first one fourth of the distance

Using concept2 from the above

T1 = $\frac{\frac{x}{4} }{8}$

T1 = $\frac{x}{32}$

Remaining distance = $x-\frac{x}{4}$
                                 = $\frac{3x}{4}$

Three fifth of the distance will be $\frac{3}{5} .\frac{3x}{4}$

T2 = Time taken to cover next three fifth of the distance

T2 =  $\frac{\frac{3}{5} .\frac{3x}{4} }{6}$

T2 = $\frac{9x}{120}$

T2 = $\frac{3x}{40}$

Now remaining distance = $x-\frac{x}{4}-\frac{9x}{20}$

                                         = $\frac{20x-5x-9x}{20}$

                                         = $\frac{3x}{10}$

T3 = Time taken to cover remaining distance

T3 = $\frac{\frac{3x}{10} }{15}$

T3 = $\frac{x}{50}$

T = Total time taken to cover the journey

T = T1+T2+T3

T = $\frac{x}{32}+ \frac{3x}{40}+ \frac{x}{50}$

T = $\frac{25x+60x+16x}{800}$

T=$\frac{101x}{800}$

Using concept3 from above,

Average speed = $\frac{x}{\frac{101x}{800} }$

                          = $\frac{800}{101}$

                          = 7.92 km per hour

Therefore, the average speed of entire journey is 7.92 km per hour.

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Thank you.