Question

SURESH STARTED HIS JOURNEY FROM P TO Q BY HIS BIKE AT THE SPEED OD 40 KM/HR AND THEN , THE SAME DISTANCE HE TRAVELLED ON HIS FOOT AT THE SPEED OF 10 KM /HR FROM Q TO R. THEN HE RETURNED FROM R TO P VIA Q AT THE SPEED OF 24 KM/ HR ..the average spped of the whole is.A)18.5KM/HR B) 19.8 C) 18.2D)19.2

Answer 1

Let the distance from P to Q and Q to R be 'x' km (since the distance from P to Q and Q and R is same)

The distance from P to R is 2x km

Total distance travelled by Suresh is 4x km (2x km for forward journey which is from P to R and 2x km for return journey which is from R to P)

Let the time travelled from P to Q be 'a' hrs, Q to R be 'b' hrs and R to P(via Q) be 'c' hrs

Time=Distance/Speed
a=x/40
b=x/10
c=2x/24

Total time taken
= a+b+c
= x/40 + x/10 + 2x/24
= x/40 + x/10 + x/12
= (3x+12x+10x)/120
= 25x/120
= 5x/25

Average speed
=Total distance/Total time
= 4x/(5x/24)
= (4x*24)/5x
= (4*24)/5
= 4*4.8
= 19.2 km/hr

The answer is 19.2 km/hr
Hope it helps

Answer 2

Solution :-

Let the distance from P to Q be 'x' km and from Q to R be also 'x' km (as the distance from P to Q and Q to R is same.

So, the distance from R to P via Q will be 2x km.

Time taken from P to Q (t1) = x/40 hr

Time taken from Q to R (t2) = x/10 hr

Time taken from R to P via Q (t3) = 2x/24 hr

Total distance = (x + x + 2x) km

= 4x km

Total time taken = x/40 + x/10 + 2x/24

Taking LCM and of the denominators and then solving it.

⇒ (3x + 12x + 10x)/120 

⇒ 5x/24

Time taken is 5x/24 hr

Speed = distance/time

⇒ 4x ÷ 5x/24

⇒ (4x*24)/5x

⇒ 96x/5x

⇒ 96/5

= 19.2 km/hr

So, the average speed is 19.2 km/hr

Option (D) is correct.