A person travels from P to Q at a speed of 40 kmph and returns at a speed 50% more than the previous speed. What is the average speed of that person?

50 kmph

No worries! We‘ve got your back. Try BYJU‘S free classes today!

48 kmph

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

45 kmph

No worries! We‘ve got your back. Try BYJU‘S free classes today!

36 kmph

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
48 kmph

Speed of the return trip = 150% of 40 kmph = 60 kmph
Average speed of the person = $\frac{T o t a l D i s t a n c e}{T o t a l T i m e} = \frac{2 \times P Q}{\frac{P Q}{40} + \frac{P Q}{60}} = \frac{2 \times 60 \times 40}{60 + 40} = 48 kmph$

Suggest Corrections

Similar questions

Q. What is the average speed if a man travels

\frac{1}{2}

the distance at speed of 40 kmph,

\frac{1}{3}

the distance at 50 kmph and rest of the distance at 60 kmph?

What is the average speed of a train that travels from one station to another station at a uniform speed of 40kmph and returns to the initial station at 60kmph.

Q. A motor boat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
OR
A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/ hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed ?

Melwin jogs 18 km at a speed of 12 km per hour. At what speed would he need to jog during the next 3 hours to have an average of 18 km per hour for the entire jogging session?

Q. A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?

Join BYJU'S Learning Program

A person travels from P to Q at a speed of 40 kmph and returns at a speed 50% more than the previous speed. What is the average speed of that person?

The correct option is B 48 kmphSpeed of the return trip = 150% of 40 kmph = 60 kmph Average speed of the person = Total DistanceTotal Time=2×PQPQ40+PQ60=2×60×4060+40=48 kmph

The correct option is B
48 kmph

Speed of the return trip = 150% of 40 kmph = 60 kmph
Average speed of the person = $\frac{T o t a l D i s t a n c e}{T o t a l T i m e} = \frac{2 \times P Q}{\frac{P Q}{40} + \frac{P Q}{60}} = \frac{2 \times 60 \times 40}{60 + 40} = 48 kmph$