Question

Priya covers a certain distance between her home and school by cycle having an average speed of 30 Km/hr she is late by 20 minutes however with a speed of 40Km/hr she reaches her school 10 minutes earlier find the distance between her home and school.​

Answer 1

Answer:

Implement concept of Inverse Variation.

If A x B = Constant, then increase in A will lead to reduction in B.

Distance = Speed x Time

If speed increase by 1/3rd, time reduces by 1/(1 + 3) th or 1/4th.

If original time is t, then t/4 = (20 + 10)

=> t = 120 minutes.

Distance = 120/60 x 30 = 60 KM's

Answer 2

Answer:

If distance is constant

Speed = 1/ time

So, speed 1 : Speed 2 = time 2 : time 1

30 :40 = T2 : T1…………….Note( 1)

Let actual time = t

Ac. To question

Lately (t1) = t +20

Early (t2) =t -10

See note 1

30 /40 = T2 /T1

30 /40 =t−10/t+20

3/4 = =t−10/t+20

3t +60 = 4t -40

T =100 min.

T = 1 hr 40 min.

Average speed = 30 km/hr

Distance = 30×(1+40/60)

=30× 5/3

=50 km

So the answer is 50km .