Question

If a man travels at a speed of 30km/h, he reaches his destination 10 minutes late and if he travels at a speed of 42 km/h, he reaches his destination 10 minutes early. What is the distance?

Answer 1

Given:

If a man travels at a speed of 30 km per hour then he reaches the destination 10 minutes late

If he travels at a speed of 42 km per hour he reaches his destination 10 minutes early.

To find:

Distance of the man from his destination.

Solution:

Let the distance of the man from his destination = x km

And he reaches is destination in 't' years

If he travels at a speed of 30 km per hour, he reaches 10 minutes late.

10 minute = $\frac{10}{60}$ hours

                = $\frac{1}{6}$ hours

Since, Speed = $\frac{\text{Distance}}{\text{Time}}$

30 = $\frac{x}{t+\frac{1}{6}}$

x = $30(t+\frac{1}{6})$

x = 30t + 5 ------- (1)

If he travels at a speed of 42 km per hour, he reaches his destination 10 minutes early.

Substitute the values in the formula,

42 = $\frac{x}{t-\frac{1}{6}}$

x = $42(t-\frac{1}{6})$

x = 42t - 7 ------- (2)

By equating the value of x from equation (1) and (2),

30t + 5 = 42t - 7

42t - 30t = 5 + 7

12t = 12

t = 1 hour

From equation (1),

x = 30(1) + 5

x = 35 km

Distance between the man and the destination is 35 km.

Answer 2

Answer:

Step-by-step explanation:

Explanation by windyyork-:    upto t=60 correct

here: t=60 means 60 min.

if he travel - 30km/h:   reaches 10 mins late,    i.e. t+10 = 60+10 =  70min

In 1 hr -  travels -  30 km  i.e

in 60 min  -30 km

in 1min -> 30/60 km

in 70 min -> (30/60 ) *70

               -> 35 km.

So the ans is: 35 km.