Question

6
44. If a man travels at a speed of 30 km/h, he
reaches his destination 10 min late and if he
travels at a speed of 42 km/h, he reaches his
destination 10min early. The distance
travelled is
(h) 26 km (ol 40 km (d) 42 km​

Answer 1

Step-by-step explanation:

Let the distance be x km.

Then, correct time at a speed of

30 km / h = x/30 - 10/60

Correct time at a speed of 42 km /h

= x/42 + 10/60

=> x/30 - 10/60 = x/42 + 10/60

=> x/30 - x/42 = 26

=> x(12)/1260 = 26

=> 72 x = 1260 × 2

=> x = 1260×272 => x = 35 km

Answer 2

Step-by-step explanation:

$\huge\bold\orange{Answer:-}$

Here we are taking two cases with different speed . Now let's let that the time taken to reach the final destination is 't'.

Let's see in case 1 where the distance covered by man to reaches his destination .

Given speed of a man = 30km/h

time taken 10min

Distance travelled =

$speed \: of \: the \: man \times time \: taken \\ = 30 \times (t + \frac{10}{60} )km \\ s = 30 \times ( \frac{1}{6} )..(1)$

Now let's see in case 2 given speed = 42 km/h then also he took 10 min to reaches his destination.

Distance travelled =

$speed \: \times time \: taken \\ = 42 \times ( t \: - \frac{10}{60} ) \\ = 42 \times (t - \frac{1}{6} )..(2)$

From equation (1) and (2)

$30(t + \frac{1}{6} ) = 42(t - \frac{1}{6} ) \\ or \\ 10t + \frac{10}{6} = 14t - \frac{14}{6} \\ \frac{10}{6} + \frac{14}{6} = 14t - 10t \\ \frac{24}{6} = 4t \\ or \: t = 1hour \\ \\ putting \: t \: in \: equation \\ s = 30(1 + \frac{1}{6} ) = 30 \times \frac{7}{6} = 35$