Question

When a person cycled at 10 km/h, he arrived at his
office 6 min late. He arrived 6 min early, when he
increased his speed by 12 km/h. The distance of his
office from the starting place is

(a) 6 km (b) 7km
(c) 12 km (d) 16 km​

Answer 1

Answer:

Correct Option: C

Let the distance be x km

∴ x\10-x\12=12\60

6x-5x/60=1/5

x=1\5×60=12 km

Second Methdod :

Here, S1 = 10, t1 = 6

S2 = 12, t2 = 6

Distance =

(S1 ×S2)(t1 + t2)/S2 −S1

= (10 × 12)(6 + 6) /12 − 10

= 120 × 12 /2

= 60 × 12km /60 km= 12 km

Answer 2

Given:

When a person cycled at 10 km/h, he arrived at his office 6 min late. He arrived 6 min early, when he increased his speed by 12 km/h.

To find: We have to find the distance of his office from the starting place.

Solution:

We know that distance=speed×time

Let the actual time to arrive in office is x hours.

When a person cycled at 10 km/h, he arrived at his office 6 min late.

Arriving time=

$x + \frac{6}{60}$

He arrived 6 min early, when he increased his speed by 12 km/h.

Arriving time

$= x - \frac{6}{60}$

The distance is same in both case so we can write-

$10(x + \frac{1}{10}) = 12(x - \frac{1}{10} ) \\ 10x + 1 = 12x - 1.2 \\ 2x = 2.2 \\ x = 1.1$

The distance of his office from starting place is

$= 10(1.1 + 0.1) \\ = 12$

So, correct option is c) 12km.