Question

Question 10 of 20
A man goes to his office by car at a speed of 20km/h and reaches 7 minutes earlier. If he goes at a speed of 15 km/h, he reaches 5 minutes
late. The distance of his office is:
O 12 km
O 18 km
O 20 km
23 km

Answer 1

Let’s assume the distance of A man to office is x , and the time at which they are supposed to arrive at office is t . Let’s call their first speed v1 and their second speed v2 . Let’s call the time it took to get to office the first time t1 and the time it took the second time t2 . So, just to be clear: the time it should take him, we call t , and the time it really took him in both cases was t1 and t2 , respectively.

The first time he is 7 minutes earlier. Let’s convert this to hours as the speed is expressed in km/h. (We could also convert the speed to km/minute, or any other unit of time and distance, but let’s stick with this.) So, 7 minutes is

$\frac{7}{60}$

of an hour

This means that, in the first attempt, A man arrived at time

$t1 = t + \frac{7}{60}$

We can now write for the distance x

$x = v1 t1 = 20(t - \frac{7}{60} )$

The second attempt, A man arrived 5 minutes late, which is

$\frac{5}{60} = \frac{1}{12}$

of an hour. So,

We can now write for distance x

$x = v2t2 = 15( t + \frac{1}{12} )$

We now have two expressions for distance x . They are both equal to x . So, we can now write

$20(t - \frac{7}{60} ) = 15( t + \frac{1}{12} )$

and solve for t , denoting the time it should take A man

$t = \frac{43}{60}$

We can now substitute this value in either expression for x to calculate the distance to office. Let’s take the first one.

$x = 12$

your answer is 12 KM

Answer 2

The distance of his office is 12 km.

Given,

When speed is 20km/h he reaches 7 minutes earlier.

When speed is 15km/h he reaches 5 minutes late.

To Find,

The distance of his office.

Solution,

Let the distance of his office and the total time taken to reach his office be d and t respectively.

We know that, 20km/h = 50/9 m/s

                         15km/h = 75/18 m/s

Using the formula, Speed = Distance/Time

For 7 minutes earlier reach,

⇒ 50/9 m/s = d/(t-7x60) m/s

⇒ 50t - 21000 = 9d

⇒ t = (9d + 21000)/50            -----------------------------------------(1)

For 5 minutes late reach,

⇒ 75/18 m/s = d/(t+5x60)

⇒ 75t + 22500 = 18d             ------------------------------------------(2)

putting value of t from (1) to (2),

⇒ 75(9d + 21000)/50 + 22500 = 18d

⇒ 54000 = 18d - 27d/2

⇒ d = 54000x2/9

⇒ d = 12000m = 12km

Hence, The distance of his office is 12km.