Question

A man reaches his office late by 5 min when he travels at the speed of 8 km/hr.
However, when he increases his speed to 12 km/hr, he reaches 5 min early. What
is the distance that he travels?​

Answer 1

Answer:

First let me define the distance as  X.  The distance,  X,  between home and work is unknown yet a constant value that we will experiment with.

So we know that

speed=distance/time  

So the distance is:

distance=speed∗time  

We know his different speeds 4 and 5 km/h respectively but we need to convert them into m/min to ease our work.

5km/h=5000m/60min=(500/6)m/min  

4km/h=4000m/60min=(400/6)m/min  

We know that:

X=(400/6)∗t1=(500/6)∗t2  

And also:

t1=t2+15  

So we will substitute the values to get the answer:

X=(400/6)∗(t2+15)=(500/6)∗t2  

A little modified:

(400/6)∗t2+(400/6)∗15=(500/6)∗t2  

Let's collect all the unknowns in one side:

(400/6)∗15=(100/6)∗t2  

To get:

400∗15=100∗t2  

t2=60minutes.

Step-by-step explanation:

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