# CAT Practice : Speed Time, Races

## Increasing Speed

Q.21: Distance between the office and the home of Alok is 100 Km. One day, he was late by an hour than the normal time to leave for the office, so he increased his speed by 5 Km/hr and reached office at the normal time. What is the changed speed of Alok?
1. 25 Km/hr
2. 20 Km/hr
3. 16 Km/hr
4. 50 Km/hr

Choice A. 25 Km/hr

## Detailed Solution

Here, again the distance is constant. However if we write the equation as we did in previous question, we get:
s x t = (s + 5) x (t - 1) = 100
In this case, we can identify other constant as the time which Alok took less to cover the distance i.e. 1 hour. So, we can write:
$\frac{100}{s} - \frac{100}{s+5} = 1$ (Difference of time take in both the cases is 1 hour)
=) 100s + 500 - 100s = s(s+5)
=) s(s+5) = 500 (One should practice to solve these kind of equations directly by factorizing 500 into 20 X 25, instead of solving the entire quadratic. However, if you are more comfortable with quadratic, keep using it. More important thing is to avoid silly mistakes once you have got the concept right)
Solving we get, s = 20 Km/hr
Therefore, Increased speed = s + 5 = 25 km/hr
We could have alternatively identified increase in speed by 5 Km/hr as a constant. Therefore, we could write the equations as:
$\frac{100}{t - 1} - \frac{100}{t} = 5$
Solving, we get t = 5 hrs, t-1 = 4hrs. Therefore, increased speed = $\frac{100}{t - 1}$ = 25 Km/hr.

## Our Online Course, Now on Google Playstore!

### Fully Functional Course on Mobile

All features of the online course, including the classes, discussion board, quizes and more, on a mobile platform.

### Cache Content for Offline Viewing

Download videos onto your mobile so you can learn on the fly, even when the network gets choppy!

## More questions from Speed Time, Races

"Life is a race ... if you don't run fast ... you will be like a broken undaa" - Veerusahasra Buddhi