The question is from the topic of Time Speed and Distance. What will be the new speed, when distance is constant and time is decreased. Time Speed and Distance is a favorite in CAT Exam and appears more often than expected in the CAT Quantitative Aptitude section in the CAT Exam

Question 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?

- 25 Km/hr
- 20 Km/hr
- 16 Km/hr
- 50 Km/hr

25 Km/hr

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Here, again the distance is constant. However if we write the equation as we did in previous question, we get:

s * t = (s + 5) * (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.

The question is **" What is the changed speed of Alok?"**

Choice A is the correct answer.

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