Thursday, April 18th, 2019
Time, Speed, and Distance is a very crucial topic in quantitative aptitude section of CAT, SNAP, XAT etc. Every year 4-5 questions are asked from this topic. We all have done questions based on Time and Distance in our high school maths but you might not be acquainted with this very significant part of the application of this concept knowns as Linear and Circular Races. You might have done some simple problems based on this concept but not enlightened with its model and theory.
In this blog, I will provide you a summarized knowledge of this concept along with itâ€™s application in forms of problems. Letâ€™s commend!
Weâ€™ll start first with familiar with what really is this topic is all about and a few expressions you will come across in problems.
A race is a competition in which contestants compete among themselves to cover some distance in shortest length of time.
Linear race: In this case participate compete in linear race track
Circular race: In this one the path of the race is circular in shape.
Starting point: Itâ€™s the point from where the race begins.
Winning Point: Itâ€™s the end point of the race.
Commonly used expressions in linear races
Expression | Meaning |
Head-Start / A gives B start of x meters | When a racer gets a start x meters ahead of the starting point itâ€™s called head-start of x meters. Here A gives B a head-start of x meters. |
Head-Start/ A can give B a start of t minutes | When a contestant gets a start by t seconds earlier than other ones, itâ€™s called head-start of t seconds. Here A gives B a head-start of t seconds. |
A beats B by x meters | When A reaches the winning point before B and is x meters away from B. Then A beats B by x meters. |
A beats B by t seconds | When A reaches the winning point t seconds before B. Then A beats B by t meters. |
Dead Heat | A dead heat is a situation of tie. When all the participants reach the winning point at the same point. |
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Given below are a few examples with the usage of terminologies mentioned above.
Example 1: In a 100 m race, A can give B 10 m and C 28 m. In the same race B can give C.
Sol: To solve this question the best way is to represent them diagrammatically like this:
A covers 100 meter in same time as B covers (100-10) meter and C covers (100-28) meter. Using unitary method, we can find out the head-start that B can give to C. Therefore, when B runs 100 meters than C runs (72/90 *100) = 80 meters. And thus, C gets head-start of 20 mtr.
Example 2: In a 500 m race, the ratio of the speeds of two contestants P and Q is 4: 5. A has a start of 140 m. Then, P wins by?
Sol) In order to reach the winning point P has to cover (500 â€“ 140) m i.e. 360 m.
Since, Ratio of Speed = Ratio of distance covered.
Therefore, P covers 4 m while Q covers 5 m.
Thus, when P covers 360 m, Q covers (^{5}/_{4} * 360) m = 450 m.
=>A win by 50 m.
Note 1: If A is n times as fast as B and A gives B a start of x meters, then the length of the race course, so that A and B reaches the winning post at the same time
x(^{n}/_{(n-1)}) meters
Note 2: If A can runÂ xÂ meters race in t_{1}Â seconds and B in t_{2}Â seconds, where t_{1}Â < t_{2}, then A beats B by a distance
x/t_{2} * (t_{2} â€“ t_{1}) meters
Â
The questions asked in this concept are usually of two types:
Before moving forth with problems letâ€™s learn some useful tips.
X and Y to meet first time anywhere on the track | X and Y to meet first time at the starting point on the track |
L/ (x – y) | LCM of L/x and L/y |
X and Y to meet first time anywhere on the track | X and Y to meet first time at the starting point on the track |
L/ (x + y) | LCM of L/x and L/y |
(Time after which theyâ€™ll meet at the starting point)/ no. of meeting points
We can now try some problems based on circular races:
Example 1: Â In a circular race of 2400m, A and B start from the same point and at the same time with speeds of 27km/hr. and 45 km/hr. Â Find when will they meet again for the first time on the track when they are running in the same direction and Opposite direction?
Sol) Length of the track = 2400 m
Theyâ€™ll meet first time on track when they are running in
Here x is 27 *^{5}/_{18 }= 7.5 m/s
y is 45*^{5}/_{18 }= 12.5 m/s
Time = 2400/ (12.5-7.5) = 480 seconds
Time = 2400/ (12.5+7.5) = 120 seconds
Example 2: Ram and Sham are running in opposite direction around a circular track of lengthÂ 20Ï€Â meters. Speed of Ram is 33.33% of the speed of Sham. Find straight line distance between their first meeting point and the second meeting point. It is given that they start simultaneously from a common starting point.
Sol) Since speed of Ram is 1/3 of Sham. Therefore, the no. of distinct points are (1+3) i.e. 4 points.
Now as all these points would be equidistant from each other. The meeting points can be represented as:
If A is the starting point then we need to find out the distance AB. And this can be done as follows:
Circumference of the circular path is 20 Ï€Â then the radius of circle will be 10 m. Thus, using Pythagoras formula we can easily calculate the length of AB i.e. 10âˆš2 m.
The above problems are few examples of the application of this concept. There can be many similar and varied problems based on both Linear and Circular race. To solve them you just need to interpret the language correctly and find the solution to the problems using the tips I have told you in this blog along with what you have learnt in your junior classes regarding Lines and circles. You will get a thorough understanding of this topic only if try different problems and practice its application.
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