In this section, we will discuss about problems on RACES which is a prat of time and distance, tips and tricks along with concept and terminologies related to it, and different cases on it.
First, we would learn the language (terminology) of races;
|1.||Race||A race is a contest of speed in running, driving, riding, sailing or rowing.|
|2.||Race Course||The ground/path on which a contest is organized in a systematic way.|
|3.||Starting Point||The exact point/place from where a race begins|
|4.||Finishing (Winning) Point||The exact point/place where a race end.|
|5.||Dead Heat Race||A race in which all the contestants reach the finishing point exactly at the same time.|
|6.||Head start (Start-up)||A runner allows another runner to stay ahead at the start.|
E.g. In a race of 100 m, A gives B a start of 10 m. What distance will be covered by B?
As ‘A’ gives a start to ‘B’ of 10 m
At the start of the race, ‘B’ would be ahead of ‘A’ by 10 m
⇒ 100 – 10 = 90 m distance would be covered by ‘B’(Ans.)
E.g. P covers 1 km in 4 min 30 s, while Q covers the same distance in 5 min. by what distance does P defeat Q?
As P covers distance faster so he would be the winner.
The extra time is taken by Q = 30 s
Distance = (1000/300) × 30 = 100 m (Ans.)
E.g. P can run 1 km in 4 min 40 s, Q can run the same distance in 5 min. How many meters start, can P give to Q in 1 km race, so that the race may end in a dead heat?
As P takes lesser time to cover he would give a start, so as we know in dead heat both crossed the finish line at the same time.
Q runs extra 20 s
Q runs in 300 seconds = 1000 m
Q runs in 1 second = 1000/300 = 10/3 m
Q runs in 20 seconds = (10/3) × 20 = 66.67 m (Ans.)
Note: In a game of x points, A scores x points and in this game, B scores y points. In that game A can give (x – y) points to B where x > y.
E.g. In a game of 100 points, A scores 100 points, while B scores only 75 points. In this game, how many points can A give to B?
A has excessive points which are 100 – 75 = 25 points, so he can give 25 points. (Ans.)
There are very fewer chances of these types of questions to be asked in paper.
Let A, B & C participate in a race where A wins by beating B by x12 m and C by x13 m.
In this condition
(L – x12) x23 = L (x13 – x12)
L = Length of race course
x12 = A beats B by distance
x23 = B beats C by distance
x13 = A beats C by distance
This is a very useful formula for questions related to races.
E.g. In a race of 100 m, A beats B by 4 m and beats C by 2 m, by how many meters would A beats C in a 100 m race?
Here Ist is A, IInd is C and IIIrd is B.
L = 100 m
x12 = 2
x13 = 4
x23 = ?
By substituting the values in the formula
⇒ (L – x12) x23 = L (x13 – x12)
⇒ (100 – 2 ) × x23 = 100 (4 – 2 )
⇒ 98 × x23 = 200
⇒ x23 = 200/98 = 2.04 m (Ans.)
E.g. A, B, and C are three contestants in a 1 km race. If A can give B a start of 40 m and A can give C a start of 64 m, how many meters start can B give C?
When A completes 1000 m in a race B would cover (1000 – 40 = 960 m) and similarly, C would cover (1000 – 64 = 936m).
We got when B covers 960m then C covers 936 m So the distance covered by C when B covers 1000 m distance,
936 /960 × 1000 = 975 m.
So B can give a start of 1000 – 75 = 25 m. (Ans.)
1. Time and Distance Part 1 – https://www.rankershubindia.com/speed-distance-and-time-part-1/
2. Time and Distance Part 2 – https://www.rankershubindia.com/speed-distance-and-time-part-2/
3. Time and Distance Part 4 – https://www.rankershubindia.com/speed-time-and-distance-part-4/
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