Question

Sagar and Akash Ran 2 kilometre race twice. Akash completed the first round 2 minutes earlier than Sagar. In second round Sagar increase his speed by 2 kilometre per hours and Aakash reduce speed by 2 kilometre per hour Sagar finished 2 minutes earlier than Aakash . Find their speed of running in the first round.

Answer 1

Solution:-

Let the speed of Sagar be considered as x km/hr.
And speed of Aakash be y km/hr.
Therefore, time taken by Sagar = 2/x hrs.
And time taken by Aakash = 2/y hrs.
Now, according to the question.
2/x - 2/y = 2/60
1/x - 1/y = 1/60 ..................(1)
(y - x)/xy = 1/60
60(y - x) = xy ...................(2)

Now, according to the second condition.
Speed of Sagar = (x + 2) km/hr.
Speed of Aakash = (y - 2) km/hr
⇒ (2/y-2) - (2/x+2) = 2/60 
⇒ (1/y-2) - (1/x+2) = 1/60
⇒ {(x+2-y+2)/(y-2)(x+2)} = 1/60
⇒ 60(x-y+4) = xy+2y-2x-4
Now, substituting the value of xy in the above, we get.
⇒ 60(x-y+4) = 60(y-x)+2y-2x-4
⇒ 30(x-y+4) = 30(y-x)+y-x-2
⇒ 30x-30y+120 = 30y-30x+y-x-2
⇒ 30x+31x-30y-31y = -2-120
⇒ 61x-61y = -122
⇒ x-y = -2
⇒ x = y-2
Substituting the value of x in (1), we get.
⇒ {1/(y-2) - (1/y) } = 1/60
⇒ (y-y-2)/(y²-2y) = 1/60
⇒ y²-2y = 120
⇒ y²-2y-120 = 0
⇒ y²+10y-12y-120 = 0
⇒ y(y+10) -12(y+10) = 0
⇒ (y+10) (y-12) = 0
⇒ y = -10 is not possible. So, y = 12 is correct value.
Substituting the value of y = 12 in x = y - 2, we get.
⇒ x = 12 - 10 
⇒ x = 10 
So, speed of Sagar in the first round was 10 km/hr and speed of Aakash was 12 km/hr.
Answer.