Question

Sagar and Aakash ran 2 km races twice. Aakash completed the first round 2 minutes earlier than Sagar. In the second round Sagar increased his speed by 2 km/hr and Aakash reduced his speed by 2 km/hr . Sagar finished 2 minutes earlier than Aakash. Find their speeds of running in the first round.

Answer 1

Hint: Take speed of Aakash and Sagar as y and x respectively and use formula;
$\text{Time = }\dfrac{\text{Distance}}{\text{Speed}}$to get those equations and solve the two obtained equations simultaneously, to get the desired result.

Complete step-by-step answer:

Now let's assume the speed of Sagar is x km/hr and the speed of Aakash is y km/hr.
So, the time taken by Sagar is $\dfrac{2}{x}$ .
The time taken by Aakash is $\dfrac{2}{y}$ .
It is given that Aakash completed the race earlier by 2 minutes, that is, by $\dfrac{2}{60}$ hours.
So now according to the question we can say that,
$\dfrac{2}{x}-\dfrac{2}{y}=\dfrac{2}{60}$
So, on simplification we get,
$2\left( \dfrac{1}{x}-\dfrac{1}{y} \right)=\dfrac{2}{60}$
Cancelling ‘2’ from both the sides, we get
$\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{60}$
By taking LCM we get,
$\dfrac{y-x}{xy}=\dfrac{1}{60}$
On cross multiplication we get,
xy = 60 (y-x) ………….(i)
Now according to the second condition,
Speed of Sagar = (x + 2) km/hr
Speed of Aakash = (y - 2)km/hr
So, the time of Sagar will be $\dfrac{2}{x+2}$ and the time of Aakash will be $\dfrac{2}{y-2}$ .
As in the second case it is given that Sagar completed the race earlier by 2 minutes, that is, by $\dfrac{2}{60}$ hours.
So we write,
$\dfrac{2}{y-2}-\dfrac{2}{x+2}=\dfrac{2}{60}$
So on simplification we get,
$\dfrac{1}{y-2}-\dfrac{1}{x+2}=\dfrac{1}{60}$
By taking LCM we get,
$\dfrac{x+2-y+2}{\left( y-2 \right)\left( x+2 \right)}=\dfrac{1}{60}$
On cross multiplication we get,
60 (x – y + 4) = (y – 2) (x + 2)
On simplification we get,
60 (x – y) + 240 = xy – 2x + 2y – 4
Now substituting equation (i), xy as 60(y – x) so we get,
60 (x – y) +240 = 60 (y –x) -2x + 2y – 4
On simplifying we get,
60x – 60y + 240 = 60y - 60x - 2x + 2y – 4
Now by taking all the variables to left hand side and all the constants to right hand side so we get,
122x – 122y = - 244
Now by dividing 122 by throughout we get,
x – y = - 2
Now adding y to both the sides we get,
x = y – 2
Now as we know that
$\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{60}$
So we can substitute x as (y – 2) we get,
$\dfrac{1}{y-2}-\dfrac{1}{y}=\dfrac{1}{60}$
By taking LCM we get,
$\dfrac{y-(y-2)}{y\left( y-2 \right)}=\dfrac{1}{60}$
Now by doing cross multiplication we get,
${{y}^{2}}-2y=120$
${{y}^{2}}-2y-120=0$
Now using middle term factor we get,
${{y}^{2}}-12y+10y-120=0$
Now by factoring we get,
(y – 12) (y + 10) = 0
So, from this we get the value of y as -10 or 12.
Now as we know that y represents the speed of Aakash and it can’t be negative so the speed of Akash is 12 km/hr.
Now as we know that
x – y = -2
Then we will substitute y = 12 we get,
x = 12 – 2 = 10.
So, ‘x’ which is the speed of Sagar is 10 km/hr.
The speed of Aakash and Sagar is 12km/hr and 10 km/hr respectively.

Note: Students must be sure about the values to know the basics on how to form an equation and solve it. They should also be careful about calculations to avoid any mistakes.
As the speeds are taken as variables then we have to compare them with the duration of time to solve this type of problem. They should use the form of time i.e., distance covered by speed and then compare it with the given value.