Question

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

Answer 1

Let Sagar run with a speed of $xkm/hr$
Let Aakash run with a speed of $ykm/hr$
According to the $1^{st}$ condition,
$\frac{2}{x}-\frac{2}{y}=\frac{2}{60}$
$\therefore \frac{1}{x}-\frac{1}{y}=\frac{1}{60}$
$\therefore \frac{y-x}{xy}=\frac{1}{60}$
$\therefore 60(y-x)=xy$ ...........$\left(1\right)$
According to the $2^{nd}$ condition,
$\frac{2}{y-2}-\frac{2}{x+2}=\frac{2}{60}$
$\therefore \frac{1}{y-2}-\frac{1}{x+2}=\frac{1}{60}$
$\therefore \frac{x+2-y+2}{(y-2)(x+2)}=\frac{1}{60}$
$\therefore \frac{x-y+4}{xy+2y-2x-4}=\frac{1}{60}$
$\therefore 60(4-y+x)=xy+2y-2x-4$
$\therefore 240-60(y-x)=xy-4+2(y-x)$
$\therefore 244=xy+2(y-x)+60(y-x)$
$\therefore 244=xy+62(y-x)$
$\therefore 244=60(y-x)+62(y-x)$ ........[from $\left(1\right)$]
$\therefore 244=122(y-x)$
$\therefore y-x=2$ ............$\left(2\right)$
$\therefore xy=60\times 2=120$ .........(from $1$)
Since, $(y+x{)}^2=(y-x{)}^2+4xy$
$=2^2+4\times 120=4+480=484$
$\therefore (y+x{)}^2=(22{)}^2$
$\therefore y+x=22$ ............$\left(3\right)$
Adding equation $\left(2\right)$ and $\left(3\right)$,
$2y=24$
$\therefore y=12$
Putting $y=12$ in equation $\left(3\right)$,
$12+x=22$
$\therefore x=10$
Sagar ran with a speed of $10km/hr$ and Aakash ran with speed of $12km/hr.$