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

Step 1. Declaring the variable.

Let the speed of Sagar be considered as $xkm/hr$.

And speed of Aakash be$ykm/hr$.

Hence, time taken by Sagar$=2/xhrs$.

And time taken by Aakash $=2/yhrs$.

Step 2. Following the instructions of the questions.

$$\begin{gathered} \frac{2}{x}–\frac{2}{y}=\frac{2}{60} \\ \frac{1}{x}-\frac{1}{y}=\frac{1}{60}\dots \dots \dots \dots \dots \dots \left(1\right) \\ \frac{(y–x)}{xy}=\frac{1}{60} \\ 60(y–x)=xy\dots \dots \dots \dots \dots \dots .\left(2\right) \end{gathered}$$

As per the second condition

Speed of Sagar$=(x+2)km/hr.$

Speed of Aakash$=(y–2)km/hr$

Step 3. On following the instruction

As per the second condition

$$\begin{gathered} \Rightarrow \frac{2}{(y-2)}–\frac{2}{(x+2)}=\frac{2}{60} \\ \Rightarrow \frac{1}{(y-2)}-\frac{1}{(x+2)}=\frac{1}{60} \\ \Rightarrow \frac{(x+2-y+2)}{(y-2)(x+2)}=1/60 \\ \Rightarrow 60(x-y+4)=xy+2y-2x-4 \end{gathered}$$

Substituting the value of $xy$ in the above equation from equation $\left(2\right)$, we get,

$$\begin{gathered} \Rightarrow \frac{1}{(y-2)}-\frac{1}{y}=\frac{1}{60} \\ \Rightarrow \frac{(y-y-2)}{(y²-2y)}=\frac{1}{60} \\ \Rightarrow y²-2y=120 \\ \Rightarrow y²-2y-120=0 \\ \Rightarrow y²+10y-12y-120=0 \\ \Rightarrow y(y+10)-12(y+10)=0 \\ \Rightarrow (y+10)(y-12)=0 \\ y=-10whichisnotpossible \end{gathered}$$

therefore, $y=12$ is the correct value.

Substituting the value of $y=12inx=y–2$, we get,

$$\begin{gathered} \Rightarrow x=12–2 \\ \Rightarrow x=10 \end{gathered}$$

Hence, Sagar ran with a speed of $10km/hr$and Aakash ran with a speed of $12km/hr$