Question

Two persons are walking along the boundary of a park with different speeds, which has a perimeter of 1.5km. If they move in the opposite directions, they meet after 15 minutes while if they move in the same direction they meet in 45 minutes. Find the speeds in km/hr.

Answer 1

Let the -

• speed of one person be x km/hr.
• speed of second (another) person be y km/hr.

Two persons are walking along the boundary of a park with different speeds and the perimeter of the park is 1.5 km.

If both the persons move in opposite directions then they meet after 15 min.

At first, convert min into hrs.

To convert min into hr. divide with 60.

=> $\sf{\frac{15}{60}}$

=> $\sf{\frac{1}{4}}$ hrs.

According to question,

$\implies$ $\sf{\frac{x}{4}\:+\:\frac{y}{4}\:=\:1.5}$

$\implies$ $\sf{x\:+\:y\:=\:1.5(4)}$

$\implies$ $\sf{x\:+\:y\:=\:6}$ ...(1)

If both person move in same direction, then they meet in 45 min.

=> $\sf{\frac{45}{60}}$

=> $\sf{0.75}$ hrs.

$\implies$ $\sf{0.75x\:-\:0.75y\:=\:1.5}$

$\implies$ $\sf{x\:-\:y\:=\:\frac{1.5}{0.75}}$

$\implies$ $\sf{x\:-\:y\:=\:2}$

$\implies$ $\sf{x\:=\:2\:+\:y}$ ...(2)

Substitute value of x = 2 + y in (eq 1)

$\implies$ $\sf{2\:+\:y\:+\:y\:=\:6}$

$\implies$ $\sf{2y\:=\:4}$

$\implies$ $\sf{y\:=\:2}$

Substitute value of y = 2 in (eq 2)

$\implies$ $\sf{x\:=\:2\:+\:2}$

$\implies$ $\sf{x\:=\:4}$

•°• Speed of one person is 4 km/hr and speed of second person is 2 k./hr.

Answer 2

Answer:

Step-by-step explanation:

$\bf\underline{Given:-}$

Perimeter of the boundary = 1.5 cm

Time taken on opposite direction = 15 min

Time taken on same direction = 45 min

$\bf\underline{To \ Find:-}$

The Speed

$\bf\underline{Solution:-}$

$\sf Let's \: Assume \: the \: speed \: of \: 1st \: person \: be \: x \: km \: per \: hour.\\And \: the \: 2nd \: person \: be \: y \: km \: per \: hour.$

$\sf Distance \: covered \: by \: 1st \: person=\frac{x}{4} \: km\\Distance \: covered \: by \: 2st \: person=\frac{y}{4} \: km$

$\sf Perimeter \: of \: the \: park=1.5 \: km\\\sf\implies \frac{x}{4}+\frac{y}{4}=1.5\\  \:\implies x+y=6  \:  ....(i)$

$\bf\implies Time \: Reqiured=\frac{Permeter \: of \: the \: Park}{Relative \: Speed}$

$\sf\implies\frac{3}{4}=\frac{1.5}{x-y}$

$\sf\implies 3(x-y)=1.5\times4$

$\sf\implies3x-3y=6$

$\sf\implies x-y=2 \: ...(ii)$

Adding Eq (i) and (ii), we get

$\sf\implies2x=8$

$\sf\implies x=\frac{8}{2}$

$\bf\implies x=4$

Putting x value, we get

$\bf\implies y =2$

$\bold{Speed \: of \: the \: 1st \: person= 4 \: km \: per \: hour.}\\\bold{Speed \: of \: the \: 2st \: person= 2 \: km \: per \: hour.}$