Question

Hello, brainlians!
Today I have a question to share. Can you figure it out?

Topic: Linear equation.

A and B walk around a circular path for once, while A goes counter-clockwise and B goes clockwise. 10 minutes after A departed, B starts walking, and both encounter at Q. After that, A arrived at P in 24 minutes, and B arrived at P in 6 minutes.
Let the speed of A and B be aa and bb (both hour/min). Find \dfrac{b}{a}
a
b
​
. (Assuming both person walks in the constant speed.)

The answer is 3. Don't spam or copy-paste, please.​

Answer 1

Given Conditions

1. A goes counter-clockwise and B goes clockwise.
2. 10 minutes after A departed, B starts walking, and both encounter at Q.
3. After that, A arrived at P in 24 minutes, and B arrived at P in 6 minutes.
4. The speed is a and b.

Keys

• $\mathrm{Speed=\dfrac{Distance}{Time} }$
• There are three speeds for each person.

Solution

Let's consider three arcs.

• arc $AQ$
• arc $BQ$
• arc $PQP$ (full-circle arc)

$a=\dfrac{AQ}{x}=\dfrac{BQ}{24} =\dfrac{PQP}{x+24}$ (hour/min)

$b=\dfrac{AQ}{6}=\dfrac{BQ}{x-10} =\dfrac{PQP}{x-4}$ (hour/min)

Thus we get equations for x.

$\therefore\dfrac{b}{a} =\dfrac{x}{6} =\dfrac{24}{x-10} =\dfrac{x+24}{x-4}$

Solving the equation gives $x=-8$ or $x=18$, then the required solution is $x=18$.

The ratio of two speed is $\dfrac{b}{a} =3$.

Hope you understood.

Answer 2

Answer:

Let us treat circular motion in degrees. 1 round = 360°

Speed of A

= 2 round/hr. = 720° in 60 min = 360° in 30 min

= (360°/30) in 1 min.

= 12° / min.

Speed of B

= 3 round/hr. = 1080° in 60 min = 540° in 30 min

= (540°/30) in 1 min.

= 18° / min.

Since A and B are moving in the opposite directions, they complete 360° together, while crossing each other.

Let it take n minutes to make a round (360°) together.

n (12°+18°) = 360°

n=360°/30°

n=12

They cross each other once in 12 minutes.

Between 8:00 am to 9:30 am, there are 90 minutes.

The number of crosses = 90/12 = 7.5

Ans: They will cross each other 7 times before 9.30 am