Question

A runs 11/3 (*Mixed fraction) as fast as B. If A gives B a start of 30 metres. How far must be the winning post, so that the race ends in a dead heat? Select one: O a. 100 m . O b. 110 m O c. 140 m . d. 120 m​

Answer 1

Answer:

A gives B a start of 30 metres. => A runs x metres and B runs (x−30) metres. That is, the distance to the winning point is 120 metres.

Answer 2

Answer:

The distance of the winning post is $120m$.

Therefore, option d) $120m$ is correct.

Step-by-step explanation:

Given,

The speed of A is $1\frac{1}{3}$ as fast as B.

The race ends in a dead heat, i.e., the time taken is the same.

Let the speed of B = $Q$ m/s

Therefore,

The speed of A = $Q \times 1\frac{1}{3}$ = $\frac{4Q}{3}$ m/s

Now,

Let the distance to the winning post for A = $x$ m

As given, the distance to the winning post for B = $(x-30)$ m

As we know,

• Speed = $\frac{Distance}{time}$
• Time = $\frac{Distance}{speed}$

For A,

• Time = $\frac{x}{\frac{4Q}{3} }$ = $\frac{3x}{4Q}$             -------equation (1)

For B,

• Time = $\frac{x-30}{Q} }$                -------equation (2)

On comparing both the equations, we get:

• $\frac{3x}{4Q} = \frac{x-30}{Q}$
• $\frac{3x}{4} = x-30$
• $3x= 4x- 120$

• x = $120m$

Therefore, the distance of the winning post is $120m$.