Sita, Geeta and Reeta decide to race a distance of y m. Reeta completes the race 28 m ahead of Sita. Geeta completes the race 40 m ahead of Reeta and 64 m ahead of Sita. What is the value of y?

Correct option is D

Given:

The distance of race is y m

Reeta completes the race 28m ahead of Sita

Geeta completes the race 40m ahead of Reeta and 64m ahead of Sita

Solution:

When Geeta completes the race she travelled y metres and Reeta travelled (y - 40)m and Sita travelled (y - 64)m

And when Reeta completes the race she travelled y metres and Sita travelled (y - 28)m

When Reeta travelled 40m Sita travelled 64 - 28 = 36m

So there is a difference of 4 m per 40m

But when Reeta crossed finish line there is gap of 28m

Hence, to make them gap of 28m, Reeta travelled = $$\frac{28}{4} \times 40 = \bf 280m$$ 

Alternate Solution: ​

​$$\frac{\text{Distance covered by slower}}{\text{Total distance}} = \frac{\text{Distance covered by even slower}}{\text{Distance when middle finishes}}$$​​

Let the total race distance (distance Geeta runs) be y metres.

Then:

Reeta covers y - 40 metres when Geeta finishes

Sita covers y - 64 metres when Geeta finishes

When Reeta finishes the race (i.e. covers y - 40 m), Sita has covered y - 68 m (since Reeta is 28 m ahead of Sita).

Using the ratio of distances:

​$$\frac{y - 40}{y} = \frac{y - 64}{y - 28}$$​​

(y - 40)(y - 28) = y(y - 64)

​$$y^2 - 68y + 1120 = y^2 - 64y$$​​

-68y + 1120 = -64y

-4y = -1120

y = 280

The race distance is 280 metres