X, Y and Z are three contestants in a race of 1000 m. Assume that all run with different uniform speeds. X gives Y a start of 40 m and X gives Z a start of 64 m. If Y and Z were to compete in a race of 1000 m, how many metres start will Y give to Z?

The correct option is 25.

Let us analyze the race step-by-step to understand the relationship between the speeds of the three contestants: $X$, $Y$, and $Z$.
Since all contestants run with uniform speeds, the ratio of the distances they cover in the same duration of time remains constant.

First, consider the relationship between $X$ and $Y$:
In a 1000 m race, $X$ gives $Y$ a start of 40 m.
This means that by the time $X$ runs the full distance of 1000 m, $Y$ only needs to cover:
$1000-40=960\text{m}$.
Thus, the ratio of the distance covered by $Y$ to that covered by $X$ in the same time interval is:
$\frac{D_Y}{D_X}=\frac{960}{1000}$

Next, consider the relationship between $X$ and $Z$:
In the same 1000 m race, $X$ gives $Z$ a start of 64 m.
This means that by the time $X$ runs 1000 m, $Z$ only needs to cover:
$1000-64=936\text{m}$.
Thus, the ratio of the distance covered by $Z$ to that covered by $X$ in the same time interval is:
$\frac{D_Z}{D_X}=\frac{936}{1000}$

Now, let us find the ratio of the distance covered by $Z$ to that covered by $Y$ in the same time interval:
We can do this by dividing the two ratios:
$\frac{D_Z}{D_Y}=\frac{D_Z}{D_X}/\frac{D_Y}{D_X}=\frac{936}{960}$

This tells us that in the time $Y$ covers 960 m, $Z$ covers 936 m.
To find out how many metres start $Y$ will give to $Z$ in a 1000 m race, we need to calculate the distance $Z$ covers when $Y$ runs exactly 1000 m:
$D_Z=1000\times \frac{936}{960}$

Let us simplify the fraction:
$\frac{936}{960}=\frac{936/24}{960/24}=\frac{39}{40}$
Substituting this back into the equation:
$D_Z=1000\times \frac{39}{40}=25\times 39=975\text{m}$

Thus, when $Y$ runs 1000 m, $Z$ runs 975 m.
The head start that $Y$ gives to $Z$ is the difference between these distances:
$1000-975=25\text{m}$.