Analyzing Work Rate Relationships

Let the speeds (work done per day) of X, Y, and Z be $S_X$, $S_Y$, and $S_Z$, respectively. The total work is considered as 1 unit.

• The first condition states X works as fast as Y and Z together:

  $S_X = S_Y + S_Z \quad (1)$

• The second condition states X and Y together can finish the task 3 times as fast as Z:

  $S_X + S_Y = 3S_Z \quad (2)$

• The third condition states all three working together finish the task in 15 days:

  $S_X + S_Y + S_Z = \frac{1}{15} \quad (3)$

Calculating Individual Speeds

We can solve the system of equations to find the speed of Z ($S_Z$).

• Substitute equation (1) into equation (2):

  $(S_Y + S_Z) + S_Y = 3S_Z$

  $2S_Y + S_Z = 3S_Z$

  $2S_Y = 2S_Z$

  $S_Y = S_Z$

• Now, substitute $S_Y = S_Z$ into equation (1) to find $S_X$ in terms of $S_Z$:

  $S_X = S_Z + S_Z$

  $S_X = 2S_Z$

• Substitute the expressions for $S_X$ and $S_Y$ (in terms of $S_Z$) into equation (3):

  $(2S_Z) + S_Z + S_Z = \frac{1}{15}$

  $4S_Z = \frac{1}{15}$

  $S_Z = \frac{1}{4 \times 15}$

  $S_Z = \frac{1}{60}$

Determining Z's Task Time

The speed of Z ($S_Z$) is $\frac{1}{60}$ of the task per day. The time Z takes to finish the task alone is the reciprocal of its speed.

Time for Z = $\frac{1}{S_Z} = \frac{1}{1/60}$

Time for Z = 60 days.