Productivity ratio of all the machine = 1 : 2 : 5
That is, in 1 hour; Slowest machine finishes just 1 work
Medium machine finishes just 2 works
Fastest machine finishes just 5 works

with 3 hours work;
Slowest machine = 1 * 3 hours =3 hours
Medium machine = 2 * 3 hours = 6 hours
Fastest machine = 5 * 3 hours = 15 hours
3 : 6 : 15
Therefore, the slowest machine breaks at the beginning of the 4th hour but was fixed at the beginning of the 7th hour,
this means that the slowest machine misses 3 hours (4th, 5th and 6th hour) but, the medium and fastest machines have
Medium machine = 2 * 3 hours = 6 works
Fastest machine = 5 * 3 hours = 15 works
Given the fact that the job is done in 9 hours we have 3 hours remaining (7th, 8th and 9th)
Therefore, slowest machine start working,
Slowest machine = 1 * 3 hours = 3 works per 3 hours
Medium machine = 2 * 3 hours = 6 works per 3 hours
Fastest machine = 5 * 3 hours = 15 works per 3 hours.
Finding the ratio of work performed by the fastest machine to the work performed by the slowest.
Therefore, Total ratio of work by fastest machine : Total ratio of work by the slowest machine.
$$\frac{\left(Total\ ratio\ of\ work\ by\ fastest\ machine\right)}{Total\ ratio\ of\ work\ by\ slowest\ machine}=\ \frac{\left(15\ +\ 15\ +\ 15\right)}{3\ +\ 3}$$
$$=\ \frac{45}{6}=\ \frac{15}{2}=\ 15\ :\ 2$$
Option C is the correct option.

swerve wrote:3 machines have a productivity ratio of 1 to 2 to 5. All 3 machines are working on a job for 3 hours. At the beginning of the 4th hour, the slowest machine breaks. It is fixed at the beginning of hour seven and begins working again. The job is done in nine hours. What was the ratio of the work performed by the fastest machine as compared to the slowest?

A. 5
B. 7
C. 15/2
D. 17/2
E. 12

Let the rate for the fastest machine = 5 units per hour and the rate for the slowest machine = 1 unit per hour.
Since the fastest machine works at a rate of 5 units per hour for all 9 hours, the work produced by the fastest machine = rt = 5*9 = 45 units.
Since the slowest machine works at a rate of 1 unit per hour for only 6 hours -- the first 3 hours and the last 3 hours -- the work produced by the slowest machine = rt = 1*6 = 6 units.
Resulting ratio:
45/6 = 15/2.

The correct answer is C.

swerve wrote:3 machines have a productivity ratio of 1 to 2 to 5. All 3 machines are working on a job for 3 hours. At the beginning of the 4th hour, the slowest machine breaks. It is fixed at the beginning of hour seven and begins working again. The job is done in nine hours. What was the ratio of the work performed by the fastest machine as compared to the slowest?

A. 5
B. 7
C. 15/2
D. 17/2
E. 12

A : 1 task/h
B: 2 tasks/h
C: 5 tasks/h
\[? = \frac{{\# \,\,{\text{tasks}}\,\,C}}{{\# \,\,{\text{tasks}}\,\,A}}\]
\[\begin{array}{*{20}{c}}
{\left( * \right)\,\,\,\# \,\,tasks}&{A\,\,\left( * \right)}&{B\,\,\left( * \right)}&{C\,\,\left( * \right)} \\
{3{\text{h}}\,\,{\text{all}}}&{3\,\,}&{6\,\,}&{15} \\
{3{\text{h}}\,\,B{\text{& C}}}&0&6&{15} \\
{3{\text{h}}\,\,{\text{all}}}&3&6&{15}
\end{array}\]
\[? = \frac{{\# \,\,{\text{tasks}}\,\,C}}{{\# \,\,{\text{tasks}}\,\,A}} = \frac{{3 \cdot 15}}{{3 \cdot 2}} = \frac{{15}}{2}\]

The above follows the notations and rationale taught in the GMATH method.