How many additional robots would be required? 
A contract is to be completed in 52 days and 125 identical robots were employed, each operational for 7 hours a day. After 39 days, five-seventh of the work was completed. How many additional robots would be required to complete the work on time, if each robot is now operational for 8 hours a day?
(A) 50 (B) 89
(C) 146 (D) 175

My trial: Given that
125 identical robots, each operational for 7 hours a day, complete work in 52 days
so the work of 1 robot per hour$=1/(125\cdot52\cdot7)$
Now, after 39 days left over work $=1-5/7=2/7$ is to be completed in left 52-39=13 days
suppose $x$ number of additional robots are required to complete the left over work then
$$\frac27=\frac{13\cdot(125+x)\cdot8}{52\cdot125\cdot7}$$
$$1=\frac{(125+x)}{125}$$
$$125+x=125\implies x=0$$
But my answer doesn't match any of given options. I don't know where I am wrong. Please give correct solution with explanation to this problem.
thanks
 A: Consider the $n$ the number of robot hours of work produced so far. We have:
$$n = 39 \times 7 \times 125 = 34125$$
In total $N$ robot hours are needed:
$$\frac{5}{7}N = n \quad \longrightarrow\quad N = 47775$$
We have $N - n$ robot hours work left, and $(52 - 39) \times8 = 104$ hours. Thus the number of additional robots needed is:
$$\frac{N - n}{104} - 125 = 6.25$$
Since we can't have fractional robots, we need $7$.
I have no idea where the multiple choice answers came from, from napkin paper math $5/7 \approx 0.71$ vs $39/52 =0.75$ even discounting that we now work more hours per day, adding $50$ robots to $125$ is ridiculous.
A: COMMENT.-A way to see none of the given possible answer is correct.
$$39\cdot125\cdot7=\frac{5T}{7}\cdots\cdots (1)\\(A)\text{ gives } 13\cdot175\cdot8=\frac{2T}{7}\\(B)\text{ gives } 13\cdot214\cdot8=\frac{2T}{7}\\(C)\text{ gives } 13\cdot271\cdot8=\frac{2T}{7}\\(D)\text{ gives } 13\cdot300\cdot8=\frac{2T}{7}$$
Taking the quotients of $(1)$ by  each of the fourth letters $A,B,C,D$ the LHS  is equal to $\dfrac 52$ but the  corresponding LHS's are $1.875,\space1.5332...,1.2107....,1.09375....$ respectively and none of them is equal to $\dfrac 52$.
The answer $6.25$ given below by orlp is correct despite its lack of meaning.
A: The number of robots is actually irrelevant. It is just a side calculation. 125x7 is 875 hours a day. Multiplied by 39 we have 34,125 work done. Since we know that 29,250 is 5/7 of the work we can divide by 5 and multiply by 2 to get 2/7 of the work. That will land us 13,650 hours of work left. 
11,700/8 divided by 125=13.65
This is not enough. The smallest amount possible is actually some number I CBA to figure out. The reason why is because it won't any of the options but it will be less than 1 of the options. Meaning every option works out fine. The smallest possible number of extra robots required out of that option is A so there is your answer.
However, if you want to be annoying to your assessor, just tell them that D is the correct answer. If D is the wrong answer then all of them are. The reason is because any less robots than D will be slower.
