F weight is 140% of Iís weight. Rís weight is 90% of Mís weight. Mís weight is twice as much as Iís. What percentage of Fís weight is Rís weight?

Iíve done like this:
Let, x be the weight of I.
So, F = 1.4x
M = 2x
R = 0.9 (2x)

Therefore required percentage:  \frac {0.9 \times 2x} {1.4x} \times 100 = 128.57

Is this right?

Another one:

A committee consists of 9 men & 4 women. In how many ways can a sub-committee of 3 men & 1 women be chosen?

So far I think,
Choosing 3 men from 9 men & 1 woman from 4 women.

So, the number of ways = ^9\mathrm{C}_3 \times ^4\mathrm{C}_1 = 336

Is that right?