F weight is 140% of I’s weight. R’s weight is 90% of M’s weight. M’s weight is twice as much as I’s. What percentage of F’s weight is R’s weight?

I’ve done like this:

Let, x be the weight of I.

So, F = 1.4x

M = 2x

R = 0.9 (2x)

Therefore required percentage: $\displaystyle \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 = $\displaystyle ^9\mathrm{C}_3 \times ^4\mathrm{C}_1 = 336 $

Is that right?