Given below are two statements:
Statement (I):
4 boys and 8 girls completed 1/3 of work in 5 days. After that 3 boys and 3 girls increased, and they completed another 1/3 of work in 3 days. If the remaining work is to be completed in 2 days, then the number of girls that should be increased is 32.
Statement (II):
The ratio of time taken by A and C to do a work is 1:2 respectively. B is 166 (2/3)% more efficient than C. Time taken by A to complete 6% of work is 6 days.
Then, the time taken by B and C together to complete the whole work is 54 (6/11) days. In light of the above statements, choose the most appropriate answer from the options given below.

Statement I

Let the efficiency of the boy be B and the girl be G.
4 boys and 8 girls completed one-third of the work in 5 days. That is, they complete the entire work in 15 days. 
Thus, Total Work = $$15\times(4B+8G)=60B+120G$$
After that, 3 boys and 3 girls increased, and they completed another one-third of the work in 3 days. This means - 
$$3\left(7B+11G\right)=\dfrac{1}{3}\left(60B+120G\right)$$
$$21B+33G=20B+40G$$
$$B=7G$$
Thus, total work = $$60(7G)+120G=540G$$
Now, two-thirds of the work is completed, and the remaining one-third, or 180G, should be finished in 2 days. For this, let's say, we need x more girls. Thus, we have 7 boys, and (11+x) girls, and B = 7G
$$2\left(7B+\left(11+x\right)G\right)=180$$
$$7\left(7G\right)+\left(11+x\right)G=90$$
$$(11+x)G=41G$$
$$x=30$$
Thus, we need 30 more girls to complete the required work in the required number of days.
Statement I is FALSE. 

Statement II

The ratio of time taken by A and C to do the work is 1:2, respectively. Thus, A takes half the time as C to do the same work. Thus, A is twice as efficient as C.
B is 166 (2/3)% more efficient than C. Thus, $$B=\left(1+\dfrac{500}{3\times100}\right)C=\dfrac{8C}{3}$$, and $$A=2C$$
The time taken by A to complete 6% of the work is 6 days. Thus, he will complete 100% work in 100 days. 
Total work = $$A\times100=2C\times100=200C$$
Let the time taken by B and C together to complete the whole work be D days. Then -
$$\left(B+C\right)\times D=200C$$
$$\left(\dfrac{8C}{3}+C\right)\times D=200C$$
$$D=\dfrac{600}{11}=54\dfrac{6}{11}$$days
Statement II is also TRUE.