A college has two departments, finance dept. and accounting

Note:
1) If order is assumed in Statement (2), this is the working:

Using (1), we do not know how many part time teachers in finance dept. (1) is not sufficient.
Using (2), we have:

# of teachers = x
# of finance teachers = 2x/3
# of accounting teachers = x/3
# of part time finance teachers = 3x/7
$ of part time accounting teachers = 4x/7

We can now compute ratio of full time: part time teachers.

(2) is sufficient.

Ans: B

However, if order cannot be assumed, then the answer is D.

ans is C.

I drew the grid, Plugged in a value for total number of employees.

B is suff here. (order assumed)

i got with B.....order assumed......wats the OA

A college has two departments, finance dept. and accounting dept. The ratio of the teachers in two dept. is 2:1. What is the ratio of the number of teachers doing a part-time job to the number of the teachers not doing?

1) 1/2 of the teachers in accounting dept. do part-time jobs.
2) The ratio of the teachers doing part-time jobs in two departments is 3:4


1) -- We don't know how many part-time teachers are there in Finance dept..--> Insuff

2) Assuming the ratio 3:4 is finance : accounting respectively, then I go with B, 2 alone is Sufficient

If this question comes in the real exam the same way as it appears here, I would pick E. may be this is one of the GMAT Traps.

I think it is C.

Explanation, lets assume the total number of teachers 24
24:- 16 teachers in Fin and 8 teachers in Acc

Stat 1. half teacher in Acc is part time, so 4 PT and 4FT - NOT SUFFICIENT

Stat2. PT job in two department is 3:4, means 3 PT teacher in Fin for every 4 PT teacher in Acc, NOT SUFFICIENT

However combined together stat 1&2.
when there are 8 teachers in Acc, and half is PT, means 4 are PT. This is same as 3 PT in Fin, so 13 FT and 3 PT.

THE RATIO OF PT/FT = (3+4)/(13+4) =7/17

both statement together are sufficient! (assuming theorder is same)

Any comments? please!

if the number of teachers is 48 ,for example, how will you go around with B only?

If order is assumed as (F:A) then its C else E.

Here I go
F is finance, A is accounting. 1 is permanent teachers and 2 is part-time teachers. So given is

F1+F2 : A1+A2 = 2:1

1) A1:A2 = 1:1
So F1+F2 : A + A = 2:1 there is no information about F2 so we can not find F2 + A2 : F1+ A1. INSUFF
2) F2: A2 = 3:4. By this we can not find the sum F2 + A2 or F1 + A1. INSUFF

Both combined

A1:A2 = 2:1 and F2:A2 = 3:4
Now assume total = x
From given F:A = 2x/3 : x/3

From 1 A1 = x/6 and A2 = x/6

From 2 F2 = (x/6)* (3/4) = x/8. So F1 = (2x/3) - x/8 = 13x/24

So (F2+A2)/(F1+A1) = [(x/8) + (x/6)] / [(13x/24) + (x/6)] = 7/17 SUFF