Who knows how to subnet? Solve this puzzle >.<

This question came up on a practice exam....
Which of the following network addresses enable you to support a network consisting of 60 subnets with up to 300 hosts on each subnet? (choose all that apply)
A. 10.48.0.0/14
B. 192.168.98.0/24
C. 172.30.0.0/25
D. 10.236.128.0/21
Which of the following network addresses enable you to support a network consisting of 60 subnets with up to 300 hosts on each subnet? (choose all that apply)
A. 10.48.0.0/14
B. 192.168.98.0/24
C. 172.30.0.0/25
D. 10.236.128.0/21
WIP: can't decide.
Comments
Therefore, you'll 15 bits total. There's only one that satisfies that.
I gave only one answer ...
The test gave 2 answers
Either I been studying too much lately and my brain is getting fried or something is wrong.
Aren't (B) and (C) wrong because they only allow 256(-2) and 128(-2) hosts per subnet?
(A) and (D) have enough (6 and 13) bits for subnets and enough (18 and 11) bits for hosts?
"Your department is given a 10.236.128.0/21 network segment [meaning you only have this segment and, therefore, have to FURTHER subnet it into subnets/hosts]."
The actual question asks us which subnets could be a PART OF an original network address that was, at some point, subnetted into answer(A) or answer(D).
What I mean is if my math teacher gives me 2*X=Y+2 and wants me to find X, I should find X, not Y.
11 bits to work with hosts.
Since the range for private 10-dot network is 10.x.x.x, you can use any amount of bits from all octets except first 8 and last 11 bits. So, 21 - 8 = 13 bits for subnets.
10.0.0.0 /21 First subnet
10.0.8.0 /21
10.0.16.0 /21
10.236.128.0 /21 Given subnet
10.255.248.0 .21 Last subnet
The former is how I'm interpreting it. It doesn't say "part of" anywhere in there; that's you reading into it. If someone told me to use those network addresses with their respective subnet masks to fulfill the requirements, I would say that A is the only valid choice. I'm not saying what you guys did is wrong. Upon knowing the answer, that seems to be what the question was after. I just don't think the question's phrased properly if that solution is what it's looking for.
Regardless kerby, I think you know what you're doing. I wouldn't get hung up over a question like this.