UMAT 2010 Question Thread

[stonecold]

And I am fairly certain that this question is wrong.  There are 2 answers, both C and D yeah?

[polky]

And I am fairly certain that this question is wrong.  There are 2 answers, both C and D yeah?

Don't know if you are supposed to know this for the UMAT... seems a bit scientific to me and requires prior knowledge (unless there was a preceding paragraph explaining ABO blood groups and you had to make inferences from the information they gave you.) (Seems like they did... did the paragraph explain that blood type AB means that the red blood cells have both A and B antigens present?)

Anyway, Option C is wrong because a person with AB blood type is a universal receiver - they can receive blood cells A, B, AB, and O.  AB red cells have both A and B antigens on the cell surface.  Patients with blood type AB have *no* anti-A or anti-B antibodies and as such they will *not* have an immune reaction to red blood cells which have A or B antigens on the cell surface (since these antigens are already naturally occurring in their blood).

The only correct answer is D.

[polky]

Can anyone explain.  I find their explanation to be plain horrible!

So Mike has either 2 or 3 of each black, blue, red and orange pencils. 

To guarantee that he gets 2 pairs of pencils, each pair the same colour, we have to consider the worst case scenario which is where he keeps on picking pencils of different colours (eg. he picks black first, then blue, then red, then orange).

So let's say for the first 4 rounds of picking he's had shocking luck and has picked 4 pencils each of different colour.  So now in his hand he has one black, one blue, one red and one orange pencil.

His next pick (for pencil #5) will give him a pair no matter what colour pencil he picks (since he already has one of each colour in his hand).  For example let's say his 5th pencil picked is red.

Pick #6 has two possible scenarios.  First, the pick could either give him his second pair of pencils (eg. he picked either black, blue or orange).  This is what he wants, and yay now he has two pairs of pencils of the same colour.  He can stop picking now.

UNFORTUNATELY, the other scenario is that he picks the same colour he picked in #5 (red, because he could have had 3 red pencils to begin with).  If he ends up picking red, the same colour in #5, he must pick again because he still doesn't have his 2 pairs.  He just has 3 red pencils and one each of black blue and orange.

Pick #7 is guaranteed to give him his other pair of pencils because he can't pick another red (max 3 of each colour).  He will have to have picked either black, blue or orange.  This will match up with what he picked during the first four rounds and he will now have 2 pairs of pencils.

Thus to guarantee he gets 2 pairs of the same, he needs to pick 7.

[shinny]

Can anyone explain.  I find their explanation to be plain horrible!

Urgh, the wording of the question really tripped me out. So you're aiming for two pairs, note, not two pencils of the same colour, but rather, two sets of pairs i.e. 4 pencils. So what they're saying is go for the worst possible case scenario to find the minimum number to guarantee two pairs. So just take it in worst possible scenario stages:
1. You draw one pencil of each colour (4 pencils drawn, 0 pairs).
2. No matter what you draw here, you will definitely make a pair. To stick with their example, pretend you pick up a black (5 pencils drawn, 1 pair).
3. However, remember they said that there could be up to 3 of each pencil. So what if you picked up another black? Let's assume that happens (6 pencils drawn, 1 pair).
4. Now finally, there's no more blacks left, so no matter what you pick up, you're bound to hit another pair (7 pencils drawn, 2 pairs).

Success!

EDIT: Goddammit =P

[polky]

Can anyone explain.  I find their explanation to be plain horrible!

Urgh, the wording of the question really tripped me out. So you're aiming for two pairs, note, not two pencils of the same colour, but rather, two sets of pairs i.e. 4 pencils. So what they're saying is go for the worst possible case scenario to find the minimum number to guarantee two pairs. So just take it in worst possible scenario stages:
1. You draw one pencil of each colour (4 pencils drawn, 0 pairs).
2. No matter what you draw here, you will definitely make a pair. To stick with their example, pretend you pick up a black (5 pencils drawn, 1 pair).
3. However, remember they said that there could be up to 3 of each pencil. So what if you picked up another black? Let's assume that happens (6 pencils drawn, 1 pair).
4. Now finally, there's no more blacks left, so no matter what you pick up, you're bound to hit another pair (7 pencils drawn, 2 pairs).

Success!

EDIT: Goddammit =P

Your explanation made alot more sense in alot less words!

[stonecold]

Thanks polky. I misinterpreted the question. Thought it was talking about the pencils left in the case.  My bad.

But with the antibody question, I am finding that they just continue to contradict themselves with the answers.  It did have a rather large stimulus, followed by 4 questions.  I'll post it up with their answers.  The problem is that they said in a previous question that because O blood had anti-A and anti-B anitbodies, then it can only be given to people with blood type O.  So I assume this would apply for all questions.

Have a read, only if you have time and it's not too much trouble.

Edit: Thanks too shinny.  I really appreciate everyone's help.

[kenhung123]

Just our of curiosity, does anyone find some medentry section 3 patterns to be very undoable? Like pick the middle when 3 identical arrows move in different patterns maybe 1 180degrees, 135 degrees and 90 degrees?

[stonecold]

I hate pick the middle.  It is a joke sometimes.  I sorta have worked out a formula for it, but am yet to try it out properly.

What I aim to do is:
-Use the 3-2 rule.  This will allow you to deduce which ones are 1st, 3rd and 5th, and which ones are 2nd and 4th.
-Now you have 3 to pick from that COULD start you off/end you off.  The thing here is the odds are in your favour, because you have a 2/3 chance of picking the first or last in the sequence. SO pick one. Then you look at both your 2nd and 4th options.  Note what has happened.  Then try and apply the rule to the next one in the  sequence.  Remember that the rule may change and could be 1 spot, 2 spots, 3 spots etc.  It doesn't have to be constant.

This is the best method I've come up with to take these annoying problems.  If anyone has anything better, please share it.

Also I find some of the number games ones to be ridiculous.

Edit: And if you can't apply the 3-2 rule and the pattern is not obvious then you are screwed.

[shinny]

Be relieved to hear that the actual UMAT has very few pick the middles, and all the ones I do remember were very simple 3-2 rule based ones. The ones in the UMAT in general were very simple compared to the random and often almost unsolvable patterns given in MedEntry.

[vexx]

Be relieved to hear that the actual UMAT has very few pick the middles, and all the ones I do remember were very simple 3-2 rule based ones. The ones in the UMAT in general were very simple compared to the random and often almost unsolvable patterns given in MedEntry.

thats such a relief, sec 3 is my worst section due to those weird as patterns.

[stonecold]

that makes me smile a little shinny.

problem solving is harder than in medentry though yeah?
if so then that sucks, because I can't even do most of the medentry ones...

[shinny]

that makes me smile a little shinny.

problem solving is harder than in medentry though yeah?
if so then that sucks, because I can't even do most of the medentry ones...

Possibly. More so because the style of the puzzles changes a little. I don't think it's necessarily more difficult. MedEntry just repeats the same stuff so you start to get a hang of it and bam, UMAT pulls out something totally different.

[stonecold]

Okay.  Thanks.  I guess I will just have to try and improve.

Also, did anyone feel on the GAT with regard to the Jose Sandwich question as if it was flash forward to the UMAT?
I swear that felt like a section 2 question haha.

[stonecold]

With problem solving, is it best to test all of the possible answers  instead of trying to solve it?  It is probably the easiest way (not that it is easy), but it just takes up a fair amount of time...

[shinny]

With problem solving, is it best to test all of the possible answers  instead of trying to solve it?  It is probably the easiest way (not that it is easy), but it just takes up a fair amount of time...

Possibly. Really depends on the question, and some questions won't give the ability to simply substitute in answers.