BMAT explanation of problem

Thank you so much dude! i know it's bad 2 ask u again but can u pls explain me q 20 same paper!
are u sitting for BMAT this year?

wow you have such a brain! how come? thanks btw

In order to fit the criteria, we have two possible scenarios:

1) YRYRYR
2) RYRYRY

Let's work with the first scenario. We've got 3 bulbs to choose from when planting the first and second bulb (since it has to be the alternating colour), 2 to choose from when planting the third and fourth bulb and then only one choice when planting the last two, so we can ignore those.

Mathematically, this can be written as: 3*3*2*2=36. However, as I said in the beginning, two scenarios fit, where we start with the other colour instead. Hence, the number of possibilites doubles, yielding 72.

Now, in total, there are 6!=720 ways of arranging the bulbs. 72/720=1/10.

where do you find the mark scheme for 2008 paper?

I get everything in bold, but I don't understand the first part. What do you mean when you say there are 3 bulbs to choose from when planting the first and second bulb? Sorry to be a pain!



http://www.admissionstests.cambridgeassessment.org.uk/adt/digitalAssets/109681_BMAT_Past_Paper_Section_1_answer_key_and_score_conversion.pdf and you can find a load of past papers/ mark schemes on here: http://www.admissionstests.cambridgeassessment.org.uk/adt/bmat/Test+Preparation

It's okay, don't worry. If you're having problems with these types of questions I recommend revising 'permutations'. The question is asking for the probability that the bulbs alternate along the whole row. In other words, they are asking for the probability that we get the order YRYRYR or RYRYRY (both show alternating colours).

Imagine you have 6 bulbs in front of you, 3 red and 3 yellow ones. Now, you want to plant them in an alternating way. This means that you have 3 bulbs to choose from when planting the first one (as you have to start with ONE of the colours). When planting the next one, you still have 3 bulbs to choose one as you are choosing one of the OTHER colour. Afterwards, you have planted one of each, so you have 2 of each left. Is it clearer now?

but the answer you get when you divide 1.59/2 and multiply by 100 is 79.5%!

hey this is probably a printing mistake. Sorry but i cud only make this out of the question. Since this is a complete combustion reaction all the carbon from the carbon containing compound must be int the CO2. I find no other way but to use the atomic ratio to find the mass of carbons. Please some one correct me if i am wrong.

Here is the solution,
First of all you know that the total carbon atoms remain same b4 and after the reaction.
so.................the ratio of carbon is to oxygen in the compound is 1:2 in CO2
so divide 4.77 by 3 which is equal to 1.59g! ( please trust the fact that in BMAT most calculation are really easy and can be done quickly)
so CO2 has 1.59g of carbon. So original 2 gram carbon containing compound, there must have been 1.59g of carbon
and 1.59 percentage of 2 is around 65% ( here they are probably testing the power of estimation) and the answer is E!!!!

Exactly! even i believe there's some printing mistake because even i dont get that answer! if anybody else think i am wrong please correct me!

You are completely wrong. That is definitely not how to solve it. What you are doing, is assume that 1 mol of carbon has the same mass as 1 mol of oxygen, which is not true. Hence, your ratio is wrong.



It's not a printing mistake, I did the question just now and got 64,5%. First of all, what you do, is calculate the percentage of carbon in 1 mol of carbon dioxide, i.e. 12/(12+32)=12/44=3/11=0,27

Now, since there are 0.27% of carbon in 1 mol, it means that 0,27% of the 4.77g of CO2 is just carbon. Hence, 0.27*4,77=1.29

Finally, as the other person said earlier, the mass of carbon (and any other element) is conserved (i.e. same before & after reaction). Hence, there are 1.29g of carbon in the 2g of the carbon compound. Thus: 1.29/2 = 64.5/100=64.5%

For question 20, this "The cylinder has the same internal diameter and length as the diameter of the sphere. " is absolutely crucial. What it is saying is that the height=diameter of sphere=2r and that the radius of the circular opening is also r.

Formula for volume of a cylinder: pi*r^2*h. Now, if h=2r, we get: 2pi*r^3.

We are told (and should know) that the volume of a sphere is 4/3pi*r^3.

To calculate the fraction the sphere takes up, we simply do (4/3pi*r^3)/(2pi*r^3). pi*r^3 cancels out on both sides, giving (4/3)/2=4/6=2/3. Hence, the answer is D.

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When it comes to 26, I actually had a bit trouble with grasping the last sentence (and what I'm supposed to find) due to, in my opinion, the poor phrasing. Nevertheless, after reading it several times, it finally made sense to me.

What they are basically asking is: out of all possible compounds of CH2BrCl, which ones will have the relative molecular mass of 128? In other words, this is a probability question within chemistry.

Now, we are told to assume that the relative atomic mass of carbon and hydrogen is 12 respectively 1 (i.e. always that) which means that we can subtract 128 by 14, giving us 114.

We have two possible masses of chlorine, 35 and 37 and two possible masses of bromine, 79 and 81. If you study these numbers, you will see that the only combination which adds up to 114 is 35 respectively 79.

Now, what is the "probability" that the chlorine in the compound has a mass of 35? We are told that the abundance of Chlorine-35 is three times that of Chlorine-37, or, in other words, that the "probability" is 3/4.

The "probability" of having bromine-79 is the same as having bromine-81, i.e. 1/2.

Hence: 3/4*1/2=3/8, hence C.

For question 20, this "The cylinder has the same internal diameter and length as the diameter of the sphere. " is absolutely crucial. What it is saying is that the height=diameter of sphere=2r and that the radius of the circular opening is also r.

Formula for volume of a cylinder: pi*r^2*h. Now, if h=2r, we get: 2pi*r^3.

We are told (and should know) that the volume of a sphere is 4/3pi*r^3.

To calculate the fraction the sphere takes up, we simply do (4/3pi*r^3)/(2pi*r^3). pi*r^3 cancels out on both sides, giving (4/3)/2=4/6=2/3. Hence, the answer is D.

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When it comes to 26, I actually had a bit trouble with grasping the last sentence (and what I'm supposed to find) due to, in my opinion, the poor phrasing. Nevertheless, after reading it several times, it finally made sense to me.

What they are basically asking is: out of all possible compounds of CH2BrCl, which ones will have the relative molecular mass of 128? In other words, this is a probability question within chemistry.

Now, we are told to assume that the relative atomic mass of carbon and hydrogen is 12 respectively 1 (i.e. always that) which means that we can subtract 128 by 14, giving us 114.

We have two possible masses of chlorine, 35 and 37 and two possible masses of bromine, 79 and 81. If you study these numbers, you will see that the only combination which adds up to 114 is 35 respectively 79.

Now, what is the "probability" that the chlorine in the compound has a mass of 35? We are told that the abundance of Chlorine-35 is three times that of Chlorine-37, or, in other words, that the "probability" is 3/4.

The "probability" of having bromine-79 is the same as having bromine-81, i.e. 1/2.

Hence: 3/4*1/2=3/8, hence C.

and wat about q11? i know the radiation is beta but half life i get 2.76!

I would have either guessed or left the question an leave it to chance. they only give a little time to find out such a easy but tricky concept.
Anyways, doppel guy you are true genius and i hope you get into Oxford or Cambridge, wherever you are applying. Would you mind if I ask you about your results?