BMAT explanation of problem
Hello,
I have a difficulty in understanding this problem's solution. It is from Past Paper 2010 Section, question 8 - http://www.admissionstests.cambridgeassessment.org.uk/adt/digitalAssets/123910_BMAT_Section_2_2010.pdf
In my opinion, the chance that a player is exactly back where they started is 1/16. Regarding the direction, after his first move, he has a 1/4 chance to go to the opposite direction (for example, if his first move is to right, he has a 1/4 chance to move to left at the second move). With respect to the number of steps, the situation is the same: he has a 1/4 chance to hit the same number of steps after the first on move on the steps counter, therefore the chance to be back exactly where he started from is 1/4 * 1/4 = 1/16, but the answer is 1/8.
I have a difficulty in understanding this problem's solution. It is from Past Paper 2010 Section, question 8 - http://www.admissionstests.cambridgeassessment.org.uk/adt/digitalAssets/123910_BMAT_Section_2_2010.pdf
In my opinion, the chance that a player is exactly back where they started is 1/16. Regarding the direction, after his first move, he has a 1/4 chance to go to the opposite direction (for example, if his first move is to right, he has a 1/4 chance to move to left at the second move). With respect to the number of steps, the situation is the same: he has a 1/4 chance to hit the same number of steps after the first on move on the steps counter, therefore the chance to be back exactly where he started from is 1/4 * 1/4 = 1/16, but the answer is 1/8.
A children's game is played on a square grid starting in the centre. Players spin two spinners to
decide how to move their counters. The first spinner decides the direction (Left, Right, Up or
Down) and the second spinner decides the distance (1, 2, 3 or 4 squares).
What are the chances that, after two moves, a player is exactly back where they started?
A 1/256
B 1/64
C 1/16
D 1/8
E 1/4
decide how to move their counters. The first spinner decides the direction (Left, Right, Up or
Down) and the second spinner decides the distance (1, 2, 3 or 4 squares).
What are the chances that, after two moves, a player is exactly back where they started?
A 1/256
B 1/64
C 1/16
D 1/8
E 1/4
Scroll to see replies
I got 1/16 as well, my method is probably bad though.
I did 0.25^4, which is the probably of any one combination that will get you back to where you start. eg, L 1 R 1 (as each spin is 0.25). Then I multiplied by 16, as I thought there were 16 ways of it happening. However, I suspect you are meant to multiply by 32, if the answer is 1/8.
Been too long since I've done probability...
EDIT: Lol, yeah, as below, the markscheme says 1/16..........
I did 0.25^4, which is the probably of any one combination that will get you back to where you start. eg, L 1 R 1 (as each spin is 0.25). Then I multiplied by 16, as I thought there were 16 ways of it happening. However, I suspect you are meant to multiply by 32, if the answer is 1/8.
Been too long since I've done probability...
EDIT: Lol, yeah, as below, the markscheme says 1/16..........
(edited 11 years ago)
Can someone explain me the problem 11?
http://www.admissionstests.cambridgeassessment.org.uk/adt/digitalAssets/119145_BMAT_Sec_2_2009.pdf
http://www.admissionstests.cambridgeassessment.org.uk/adt/digitalAssets/119145_BMAT_Sec_2_2009.pdf
Original post by dornam
Can someone explain me the problem 11?
http://www.admissionstests.cambridgeassessment.org.uk/adt/digitalAssets/119145_BMAT_Sec_2_2009.pdf
http://www.admissionstests.cambridgeassessment.org.uk/adt/digitalAssets/119145_BMAT_Sec_2_2009.pdf
What are your thoughts on it?
Original post by nexttime
What are your thoughts on it?
I'd say D. Not completely sure though.
I got the half life to be 2.76 as half of 220 is 110 which corresponds to slightly more than 2.5.
The fact that the radiation reaches the detector at 30cm with high energy but not at 100cm makes me think beta, as alpha wouldn't reach the first detector and gamma would reach both with fairly high energy.
Can anyone confirm or refute?
Original post by RexRemedium
I'd say D. Not completely sure though.
I got the half life to be 2.76 as half of 220 is 110 which corresponds to slightly more than 2.5.
The fact that the radiation reaches the detector at 30cm with high energy but not at 100cm makes me think beta, as alpha wouldn't reach the first detector and gamma would reach both with fairly high energy.
Can anyone confirm or refute?
I got the half life to be 2.76 as half of 220 is 110 which corresponds to slightly more than 2.5.
The fact that the radiation reaches the detector at 30cm with high energy but not at 100cm makes me think beta, as alpha wouldn't reach the first detector and gamma would reach both with fairly high energy.
Can anyone confirm or refute?
The radiation is beta, but you have the wrong half life.
Spoiler
Original post by Glacier
The radiation is beta, but you have the wrong half life.
Spoiler
Aha ok that makes sense
Original post by dornam
Can someone explain me this problem?
What have you done so far?
Original post by nexttime
What have you done so far?
Answer to this question will be:
First calculate the average speed for the first 48 m which is 4 m/s.
Then calculate the average speed from the last 84 m which seems 7 m/s.
Then calculate the the acceleration which is (7 - 4)/12 => notice that there are 3 equal spaces and the answer is 0.25 which is D is think.
Original post by dornam
I am stuck, I don't know what to begin with. I tried the equation final velocity = initial velocity + acceleration * time , didn't work...
All I know is that practically he is accelerating for 4 seconds.
All I know is that practically he is accelerating for 4 seconds.
The bolded bit is exactly right. How could you work out initial velocity and final velocity? hint: the rate of the oil dropping is very important.
Original post by dornam
How do you get the speed from the last 84 metres?
Do those points represent the drips? I mean, between each drip it passed 4 seconds? I didn't think about it this way...
Do those points represent the drips? I mean, between each drip it passed 4 seconds? I didn't think about it this way...
You know the distance, and since one drop is every 4 seconds, you know the time.
Speed = distance / time. So you can get initial speed and final speed, subtract them to get 3.
Then divide that by the time from K to L to get the acceleration.
Well, that's what I did. That gives Answer D, but I could be wrong.
(edited 11 years ago)
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