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# Trinity Admissions Test Solutions watch

1. (Original post by JakeThomasLee)
Specimen Paper III, Question 8:
Spoiler:
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In a grid of 8 dots, we have an area of 7 x 7 = 49 unit squares

It follows that there are 49 unit squares (1x1 squares)

There are 6 double squares on each row (2x2 squares) and hence 6x6 = 36 double squares in the whole grid

There are 5 triple squares on each row (3x3 squares) and hence 5x5 = 25 triple squares in the whole grid

There are 4 quadruple squares on each row (4x4 squares) and hence 4x4=16 quadruple squares in the whole grid

There are 3 quintuple squares on each row (5x5 squares) and hence 3x3=9 quintuple squares in the whole grid

There are 2 sixfold squares on each row (6x6 squares) and hence 2x2=4 sixfold squares in the whole grid

There is 1 sevenfold square in the grid only

Hence there are squares in the grid who have sides parallel to the dots in the grid

It's clear that this may be applicable to any grid of n dots such that the number of squares with sides parallel to the dots is equal to the sum of all square numbers less than n

Assuming a square must have dots for all four of it's vertices:

Considering diagonal squares:

When I say a 1x1 diagonal square I mean a x square but have written it this way for simplicity.

1x1 diagonal squares take up a 2x2 parallel area in the grid. Hence there are the same number of 1x1 diagonal squares as 2x2 parallel squares in the grid = 36

2x2 diagonal squares take up a 4x4 parallel area in the grid. Hence there are the same number of 2x2 diagonal squares in the grid as 4x4 parallel squares = 16

3x3 diagonal squares take up a 6x6 parallel area in the grid. Hence there are the same number of 3x3 diagonal squares in the grid as 6x6 parallel squares = 4

There can be no 4x4 diagonal squares contained in the grid.

Therefore there are total squares
I disagree with your answer. Squares do not have to be diagonal, they can be just slightly slanted.
2. (Original post by Renzhi10122)
I disagree with your answer. Squares do not have to be diagonal, they can be just slightly slanted.
Ah, yes you're right
3. (Original post by JakeThomasLee)
I currently have this integral, not too sure how I'm going to evaluate it, although it may just be completely obvious and I can't see it

Use the same sub as for the other one. . .
I hope the others made it clear what I did
4. (Original post by JakeThomasLee)
I currently have this integral, not too sure how I'm going to evaluate it, although it may just be completely obvious and I can't see it

Consider then look at and .
5. (Original post by JakeThomasLee)
I currently have this integral, not too sure how I'm going to evaluate it, although it may just be completely obvious and I can't see it

Can't you just divide top and bottom by cos (theta)

Giving 1/(1+tan)

Let u = tan

And using sec ^2 = tan ^2 +1

Gives,

1/(1+u)(1+u^2) which can be done by partial fractions?

Not the most elegant though

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6. (Original post by Zacken)
Consider then look at and .
Deja vu, Siklos somewhere!

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7. (Original post by I am Ace)
Deja vu, Siklos somewhere!

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I will forever remember that STEP question.
8. (Original post by I am Ace)
Deja vu, Siklos somewhere!

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9. (Original post by physicsmaths)
i don't agree with part 1, surely it is P(heads/ 3 heads) which is an 1/8?
then again if he is right can someone explain since i am **** at probability.
got it now my division of numbers was wrong
The probability of and all being heads is though, by definition . So the probability that is just the probability the coin is flipped heads at .
The probability of and all being heads is though, by definition . So the probability that is just the probability the coin is flipped heads at .
p( h|3 heads)=1/2 I had an error with numbers causing me to get 1/8

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11. (Original post by joostan)
Specimen Paper 2, Question 9:
Caveat - my probability is a bit weak and I'm not convinced about the last part.
Also succinct notation was a pain.
Spoiler:
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Notational stuff:
Let be the number of cards of type i.e
Then, assuming :

a)

b)

c) I claim (perhaps erroneously) that:

i.e I'm saying that we need Queens from cards and Kings from cards - the full pack without the Queens, and that these are independent.

Simplifying a little we obtain:
.
for part c why is it 48 and not 52-i for the number of cards in which to draw a king?
12. Is there a thread for the Natural Sciences solutions?
13. (Original post by ak33m98)
Is there a thread for the Natural Sciences solutions?
Don't think so. U might aswell make one then

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14. (Original post by mattyd375)
for part c why is it 48 and not 52-i for the number of cards in which to draw a king?
Because you want to pull a King from the pack, without pulling a Queen as this would change though as I said - I'm not entirely certain I'm right.
15. (Original post by ak33m98)
Is there a thread for the Natural Sciences solutions?
NatSci's aren't as cool as Mathmos, it seems.

jk, I don't think they have a thread - you might want to make one yourself, but it seems a little late for this years interview session.
16. Test 4, Question 5

Hint
Spoiler:
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You have two unknowns - so you need two equations. Use the trigonometry the question suggests to find them.

Solution
Spoiler:
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I attach a diagram. I've used A and B instead of theta and phi because latex is effort.

Consider triangle XX'P (where P is the location of the deer), we can write down

Similarly from YY'P,

Substituting in the values of Y' from the first equation into the second, we obtain (after messy algebra)

and substituting the second equation into the first:

Someone should probably check my algebra.
17. Test 5, Question 1 (Maths and Physics, Paper 1)

Hint:
Spoiler:
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A nice sub then Integration by parts.
Solution:
Spoiler:
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Using the sub we have .

Integration by parts with and gives us .

Repeated integration by parts yields (which is the integral we were given, with some fiddling with limits) so, finally:

N.B: .
18. Test 5, Question 5 (Maths and Physics, Paper 1)

Thoroughly enjoyed this question.

Hint:
Spoiler:
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Solution:
(a)
Spoiler:
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This part is fairly trivial:

(b)
Spoiler:
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(Without discussing convergence, as per the question, we then have):

Hence:

Where we used integration by parts with and .

as required.
19. (Original post by Zacken)
Test 5, Question 5 (Maths and Physics, Paper 1)

Thoroughly enjoyed this question.

Hint:
Spoiler:
Show
Solution:
(a)
Spoiler:
Show

This part is fairly trivial:

(b)
Spoiler:
Show

(Without discussing convergence, as per the question, we then have):

Hence:

Where we used integration by parts with and .

as required.
How did you manage to get Test 5 out of 4 numbered tests?
20. (Original post by FutureOxford3)
How did you manage to get Test 5 out of 4 numbered tests?
(Original post by Zacken)
(Maths and Physics, Paper 1)
...

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