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username1763791
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:king1:


Welcome to the Trinity Admissions Test Solutions Thread.This is the place to discuss, post problems, or ask any questions you may have regarding the exam before your interview!

Here we are going to post solutions to all of the sample Trinity College Cambridge Admissions Tests available online, in the hope to help those preparing for interviews in the coming years. These tests will be useful preparation for all Cambridge Colleges, and Oxford for any mathematical interview.

We do ask that any solutions are spoilers, and, if you are stuck with any questions to first ask for any hints or look at the hints given before the TSR solutions.

When preparing for your interview at Trinity College specifically, please remember that you are not expected to answer all of the questions, or necessarily to give complete answers to questions. You should just attempt those that appeal to you, and they will be used as a basis for discussion in the interview that follows. It has also been mentioned that they will most likely ignore any full solutions you have done in the test, and have you attempt the questions you have not done.

Once finished, each solution will be linked below to the relevant post!



SPECIMEN TEST I
1. Solution by Edothero
2. Solution by Zacken
3. Solution by Zacken
4. (1)Solution by Number Nine(2) DFranklin
5. Solution by Jordan\
6. Solution by 16Characters...
7. Solution by 16Characters...
8. Solution by 16Characters...
9. Solution by Joostan
10. Solution by Edothero;Part C by DFranklin

SPECIMEN TEST II
1. (1) Solution by Zacken(2) DFrankin
2. Solution by Zacken
3. Solution by Krollo
4. Solution by Zacken
5. Solution by MadChickenMan
6. Solution by Jordan\
7. Solution by Krollo
8. Solution by Renzhi10122
9. Solution by Joosten
10. Solution by Krollo /with some further comments (1) astruser (2) DFranklin


SPECIMEN TEST III
1. Solution by Joostan
2. Solution by Joostan
3. Solution by Hauss
4. Solution by ManChickenMan
5. Solution by Krollo
6. Solution by Joostan
7. Solution by Krollo
8. Solution by Euclidean
9. Solution by MadChickenMan
10. Solution by Krollo

SPECIMEN TEST IV
1. Solution by Krollo
2. Solution by 16Characters
3. Solution by Zacken
4.
5. Solution by Krollo
6. Solution by joostan
7. Solution by Krollo
8. Solution by Mop3476
9. Solution by Krollo
10. Solution by Krollo

Maths and Physics Paper I

We have also came across the Downing Admissions Paper,
where you can also find the solutions to below:
1.
2.
3.
4. Solution by Number Nine
5.
6.
7.
8.
GOOD LUCK!
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Zacken
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Specimen Test 1, Question 2:

Hints:
Spoiler:
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Draw a diagram.
Label the angles the lines make with the positive x-axis, how can you relate these angles to the tangents of the slope?
Full solution:
Spoiler:
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Let the angle that L_1 makes with the positive x-axis be \alpha and the angle that L_2 makes be \beta, then we know that m_1 = \tan \alpha and m_2 = \tan \beta.

A quick sketch shows us that \theta = \alpha - \beta and hence, applying the tangent to both sides yields:

\displaystyle \tan \theta = \tan(\alpha - \beta) =  \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = \frac{m_1 - m_2}{1 + m_1 m_2}, since we only want the acute angle, then we need \tan \theta > 0, so we have, finally:

\displaystyle \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|

We note that our lines cannot be perpendicular since 1 + m_1 m_2 \neq 0 \implies m_1 m_2 \neq -1 and that L_1 and L_2 cannot be vertical.

2
16Characters....
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So we post solutions for them to be put in? If so I have one for Q2. Also I am not a Trinity applicant, but do you think these questions are worthwhile doing to prepare for my St John's interview (if I get one)?

Edit: Whoops Zacken beat me to it anyway :-)
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Zacken
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(Original post by 16Characters....)
So we post solutions for them to be put in? If so I have one for Q2. Also I am not a Trinity applicant, but do you think these questions are worthwhile doing to prepare for my St John's interview (if I get one)?

Edit: Whoops Zacken beat me to it anyway :-)
Very much so, the questions test general maths and will be useful practice for any college including Oxford.
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username1763791
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(Original post by 16Characters....)
So we post solutions for them to be put in? If so I have one for Q2. Also I am not a Trinity applicant, but do you think these questions are worthwhile doing to prepare for my St John's interview (if I get one)?

Edit: Whoops Zacken beat me to it anyway :-)
Definitely - these will be helpful for any college interview. I myself have applied to Pembroke, and Zacken^ for Kings I believe

Also, If you're solution is any different in the approach it'll still be useful to post it!
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16Characters....
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(Original post by Jordan\)
Definitely - these will be helpful for any college interview. I myself have applied to Pembroke, and Zacken^ for Kings I believe

Also, If you're solution is any different in the approach it'll still be useful to post it!
Nope it's pretty much identical. And best of luck to you all :-)

Speciment Test 1, Question 6

Hints
Spoiler:
Show

For the first part consider the necessary condition for the object to decelerate..
The second part can either be approached using energy considerations or a combination of Newton's Second Law and the SUVAT equations.

Solution
Spoiler:
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Let m be the mass of the packing case. For the sliding packing case to eventually come to a rest there must be a resultant force acting up the slope. Hence resolving parallel up the slope:

\mu R - mg\sin \theta \geq \implies \mu mg\cos \theta - mg\sin\theta \geq 0
Since  0 < \theta < 90 I can safely divide by mg\cos \theta without altering the direction of the inequalities hence
 \mu \geq \tan \theta

The work done by the force up the slope in decelerating the packing case is equal to the initial KE of the object,  \frac{1}{2}mu^2 where u is the intitial speed. Hence:

 (\mu mg\cos\theta - mg\sin\theta)d = \frac{1}{2}mu^2
 \implies 2d(\mu g\cos\theta - g\sin\theta) = u^2 

\implies u = (2dg(\mu \cos\theta - \sin\theta))^{\frac{1}{2}}
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Specimen Test 1, Question 3:

Hints:
Spoiler:
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Integration by parts, although I'm sure there's a more elegant method that I can't spot.
Full solution:
Spoiler:
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Using IBP with u = x^2 and \mathrm{d}v = \sin^2 x yields:


\displaystyle I = \int x^2 \sin^2 x \, \mathrm{d}x =  \bigg[\frac{x^2}{4}\left(2x - \sin 2x\right)\bigg]_0^{\pi} - \frac{1}{2}\int_{0}^{\pi}2x^2 - x \sin 2x \, \mathrm{d}x

\displaystyle I = \frac{\pi^3}{2} - \bigg[\frac{x^3}{3}\bigg]_0^{\pi} + \frac{1}{2}\int_{0}^{\pi}x \sin 2x \, \mathrm{d}x

Where another use of IBP on the last integral gives us \displaystyle \int_0^{\pi} x \sin 2x \, \mathrm{d}x = -\frac{\pi}{2} so we get:

\displaystyle I = \frac{\pi^3}{2} - \frac{\pi^3}{3} - \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi^3}{6} - \frac{\pi}{4}.
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Number Nine
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Specimen Test 1 Question 4
Hint:
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Try finding numbers that aren't multiples of 2,3,5 for numbers not quite as big as 6000
Solution:
Spoiler:
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Using 2 x 3 x 5 = 30 and 6000 ÷ 30 = 200:
Numbers that aren't multiples of 2,3,5 between 1 and 30 are as follows
1, 7, 11, 13, 17, 19, 23, 29
There are 8 numbers here, 8 * 200 = 1600
Therefore there are 1600 numbers between 1-6000 that aren't multiples of 2, 3 and 5
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DFranklin
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(Original post by Number Nine)
Specimen Test 1 Question 4
Hint:
Spoiler:
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Try finding numbers that aren't multiples of 2,3,5 for numbers not quite as big as 6000
Solution:
Spoiler:
Show
Using 2 x 3 x 5 = 30 and 6000 ÷ 30 = 200:
Numbers that aren't multiples of 2,3,5 between 1 and 30 are as follows
1, 7, 11, 13, 17, 19, 23, 29
There are 8 numbers here, 8 * 200 = 1600
Therefore there are 1600 numbers between 1-6000 that aren't multiples of 2, 3 and 5
Shorter:

Spoiler:
Show
Since 2,3,5 are prime, removing multiples of one number doesn't affect the proportion of multiples of the others. So the answer is simply 6000 * (1/2) * (2/3) * (4/5) = 1600.
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16Characters....
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Speciment Test 1, Question 8

Hint
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Again conservation of energy, but rotational KE must be considered as opposed to regular KE.
Solution
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Let m be the mass of the rod. The loss of GPE is given by  \frac{mgl}{2} and the rotational kinetic energy is given by  KE = \frac{1}{2}I w^2 where I is the moment of inertia of the rod about its end and w is its angular speed. These two quantities are equal hence

 \frac{mgl}{2} = \frac{ml^2}{6}w^2 \implies w = \sqrt{\frac{3g}{l}}

Hence the speed v of a point on the unhinged end of the rod is given by

 v = lw = \sqrt{3gl}
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username1763791
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SPECIMEN TEST 1, QUESTION 5

Hints
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Using the formula  P(A|B) = \frac{P(B|A)P(A)}{ P(B|A)P(A) + P(B|A' )P(A' )} will help.
Solution
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P ( biased | 8 heads ) =  \displaystyle \frac{\frac{1}{129}}{\frac{1}{12  9} + \frac{128}{129} \left ( \frac{1}{2} \right )^8} = \frac{2}{3}

P ( 'biased | 8 heads) = \frac{1}{3}

P (heads) =  \frac{1}{3}* \frac{1}{2} + \frac{2}{3} = \frac{5}{6}
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tiny hobbit
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(Original post by Jordan\)
[center]:king1:

When preparing for your interview at Trinity College specifically, please remember that you are not expected to answer all of the questions, or necessarily to give complete answers to questions. You should just attempt those that appeal to you, and they will be used as a basis for discussion in the interview that follows.
They may also just ignore any complete solutions that you have done and ask you about the questions you hadn't attempted. This is what happened to my son (several years back).
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DFranklin
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(Original post by atsruser)
Question 3

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1. Prove that \int_0^\pi x f(\sin x) dx = \frac{\pi}{2}\int_0^\pi f(\sin x) dx

Spoiler:
Show
Hint 1: over the range [0,\pi], \sin x has a line of symmetry at \pi/2.
Spoiler:
Show
Hint 2: Find a nice substitution that makes use of Hint 1

2. Reduce the given integral to something trivial using point 1. above.
Spoiler:
Show
Not convinced. To use your suggestion requires f(sin x) = x sin^2 x. Which even if you allow arcsin isn't possible in the range [0,pi] since sin x isn't monotonic.

And f we ignore than and just plow on, we get

\int_0^\pi x^2 \sin^2 x\,dx = \dfrac{\pi}{2}\int_0^\pi x \sin^2 x\.dx = \dfrac{\pi^2}{4} \int_0^\pi \sin^2x = \dfrac{\pi^3}{8}, which is definitely incorrect.
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-Gifted-
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Could anyone help me with question 10 of this paper:

https://share.trin.cam.ac.uk/sites/p...ampletest3.pdf

Do I assume that volume of the dust is equal to the volume of the spaceship? Do I also assume that the spaceship has unit length, so the volume equals the cross sectional area? And also my answer has the constant velocity,'u', in it- is it not meant to?

Thank you.
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username1763791
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(Original post by tiny hobbit)
They may also just ignore any complete solutions that you have done and ask you about the questions you hadn't attempted. This is what happened to my son (several years back).
Ah, ok thank you I'll put that in the OP now!
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(Original post by DFranklin)
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Not convinced. To use your suggestion requires f(sin x) = x sin^2 x. Which even if you allow arcsin isn't possible in the range [0,pi] since sin x isn't monotonic.

And f we ignore than and just plow on, we get

\int_0^\pi x^2 \sin^2 x\,dx = \dfrac{\pi}{2}\int_0^\pi x \sin^2 x\.dx = \dfrac{\pi^2}{4} \int_0^\pi \sin^2x = \dfrac{\pi^3}{8}, which is definitely incorrect.
You're right. Damn. sin^2 breaks this. Haven't got time to think about it in detail now. I'll delete that post forthwith.
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username1258398
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(Original post by -Gifted-)
Could anyone help me with question 10 of this paper:

https://share.trin.cam.ac.uk/sites/p...ampletest3.pdf

Do I assume that volume of the dust is equal to the volume of the spaceship? Do I also assume that the spaceship has unit length, so the volume equals the cross sectional area? And also my answer has the constant velocity,'u', in it- is it not meant to?

Thank you.
I don't think you need the length of the spaceship, or to assume anything about the volume of the dust, but, I do think you have to assume that dust only sticks to the front of the ship. I'm not sure if my solution is correct or not, but here it is:

Spoiler:
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Let u be the constant velocity it travels at initially. After a distance x, a mass of dust pAx has stuck to the front of the ship. Momentum is conserved, so when it's travelling at half its initial velocity,

Mu = (M+pAx)\frac{u}{2}

Cancelling u from both sides,

2M = M+pAx

And so,

x = \frac{M}{pA}
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username1258398
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(Original post by Jordan\)
Zacken^ for Kings I believe
ah, some competition for me, then :P

What made you choose Pembroke? I hear the food's fantastic.

(Original post by Zacken)
x
We may see each other at the interviews. ^^ Why King's, if you don't mind me asking?
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(Original post by StrangeBanana)
I don't think you need the length of the spaceship, or to assume anything about the volume of the dust, but, I do think you have to assume that dust only sticks to the front of the ship. I'm not sure if my solution is correct or not, but here it is:
Spoiler:
Show

Let u be the constant velocity it travels at initially. After a distance x, a mass of dust pAx has stuck to the front of the ship. Momentum is conserved, so when it's travelling at half its initial velocity,

Mu = (M+pAx)\frac{u}{2}

Cancelling u from both sides,

2M = M+pAx

And so,

x = \frac{M}{pA}
Thanks a ton, it makes sense to me. However, I used conservation of mechanical energy, but that yielded a different result where the velocity doesn't cancel out. Let me scan it.
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Specimen Test 1, Question 7
Hint
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For the first part consider what each of the two possible cases means with regards to what set the other elements in the subset come from, and remember that
Total number of subsets = Number of subsets containing 1 + Number of subsets NOT containing 1
The second part is similar to the first but by considering another element in the set {1, 2,... , n}
Solution
Spoiler:
Show

Consider a subset of S = {1,2,....,n}.There are  \binom{n}{r} possible subsets containing r elements (since this is the number of combinations of r elements).

Each possible subset can either contain 1 or not contain 1. If it contains 1 then the other (r-1) elements must come from {2, 3, ... , n} of which there are  \binom{n-1}{r-1} combinations. If it does not contain 1 then all r elements come from {2, 3, ..., n}, of which there are  \binom{n-1}{r} combinations. Hence
 \binom{n-1}{r-1} + \binom{n-1}{r} = \binom{n}{r}

Equally, a subset of S with r elements either contains 2 or does not contain 2. This gives rise to 4 possibilities:

- A given subset of S contains 1 and 2
- A given subset of S contains 1 but NOT 2
- A given subset of S contains 2 but NOT 1
- A given subset of S contains neither 1 or 2

In the first case the other (r-2) elements of the subset come from {3, 4,...n} of which there are \binom {n-2}{r-2} combinations.

In the second case the other (r-1) elements of the subset come from {2, 3,..., n} of which there are  \binom{n-1}{r-1} combinations. The third case can be derived by replacing the "2" in {2, 3,..., n} with a "1" hence there are  \binom{n-1}{r-1} subsets for the third case.

In the final case all r elements of the subset come from {3, 4,..., n}, of which there are  \binom {n-2}{r} combinations.

Hence the total number of subsets is  \binom{n}{r} = \binom{n-2}{r-2} + 2\binom{n-1}{r-1} + \binom{n-2}{r}

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