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Trinity Admissions Test Solutions

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Reply 60

username1162972

is paper 3 getting done?

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Reply 61

Krollo

Original post by physicsmaths

is paper 3 getting done?

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At some point. I'll start now in fact...

Paper 3, Q5

Hint

Consider parity (even/odd). What can we apply it to?

Solution

Initially, the sum of all the cards is 5050 (from arithmetic series formula), which is even.

Suppose the two cards you choose are even. Then their sum is even, and their difference is even. The sum of all cards thus remains even.

Suppose they are both odd. Then their sum is even, and their difference is even. The sum of all cards thus remains even.

Suppose one is even and one is odd. Then their sum is odd, and their difference is odd. The sum of all cards thus remains even.

Thus the sum of all cards will always remain even, which remains true when a single card remains. That card must hence be even.

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Reply 62

Hauss

Paper 3 Q3 Solution

Spoiler

(edited 8 years ago)

Reply 63

DFranklin

Original post by atsruser

Some further thoughts (I don't have a complete solution):

Spoiler

You're pretty much there,

Spoiler

Reply 64

joostan

Paper 3 Question 1:

Spoiler

Paper 3 Question 2:

Spoiler

Paper 3 Question 6:

Spoiler

(edited 8 years ago)

Reply 65

Zacken

Test 3, Question 7:
Thanks to Krollo for this solution.

Hints:

Spoiler

Solution:

Spoiler

Reply 66

atsruser

Original post by DFranklin

You're pretty much there,

Spoiler

Yes, thanks for this. I've gone through this in detail now, and I've got a complete solution, I think. This is a pretty interesting result, and I can see why I was baffled last night (I didn't take the equations seriously enough and tried to rely on my intuition - my intuition was wrong).

So, in full:

Spoiler

(edited 8 years ago)

Reply 67

DFranklin

Original post by atsruser

Spoiler

Yes all looks good.

One thing I'll add:

If you're doing mechanics and have something like:

\ddot{x} + kx = A

, if you write u = x-A/k, you find

\ddot{u} + ku = 0

.

That is, looking at u immediately shows we have SHM about u = 0.

I don't know that it's objectively "better" than solving via CF+PI, but to me it shows more clearly what's going on.

Reply 68

atsruser

Original post by DFranklin

Yes all looks good.

One thing I'll add:

If you're doing mechanics and have something like:

\ddot{x} + kx = A

, if you write u = x-A/k, you find

\ddot{u} + ku = 0

.

That is, looking at u immediately shows we have SHM about u = 0.

I don't know that it's objectively "better" than solving via CF+PI, but to me it shows more clearly what's going on.

Yes, I think that tends to crop up in questions where you have a weight hung from an spring vertically and you extend the system from its equilibrium point. It's objectively *quicker* than CF+PI, for sure, but it didn't occur to me immediately last night when I first wrote it up - still, it's just as well to demonstrate a variety of approaches, I guess.

Reply 69

Krollo

Paper 3, Question 10

Hint

Spoiler

Solution

Spoiler

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Reply 70

shamika

Why aren't people touching the probability ones? What about paper 4 (from Krollo's interview prep thread?)

Reply 71

username1763791

Original post by shamika

Why aren't people touching the probability ones? What about paper 4 (from Krollo's interview prep thread?)

I can't do any of them unfortunately and I think some people, especially those who can do them easily, have no need to come to this thread so they just aren't being done. If you have any you would like to do please go ahead it would be a big help :smile:

I'll add Paper 4 now.

Reply 72

Krollo

Original post by Jordan\

I'll add Paper 4 now.

Shotgun all the non probability ones

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Reply 73

username1763791

Original post by Krollo

Shotgun all the non probability ones

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Go ahead mate :lol:

If I could do any of the others I would have them done already :wink:

Reply 74

Krollo

Test 4 question 1

Hint

When you need to approximate a probability p think about how you might approximate firstly lnp. Can you work back from there?

Solution

If the probability of one bond winning is 1/14000, then the probability of it not winning is 13999/14000. So the probability of at least one winning is 1-(13999/14000)^14000, evidently not 1.

We need to approximate (13999/14000)^14000. Take natural log of it, giving 14000ln(1-1/14000) req 14000 * -1/14000 req -1 by the taylor expansion.

So the probability required is about 1 - e^-1 req 0.63.

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Reply 75

username1763791