STEP Maths I, II, III 1993 Solutions

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Reply 40

insparato

I wouldnt even know about matrices if i didnt do FP3.

Reply 41

Swayum

DFranklin

Huh? Why do you think there's an r^(3n+1) term in S_n(r)?

Well their series was:

S_n(r) = r + r^2 + r^4 + r^5 + r^7 + r^8 + r^10 + ... + r^(3n-1)

Let n = 3. Then the last term would be r^(3*3 - 1) = r^8

Consider the series U_n = 3n - 1, it gives:

u_1 = 2
u_2 = 5
u_3 = 8

If you then take r^(U_n) (so basically r^2, r^5, r^8), the terms left out from S_n(r) are r, r^4 and r^7.

Consider the series U2_n = 1, 4, 7, 10.... Its nth term would be 3n + 1 (if you include n = 0). If you then took r^(U_n) (which would give r, r^4, r^7) you'd get all the terms that the r^(U_n) series missed out.

Which is why I think S_n(r) can be expressed as r^(U_n) + r^(U2_n).

And the last term in the series is given by r^(3n-1). So if n was 3, S_n(r) would be everything up to and including r^8 which is essentially:

S_3(r) = r + r^2 + r^4 + r^5 + r^7 + r^8.

And that sum is equal to the formula I gave.

Reply 42

Swayum

nota bene

edit: DeathAwaitsU, S_n(r) is just a silly way of writing $\displaystyle\sum_1^n r^{3n-1}$ (like f(x) for a function of x this is a sum of r) as far as I know...

I don't think it is though. If you do the iterations from n = 1 to n = 3, using the formula r^(3n - 1), you get:

r^(3*1 - 1) = r^2
r^(3*2 - 1) = r^5
r^(3*3 - 1) = r^8

Where as their series suggests that it should give r + r^2 + r^4 + r^5 + r^7 + r^8 or maybe r + r^2 + r^4.

Although actually I think I'm totally misunderstanding the question.

Reply 43

DFranklin

nota bene

Okay, do you think it is better if I prove it is impossible (which I accidentally did in my working as I slipped on an argument and went two steps wrong which I quickly realised). Because I tried to see an obvious way of dismissing the possibility of 2+2 but didn't find any information under the points that were dealt with . (Well if you chose 2+2 strategy you'll end up with 112 on one die and 2 on the other and then trying to move on with this proves impossible when we add the 3 to 112 (can't add a 4 etc.) which makes it impossible to get what we want).Having pondered this a bit, I couldn't find a nice way without basically solving the whole thing at the same time. (The big problem I had was showing you can't have two 1s - maybe I missed something).

First make the easy progress at the "edges" of the freq table:

Since we can't get a 2, we know that there can't be a 1 on both dice. But since we can get a 3, we know exactly one die has a 1. Call that dice A and the other one B. Then B must have a 2, so we know 1 \in A and 2 \in B.

Now, there are 6 ways of getting 11 and no ways of getting 12. So there isn't a 6 on both dice. So we either have 2 5's on 1 die and 3 6's on the other, or 2 6's on one die and 3 5's on the other. (Where the die with 6's can also have 5's).

Now it gets harder. Since we have 2 ways of getting 3, either 1,1 \in A, 2 \in B or 1 \in A, 2,2 \in B.

Suppose 1,1 \in A, 2 \in B. We know there's exactly one way of getting a 4, so we must have 2 \in A (as if 3\in B we get 2 ways). So in fact we have 1,1,2 \in A, 2 \in B. Then there's exactly one way of getting a 5, so either 3 \in A or 3\in B (but not both). Now consider how we can find 4 ways of making 6. We can't combine 2 3's. We can have at most one 4 \in A (because A must contain at least 2 sixes or 2 fives). So we need some 5's \in B. But then each 5 combines with 2 1's, so each 5 contributes 2 ways. So as the number of ways is even, 4 \notin A. So then B contains two 5's, and so A must contain 3 sixes. So A = 1,1,2,6,6,6 and 2,5,5 \in B (and 1,6 \notin B). But then there are only 2 ways of making 7.

So we must have 1 \in A, 2,2 \in B. So then there's one way of getting a 4, so 3 \in B, 2 \notin A. And then 1 way of getting a 5, so 4 \in B, 3 \notin A. So 1\in A, 2,3 \notin A, 2,2,3,4 \in B. Suppose B is the die with the sixes, then B = 2,2,3,4,6,6. Then 1,5,5,5 \in A, 2,3,6 \notin A. So A = 1,4,4,5,5,5. But then we choosing 2 from B and 5 from A gives 6 ways of making 7.

So B must contain the 5's, so B = 2,2,3,4,5,5. So 1,6,6,6\in A, 2,3 \not in A. So we can now reduce to the possibilities A = {1,4,4,6,6,6}, A = {1,4,5,6,6,6} or A = {1,5,5,6,6,6}. If A = {1,4,4,6,6,6} there are only 2 ways of making 7, if A = {1,5,5,6,6,6] there are 4 ways. So the only possibility is A = {1,4,5,6,6,6}.

(I've still been a bit sloppy about explicitly saying what can't be in the sets).

Reply 44

DFranklin

DeathAwaitsU

Well their series was:

S_n(r) = r + r^2 + r^4 + r^5 + r^7 + r^8 + r^10 + ... + r^(3n-1)

Let n = 3. Then the last term would be r^(3*3 - 1) = r^8

Consider the series U_n = 3n - 1, it gives:

u_1 = 2
u_2 = 5
u_3 = 8

If you then take r^(U_n) (so basically r^2, r^5, r^8), the terms left out from S_n(r) are r, r^4 and r^7.

Consider the series U2_n = 1, 4, 7, 10.... Its nth term would be 3n + 1 (if you include n = 0).

If you then took r^(U_n) (which would give r, r^4, r^7) you'd get all the terms that the r^(U_n) series missed out.No, you'd get r^4, r^7, r^10. You have to start from n=1 if the r^(3n-1) part of what you're doing is going to make sense.

Reply 45

generalebriety

STEP I question 11.

Introduce cartesian axes into the problem, such that the x-axis lies along BC and the y-axis is perpendicular and passes through the midpoint. Let BC = 2a, such that A is the point (-a, 2a), B is the point (-a, 0), C is the point (a, 0) and D is the point (a, 2a). Model the wire as three particles, each of mass m, located at (-a, a), (0, 0) and (a, a). The centre of mass has coordinates (0, y) (by symmetry). Then the total mass M = 3m, and My = ma + ma = 2ma, so y = 2/3 a.

(See diagram for the next part.)

The angle required (between directions of BC) is θ. Since γ = 180 - α - β, and θ = 180 - γ (or simply by vertically opposite angles), θ = α + β. From the diagram,
tan α = 4/3
tan β = 2/3
and so tan θ = (4/3 + 2/3) / (1 - (4/3)(2/3)) = 2 / (1 - 8/9) = 2 / (1/9) = 18.

q11.gif6.6KB

Reply 46

generalebriety

Also Q12...

Let the angle between the vertical and the rod be θ. Take GPE=0 at the horizontal line through B.

Let the extension of the spring CD be x. Then the length of CD is (a+x).
a+x = 2a sin(θ/2) --> x = a(2 sin (θ/2) - 1).
$\displaystyle \text{EPE} = \frac{\lambda a^2 (2 \sin \frac{\theta}{2} - 1)^2}{2a} = \frac{1}{2} \lambda a(2 \sin \frac{\theta}{2} - 1)^2. \\[br]\text{GPE} = 2mga \cos \theta. \\ \\[br][br]\text{Potential energy of the system:} \\[br]V = 2mga \cos \theta + \frac{1}{2} \lambda a(4 \sin^2 \frac{\theta}{2} - 4 \sin \frac{\theta}{2} + 1) \\[br]= mga(2 - 4 \sin^2 \frac{\theta}{2}) + 2\lambda a \sin^2 \frac{\theta}{2} - 2\lambda a\sin \frac{\theta}{2} + \frac{1}{2}\lambda a \\[br]= 2(\lambda a - 2mga)\sin^2 \frac{\theta}{2} - 2 \lambda a \sin \frac{\theta}{2} + \text{constant} \\[br]\frac{\text{d}V}{\text{d}\theta} = 2(\lambda a - 2mga) \sin \frac{\theta}{2} \cos \frac{\theta}{2} - \lambda a\cos \frac{\theta}{2} \\[br]= a\cos \frac{\theta}{2} (2(\lambda - 2mg)\sin \frac{\theta}{2} - \lambda). \\[br]\frac{\text{d}V}{\text{d}\theta} = 0 \Rightarrow \cos \frac{\theta}{2} = 0 \text{ (n/a)}, \; \sin \frac{\theta}{2} = \frac{\lambda}{2(\lambda - 2mg)}. \\[br]\lambda > 4mg \Rightarrow \lambda - 2mg > 2mg \\[br]\Rightarrow 0 < \frac{mg}{\lambda - 2mg} < \frac{1}{2} \\[br]\Rightarrow \frac{1}{2} < \sin \frac{\theta}{2} < 1 \\[br]\Rightarrow 30 < \frac{\theta}{2} < 90 \\[br]\Rightarrow 60 < \theta < 180.$

Reply 47

generalebriety

And STEP II Q7:

a (mod 5) | 2a^2 (mod 5)
0 | 0
1 | 2
2 | 3
3 | 3
4 | 2

b (mod 5) | b^2 (mod 5)
0 | 0
1 | 1
2 | 4
3 | 4
4 | 1

So 2a^2 + b^2 is divisible by 5 if and only if 2a^2 + b^2 = 0 (mod 5), which happens if and only if a = 0 (mod 5), b = 0 (mod 5).

2a^2 + b^2 = 5c^2. (*)

Assume a = 0 (mod 5), b = 0 (mod 5). Then a = 5p, b = 5q.
Then 2a^2 + b^2 = 5c^2
$\Rightarrow$ 25(2p^2 + q^2) = 5c^2
$\Rightarrow$ 5(2p^2 + q^2) = c^2
therefore c^2 = 0 (mod 5), and so c = 0 (mod 5). Then c = 5r.
Then 5(2p^2 + q^2) = c^2
$\Rightarrow$ 5(2p^2 + q^2) = 25r^2
$\Rightarrow$ 2p^2 + q^2 = 5r^2.
But this is equivalent to (*).

Hence, by the method of infinite descent, a, b and c can be divided by 5 infinitely many times and still remain integers.

Therefore the only solution is a = b = c = 0.

Ok, that's me done for the day. :biggrin:

Are these solutions going on the wiki? If not, would it be a nice idea to do so? :smile:

Reply 48

Speleo

Did that last one already g.e. :tongue:

DFranklin, is that question finished now (the dice one)?

Reply 49

DFranklin

Speleo

Did that last one already g.e. :tongue:

DFranklin, is that question finished now (the dice one)?

Unless someone finds a mistake!

Reply 50

*bobo*

10 STEP I)modulous of elascticity=x and extension=e
<BCD=<BDC= Y
tension in wire= xe/a

for equib: moments about B:
xesinY= 2amgsin(180-2Y)=2amgsin2Y

e=a(2-2cos(180-2Y))^0.5 -a
=a(2+2cos2Y)^0.5 -a
=a(2 + 4cos^2 Y -2)^0.5 -a
=a(2cosY -1)

a(2cosY-1)xsinY=4amgsinYcosY
(2cosY-1)x=4mgcosY
cosY= x/(2x-4mg)

if rod is in equib away from wall (ie not against it):
Y>0 so cosY<1
x/(2x-4mg)<1

=> x>4mg

Reply 51

*bobo*

13 STEP I)
i)time taken to reach speed u:
dv/dt= F/M
v=Ft/M => t= uM/F

after it reaches speed u:
Mv dv/dt= Fu
=> (M/2)v^2= Fut +C
when t=0, v=u
(M/2)(v^2 -u^2)=Fut
=> t= (M/2Fu)(v^2-u^2)

so time taken to reach speed v=
(M/2Fu)(v^2-u^2) + uM/F (P=Fu)
=(M/2P)(v^2-u^2) + Mu^2/P
=(M/2P)(v^2+u^2)

Reply 52

*bobo*

13 ii)

distance travelled before speed reaches u:
initial speed=0, v=u, a=F/M
using v^2= u^2 +2as
s= u^2M/2F= Mu^3/2P

distance travelled after speed reaches u:
Mv^2 dv/dx=Fu

(M/3)( v^3- u^3)= Fux (if x=0 when v=u)

=> x= (M/3P)(v^3-u^3)

total distance travelled= (M/3P)(v^3- u^3) + Mu^3/2P
=(M/6P)(2v^3 + u^3)

Reply 53

Rabite

I think I did Q2 in the Further Maths A.
But it seems quite easy so I've probably made a mistake. It's still red on the front page, so if no one else has typed it out already, I'll do so~

[edit] Here it is anyway.
$\int \cos{mx}\cos{nx} dx = ½ \int \cos (m+n)x + \cos (m-n)x dx$
By the product/sum formulae that no one remembers.
$= ½[\frac{1}{m+n} \sin(m+n)x + \frac{1}{m-n} \sin(m-n)x]_0 ^{2\pi}$

But if m=±n, one of the fractions explodes. So in that case the question is:

$\int \cos^2 {mx} dx$

$=½ \int 1+ cos{2mx}dx$

$= ½[x+\frac{1}{2m} \sin{2mx}]_0 ^{2\pi}$

$=\pi$

If m=n=0, the integral turns to $2\pi$ .

As for the second bit.
Let x = sinh²t

dx = 2sinhtcosht dt

$I = \int \sqrt{\frac{x+1}{x}}dx$

$= \int \sqrt{\frac{\sinh^2 t +1}{\sinh^2 t}}(2\sinh t\cosh t)dt$

$= \int 2 \frac{\cosh t}{\sinh t} (\sinh t \ cosh t) dt$

$= \int 2 \cosh^2 t dt$

$= \int 1+ \cosh 2t dt = t + ½ \sinh 2t$ Ignoring the +c for now

$= t + \sinh t \cosh t = \sinh ^{-1}(\sqrt{x}) + \sqrt(x(x+x)) +c$
Which you can rewrite using the log form of arsinh.

Reply 54

*bobo*

12 STEP III)
i)T= (kmg/2a)e
e= 2(9a^2 + a^2/16)^0.5= a(145)^0.5/2

2Tsin(PDA)=mg
sin(PDA)=(145)^-0.5

=>2k(145)^0.5x0.25(145)^-0.5=1
=> k= 2

conservation of energy:
0.5(2mg/2a)(4a^2(p^2 +9))= (p+3)mga + 0.5(2mg/2a)64a^2

=>2p^2 + 18= p+3 +32
=>2p^2 -p -17=0

Reply 55

*bobo*

12 ii)
at Q: e=2a(p^2 + 9)^0.5
at R: e= a(4^2 +3^2)^0.5 - 2= 8a

2p^2 -p -17=0
p= (1 + 137^0.5)/4
applying p it can be seen that e at Q is greater than at R
hence tension in string is greater at Q than at R

diagram.doc78.3KB

Reply 56

nota bene

DFranklin

Having pondered this a bit, I couldn't find a nice way without basically solving the whole thing at the same time. (The big problem I had was showing you can't have two 1s - maybe I missed something).

The same problem I had, so I don't think we are missing anything. In your solution you had to disprove the (1,1,2) as well, although you did it in a much better way by starting to consider the ways to combine the sum of 11, which cuts down a bit on the testing of possibilities.

(I think I'll stay away from any further tedious questions on testing possibilities until I learn to explain, or I'll scan up scribbled unreadable notes...)

Now to the next STEP question...
STEP I Q 15
Probability an Ascii gives the same answer twice is TT+FF (T=True, F=False) Hence $\frac{3}{4}\frac{3}{4}+\frac{1}{4}\frac{1}{4}= \frac{10}{16}= \frac{5}{8}$
So, given that an Ascii tells the same answer twice in succession the probability of lying is $\displaystyle\frac{\frac{1}{16}}{\frac{10}{16}}= \frac{1}{10}$ and therefore the probability of truth is 1-P(lying) = 9/10.

Given that a tribe member tells the same answer twice (as in the scenario) the probability of him being Ascii is possible to calculate via Bayes theorem, and Biscii is 1-P(A). By Bayes, P(A|two consequtive answers)= $\displaystyle\frac{P(A)P(two|A)}{P(A)P(two|A)+P(A')P(two|A')}$
Now we know P(A)=11/16, P(two|A)=5/8, P(A')=5/16 and P(two|A')=1
Calculating this gives $P(A|two)=\frac{11}{19}$ and therefore Probability of Biscii given two consecutive answers is $\frac{8}{19}$ . Now the probability of lie is $\frac{11}{19}\frac{1}{10}+\frac{8}{19}\times 1=\frac{91}{190}$ Therefore probability of truth is 1-P(lie) and thus $\frac{99}{190}$ as desired.

Following the scenario we are now given three consecutive answers, working with the same principle as above we calculate P(three|A) to be FFF+TTT = $\frac{3}{4}\frac{3}{4}\frac{3}{4}+\frac{1}{4}\frac{1}{4}\frac{1}{4}=\frac{28}{64}=\frac{7}{16}$
(The other components of the earlier calculation stays the same)
So, by Bayes, P(A|three consequtive answers)= $\displaystyle\frac{P(A)P(three|A)}{P(A)P(three|A)+P(A')P(three|A')}$
$\displaystyle\frac{\frac{11}{16}\frac{7}{16}}{ \frac{11}{16}\frac{7}{16}+\frac{5}{16}\times1}= \frac{77}{157}$
Therefore probability of Biscii giving three consecutive answers is $\frac{80}{157}$ .
Now we need the probability for lie given it is an Ascii giving three consecutive answers. This is given by $\frac{\frac{1}{64}}{\frac{1}{64}+\frac{27}{64}}= \frac{1}{28}$

So, now P(truth)=1-P(lie) and P(lie)= $\frac{1}{28}\frac{77}{157}+\frac{80}{157}= \frac{331}{628}$
So P(truth) is $\frac{297}{628}$ . Therefore the man shall not follow the given answer (i.e. left); he shall go to the right, with the probability $\frac{331}{628}$ of coming to the town.

Reply 57

DFranklin

nota bene

Now to the next STEP question...

It's so much nicer when you get a "show..." question isn't it? :smile:

Reply 58

*bobo*

12 Stepii:

12step2page1.doc530.4KB

page2.doc383.5KB

Reply 59

*bobo*

part 2 to question12 STEP II:
take moments about Q(2), (where A is the angle bwteen the two walls):
rMgcosA= mg sin0.5A. r(2-2cosA)^0.5

(McosA)^2=m^2sin^2 0.5A(2-2cosA)=m^2(2-2cosA)(0.5-0.5cosA)m^2(1-cosA)^2
=> McosA=m(1-cosA)
m= McosA/(1-cosA)= M/(secA-1)

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