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STEP 2006 Solutions Thread

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Reply 40
Avatar for SimonM
SimonM
OP
18
STEP I, Question 10

Spoiler

Reply 41
Avatar for SimonM
SimonM
OP
18
STEP II, Question 7

Spoiler

Reply 42
Avatar for jj193
jj193
14
STEP II, Question 5.

LaTeX shows error messeges in the solution.
For i) I get 3(n-1)/2
The mark scheme says it's 3n(n-1)/2

My working:
The second curve (with the integer part in) can be viewed as rectangles accounting for the 3[x], The area of this part is (Sum of (3r) from (1 to (n-1)). The other part of the curve is simply (x^2 - 1).
This has an area of
Integral(x^2 -1)from(1 to n) +Sum of (3r)from(1 to (n-1))

The other curve has area Integral(x^2+3x-1)from(1 to n)

So the area between the curves is:
Integral(3x)from(1 to n) -Sum of (3r)(1 to (n-1))

I can't see my mistake.
jj193
..
I've fixed the LaTeX. It agrees with you, and glancing at the question, the answer is definitely <= 3(n-1) and I would expect it to be of order 3(n-1)/2. So I'm sure you're right and the markscheme is wrong.
Reply 44
Avatar for jj193
jj193
14
DFranklin
I've fixed the LaTeX. It agrees with you, and glancing at the question, the answer is definitely <= 3(n-1) and I would expect it to be of order 3(n-1)/2. So I'm sure you're right and the markscheme is wrong.

Thanks :yes:
Nice to know i'm not going mad...
Sorry but how did you do Part 2, how can you just know m = 2? I thought you needed to work it out ...

I did Part 1:

Let x^2 = 33127 = 40, 000(1 - 6873/40, 000)
=> x = 200Sqrt(1 - 6873/40, 000)
~ 200Sqrt(1 - 6870/40, 000)
= 200Sqrt(1 - 17/100)
= 20Sqrt(83)
= 20Sqrt[81(1 + 2/81)]
~ 180Sqrt(1 + 1/40)
= 180Sqrt(1.025)
~ 180 x 1.1
~ 182

Testing 182^2 gives a value just under 33127. Hence n = 182.

How do you do the 2nd part? It's impossible.
(n+m)^2 - 33127 = 182^2 + 364m +m^2 -33127 = 32400 + 720 + 4 + 364m +m^2-33127 = 33124 + 364m + m^2 -33127 = m^2 + 364m - 3.

So you're looking for m such that m^2+364m-3 is a perfect square. It's obvious m can't be -ve (and small), then by trial and error we find m = 2. (After trying 0, 1, 2).
DFranklin
(n+m)^2 - 33127 = 182^2 + 364m +m^2 -33127 = 32400 + 720 + 4 + 364m +m^2-33127 = 33124 + 364m + m^2 -33127 = m^2 + 364m - 3.

So you're looking for m such that m^2+364m-3 is a perfect square. It's obvious m can't be -ve (and small), then by trial and error we find m = 2. (After trying 0, 1, 2).

Bingo! Thanks so much. I was getting something similar last night but decided to hit the bed at 4am. I got a k^2 + 3 I remember, seems like you brought that 3 across. I'll take another look.
Reply 48
Avatar for IDGAF
IDGAF
Daniel Freedman
STEP II 2006, Question 8

Spoiler




When I was doing this question, I was unable to understand what was meant by 'the point D is the intersection of OC produced and AB produced'. Specifically the use of the word 'produced'. Could someone just clarify what it means?

Thanks in advance
Without looking at the paper for context, I'd assume it means "extended".
Hey guys, I'm having a brain freeze here.

In STEP 1 Q11 Part 3, in the markscheme they say p < q < r at the start, where p, q and r are the velocities of A_(n-2), A_(n-1) and A_n.

The way it's stated so casually at the start, assumes it's something obvious. I don't get why though, and it's frustrating, as this assumption at the start will prove the contradiction needed.

DFranklin in particular, any help?
I can easily justify p <= q <= r (if p > q, then obviously there will be a subsequent collision between A_n-2 and A_n-1, contradicting the assumption that all collisions have taken place, etc.)

It's not obvious to me that you can't have p = q or q = r, but I think the assumption p<=q<=r is sufficient to get the same contradiction and the end of (iii).
DFranklin
I can easily justify p <= q <= r (if p > q, then obviously there will be a subsequent collision between A_n-2 and A_n-1, contradicting the assumption that all collisions have taken place, etc.)

Perfect! If all collisions took place, yet p > q, then you'd get another collision as A_(n-2) would catch A_(n-1), which is a contradiction.

The God
It's not obvious to me that you can't have p = q or q = r, but I think the assumption p<=q<=r is sufficient to get the same contradiction and the end of (iii).

Yes they could keep trotting along at the same speed without colliding, but don't worry, all I need at the end is to justify p <= q + (lambda)r which is a formality now.

Thanks so much mate, :smile:
Original post by SimonM
...

STEP II Q10

(i)


(ii)

Original post by SimonM
...

STEP I Q12

First part



Second part

(edited 13 years ago)
Reply 55
Avatar for kiddey
kiddey
I 2006 q1

1802=32400180^2 = 32400
1812=32761181^2 = 32761
1822=33124182^2 = 33124
1832=33489183^2 = 33489

hence n=182n=182
satisfies n2<33127<(n+1)2n^2<33127<(n+1)^2

if m=2m=2, then (n+m)2=1842=33856(n+m)^2 = 184^2 = 33856

33856−33127=729=27233856-33127=729=27^2

so n=182n=182, m=2m=2
satisfies (n+m)2−33127=x2(n+m)^2 - 33127 = x^2
where x=27x=27


33127=1842−27233127=184^2 - 27^2

⇒33127=(184−27)(184+27)\Rightarrow 33127=(184-27)(184+27)

⇒33127=211×157\Rightarrow 33127=211 \times 157

as 211211 and 157157 are prime
33127=211×15733127=211 \times 157 or 33127×133127 \times 1

so 33127=(16564+16563)(16564−16563)33127= (16564+16563)(16564-16563)

⇒33127=165642−165632\Rightarrow 33127 = 16564^2 - 16563^2

⇒165642−33127=165632\Rightarrow 16564^2 - 33127 = 16563^2

hence other value of n+m is 1656416564, and n=182⇒m=16382n=182 \Rightarrow m=16382
(edited 13 years ago)
STEP I Q14
Whilst this has been discussed earlier I haven't seen anyone post the actual solution so I'll post mine.

i) P(choosing red on rth go)=nr−1(n+1)r\frac{n^{r-1}}{(n+1)^r}
(usually I would draw a tree diagram but here it just isn't necessary)
To find the maximum we should differentiate and equate the derivative to zero.
let p=P(choosing red on rth go):
p=nr−1(n+1)rp=\frac{n^{r-1}}{(n+1)^r}
take logs of both sides:
ln[f(n)]=ln(nr−1)−ln[(n+1)r]ln[f(n)]=ln(n^{r-1})-ln[(n+1)^{r}]
ln[f(n)]=(r−1)ln(n)−rln(n+1)ln[f(n)]=(r-1)ln(n)-rln(n+1)
Now differentiate both sides:
1f(n)f′(n)=(r−1)1n−r1n+1\frac{1}{f(n)}f'(n)=(r-1)\frac{1}{n}-r\frac{1}{n+1}
f′(n)=[(r−1)1n−r1n+1]f(n)f'(n)=[(r-1)\frac{1}{n}-r\frac{1}{n+1}]f(n)
To find stationary points, let f′(x)=0f'(x)=0


[(r−1)1n−r1n+1]f(n)=0[(r-1)\frac{1}{n}-r\frac{1}{n+1}]f(n)=0. Since f(n) can not be 0:
[(r−1)1n−r1n+1]=0[(r-1)\frac{1}{n}-r\frac{1}{n+1}]=0
Rearranging we see that:
(n+1)(r−1)−rn=0(n+1)(r-1)-rn=0 and so:n=r−1n=r-1

ii) A trick in these type of questions where the denominator and the numerator of the probabilities go down by one every time is to notice that the k+1th denominator is canceled by the kth numerator. Therefore all we care about is the first denominator and the last numerator as they are the only ones which don't cancel.
P(choosing red on rth go)=1n+1\frac{1}{n+1}
so n should be the smallest it can be.
n+1 is greater than or equal to r so the smallest value of n is:
n=r-1
(edited 13 years ago)
Original post by SimonM
...
STEP II Q14

Solution

(edited 13 years ago)
2005 STEP III question 12 \text {2005 STEP III question 12}

X=aT+b(T1+T2+T3+T4) X=aT+b(T_1+T_2+T_3+T_4)
  ⟹  E[X]=at+b(t1+t2+t3+t4)=at+bt=(a+b)t=tbifa+b=1 \implies \text {E}[X]=at+b(t_1+t_2+t_3+t_4)=at+bt=(a+b)t=t bif a+b=1
Var[X]=a2σ2+4b2σ2=(a2+4b2)σ2 which is as small as possible if a2+b2 is as small as possible \text {Var}[X]=a^2 \sigma^2+4b^2 \sigma^2=(a^2+4b^2) \sigma^2 \text { which is as small as possible if }a^2+b^2 \text { is as small as possible}
Now a2+4b2=a2+4(1−a)2=5a2−8a+4=5(a2−85a+45)=5[(a−45)2+425] \text {Now }a^2+4b^2=a^2+4(1-a)^2=5a^2-8a+4=5 \left(a^2- \dfrac{8}{5}a+ \dfrac{4}{5} \right)=5 \left[ \left(a-\dfrac{4}{5} \right)^2+\dfrac{4}{25} \right]
Unparseable latex formula:

\text {which has a smallest value of }\dfrac{4}{5} \text { when }a=\dfrac{4}{5} \text { so } b= \dfrac{1}{5} \text { and } X= \dfrac{}1}{5} \left[4T+T_1+T_2+T_3+TR_4 right]


Now let Y=cT+d(T1+T2+T3+T4),E[Y]=(c+d)t as above and Var[Y]=(c2+4d2)σ2 \text {Now let }Y=cT+d(T_1+T_2+T_3+T_4), \text{E}[Y]=(c+d)t \text { as above and Var}[Y]=(c^2+4d^2) \sigma^2
So E[Y2]=Var[Y]+(E[Y])2=(c2+4d2)σ2+(c+d)2t2=σ2 independent of lap times if c+d=0 \text {So E}[Y^2]= \text{Var}[Y]+ (\text {E}[Y])^2=(c^2+4d^2) \sigma^2+(c+d)^2t^2= \sigma^2 \text { independent of lap times if }c+d=0
and c2+4d2=1  ⟹  5c2=1  ⟹  c=15,d=−15 so Y=15[T−(T1+T2+T3+T4)]\text {and }c^2+4d^2=1 \implies 5c^2=1 \implies c= \dfrac{1}{\sqrt5}, d=- \dfrac{1}{ \sqrt5} \text { so } Y= \dfrac{1}{ \sqrt5}[T-(T_1+T_2+T_3+T_4)]
T=220 and (T1+T2+T3+T4)=220.5  ⟹  E[X]=15(880+220.5)=220.1 T=220 \text { and }(T_1+T_2+T_3+T_4)=220.5 \implies \text {E}[X]= \dfrac{1}{5}(880+220.5)=220.1
and Var[X]=45σ2=45×15[T−(T1+T2+T3+T4)]2=425(220−220.5)2=0.04 \text {and Var}[X]=\dfrac{4}{5} \sigma^2=\dfrac{4}{5} \times \dfrac{1}{5}[T-(T_1+T_2+T_3+T_4)]^2=\dfrac{4}{25}(220-220.5)^2=0.04
we can reasonably expect the true value to be within 2 standard errors of the expected value \text{we can reasonably expect the true value to be within 2 standard errors of the expected value}
i.e. between 220.1+0.4 and 2−0.4=219.7 to 220.5 \text {i.e. between }220.1+0.4 \text { and } 2-0.4=219.7 \text { to }220.5
2006 STEP I question 11 \text {2006 STEP I question 11}

(i) If only one particle is moving after all collisions it would have to be An with velocity v \text {(i) If only one particle is moving after all collisions it would have to be }A_n \text { with velocity }v
overall conservation of momentum then gives mu=λmv and conservation of energy gives  \text {overall conservation of momentum then gives }mu= \lambda mv \text { and conservation of energy gives }
mu2=λmv2 mu^2= \lambda mv^2
This would therefore require that u=λv  ⟹  u2=λ2v2, but u2=λv2 so λ=0 or 1, neither of which is \text {This would therefore require that }u= \lambda v \implies u^2= \lambda^2 v^2, \text { but } u^2= \lambda v^2 \text { so } \lambda = 0 \text { or }1, \text { neither of which is}
possible so we cannot have exactly one particle moving after all collisions \text {possible so we cannot have exactly one particle moving after all collisions}
(ii) Let final speeds of An−1 and An be v and w respectively \text {(ii) Let final speeds of }A_{n-1} \text { and }A_n \text { be }v \text { and } w \text { respectively}
Unparseable latex formula:

\text {conservation of momentum gives }mu=mu+ \lambda mw \text { \nd of energy } mu^2=mv^2+ \lambda mw^2


Unparseable latex formula:

\text {hence, }v+ \lambda w)^2=v^2+ \lambda w^2 \implies 2v \lambda w+ \lambda^2 w^2= \lambda w^2 \impliea 2v=w(1-\lambda) \lambda \implies v<0 \text { since } \lambda>1


this is clearly not possible so An−1 and An cannot be the only two particles moving \text {this is clearly not possible so }A_{n-1} \text { and }A_n \text { cannot be the only two particles moving}
(iii) Let final velocities of An−2,An−1 and An be p,q and r respectively, then\text {(iii) Let final velocities of }A_{n-2},A_{n-1} \text { and }A_n \text { be } p,q \text { and }r \text { respectively, then}
Unparseable latex formula:

\text {conservation of momentum and energy gives }mu= kmp+mq+ \lambda mr \text { and }mu^2=kmp^2+mq^2+ \lambdamr^2


where km is the unknown mass of An−2 \text {where }km \text { is the unknown mass of }A_{n-2}
  ⟹  (kp+q+λr)2=kp2+q2+λr2  ⟹  2kpq+2qλr+2kλpr+k2p2+λ2r2=kp2+λr2\implies (kp+q+ \lambda r)^2=kp^2+q^2+ \lambda r^2 \implies 2kpq+2q \lambda r+2k \lambda pr+k^2p^2+ \lambda^2 r^2=kp^2+ \lambda r^2
Unparseable latex formula:

\implies kp(2q+2 \lambda r+kp-p)= \lambda r^2-2 \lambda qr- \lambda^2 r^2=r^2( \lambdea- \lambda^2)-2 \lambda qr<0 \text { since }k>1


i.e. kp(2q+2λr+kp−p)<0 but q>p  ⟹  2q+kp−p>2p+kp−p=p(1+k)>0 \text {i.e. }kp(2q+2 \lambda r+kp-p)<0 \text { but } q>p \implies 2q+kp-p>2p+kp-p=p(1+k)>0
so kp(2q+2λr+kp−p)>0 a contradiction \text {so }kp(2q+2 \lambda r+kp-p)>0 \text { a contradiction}
(iv) If there are only two particles moving they must be A0 and An with velocities x and y say \text {(iv) If there are only two particles moving they must be }A_0 \text { and }A_n \text { with velocities } x \text { and }y \text { say}
so mu=mx+λmy and mu2=mx2+λmy2  ⟹  (u−λy)2=u2−λy2 \text {so }mu=mx+ \lambda my \text { and } mu^2=mx^2+ \lambda my^2 \implies (u- \lambda y)^2=u^2- \lambda y^2
Unparseable latex formula:

\implies \lambda ^y^2-2u \lambda y=- \lambda y^2 \implies \lambda y-2u=-y \implies y= \dfrac{2u}{1+ \lambda} \text { and } x=u- \lambda y=u-\dfrac{2 \lambda u}{1+ \lambda}= \dfrac{u(1- \lambda)}{1+ \lambda}

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