STEP I, II, III 2002 Solutions

What was the difference between yours and Unbounded? As described it looked similar, but from what you're saying, his turned out a bit nastier.

Reply 121

ghostwalker

17

DFranklin

What was the difference between yours and Unbounded? As described it looked similar, but from what you're saying, his turned out a bit nastier.

You disagree then?

Well it avoided the

(\theta+2\phi)

, avoided the product-sum, seemed, to me, a hell of an lot simpler with the trig manipulations, and really reduced the problem to more standard A-level, rather than STEP, IMHO.

P.S. I did say I was impartial (not)!

Reply 122

DFranklin

18

Wasn't disagreeing, just trying to ascertain where the big "don't do this, do that" point of decision was. It wasn't obvious to me on a cursory glance why your algebra was so much nicer - largely because I wasn't sure what was involved in your "sub (4) into (3) and fiddle" comment. Looking a bit more carefully, I see that there is very little fiddling required, but I didn't notice that at the time.

Reply 123

ghostwalker

17

DFranklin

...

Yep, "fiddle" with me is a small thing (though someone else may use the term compeletely differently), as opposed to the "after some nasty algebra" in my second solution which involved 8 terms each comprising a combination of three, sines and cosines; yuck!

I guess the main difference was that I didn't square the "momentum in the direction of the initial velocity" equation early on; and later didn't need to.

After stating the initial equations, putting v2 in terms of v1 seemed a natural first step, which would almost certainly come in handy later.

Then "putting (4) into (3) and fiddle" so u is just in terms of v1 produced something akin to the one side of the desired equation (fortuitously perhaps), and the next step was obvious.

Reply 124

anonanon123123

14

SimonM

STEP I, Question 4

\displaystyle y' = - \frac{2x}{(1+x^2)^2}

Therefore the equation of the tangent is

\displaystyle y = - \frac{2a}{(1+a^2)^2} x + \frac{1}{1+a^2} +\frac{2a^2}{(1+a^2)^2}

Where did that part on the end (not the coefficient of x, everything else) come from? :frown:

Reply 125

miml

17

ziedj

Where did that part on the end (not the coefficient of x, everything else) come from? :frown:

For what it's worth I set it out a slightly different way
Tangent is at x=a (>0) and passes through (0,1)

y - 1 = \dfrac{-2a}{(1+a^2)^2}x

and

\dfrac{y-1}{x} = \dfrac{\dfrac{1}{1+a^2} - 1}{a} = \dfrac{-a}{1+a^2}

So we must have

\dfrac{-2a}{(1+a^2)^2} = \dfrac{-a}{1+a^2}

Which quickly cancels down to give

a^2 = 1

Reply 126

grusini

My solution to question 14 STEP II

STEP II.pdf43.7KB

Reply 127

grusini

Completing Dadeyemi's work on STEP III Q2, last part

STEP III Q2.pdf33.2KB

Reply 128

disco1000

7

SimonM

STEP III, Question 1

Spoiler

Sorry could you explain how you get to 2 divided by pi? I got the -(ln^2 (a) +2 ln(a) +2)/a but could not justify whether that expression would tend to 0 or 1, im thinking now probably 0 as if you sub in x = e^2 then e^5 that gets progressively smaller and smaller, so did you mean 2pi?

Also a little suggestion maybe it would have been easier to use the substitution u = ln x.

Reply 129

SimonM

OP

18

Diego Granziol

Sorry could you explain how you get to 2 divided by pi? I got the -(ln^2 (a) +2 ln(a) +2)/a but could not justify whether that expression would tend to 0 or 1, im thinking now probably 0 as if you sub in x = e^2 then e^5 that gets progressively smaller and smaller, so did you mean 2pi?

Also a little suggestion maybe it would have been easier to use the substitution u = ln x.

It's 2 pi, in latex that is 2\pi. Perhaps I forgot the tags?

Reply 130

welshenglish

Solution to 2002 STEP II Q6
Draw a decent diagram so you can see the lines and angles clearly.
Let l1, l2 and l3 be the lengths of lines l1, l2 and l3 from a horizontal plane to the point of intersection. Let vertical height of point of intersection above the plane be h. Then: -
l3 sin

/4 = l1 sin

/6 = l2 sin

= h
l3 sin

= l1 cos

= l2
Therefore
l1 cos

sin

= l1 sin

/6
cos

sin

= 1/2
l3 sin

sin

= l3 sin

/4
sin

sin

= 1/

(sin

sin

)^2 + (cos

sin

)^2 = 3/4
sin

=

/2

=

/3
Same diagram for second part but
replace

with

and

/4,

and

/6 with 2

,

/3 and

l3 sin 2

= l1 sin

= l2 sin

/3 = h
l3 sin

= l1 cos

= l2
l1 / l3 = tan

= sin 2

/ sin

tan

= 2 cos

l1 sin

= l1 cos

sin

/3
(tan

) ^ 2 + 3 (cos

) ^ 2 = 4
(tan

) ^ 4 - 3 (tan

) ^ 2 - 1= 0
(tan

) ^ 2 = (3 +

)/2
(3 -

)/2 is not a solution as (tan

) ^ 2

Reply 131

welshenglish

SimonM

STEP I:
1: Solution by Unbounded
2: Solution by Unbounded
3: Solution by nota bene
4: Solution by SimonM
5: Solution by nota bene
6: Solution by Unbounded
7: Solution by sonofdot
8: Solution by Unbounded
9: Solution by cliverlong
10: Solution by Dadeyemi
11:Solution by Unbounded
12: Solution by Robbie10538
13: Solution by Unbounded
14:

STEP II:
1: Solution by sonofdot
2: Solution by dadeyemi
3: Solution by sonofdot
4: Solution by dadeyemi
5: Solution by dadeyemi
6:
7: Solution by Glutamic Acid
8: Solution by Glutamic Acid
9: Solution by tommm
10: Solution by tommm
11: Solution by tommm
12:
13:
14:

STEP III:
1: Solution by SimonM
2: Solution by Dadeyemi
3: Solution by Daniel Freedman
4: Solution by Dadeyemi
5: Solution by Dadeyemi
6: Solution by Elongar
7: Solution by Dadeyemi
8: Solution by Dadeyemi
9:
10:
11:
12:
13:
14:

Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007

Hi Simon,
I'm working my way through the gaps in the solutions. I've put in an answer for 2002, STEP II Q6. Was your list up to date? If so, can you include this answer.

Cheers, Peter

Reply 132

odp04y

Answer to Step III 2002 Q9.

For the case

k>\frac{1}{2}\tan\theta

.
Where

\theta

is the angle of the plane.

The area of the fluid =

ka^2

and forms a trapezium.
If you cut the trapezium into two parts, a rectangle and a square angled triangle with angle

\theta

, the dimensions of the parts are:

Triangle: Width

= a

Height

= a\tan\theta

Rectangle: width

= a

let height

= z

but

ka^2 = za+\frac{a^2 \tan\theta}{2}

ka = z+\frac{a \tan\theta}{2}

z = ka-\frac{a \tan\theta}{2}

z = a(k-\frac{\tan\theta}{2})

Therefore the centre of mass of the triangle is

\frac{1}{3}a

from the side of the container and

a(k-\frac{\tan\theta}{2})+\frac{a\tan\theta}{3}

from the base. The mass of the triangle is

\frac{a^2 \tan\theta}{2}

units.

The centre of mass of the rectangle is

\frac{1}{2}a

from the side of the container and

\frac{a}{2} (k-\frac{\tan\theta}{2})

from the base. Its mass is

a^2(k-\frac{\tan\theta}{2})

units.

Using: Moments of whole = Sum moments of parts we can find x and y
To find x:

ka^2 x= (\frac{1}{3}a)(\frac{a^2 \tan\theta}{2})+(\frac{1}{2}a)(a^2(k-\frac{\tan\theta}{2}))

ka^2 x= \frac{a^3 \tan\theta}{6}+\frac{a^3}{2}(k-\frac{\tan\theta}{2})

\frac{kx}{a}= \frac{ \tan\theta}{6}+\frac{k}{2}-\frac{\tan\theta}{4}

\frac{kx}{a}=\frac{k}{2}-\frac{ \tan\theta}{12}

\frac{x}{a}=\frac{1}{2}-\frac{ \tan\theta}{12k}

To find y:

ka^2 y= (\frac{a^2 \tan\theta}{2})(ka-\frac{a \tan\theta}{6})+(ka^2-\frac{a^2 \tan\theta}{2})(\frac{ka}{2}-\frac{a\tan\theta}{4})

ka^2 y= \frac{ka^3 \tan\theta}{2}-\frac{a^3 \tan^2\theta}{12}+\frac{k^2a^3}{2}-\frac{ka^3 \tan\theta}{4} -\frac{ka^3 \tan\theta}{4}+\frac{a^3 \tan^2\theta}{8}

\frac{ky}{a}= \frac{k^2}{2}-\frac{\tan^2\theta}{12}+\frac{\tan^2\theta}{8}

\frac{y}{a}= \frac{k}{2}+\frac{\tan^2\theta}{24k}

For the case

k<\frac{1}{2}\tan\theta

.
The area of the fluid =

ka^2

and forms a square angled triangle with angle

\theta

.

The area can be found using

\frac{1}{2} bc \sin\theta=ka^2

where b and c are the hypotenuse and base of the triangle respectively.
Also

\cos\theta = \frac{c}{b}

b=\frac{c}{\cos\theta}

Therefore by substitution:

\frac{1}{2} c^2 \tan\theta = ka^2

c^2 = \frac{2ka^2}{\tan\theta}

The distance

x

of the centre of mass from the side is

\frac{1}{3} c

.

Therefore

x=\frac{1}{3} c

x^2=\frac{1}{9} c^2

9x^2= c^2

Therefore:

9x^2 = \frac{2ka^2}{\tan\theta}

\frac{x^2}{a^2} = \frac{2k}{9\tan\theta}

\frac{x}{a} = \sqrt{\frac{2k}{9\tan\theta}}

To find y:

\sin\theta = \frac{d}{b}

\tan\theta = \frac{d}{c}

Where d is the side opposite the angle theta.

b= \frac{d}{\sin\theta}

c= \frac{d}{\tan\theta}

By substitution

\frac{d^2}{2\tan\theta\sin\theta} \sin\theta = ka^2

\frac{d^2}{a^2} = 2k\tan\theta

The distance

y

of the centre of mass from the base is

\frac{1}{3} d

.

Therefore

9y^2= d^2

\frac{y}{a} = \sqrt{\frac{2k\tan\theta}{9}}

For the case when

k<\frac{1}{2}

and

\theta>45

2k<1

and

\tan\theta>1

Therefore

\frac{1}{2} \tan\theta>k

the case where the fluid forms a triangle.

Also

\tan \phi=\frac{y}{x}

Where

\phi

is the angle between the centre of mass and the base of the container. and

x

and

y

are the distances of the centre of mass from the side and base of the container.

\tan \phi=\frac{y}{x}=\frac{\sqrt{\frac{2k\tan\theta}{9}}}{\sqrt{\frac{2k}{9\tan\theta}}}

\frac{y^2}{x^2}=(\frac{2k \tan\theta}{9})(\frac {9\tan\theta}{2k})

\frac{y^2}{x^2}=\tan^2\theta

\tan \phi=\frac{y}{x}=\tan\theta

\phi=\theta

Container will topple when

\theta+\phi>90

In the case when

\theta>45

\theta+\phi>90

Therefore the container will topple.

Reply 133

welshenglish

Well done odp04y,

You've obviously got a better command of LaTex than me.

Reply 134

theOldBean

15

Here's an alternative for the last part of Q5

Let the roots be a,b,c,d. We know that

abcd=576=2^6 * 3^2
(a+1)(b+1)(c+1)(d+1)=1+22+172...=3^3 * 7^2
(a-1)(b-1)(c-1)(d-1)=1-22+172-552+576=7 * 5^2.

All the roots are negative (Since all the signs in the equation are "+".)

If the "+1" equation has a factor of 7^2 and the "-1" has a factor 5^2, 6 must be a duplicated root. Also since the "-1" has a factor of 7, 8 must be a root. It is now easy to deduce that the other root is 2.

So the roots are -2, -6 (twice) and -8.

By the fundamental theorem of algebra, this is the only solution set.

(edited 13 years ago)

Reply 135

AnonyMatt

15

STEP III Q14

Spoiler

Very much unfinished, and I don't think I've done part i) correctly...

Reply 136

Asboob

OK, I know this is an old thread, but I would be forever grateful if someone could clarify a tiny bit of the solution for II/7:

Original post by Glutamic Acid

II/7:

So

1/3 + \alpha(\lambda - 1/3) = 1 + \beta(\lambda - 1)

2/3 + \alpha(\lambda - 2/3) = 0

2/3 - \alpha 2/3 = \lambda \beta

The second equation gives

\alpha = \dfrac{2}{2 - 3 \lambda}

. Substituting into the third gives

\beta = \dfrac{2}{2 - 3 \lambda}

. Substituting these into the first:

1/3 + \dfrac{2}{2 - 3 \lambda}(\lambda - 1/3) = 1 + \dfrac{2}{3 \lambda - 2}(\lambda - 1)

; multiplying out gives

2 \lambda = 0 \Rightarrow \lambda = 0

, so no non-zero solutions.

I have the same three equations. From the second equation, I get the same value of

\alpha

(\dfrac{2}{2 - 3 \lambda})

. However, when I substitute it into the third equation, I get

\beta = \dfrac{2}{3 \lambda - 2}

, the negative of what Glutamic Acid got (which is the answer that works, putting

\lambda = 0

into

1/3 + \dfrac{2}{2 - 3 \lambda}(\lambda - 1/3) = 1 + \dfrac{2}{3 \lambda - 2}(\lambda - 1)

gives a false equation). I'm probably just being an idiot, but I'm tearing my hair out over this, can someone put me out of my misery?

Reply 137

ghostwalker

17

Original post by Asboob

I have the same three equations. From the second equation, I get the same value of

\alpha

(\dfrac{2}{2 - 3 \lambda})

. However, when I substitute it into the third equation, I get

\beta = \dfrac{2}{3 \lambda - 2}

, the negative of what Glutamic Acid got (which is the answer that works, putting

\lambda = 0

into

1/3 + \dfrac{2}{2 - 3 \lambda}(\lambda - 1/3) = 1 + \dfrac{2}{3 \lambda - 2}(\lambda - 1)

gives a false equation). I'm probably just being an idiot, but I'm tearing my hair out over this, can someone put me out of my misery?

I suspect it's just a typo on his part, and he meant to put

\beta = \dfrac{-2}{2-3 \lambda }

since he's used that value in the following line (having multiplied top and bottom by -1).

Reply 138

brianeverit

9

2002 STEP I question 14

\text{P(I get money at first attempt) is }\dfrac{1}{n}, \text{P(at second attempt) is }\dfrac{n-1}{n} \times \dfrac{1}{n-1} \times \text{e}^{-\lambda /r}

Unparseable latex formula:

\text{P(at third attempt} is }\dfrac{n-1}{n}\times\dfrac{n-2}{n-1}\times\dfrac{1}{n-2}\times \text{e}^{-2\lambda/r} \text{ etc}

\text{hence, P(I get my money) is}\dfrac{1}{n}+\dfrac{1}{n} \displaystyle \sum_{k=1}^n \text{e}^{-k\lambda/r}=\dfrac{1}{n} \displaystyle \sum_{k=1}^n\text{e}^{-\lambda(k-1)/r}

\text{so Probability that police arrive before I get my mone is }1-\dfrac{1}{n} \displaystyle \sum_{k=1}^n\text{e}^{-\lambda(k-1)r}=1-\dfrac{1}{n} \left( \dfrac{1-\text{e}^{-n\lambda/r}}{1-\text{e}^{-\lambda/r}}\right)

\text{If I punch in numbers at random, the probability of getting the correct one or an incorrect one }

\text {remain constant at } \dfrac{1}{n} \text{ and } \dfrac{n-1}{n} \text{ respectively, hence Prob(I get my money) is now }

\dfrac{1}{n}+ \dfrac{n-1}{n}\times\dfrac{1}{n}\text{e}^{-\lambda/r}+ \left(\dfrac{n-1}{n}\right)^2\times \dfrac{1}{n}\text{e}^{-2\lambda/r} + \dots

= \dfrac{1}{n} \left(1+ \displaystyle \sum_{k=1}^{\infty} \left( \dfrac{n-1}{n} \right)^k \text{e}^{-k \lambda/r} \right)= \dfrac{1}{n} \left( \dfrac{1}{1- \frac{n-1}{n} \text{e}^{- \lambda/r}} \right)=1- \dfrac{1}{n-(n-1)\text{e}^{-\lambda/r}}

Reply 139

brianeverit

9

2002 STEP II question 12

\text{Consider a single coin thrown }M\text{ times, then number of heads }X\text{ say is a binomial variable}Bin(M,p)

\text{We use the normal approximation }X \approx(1-p))

\text{so }P(X<m)=\Phi\left(\dfrac{m-\frac{1}{2}-Mp}{\sqrt{Mp(1-p)}}\right)=\Phi\left(\dfrac{2m-1-2MP}{2\sqrt{Mp(1-p)}}\right)= \dfrac{h}{L} h\text{ defined as in question}

\text{Now consider }L\text{ such trials. The number of heads }Y \text{ say will have a }Bin\left(L,\dfrac{h}{L}\right) \text{ distribution}

\text{it is reasonable to assume that }\dfrac{h}{L} \text{ is small, so we use a Poisspon approximation }Po(h)

\text{if }h \text{ is not too small we may approximate this with }N(h.h)

Unparseable latex formula:

\text{so we now have }P(Y>l)=1- \Phi \left( \dfrac{1-0.5-h}{\sqrt{h}}\right)=1-\Phi\left(\dfrac{2l-1-2h}{2\sqrth}\right)= \Phi \left( \dfrac{2h-2l-1}{2\sqrt{h}}\right)=q

\text{If each such set of trials is performed }K\text{ times then the number of successful outcomes is }B(K,q)

\text{hence, P(}k\text{ successful outcomes })=\displaystyle\binom{K}{k}q^k(1-q)^K-k \text{ as required}

\text{With }K=7,k=2,L=500,l=4,M=1)),m=48 \text{ and }p=0.6

M\text{ is large and }p\text{ is close to }0.5 \text{ so first normal approximation is acceptable}

h=500\Phi\left(\dfrac{96-1-120}{2\sqrt24}\right)=500\Phi \left(- \dfrac{25}{2\sqrt24}\right)=2.7

\dfrac{h}{L}=0.0054 \text{ so Poisson approximation }Po(h) \text{ is acceptable but the normal approximation to this is not}

\text{so I would not expect theresult to be very accurate}

STEP I, II, III 2002 Solutions

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