1987 Specimen STEP solutions thread

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Found this paper the other day and realised there wasn't a solutions thread up and running so I figured it should be done. Unfortunately, I've only been able to locate the STEP III paper so if anyone could locate the others it would be awesome.
EDIT: Got them. Courtesy of Professor Siklos himself!

STEP I
1.
2.Solution by ben-smith
3.Solution by ben-smith
4.Solution by ben-smith
5.
6.Solution by ben-smith
7.Solution by Farhan.Hanif93
8.
9.
10.
11.Solution by Farhan.Hanif93
12.Solution by Farhan.Hanif93
13.Solution by Farhan.Hanif93
14.
15.
16.
STEP II
-
STEP III
1.Solution by ben-smith.
2.Solution by DFranklin
3.Solution by ben-smith
4.Solution by ben-smith
5.
6.Solution by Farhan.Hanif93
7.Solution by Farhan.Hanif93
8.Solution by DFranklin
9.Solution by Farhan.Hanif93
10.Solution by Kennethx
11.solution by ben-smith
12.Solution by ben-smith
13.Solution by ben-smith
14.Solution by ben-smith
15.Solution by DFranklin
16.Solution by Dfranklin

Attached files

9465.pdf (128.3 KB)
9470.pdf (154.9 KB)

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Dirac Spinor

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STEP III Q1
a)let

denote the product up to n.
$I_2=\frac{3}{4},I_3=\frac{2}{3},I_4=\frac{5}{8}, I_5=\frac{3}{5}, I_6=\frac{7}{12},...$
By inspection, these numbers seem to be of the form $I_n=\frac{n+1}{2n}$ so let's guess that.
We have already proved this to be the case for n=2 so let us presume, for the purpose of induction, that $I_k=\frac{k+1}{2k}$ . Now let n=k+1:
$I_{k+1}=I_k*(1-\frac{1}{(k+1)^2})=\frac{k+1}{2(k)}*\frac{(k+1)^2-1}{(k+1)^2}=\frac{k+2}{2(k+1)}$ So our guess is thus true by mathematical induction.
b) Consider the binomial expansion of

:
$(1+x)^n=\displaystyle\sum_{r=0}^n \displaystyle \binom{n}{r}x^r$ . Letting x=-1 gives us:
$\displaystyle\sum_{r=0}^n \displaystyle \binom{n}{r}(-1)^r=0 \Rightarrow \displaystyle\sum_{r=0}^k (-1)^r\displaystyle \binom{n}{r}=-\displaystyle\sum_{r=k+1}^n(-1)^r \displaystyle \binom{n}{r} =\displaystyle\sum_{r=k+1}^n(-1)^{r-1} \displaystyle \binom{n}{r} =\displaystyle\sum_{r=k+1}^n(-1)^{r-1} \displaystyle \binom{n-1}{r-1}+\displaystyle\sum_{r=k+1}^n(-1)^{r-1} \displaystyle \binom{n-1}{r}$
Notice that the i+1 th element in the first summation cancels with the ith element in the second summation leaving the first element in the first summation so:
$\displaystyle\sum_{r=0}^k (-1)^r\displaystyle \binom{n}{r}=(-1)^k \displaystyle \binom{n-1}{k}$
As required.

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DFranklin

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At a brief glance, the specimen questions are quite a lot easier than the actual exam questions were that year... {Tut tut}.

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Dirac Spinor

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(Original post by DFranklin)
At a brief glance, the specimen questions are quite a lot easier than the actual exam questions were that year... {Tut tut}.

Aren't all STEP questions relatively easy for you though?
Shall I continue uploading solutions?

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(Original post by ben-smith)
Aren't all STEP questions relatively easy for you though?

Not at all.

Shall I continue uploading solutions?

Sure. (My tut-tut was at the examiners. I can well imagine what it would be like to prepare based on this paper and then find you can only do a third as many questions as you expect in the actual exam

).

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Question 10 was quite easy, managing attachments is alas a bit more troublesome. Enjoy reading down to up instead of left to right.

Attached files

photo (4).JPG (95.6 KB)
photo (5).JPG (83.5 KB)
photo (7).JPG (86.8 KB)

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Farhan.Hanif93

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Oh why not.

STEP III Q6

Solution

Note $J_n=\displaystyle\int ^{\infty}_0 sech ^{n+2}u \sinh ^2u du = \displaystyle\int ^{\infty}_0 sech ^{n}u (sech u \tanh u) \times \sinh u du$

Integrating by parts with $p=\sinh u$ and $\dfrac{dq}{du}=sech ^{n}u (sech u \tanh u)$ :

$J_n=\left[\dfrac{1}{n+1}sech ^{n+1}u \sinh u\right]^{\infty}_0 + \dfrac{1}{n+1}\displaystyle\int ^{\infty}_0 sech ^{n+1}u \cosh u du$
$=0+\dfrac{1}{n+1}I_n$

It follows that $\boxed{\displaystyle\int ^{\infty}_0 sech ^{n+2}u \sinh ^2u du = \dfrac{1}{n+1}I_n}$ as required.

Similarly $J_n = \displaystyle\int ^{\infty}_0 sech ^{n-1}u (sech u \tanh u) \times \tanh u du$

Integrating by parts with $p=\tanh u$ and $\dfrac{dq}{du}= sech ^{n-1}u (sech u \tanh u)$ :

$J_n = \left[\dfrac{1}{n}sech ^{n}u \tanh u \right] ^{\infty}_0 + \dfrac{1}{n}\displaystyle\int ^{\infty}_0 sech ^{n+2}u du$

$= 0 + \dfrac{1}{n}I_{n+2}$

It follows that $\dfrac{1}{n+1}I_n = \dfrac{1}{n}I_{n+2}$ and hence it can be deduced that $\boxed{(n+1)I_{n+2} = nI_n}$ as required.

Note that

$I_6 = \dfrac{4}{5}I_4 =\dfrac{4}{5} \times \dfrac{2}{3} I_2 = \dfrac{8}{15} [\tanh u]^{\infty}_0 = \boxed{\dfrac{8}{15}}$

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Kennethx

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Dibs on question no. 4 btw.

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DFranklin

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STEP III, Q15: It's easy to convince yourself that the villages all remain connected if less than 3 roads are down (draw diagrams if you wish to illustrate the possible cases).
If 3 roads are down and they all go to the same village, that village is cut off. Otherwise, the villages remain connected. Again, draw diagrams if you wish.
If 4 roads are down, then it's clearly impossible for 2 roads to connect all 4 villages. (The roads must connect at a village, and that only leaves 2 ends to connect 3 villages).
Even more obvious for 5 and 6 roads.

P(more than 3 roads down) =

(binomial)
P(A cut off by having exactly 3 roads down) =

. So P(any village cut off by exactly 3 roads down) =

So P(villages are cut off) =

So P(villages are connected) =

When p = 1/2, this equals $\dfrac{64 - 4 - 15 - 6 - 1}{64} = \dfrac{38}{64} = \dfrac{19}{32}$

Wow - only 2 probablity questions out of 16. Hard times for probability specialists...

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Dirac Spinor

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STEP III Q3
Consider the complex numbers

which forms a triangle of side

. By the triangle inequality, $|z_1|+|z_2|\geq|z_1+z_2|$ so the given inequality is just a restatement of the triangle inequality.
$\Sigma a_i z^i=1$ . Taking the modulus of both sides we get:
$|\Sigma a_iz^i|=1 \leq \Sigma|a_iz^i|=\Sigma |a_i||z|^i$ .
We can maximise the RHS summation by letting

so:
$\Sigma2|z|^i=2\frac{|z|(1-|z|^n)}{1-|z|}\Rightarrow \frac{1}{2}\leq \frac{|z|(1-|z|^n)}{1-|z|}$ which must hold if there are to be solutions.
notice that this does not hold for |z|=1/3 and therefore does not hold for values less than a third (this is obvious as the sum is just a power series with positive coefficients and so must have a +ve derivative). Furthermore since we chose |a| such that the sum was maximised, the result must hold for smaller values of it. Therefore there must be no solutions for |z| less than or equal to a third.

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Dirac Spinor

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(Original post by DFranklin)
STEP III, Q15: It's easy to convince yourself that the villages all remain connected if less than 3 roads are down (draw diagrams if you wish to illustrate the possible cases).
If 3 roads are down and they all go to the same village, that village is cut off. Otherwise, the villages remain connected. Again, draw diagrams if you wish.
If 4 roads are down, then it's clearly impossible for 2 roads to connect all 4 villages. (The roads must connect at a village, and that only leaves 2 ends to connect 3 villages).
Even more obvious for 5 and 6 roads.

P(more than 3 roads down) =

(binomial)
P(A cut off by having exactly 3 roads down) =

. So P(any village cut off by exactly 3 roads down) =

So P(villages are cut off) =

So P(villages are connected) =

When p = 1/2, this equals $\dfrac{64 - 4 - 15 - 6 - 1}{64} = \dfrac{38}{64} = \dfrac{19}{32}$

Wow - only 2 probablity questions out of 16. Hard times for probability specialists...

Yeah, the same thing happened in 88 except that year imo the two questions were very hard.

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STEP III, Q16:

(i) If

is the number of extra children needed for the ith male child (that is, the number of children since the i-1th male child), then C = X_1 + ... + X_r, and so E[C] = E[X_1]+...+E[X_r]. The X_i are iid geometric distributions with p = 1/2, so E[X_i] = 2 and C = 2r.

(ii) Suppose the king ignores the rules and just has 2r-1 children. If he has r or more boys, clearly C < 2r, and conversely, if he has < r boys, C>=2r.
So P(C < 2r) = P(there are at least r boys from 2r-1 children). Since boys and girls are equally likely, P(there are at least r girls from 2r-1 children) = P(there are at least r boys from 2r-1 children).
But these two events are mutually exclusive, and 1 of them must occur. So P(there are at least r girls from 2r-1 children) = 1 - P(there are at least r boys from 2r-1 children).
So P(there are at least r boys from 2r-1 children) = 1 - P(there are at least r boys from 2r-1 children), and so P(there are at least r boys from 2r-1 children) = 1/2. That is, P(C < 2r) = 1/2.

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STEP III, Q8.

(i) I'm going to be lazy and use capitals to denote vectors. Let the vertices of the tetrahedron be A, B, C, D. Sum of squares of edges = |A-B|^2+|A-C|^2+|A-D|^2+|B-C|^2+|B-D|^2+|C-D|^2.
Sum of squares of midpoints = [(A+B-C-D)^2 + (A+C-B-D)^2 + (A+D-B-C)^2] / 4.
So 4 x Sum of squares of midpoints = (A+B-C-D)^2 + (A+C-B-D)^2 + (A+D-B-C)^2 = ((A-D)+(B-C))^2 + ((A-B)+(C-D))^2 +((A-C)+(D-B))^2
= (A-D)^2+(B-C)^2 + 2(A-D).(B-C)+(A-B)^2+(C-D)^2+2(A-B).(C-D)+(A-C)^2+(D-B)^2 + (A-C).(D-B).
So, sufficient to show (A-D).(B-C) + (A-B).(C-D) + (A-C).(D-B) = 0. Some tedious expansion gives us A.B-A.C+C.D-B.D+A.C-A.D+B.D-B.C+A.D-A.B+B.C-C.D and everything cancels.

(ii) Take a = 3I + 2J + 6K, b = xI + yJ + K.
Then we have $(3x+2y+6) \leq \sqrt{3^2+4^2+6^2} \sqrt{x^2+y^2+1} = 7\sqrt{x^2+y^2+1}$ .
a.b = cos t |a| |b, where cos t is the angle between the vectors.

So a.b = |a| |b if and only if cos t = 1.
So we require 3I+2J+6K and xI+yJ+1 to be in the same direction. So x = 3/6, y=2/6, or x = 1/2, y=1/3.

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STEP III Q7

Solution

Setting $y=kx^{\alpha}, y'= \alpha k x^{\alpha -1} , y'' =\alpha(\alpha -1)kx^{\alpha -2}$ in the differential equation:

$\alpha (\alpha -1)kx^{\alpha} + (x-2)(\alpha kx^{\alpha} -kx ^{\alpha}) = 0$
$\implies kx^{\alpha} (\alpha -1)(x+\alpha -2) =0$

Clearly this holds if $\alpha =1$ and hence there exists a solution of the form $x^\alpha$ as required.

Setting

in the DE:

$\implies v'' + v' =0$

$\implies v'=Ce^{-x}$

$\implies \boxed{y=Axe^{-x}+Bx}$ (where

)

They call that a STEP question...

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Dirac Spinor

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STEP III Q14
v is the velocity of the particle after the collision.
Conservation of KE: $\frac{1}{2}mu^2=\frac{1}{2}I \omega^2+\frac{1}{2}mv^2$
I is the moment of inertia of the rod.
Conservation of angular momentum:
$ma^2\frac{u}{a}=mav+I \omega I= 1/3 m4a^2 \Rightarrow v=u-\frac{4}{3}a\omega \Rightarrow 0.5mu^2=0.5(\frac{1}{3}m4a^2)\omega^2+0.5m(u-\frac{4}{3}a\omega)^2 \Rightarrow \frac{28a^2}{9}\omega^2-\frac{8ua}{3}\omega=0 \Rightarrow \frac{28a^2}{9}\omega=\frac{8u}{3}\Rightarrow a\omega=\frac{6u}{7}$
As required.
The rod forms a circle when it rotates. The particle and the rod can only collide when the particle is within it. By pythagoras, the particle will travel $\sqrt{3}$ before leaving the circle and so the time it takes is:
$T_1=\frac{\sqrt{3}a}{-v}=\frac{\sqrt{3}a}{(4a\omega /3)-(7a\omega /6)}= \frac{6\sqrt3}{\omega}$
The time for the rod rotate through $\frac{5\pi}{3}$ radians such that the tip touches the point where the particle leaves the circle is $\frac{5\pi}{3}/ \omega$
Note that $\frac{5\pi}{3}<6sqrt3$ which can easily be seen if we overestimate pi to 3.2 and underestimate root 3 to 1.5 where the inequality holds. If it holds for those values then it certainly must hold for their actual values. Therefore the tip of the rod reaches the leaving point before the particle so they must collide again.

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Dirac Spinor

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STEP III Q11
$\frac{dx}{dt}=k/\alpha z ,\frac{dz}{dt}=k \Rightarrow z=kt+h \Rightarrow \frac{dx}{dt}=\frac{k}{\alpha(kt+h)} \Rightarrow x=k/ \alpha(\frac{1}{k}ln(kt+h)+C). t=0, x=0 \Rightarrow C=-\frac{lnh}{k} \therefore \alpha x=ln(\frac{kt}{h}+1) \Rightarrow t=\frac{h}{k}(e^{\alpha x}-1)$
as required.
For the second plough:
z=k(time it takes for 2nd plough to go y metres-time taken for first plough to reach y)
= $k(t-(e^{\alpha y}-1)\frac{h}{k}$
And since
$\frac{dy}{dt}=k/\alpha z \Rightarrow 1/ \alpha \frac{dt}{dz}=z/k= t-(e^{\alpha y}-1)\frac{h}{k}$
As required.
Differentiating this with respect to time we get:
$1/ \alpha \frac{dt}{dz} =1/ \alpha (\frac{h\alpha}{k}e^{\alpha y}+\alphae^{\alpha y}(T-\frac{\alpha hy}{k})-\frac{h\alpha}{k}e^{\alpha y})=e^{\alpha y}(T-\frac{\alpha hy}{k})=(e^{\alpha y}-1)\frac{h}{k}+e^{\alpha y}(T-\frac{\alpha hy}{k})-e^{\alpha y}(T-\frac{\alpha hy}{k})=t-(e^{\alpha y}-1)\frac{h}{k}$
Which means it is a solution as it satisfies the equation.
They will collide when their times are equal and x=y. Notice how the time for second plough is the same as the one for the 1st $+e^{\alpha y}(T-\frac{\alpha hy}{k})$
This should disappear when they collide so:
$T-\frac{\alpha hx}{k}=0 \Rightarrow x=\frac{Tk}{\alpha h}$
As required.

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Dirac Spinor

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STEP III Q4
The parametric form of a parabola is

.
$y^2=2ax \Rightarrow \frac{dy}{dx}=\sqrt{a/x}=1/t \therefore 1/t=\frac{y-2at}{x-at^2} \Rightarrow yt=x+at^2$
So the given line touches the paabola, as required, at the point

.
Lets assume the tangents intersect at the point (x_1,y_1) which would mean:
$ty_1=x_1+at^2 \Rightarrow t=\frac{y_1\pm \sqrt{y_1^2-4ax_1}}{2a}$
So the equation of the tangent that goes through (x_1,y_1) is:
$y=\frac{2ax}{y_1\pm \sqrt{y_1^2-4ax_1}}+a\frac{y_1\pm \sqrt{y_1^2-4ax_1}}{2a}$

Substituing in x=-a and subtracting the y values we get:
$l= -2a^2(\frac{2 \sqrt{y_1^2-4ax_1}}{y_1^2-y_1^2+4ax_1})+a(\frac{-2 \sqrt{y_1^2-4ax_1}}{2a}) =(-4a^2 \sqrt{y_1^2-4ax_1})/{4ax_1}-\sqrt{y_1^2-4ax_1} =(-\sqrt{y_1^2-4ax_1}(a+x_1))/{x_1} \Rightarrow l^2x^2=(y^2-4ax)(a+x)^2$
As required.
Note: at the end I removed the suffixes, they were only there in the first place for notational purposes. They are just generalised coordinates.

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STEP III, Q2. This is completely standard once you've done an analysis course...

(1+x)^n = 1 + nx + n(n-1)/2 x^2 + ... + x^n. All terms are +ve, so discarding all but the first two terms on the RHS we have (1+x)^n >= 1+nx > nx. Now suppose y > 1, write y = 1 + x, where x > 0. y^n > nx and so if n > K/x, y^n > K. So we can take N = K/x = K/(y-1).

Similarly, for n >=2, by discarding all but the n(n-1)/2 term we have (1+x)^n >= n(n-1)/2 x^2. If y > 1, again write y = 1+x, then y^n > n(n-1)/2 x^2 and so y^n / n > (n-1)/2 x^2. If n > (2K/x^2) + 1, then (n-1)/2 > K / x^2 and so (n-1)/2 x^2 > K. So we can take N = (2K/(y-1)^2) + 1.

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Dirac Spinor

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Q12
A, B and C form a triangle. Let the angle at A be theta and the angle a C be phi. Let us further denote D to be the point on CB that also lies on the line perpendicular to CB and goes through A and T is the tension in the string.
The rod is in equilibrium so the forces vertically and the moments about A are balanced:
$Mg=Tcos\phi T(AD)=(L/2)sin\thetaMg$
To find AD, notice that the are of triangle is $1/2(\alpha L (AD))$ and also $\frac{1}{2}L^2\alpha sin(\pi-(\theta+\phi))$ .
Equating the two, we get:
$AD=Lsin(\theta+\phi)$
Substituting in from our first equations:
$TLsin(\theta+\phi)=(L/2)sin\thetaMg \Rightarrow sin(\theta+\phi)=\frac{sin\thetacos\phi}{2} \Rightarrow sin\theta cos\phi+cos\theta sin\phi=\frac{sin\theta cos\phi}{2} \Rightarrow cos\theta sin\phi=-\frac{sin\theta cos\phi}{2}$
By the sine rule:
$\frac{sin\phi}{L}=\frac{sin\theta}{\alpha L}\Rightarrow sin\phi=\frac{sin\theta}{\alpha}$
So:
$\frac{sin\theta cos\theta }{\alpha}=-\frac{sin\theta \sqrt{\alpha^2-sin^2\theta}}{2\alpha} \Rightarrow 2cos\theta=-\sqrt{\alpha^2-sin^2\theta} \Rightarrow 4cos^2\theta=\alpha^2-sin^2\theta \Rightarrow cos^2\theta=\frac{\alpha^2-1}{3}$
The function $cos^2\theta$ is greater that or equal to 0 and less than or equal to 1 but, in this case, we don't want the equality case as that would mean A, B and C would be collinear so:
$1<\alpha<2$
To find the tension:
$T=\frac{Mg}{cos\phi} sin\phi=\frac{sin\theta}{\alpha} \Rightarrow cos\phi = 1/ \alpha(\sqrt{\alpha^2-\frac{4-\alpha^2}{3}})=\frac{2}{\alpha}(\frac{\alpha^2-1}{3})^{1/2}$
So:
$T=\frac{Mg}{cos\phi)}=\frac{Mg \alpha}{2} (\frac{3}{\alpha^2-1})^{1/2}$
As required.

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STEP III Q13
By conservation of energy:
$\Delta P.E.= \Delta K.E. \Rightarrow mga(1-cos2\theta)=\frac{1}{2}mv^2 \Rightarrow v=\sqrt{2ga(1-cos2\theta)}=2(ga)^{1/2}sin\theta$
As required
When the ring has rotated 2theta radians, the particle has moved $asin2\theta$ horizontally relative to the ring but the ring itself has moved $2a\theta horizontally$ so:
$x=2a\theta+asin2\theta$ and similarly for y
$y=a+acos2\theta$

$\mathbf{v}=\frac{d}{dt}[(2a\theta+asin2\theta)\mathbf{i}+(a+acos2\theta)\mathbf{j}] = (2a \theta'+2a\theta'cos2\theta) \mathbf{i}+(-ra\theta'sin2\theta)\mathbf{j} \therefore v=2a\theta'\sqrt{(1+cos2\theta)^2+sin^22\theta}=2a\theta'\sqrt{4cos^2\theta}=4a\theta'cos\theta$
And by our first result:
$v=2(ga)^{1/2}sin\theta=4a\theta'cos\theta \Rightarrow \theta'=\frac{1}{2}(g/a)^{1/2}Tan\theta$
as required.
Differentiating w.r.t time:
$\theta'=\frac{1}{2}(g/a)^{1/2}Tan\theta \Rightarrow \theta''=\frac{1}{2}(g/a)^{1/2}sec^2\theta \theta'=\frac{1}{4}(g/a)sec^2\theta Tan\theta$
Similarly:
$y'=-2a\theta' sin2\theta \Rightarrow y''=-2a(\theta''sin2\theta+cos2\theta*2(\theta')^2) =-2a[(g/4a)sec^2\theta Tan\theta sin2\theta + cos2\theta*\frac{g}{2a}Tan^2\theta] =-2g[(1/2)sec^2\thetaTan\theta sin\theta cos\theta +\frac{1}{2}cos2\theta Tan^2\theta] =-2g[(1/2)Tan^2\theta +\frac{1}{2}Tan^2\theta(2cos^2 \theta-1)] =-2g[(Tan^2\theta/2)(1+2cos^\theta-1)]=-2gsin^2\theta$
As required.
At theta=pi/4, from our previous results, we can see that the acceleration vertically is g which would imply the reaction force is 0 but at that point the hoop ha left the table so it can't roll past theta=pi/4.

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