post in another thread, Ive gone through this years smc
Model solutions: https://ukmt.org.uk/wp-content/uploads/2023/10/SMC-2023-Solutions.pdf
and tried to do it in a simple problem solving / elementary way. I doubt Ive done an "optimal" approach in some cases, but the usual points that
* There is little A level content. Up to (and including) 22 was reasonably straightforward.
* Know your pythagorean triples, and 45-45-90 and 30-60-90 side ratios and spot them in questions/answers
* Round stuff if the answers are sufficiently different
* Subbing answers is a reasonable solution in some questions
* hard questionns 20-22 were fairly write down answers with a bit of simple reasoning.
* completing the square and dots are popular
* a decent sketch with tangents .... added is very useful.
* exam time shouldnt be too much of an issue, though doing full algebraic working for each question will eat up time.
1) Either note its a ukmt “year - prime factors” question so 2023 = 7*17^2, or note the answers are reasonably different (square of the values) so 2023/7 ~ 2100/7 ~ 300 and 17^2 = 289 ~ 300. So C)
2)The difference is a recurring decimal so C) or E) and the latter is too small. So C)
3) The multiplier is 1.2*0.85 > 1. So D) or E). Simple enough to evaluate so 0.85+0.17. So D)
4) Answers are reasonably different so 5km in ⅓ hr so 15km/h. So C)
5) The possible areas are integers, so pythagorean triples are likely involved. Using the given 5-5-6 isosceles triangle, so bisect 6 to form 2 congruent right triangles which are 3-4-5, so area 12, so B).
6) After trying a few examples, you should note theyre all c-d + c + c+d = 3c. So 8+8+8, so B). Really you just note that there are 2 examples per row/col and a few on the diagonal as the answers are sufficiently different and it must be > 16 and < 30.
7) Just plugging in a few more values, it goes down 2 every 3 (as the model solution notes) so 2023-16=2007, so D).
8) The difference is 0.050505. Multiply by 100 gives ~ 5. So 5, so A)
9) Starting with 1 down, the only possibility is 2*243 = 486. 1 across must be 21^2=441 as 20^2 would mean 2 down would start with a 0 digit and 22^2 would need an 8 option in the answers so D)
10) Simple factorisation. See ukmt model solution.
11) See ukmt model solution
12) Simple factorial counting. See ukmt model solution
13) Simple sketch/angle chasing. See ukmt model solution.
14) Pythagoras / ukmt model solution isnt hard but you could reason d cant be negative as 2d-6 would be too large (magnitude) so not A) or B) and not C) or D) as d<=4 is too small as the distances of the two points from the origin would be too different. So E).
15) Sub answers in, start small. Overlap 0. Then neither : football : basketball = 1:2:4 but 30 isnt divisible by 7. Overlap 5, then 30+5 is divisible by 7 so 5:10:20. So D). Note that overlap 19 A) is also divisible by 7, but the numbers dont work (7:14:28).
16) Fairly simply the base is 4 (square length to diagonal is 1:sqrt(2)) and top is half that so the average is 3. So A) is looking likely. Putting another cube on the top, the perp height is half of sqrt (2*2sqrt(2))^2 + 2^2) = 3, so A)
17) See ukmt model solution
18) Negate and sub z=2sin(x) so z^3+z^2-z-1=0, so a cubic in z and has half the number of “x” solutions as z=+/-2 are not roots and for |z|>2, the cubic term dominates, so all solutions |z|<2. z=1 is one solution as the coeffs sum to zero so (z-1)(z^2+2z +1) = (z-1)(z+1)^2, so 2 (distinct) “z” roots and 4 “x” solutions, so C.
19) Alternative to model solution, so m = (7n+12)/(2n+3) so (rearrange and multiply by 2
4nm + 6m - 14n = 24
Factorise and subtract 21 from both sides
(2n+3)(2m-7) = 3
The potential values of (2m-7) are +/-factors of 3 and as they’re symmetric about 0 the sum of the 4 values of m is (7/2)*4 = 14, so E). m,n are integers for all these values as 2# = odd - odd.
20) Shrink r to zero, then the only option for which q=0 is C)
21) See the model solution, but the key thing is to know the 30-60-90 side ratios and “always” connect tangent points with circle centres (OX). Doing that, the question is straightforward
22) Sub x=0 gives y=1 and its the max value so C).
23) See the model solution. Note it reduces to finding the area of the triangular overlap as base * perpendicular height / 2. A good sketch is key, the actual algrebra is relatively simple.
24) Similar to the ukmt model solution, the point where the shapes touch corresponds to the intersection of the lines y=sqrt(3)x and y=-x+1, where the origin is the left most point of the hexagon. Elimminate x and get y = sqrt(3)/(sqrt(3)+1). Rationalise and double. A longer alternative would be trig heavy using sin(75) as a half angle of sin(30) and sin rule on the 60-45-75 triangle.
25) Similar to a question a few years ago, once youve factored out xy^2 it should be clear you want to complete the square in x and y as they are very similar (constant terms cancel) so you can root it as youll get an area bounded by lines. The model solution does dots which is similar. Writing down the first factorisation is really the hard part and then completing the square or dots should be clear.