arthham

Exam

UKMT Senior Mathematical Challenge

Conducted by
UK Mathematics Trust (UKMT)
For
Students aged 16-18 (Year 13/S6 or below)
Subject
Mathematics
Format
25 multiple-choice questions
Duration
90 minutes
Marking
4 marks per correct answer; penalty for wrong answers on later questions
Frequency
Annually
Qualifies for
Senior Kangaroo, British Mathematical Olympiad

The Senior Mathematical Challenge (SMC) is a national competition run by the UK Mathematics Trust for students in their final two years of secondary school. It aims to stimulate interest in mathematics and encourage problem solving. Participants are typically those studying A-level Mathematics or equivalent qualifications, though younger students can also enter. High scores can lead to invitations to follow-on rounds like the Senior Kangaroo and the British Mathematical Olympiad.

The paper consists of 25 multiple-choice questions to be answered in 90 minutes. The questions are arranged in order of increasing difficulty, with the first fifteen being more straightforward applications of school mathematics. The real test lies in the final ten questions, which demand deeper insight and creative thinking. These problems often require you to see familiar concepts in a new light.

The challenge of the SMC is not the advanced level of the mathematics, which is largely based on the GCSE and early A-level syllabus. The difficulty comes from the clever and unfamiliar ways the questions are framed. A key feature is the marking scheme: while all correct answers are worth four marks, a mark is deducted for an incorrect answer to any of the last ten questions. This introduces a strategic element, as you must decide whether to risk guessing or leave a question unanswered.

How a question gets worked here

From a mathematics paper rather than UKMT Senior Mathematical Challenge, worked the way every question on this site is. UKMT Senior Mathematical Challenge questions are being added.

JEE Advanced 2013

Two lines L1:x=5,y3−α=z−2L_{1} : x = 5, \frac{y}{3-\alpha} = \frac{z}{-2} and L2:x=α,y−1=z2−αL_{2} : x = \alpha, \frac{y}{-1} = \frac{z}{2-\alpha} are coplanar. Then α\alpha can take value(s)
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Two lines L₁ : x = 5, y/(3-α) = z/(-2) and L₂ : x = α, y/(-1) = z/(2-α) are coplanar. Then α can take value(s) Given: L1: x=5, y3−α=z−2\text{Given: } L_1:\ x = 5,\ \dfrac{y}{3-\alpha} = \dfrac{z}{-2} L2: x=α, y−1=z2−αL_2:\ x = \alpha,\ \dfrac{y}{-1} = \dfrac{z}{2-\alpha} To find: α such that L1,L2 are coplanar\text{To find: } \alpha \text{ such that } L_1, L_2 \text{ are coplanar} 1. point and direction of L1,L21.\ \text{point and direction of } L_1, L_2 2. coplanar⇔[A2⃗−A1⃗, d1⃗, d2⃗]=02.\ \text{coplanar} \Leftrightarrow [\vec{A_2}-\vec{A_1},\ \vec{d_1},\ \vec{d_2}] = 0 3. solve for α, then check the options3.\ \text{solve for } \alpha, \text{ then check the options} L1: A1=(5,0,0),d1⃗=(0, 3−α, −2)L_1:\ A_1 = (5, 0, 0), \quad \vec{d_1} = (0,\ 3-\alpha,\ -2) L2: A2=(α,0,0),d2⃗=(0, −1, 2−α)L_2:\ A_2 = (\alpha, 0, 0), \quad \vec{d_2} = (0,\ -1,\ 2-\alpha) Coplanar⇔[A2⃗−A1⃗, d1⃗, d2⃗]=0\text{Coplanar} \Leftrightarrow [\vec{A_2} - \vec{A_1},\ \vec{d_1},\ \vec{d_2}] = 0 ⇔(A2⃗−A1⃗)⋅(d1⃗×d2⃗)=0\Leftrightarrow (\vec{A_2} - \vec{A_1}) \cdot (\vec{d_1} \times \vec{d_2}) = 0 A2⃗−A1⃗=(α−5, 0, 0)\vec{A_2} - \vec{A_1} = (\alpha - 5,\ 0,\ 0) d1⃗×d2⃗=∣i^j^k^03−α−20−12−α∣\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[2pt] 0 & 3-\alpha & -2 \\[2pt] 0 & -1 & 2-\alpha \end{vmatrix} =[(3−α)(2−α)−2]i^=(α2−5α+4) i^= \left[(3-\alpha)(2-\alpha) - 2\right]\hat{i} = (\alpha^{2} - 5\alpha + 4)\,\hat{i} α2−5α+4=(α−1)(α−4)\alpha^{2} - 5\alpha + 4 = (\alpha - 1)(\alpha - 4) d1⃗×d2⃗=(α−1)(α−4) i^\vec{d_1} \times \vec{d_2} = (\alpha-1)(\alpha-4)\,\hat{i} (A2⃗−A1⃗)⋅(d1⃗×d2⃗)=(α−5)(α−1)(α−4)(\vec{A_2} - \vec{A_1}) \cdot (\vec{d_1} \times \vec{d_2}) = (\alpha-5)(\alpha-1)(\alpha-4) (α−5)(α−1)(α−4)=0⇒α=1, 4, 5(\alpha-5)(\alpha-1)(\alpha-4) = 0 \Rightarrow \alpha = 1,\ 4,\ 5 Given options are 1,2,3,4⇒α=1 or α=4\text{Given options are } 1, 2, 3, 4 \Rightarrow \alpha = 1 \text{ or } \alpha = 4 α=1: d1⃗=(0,2,−2), d2⃗=(0,−1,1)⇒d1⃗∥d2⃗\alpha = 1:\ \vec{d_1} = (0, 2, -2),\ \vec{d_2} = (0, -1, 1) \Rightarrow \vec{d_1} \parallel \vec{d_2} α=4: d1⃗=d2⃗=(0,−1,−2)⇒d1⃗∥d2⃗\alpha = 4:\ \vec{d_1} = \vec{d_2} = (0, -1, -2) \Rightarrow \vec{d_1} \parallel \vec{d_2} α=2: (α−5)(α−1)(α−4)=(−3)(1)(−2)=6≠0\alpha = 2:\ (\alpha-5)(\alpha-1)(\alpha-4) = (-3)(1)(-2) = 6 \neq 0 α=3: (−2)(2)(−1)=4≠0\alpha = 3:\ (-2)(2)(-1) = 4 \neq 0 Answer: α=1 and α=4⇒options (1) and (4)\boxed{\text{Answer: } \alpha = 1 \text{ and } \alpha = 4 \Rightarrow \text{options (1) and (4)}}

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