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Admissions tests / TMUA / Paper 2 / Reasoning with algebra and number

Demanding. 12 questions, 12 marks, about 50 minutes.

TMUA Paper 2: Reasoning with algebra and number, set 3

Section 2 reasoning applied to the Section 1 content: statements about divisibility, parity, inequalities, sequences and functions, where the work is deciding what follows rather than computing.

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  • Answer all questions. No calculator is allowed.
  • Each question has exactly one correct answer.
  • This set is pitched above standard exam difficulty: expect multi-step routes, less signposting, and numbers designed to punish a careless line of working.
  1. 11 mark

    Solve the equation 9^(x+1) = 27^(x-1) for x.

    1. A x = 3
    2. B x = 4
    3. C x = 5
    4. D x = -1
  2. 21 mark

    Which of the following is the fully simplified form of 5/(3 - sqrt(2)), with a rational denominator?

    1. A (15 + 5sqrt(2))/7
    2. B (15 - 5sqrt(2))/7
    3. C (15 + 5sqrt(2))/11
    4. D 15/7
  3. 31 mark

    For how many integer values of k, where -10 <= k <= 10, does the equation x^2 - kx + 9 = 0 have two distinct real roots?

    1. A 4
    2. B 10
    3. C 21
    4. D 8
  4. 41 mark

    The curve y = x^2 - 4x + 1 and the line y = 3x - 5 intersect at two points, A and B. What is the sum of the x-coordinates of A and B?

    1. A -7
    2. B 7
    3. C 6
    4. D 1
  5. 51 mark

    The polynomial f(x) = x^3 + ax^2 + bx - 6 has (x - 1) and (x + 2) as factors, where a and b are constants. What is the remainder when f(x) is divided by (x - 2)?

    1. A 0
    2. B 26
    3. C 20
    4. D 18
  6. 61 mark

    For which values of x is sqrt(6 + x - x^2) a real number?

    1. A x <= -2 or x >= 3
    2. B x <= -3 or x >= 2
    3. C -3 <= x <= 2
    4. D -2 <= x <= 3
  7. 71 mark

    A quadratic sequence begins 2, 7, 16, 29, ... . What is the next term?

    1. A 46
    2. B 42
    3. C 45
    4. D 70
  8. 81 mark

    The sum of the first 10 terms of an arithmetic series is 155, and the sum of the first 20 terms is 610. What is the first term of the series?

    1. A 17
    2. B 4
    3. C 2
    4. D 29
  9. 91 mark

    A geometric series has a positive first term a and common ratio r, where 0 < r < 1. Its sum to infinity is 3 times its first term. What is the value of r?

    1. A 1/3
    2. B 2/3
    3. C -2/3
    4. D -2
  10. 101 mark

    In the expansion of (2 - x)^5, what is the coefficient of x^3?

    1. A 40
    2. B -80
    3. C -60
    4. D -40
  11. 111 mark

    A length is measured as 3.5 x 10^4 m, correct to 2 significant figures. What is the upper bound of this length?

    1. A 3.55 x 10^4 m
    2. B 3.6 x 10^4 m
    3. C 3.45 x 10^4 m
    4. D 4.0 x 10^4 m
  12. 121 mark

    How many ordered pairs (m, n) of positive integers satisfy m x n = 24 and HCF(m, n) = 2?

    1. A 8
    2. B 4
    3. C 2
    4. D 1

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: C

    1. Rewrite both sides with base 3: since 9 = 3^2, 9^(x+1) = 3^(2(x+1)) = 3^(2x+2). Since 27 = 3^3, 27^(x-1) = 3^(3(x-1)) = 3^(3x-3).
    2. The bases now match, so equate the exponents: 2x+2 = 3x-3.
    3. Solve: 2+3 = 3x-2x, so x = 5.
    4. Check: with x = 5, 9^6 = 3^12 and 27^4 = 3^12, so both sides agree. Answer C.
    • Why not A: Expands 27^(x-1) as 3^(3x-1) instead of 3^(3(x-1)) = 3^(3x-3), multiplying only the x-term by 3 and forgetting to also multiply the -1, giving 2x+2 = 3x-1 and so x = 3.
    • Why not B: Expands 9^(x+1) as 3^(2x+1) instead of 3^(2(x+1)) = 3^(2x+2), multiplying only the x-term by 2 and forgetting to also multiply the +1, giving 2x+1 = 3x-3 and so x = 4.
    • Why not D: Misreads 27^(x-1) as 27^(x+1), flipping the sign of the exponent's constant term, giving 2x+2 = 3x+3 and so x = -1.

  2. Question 2Answer: A

    1. To rationalise 5/(3 - sqrt(2)), multiply top and bottom by the conjugate of the denominator, (3 + sqrt(2)).
    2. Denominator: (3 - sqrt(2))(3 + sqrt(2)) = 3^2 - (sqrt(2))^2 = 9 - 2 = 7.
    3. Numerator: 5(3 + sqrt(2)) = 15 + 5sqrt(2).
    4. So 5/(3 - sqrt(2)) = (15 + 5sqrt(2))/7, which cannot be simplified further. Answer A.
    • Why not B: Multiplies numerator and denominator by (3 - sqrt(2)) again instead of the conjugate (3 + sqrt(2)), so the denominator is not rationalised correctly and the numerator's surd term keeps the wrong sign, giving (15 - 5sqrt(2))/7.
    • Why not C: Computes (3 - sqrt(2))(3 + sqrt(2)) as 9 + 2 = 11 instead of 9 - 2 = 7, treating the product of conjugates as a sum of squares rather than a difference of squares.
    • Why not D: Multiplies the denominator by the conjugate correctly to get 7, but forgets to distribute 5 x sqrt(2) in the numerator, keeping only the 5 x 3 = 15 term and dropping the surd entirely.

  3. Question 3Answer: D

    1. For x^2 - kx + 9 = 0 to have two distinct real roots, its discriminant must be strictly positive: k^2 - 4(1)(9) > 0, i.e. k^2 > 36.
    2. Since k is an integer, k^2 > 36 means |k| >= 7 (because 6^2 = 36 is not greater than 36, but 7^2 = 49 is).
    3. Within -10 <= k <= 10, the values satisfying |k| >= 7 are k = -10, -9, -8, -7 and k = 7, 8, 9, 10: eight values in total.
    4. So there are 8 integer values of k giving two distinct real roots. Answer D.
    • Why not A: Only counts the positive values of k (7, 8, 9, 10), missing that k^2 > 36 is satisfied symmetrically by negative k as well, since the discriminant depends only on k^2.
    • Why not B: Uses k^2 >= 36 instead of the strict inequality k^2 > 36, wrongly including the boundary values k = -6 and k = 6, where the discriminant is exactly zero and the equation has a repeated root, not two distinct ones.
    • Why not C: Computes the discriminant with the wrong sign, working out k^2 - 4(1)(9) as k^2 + 36 (as if -4ac contributed positively) instead of k^2 - 36, so every integer k in the range wrongly appears to give two distinct real roots.

  4. Question 4Answer: B

    1. Set the curve equal to the line: x^2 - 4x + 1 = 3x - 5.
    2. Rearrange to one side: x^2 - 4x + 1 - 3x + 5 = 0, i.e. x^2 - 7x + 6 = 0.
    3. By Vieta's formulas, for ax^2 + bx + c = 0 the sum of the roots is -b/a; here that is -(-7)/1 = 7.
    4. So the sum of the x-coordinates of A and B is 7 (the roots are x = 1 and x = 6, and 1 + 6 = 7). Answer B.
    • Why not A: Uses Vieta's formula for the sum of roots as b/a instead of the correct -b/a; from x^2 - 7x + 6 = 0, this gives sum = -7 instead of the correct 7.
    • Why not C: Confuses the sum of the roots with their product; the product of the roots of x^2 - 7x + 6 = 0 is c/a = 6, not the sum.
    • Why not D: Makes a sign slip rearranging the equation, adding 3x instead of subtracting it when moving all terms to one side: x^2 - 4x + 1 + 3x + 5 = 0 instead of x^2 - 4x + 1 - 3x + 5 = 0, giving x^2 - x + 6 = 0 and a wrong sum of roots of 1.

  5. Question 5Answer: C

    1. By the factor theorem, since (x - 1) is a factor, f(1) = 0: 1 + a + b - 6 = 0, so a + b = 5.
    2. Since (x + 2) is a factor, f(-2) = 0: -8 + 4a - 2b - 6 = 0, so 4a - 2b = 14, i.e. 2a - b = 7.
    3. Adding a + b = 5 and 2a - b = 7 gives 3a = 12, so a = 4, and then b = 5 - 4 = 1.
    4. So f(x) = x^3 + 4x^2 + x - 6. By the remainder theorem, the remainder on division by (x - 2) is f(2) = 2^3 + 4(2)^2 + 1(2) - 6 = 8 + 16 + 2 - 6 = 20. Answer C.
    • Why not A: Substitutes x = -2 rather than x = 2 into f(x), confusing the remainder on division by (x - 2) with the value at the already-known factor (x + 2); f(-2) = 0 is the remainder for division by (x + 2), not (x - 2).
    • Why not B: Correctly finds a = 4 and b = 1, but omits the constant term -6 when evaluating f(2) = 2^3 + 4(2)^2 + 1(2) - 6, computing 8 + 16 + 2 = 26 instead of 8 + 16 + 2 - 6 = 20.
    • Why not D: Correctly finds a = 4 and b = 1, but drops the linear term when evaluating f(2), computing 8 + 16 - 6 = 18 instead of including 1(2) = 2, i.e. 8 + 16 + 2 - 6 = 20.

  6. Question 6Answer: D

    1. sqrt(6 + x - x^2) is a real number exactly when the expression under the root is non-negative: 6 + x - x^2 >= 0.
    2. Multiplying both sides by -1 and flipping the inequality gives x^2 - x - 6 <= 0.
    3. Factorise: x^2 - x - 6 = (x - 3)(x + 2), so (x - 3)(x + 2) <= 0.
    4. This product is non-positive between its roots, so -2 <= x <= 3. Answer D.
    • Why not A: Applies the 'outside the roots' rule appropriate for an upward-opening parabola, forgetting that 6 + x - x^2 has a negative leading coefficient (it opens downward), so the region where it is non-negative lies between the roots, not outside them.
    • Why not B: Drops the negative sign on the x^2 term when identifying the quadratic to solve, effectively solving x^2 + x - 6 >= 0 instead of the correct 6 + x - x^2 >= 0; this gives the wrong roots -3 and 2, and then applies the outside-the-roots rule on top of that.
    • Why not C: Makes the same sign-dropping error, finding roots -3 and 2 from x^2 + x - 6 instead of the correct 6 + x - x^2, but this time correctly applies the 'between the roots' rule for a downward parabola: the direction of reasoning is right, but the roots themselves are wrong.

  7. Question 7Answer: A

    1. Find the first differences of 2, 7, 16, 29: 7-2=5, 16-7=9, 29-16=13.
    2. Find the second differences: 9-5=4, 13-9=4, which are constant, confirming this is a quadratic sequence.
    3. Since the second difference is constant at 4, the next first difference continues the pattern by increasing by 4 again: 13+4=17.
    4. So the next term is 29+17=46. Answer A.
    • Why not B: Assumes the run of first differences (5, 9, 13) should continue by simply repeating the most recent difference (13) rather than increasing by the constant second difference of 4, giving the next term as 29 + 13 = 42.
    • Why not C: Correctly recognises the second difference should be constant, but miscalculates it as 3 instead of 4 (a slip in 13 - 9), extending the first differences as ..., 13, 16, ... and giving the next term as 29 + 16 = 45.
    • Why not D: Sets the coefficient of n^2 in the nth-term formula equal to the constant second difference itself (4) rather than half of it (2), since the correct rule is that the second difference equals 2a; this wrong coefficient propagates through solving for the remaining coefficients and evaluating the 5th term.

  8. Question 8Answer: C

    1. Using S_n=(n/2)(2a+(n-1)d): S_10=5(2a+9d)=155, so 2a+9d=31.
    2. S_20=10(2a+19d)=610, so 2a+19d=61.
    3. Subtracting the first equation from the second: 10d=30, so d=3.
    4. Substituting into 2a+9d=31: 2a+27=31, so 2a=4 and a=2. Answer C.
    • Why not A: Correctly solves the pair of equations to find the common difference d=3, but then substitutes it back using the constant from the wrong equation (61, from the S_20 equation) instead of the S_10 equation's constant (31): 2a+9(3)=61 gives a=17.
    • Why not B: Uses the formula S_n=(n/2)(a+(n-1)d), omitting the factor of 2 on the first term a, instead of the correct S_n=(n/2)(2a+(n-1)d); solving the resulting pair a+9d=31 and a+19d=61 gives d=3 and a=4.
    • Why not D: Correctly reaches 2a+9d=31 but makes a sign error isolating a, adding 27 to both sides instead of subtracting it: 2a=31+27=58, giving a=29 instead of the correct a=(31-27)/2=2.

  9. Question 9Answer: B

    1. Sum to infinity: S=a/(1-r). The condition given is a/(1-r)=3a.
    2. Since a is positive (so nonzero), divide both sides by a: 1/(1-r)=3.
    3. Take the reciprocal of both sides: 1-r=1/3.
    4. Solve for r: r=1-1/3=2/3, which does satisfy 0<r<1. Answer B.
    • Why not A: Correctly derives 1-r=1/3 but then reads off r=1/3 directly rather than subtracting from 1, missing the final step r=1-1/3=2/3.
    • Why not C: Uses the sum-to-infinity formula with the wrong sign in the denominator, a/(1+r) instead of the correct a/(1-r), so solves 1+r=1/3, giving r=1/3-1=-2/3.
    • Why not D: Treats the equation 1/(1-r)=3 as though it can be solved by setting 1-r=3 directly, rather than correctly inverting both sides to get 1-r=1/3, giving r=1-3=-2.

  10. Question 10Answer: D

    1. The general term in the expansion of (2-x)^5 is C(5,k) * 2^(5-k) * (-x)^k.
    2. For the x^3 term, take k=3: C(5,3) * 2^2 * (-x)^3.
    3. C(5,3)=10, 2^2=4, and (-x)^3=-x^3.
    4. So the term is 10 x 4 x (-1) x^3 = -40x^3, so the coefficient is -40. Answer D.
    • Why not A: Correctly computes C(5,3)=10 and 2^2=4, but drops the sign from (-x)^3=(-1)^3 x^3=-x^3, treating (-1)^3 as +1 rather than -1, giving +40 instead of -40.
    • Why not B: Pairs the power of 2 with k instead of (n-k) in the general term C(n,k) a^(n-k) b^k, using 2^3 instead of the correct 2^(5-3)=2^2 (with a=2, b=-x); this gives 10 x 8 x (-1) = -80.
    • Why not C: Miscalculates the binomial coefficient C(5,3) as 5x3=15 (multiplying n and k directly) instead of the correct value 10 (=5!/(3!2!)), giving 15 x 4 x (-1) = -60.

  11. Question 11Answer: A

    1. '3.5 x 10^4 m, correct to 2 significant figures' means the true length could lie anywhere within half a unit of the last significant figure (the 1st decimal place of 3.5) above or below the stated value.
    2. Half of that unit is 0.05 (in units of 10^4).
    3. The upper bound is therefore 3.5+0.05=3.55, i.e. 3.55 x 10^4 m.
    4. (Equivalently, 35500 m is the largest value that would round to 35000 at 2 significant figures.) Answer A.
    • Why not B: Mistakes the bound as a full unit above the last significant figure rather than half a unit, using 3.5+0.1=3.6 instead of the correct half-unit adjustment of 3.5+0.05=3.55.
    • Why not C: Computes the correct half-unit adjustment (0.05) but applies it in the wrong direction, giving the lower bound (3.5-0.05=3.45) while mislabelling it as the upper bound.
    • Why not D: Misjudges how many significant figures are being rounded to, treating the precision as if only 1 significant figure were given (rounding to the nearest 10^4, +-0.5x10^4) rather than the correct 2 significant figures (+-0.05x10^4), giving 3.5+0.5=4.0.

  12. Question 12Answer: B

    1. If HCF(m,n)=2, write m=2p and n=2q where HCF(p,q)=1.
    2. Then m x n=4pq=24, so pq=6.
    3. Since 6=2x3 has only two distinct prime factors, every ordered pair of positive divisors (p,q) with pq=6 automatically has HCF(p,q)=1: these are (1,6),(2,3),(3,2),(6,1) - four ordered pairs.
    4. Converting back, (m,n)=(2p,2q) gives (2,12),(4,6),(6,4),(12,2) - four ordered pairs in total, each with product 24 and HCF exactly 2. Answer B.
    • Why not A: Lists every ordered factor pair of 24 - (1,24),(2,12),(3,8),(4,6),(6,4),(8,3),(12,2),(24,1) - without checking the HCF(m,n)=2 condition at all, counting all 8 ordered factor pairs of 24 rather than only those with HCF exactly 2.
    • Why not C: Correctly identifies the two valid unordered pairs, {2,12} and {4,6}, but treats (m,n) and (n,m) as the same pair rather than as distinct ordered pairs, missing (12,2) and (6,4) and undercounting by half.
    • Why not D: Finds only one qualifying pair, (4,6), by spotting it directly rather than systematically listing every factor pair of 24 and checking each one's HCF, missing (2,12), (6,4) and (12,2) entirely.

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