Question 1Algebra
TMUA Algebra — Practice Questions by Topic
These are SummitPapers original questions, not official past paper questions. Official TMUA questions sorted by topic are on TMUA past papers by topic.
- M4Algebra
- 25 questions11 on Paper 1 · 14 on Paper 2
- 8 free solutionsThe rest show the correct letter only
Covers: inequalities in one variable, remainders and factors, equations that include a modulus, counting the integer solutions of an inequality.
What this topic tests
The Part 2 algebra topic is narrower than algebra and functions. The questions here are inequalities in one variable, remainders and factors, equations with a modulus, and counting the integer solutions of an inequality.
How it is assessed
This is Section 1, on both papers in this set. Each question has five options. A paper is 20 questions in 75 minutes, with no calculator.
This note is written for SummitPapers. The official list of what can be examined is the specification, together with the Notes on Mathematics and, for Paper 2, the Notes on Logic and Proof.
Key methods
An inequality
Multiplying by a negative number reverses the inequality. Multiplying by a quantity that might be negative splits into cases. A squared expression is at least zero, which often settles a comparison.
Integer solutions
List the integers inside the interval, and check an endpoint if the inequality is not strict. The count is not the length of the interval.
Common mistakes
Dividing by the variable
Dividing an inequality by x assumes x is positive. The case x < 0 reverses the sign, and x = 0 has to be checked on its own.
Treating a modulus as always positive in the wrong place
The modulus of an expression is non-negative. The expression inside can still be negative, and the equation splits there.
Worked example
This question is also in the list below, with the solution folded. It is opened here so the method is on the page before the other questions.
Suppose x and y are real, x² + y² = 73, x + y = 11 and x > y. Find x − y.
Answer: D. 5
Worked solution. Square the sum: (x + y)² = 121 = x² + 2xy + y² = 73 + 2xy, so 2xy = 48. Then (x − y)² = x² − 2xy + y² = 73 − 48 = 25. Since x > y, x − y = 5.
Why the other options look right. A reports the given sum x + y = 11 instead of working out x − y. B finds (x − y)² = 73 − 48 = 25 and forgets to take the square root. C solves x + y = 11 and x − y = 5 to get x = 8 and reports x instead of x − y. E stops after finding 2xy = 121 − 73 = 48 and reports it instead of going on to (x − y)².
Paper 1 questions
Paper 1 is Applications of Mathematical Knowledge. Calculators are not allowed.
Q1. Suppose x and y are real, x² + y² = 73, x + y = 11 and x > y. Find x − y.
Free · worked solution included
Answer and worked solution
Answer: D. 5
Worked solution. Square the sum: (x + y)² = 121 = x² + 2xy + y² = 73 + 2xy, so 2xy = 48. Then (x − y)² = x² − 2xy + y² = 73 − 48 = 25. Since x > y, x − y = 5.
Why the other options look right. A reports the given sum x + y = 11 instead of working out x − y. B finds (x − y)² = 73 − 48 = 25 and forgets to take the square root. C solves x + y = 11 and x − y = 5 to get x = 8 and reports x instead of x − y. E stops after finding 2xy = 121 − 73 = 48 and reports it instead of going on to (x − y)².
Q2. What is the sum of all positive real solutions of √x + 12/√x = 7?
Free · correct letter only
Answer
Answer: D. 25
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Q3. What is the coefficient of x2 y in the expansion of (1 + x + 2y)5?
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Answer
Answer: D. 60
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Q4. Which of the following is the complete set of real values of x for which x − 2 < 3/x?
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Answer
Answer: C. x < −1 or 0 < x < 3
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Q5. The product (3x2 − 5x + 4)(a x + 2) is divided by x − 1, and the remainder is 30. What is a?
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Answer
Answer: E. 13
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Q6. The real numbers x and y satisfy y + |x − 3| ≤ 2 + |x| and 3|x| ≤ y + 6. What is the greatest possible value of |x y|?
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Answer
Answer: D. 55/3
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Q7. The numbers a and b are positive integers. When x2 − 2 a x − a2 is divided by x − b, the remainder is −4. Also, 2x − a is a factor of 4 b x2 − 6x − 10. What is a + b?
Free · correct letter only
Answer
Answer: D. 6
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Q8. For each positive integer n, the polynomial qn(x) is the sum of the n terms k(x + 2) − 3n for k = 1, 2, …, n. For n ≥ 2, what is the remainder when qn(x) is divided by qn−1(x)?
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Answer
Answer: E. −3
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Q9. Positive integers a, b and c satisfy abc + ab + bc + ca + a + b + c = 29. What is (a + 1)² + (b + 1)² + (c + 1)²?
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Answer
Answer: D. 38
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Q10. Simplify √(11 − 6√2).
Free · correct letter only
Answer
Answer: C. 3 − √2
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Q11. Positive numbers a and d satisfy a² + d² = 13 and ad = 6. What is a + d?
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Answer
Answer: A. 5
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Paper 2 questions
Paper 2 is Mathematical Reasoning. It can test this same topic. Argument, proof, and identifying errors are the topics that appear on Paper 2 only.
Q1. Which statement is true at every point of the region y > x2 and y < 2x + 3?
Free · worked solution included
Answer and worked solution
Answer: B. y < 9
Worked solution. At every point of the region x2 < y < 2x + 3, so x2 < 2x + 3, that is (x − 3)(x + 1) < 0, and −1 < x < 3. Then y < 2x + 3 < 2(3) + 3 = 9, so y < 9 everywhere in the region. The point (−0.5, 0.3) is in the region, because 0.3 > 0.25 and 0.3 < 2, but it has x < 0 and x + y = −0.2 < 0. The point (0, 0.5) is in the region, because 0.5 > 0 and 0.5 < 3, but y < 1 there. Only y < 9 holds at every point.
Why the other options look right. A fails at (−0.5, 0.3): the region extends to the left of the y-axis, as far as x = −1, where the parabola meets the line. C fails at (0, 0.5): near the vertex of the parabola y can be close to 0. D fails at (−0.5, 0.3), where x + y = −0.2. E says none of the statements holds everywhere, but B does, because x < 3 forces y < 2x + 3 < 9.
Q2. The real numbers p, q and r satisfy p < q < r. Which of the following must be true? I. p2 < q2 < r2. II. p + q + r > 0. III. pr < qr.
Free · worked solution included
Answer and worked solution
Answer: A. none of them
Worked solution. Take p = −4, q = −2 and r = −1. Then I fails because 16 > 4. II fails because the sum is −7. III fails because pr = 4 and qr = 2, so pr < qr is false. None of the three statements holds for every triple with p < q < r.
Why the other options look right. B relies on I, but squares reverse order for these negative numbers. C relies on II, but the sum of three negative numbers is negative. D relies on III, but multiplying the inequality p < q by the negative number r reverses it. E combines I and III, and the same triple is a counterexample to both.
Q3. How many positive integers n satisfy n² < 6n + 16?
Free · worked solution included
Answer and worked solution
Answer: D. 7
Worked solution. Rearrange to n² − 6n − 16 < 0, which factors as (n − 8)(n + 2) < 0. The product is negative for −2 < n < 8. The positive integers in that range are 1, 2, 3, 4, 5, 6 and 7, so there are 7. At n = 8 the two sides are equal, so the strict inequality fails.
Why the other options look right. A replaces the strict inequality by ≤ and includes n = 8. B counts every integer from −1 to 7, which is nine integers, and drops the condition that n is positive. C drops +16 and solves n² < 6n, so 1 ≤ n ≤ 5. E drops the term 6n and solves n² < 16, so n = 1, 2 and 3.
Q4. Which statement is not true for all real numbers x and y?
Free · worked solution included
Answer and worked solution
Answer: E. x < y implies x² < xy
Worked solution. The statement x < y implies x² < xy fails when x = −2 and y = 1: here −2 < 1, but x² = 4 and xy = −2, and 4 < −2 is false. (In general x² − xy = x(x − y), which is negative for x < y only when x > 0.) The other four statements hold for all real x and y. In particular, (x + y)² ≥ 4xy rearranges to (x − y)² ≥ 0, and x > y > 0 does force x² > y² because both numbers are positive.
Why the other options look right. A is equivalent to (x − y)² ≥ 0. B is the reverse triangle inequality. C holds because a sum of squares of real numbers is zero only when each square is zero. D holds because squaring preserves order for positive real numbers.
Q5. For how many integers n, positive, negative or zero, is (n² + 11)/(n + 1) an integer?
Free · worked solution included
Answer and worked solution
Answer: C. 12
Worked solution. Divide first: n² + 11 = (n + 1)(n − 1) + 12, so (n² + 11)/(n + 1) = n − 1 + 12/(n + 1). Since n − 1 is an integer, the fraction is an integer exactly when n + 1 divides 12. The integer divisors of 12 are ±1, ±2, ±3, ±4, ±6 and ±12, which is 12 values, and each gives one integer n = d − 1 (none of them is n = −1, because d ≠ 0). So there are 12 such integers n, from n = −13 up to n = 11.
Why the other options look right. A uses only the positive divisors 1, 2, 3, 4, 6 and 12 of 12, forgetting that n + 1 can be negative. B expands (n + 1)(n − 1) as n² + 1, so it takes the remainder as 10 and counts the 8 integer divisors of 10. D treats n² as if n + 1 divided it, so it requires n + 1 to divide 11 and counts the 4 divisors ±1 and ±11. E counts only positive n, which drops n = 0 from the six positive divisors and leaves 5.
Q6. How many positive integers n satisfy n² < 5n + 6?
Free · worked solution included
Answer and worked solution
Answer: D. 5
Worked solution. Rearrange to n² − 5n − 6 < 0, or (n − 6)(n + 1) < 0. The quadratic is negative strictly between the roots, so −1 < n < 6. The positive integers in that range are 1, 2, 3, 4 and 5, five numbers in all. At n = 6 the two sides are equal (36 = 36), so the strict inequality excludes 6.
Why the other options look right. A drops the constant 6 and solves n² < 5n, so n < 5 and n = 1, 2, 3, 4. B includes n = 6, where the two sides are equal. C counts every integer from −1 to 5 inclusive, keeping the root −1 and the value 0, which are not positive. E drops the 5n term and solves n² < 6, giving only n = 1 and 2.
Q7. Find the number of positive integers n for which n² is strictly less than 4n + 5.
Free · worked solution included
Answer and worked solution
Answer: E. 4
Worked solution. Rearrange to n² − 4n − 5 < 0, so (n − 5)(n + 1) < 0. The product is negative when −1 < n < 5. The positive integers in this range are 1, 2, 3 and 4, so there are 4. At n = 5 the two sides of the original inequality are equal, and the inequality is strict.
Why the other options look right. A treats the inequality as n² ≤ 4n + 5 and includes n = 5, where the two sides are equal. B drops the 4n term and solves n² < 5, which gives only n = 1 and n = 2. C drops the constant 5 and solves n² < 4n, that is n < 4, which gives n = 1, 2 and 3. D factorises n² − 4n − 5 as (n + 5)(n − 1), so it gets −5 < n < 1, which contains no positive integer.
Q8. The real numbers p, q and r satisfy 0 < p < q < r < 1. Which of the following inequalities must be true? I: pq < pr. II: p + q < r + 1. III: q2 > pr.
Free · correct letter only
Answer
Answer: D. I and II only
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Q9. Positive integers p and q have product 360 and greatest common divisor 6. What is the least possible value of p + q?
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Answer
Answer: C. 42
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Q10. The polynomial p(x) = x4 + a x3 + b x2 + c x − 2, where a, b and c are constants, leaves remainder 3x + 5 when it is divided by x2 − 1. What is b?
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Answer
Answer: A. 6
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Q11. A region R consists of the points (x, y) satisfying both y < x + 2 and x + 2y < 8. Consider the four statements: (1) x < 8; (2) y < 4; (3) x + y < 10; (4) y < x + 3. Which of these statements are true for every point of R?
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Answer
Answer: C. 2 and 4 only
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Q12. How many integer pairs (x, y) satisfy −2 ≤ x ≤ 3, 1 ≤ y ≤ 4 and (x − 1)(y − 2) ≤ 2?
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Answer
Answer: D. 22
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Q13. Let n be a positive integer. For which n is x2 + 3 a factor of (x2 + 1)n + (x2 + 5)n?
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Answer
Answer: A. every odd positive integer n
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Q14. The positive integers x and y satisfy x² − y = 2 and y² − x = 2. What is x + y?
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Answer
Answer: C. 4
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