Solved Questions

How to Use These Questions

There are 36 questions here in no fixed order, so an $x + \frac{1}{x}$ question can come right after one on linear equations. The first job every time is to name the type. Once you know it, most answers take two or three lines.

Work each one on paper before reading the answer, and give a normal Tier 1 question about 45 seconds. If a question runs well past that, find its fast route in Shortcuts & Speed Methods and redo it the next day.

About a third of the questions are on identities: squares and cubes of two letters, factorising, the three-letter identities, $a + b + c = 0$ and the special conditions such as a sum of squares equal to zero. About a quarter are on $x + \frac{1}{x}$, its special values and surd values of $x$. The rest are on polynomials, quadratic equations, linear equations, word problems and least or greatest values.

A question marked [Tier 2] needs one more step than Tier 1 usually asks for, but it uses the same identities, so do not skip it. Whenever the answer is an expression, you can test it by putting small numbers such as $x = 2$ and $y = 1$ into both the question and your answer.

Solved Examples

Question 1

If $a + b = 10$ and $a^2 + b^2 = 58$, find $a - b$, given that $a > b$.

Answer. First get $ab$. Since $(a + b)^2 = a^2 + b^2 + 2ab$, we have $100 = 58 + 2ab$, so $ab = 21$. Then $(a - b)^2 = a^2 + b^2 - 2ab = 58 - 42 = 16$. As $a > b$, take the positive root: $a - b$ = 4. (The numbers are 7 and 3.)

Question 2

If $x + \frac{1}{x} = 4$, find $x^5 + \frac{1}{x^5}$.

Answer. Build up step by step. $x^2 + \frac{1}{x^2} = 16 - 2 = 14$ and $x^3 + \frac{1}{x^3} = 64 - 12 = 52$. Now multiply these two:

$$\left(x^2 + \frac{1}{x^2}\right)\left(x^3 + \frac{1}{x^3}\right) = x^5 + \frac{1}{x^5} + x + \frac{1}{x}$$

So $x^5 + \frac{1}{x^5} = 14 \times 52 - 4 = 728 - 4$ = 724.

Question 3

When $2x^3 + ax^2 - 5x + 6$ is divided by $x - 2$, the remainder is 20. Find $a$.

Answer. By the remainder theorem, the remainder is the value of the polynomial at $x = 2$: $16 + 4a - 10 + 6 = 12 + 4a$. Setting $12 + 4a = 20$ gives $a$ = 2.

Question 4

If $x = \sqrt{7} + \sqrt{6}$, find $x^3 + \frac{1}{x^3}$.

Answer. $(\sqrt{7} + \sqrt{6})(\sqrt{7} - \sqrt{6}) = 7 - 6 = 1$, so $\frac{1}{x} = \sqrt{7} - \sqrt{6}$ and $x + \frac{1}{x} = 2\sqrt{7}$. Then

$$x^3 + \frac{1}{x^3} = (2\sqrt{7})^3 - 3 \times 2\sqrt{7} = 56\sqrt{7} - 6\sqrt{7}$$

which is $50\sqrt{7}$.

Question 5

Solve $\frac{2x - 1}{3} - \frac{x - 2}{4} = \frac{x + 4}{6}$.

Answer. Clear the fractions by multiplying everything by 12, the LCM of 3, 4 and 6: $4(2x - 1) - 3(x - 2) = 2(x + 4)$. That is $8x - 4 - 3x + 6 = 2x + 8$, so $5x + 2 = 2x + 8$, $3x = 6$ and $x$ = 2. Check: the left side is $1 - 0 = 1$, and the right side is $\frac{6}{6} = 1$.

Question 6

Simplify $\frac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y - z)^3 + (z - x)^3}$.

Answer. On top, the three brackets add up to 0, so their cubes add up to $3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)$. The bottom brackets also add up to 0, so the bottom is $3(x - y)(y - z)(z - x)$. Each $x^2 - y^2$ is $(x - y)(x + y)$, so after cancelling the 3 and the three differences we are left with $(x + y)(y + z)(z + x)$.

Question 7

Find the positive value of $k$ for which $2x^2 + kx + 18 = 0$ has equal roots, and find that root.

Answer. Equal roots means $b^2 = 4ac$: $k^2 = 4 \times 2 \times 18 = 144$, so $k = 12$ (taking the positive value). The equation becomes $2x^2 + 12x + 18 = 2(x + 3)^2 = 0$, so the root is $x = -3$.

Question 8

If $a + b = 11$ and $a^3 + b^3 = 737$, find $ab$ and $a^2 + b^2$.

Answer. Use $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$: $1331 = 737 + 33ab$, so $33ab = 594$ and $ab$ = 18. Then $a^2 + b^2 = 121 - 36$ = 85. (The numbers are 2 and 9, and $8 + 729 = 737$.)

Question 9

If $x + \frac{1}{x} = 1$, find $x^{2026} + \frac{1}{x^{2026}}$.

Answer. $x + \frac{1}{x} = 1$ means $x^2 - x + 1 = 0$, and multiplying by $x + 1$ gives $x^3 + 1 = 0$. So $x^3 = -1$ and $x^6 = 1$. Divide 2026 by 6: $2026 = 6 \times 337 + 4$, so $x^{2026} = x^4 = x^3 \times x = -x$. In the same way $\frac{1}{x^{2026}} = -\frac{1}{x}$. Adding, the value is $-\left(x + \frac{1}{x}\right)$ = $-1$.

Question 10

Simplify $\frac{x^2 - 9}{x^2 + x - 12} \div \frac{x^2 + 6x + 9}{x^2 + 2x - 8}$.

Answer. Factorise everything first: $x^2 - 9 = (x - 3)(x + 3)$, $x^2 + x - 12 = (x + 4)(x - 3)$, $x^2 + 6x + 9 = (x + 3)^2$ and $x^2 + 2x - 8 = (x + 4)(x - 2)$. Dividing means turning the second fraction over:

$$\frac{(x - 3)(x + 3)}{(x + 4)(x - 3)} \times \frac{(x + 4)(x - 2)}{(x + 3)^2} = \frac{x - 2}{x + 3}$$

So the answer is $\frac{x - 2}{x + 3}$.

Question 11

If $a + b + c = 11$, $ab + bc + ca = 38$ and $abc = 40$, find $a^2 + b^2 + c^2$ and $a^3 + b^3 + c^3$.

Answer. $a^2 + b^2 + c^2 = 11^2 - 2 \times 38 = 121 - 76$ = 45. For the cubes,

$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 11 \times (45 - 38) = 77$$

So $a^3 + b^3 + c^3 = 77 + 3 \times 40$ = 197. (The numbers are 2, 4 and 5.)

Question 12

Find the least value of $x^2 + y^2 - 6x + 8y + 30$.

Answer. Complete the square for each letter: $x^2 - 6x = (x - 3)^2 - 9$ and $y^2 + 8y = (y + 4)^2 - 16$. So the expression is $(x - 3)^2 + (y + 4)^2 + 30 - 25 = (x - 3)^2 + (y + 4)^2 + 5$. Squares cannot be negative, so the least value is 5, at $x = 3$ and $y = -4$.

Question 13

A two-digit number is 6 times the sum of its digits. If 9 is taken away from it, the digits swap places. Find the number.

Answer. Let the tens digit be $a$ and the units digit $b$. The first fact gives $10a + b = 6(a + b)$, so $4a = 5b$. Taking 9 away swaps the digits, so the number falls by $9(a - b) = 9$, and $a - b = 1$. Put $a = b + 1$ into $4a = 5b$: $4b + 4 = 5b$, so $b = 4$ and $a = 5$, and the number is 54. Check: $6 \times (5 + 4) = 54$, and $54 - 9 = 45$.

Question 14

If $2x + \frac{1}{3x} = 4$, find $36x^2 + \frac{1}{x^2}$.

Answer. The two terms are not reciprocals, so square and watch the middle term:

$$\left(2x + \frac{1}{3x}\right)^2 = 4x^2 + \frac{1}{9x^2} + 2 \times 2x \times \frac{1}{3x} = 4x^2 + \frac{1}{9x^2} + \frac{4}{3}$$

So $4x^2 + \frac{1}{9x^2} = 16 - \frac{4}{3} = \frac{44}{3}$. The question wants 9 times this: $36x^2 + \frac{1}{x^2} = 9 \times \frac{44}{3}$ = 132.

Question 15

If $a^2 + b^2 + c^2 = ab + bc + ca$, and $a$, $b$, $c$ are not all zero, find $\frac{(a + b + c)^2}{a^2 + b^2 + c^2}$.

Answer. Double both sides and move everything to one side: $(a - b)^2 + (b - c)^2 + (c - a)^2 = 0$. A sum of squares is 0 only when each square is 0, so $a = b = c$. Then the top is $(3a)^2 = 9a^2$ and the bottom is $3a^2$, and the value is 3.

Question 16 [Tier 2]

If $\alpha$, $\beta$ and $\gamma$ are the zeros of $x^3 - 7x^2 + 14x - 8$, find $\alpha^2 + \beta^2 + \gamma^2$ and $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$.

Answer. For $ax^3 + bx^2 + cx + d$, the sum of the zeros is $-\frac{b}{a} = 7$, the sum of their products in pairs is $\frac{c}{a} = 14$, and their product is $-\frac{d}{a} = 8$. So $\alpha^2 + \beta^2 + \gamma^2 = 7^2 - 2 \times 14 = 49 - 28$ = 21, and $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{14}{8}$ = $\frac{7}{4}$. (The zeros are 1, 2 and 4.)

Question 17

If $x = \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} - \sqrt{5}}$ and $y = \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} + \sqrt{5}}$, find $x^2 - xy + y^2$.

Answer. $y$ is just $x$ turned upside down, so $xy = 1$. Rationalising, $x = \frac{(\sqrt{7} + \sqrt{5})^2}{7 - 5} = \frac{12 + 2\sqrt{35}}{2} = 6 + \sqrt{35}$, and so $y = 6 - \sqrt{35}$. Then $x + y = 12$ and $x^2 + y^2 = 144 - 2 = 142$, so $x^2 - xy + y^2 = 142 - 1$ = 141.

Question 18

For what value of $k$ do the equations $kx + 4y = 6$ and $9x + ky = 9$ have no solution?

Answer. No solution needs $\frac{k}{9} = \frac{4}{k}$, so $k^2 = 36$ and $k = 6$ or $k = -6$. Now check the constants. For $k = 6$: $\frac{6}{9} = \frac{4}{6} = \frac{6}{9}$, so all three ratios are equal, the two lines are the same, and there are infinitely many solutions. For $k = -6$: $\frac{-6}{9} = \frac{4}{-6} = -\frac{2}{3}$, but $\frac{6}{9} = \frac{2}{3}$, so there is no solution. The answer is $k = -6$.

Question 19

If $a - b = 3$ and $a^3 - b^3 = 117$, find $a + b$, given that $a$ and $b$ are positive.

Answer. Use $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$: $27 = 117 - 9ab$, so $ab = 10$. Then $(a + b)^2 = (a - b)^2 + 4ab = 9 + 40 = 49$, and since both numbers are positive, $a + b$ = 7. (They are 5 and 2.)

Question 20

If $x + \frac{1}{x} = 9$ and $x > 1$, find $x^2 - \frac{1}{x^2}$.

Answer. First get $x - \frac{1}{x}$. Its square is $\left(x + \frac{1}{x}\right)^2 - 4 = 81 - 4 = 77$. As $x > 1$, $x - \frac{1}{x}$ is positive, so it is $\sqrt{77}$. Then $x^2 - \frac{1}{x^2} = \left(x + \frac{1}{x}\right)\left(x - \frac{1}{x}\right)$ = $9\sqrt{77}$.

Question 21

If $a^3 + 8b^3 + 125c^3 = 30abc$ and $a + 2b + 5c \neq 0$, find $a : b : c$.

Answer. Write it as $a^3 + (2b)^3 + (5c)^3 = 3 \times a \times 2b \times 5c$; this works because $3 \times 2 \times 5 = 30$. That is $p^3 + q^3 + r^3 = 3pqr$, so either $p + q + r = 0$ or $p = q = r$. The question rules out the first, so $a = 2b = 5c$. Take this common value as 10: $a = 10$, $b = 5$, $c = 2$. So $a : b : c$ = 10 : 5 : 2.

Question 22

If $a^3 + 3a^2 + 3a = 26$ and $a$ is real, find $a^2 + 2a$.

Answer. The left side is one short of a cube: $a^3 + 3a^2 + 3a + 1 = (a + 1)^3$. Adding 1 to both sides, $(a + 1)^3 = 27$, so $a + 1 = 3$ and $a = 2$. Then $a^2 + 2a = 4 + 4$ = 8. Quicker still: $a^2 + 2a = (a + 1)^2 - 1 = 9 - 1$.

Question 23 [Tier 2]

If $\alpha$ and $\beta$ are the roots of $x^2 - 8x + 3 = 0$, find the quadratic equation whose roots are $\alpha^2$ and $\beta^2$.

Answer. $\alpha + \beta = 8$ and $\alpha\beta = 3$. The new sum is $\alpha^2 + \beta^2 = 64 - 6 = 58$, and the new product is $\alpha^2\beta^2 = 9$. An equation with a given sum and product of roots is $x^2 - (\text{sum})x + \text{product} = 0$, so the answer is $x^2 - 58x + 9 = 0$.

Question 24

The perimeter of a rectangle is 46 cm and its area is 120 square cm. Find the length of its diagonal.

Answer. Call the sides $l$ and $b$. Then $l + b = 23$ and $lb = 120$. The diagonal is $\sqrt{l^2 + b^2}$, and $l^2 + b^2 = 23^2 - 2 \times 120 = 529 - 240 = 289$. So the diagonal is $\sqrt{289}$ = 17 cm. (The sides are 15 cm and 8 cm.)

Question 25

If $x^4 + \frac{1}{x^4} = 3842$ and $x > 0$, find $x^3 + \frac{1}{x^3}$.

Answer. Go backwards. $\left(x^2 + \frac{1}{x^2}\right)^2 = 3842 + 2 = 3844 = 62^2$, so $x^2 + \frac{1}{x^2} = 62$. Then $\left(x + \frac{1}{x}\right)^2 = 62 + 2 = 64$, and since $x > 0$, $x + \frac{1}{x} = 8$. Finally $x^3 + \frac{1}{x^3} = 512 - 24$ = 488.

Question 26

If $p + q + r = 0$, and they are not all zero, find $\frac{p^2 + q^2 + r^2}{pq + qr + rp}$.

Answer. Square the condition: $p^2 + q^2 + r^2 + 2(pq + qr + rp) = 0$. So $p^2 + q^2 + r^2 = -2(pq + qr + rp)$, and the fraction is $-2$. A quick check with $p = 1$, $q = 1$, $r = -2$: the top is 6 and the bottom is $1 - 2 - 2 = -3$.

Question 27 [Tier 2]

If $x = \sqrt{5} - 1$, find $x^3 + 3x^2 - 2x + 4$.

Answer. Make the equation first: $x + 1 = \sqrt{5}$, so $x^2 + 2x + 1 = 5$, that is $x^2 + 2x - 4 = 0$. Now split the cubic using this: $x^3 + 3x^2 - 2x + 4 = (x^2 + 2x - 4)(x + 1) + 8$. You can check it by multiplying out: $(x^2 + 2x - 4)(x + 1) = x^3 + 3x^2 - 2x - 4$. The bracket is 0, so the value is 8.

Question 28 [Tier 2]

2 notebooks, 3 pens and 1 file cost ₹95, and 1 notebook, 2 pens and 3 files cost ₹90. What do 5 notebooks, 8 pens and 5 files cost?

Answer. There are three prices and only two lines, so we cannot find each price. But we may not need to. Try to build the asked line from the given ones: twice the first line is 4 notebooks, 6 pens and 2 files, and adding the second line gives 5 notebooks, 8 pens and 5 files. That is exactly what is asked, so the cost is $2 \times 95 + 90$ = ₹280.

Question 29

Simplify $(x + y)^3 - (x - y)^3 - 6y(x^2 - y^2)$.

Answer. In $(x + y)^3 - (x - y)^3$, the $x^3$ terms cancel and so do the $3xy^2$ terms, while the other two terms double: it is $6x^2y + 2y^3$. Taking away $6y(x^2 - y^2) = 6x^2y - 6y^3$ leaves $2y^3 + 6y^3$ = $8y^3$. A quick check with $x = 1$, $y = 1$: $8 - 0 - 0 = 8$.

Question 30 [Tier 2]

If $x$ and $y$ are positive and $3x + 4y = 48$, find the greatest value of $xy$.

Answer. For a fixed sum, a product is greatest when the two parts are equal. Here the sum $3x + 4y$ is fixed, so $(3x)(4y)$ is greatest when $3x = 4y = 24$. Then $12xy \le 24 \times 24 = 576$, so $xy \le 48$. The greatest value is 48, at $x = 8$ and $y = 6$.

Question 31 [Tier 2]

If $x + \frac{1}{x} = \sqrt{2}$, find $x^{10} + \frac{1}{x^{10}}$.

Answer. Square the given line: $x^2 + \frac{1}{x^2} = 2 - 2 = 0$. Multiplying by $x^2$ gives $x^4 + 1 = 0$, so $x^4 = -1$ and $x^8 = 1$. Then $x^{10} = x^8 \times x^2 = x^2$, and in the same way $\frac{1}{x^{10}} = \frac{1}{x^2}$. So the value is $x^2 + \frac{1}{x^2}$, which we found is 0.

Question 32

Simplify $\frac{a^4 - b^4}{a^3 + a^2b + ab^2 + b^3}$.

Answer. Factorise the top as a difference of squares twice: $a^4 - b^4 = (a^2 + b^2)(a + b)(a - b)$. Group the bottom in pairs: $a^2(a + b) + b^2(a + b) = (a + b)(a^2 + b^2)$. Cancelling the common factors, the answer is $a - b$.

Question 33

The product of two consecutive positive even numbers is 168. Find their sum.

Answer. Call them $n$ and $n + 2$. Then $n(n + 2) = 168$, so $n^2 + 2n - 168 = 0$. Two numbers that multiply to $-168$ and add up to 2 are 14 and $-12$, so $(n + 14)(n - 12) = 0$. As the numbers are positive, $n = 12$, so they are 12 and 14, and their sum is 26.

Question 34 [Tier 2]

If $a + b + c = 12$, $a^2 + b^2 + c^2 = 66$ and $a^3 + b^3 + c^3 = 408$, find $abc$.

Answer. First $ab + bc + ca = \frac{144 - 66}{2} = 39$. Then $a^3 + b^3 + c^3 - 3abc = 12 \times (66 - 39) = 324$. So $3abc = 408 - 324 = 84$, and $abc$ = 28. (The numbers are 1, 4 and 7.)

Question 35

If $x = 3 + \sqrt{5}$ and $y = 3 - \sqrt{5}$, find $\frac{x}{y} + \frac{y}{x}$.

Answer. Here $x$ and $y$ are not reciprocals, because $xy = 9 - 5 = 4$, not 1. So work with the sum and the product: $x + y = 6$ and $xy = 4$. Then $x^2 + y^2 = 36 - 8 = 28$, and $\frac{x}{y} + \frac{y}{x} = \frac{x^2 + y^2}{xy} = \frac{28}{4}$ = 7.

Question 36

If $x + 2y = 12$, $y + 2z = 11$ and $z + 2x = 7$, find $x^2 + y^2 + z^2$.

Answer. Each letter appears 3 times in all, so add the three lines: $3(x + y + z) = 30$ and $x + y + z = 10$. Now take the total away from the lines: $(x + 2y) - 10$ gives $y - z = 2$, and $(y + 2z) - 10$ gives $z - x = 1$. So $z = x + 1$ and $y = z + 2 = x + 3$. Then $x + (x + 3) + (x + 1) = 10$ gives $x = 2$, so $y = 5$ and $z = 3$. The answer is $4 + 25 + 9$ = 38.