Elementary Mathematics 2017 (I)

Set A

Questions 1–20 of 100

1Algebra • Surds • Simplification

ModerateIdeal time: about 1.5 min

What is $\frac{0.064 \times 6.25}{0.081 \times 4.84}$ equal to?

\frac{10}{99}

\frac{100}{99}

9

99

Show answer & solution

Option B

\frac{100}{99}

Approach

Write every decimal as a fraction, reduce the quotient to a single fraction, and take the square root of numerator and denominator separately.

Prime factorisation

Step-by-step solution

1
$0.064 \times 6.25 = \frac{64}{1000} \times \frac{625}{100} = \frac{40000}{100000}$
Why: Decimals stay exact as fractions, and the powers of $10$ can then be cancelled at the end rather than tracked through a decimal multiplication.
Prime factorisation
2
$0.081 \times 4.84 = \frac{81}{1000} \times \frac{484}{100} = \frac{39204}{100000}$
Why: The same treatment for the denominator, so both products carry the same power of $10$ .
3
$\frac{0.064 \times 6.25}{0.081 \times 4.84} = \frac{40000}{39204} = \frac{10000}{9801}$
Why: The factor $1 0^{5}$ cancels; dividing both parts by $4$ leaves the fraction in lowest terms.
4
$10000 = 10 0^{2}$ and $9801 = 3^{4} \times 1 1^{2} = 9 9^{2}$ , so $\frac{10000}{9801} = \frac{100}{99}$
Why: Both parts are perfect squares, so the root is exact and no decimal approximation is needed.
denotes the non-negative root $ab = a b$ and $a / b = a / b$

Common mistakes

Option A

Lost one power of $10$ when converting $0.064$ , leaving $\frac{100}{9801}$ under the root and $\frac{10}{99}$ outside it.

Option C

No route reaches $9$ ; the quotient inside the root is barely above $1$ .

Option D

Took the root of the denominator only, $9801 = 99$ , and reported it as the whole value.

Skills used

Prime factorisation

denotes the non-negative root

ab = a b

and

a / b = a / b

2Algebra • Polynomial • Factor and Remainder Theorem

EasyIdeal time: about 55 s

$(x + 4)$ is a factor of which one of the following expressions?

x^{2} - 7 x + 44

x^{2} + 7 x - 44

x^{2} - 7 x - 44

x^{2} + 7 x + 44

Show answer & solution

Option C

x^{2} - 7 x - 44

Approach

Any quadratic with $(x + 4)$ as a factor is $(x + 4) (x + m)$ , which ties its constant term to its middle coefficient.

Factorise a quadratic by splitting the middle term

Step-by-step solution

1
$(x + 4) (x + m) = x^{2} + (4 + m) x + 4 m$
Why: This is the shape every option must have if $(x + 4)$ divides it, so the constant term fixes $m$ and $m$ then fixes the coefficient of $x$ .
Factorise a quadratic by splitting the middle term
2
constant $= - 44 \Rightarrow 4 m = - 44 \Rightarrow m = - 11 \Rightarrow 4 + m = - 7$
Why: Of the two constants on offer, $- 44$ forces the middle term $- 7 x$ .
3
constant $= + 44 \Rightarrow m = 11 \Rightarrow 4 + m = 15$ , giving $x^{2} + 15 x + 44$
Why: The other constant forces a middle term of $+ 15 x$ , which is not printed on any option.
4
$x^{2} - 7 x - 44 = (x + 4) (x - 11)$
Why: Only this expression pairs the constant $- 44$ with the middle coefficient $- 7$ that $(x + 4)$ demands.

Shortcut

Apply the factor theorem directly: $(x + 4)$ is a factor exactly when substituting $x = - 4$ gives $0$ , and only $x^{2} - 7 x - 44$ does.

Common mistakes

Option A

Kept the middle term $- 7 x$ but flipped the constant's sign; $4 m = 44$ needs $m = 11$ , which would give $+ 15 x$ .

Option B

$x^{2} + 7 x - 44 = (x - 4) (x + 11)$ — the sign of the root was misread, so $(x - 4)$ came out as the factor.

Option D

$x^{2} + 7 x + 44$ has discriminant $49 - 176 < 0$ and no real linear factor at all.

Skills used

Factorise a quadratic by splitting the middle term

3Algebra • Quadratic Equations • Roots

ModerateIdeal time: about 1.5 min

If $α$ and $β$ are the roots of the quadratic equation $2 x^{2} + 6 x + k = 0$ , where $k < 0$ , then what is the maximum value of $\frac{α}{β} + \frac{β}{α}$ ?

2

- 2

9

- 9

Show answer & solution

Option B

- 2

Approach

Rewrite $\frac{α}{β} + \frac{β}{α}$ in terms of the sum and product of the roots, express it in $k$ , and see how large it can get for $k < 0$ .

Sum and product of roots (Vieta)

Step-by-step solution

1
$α + β = - \frac{6}{2} = - 3$ and $α β = \frac{k}{2}$
Why: Vieta's relations give both symmetric functions without solving the equation.
Sum and product of roots (Vieta)
2
$\frac{α}{β} + \frac{β}{α} = \frac{α ^{2} + β ^{2}}{α β} = \frac{( α + β ) ^{2} - 2 α β}{α β} = \frac{( α + β ) ^{2}}{α β} - 2$
Why: The expression is symmetric, so it can be written using only $α + β$ and $α β$ .
Square of a sum and of a difference
3
$\frac{( α + β ) ^{2}}{α β} - 2 = \frac{9}{k /2} - 2 = \frac{18}{k} - 2$
Why: Everything is now a function of the single parameter $k$ .
4
$k < 0 \Rightarrow \frac{18}{k} < 0$ , and $\frac{18}{k} \to 0$ as $k \to - \infty$ , so the value runs over $(- \infty, - 2)$
Why: The expression increases towards $- 2$ as $k$ becomes more negative, so $- 2$ is the least upper bound and the largest value on offer.

Common mistakes

Option A

Quoted $\frac{α}{β} + \frac{β}{α} \geq 2$ from memory; that bound needs $α β > 0$ , which $k < 0$ rules out.

Option C

Stopped at $(α + β)^{2} = 9$ without dividing by $α β$ or subtracting $2$ .

Option D

Computed $9$ as above and then attached a minus sign on seeing $k < 0$ .

Skills used

Sum and product of roots (Vieta)Square of a sum and of a difference

4Arithmetic • Number System • LCM and HCF

HardIdeal time: about 2 min

Consider the following statements:

1: If $a = b c$ with HCF $(b, c) = 1$ , then HCF $(c, b d) =$ HCF $(c, d)$ .

2: If $a = b c$ with HCF $(b, c) = 1$ , then LCM $(a, d) =$ LCM $(c, b d)$ .

Which of the above statements is/are correct?

A1 onlyB2 onlyCBoth 1 and 2DNeither 1 nor 2

Show answer & solution

Option A

1 only

Approach

Compare prime exponents on the two sides of statement 1, and test statement 2 on the smallest coprime pair.

Prime factorisation

Step-by-step solution

1
$HCF (b, c) = 1$ means no prime divides both $b$ and $c$
Why: This is the only use made of $a = b c$ : it is what lets $b$ be removed from a computation involving $c$ .
Coprime means no shared prime factor
2
For each prime $p$ : $min (v_{p} (c), v_{p} (b) + v_{p} (d)) = min (v_{p} (c), v_{p} (d))$ , since $v_{p} (c) > 0$ forces $v_{p} (b) = 0$
Why: The HCF takes the minimum exponent of each prime; where $c$ contributes, $b$ contributes nothing, and where $c$ does not, both minima are $0$ . Hence $HCF (c, b d) = HCF (c, d)$ and statement 1 is correct.
Prime factorisation
3
Statement 2 with $b = 2$ , $c = 3$ (so $a = 6$ ) and $d = 2$ : $LCM (a, d) = LCM (6, 2) = 6$
Why: $HCF (2, 3) = 1$ , so this is an admissible case of the hypothesis.
4
$LCM (c, b d) = LCM (3, 4) = 12 \neq = 6$
Why: One counterexample is enough to refute a universal claim, so statement 2 is not correct.

Common mistakes

Option B

Accepted the LCM claim by analogy with the HCF one and rejected statement 1 as too strong.

Option C

Assumed statement 2 mirrors statement 1, since the coprime splitting of $a$ does work for the HCF.

Option D

Rejected statement 1 as well, typically by testing a $d$ sharing a factor with $b$ and misreading the result.

Skills used

Prime factorisation Coprime means no shared prime factor

5Algebra • Surds • Simplification

EasyIdeal time: about 1 min

What is the square root of $\frac{( 0.35 ) ^{2} + 0.70 + 1}{2.25} + 0.19$ ?

1

2

3

4

Show answer & solution

Option A

1

Approach

Recognise the numerator as a perfect square, divide, then add and take the root.

Square of a sum and of a difference

Step-by-step solution

1
$(0.35)^{2} + 0.70 + 1 = (0.35)^{2} + 2 (0.35) + 1 = (0.35 + 1)^{2} = (1.35)^{2}$
Why: $0.70$ is exactly twice $0.35$ , which is the signature of $(a + 1)^{2}$ .
Square of a sum and of a difference
2
$\frac{( 1.35 ) ^{2}}{2.25} = \frac{( 1.35 ) ^{2}}{( 1.5 ) ^{2}} = (\frac{1.35}{1.5})^{2} = (0.9)^{2} = 0.81$
Why: $2.25$ is also a perfect square, so the quotient stays a square and no long division is needed.
3
$0.81 + 0.19 = 1$
Why: The stray $0.19$ is there to complete the sum to $1$ .
denotes the non-negative root
4
$1 = 1$
Why: The square root of $1$ is $1$ .

Common mistakes

Option B

No route reaches $2$ ; the quantity under the root is exactly $1$ .

Option C

No route reaches $3$ .

Option D

No route reaches $4$ .

Skills used

Square of a sum and of a difference

denotes the non-negative root

6Algebra • Logarithm • Characteristic and Mantissa

EasyIdeal time: about 45 s

What is the number of digits in $2^{40}$ ? (Given that $lo g_{10} 2 = 0.301$ )

14

13

12

11

Show answer & solution

Option B

13

Approach

Take $lo g_{10}$ of the number and use that the digit count is the characteristic plus one.

Number of digits from a common logarithm

Step-by-step solution

1
$lo g_{10} 2^{40} = 40 lo g_{10} 2 = 40 \times 0.301 = 12.04$
Why: The power rule brings the exponent down so the tabulated value of $lo g_{10} 2$ can be used.
$lo g a^{n} = n lo g a$
2
$12 < 12.04 < 13 \Rightarrow 1 0^{12} < 2^{40} < 1 0^{13}$
Why: A number between $1 0^{12}$ and $1 0^{13}$ has thirteen digits.
Number of digits from a common logarithm
3
digits $= ⌊ 12.04 ⌋ + 1 = 12 + 1 = 13$
Why: The characteristic is the integer part, and the digit count is one more.

Common mistakes

Option A

Rounded $12.04$ up to $13$ and then still added $1$ .

Option C

Took the characteristic $12$ itself as the number of digits, dropping the $+ 1$ .

Option D

No route reaches $11$ .

Skills used

Number of digits from a common logarithm

lo g a^{n} = n lo g a

7Algebra • Quadratic Equations • Roots

ModerateIdeal time: about 2 min

If one root of $(a^{2} - 5 a + 3) x^{2} + (3 a - 1) x + 2 = 0$ is twice the other, then what is the value of 'a'?

\frac{2}{3}

- \frac{2}{3}

\frac{1}{3}

- \frac{1}{3}

Show answer & solution

Option A

\frac{2}{3}

Approach

Name the roots $α$ and $2 α$ , write Vieta's two relations, and eliminate $α$ by squaring the sum.

Sum and product of roots (Vieta)

Step-by-step solution

1
Write $A = a^{2} - 5 a + 3$ . With roots $α$ and $2 α$ : $3 α = - \frac{3 a - 1}{A}$ and $2 α^{2} = \frac{2}{A}$
Why: 'One root is twice the other' is imposed by naming the roots that way, so Vieta's relations carry the condition.
Sum and product of roots (Vieta)
2
From the product: $α^{2} = \frac{1}{A}$
Why: The factor $2$ cancels, leaving $α^{2}$ in terms of $a$ alone.
3
Squaring the sum: $9 α^{2} = \frac{( 3 a - 1 ) ^{2}}{A ^{2}}$ , so $\frac{9}{A} = \frac{( 3 a - 1 ) ^{2}}{A ^{2}}$ , i.e. $9 A = (3 a - 1)^{2}$
Why: Squaring is what makes the two relations comparable; $α$ then disappears.
Square of a sum and of a difference
4
$9 (a^{2} - 5 a + 3) = 9 a^{2} - 6 a + 1 \Rightarrow - 45 a + 27 = - 6 a + 1 \Rightarrow 39 a = 26 \Rightarrow a = \frac{2}{3}$
Why: The $9 a^{2}$ terms cancel, leaving a linear equation in $a$ .

Common mistakes

Option B

Dropped the sign when transposing $- 45 a$ and $- 6 a$ , giving $39 a = - 26$ .

Option C

Expanded $(3 a - 1)^{2}$ as $9 a^{2} - 6 a + 1$ but $9 A$ as $9 a^{2} - 45 a + 3$ , leaving $39 a = 2$ mis-solved to $\frac{1}{3}$ .

Option D

Both slips together: a sign error on top of the mis-expansion.

Skills used

Sum and product of roots (Vieta)Square of a sum and of a difference

8Arithmetic • Number System • Divisibility

ModerateIdeal time: about 1.5 min

What is the remainder when the number $(4444)^{4444}$ is divided by $9$ ?

4

6

7

8

Show answer & solution

Option C

7

Approach

Reduce the base modulo $9$ by its digit sum, find the short cycle of that residue's powers, and reduce the exponent.

Reduce a large power by the cycle of its base modulo

n

Step-by-step solution

1
$4 + 4 + 4 + 4 = 16$ and $1 + 6 = 7$ , so $4444 \equiv 7 (mod 9)$
Why: Since $10 \equiv 1 (mod 9)$ , every power of $10$ is $\equiv 1$ , so a number is congruent to the sum of its digits.
A number is congruent to its digit sum modulo $9$
2
$7^{1} \equiv 7$ , $7^{2} = 49 \equiv 4$ , $7^{3} = 343 = 38 \times 9 + 1 \equiv 1 (mod 9)$
Why: The powers of $7$ close a cycle of length $3$ modulo $9$ , so only the exponent's remainder on division by $3$ matters.
Reduce a large power by the cycle of its base modulo $n$
3
$4444 = 3 \times 1481 + 1$ , so $7^{4444} = (7^{3})^{1481} \times 7$
Why: Splitting the exponent isolates a block that is $\equiv 1$ .
Power of a power, and the move to a common base
4
$(7^{3})^{1481} \times 7 \equiv 1^{1481} \times 7 = 7 (mod 9)$
Why: The remainder is therefore $7$ .

Common mistakes

Option A

Reduced the base to its unit digit $4$ instead of its digit sum: $4^{4444} \equiv 4 (mod 9)$ by the same cycle argument.

Option B

No route reaches $6$ .

Option D

No route reaches $8$ ; $8$ would need the base $\equiv - 1$ and an odd exponent.

Skills used

Power of a power, and the move to a common base A number is congruent to its digit sum modulo

9

Reduce a large power by the cycle of its base modulo

n

9Algebra • Surds • Simplification

ModerateIdeal time: about 1.5 min

If $x = \frac{a + b - a - b}{a + b + a - b}$ , then what is $b x^{2} - 2 a x + b$ equal to $(b \neq = 0)$ ?

0

1

ab

2 ab

Show answer & solution

Option A

0

Approach

Rationalise $x$ , isolate the remaining surd, and square to get a polynomial relation in $x$ .

Rationalise a denominator using the conjugate

Step-by-step solution

1
$x = \frac{( a + b - a - b ) ^{2}}{( a + b + a - b ) ( a + b - a - b )} = \frac{( a + b - a - b ) ^{2}}{( a + b ) - ( a - b )}$
Why: Multiplying above and below by the conjugate turns the denominator into a difference of squares, which is free of surds.
Rationalise a denominator using the conjugate Difference of two squares
2
numerator $= (a + b) - 2 a^{2} - b^{2} + (a - b) = 2 a - 2 a^{2} - b^{2}$ , denominator $= 2 b$ , so $b x = a - a^{2} - b^{2}$
Why: Expanding the square leaves a single surd, and clearing the $2 b$ puts the expression in the form the question's target suggests.
Square of a sum and of a difference
3
$a - b x = a^{2} - b^{2} \Rightarrow (a - b x)^{2} = a^{2} - b^{2} \Rightarrow a^{2} - 2 ab x + b^{2} x^{2} = a^{2} - b^{2}$
Why: With the surd alone on one side, squaring removes it without introducing cross terms.
4
$b^{2} x^{2} - 2 ab x + b^{2} = 0$ , and dividing by $b \neq = 0$ : $b x^{2} - 2 a x + b = 0$
Why: Every term carries a factor $b$ , which is why the question insists $b \neq = 0$ .

Shortcut

The target $b x^{2} - 2 a x + b$ is the quadratic with roots $x$ and $\frac{1}{x}$ and product $1$ ; since $\frac{1}{x}$ is the same expression with the surds swapped, $x$ is a root and the value is $0$ .

Common mistakes

Option B

Divided the relation $b^{2} x^{2} - 2 ab x + b^{2} = 0$ by its own left side, or read the target as an evaluation that must be non-zero.

Option C

No route reaches $ab$ ; the expression is a multiple of the vanishing quadratic.

Option D

Carried the $2 ab$ from the cross term $- 2 ab x$ across as a constant.

Skills used

Rationalise a denominator using the conjugate Difference of two squares Square of a sum and of a difference

10Algebra • Expressions • Simplification

EasyIdeal time: about 45 s

What is the value of $\frac{( 443 + 547 ) ^{2} + ( 443 - 547 ) ^{2}}{443 \times 443 + 547 \times 547}$ ?

0

1

2

3

Show answer & solution

Option C

2

Approach

Use $(m + n)^{2} + (m - n)^{2} = 2 (m^{2} + n^{2})$ so that no arithmetic with $443$ and $547$ is needed.

Square of a sum and of a difference

Step-by-step solution

1
With $m = 443$ and $n = 547$ : $(m + n)^{2} + (m - n)^{2} = (m^{2} + 2 mn + n^{2}) + (m^{2} - 2 mn + n^{2}) = 2 (m^{2} + n^{2})$
Why: The cross terms cancel, which is the whole point of pairing a sum with a difference.
Square of a sum and of a difference
2
$\frac{2 ( m ^{2} + n ^{2} )}{m ^{2} + n ^{2}} = 2$
Why: The denominator is exactly $m^{2} + n^{2}$ , so everything cancels.

Common mistakes

Option A

Cancelled the two squares against each other as if the numerator were a difference.

Option B

Saw the cancellation but dropped the factor $2$ that the two copies of $m^{2} + n^{2}$ produce.

Option D

No route reaches $3$ ; only two copies of $m^{2} + n^{2}$ appear.

Skills used

Square of a sum and of a difference

11Algebra • Expressions • Simplification

ModerateIdeal time: about 1.5 min

If $x = t^{\frac{1}{t - 1}}$ and $y = t^{\frac{t}{t - 1}}$ , $t > 0$ , $t \neq = 1$ , then what is the relation between $x$ and $y$ ?

y^{x} = x^{1/ y}

x^{1/ y} = y^{1/ x}

x^{y} = y^{x}

x^{y} = y^{1/ x}

Show answer & solution

Option C

x^{y} = y^{x}

Approach

Show that $y = t x$ and $y = x^{t}$ ; together they force $x^{y} = y^{x}$ .

Power of a power, and the move to a common base

Step-by-step solution

1
$\frac{y}{x} = t^{\frac{t}{t - 1} - \frac{1}{t - 1}} = t^{\frac{t - 1}{t - 1}} = t$ , so $y = t x$
Why: Dividing powers of the same base subtracts the indices, and the exponents here differ by exactly $1$ over $t - 1$ .
Power of a power, and the move to a common base
2
$x^{t} = (t^{\frac{1}{t - 1}})^{t} = t^{\frac{t}{t - 1}} = y$
Why: A power of a power multiplies the indices, which turns $x$ into $y$ .
3
$y^{x} = (x^{t})^{x} = x^{t x} = x^{y}$
Why: Substituting $y = x^{t}$ and then $t x = y$ closes the loop.

Shortcut

Test $t = 2$ : $x = 2$ , $y = 4$ , and $2^{4} = 4^{2}$ is the only printed relation that holds.

Common mistakes

Option A

Inverted one exponent, reading $y = x^{t}$ as $y = x^{1/ t}$ .

Option B

Took reciprocals of both exponents, which does not preserve the relation: at $t = 2$ it reads $2^{1/4} = 4^{1/2}$ .

Option D

Mixed the two forms, keeping $x^{y}$ on the left but inverting the exponent on the right.

Skills used

Power of a power, and the move to a common base

12Arithmetic • Ratio and Proportion

ModerateIdeal time: about 1.5 min

If $A : B = 3 : 4$ , then what is the value of the expression $(\frac{3 A ^{2} + 4 B}{3 A - 4 B ^{2}})$ ?

\frac{43}{55}

- \frac{43}{55}

\frac{47}{55}

DCannot be determined

Show answer & solution

Option D

Cannot be determined

Approach

Write $A = 3 k$ , $B = 4 k$ and check whether $k$ survives the substitution.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
$A = 3 k$ , $B = 4 k$ for some $k > 0$
Why: A ratio fixes $A$ and $B$ only up to a common factor.
Introduce a common factor $k$ for a ratio
2
$\frac{3 A ^{2} + 4 B}{3 A - 4 B ^{2}} = \frac{27 k ^{2} + 16 k}{9 k - 64 k ^{2}} = \frac{27 k + 16}{9 - 64 k}$
Why: Only one power of $k$ cancels, because the numerator and the denominator each mix a degree- $2$ term with a degree- $1$ term.
3
$k = 1$ gives $\frac{43}{- 55}$ ; $k = 2$ gives $\frac{70}{- 119} = - \frac{10}{17}$
Why: The expression is not homogeneous, so $k$ does not cancel and the value genuinely changes with the actual sizes of $A$ and $B$ .
4
The ratio $A : B = 3 : 4$ alone does not determine the expression
Why: Two admissible pairs give two different values, so no single value can be reported.

Common mistakes

Option A

Substituted $A = 3$ , $B = 4$ and then lost the sign of the denominator $9 - 64 = - 55$ .

Option B

Substituted $A = 3$ , $B = 4$ as if the ratio fixed the values themselves — the intended trap.

Option C

No route reaches $\frac{47}{55}$ .

Skills used

Introduce a common factor

k

for a ratio

13Algebra • Sets • Operations

EasyIdeal time: about 1.5 min

If $A = {x : x$ is a multiple of $7}$ , $B = {x : x$ is a multiple of $5}$ and $C = {x : x$ is a multiple of $35}$ , then which one of the following is a null set?

(A - B) \cup C

(A - B) - C

(A \cap B) \cap C

(A \cap B) - C

Show answer & solution

Option D

(A \cap B) - C

Approach

Identify $A \cap B$ with $C$ , then evaluate each of the four combinations.

Set difference

A - B

Step-by-step solution

1
$x \in A \cap B$ iff $7 ∣ x$ and $5 ∣ x$ ; as $HCF (7, 5) = 1$ , $LCM (7, 5) = 35$ , so $A \cap B = C$
Why: A number divisible by two coprime numbers is divisible by their product, so the common multiples of $7$ and $5$ are exactly the multiples of $35$ .
Coprime means no shared prime factor
2
$(A \cap B) - C = C - C = \emptyset$
Why: Removing from a set exactly the elements of that same set leaves nothing.
Set difference $A - B$
3
$(A \cap B) \cap C = C$ and $(A - B) \cup C \supseteq C$ , both containing $35$
Why: Neither of these can be empty, since $35$ lies in $C$ .
4
$A - B$ is the multiples of $7$ that are not multiples of $5$ ; no such number is a multiple of $35$ , so $(A - B) - C = A - B$ , which contains $7$
Why: $C$ and $A - B$ are disjoint, so subtracting $C$ removes nothing at all.

Common mistakes

Option A

Read $\cup$ as $\cap$ : $(A - B) \cap C$ is indeed empty, since $A - B$ and $C$ are disjoint.

Option B

Assumed $C$ empties $A - B$ , forgetting that a multiple of $35$ is a multiple of $5$ and so was never in $A - B$ .

Option C

Read the triple overlap as needing an element outside $C$ , though $A \cap B \cap C = C$ .

Skills used

Set difference

A - B

Coprime means no shared prime factor

14Algebra • Surds • Simplification

HardIdeal time: about 2.5 min

If $x = 2 + 2^{2/3} + 2^{1/3}$ , then what is the value of $x^{3} - 6 x^{2} + 6 x$ ?

3

2

1

0

Show answer & solution

Option B

2

Approach

Put $t = 2^{1/3}$ , compute $x^{2}$ and $x^{3}$ using $t^{3} = 2$ , and combine the three terms.

Power of a power, and the move to a common base

Step-by-step solution

1
$t = 2^{1/3}$ , so $x = 2 + t + t^{2}$ , with $t^{3} = 2$ and $t^{4} = t \cdot t^{3} = 2 t$
Why: Every power of $t$ above the second collapses back to a linear expression in $t$ , which keeps the algebra finite.
Power of a power, and the move to a common base
2
$x^{2} = (2 + (t + t^{2}))^{2} = 4 + 4 (t + t^{2}) + (t^{2} + 2 t^{3} + t^{4}) = 8 + 6 t + 5 t^{2}$
Why: Group $x$ as $2$ plus $t + t^{2}$ and square; the inner square gives $t^{2} + 4 + 2 t$ once $t^{3}$ and $t^{4}$ are reduced.
Square of a sum and of a difference
3
$x^{3} = x \cdot x^{2} = (2 + t + t^{2}) (8 + 6 t + 5 t^{2}) = 38 + 30 t + 24 t^{2}$
Why: Multiply out and reduce $t^{3} = 2$ , $t^{4} = 2 t$ : the constants give $16 + 10 + 12 = 38$ , the $t$ terms $12 + 8 + 10 = 30$ , the $t^{2}$ terms $10 + 6 + 8 = 24$ .
4
$x^{3} - 6 x^{2} + 6 x = (38 + 30 t + 24 t^{2}) - 6 (8 + 6 t + 5 t^{2}) + 6 (2 + t + t^{2}) = (38 - 48 + 12) + (30 - 36 + 6) t + (24 - 30 + 6) t^{2} = 2$
Why: Both surd terms cancel, which is the point of the particular coefficients $- 6$ and $6$ .

Shortcut

With $y = x - 2 = t + t^{2}$ : $y^{2} = t^{2} + 2 t + 4$ and $y^{3} = y \cdot y^{2} = 6 (t + t^{2}) + 6 = 6 y + 6$ , so $(x - 2)^{3} = 6 (x - 2) + 6$ , which expands directly to $x^{3} - 6 x^{2} + 6 x = 2$ .

Common mistakes

Option A

Dropped one unit somewhere in the constant tally $38 - 48 + 12$ .

Option C

No route reaches $1$ .

Option D

Assumed the expression was built to vanish, as a rearranged cube of $x - 2$ would.

Skills used

Power of a power, and the move to a common base Square of a sum and of a difference

15Algebra • Equations

ModerateIdeal time: about 1.5 min

If $\frac{x}{y} = \frac{24}{5} + \frac{y}{x}$ and $x + y = 26$ , then what is the value of $x y$ ?

5

15

25

30

Show answer & solution

Option C

25

Approach

Set $u = x / y$ ; the equation becomes a quadratic in $u$ , and $u$ fixes the ratio $x : y$ .

Factorise a quadratic by splitting the middle term

Step-by-step solution

1
$u = \frac{x}{y} > 0 \Rightarrow \frac{y}{x} = \frac{1}{u}$ , so $u - \frac{1}{u} = \frac{24}{5}$
Why: The two radicals are reciprocals, so one letter carries both.
Negative exponent is a reciprocal denotes the non-negative root
2
$5 u^{2} - 24 u - 5 = 0 \Rightarrow (5 u + 1) (u - 5) = 0$
Why: Clearing the fraction gives a quadratic; $- 25 u + u$ is the split of the middle term with product $5 \times (- 5) = - 25$ .
Factorise a quadratic by splitting the middle term
3
$u > 0$ discards $u = - \frac{1}{5}$ , leaving $u = 5$ , so $\frac{x}{y} = u^{2} = 25$
Why: A square root is non-negative, so only the positive root survives.
4
$x = 25 y$ and $x + y = 26 \Rightarrow 26 y = 26 \Rightarrow y = 1, x = 25, x y = 25$
Why: The ratio plus the given sum pins down both numbers.

Common mistakes

Option A

Stopped at $u = 5$ and reported it as the answer.

Option B

No route reaches $15$ .

Option D

No route reaches $30$ ; $x$ and $y$ are forced to be $25$ and $1$ .

Skills used

Factorise a quadratic by splitting the middle term Negative exponent is a reciprocal

denotes the non-negative root

16Algebra • Logarithm • Equations

HardIdeal time: about 2 min

What is the solution of the equation $x lo g_{10} (\frac{10}{3}) + lo g_{10} 3 = lo g_{10} (2 + 3^{x}) + x$ ?

10

3

1

0

Show answer & solution

Option D

0

Approach

Split $lo g_{10} \frac{10}{3}$ , cancel the free $x$ on both sides, and solve the resulting quadratic in $3^{x}$ .

Logarithm to exponential form

Step-by-step solution

1
$lo g_{10} \frac{10}{3} = lo g_{10} 10 - lo g_{10} 3 = 1 - lo g_{10} 3$ , so the left side is $x - x lo g_{10} 3 + lo g_{10} 3$
Why: Splitting the quotient is what exposes the bare $x$ that will cancel against the $x$ on the right.
$lo g a^{n} = n lo g a$ $lo g (ab) = lo g a + lo g b$ and $lo g \frac{a}{b} = lo g a - lo g b$
2
$lo g_{10} 3 - x lo g_{10} 3 = lo g_{10} (2 + 3^{x})$ , i.e. $lo g_{10} 3^{1 - x} = lo g_{10} (2 + 3^{x})$
Why: Cancelling $x$ and folding the coefficient back into the exponent with the power rule leaves one logarithm on each side.
3
$3^{1 - x} = 2 + 3^{x}$ ; with $u = 3^{x} > 0$ : $\frac{3}{u} = 2 + u \Rightarrow u^{2} + 2 u - 3 = 0 \Rightarrow (u + 3) (u - 1) = 0$
Why: Equal logarithms with the same base mean equal arguments, and the substitution turns the exponential equation into a quadratic.
Logarithm to exponential form Factorise a quadratic by splitting the middle term
4
$u > 0$ rules out $u = - 3$ , so $u = 1$ , i.e. $3^{x} = 1$ and $x = 0$
Why: A power of $3$ is always positive, and $3^{x} = 1$ only at $x = 0$ .

Common mistakes

Option A

Read the base $10$ inside $lo g_{10} \frac{10}{3}$ as the solution.

Option B

Read the $3$ in the same place, or took $u = 3$ from the quadratic.

Option C

Took the root of $u^{2} + 2 u - 3 = 0$ as $u = 3$ rather than $u = 1$ , giving $3^{x} = 3$ .

Skills used

Logarithm to exponential form

lo g a^{n} = n lo g a

Factorise a quadratic by splitting the middle term

lo g (ab) = lo g a + lo g b

and

lo g \frac{a}{b} = lo g a - lo g b

17Algebra • Quadratic Equations • Roots

Very easyIdeal time: about 30 s

If $α$ and $β$ are the roots of the equation $x^{2} + p x + q = 0$ , then what is $α^{2} + β^{2}$ equal to?

p^{2} - 2 q

q^{2} - 2 p

p^{2} + 2 q

q^{2} - q

Show answer & solution

Option A

p^{2} - 2 q

Approach

Write $α^{2} + β^{2}$ in terms of the sum and product of the roots, then read those off the coefficients.

Sum and product of roots (Vieta)

Step-by-step solution

1
$α + β = - p$ and $α β = q$
Why: For $x^{2} + p x + q = 0$ the leading coefficient is $1$ , so the sum is $- p$ and the product is $q$ .
Sum and product of roots (Vieta)
2
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = (- p)^{2} - 2 q = p^{2} - 2 q$
Why: The square of the sum overshoots by exactly $2 α β$ , and the minus sign on $p$ disappears on squaring.
Square of a sum and of a difference

Common mistakes

Option B

Interchanged $p$ and $q$ , using the product where the sum belongs.

Option C

Used $(α + β)^{2} + 2 α β$ , adding the correction instead of subtracting it.

Option D

No route reaches $q^{2} - q$ .

Skills used

Sum and product of roots (Vieta)Square of a sum and of a difference

18Algebra • Expressions • Simplification

ModerateIdeal time: about 1.5 min

If $a^{3} = 335 + b^{3}$ and $a = 5 + b$ , then what is the value of $a + b$ (given that $a > 0$ and $b > 0$ )?

7

9

16

49

Show answer & solution

Option B

9

Approach

Read the two givens as $a - b = 5$ and $a^{3} - b^{3} = 335$ , extract $ab$ from the cube factorisation, then get $a + b$ from $(a + b)^{2} = (a - b)^{2} + 4 ab$ .

Sum and difference of two cubes

Step-by-step solution

1
$a = 5 + b \Rightarrow a - b = 5$ , and $a^{3} = 335 + b^{3} \Rightarrow a^{3} - b^{3} = 335$
Why: Both givens are differences, which is what the cube identity needs.
Sum and difference of two cubes
2
$a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2}) \Rightarrow 5 (a^{2} + ab + b^{2}) = 335 \Rightarrow a^{2} + ab + b^{2} = 67$
Why: Dividing out the known factor $a - b = 5$ leaves a symmetric expression.
3
$(a - b)^{2} = a^{2} - 2 ab + b^{2} = 25$ ; subtracting from $a^{2} + ab + b^{2} = 67$ gives $3 ab = 42$ , so $ab = 14$
Why: The two symmetric expressions differ by exactly $3 ab$ .
Square of a sum and of a difference
4
$(a + b)^{2} = (a - b)^{2} + 4 ab = 25 + 56 = 81 \Rightarrow a + b = 9$
Why: Both $a$ and $b$ are positive, so the positive root is the one to take.

Shortcut

Guess-and-check on $a - b = 5$ with small positive integers reaches $7^{3} - 2^{3} = 335$ in a few trials, giving $a + b = 9$ .

Common mistakes

Option A

Solved for $a$ and $b$ but reported $a = 7$ instead of $a + b$ .

Option C

No route reaches $16$ .

Option D

Reported $a^{2} = 49$ , or stopped one square-root short having found $(a + b)^{2}$ wrongly.

Skills used

Sum and difference of two cubes Square of a sum and of a difference

19Algebra • Equations • Exponential Equations

ModerateIdeal time: about 2 min

If $9^{x} 3^{y} = 2187$ and $2^{3 x} 2^{2 y} - 4^{x y} = 0$ , then what can be the value of $(x + y)$ ?

1

3

5

7

Show answer & solution

Option C

5

Approach

Bring each equation to a single base, equate exponents, and solve the resulting system.

Power of a power, and the move to a common base

Step-by-step solution

1
$9^{x} 3^{y} = 3^{2 x + y}$ and $2187 = 3^{7}$ , so $2 x + y = 7$
Why: With both sides written to base $3$ , equal powers force equal exponents.
Power of a power, and the move to a common base
2
$2^{3 x} 2^{2 y} = 4^{x y} \Rightarrow 2^{3 x + 2 y} = 2^{2 x y} \Rightarrow 3 x + 2 y = 2 x y$
Why: The same move to base $2$ ; the difference being zero means the two powers are equal.
3
$y = 7 - 2 x \Rightarrow 3 x + 2 (7 - 2 x) = 2 x (7 - 2 x) \Rightarrow 4 x^{2} - 15 x + 14 = 0 \Rightarrow (x - 2) (4 x - 7) = 0$
Why: Substituting the linear relation into the second equation leaves a quadratic in $x$ alone.
Factorise a quadratic by splitting the middle term
4
$x = 2 \Rightarrow y = 3 \Rightarrow x + y = 5$ ; the other root $x = \frac{7}{4}$ gives $y = \frac{7}{2}$ and $x + y = \frac{21}{4}$
Why: The question asks what $x + y$ *can* be, and only $5$ is printed.

Common mistakes

Option A

No route reaches $1$ .

Option B

Reported $y = 3$ alone rather than $x + y$ .

Option D

Reported $2 x + y = 7$ , the first equation's right-hand side, as if it were $x + y$ .

Skills used

Power of a power, and the move to a common base Factorise a quadratic by splitting the middle term

20Algebra • Equations • Linear Equations

EasyIdeal time: about 50 s

The pair of linear equations $k x + 3 y + 1 = 0$ and $2 x + y + 3 = 0$ intersect each other, if

k = 6

k \neq = 6

k = 0

k \neq = 0

Show answer & solution

Option B

k \neq = 6

Approach

Eliminate $y$ and ask when the resulting equation in $x$ has a solution.

When a pair of linear equations meets, is parallel, or coincides

Step-by-step solution

1
$2 x + y + 3 = 0 \Rightarrow y = - 2 x - 3$
Why: The second equation is the easier one to solve for a variable.
A product is zero only when a factor is zero
2
$k x + 3 (- 2 x - 3) + 1 = 0 \Rightarrow k x - 6 x - 8 = 0 \Rightarrow (k - 6) x = 8$
Why: Substituting collapses the pair to one equation in $x$ .
3
$(k - 6) x = 8$ has a solution exactly when $k - 6 \neq = 0$ ; at $k = 6$ it reads $0 = 8$
Why: A product can equal a non-zero number only if neither factor is zero, so $k = 6$ leaves the system inconsistent — the lines are parallel and never meet.
When a pair of linear equations meets, is parallel, or coincides
4
The lines intersect iff $k \neq = 6$
Why: For every other $k$ there is exactly one $x$ , and then exactly one $y$ .

Shortcut

Two lines meet in a point exactly when $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ ; here $\frac{k}{2} \neq = \frac{3}{1}$ , i.e. $k \neq = 6$ .

Common mistakes

Option A

$k = 6$ is precisely the value that makes the lines parallel — the condition read the wrong way round.

Option C

$k = 0$ still gives an intersection, so it is neither necessary nor sufficient.

Option D

Compared the constant terms instead of the coefficients of $x$ and $y$ .

Skills used

A product is zero only when a factor is zero When a pair of linear equations meets, is parallel, or coincides