Elementary Mathematics 2017 (II)

Set A

Questions 1–20 of 100

1Algebra • Expressions • Factorisation

ModerateIdeal time: about 1.5 min

What is the value of $α$ ( $α \neq = 0$ ) for which $x^{2} - 5 x + α$ and $x^{2} - 7 x + 2 α$ have a common factor ?

6

4

3

2

Show answer & solution

Option A

6

Approach

A common factor means a common root; subtracting the two quadratics pins that root down in terms of $α$ .

Common root of two quadratics by elimination

Step-by-step solution

1
$x^{2} - 5 x + α = 0$ and $x^{2} - 7 x + 2 α = 0$ at $x = r$
Why: A common linear factor $(x - r)$ makes $r$ a root of both quadratics, by the factor theorem.
Factor and remainder theorem
2
$(x^{2} - 5 x + α) - (x^{2} - 7 x + 2 α) = 2 x - α$ , so $2 r - α = 0$ and $r = \frac{α}{2}$
Why: The $x^{2}$ terms cancel on subtraction, leaving a linear equation that the common root must also satisfy.
Common root of two quadratics by elimination
3
$\frac{α ^{2}}{4} - \frac{5 α}{2} + α = 0 \Rightarrow \frac{α}{4} (α - 6) = 0$
Why: Substituting $r = \frac{α}{2}$ back into the first quadratic turns it into an equation in $α$ alone.
A product is zero only when a factor is zero
4
$α \neq = 0 \Rightarrow α = 6$
Why: The question excludes $α = 0$ , so the other factor must vanish.

Shortcut

Subtracting kills the $x^{2}$ term at once: $2 x = α$ , so the common root is $\frac{α}{2}$ , and one substitution finishes it.

Common mistakes

Option B

Dropped the $2$ when solving $2 x - α = 0$ , taking $r = α$ ; then $α^{2} - 5 α + α = 0$ gives $α = 4$ .

Option C

Reached $\frac{α}{4} = \frac{3}{2}$ and multiplied by $2$ instead of $4$ .

Option D

No route reaches $2$ ; it echoes the coefficient of $α$ in the second quadratic.

Skills used

Common root of two quadratics by elimination Factor and remainder theorem A product is zero only when a factor is zero

2Arithmetic • Number System • LCM and HCF

Very easyIdeal time: about 40 s

How many numbers from $1$ to $1000$ are divisible by $2, 3, 4$ and $5$ ?

16

17

32

DNone of the above

Show answer & solution

Option A

16

Approach

A number divisible by each of $2, 3, 4, 5$ is exactly a multiple of their LCM, so count the multiples of that LCM up to $1000$ .

Prime factorisation

Step-by-step solution

1
$2 = 2, 3 = 3, 4 = 2^{2}, 5 = 5 \Rightarrow LCM = 2^{2} \cdot 3 \cdot 5 = 60$
Why: The LCM takes the highest power of each prime that occurs, and the common multiples of the four numbers are precisely the multiples of this LCM.
Prime factorisation
2
$60 k \leq 1000 \Rightarrow k \leq \frac{1000}{60} = 16. 6$
Why: The multiples of $60$ form an arithmetic progression $60, 120, \dots$ ; bounding it by $1000$ bounds $k$ .
Enumerate a residue class within a bound
3
$k = 1, 2, \dots, 16$ , so there are $16$ such numbers
Why: $k$ is a positive integer, and the largest admissible value is $⌊ 16. 6 ⌋ = 16$ .

Shortcut

Divisibility by $4$ already forces divisibility by $2$ , so the test is just $LCM (3, 4, 5) = 60$ ; then $⌊ 1000/60 ⌋ = 16$ .

Common mistakes

Option B

Rounded $\frac{1000}{60} = 16. 6$ up to $17$ instead of taking the floor.

Option C

No route reaches $32$ from a correct LCM of $60$ .

Option D

Chosen only when the count is mis-derived; $16$ is printed, so this is never correct.

Skills used

Prime factorisation Enumerate a residue class within a bound

3Algebra • Expressions • Factorisation

EasyIdeal time: about 1 min

What are the factors of $x^{3} + 4 x^{2} - 11 x - 30$ ?

(x - 2), (x + 3)

and

(x + 5)

(x + 2), (x + 3)

and

(x - 5)

(x + 2), (x - 3)

and

(x + 5)

(x + 2), (x - 3)

and

(x - 5)

Show answer & solution

Option C

(x + 2), (x - 3)

and

(x + 5)

Approach

Find one root by the factor theorem, divide it out, and factorise the remaining quadratic.

Factor and remainder theorem

Step-by-step solution

1
$p (- 2) = - 8 + 16 + 22 - 30 = 0$ , so $(x + 2)$ is a factor
Why: The factor theorem says $(x + 2)$ divides $p (x)$ exactly when $p (- 2) = 0$ ; the divisors of $30$ are the candidates worth testing.
Factor and remainder theorem
2
$x^{3} + 4 x^{2} - 11 x - 30 = (x + 2) (x^{2} + 2 x - 15)$
Why: Dividing the cubic by the known linear factor leaves a quadratic.
3
$x^{2} + 2 x - 15 = (x + 5) (x - 3)$
Why: Split the middle term using $5 \times (- 3) = - 15$ and $5 + (- 3) = 2$ .
Factorise a quadratic by splitting the middle term

Shortcut

Check the constant term: the three roots must multiply to $+ 30$ and add to $- 4$ ; only $- 2, 3, - 5$ does both.

Common mistakes

Option A

Sign error on the $(x + 2)$ factor: $(x - 2) (x + 3) (x + 5)$ has the right constant $- 30$ but its roots sum to $- 6$ , not $- 4$ .

Option B

Sign error on the last factor: $(x + 2) (x + 3) (x - 5)$ gives constant $- 30$ but roots summing to $0$ .

Option D

Signs flipped on both quadratic factors, giving constant $+ 30$ instead of $- 30$ .

Skills used

Factor and remainder theorem Factorise a quadratic by splitting the middle term

4Algebra • Expressions • Simplification

HardIdeal time: about 2.5 min

If $x = 111 \dots 1$ ( $20$ digits), $y = 333 \dots 3$ ( $10$ digits) and $z = 222 \dots 2$ ( $10$ digits), then what is $\frac{x - y ^{2}}{z}$ equal to ?

\frac{1}{2}

1

2

3

Show answer & solution

Option B

1

Approach

Write everything in terms of the $10$ -digit repunit $a = 10 11 \dots 1$ , using $1 0^{10} = 9 a + 1$ .

Closed form of a repeated-digit number

Step-by-step solution

1
$a = 10 11 \dots 1, y = 3 a, z = 2 a$
Why: $10 33 \dots 3$ and $10 22 \dots 2$ are $3$ and $2$ times the $10$ -digit repunit, by place value.
Place-value expansion of a numeral Closed form of a repeated-digit number
2
$9 a + 1 = 1 0^{10}$
Why: $9 \times 10 11 \dots 1 = 10 99 \dots 9 = 1 0^{10} - 1$ .
3
$x = 20 11 \dots 1 = a \times 1 0^{10} + a = a (1 0^{10} + 1) = a (9 a + 2) = 9 a^{2} + 2 a$
Why: Writing twenty $1$ s is the same as writing ten $1$ s, shifting them ten places, and adding ten more $1$ s.
4
$x - y^{2} = (9 a^{2} + 2 a) - 9 a^{2} = 2 a$
Why: $y^{2} = (3 a)^{2} = 9 a^{2}$ cancels the quadratic part of $x$ exactly.
5
$\frac{x - y ^{2}}{z} = \frac{2 a}{2 a} = 1$
Why: The remaining numerator is precisely $z$ .

Shortcut

Test the pattern on the smallest case: $x = 1111$ , $y = 33$ , $z = 22$ gives $\frac{1111 - 1089}{22} = 1$ , and the algebra shows it is not a coincidence.

Common mistakes

Option A

Halved the result by taking the denominator as $4 a$ , or by getting $x - y^{2} = a$ instead of $2 a$ .

Option C

Read $z$ as the repunit $a$ rather than $2 a$ , giving $\frac{2 a}{a} = 2$ .

Option D

No route reaches $3$ ; it echoes the digit of $y$ .

Skills used

Closed form of a repeated-digit number Place-value expansion of a numeral

5Algebra • Quadratic Equations • Roots

ModerateIdeal time: about 1.5 min

What is the positive value of $m$ for which the roots of the equation $12 x^{2} + m x + 5 = 0$ are in the ratio $3 : 2$ ?

510

\frac{5 10}{12}

\frac{5}{12}

\frac{12}{5}

Show answer & solution

Option A

510

Approach

Name the roots $3 k$ and $2 k$ , then read their sum and product off the coefficients.

Sum and product of roots (Vieta)

Step-by-step solution

1
roots $= 3 k$ and $2 k$
Why: A ratio $3 : 2$ fixes the roots only up to a common factor $k$ .
Introduce a common factor $k$ for a ratio
2
$3 k \cdot 2 k = \frac{5}{12} \Rightarrow 6 k^{2} = \frac{5}{12} \Rightarrow k^{2} = \frac{5}{72}$
Why: The product of the roots of $a x^{2} + b x + c = 0$ is $\frac{c}{a}$ , here $\frac{5}{12}$ .
Sum and product of roots (Vieta)
3
$k = \pm \frac{5}{72} = \pm \frac{5}{6 2} = \pm \frac{10}{12}$
Why: $72 = 62$ , and rationalising gives $\frac{10}{12}$ .
$ab = a b$ and $a / b = a / b$
4
$3 k + 2 k = - \frac{m}{12} \Rightarrow m = - 60 k$
Why: The sum of the roots is $- \frac{b}{a} = - \frac{m}{12}$ .
5
$m > 0 \Rightarrow k = - \frac{10}{12} \Rightarrow m = 60 \cdot \frac{10}{12} = 510$
Why: Only the negative value of $k$ makes $m$ positive, as the question demands.

Common mistakes

Option B

Stopped at the sum of the roots $5 k = \frac{5 10}{12}$ instead of going on to $m$ .

Option C

Quoted the product of the roots $\frac{c}{a} = \frac{5}{12}$ as the answer.

Option D

Inverted $\frac{c}{a}$ , giving $\frac{12}{5}$ .

Skills used

Sum and product of roots (Vieta)Introduce a common factor

k

for a ratio

ab = a b

and

a / b = a / b

6Algebra • Polynomial

Very easyIdeal time: about 20 s

Let $f (x)$ and $g (x)$ be two polynomials (with real coefficients) having degrees $3$ and $4$ respectively. What is the degree of $f (x) g (x)$ ?

12

7

4

3

Show answer & solution

Option B

7

Approach

Multiply the leading terms: their product is non-zero, so its exponent is the degree.

Degree of a product of polynomials

Step-by-step solution

1
$f (x) = a_{0} x^{3} + \dots$ with $a_{0} \neq = 0$ , $g (x) = b_{0} x^{4} + \dots$ with $b_{0} \neq = 0$
Why: A polynomial of degree $3$ has a non-zero coefficient on $x^{3}$ , and likewise for degree $4$ .
2
$f (x) g (x) = a_{0} b_{0} x^{7} + (lower powers)$
Why: The highest power in the product comes from multiplying the two highest powers, and $a_{0} b_{0} \neq = 0$ because a product of non-zero reals is non-zero, so no cancellation can lower the degree.
A product is zero only when a factor is zero Degree of a product of polynomials

Shortcut

Degrees add under multiplication: $3 + 4 = 7$ .

Common mistakes

Option A

Multiplied the degrees, $3 \times 4 = 12$ , instead of adding them.

Option C

Quoted the larger of the two degrees.

Option D

Quoted the smaller of the two degrees.

Skills used

Degree of a product of polynomials A product is zero only when a factor is zero

7Algebra • Polynomial • Factor and Remainder Theorem

Very easyIdeal time: about 45 s

If $5 x^{3} + 5 x^{2} - 6 x + 9$ is divided by $(x + 3)$ , then the remainder is

135

- 135

63

- 63

Show answer & solution

Option D

- 63

Approach

The remainder on dividing by $(x + 3)$ is the value of the polynomial at $x = - 3$ .

Factor and remainder theorem

Step-by-step solution

1
$p (x) = 5 x^{3} + 5 x^{2} - 6 x + 9$ ; remainder $= p (- 3)$
Why: By the remainder theorem, dividing by $(x - c)$ leaves $p (c)$ ; here $x + 3 = 0$ gives $c = - 3$ .
Factor and remainder theorem
2
$p (- 3) = 5 (- 27) + 5 (9) - 6 (- 3) + 9$
Why: Substitute $x = - 3$ term by term, keeping the sign of the odd power.
3
$= - 135 + 45 + 18 + 9 = - 63$
Why: Collect: $- 135 + 45 = - 90$ , and $- 90 + 18 + 9 = - 63$ .

Shortcut

No long division is needed: evaluate at $x = - 3$ .

Common mistakes

Option A

Sign lost on $(- 3)^{3}$ , so the first term was taken as $+ 135$ and the rest discarded.

Option B

Stopped after the leading term, $5 (- 3)^{3} = - 135$ .

Option C

Correct magnitude but the final sign flipped.

Skills used

Factor and remainder theorem

8Arithmetic • Number System • LCM and HCF

EasyIdeal time: about 45 s

The product of two non-zero expressions is $(x + y + z) p^{3}$ . If their HCF is $p^{2}$ , then their LCM is

(x + y + z)

(x + y + z) p^{2}

(x + y + z) p^{5}

(x + y + z) p

Show answer & solution

Option D

(x + y + z) p

Approach

For two expressions, $HCF \times LCM$ equals their product, so divide the product by the HCF.

HCF

\times

LCM = product of the two numbers

Step-by-step solution

1
$HCF \times LCM = product = (x + y + z) p^{3}$
Why: The product of two expressions equals the product of their HCF and LCM, exactly as for two integers.
HCF $\times$ LCM = product of the two numbers
2
$LCM = \frac{( x + y + z ) p ^{3}}{p ^{2}}$
Why: Three of the four quantities are known, so solve for the fourth.
3
$LCM = (x + y + z) p$
Why: $\frac{p ^{3}}{p ^{2}} = p$ , and $(x + y + z)$ carries through untouched.
HCF and LCM of polynomials from their complete factorisations

Common mistakes

Option A

Divided the product by $p^{3}$ rather than by the HCF $p^{2}$ .

Option B

Divided by $p$ instead of $p^{2}$ , leaving $p^{2}$ in the LCM.

Option C

Multiplied the product by the HCF instead of dividing by it: $p^{3} \times p^{2} = p^{5}$ .

Skills used

HCF

\times

LCM = product of the two numbers HCF and LCM of polynomials from their complete factorisations

9Arithmetic • Number System • Rational Numbers

ModerateIdeal time: about 1.5 min

If the points $P$ and $Q$ represent the real numbers $0.8 3$ and $0.6 2$ on the number line, then the distance between $P$ and $Q$ is

\frac{21}{90}

\frac{19}{90}

\frac{21}{100}

\frac{56}{90}

Show answer & solution

Option B

\frac{19}{90}

Approach

Turn each mixed recurring decimal into a fraction over $90$ , then subtract.

Convert a recurring decimal to a fraction

Step-by-step solution

1
$N = 0.8 3 \Rightarrow 100 N = 83. 3, 10 N = 8. 3 \Rightarrow 90 N = 75 \Rightarrow N = \frac{75}{90}$
Why: Multiplying by $100$ and by $10$ puts the same recurring tail after the point in both, so subtracting removes it.
Convert a recurring decimal to a fraction
2
$M = 0.6 2 \Rightarrow 100 M = 62. 2, 10 M = 6. 2 \Rightarrow 90 M = 56 \Rightarrow M = \frac{56}{90}$
Why: The same move on the second decimal; one digit before the bar and one under it again give the denominator $90$ .
3
$P Q = ∣ N - M ∣ = \frac{75}{90} - \frac{56}{90} = \frac{19}{90}$
Why: The distance between two points of the number line is the absolute difference of the numbers they represent, and the two fractions already share a denominator.
Comparing two fractions

Shortcut

For a decimal with one non-recurring and one recurring digit, the fraction is $\frac{whole digits - non-recurring digits}{90}$ : $\frac{83 - 8}{90}$ and $\frac{62 - 6}{90}$ , so the difference is $\frac{83 - 62 - ( 8 - 6 )}{90} = \frac{19}{90}$ .

Common mistakes

Option A

Subtracted the two-digit numerators as $83 - 62 = 21$ and kept the denominator $90$ , forgetting the non-recurring digits.

Option C

Treated both decimals as terminating, $0.83 - 0.62 = 0.21 = \frac{21}{100}$ .

Option D

Quoted $0.6 2 = \frac{56}{90}$ itself instead of the difference.

Skills used

Convert a recurring decimal to a fraction Comparing two fractions

10Arithmetic • Profit and Loss • Discount

EasyIdeal time: about 1 min

Sudhir purchased a chair with three consecutive discounts of $20%$ , $12.5%$ and $5%$ . The actual deduction will be

33.5%

30%

32%

35%

Show answer & solution

Option A

33.5%

Approach

Chain the three discounts as multipliers and read the total deduction off the surviving fraction.

Percentage change as a multiplier

Step-by-step solution

1
price factor $= (1 - 0.20) (1 - 0.125) (1 - 0.05) = 0.8 \times 0.875 \times 0.95$
Why: Each discount multiplies the current price by $(1 - \frac{p}{100})$ , and successive discounts multiply their factors.
Percentage change as a multiplier
2
$0.8 \times 0.875 = 0.7$
Why: $0.875 = \frac{7}{8}$ , and $\frac{4}{5} \times \frac{7}{8} = \frac{7}{10}$ .
3
$0.7 \times 0.95 = 0.665$
Why: $0.7 - 0.05 \times 0.7 = 0.7 - 0.035 = 0.665$ is what the buyer still pays.
4
deduction $= 1 - 0.665 = 0.335 = 33.5%$
Why: The actual deduction is the fraction of the marked price that was NOT paid.

Shortcut

In fractions: $\frac{4}{5} \times \frac{7}{8} \times \frac{19}{20} = \frac{133}{200} = 0.665$ , so $33.5%$ comes off.

Common mistakes

Option B

Stopped after two discounts: $0.8 \times 0.875 = 0.7$ gives exactly $30%$ .

Option C

No clean route reaches $32%$ from these three factors.

Option D

Applied the final $5%$ to the original price rather than to the reduced one: $100 \to 80 \to 70 \to 65$ .

Skills used

Percentage change as a multiplier

11Arithmetic • Percentage

EasyIdeal time: about 1 min

A fruit seller has a certain number of mangoes of which $5%$ are rotten. He sells $75%$ of the remainder and he is left with $95$ mangoes. How many mangoes did he have originally ?

500

450

400

350

Show answer & solution

Option C

400

Approach

Track the surviving fraction of the original stock through both stages and equate it to $95$ .

Percentage change as a multiplier

Step-by-step solution

1
let $N$ be the original number; good mangoes $= 0.95 N$
Why: $5%$ are rotten, so $95%$ of the stock is left before any sale.
$x %$ of a quantity
2
unsold $= (1 - 0.75) \times 0.95 N = 0.25 \times 0.95 N = 0.2375 N$
Why: He sells $75%$ OF THE REMAINDER, so a quarter of the good mangoes is what is left.
Percentage change as a multiplier
3
$0.2375 N = 95 \Rightarrow N = \frac{95}{0.2375} = 400$
Why: Solve the single linear equation for the original number.

Shortcut

Work backwards: $95$ is a quarter of the good stock, so $380$ were good; $380$ is $95%$ of the original, so $N = \frac{380 \times 100}{95} = 400$ .

Common mistakes

Option A

Took the unsold share as $20%$ rather than $25%$ of the remainder, giving $\frac{95}{0.19} = 500$ .

Option B

No route reaches $450$ .

Option D

No route reaches $350$ .

Skills used

Percentage change as a multiplier

x %

of a quantity

12Arithmetic • Time, Speed and Distance • Trains

HardIdeal time: about 2 min

If a train crosses a km-stone in 12 seconds, how long will it take to cross 91 km-stones completely if its speed is $60 km/hr$ ?

A1 hr 30 minB1 hr 30 min 12 secC1 hr 51 minD1 hr 1 min 3 sec

Show answer & solution

Option B

1 hr 30 min 12 sec

Approach

Get the train's length from the km-stone crossing, then add it to the $90$ km that separates the first and the $91$ st stone.

What a train covers when it crosses something

Step-by-step solution

1
$60 km/hr = 60 \times \frac{5}{18} = \frac{50}{3} m/s$
Why: Convert to metres per second so the $12$ second crossing can be used.
km/hour to m/second
2
length $= \frac{50}{3} \times 12 = 200 m$
Why: A km-stone has no length of its own, so crossing it means the train covers exactly its own length.
What a train covers when it crosses something
3
distance between the $1$ st and the $91$ st stone $= (91 - 1) \times 1 = 90 km$
Why: Consecutive km-stones are $1$ km apart, so $91$ stones span $90$ gaps.
4
total distance $= 90 km + 200 m = 90.2 km$
Why: To cross the last stone COMPLETELY the rear of the train must pass it, which costs one extra train length.
5
time $= \frac{90.2}{60} hr = 1.5033 3 hr = 1 hr 30 min 12 sec$
Why: $0.50333 3 \times 60 = 30.2$ minutes, and $0.2$ minute is $12$ seconds.
Speed, distance and time

Shortcut

In seconds: $\frac{90200}{50/3} = 5412$ s $= 5400 + 12$ s $= 1$ hr $30$ min $12$ sec.

Common mistakes

Option A

Ignored the train's own length, timing only the $90$ km between the first and last stone.

Option C

No route reaches $1$ hr $51$ min from a speed of $60$ km/hr.

Option D

No route reaches $1$ hr $1$ min $3$ sec.

Skills used

What a train covers when it crosses something km/hour to m/second Speed, distance and time

13Arithmetic • Time, Speed and Distance • Races

ModerateIdeal time: about 1.5 min

In a 100 m race, A runs at $6 km/hr$ . If A gives B a start of $8 m$ and still beats him by 9 seconds, what is the speed of B ?

4.6 km/hr

4.8 km/hr

5.2 km/hr

5.4 km/hr

Show answer & solution

Option B

4.8 km/hr

Approach

Compute A's time over $100$ m, add the $9$ second margin to get B's time, and divide B's shorter distance by it.

Head starts and winning margins in a race

Step-by-step solution

1
$6 km/hr = 6 \times \frac{5}{18} = \frac{5}{3} m/s$
Why: The race is in metres and seconds, so convert A's speed first.
km/hour to m/second
2
A's time $= \frac{100}{5/3} = 60 s$
Why: Time is distance divided by speed.
Speed, distance and time
3
B's distance $= 100 - 8 = 92 m$ ; B's time $= 60 + 9 = 69 s$
Why: An $8$ m start shortens B's run, and 'beats him by $9$ seconds' means B is still running $9$ s after A finishes.
Head starts and winning margins in a race
4
B's speed $= \frac{92}{69} = \frac{4}{3} m/s = \frac{4}{3} \times \frac{18}{5} = 4.8 km/hr$
Why: Convert back to km/hr by multiplying by $\frac{18}{5}$ .

Common mistakes

Option A

No route reaches $4.6$ km/hr.

Option C

Ignored the $8$ m start and used $\frac{100}{69}$ m/s, which is $5.2$ km/hr.

Option D

No route reaches $5.4$ km/hr.

Skills used

Head starts and winning margins in a race km/hour to m/second Speed, distance and time

14Algebra • Expressions • Factorisation

EasyIdeal time: about 40 s

The quotient of $8 x^{3} - y^{3}$ when divided by $2 x y + 4 x^{2} + y^{2}$ is

2 x + y

x + 2 y

2 x - y

4 x - y

Show answer & solution

Option C

2 x - y

Approach

Recognise the divisor as the quadratic factor in the difference-of-cubes identity.

Sum and difference of two cubes

Step-by-step solution

1
$8 x^{3} - y^{3} = (2 x)^{3} - y^{3}$
Why: Both terms are perfect cubes, with $a = 2 x$ and $b = y$ .
Sum and difference of two cubes
2
$(2 x)^{3} - y^{3} = (2 x - y) (4 x^{2} + 2 x y + y^{2})$
Why: The identity $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ gives $a^{2} = 4 x^{2}$ , $ab = 2 x y$ and $b^{2} = y^{2}$ - exactly the printed divisor, reordered.
3
quotient $= \frac{( 2 x - y ) ( 4 x ^{2} + 2 x y + y ^{2} )}{4 x ^{2} + 2 x y + y ^{2}} = 2 x - y$
Why: The divisor cancels, leaving the linear factor.

Shortcut

The divisor is $a^{2} + ab + b^{2}$ with $a = 2 x$ , $b = y$ ; the quotient of $a^{3} - b^{3}$ by it is always $a - b$ .

Common mistakes

Option A

Used the sum-of-cubes form, whose quadratic factor is $4 x^{2} - 2 x y + y^{2}$ , and took the linear factor as $2 x + y$ .

Option B

Swapped the roles of $2 x$ and $y$ when reading the identity.

Option D

Took the linear factor's first term from $4 x^{2}$ rather than from $3 8 x^{3} = 2 x$ .

Skills used

Sum and difference of two cubes

15Algebra • Polynomial • Factor and Remainder Theorem

EasyIdeal time: about 1 min

If $(x + 2)$ is a common factor of $x^{2} + a x + b$ and $x^{2} + b x + a$ , then the ratio $a : b$ is equal to

1

2

3

4

Show answer & solution

Option A

1

Approach

Both quadratics vanish at $x = - 2$ ; the two resulting equations force $a = b$ .

Factor and remainder theorem

Step-by-step solution

1
$(- 2)^{2} + a (- 2) + b = 0 \Rightarrow 4 - 2 a + b = 0$
Why: $(x + 2)$ is a factor of $x^{2} + a x + b$ exactly when the quadratic vanishes at $x = - 2$ .
Factor and remainder theorem
2
$(- 2)^{2} + b (- 2) + a = 0 \Rightarrow 4 - 2 b + a = 0$
Why: The same test applied to the second quadratic.
3
subtracting: $- 2 a + b + 2 b - a = 0 \Rightarrow 3 (b - a) = 0 \Rightarrow a = b$
Why: The constant $4$ cancels, leaving a relation between $a$ and $b$ alone.
4
$4 - 2 a + a = 0 \Rightarrow a = 4$ , so $a = b = 4$ and $a : b = 1$
Why: With $a = b$ , either equation gives the common value, and the ratio of two equal non-zero numbers is $1$ .

Shortcut

The two conditions are the same equation with $a$ and $b$ swapped, so symmetry forces $a = b$ and the ratio is $1$ without finding either value.

Common mistakes

Option B

No route reaches $2$ .

Option C

No route reaches $3$ .

Option D

Quoted the common value $a = b = 4$ instead of the ratio $a : b$ .

Skills used

Factor and remainder theorem

16Algebra • Polynomial • Factor and Remainder Theorem

Very easyIdeal time: about 25 s

Let $f (x) = a_{0} x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots + a_{n - 1} x + a_{n}$ , where $a_{0}, a_{1}, a_{2}, \dots, a_{n}$ are real numbers. If $f (x)$ is divided by $(a x - b)$ , then the remainder is

f (\frac{b}{a})

f (- \frac{b}{a})

f (\frac{a}{b})

f (- \frac{a}{b})

Show answer & solution

Option A

f (\frac{b}{a})

Approach

Set the divisor to zero and evaluate $f$ there.

Factor and remainder theorem

Step-by-step solution

1
$a x - b = 0 \Rightarrow x = \frac{b}{a}$
Why: The remainder theorem is read at the zero of the divisor, whatever its leading coefficient.
Factor and remainder theorem
2
$f (x) = (a x - b) q (x) + R \Rightarrow f (\frac{b}{a}) = 0 \cdot q (\frac{b}{a}) + R = R$
Why: A linear divisor leaves a constant remainder, and substituting the divisor's root kills the quotient term.

Shortcut

Write $a x - b = a (x - \frac{b}{a})$ : the root is $\frac{b}{a}$ , so the remainder is $f (\frac{b}{a})$ .

Common mistakes

Option B

Used the root of $a x + b$ instead of $a x - b$ .

Option C

Inverted the root, reading $x = \frac{a}{b}$ .

Option D

Both inverted and sign-flipped the root.

Skills used

Factor and remainder theorem

17Arithmetic • Number System • Perfect Squares

ModerateIdeal time: about 1.5 min

Consider the following numbers :

1: 2222

2: 11664

3: 343343

4: 220347

Which of the above are not perfect squares ?

A1, 2 and 3B1, 2 and 4C2, 3 and 4D1, 3 and 4

Show answer & solution

Option D

1, 3 and 4

Approach

Screen by last digit first, then settle the survivors by prime factorisation.

Prime factorisation

Step-by-step solution

1
$2222$ ends in $2$ , and $220347$ ends in $7$
Why: A perfect square can only end in $0, 1, 4, 5, 6$ or $9$ , so both are ruled out at sight; statements $1$ and $4$ are not perfect squares.
Last digit of a perfect square
2
$11664 = 2^{4} \times 3^{6}$
Why: Halving four times gives $729 = 3^{6}$ , so every prime exponent is even.
Prime factorisation
3
$11664 = (2^{2} \times 3^{3})^{2} = 10 8^{2}$
Why: All exponents even means the number IS a perfect square, so statement $2$ must be excluded from the answer.
4
$343343 = 343 \times 1001 = 7^{3} \times 7 \times 11 \times 13 = 7^{4} \times 11 \times 13$
Why: The repeated block $343343$ factorises as $343 \times 1001$ , and $1001 = 7 \times 11 \times 13$ .
5
$11$ and $13$ appear to the first power, so $343343$ is not a perfect square
Why: A perfect square has every prime to an even power.

Shortcut

Only statement $2$ is a perfect square ( $10 8^{2} = 11664$ ), so the answer must be the option that omits $2$ - and just one does.

Common mistakes

Option A

Missed that $11664 = 10 8^{2}$ and also judged $220347$ a square.

Option B

Missed that $11664 = 10 8^{2}$ and also judged $343343$ a square.

Option C

Missed that $11664 = 10 8^{2}$ and also judged $2222$ a square.

Skills used

Prime factorisation Last digit of a perfect square

18Algebra • Expressions • Simplification

ModerateIdeal time: about 1.5 min

The product of the polynomials $(x + 2)$ , $(x - 2)$ , $(x^{3} - 2 x^{2} + 4 x - 8)$ and $(x^{3} + 2 x^{2} + 4 x + 8)$ is

x^{8} - 256

(x^{4} - 16)^{2}

(x^{4} + 16)^{2}

(x^{2} - 4)^{4}

Show answer & solution

Option B

(x^{4} - 16)^{2}

Approach

Factorise each cubic by the factor theorem; the four factors pair into two copies of $x^{4} - 16$ .

Factor and remainder theorem

Step-by-step solution

1
$x^{3} - 2 x^{2} + 4 x - 8$ vanishes at $x = 2$ , so $x^{3} - 2 x^{2} + 4 x - 8 = (x - 2) (x^{2} + 4)$
Why: $8 - 8 + 8 - 8 = 0$ , so $(x - 2)$ is a factor; dividing leaves $x^{2} + 4$ .
Factor and remainder theorem
2
$x^{3} + 2 x^{2} + 4 x + 8$ vanishes at $x = - 2$ , so $x^{3} + 2 x^{2} + 4 x + 8 = (x + 2) (x^{2} + 4)$
Why: $- 8 + 8 - 8 + 8 = 0$ , so $(x + 2)$ is a factor; dividing leaves $x^{2} + 4$ .
3
product $= (x + 2) (x - 2) \cdot (x - 2) (x^{2} + 4) \cdot (x + 2) (x^{2} + 4)$
Why: Collect all four printed polynomials in factored form.
4
$= [(x + 2) (x - 2) (x^{2} + 4)]^{2} = [(x^{2} - 4) (x^{2} + 4)]^{2} = (x^{4} - 16)^{2}$
Why: The six factors pair off into two identical triples, and each triple collapses by the difference of two squares, twice.
Difference of two squares

Shortcut

Each cubic is $(x \mp 2) (x^{2} + 4)$ , so the whole product is $[(x - 2) (x + 2) (x^{2} + 4)]^{2} = (x^{4} - 16)^{2}$ .

Common mistakes

Option A

Multiplied $(x^{4} - 16)$ by $(x^{4} + 16)$ instead of squaring, as if the two triples were different.

Option C

Sign lost in $(x^{2} - 4) (x^{2} + 4)$ , giving $x^{4} + 16$ inside the square.

Option D

Took the $(x^{2} + 4)$ factors as $(x^{2} - 4)$ , collapsing everything to $(x^{2} - 4)^{4}$ .

Skills used

Factor and remainder theorem Difference of two squares

19Algebra • Expressions • Factorisation

HardIdeal time: about 2 min

The factors of $x (x + 2) (x + 3) (x + 5) - 72$ are

x

(x + 3)

(x + 4)

and

(x - 6)

(x - 1)

(x + 6)

and

(x^{2} - 2 x - 12)

(x - 1)

(x + 6)

and

(x^{2} + 5 x + 12)

(x + 1)

(x - 6)

and

(x^{2} - 5 x - 12)

Show answer & solution

Option C

(x - 1)

(x + 6)

and

(x^{2} + 5 x + 12)

Approach

Pair the four linear factors so both products share $x^{2} + 5 x$ , then substitute $t = x^{2} + 5 x$ .

Pair four linear factors so one quadratic repeats

Step-by-step solution

1
$x (x + 5) = x^{2} + 5 x$ and $(x + 2) (x + 3) = x^{2} + 5 x + 6$
Why: Pairing the outer two and the inner two makes both quadratics agree except for a constant - this is the whole trick.
Pair four linear factors so one quadratic repeats
2
with $t = x^{2} + 5 x$ , the expression is $t (t + 6) - 72 = t^{2} + 6 t - 72$
Why: The substitution turns a quartic into a quadratic in $t$ .
3
$t^{2} + 6 t - 72 = (t + 12) (t - 6)$
Why: Split the middle term with $12 \times (- 6) = - 72$ and $12 + (- 6) = 6$ .
Factorise a quadratic by splitting the middle term
4
$= (x^{2} + 5 x + 12) (x^{2} + 5 x - 6)$
Why: Substitute $t = x^{2} + 5 x$ back into both factors.
5
$x^{2} + 5 x - 6 = (x + 6) (x - 1)$ , while $x^{2} + 5 x + 12$ has discriminant $25 - 48 < 0$ and does not factorise further
Why: Split the middle term with $6 \times (- 1) = - 6$ and $6 + (- 1) = 5$ ; the other quadratic is irreducible over the reals.

Shortcut

Check the constant term: at $x = 0$ the expression is $- 72$ , and only an option whose factors multiply to $- 72$ at $x = 0$ can be right.

Common mistakes

Option A

Kept the leading $x$ as a factor, but $x = 0$ gives $- 72 \neq = 0$ , so $x$ cannot divide the expression.

Option B

Got the linear factors right but mis-split $t^{2} + 6 t - 72$ , landing on $x^{2} - 2 x - 12$ for the irreducible quadratic.

Option D

Signs flipped throughout: $(x + 1)$ and $(x - 6)$ in place of $(x - 1)$ and $(x + 6)$ .

Skills used

Pair four linear factors so one quadratic repeats Factorise a quadratic by splitting the middle term

20Algebra • Polynomial • HCF and LCM

ModerateIdeal time: about 1.5 min

If the HCF of polynomials $f (x) = (x - 1) (x^{2} + 3 x + a)$ and $g (x) = (x + 2) (x^{2} + 2 x + b)$ is $(x^{2} + x - 2)$ , then what are the values of $a$ and $b$ respectively ?

2, 2

2, - 3

- 1, - 3

- 2, - 1

Show answer & solution

Option B

2, - 3

Approach

Factorise the stated HCF; each of its factors must divide the polynomial that does not already carry it.

HCF and LCM of polynomials from their complete factorisations

Step-by-step solution

1
$x^{2} + x - 2 = (x + 2) (x - 1)$
Why: Split the middle term with $2 \times (- 1) = - 2$ and $2 + (- 1) = 1$ .
Factorise a quadratic by splitting the middle term
2
$(x + 2)$ and $(x - 1)$ must both divide $f (x)$ and $g (x)$
Why: The HCF is the product of the factors common to both, so each of its factors divides each polynomial.
HCF and LCM of polynomials from their complete factorisations
3
$f (x) = (x - 1) (x^{2} + 3 x + a)$ already has $(x - 1)$ , so $(x + 2)$ must divide $x^{2} + 3 x + a$ : $4 - 6 + a = 0 \Rightarrow a = 2$
Why: Apply the factor theorem at $x = - 2$ to the quadratic factor of $f$ .
Factor and remainder theorem
4
$g (x) = (x + 2) (x^{2} + 2 x + b)$ already has $(x + 2)$ , so $(x - 1)$ must divide $x^{2} + 2 x + b$ : $1 + 2 + b = 0 \Rightarrow b = - 3$
Why: Apply the factor theorem at $x = 1$ to the quadratic factor of $g$ .

Common mistakes

Option A

Found $a = 2$ correctly, then assumed $b = a$ by symmetry instead of testing $g$ .

Option C

Sign slip in $4 - 6 + a = 0$ , reading $a = - 1$ rather than $a = 2$ .

Option D

No route reaches $- 2$ and $- 1$ .

Skills used

HCF and LCM of polynomials from their complete factorisations Factorise a quadratic by splitting the middle term Factor and remainder theorem