Elementary Mathematics 2020 (I)

Set A

Questions 41–60 of 100

41Arithmetic • Number System • Divisibility

HardIdeal time: about 3 min

The number of three digit numbers (all digits are different) which are divisible by $7$ and also divisible by $7$ on reversing the order of the digits, is

ASixBFiveCFourDThree

Show answer & solution

Option C

Four

Approach

Subtract the number from its reversal: the difference is $99 (a - c)$ , which forces $a - c$ to be a multiple of $7$ , leaving only a handful of digit pairs to test.

Place-value expansion of a numeral

Step-by-step solution

1
$N = 100 a + 10 b + c$ , reversed $N^{'} = 100 c + 10 b + a$
Why: Working with the digits as letters is the only way to say anything about both the number and its reversal at once.
Place-value expansion of a numeral
2
$N - N^{'} = 99 (a - c)$
Why: If $7$ divides $N$ and $N^{'}$ it divides their difference, and the $10 b$ term cancels — so the middle digit drops out of this condition entirely.
3
$99 = 7 \times 14 + 1$ , so $7$ must divide $a - c$
Why: $7$ is prime and does not divide $99$ , so the factor of $7$ has to sit in $a - c$ .
Euclid's lemma Dividend $=$ divisor $\times$ quotient $+$ remainder
4
$a \neq = c$ (digits all different), so $a - c = \pm 7$ : $(a, c) = (7, 0), (8, 1), (9, 2), (1, 8), (2, 9)$
Why: Digits run $0$ to $9$ and $a \neq = 0$ , so a difference of $\pm 14$ is impossible; this is the whole list.
Enumerate a residue class within a bound
5
$N = 7 (14 a + b) + (2 a + 3 b + c)$ , so $7$ must divide $2 a + 3 b + c$
Why: Now impose divisibility on $N$ itself. Reducing each place value ( $100 \to 2$ , $10 \to 3$ ) turns the test into one small sum.
6
$(7, 0) : 3 b + 14$ ; $(8, 1) : 3 b + 17$ ; $(9, 2) : 3 b + 20$ ; $(1, 8) : 3 b + 10$ ; $(2, 9) : 3 b + 13$
Why: One linear expression in $b$ per pair; run $b$ from $0$ to $9$ and keep the multiples of $7$ .
7
$b = 0, 7$ for $(7, 0)$ ; $b = 6$ for $(8, 1)$ and $(1, 8)$ ; $b = 5$ for $(9, 2)$ and $(2, 9)$
Why: $3 b + 14 = 14, 35$ at $b = 0, 7$ ; $3 b + 17 = 35$ at $b = 6$ ; $3 b + 20 = 35$ at $b = 5$ ; $3 b + 10 = 28$ at $b = 6$ ; $3 b + 13 = 28$ at $b = 5$ .
8
$700$ and $770$ have repeated digits; $861, 168, 952, 259$ survive, so the count is $4$
Why: The condition 'all digits are different' is what kills the two numbers coming from $(7, 0)$ , and it is the easiest condition in the question to forget.

Shortcut

Multiples of $7$ whose reversal is also a multiple of $7$ need $7 ∣ 99 (a - c)$ , i.e. $7 ∣ a - c$ ; among distinct digits that is only ${1, 8}$ and ${2, 9}$ , and each pair admits exactly one middle digit, giving $168, 861, 259, 952$ .

Common mistakes

Option A

Kept both $700$ and $770$ — each is a multiple of $7$ whose reversal is too — forgetting that all three digits must differ.

Option B

Kept one of $700$ or $770$ , overlooking its repeated digit.

Option D

Found $168$ , $259$ and one reversal, then stopped — the second reversal $952$ was never written down.

Skills used

Place-value expansion of a numeral Euclid's lemma Dividend

=

divisor

\times

quotient

+

remainder Enumerate a residue class within a bound

42Algebra • Equations • Linear Equations

HardIdeal time: about 2.5 min

How many integral values of $x$ and $y$ satisfy the equation $5 x + 9 y = 7$ , where $- 500 < x < 500$ and $- 500 < y < 500$ ?

110

111

112

DNone of the above

Show answer & solution

Option B

111

Approach

Pin down which $x$ can work by making $7 - 5 x$ a multiple of $9$ , write the whole family in one parameter, then count the parameter values both bounds allow.

Enumerate a residue class within a bound

Step-by-step solution

1
$9 y = 7 - 5 x$ , so $9$ must divide $7 - 5 x$
Why: $y$ is an integer only when the right-hand side is a multiple of $9$ ; this converts the two-variable problem into one condition on $x$ .
Dividend $=$ divisor $\times$ quotient $+$ remainder
2
$x = 5$ gives $7 - 25 = - 18 = 9 \times (- 2)$ , so $x = 5, y = - 2$ is one solution
Why: Testing $x = 0, 1, 2, \dots$ finds the first fit quickly; one solution is all that is needed to generate the rest.
Enumerate a residue class within a bound
3
$x = 9 t + 5, y = - 5 t - 2$ for every integer $t$
Why: Moving $x$ by $9$ changes $5 x$ by $45$ , which $9 y$ absorbs by moving $y$ by $5$ the other way; $5$ and $9$ share no factor, so no smaller step works and nothing is missed.
Integer solutions of $a x + b y = c$
4
$- 500 < 9 t + 5 < 500 \Rightarrow - 56 \leq t \leq 54$
Why: $9 t < 495$ gives $t \leq 54$ , and $9 t > - 505$ gives $t \geq - 56$ since $- 505/9 = - 56.1$ .
Multiplying an inequality by a negative number reverses it
5
$- 500 < - 5 t - 2 < 500 \Rightarrow - 100 \leq t \leq 99$
Why: The $y$ -bound is the looser of the two, so it removes nothing; checking it is still necessary before counting.
6
$t = - 56, \dots, 54$ , a count of $54 - (- 56) + 1 = 111$
Why: Both endpoints are included, so the count is the difference plus one — the step this question is built to catch.

Shortcut

$x$ must leave remainder $5$ on division by $9$ , so $x$ runs over $- 499, - 490, \dots, 491$ : an arithmetic run of common difference $9$ from $- 499$ to $491$ , which has $\frac{491 - ( - 499 )}{9} + 1 = 111$ terms.

Common mistakes

Option A

Counted $54 - (- 56) = 110$ , forgetting that both endpoints are themselves solutions.

Option C

Allowed $t = 55$ , which gives $x = 500$ — the bound $x < 500$ is strict.

Option D

No plausible route reaches 'None of the above'; $111$ is on the list.

Skills used

Dividend

=

divisor

\times

quotient

+

remainder Enumerate a residue class within a bound Multiplying an inequality by a negative number reverses it Integer solutions of

a x + b y = c

43Arithmetic • Number System • Divisibility

ModerateIdeal time: about 1.5 min

Let $X Y Z$ be a $3$ -digit number. Let $S = X Y Z + Y Z X + Z X Y$ . Which of the following statements is/are correct?

1: $S$ is always divisible by $3$ and $(X + Y + Z)$

2: $S$ is always divisible by $9$

3: $S$ is always divisible by $37$

Select the correct answer using the code given below:

A1 onlyB2 onlyC1 and 2D1 and 3

Show answer & solution

Option D

1 and 3

Approach

Expand all three cyclic numerals by place value; every digit ends up with the same coefficient, so $S = 111 (X + Y + Z)$ , and the three claims are read off from $111 = 3 \times 37$ .

Place-value expansion of a numeral

Step-by-step solution

1
$X Y Z = 100 X + 10 Y + Z, Y Z X = 100 Y + 10 Z + X, Z X Y = 100 Z + 10 X + Y$
Why: The three numerals are the same digits cycled, so each digit occupies each place exactly once.
Place-value expansion of a numeral
2
$S = (100 + 10 + 1) (X + Y + Z) = 111 (X + Y + Z)$
Why: Adding the three expansions gives each digit the coefficient $100 + 10 + 1$ — this one line settles the whole question.
3
$111 = 3 \times 37$
Why: $111$ is not prime, and the factor $37$ is the part a candidate is most likely to miss.
Prime factorisation
4
Statement 1 holds: $S = 3 \times 37 \times (X + Y + Z)$ is divisible by $3$ and by $(X + Y + Z)$
Why: Both divisors appear explicitly as factors, whatever the digits are.
Dividend $=$ divisor $\times$ quotient $+$ remainder
5
Statement 2 fails: $X Y Z = 100$ gives $S = 100 + 1 + 10 = 111$ , and $9 ∤ 111$
Why: $111$ carries only one factor of $3$ , so $9$ divides $S$ only when $X + Y + Z$ is itself a multiple of $3$ — not something the question guarantees.
A number is congruent to its digit sum modulo $9$
6
Statement 3 holds: $37$ is a factor of $111$ , hence of $S$
Why: Same reason as statement 1; the factor is there for every choice of digits.

Shortcut

$S = 111 (X + Y + Z)$ and $111 = 3 \times 37$ : statements 1 and 3 are immediate, and statement 2 needs a second factor of $3$ that nothing supplies.

Common mistakes

Option A

Reached $S = 111 (X + Y + Z)$ but left $111$ unfactorised, so the factor $37$ was never seen.

Option B

No route makes statement 2 the only correct one; statement 1 is visible the moment $S$ is written as $111 (X + Y + Z)$ .

Option C

Assumed $X + Y + Z$ supplies the second factor of $3$ needed for $9$ ; $X Y Z = 100$ , where $S = 111$ , shows it need not.

Skills used

Place-value expansion of a numeral Prime factorisation Dividend

=

divisor

\times

quotient

+

remainder A number is congruent to its digit sum modulo

9

44Arithmetic • Time, Speed and Distance

EasyIdeal time: about 55 s

In covering certain distance, the average speeds of $X$ and $Y$ are in the ratio $4 : 5$ . If $X$ takes $45$ minutes more than $Y$ to reach the destination, then what is the time taken by $Y$ to reach the destination?

135

minutesB

150

minutesC

180

minutesD

225

minutes

Show answer & solution

Option C

180

minutes

Approach

Over the same distance time runs inversely to speed, so the times are in the ratio $5 : 4$ ; the $45$ minutes is the single part by which they differ.

Over a fixed distance, time varies inversely with speed

Step-by-step solution

1
Speeds $X : Y = 4 : 5 \Rightarrow$ times $X : Y = 5 : 4$
Why: The distance is the same for both, so whoever is slower takes proportionally longer — the ratio simply turns over.
Over a fixed distance, time varies inversely with speed
2
Times $= 5 k$ and $4 k$ minutes
Why: A ratio fixes the two times only up to a common factor; naming it $k$ lets the one given difference pin it down.
Introduce a common factor $k$ for a ratio
3
$5 k - 4 k = 45 \Rightarrow k = 45$
Why: $X$ takes $45$ minutes more, and in these units the gap is exactly one part.
Turn a worded condition into a linear equation in one unknown
4
$Y$ 's time $= 4 k = 180$ minutes
Why: The question asks for $Y$ , the faster of the two, so it is the $4$ -part time, not the $5$ -part one.

Shortcut

Times are in the ratio $5 : 4$ , the difference is $1$ part $= 45$ min, so $Y$ takes $4 \times 45 = 180$ min.

Common mistakes

Option A

No plausible route reaches $135$ minutes.

Option B

No plausible route reaches $150$ minutes.

Option D

Gave $X$ 's time, $5 \times 45 = 225$ minutes, instead of $Y$ 's.

Skills used

Over a fixed distance, time varies inversely with speed Introduce a common factor

k

for a ratio Turn a worded condition into a linear equation in one unknown

45Statistics • Mean

EasyIdeal time: about 50 s

For two observations, the sum is $S$ and product is $P$ . What is the harmonic mean of these two observations?

\frac{2 S}{P}

\frac{S}{( 2 P )}

\frac{2 P}{S}

\frac{P}{( 2 S )}

Show answer & solution

Option C

\frac{2 P}{S}

Approach

Write the harmonic mean of two numbers from its definition — the reciprocal of the average of the reciprocals — and the sum and product appear on their own.

Harmonic mean:

n

over the sum of the reciprocals

Step-by-step solution

1
Let the observations be $a$ and $b$ , so $a + b = S$ and $ab = P$
Why: Naming them lets the definition be written out; $S$ and $P$ are the only combinations that will survive.
Sum and product of roots (Vieta)
2
$HM = \frac{2}{\frac{1}{a} + \frac{1}{b}}$
Why: The harmonic mean of $n$ values is $n$ divided by the sum of the reciprocals; for two values that is $2$ over two reciprocals.
Average speed over two equal distances Harmonic mean: $n$ over the sum of the reciprocals
3
$\frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab} = \frac{S}{P}$
Why: Combining the reciprocals over a common denominator produces exactly the sum over the product.
Clear the denominators of an equation
4
$HM = \frac{2}{S / P} = \frac{2 P}{S}$
Why: Dividing by a fraction multiplies by its reciprocal, and the $2$ stays on top.
Negative exponent is a reciprocal

Shortcut

For two numbers the harmonic mean is $\frac{2 ab}{a + b}$ , which is $\frac{2 P}{S}$ on sight.

Common mistakes

Option A

Turned the formula upside down, writing $\frac{2 ( a + b )}{ab}$ in place of $\frac{2 ab}{a + b}$ .

Option B

Averaged the reciprocals and stopped: $\frac{1}{2} (\frac{1}{a} + \frac{1}{b}) = \frac{S}{2 P}$ is the mean of the reciprocals, not its reciprocal.

Option D

Divided $\frac{P}{S}$ by $2$ instead of multiplying by $2$ .

Skills used

Harmonic mean:

n

over the sum of the reciprocals Sum and product of roots (Vieta)Average speed over two equal distances Clear the denominators of an equation Negative exponent is a reciprocal

46Arithmetic • Percentage

EasyIdeal time: about 50 s

If the annual income of $X$ is $20%$ more than that of $Y$ , then the income of $Y$ is less than that of $X$ by $p %$ . What is the value of $p$ ?

10

16 \frac{2}{3}

17 \frac{1}{3}

20

Show answer & solution

Option B

16 \frac{2}{3}

Approach

Give $Y$ a convenient income of $100$ , compute $X$ , then measure the shortfall as a percentage of $X$ — because 'less than $X$ ' makes $X$ the base.

x %

of a quantity

Step-by-step solution

1
$Y = 100 \Rightarrow X = 120$
Why: $X$ is $20%$ more than $Y$ , so the increase is measured on $Y$ ; choosing $Y = 100$ makes that arithmetic trivial.
Percentage change as a multiplier
2
Shortfall $= X - Y = 20$
Why: The absolute gap is the same whichever direction it is read; only the base changes.
$x %$ of a quantity
3
$p = \frac{20}{120} \times 100$
Why: ' $Y$ is less than $X$ by $p %$ ' measures the gap against $X$ , which is $120$ , not against $Y$ .
4
$p = \frac{50}{3} = 16 \frac{2}{3}$
Why: $\frac{20}{120} = \frac{1}{6}$ and $\frac{100}{6} = \frac{50}{3}$ .
Comparing two fractions

Shortcut

A rise of $r %$ one way is a fall of $\frac{100 r}{100 + r} %$ the other: $\frac{100 \times 20}{120} = 16 \frac{2}{3}$ .

Common mistakes

Option A

No plausible route reaches $10$ .

Option C

No plausible route reaches $17 \frac{1}{3}$ ; it is placed next to the key to catch a guess.

Option D

Assumed the percentage reads the same both ways and reused the given $20$ ; but the base shifts from $Y = 100$ to $X = 120$ .

Skills used

Percentage change as a multiplier

x %

of a quantity Comparing two fractions

47Arithmetic • Number System • LCM and HCF

ModerateIdeal time: about 1.5 min

What is the least perfect square which is divisible by $3$ , $4$ , $5$ , $6$ and $7$ ?

1764

17640

44100

176400

Show answer & solution

Option C

44100

Approach

Any common multiple of $3, 4, 5, 6, 7$ is a multiple of their LCM; make that LCM a square by topping up the primes whose exponents are odd.

LCM and HCF of several numbers from prime powers

Step-by-step solution

1
$3, 4 = 2^{2}, 5, 6 = 2 \cdot 3, 7$
Why: Prime factorisations are what the LCM and the square condition are both read off.
Prime factorisation
2
$LCM = 2^{2} \cdot 3 \cdot 5 \cdot 7 = 420$
Why: Take each prime to the highest power appearing; $420$ is the smallest number divisible by all five.
LCM and HCF of several numbers from prime powers
3
A perfect square has every prime exponent even; in $2^{2} \cdot 3 \cdot 5 \cdot 7$ the odd ones are $3, 5, 7$
Why: The required number must be a multiple of $420$ , so it carries at least these primes; squaring demands one more of each of the three that appear singly.
A perfect square has every prime exponent even
4
$420 \times (3 \cdot 5 \cdot 7) = 420 \times 105 = 44100$
Why: Multiplying by exactly $105$ is the least top-up: anything smaller leaves some exponent odd.
5
$44100 = 2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 7^{2} = (2 \cdot 3 \cdot 5 \cdot 7)^{2} = 21 0^{2}$
Why: Confirms it is a square and shows the answer in the compact form $21 0^{2}$ .
Least perfect square not below a given bound

Shortcut

The number must be $(2 \cdot 3 \cdot 5 \cdot 7)^{2} = 21 0^{2} = 44100$ : every prime in $LCM = 420$ has to appear to an even power, and $2, 3, 5, 7$ each need to appear at least once.

Common mistakes

Option A

$1764 = 4 2^{2}$ is the least square divisible by $3, 4, 6, 7$ — the $5$ was dropped from the LCM.

Option B

Multiplied $420$ by $42$ to 'balance' it, but $17640 = 2^{3} \cdot 3^{2} \cdot 5 \cdot 7^{2}$ still has odd exponents and is not a square.

Option D

Squared the LCM, $42 0^{2} = 176400$ ; that works but is four times larger than needed, because the $2^{2}$ in $420$ was already even.

Skills used

Prime factorisation LCM and HCF of several numbers from prime powers Least perfect square not below a given bound A perfect square has every prime exponent even

48Arithmetic • Time and Work • Pipes and Cisterns

EasyIdeal time: about 40 s

In a water tank there are two outlets. It takes $20$ minutes to empty the tank if both the outlets are opened. If the first outlet is opened, the tank is emptied in $30$ minutes. What is the time taken to empty the tank by second outlet?

30

minutesB

40

minutesC

50

minutesD

60

minutes

Show answer & solution

Option D

60

minutes

Approach

Outlets add their rates, not their times: subtract the first outlet's rate from the combined rate to get the second's.

Take total work as the LCM of the times

Step-by-step solution

1
Total work $= 60$ units (the LCM of $20$ and $30$ )
Why: Choosing the LCM as the tankful keeps every rate a whole number and avoids fractions entirely.
Take total work as the LCM of the times
2
Both together: $\frac{60}{20} = 3$ units/min; first alone: $\frac{60}{30} = 2$ units/min
Why: Rate is work divided by time, and rates of outlets working at the same moment add.
3
Second outlet $= 3 - 2 = 1$ unit/min
Why: The combined rate is the sum of the two, so the unknown rate is the difference.
4
Time $= \frac{60}{1} = 60$ minutes
Why: Back from rate to time by dividing the tankful by the rate.

Shortcut

$\frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}$ , so $60$ minutes.

Common mistakes

Option A

Copied the first outlet's time, $30$ minutes.

Option B

No plausible route reaches $40$ minutes.

Option C

Added the two given times, $20 + 30$ , instead of subtracting rates.

Skills used

Take total work as the LCM of the times

49Algebra • Polynomial • Factor and Remainder Theorem

ModerateIdeal time: about 1.5 min

If $(x^{2} - 1)$ is a factor of $a x^{4} + b x^{3} + c x^{2} + d x + e$ , then which one of the following is correct?

a + b + c = d + e

a + b + e = c + d

b + c + d = a + e

a + c + e = b + d

Show answer & solution

Option D

a + c + e = b + d

Approach

$x^{2} - 1 = (x - 1) (x + 1)$ , so both $x = 1$ and $x = - 1$ are roots; add and subtract the two resulting equations.

Factor and remainder theorem

Step-by-step solution

1
$x^{2} - 1 = (x + 1) (x - 1)$
Why: A quadratic factor splits into two linear ones, which is what lets the factor theorem be used twice.
Difference of two squares
2
$f (1) = a + b + c + d + e = 0$
Why: If $(x - 1)$ is a factor then $f (1) = 0$ ; substituting $x = 1$ just adds the coefficients.
Factor and remainder theorem
3
$f (- 1) = a - b + c - d + e = 0$
Why: $(- 1)^{4} = 1$ , $(- 1)^{3} = - 1$ , $(- 1)^{2} = 1$ , $(- 1)^{1} = - 1$ : the odd-power coefficients change sign, the even ones do not.
4
Adding: $2 (a + c + e) = 0$ ; subtracting: $2 (b + d) = 0$
Why: The two equations have the same terms with reversed signs on $b$ and $d$ , so adding isolates the even part and subtracting the odd part.
Add and subtract when the coefficients are reversed
5
$a + c + e = 0$ and $b + d = 0$ , hence $a + c + e = b + d$
Why: Both sides are zero, so the stated equality holds — and it is the only one of the four that follows without extra assumptions.
Eliminate one variable from a linear pair

Shortcut

$f (1) = f (- 1) = 0$ means the even-power coefficients sum to zero and the odd-power ones sum to zero: $a + c + e = 0 = b + d$ .

Common mistakes

Option A

From $f (1) = 0$ , $a + b + c = - (d + e)$ ; the minus sign was dropped on the way across.

Option B

Alternated the signs in $f (- 1)$ starting from the wrong end, giving $a + b - c - d + e = 0$ .

Option C

Took $(- 1)^{2}$ as $- 1$ in $f (- 1)$ , giving $a - b - c - d + e = 0$ and hence $b + c + d = a + e$ .

Skills used

Difference of two squares Factor and remainder theorem Add and subtract when the coefficients are reversed Eliminate one variable from a linear pair

50Algebra • Expressions • Algebraic Identities

ModerateIdeal time: about 1.5 min

If $(x^{8} + \frac{1}{x ^{8}}) = 47$ , what is the value of $(x^{6} + \frac{1}{x ^{6}})$ ?

36

27

18

9

Show answer & solution

Option C

18

Approach

Treat $u = x^{2}$ and walk down the ladder $u^{4} + \frac{1}{u ^{4}} \to u^{2} + \frac{1}{u ^{2}} \to u + \frac{1}{u}$ with $t^{2} - 2$ , then climb to $u^{3} + \frac{1}{u ^{3}}$ with the cube identity.

Square of a sum and of a difference

Step-by-step solution

1
Put $u = x^{2}$ , so the given is $u^{4} + \frac{1}{u ^{4}} = 47$ and the target is $u^{3} + \frac{1}{u ^{3}}$
Why: Every exponent in the question is even, so working in $u = x^{2}$ removes the clutter and halves the ladder.
Power of a power, and the move to a common base
2
$u^{2} + \frac{1}{u ^{2}} = 47 + 2 = 7$
Why: $(u^{2} + \frac{1}{u ^{2}})^{2} = u^{4} + 2 + \frac{1}{u ^{4}}$ , so the square of the smaller expression is $47 + 2 = 49$ . The negative root is rejected because $u^{2} + \frac{1}{u ^{2}} \geq 2$ for real $u$ .
Square of a sum and of a difference A positive number plus its reciprocal is at least $2$
3
$u + \frac{1}{u} = 7 + 2 = 3$
Why: The same identity one rung down: $(u + \frac{1}{u})^{2} = u^{2} + 2 + \frac{1}{u ^{2}} = 9$ , and again only the positive root can occur.
4
$u^{3} + \frac{1}{u ^{3}} = (u + \frac{1}{u})^{3} - 3 (u + \frac{1}{u}) = 27 - 9 = 18$
Why: $(a + b)^{3} = a^{3} + b^{3} + 3 ab (a + b)$ with $ab = 1$ gives $a^{3} + b^{3} = (a + b)^{3} - 3 (a + b)$ — the correction term is what the wrong options drop.
Cube of a sum and of a difference

Shortcut

$47 = 7^{2} - 2$ and $7 = 3^{2} - 2$ , so $x^{2} + \frac{1}{x ^{2}} = 3$ ; then $x^{6} + \frac{1}{x ^{6}} = 3^{3} - 3 \cdot 3 = 18$ .

Common mistakes

Option A

Added the correction term instead of subtracting it: $27 + 9 = 36$ .

Option B

Cubed $x^{2} + \frac{1}{x ^{2}} = 3$ and stopped at $27$ , forgetting the $- 3 (x^{2} + \frac{1}{x ^{2}})$ term.

Option D

Squared $3$ in place of cubing, as if $x^{6} + \frac{1}{x ^{6}}$ came from $(x^{2} + \frac{1}{x ^{2}})^{2}$ .

Skills used

Power of a power, and the move to a common base Square of a sum and of a difference A positive number plus its reciprocal is at least

2

Cube of a sum and of a difference

51Trigonometry • Trigonometric Identity • Degree and Radian

EasyIdeal time: about 45 s

A wheel makes $360$ revolutions in one minute. What is the number of radians it turns in one second?

4 π

6 π

12 π

16 π

Show answer & solution

Option C

12 π

Approach

Convert revolutions per minute to revolutions per second, then one revolution to $2 π$ radians.

Converting between degrees and radians

Step-by-step solution

1
$\frac{360 rev}{60 s} = 6$ revolutions per second
Why: One minute is $60$ seconds, so the rate per second is the rate per minute divided by $60$ .
Unitary method: find one unit, then scale
2
One revolution $= 36 0^{\circ} = 2 π$ radians
Why: $π$ radians $= 18 0^{\circ}$ , so a full turn is $2 π$ — a half-turn would be only $π$ .
Converting between degrees and radians
3
$6 \times 2 π = 12 π$ radians per second
Why: Multiply the number of turns by the radian measure of one turn.

Shortcut

$360$ rev/min $= 6$ rev/s $= 12 π$ rad/s.

Common mistakes

Option A

No plausible route reaches $4 π$ .

Option B

Found $6$ revolutions per second but took one revolution as $π$ radians instead of $2 π$ .

Option D

No plausible route reaches $16 π$ .

Skills used

Unitary method: find one unit, then scale Converting between degrees and radians

52Trigonometry • Trigonometric Identity • Maximum and Minimum Values

HardIdeal time: about 2 min

What is the least value of $(25 cosec^{2} x + sec^{2} x)$ ?

40

36

26

24

Show answer & solution

Option B

36

Approach

Use the Pythagorean identities to split off a constant, leaving a term plus a multiple of its reciprocal, which has a known least value.

A positive number plus its reciprocal is at least

2

Step-by-step solution

1
$25 cosec^{2} x + sec^{2} x = 25 (1 + cot^{2} x) + (1 + tan^{2} x)$
Why: $1 + cot^{2} x = cosec^{2} x$ and $1 + tan^{2} x = sec^{2} x$ ; rewriting this way pulls out the part that cannot vary.
Pythagorean trigonometric identities
2
$= 26 + 25 cot^{2} x + tan^{2} x$
Why: The $26$ is fixed, so minimising the whole expression means minimising $25 cot^{2} x + tan^{2} x$ .
3
With $t = tan^{2} x > 0$ : $\frac{25}{t} + t = 5 (\frac{5}{t} + \frac{t}{5}) \geq 5 \times 2 = 10$
Why: $cot^{2} x = \frac{1}{t a n ^{2} x}$ , so the two terms are a number and a multiple of its reciprocal; $z + \frac{1}{z} \geq 2$ for $z > 0$ with $z = \frac{t}{5}$ .
Reciprocal and quotient relations between the ratios A positive number plus its reciprocal is at least $2$
4
Least value $= 26 + 10 = 36$ , attained when $t = 5$ , i.e. $tan^{2} x = 5$
Why: Equality in $z + \frac{1}{z} \geq 2$ needs $z = 1$ , i.e. $t = 5$ , and $tan^{2} x = 5$ is reachable, so the bound is actually achieved.

Shortcut

For positive $a, b$ the least value of $a^{2} cosec^{2} x + b^{2} sec^{2} x$ is $(a + b)^{2}$ ; here $(5 + 1)^{2} = 36$ .

Common mistakes

Option A

No plausible route reaches $40$ .

Option C

Read $26$ off the constant term, as if $25 cot^{2} x$ and $tan^{2} x$ could both be made zero; they are reciprocals, so they cannot.

Option D

Sign slip on the two $1$ s from the identities, giving $25 - 1$ as the constant part.

Skills used

Pythagorean trigonometric identities Reciprocal and quotient relations between the ratios A positive number plus its reciprocal is at least

2

53Trigonometry • Trigonometric Identity • Complementary Angles

HardIdeal time: about 2 min

Let $0 < θ < 9 0^{\circ}$ and $100 θ = 9 0^{\circ}$ . If $α = \prod_{n = 1}^{99} cot n θ$ , then which one of the following is correct?

α = 1

α = 0

α > 1

0 < α < 1

Show answer & solution

Option A

α = 1

Approach

$θ = 0. 9^{\circ}$ , so the angles $n θ$ and $(100 - n) θ$ are complementary; pair the factors from the two ends and each pair multiplies to $1$ .

Ratios of complementary angles

Step-by-step solution

1
$100 θ = 9 0^{\circ} \Rightarrow θ = 0. 9^{\circ}$ , and the factors run $cot 0. 9^{\circ}, cot 1. 8^{\circ}, \dots, cot 89. 1^{\circ}$
Why: $n$ runs from $1$ to $99$ , so the largest angle is $99 θ = 89. 1^{\circ}$ — every factor is a cotangent of an acute angle, so all $99$ are defined and positive.
Converting between degrees and radians
2
$n θ + (100 - n) θ = 100 θ = 9 0^{\circ}$
Why: This is why pairing from the two ends is the move: the $n$ th factor and the $(100 - n)$ th sit on complementary angles.
Ratios of complementary angles
3
$cot (9 0^{\circ} - n θ) = tan n θ$ , so $cot n θ \cdot cot (100 - n) θ = cot n θ \cdot tan n θ = 1$
Why: Cotangent and tangent of the same angle are reciprocals, so each pair collapses to $1$ .
Reciprocal and quotient relations between the ratios
4
$n = 1, \dots, 99$ gives $49$ such pairs plus the self-paired $n = 50$ , where $50 θ = 4 5^{\circ}$ and $cot 4 5^{\circ} = 1$
Why: $99$ is odd, so one factor is left over; it happens to be the one at $4 5^{\circ}$ , whose cotangent is $1$ .
Exact trigonometric ratios of the standard angles
5
$α = 1^{49} \times 1 = 1$
Why: Every pair contributes $1$ and the middle factor is $1$ , so the whole product is $1$ .

Shortcut

Pair $cot n θ$ with $cot (100 - n) θ$ : complementary angles make each pair $1$ , and the leftover $cot 4 5^{\circ}$ is also $1$ , so $α = 1$ .

Common mistakes

Option B

Expected a factor of $cot 9 0^{\circ} = 0$ , but $n θ$ stops at $89. 1^{\circ}$ and never reaches $9 0^{\circ}$ .

Option C

Noticed the early factors $cot 0. 9^{\circ}, cot 1. 8^{\circ}, \dots$ are very large and concluded the product must exceed $1$ , without pairing them against the small factors at the other end.

Option D

Paired correctly but inverted one factor in each pair, reading every pair as less than $1$ .

Skills used

Converting between degrees and radians Ratios of complementary angles Reciprocal and quotient relations between the ratios Exact trigonometric ratios of the standard angles

54Trigonometry • Trigonometric Equation

EasyIdeal time: about 55 s

If $tan 6 θ = cot 2 θ$ , where $0 < 6 θ < \frac{π}{2}$ , then what is the value of $sec 4 θ$ ?

2

2

\frac{2}{3}

\frac{4}{3}

Show answer & solution

Option A

2

Approach

Turn the cotangent into a tangent of the complementary angle; the two tangents then force their angles to be equal, which pins $θ$ .

Ratios of complementary angles

Step-by-step solution

1
$cot 2 θ = tan (9 0^{\circ} - 2 θ)$
Why: Putting both sides in terms of tangent is what makes the equation solvable without any extra machinery.
Ratios of complementary angles
2
$tan 6 θ = tan (9 0^{\circ} - 2 θ) \Rightarrow 6 θ = 9 0^{\circ} - 2 θ$
Why: $0 < 6 θ < 9 0^{\circ}$ and $0 < 9 0^{\circ} - 2 θ < 9 0^{\circ}$ , and tangent is strictly increasing on that range, so equal tangents mean equal angles.
How the ratios move across the first quadrant
3
$8 θ = 9 0^{\circ} \Rightarrow θ = 11.2 5^{\circ}$ , so $4 θ = 4 5^{\circ}$
Why: The question asks about $4 θ$ , which is exactly half of $8 θ = 9 0^{\circ}$ — no need to carry the awkward $11.2 5^{\circ}$ any further.
Turn a worded condition into a linear equation in one unknown
4
$sec 4 θ = sec 4 5^{\circ} = 2$
Why: $cos 4 5^{\circ} = \frac{1}{2}$ , and secant is its reciprocal.
Exact trigonometric ratios of the standard angles Reciprocal and quotient relations between the ratios

Shortcut

$tan A = cot B$ means $A + B = 9 0^{\circ}$ , so $8 θ = 9 0^{\circ}$ and $4 θ = 4 5^{\circ}$ : $sec 4 θ = 2$ .

Common mistakes

Option B

Solved $6 θ = 9 0^{\circ}$ , i.e. $θ = 1 5^{\circ}$ , so $sec 4 θ$ was read as $sec 6 0^{\circ} = 2$ .

Option C

Solved $8 θ = 6 0^{\circ}$ instead of $9 0^{\circ}$ , giving $4 θ = 3 0^{\circ}$ and $sec 3 0^{\circ} = \frac{2}{3}$ .

Option D

No plausible route reaches $\frac{4}{3}$ .

Skills used

Ratios of complementary angles How the ratios move across the first quadrant Turn a worded condition into a linear equation in one unknown Exact trigonometric ratios of the standard angles Reciprocal and quotient relations between the ratios

55Trigonometry • Heights and Distances

EasyIdeal time: about 1 min

A tree of height $15$ m is broken by wind in such a way that its top touches the ground and makes an angle $3 0^{\circ}$ with the ground. What is the height from the ground to the point where tree is broken?

10

7

5

3

Show answer & solution

Option C

5

Approach

The broken piece is the hypotenuse of a right triangle whose vertical leg is the standing stump; the two pieces together are $15$ m.

Sine, cosine and tangent in a right triangle

Step-by-step solution

1
Stump $= h$ , broken piece $= 15 - h$
Why: The tree snaps but nothing is lost, so the two pieces account for the whole $15$ m.
Turn a worded condition into a linear equation in one unknown
2
The broken piece leans from the break point to the ground at $3 0^{\circ}$ , so it is the hypotenuse and $h$ is the side opposite the $3 0^{\circ}$
Why: The ground is horizontal and the stump vertical, so the triangle is right-angled at the foot of the stump.
Sine, cosine and tangent in a right triangle
3
$sin 3 0^{\circ} = \frac{h}{15 - h} \Rightarrow \frac{1}{2} = \frac{h}{15 - h}$
Why: Opposite over hypotenuse — and crucially the hypotenuse is the broken piece, not the original $15$ m.
Exact trigonometric ratios of the standard angles
4
$15 - h = 2 h \Rightarrow h = 5$ m
Why: Cross-multiplying clears the fraction and leaves a one-line linear equation.
Clear the denominators of an equation

Shortcut

At $3 0^{\circ}$ the hypotenuse is twice the opposite side, so the broken piece is twice the stump: $h + 2 h = 15$ , $h = 5$ m.

Common mistakes

Option A

Gave the length of the broken piece, $15 - h = 10$ m, rather than the height of the break.

Option B

Measured the $3 0^{\circ}$ against the full height, $sin 3 0^{\circ} = \frac{h}{15}$ , giving $7.5$ and rounding to $7$ .

Option D

No plausible route reaches $3$ m.

Skills used

Turn a worded condition into a linear equation in one unknown Sine, cosine and tangent in a right triangle Exact trigonometric ratios of the standard angles Clear the denominators of an equation

56Geometry • Triangle • Similarity

HardIdeal time: about 3 min

On a plane area there are two vertical towers separated by $100$ feet apart. The shorter tower is $40$ feet tall. A pole of length $6$ feet stands on the line joining the base of two towers so that the tip of the towers and tip of the pole are also on the same line. If the distance of the pole from the shorter tower is $75$ feet, then what is the height of the taller tower (approximately)?

85

feetB

110

feetC

125

feetD

140

feet

Show answer & solution

Option A

85

feet

Approach

Three collinear tips means one straight line of sight; measure heights above the pole's tip and use similar triangles on the two horizontal distances.

Step-by-step solution

1
Shorter tower $40$ ft and taller tower $H$ ft stand $100$ ft apart; the pole is $6$ ft tall and $75$ ft from the shorter tower
Why: Write the configuration down before drawing: the three tips lie on one line, so two of the three right triangles under that line are similar.
Angles of elevation and depression
2
The pole cannot lie between the towers: a line through $(0, 40)$ and $(75, 6)$ falls to $- 5 \frac{1}{3}$ at the far tower, which is below ground
Why: A straight line cannot drop from $40$ to $6$ and then climb to a height above $40$ , so the pole must sit beyond the shorter tower, $75$ ft out on the same straight line.
Parallel lines cut transversals proportionally
3
Measuring from the pole's tip: the shorter tower rises $40 - 6 = 34$ ft over $75$ ft, the taller rises $H - 6$ ft over $75 + 100 = 175$ ft
Why: Taking the pole's tip as the eye-point makes the two triangles share the angle of elevation, so they are similar by AA.
Similar triangles have proportional sides
4
$\frac{34}{75} = \frac{H - 6}{175} \Rightarrow H - 6 = 34 \times \frac{7}{3} = \frac{238}{3}$
Why: Corresponding sides of similar triangles are proportional; $\frac{175}{75} = \frac{7}{3}$ keeps the arithmetic in small numbers.
5
$H = 6 + \frac{238}{3} = \frac{256}{3} = 85 \frac{1}{3} \approx 85$ ft
Why: The $6$ ft has to be added back because the ratio was measured from the pole's tip, not from the ground.
Unitary method: find one unit, then scale

Shortcut

From the pole's tip the sight line climbs $34$ ft in $75$ ft; over $175$ ft it climbs $\frac{7}{3} \times 34 = 79 \frac{1}{3}$ , so the taller tower is $6 + 79 \frac{1}{3} \approx 85$ ft.

Common mistakes

Option B

No plausible route reaches $110$ feet.

Option C

No plausible route reaches $125$ feet.

Option D

Put the pole between the towers, $25$ ft from the taller one, and solved $\frac{40 - 6}{25} = \frac{H - 6}{100}$ to get $142 \approx 140$ .

Skills used

Angles of elevation and depression Parallel lines cut transversals proportionally Similar triangles have proportional sides Unitary method: find one unit, then scale

57Trigonometry • Heights and Distances

EasyIdeal time: about 55 s

The angles of elevation of the top of a tower from two points at distances $p$ and $q$ from the base and on the same straight line are $2 7^{\circ}$ and $6 3^{\circ}$ respectively. What is the height of the tower?

pq

pq

\frac{pq}{2}

\frac{pq}{2}

Show answer & solution

Option B

pq

Approach

Write the two tangents from the right triangles; since $2 7^{\circ}$ and $6 3^{\circ}$ are complementary, their product is $1$ and the height drops out as a geometric mean.

Ratios of complementary angles

Step-by-step solution

1
$tan 2 7^{\circ} = \frac{h}{p}$ and $tan 6 3^{\circ} = \frac{h}{q}$
Why: Each point of observation, the base and the top form a right triangle with the tower as the opposite side.
Angles of elevation and depression Sine, cosine and tangent in a right triangle
2
$2 7^{\circ} + 6 3^{\circ} = 9 0^{\circ}$ , so $tan 6 3^{\circ} = cot 2 7^{\circ} = \frac{1}{tan 2 7 ^{\circ}}$
Why: Spotting that the two angles are complementary is what makes the unknown angles cancel; without it the question needs tables.
Ratios of complementary angles Reciprocal and quotient relations between the ratios
3
$tan 2 7^{\circ} tan 6 3^{\circ} = 1 \Rightarrow \frac{h}{p} \cdot \frac{h}{q} = 1$
Why: Multiplying the two equations removes both tangents at once.
4
$h^{2} = pq \Rightarrow h = pq$
Why: A height is positive, so only the positive root is taken.
denotes the non-negative root

Shortcut

Elevations that add to $9 0^{\circ}$ from distances $p$ and $q$ always give $h = pq$ .

Common mistakes

Option A

Reached $h^{2} = pq$ and forgot to take the square root.

Option C

No plausible route reaches $\frac{pq}{2}$ .

Option D

No plausible route reaches $\frac{pq}{2}$ .

Skills used

Angles of elevation and depression Sine, cosine and tangent in a right triangle Ratios of complementary angles Reciprocal and quotient relations between the ratios

denotes the non-negative root

58Trigonometry • Trigonometric Identity • Complementary Angles

ModerateIdeal time: about 1.5 min

What is the value of $sin^{2} 6^{\circ} + sin^{2} 1 2^{\circ} + sin^{2} 1 8^{\circ} + \dots + sin^{2} 8 4^{\circ} + sin^{2} 9 0^{\circ}$ ?

1

2

4

8

Show answer & solution

Option D

8

Approach

List the angles, pair each with its complement so that the two squared sines add to $1$ , and handle the one term left over.

Ratios of complementary angles

Step-by-step solution

1
The angles are $6^{\circ}, 1 2^{\circ}, \dots, 9 0^{\circ}$ : $\frac{90}{6} = 15$ terms
Why: Counting the terms first is what makes the pairing bookkeeping safe.
An evenly spaced run averages to its middle term
2
$sin^{2} θ + sin^{2} (9 0^{\circ} - θ) = sin^{2} θ + cos^{2} θ = 1$
Why: $sin (9 0^{\circ} - θ) = cos θ$ , so complementary angles make each pair collapse to $1$ .
Ratios of complementary angles Pythagorean trigonometric identities
3
$(6^{\circ}, 8 4^{\circ}), (1 2^{\circ}, 7 8^{\circ}), (1 8^{\circ}, 7 2^{\circ}), (2 4^{\circ}, 6 6^{\circ}), (3 0^{\circ}, 6 0^{\circ}), (3 6^{\circ}, 5 4^{\circ}), (4 2^{\circ}, 4 8^{\circ})$ — $7$ pairs
Why: Every multiple of $6^{\circ}$ from $6^{\circ}$ to $8 4^{\circ}$ pairs with another; $4 5^{\circ}$ is not a multiple of $6^{\circ}$ , so no term is its own partner.
4
$7$ pairs use $14$ terms, leaving $sin^{2} 9 0^{\circ} = 1$
Why: $15$ is odd, so exactly one term is unpaired — and it is the last one, whose sine is $1$ .
Exact trigonometric ratios of the standard angles
5
Sum $= 7 \times 1 + 1 = 8$
Why: Seven pairs contribute $1$ each and the leftover contributes $1$ .

Shortcut

Fifteen terms, seven complementary pairs each worth $1$ , plus $sin^{2} 9 0^{\circ} = 1$ : total $8$ .

Common mistakes

Option A

Treated the series as collapsing entirely to its last term.

Option B

No plausible route reaches $2$ .

Option C

No plausible route reaches $4$ .

Skills used

An evenly spaced run averages to its middle term Ratios of complementary angles Pythagorean trigonometric identities Exact trigonometric ratios of the standard angles

59Trigonometry • Trigonometric Identity

EasyIdeal time: about 1 min

What is $\frac{c o s θ}{1 + s i n θ} + \frac{1}{c o t θ}$ equal to?

cosec θ

sec θ

sec θ + cosec θ

cosec θ - cot θ

Show answer & solution

Option B

sec θ

Approach

Rewrite $\frac{1}{c o t θ}$ as $\frac{s i n θ}{c o s θ}$ , put both terms over the common denominator $cos θ (1 + sin θ)$ , and use $sin^{2} θ + cos^{2} θ = 1$ .

Pythagorean trigonometric identities

Step-by-step solution

1
$\frac{1}{cot θ} = tan θ = \frac{sin θ}{cos θ}$
Why: Both terms must be in sine and cosine before they can be combined.
Reciprocal and quotient relations between the ratios
2
$\frac{cos θ}{1 + sin θ} + \frac{sin θ}{cos θ} = \frac{cos ^{2} θ + sin θ ( 1 + sin θ )}{cos θ ( 1 + sin θ )}$
Why: The common denominator is just the product of the two denominators; nothing cancels yet.
Clear the denominators of an equation
3
Numerator $= cos^{2} θ + sin^{2} θ + sin θ = 1 + sin θ$
Why: $sin^{2} θ + cos^{2} θ = 1$ turns the numerator into a copy of the factor $(1 + sin θ)$ sitting in the denominator — that is the whole trick.
Pythagorean trigonometric identities
4
$\frac{1 + sin θ}{cos θ ( 1 + sin θ )} = \frac{1}{cos θ} = sec θ$
Why: $1 + sin θ$ cancels (it is non-zero except at $θ = - 9 0^{\circ}$ , where the expression is undefined anyway).

Common mistakes

Option A

Interchanged sine and cosine at the last step, so $\frac{1}{c o s θ}$ was written as $\frac{1}{s i n θ}$ .

Option C

Read $\frac{1}{c o t θ}$ as $cosec θ$ and the first fraction as $sec θ$ , then added the two separately instead of combining them.

Option D

Rationalised the first fraction with sine and cosine interchanged, giving $\frac{1 - c o s θ}{s i n θ} = cosec θ - cot θ$ , and dropped the second term.

Skills used

Reciprocal and quotient relations between the ratios Clear the denominators of an equation Pythagorean trigonometric identities

60Trigonometry • Trigonometric Identity

ModerateIdeal time: about 1.5 min

What is $\frac{s i n θ - c o s θ + 1}{s i n θ + c o s θ - 1} - \frac{s i n θ + 1}{c o s θ}$ equal to ?

0

1

2 sin θ

2 cos θ

Show answer & solution

Option A

0

Approach

Show the first fraction equals $\frac{1 + s i n θ}{c o s θ}$ — the second term exactly — so the difference is zero.

Difference of two squares

Step-by-step solution

1
Multiply the first fraction above and below by $(sin θ + cos θ + 1)$
Why: The denominator $sin θ + cos θ - 1$ is one half of a difference of squares; supplying the other half is what clears it.
Rationalise a denominator using the conjugate
2
Denominator $= (sin θ + cos θ)^{2} - 1 = 2 sin θ cos θ$
Why: $(a - b) (a + b) = a^{2} - b^{2}$ with $a = sin θ + cos θ$ and $b = 1$ , then $sin^{2} θ + cos^{2} θ = 1$ leaves only the cross term.
Difference of two squares Pythagorean trigonometric identities
3
Numerator $= (1 + sin θ - cos θ) (1 + sin θ + cos θ) = (1 + sin θ)^{2} - cos^{2} θ = 2 sin θ (1 + sin θ)$
Why: Grouping as $(1 + sin θ) \mp cos θ$ is another difference of squares; $(1 + sin θ)^{2} - cos^{2} θ = 1 + 2 sin θ + sin^{2} θ - cos^{2} θ = 2 sin θ + 2 sin^{2} θ$ .
Square of a sum and of a difference
4
First fraction $= \frac{2 sin θ ( 1 + sin θ )}{2 sin θ cos θ} = \frac{1 + sin θ}{cos θ}$
Why: $2 sin θ$ cancels, leaving exactly the second term of the question.
Reciprocal and quotient relations between the ratios
5
$\frac{1 + sin θ}{cos θ} - \frac{sin θ + 1}{cos θ} = 0$
Why: The two expressions are identical, so the whole expression vanishes for every $θ$ where it is defined.
Evaluate a nested fraction from the inside out

Shortcut

The standard identity $\frac{s i n θ - c o s θ + 1}{s i n θ + c o s θ - 1} = sec θ + tan θ = \frac{1 + s i n θ}{c o s θ}$ makes the subtraction immediate.

Common mistakes

Option B

Simplified the first fraction correctly but then divided by the second term instead of subtracting it.

Option C

No plausible route reaches $2 sin θ$ .

Option D

No plausible route reaches $2 cos θ$ .

Skills used

Rationalise a denominator using the conjugate Difference of two squares Pythagorean trigonometric identities Square of a sum and of a difference Reciprocal and quotient relations between the ratios Evaluate a nested fraction from the inside out