Elementary Mathematics 2020 (II)

Set A

Questions 41–60 of 100

41Arithmetic • Time, Speed and Distance

ModerateIdeal time: about 2 min

A car did a journey in t hours. Had the average speed been x kmph greater, the journey would have taken y hours less. How long was the journey?

x (t - y) t y

x (t - y) t y^{- 1}

x (t - y) t y^{- 2}

x (t + y) t y

Show answer & solution

Option B

x (t - y) t y^{- 1}

Approach

Write the distance as speed $\times$ time for both journeys, then eliminate the unknown original speed.

Speed, distance and time

Step-by-step solution

1
$D = v t$ , so $v = \frac{D}{t}$
Why: The only given about the actual journey is that it lasted $t$ hours, so call the average speed $v$ and the distance $D$ ; the distance is what is asked for, and $v$ is the quantity we must get rid of.
Speed, distance and time
2
$D = (v + x) (t - y)$
Why: The hypothetical journey covers the same distance $D$ at speed $x$ kmph greater, taking $y$ hours less. Same distance is the equation that links the two cases.
Turn a worded condition into a linear equation in one unknown
3
$D = (\frac{D}{t} + x) (t - y) = D - \frac{D y}{t} + x (t - y)$
Why: Substituting $v = \frac{D}{t}$ leaves only $D$ unknown. Expanding, the term $\frac{D}{t} \cdot t = D$ cancels against the $D$ on the left.
Clear the denominators of an equation
4
$\frac{D y}{t} = x (t - y) \Rightarrow D = \frac{x ( t - y ) t}{y} = x (t - y) t y^{- 1}$
Why: What survives the cancellation is a single linear equation in $D$ . Dividing by $y$ is the same as multiplying by $y^{- 1}$ , which is how the options are printed.
Negative exponent is a reciprocal

Shortcut

Difference of times at a fixed distance: $\frac{D}{v} - \frac{D}{v + x} = y$ with $\frac{D}{v} = t$ gives $t - \frac{D}{v + x} = y$ , so $\frac{D}{v + x} = t - y$ and dividing the two expressions for $D$ gives $\frac{v + x}{v} = \frac{t}{t - y}$ , hence $v = \frac{x ( t - y )}{y}$ and $D = v t$ .

Common mistakes

Option A

Cleared the fraction $\frac{D y}{t}$ by multiplying by $t$ but also by $y$ instead of dividing, so $y$ ended up in the numerator.

Option C

Divided by $y$ twice, once when clearing $\frac{D y}{t}$ and again at the end.

Option D

Read " $y$ hours less" as $t + y$ , and misplaced $y$ as in option A.

Skills used

Speed, distance and time Turn a worded condition into a linear equation in one unknown Clear the denominators of an equation Negative exponent is a reciprocal

42Arithmetic • Ratio and Proportion • Direct and Inverse Variation

EasyIdeal time: about 55 s

When a ball is allowed to fall, the time it takes to fall any distance varies as the square root of the distance and it takes $4$ seconds to fall $78.40$ m. How long would it take to fall $122.50$ m?

5

secondsB

5.5

secondsC

6

secondsD

6.5

seconds

Show answer & solution

Option A

5

seconds

Approach

Use the ratio form of direct variation so the constant never has to be found.

Direct variation as a power:

y \propto x^{n}

Step-by-step solution

1
$t \propto d$ , so $\frac{t _{2}}{t _{1}} = \frac{d _{2}}{d _{1}}$
Why: "Varies as the square root" is direct variation with power $\frac{1}{2}$ . Writing it as a ratio of the two cases kills the constant of proportionality, which the question never asks for.
Direct variation as a power: $y \propto x^{n}$
2
$\frac{d _{2}}{d _{1}} = \frac{122.50}{78.40} = \frac{1225}{784}$
Why: Multiplying both decimals by $10$ turns the ratio into whole numbers, and $1225$ and $784$ are both recognisable squares ( $3 5^{2}$ and $2 8^{2}$ ) -- which is the hint that the numbers were chosen to be clean.
Comparing two fractions
3
$\frac{1225}{784} = \frac{35}{28} = \frac{5}{4}$
Why: The square root of a quotient is the quotient of the square roots, so no decimal square root is needed.
$ab = a b$ and $a / b = a / b$
4
$t_{2} = 4 \times \frac{5}{4} = 5$ seconds
Why: Multiply the known time by the ratio just found.
Unitary method: find one unit, then scale

Shortcut

$\frac{122.50}{78.40} = 1.5625 = 1.2 5^{2}$ , so the time scales by $1.25$ : $4 \times 1.25 = 5$ s.

Common mistakes

Option B

No route reaches $5.5$ s; it sits between the correct $5$ and the linear-variation value.

Option C

Took the time proportional to the distance itself: $4 \times \frac{122.50}{78.40} = 6.25$ s, then rounded to the nearest option.

Option D

No route reaches $6.5$ s.

Skills used

Direct variation as a power:

y \propto x^{n}

Comparing two fractions

ab = a b

and

a / b = a / b

Unitary method: find one unit, then scale

43Algebra • Logarithm • Equations

ModerateIdeal time: about 2 min

If $6^{3 - 4 x} 4^{x + 5} = 8$ (Given $lo g_{10} 2 = 0.301$ and $lo g_{10} 3 = 0.477$ ), then which one of the following is correct?

0 < x < 1

1 < x < 2

2 < x < 3

3 < x < 4

Show answer & solution

Option B

1 < x < 2

Approach

Take $lo g_{10}$ of both sides, which turns the equation into a linear one in $x$ , and solve it with the two given logarithm values.

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

Step-by-step solution

1
$(3 - 4 x) lo g 6 + (x + 5) lo g 4 = lo g 8$
Why: The unknown sits only in exponents, so logarithms are the move: $lo g (a^{m} b^{n}) = m lo g a + n lo g b$ brings every exponent down into a linear expression.
$lo g a^{n} = n lo g a$ $lo g (ab) = lo g a + lo g b$ and $lo g \frac{a}{b} = lo g a - lo g b$
2
$lo g 6 = 0.301 + 0.477 = 0.778$ , $lo g 4 = 2 (0.301) = 0.602$ , $lo g 8 = 3 (0.301) = 0.903$
Why: Only $lo g 2$ and $lo g 3$ are given, so every logarithm in the equation has to be rebuilt from them: $6 = 2 \times 3$ , $4 = 2^{2}$ , $8 = 2^{3}$ .
Power of a power, and the move to a common base
3
$(3 - 4 x) (0.778) + (x + 5) (0.602) = 0.903$
Why: Substitute the three values. Everything is now ordinary arithmetic in $x$ .
Turn a worded condition into a linear equation in one unknown
4
$2.334 - 3.112 x + 0.602 x + 3.010 = 0.903 \Rightarrow 5.344 - 2.510 x = 0.903$
Why: Expand and collect: the $x$ terms give $- 3.112 x + 0.602 x = - 2.510 x$ and the constants give $2.334 + 3.010 = 5.344$ .
Eliminate one variable from a linear pair
5
$x = \frac{4.441}{2.510} \approx 1.77$ , so $1 < x < 2$
Why: A single division finishes it. Since $2.510 \times 1.77 \approx 4.44$ , the quotient is just under $1.8$ -- there is no need to carry the division further than one decimal place, because the options are whole-number intervals.
Comparing two fractions

Shortcut

Only the interval is asked, so test the endpoints: substitute $x = 1$ and $x = 2$ into $5.344 - 2.510 x$ and see where it crosses $0.903$ .

Common mistakes

Option A

Dropped the $5$ when distributing $lo g 4$ over $x + 5$ , giving $2.334 - 2.510 x = 0.903$ and $x \approx 0.57$ .

Option C

Sign slip when moving $0.903$ across: solved $2.510 x = 5.344 + 0.903$ to get $x \approx 2.49$ .

Option D

No route reaches an $x$ above $3$ .

Skills used

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

Power of a power, and the move to a common base Turn a worded condition into a linear equation in one unknown Eliminate one variable from a linear pair Comparing two fractions

44Arithmetic • Number System • LCM and HCF

Very easyIdeal time: about 15 s

The Euclidean algorithm is used to calculate the

Asquare root of an integerBcube root of an integerCsquare of an integerDHCF of two integers

Show answer & solution

Option D

HCF of two integers

Approach

Recall what the algorithm does: repeated division with remainder, which terminates at the greatest common divisor.

HCF by Euclid's algorithm

Step-by-step solution

1
$a = b q + r$ , $0 \leq r < b$
Why: The Euclidean algorithm is the division algorithm applied over and over: divide, keep the remainder, then divide the old divisor by that remainder.
Dividend $=$ divisor $\times$ quotient $+$ remainder
2
$g cd (a, b) = g cd (b, r)$ , and the process stops when $r = 0$
Why: Every common divisor of $a$ and $b$ also divides $r = a - b q$ , and conversely, so the pair's common divisors never change. The last non-zero remainder is therefore the HCF.
HCF by Euclid's algorithm
3
$g cd (48, 18)$ : $48 = 2 (18) + 12$ , $18 = 1 (12) + 6$ , $12 = 2 (6) + 0$ , so HCF $= 6$
Why: A two-line example confirms the reading: the algorithm produces $6$ , which is the HCF of $48$ and $18$ , not a root or a square.

Common mistakes

Option A

Confused with the long-division method for square roots, which is a different algorithm.

Option B

No route; cube roots have no classical division algorithm in the syllabus.

Option C

No route; squaring needs no algorithm.

Skills used

HCF by Euclid's algorithm Dividend

=

divisor

\times

quotient

+

remainder

45Arithmetic • Number System • Rational Numbers

ModerateIdeal time: about 1 min

If radius of a sphere is rational, then which of the following is/are correct?

1: Its surface area is rational.

2: Its volume is rational.

Select the correct answer using the code given below:

1

onlyB

2

onlyCBoth

1

and

2

DNeither

1

nor

2

Show answer & solution

Option D

Neither

1

nor

2

Approach

Write both quantities out and ask where the irrationality comes from: both carry a factor of $π$ , which no rational radius can remove.

π

is irrational

Step-by-step solution

1
Surface area $= 4 π r^{2}$ , volume $= \frac{4}{3} π r^{3}$ , with $r$ rational and $r > 0$
Why: Both standard formulae are needed, since the two statements are about one each.
Volume and surface area of a sphere
2
$4 r^{2}$ and $\frac{4}{3} r^{3}$ are rational and non-zero
Why: Rationals are closed under multiplication, so any power or rational multiple of a rational is rational. The radius being rational settles everything except the $π$ .
Comparing two fractions
3
$π$ is irrational, and a non-zero rational multiple of an irrational number is irrational
Why: This is the step that decides the question. If $4 π r^{2}$ were rational, dividing by the rational $4 r^{2}$ would make $π$ rational.
$π$ is irrational $n$ is irrational unless $n$ is a perfect square
4
Both $4 π r^{2}$ and $\frac{4}{3} π r^{3}$ are irrational, so neither statement holds
Why: The same argument applies unchanged to the volume, so both statements fail together.

Common mistakes

Option A

Treated the irrationality as coming from the cube, so $r^{3}$ looked suspect while $4 π r^{2}$ looked safe -- overlooking that $π$ is in both.

Option B

No route singles out the volume as rational while rejecting the surface area.

Option C

Took $π = \frac{22}{7}$ , the value used for computation throughout the paper, as the actual value of $π$ , making both quantities rational.

Skills used

π

is irrational Volume and surface area of a sphere Comparing two fractions

n

is irrational unless

n

is a perfect square

46Trigonometry • Trigonometric Identity • Elimination of the Angle

HardIdeal time: about 2.5 min

If $cosec θ - sin θ = m$ and $sec θ - cos θ = n$ , then what is $m^{\frac{4}{3}} n^{\frac{2}{3}} + m^{\frac{2}{3}} n^{\frac{4}{3}}$ equal to?

0

1

mn

m^{2} n^{2}

Show answer & solution

Option B

1

Approach

Reduce $m$ and $n$ to single trigonometric ratios, then notice that $m^{2} n$ and $m n^{2}$ are perfect cubes of $cos θ$ and $sin θ$ .

Reciprocal and quotient relations between the ratios

Step-by-step solution

1
$m = cosec θ - sin θ = \frac{1 - sin ^{2} θ}{sin θ} = \frac{cos ^{2} θ}{sin θ}$
Why: Write the reciprocal ratio as a fraction and combine over one denominator; $1 - sin^{2} θ = cos^{2} θ$ then collapses the numerator.
Reciprocal and quotient relations between the ratios Pythagorean trigonometric identities
2
$n = sec θ - cos θ = \frac{1 - cos ^{2} θ}{cos θ} = \frac{sin ^{2} θ}{cos θ}$
Why: Exactly the same move on the second given, with $sin$ and $cos$ interchanged.
3
$m^{2} n = \frac{cos ^{4} θ}{sin ^{2} θ} \cdot \frac{sin ^{2} θ}{cos θ} = cos^{3} θ$ and $m n^{2} = \frac{cos ^{2} θ}{sin θ} \cdot \frac{sin ^{4} θ}{cos ^{2} θ} = sin^{3} θ$
Why: The fractional powers $\frac{4}{3}$ and $\frac{2}{3}$ in the target are the signal to look for cube combinations of $m$ and $n$ . These two products are the only combinations in which both denominators cancel.
Power of a power, and the move to a common base
4
$m^{\frac{4}{3}} n^{\frac{2}{3}} = (m^{2} n)^{\frac{2}{3}} = (cos^{3} θ)^{\frac{2}{3}} = cos^{2} θ$ and $m^{\frac{2}{3}} n^{\frac{4}{3}} = (m n^{2})^{\frac{2}{3}} = sin^{2} θ$
Why: Each target term is exactly one of those cubes raised to the power $\frac{2}{3}$ , which is why the exponents were chosen as they were.
5
$m^{\frac{4}{3}} n^{\frac{2}{3}} + m^{\frac{2}{3}} n^{\frac{4}{3}} = cos^{2} θ + sin^{2} θ = 1$
Why: The two terms are the two squares of the Pythagorean identity, so the sum is $1$ for every $θ$ -- a constant, which is why the options contain a bare number.

Shortcut

Having seen $m = \frac{cos ^{2} θ}{sin θ}$ and $n = \frac{sin ^{2} θ}{cos θ}$ , read off $m^{2/3} n^{1/3} = cos θ$ and $m^{1/3} n^{2/3} = sin θ$ ; the asked expression is the sum of their squares.

Common mistakes

Option A

Read the two terms as a difference and cancelled them against each other.

Option C

Stopped at $m^{2} n = cos^{3} θ$ and $m n^{2} = sin^{3} θ$ , multiplied them to get $mn = sin θ cos θ$ , and quoted that instead of the expression asked for.

Option D

No route produces $m^{2} n^{2}$ ; it is a plausible-looking power of the givens.

Skills used

Reciprocal and quotient relations between the ratios Pythagorean trigonometric identities Power of a power, and the move to a common base

47Trigonometry • Trigonometric Identity

ModerateIdeal time: about 1.5 min

If $cos θ + sec θ = k$ , then what is the value of $sin^{2} θ - tan^{2} θ$ ?

4 - k

4 - k^{2}

k^{2} - 4

k^{2} + 2

Show answer & solution

Option B

4 - k^{2}

Approach

Rewrite the target so that it contains only $cos^{2} θ + sec^{2} θ$ , which is what squaring the given produces.

Pythagorean trigonometric identities

Step-by-step solution

1
$sin^{2} θ - tan^{2} θ = (1 - cos^{2} θ) - (sec^{2} θ - 1) = 2 - cos^{2} θ - sec^{2} θ$
Why: The given involves $cos θ$ and $sec θ$ only, so both parts of the target must be converted into those. $sin^{2} θ = 1 - cos^{2} θ$ and $tan^{2} θ = sec^{2} θ - 1$ are the two Pythagorean forms that do it.
Pythagorean trigonometric identities Reciprocal and quotient relations between the ratios
2
$k^{2} = (cos θ + sec θ)^{2} = cos^{2} θ + 2 cos θ sec θ + sec^{2} θ$
Why: Squaring a sum is the standard way to turn a sum of two reciprocal quantities into a sum of their squares.
Square of a sum and of a difference
3
$cos θ sec θ = 1$ , so $cos^{2} θ + sec^{2} θ = k^{2} - 2$
Why: The cross term collapses because $sec θ$ is the reciprocal of $cos θ$ -- this is the whole reason the question pairs them.
4
$sin^{2} θ - tan^{2} θ = 2 - (k^{2} - 2) = 4 - k^{2}$
Why: Substitute the bracket found in step 3 into the expression from step 1.
Eliminate one variable from a linear pair

Shortcut

$sin^{2} θ - tan^{2} θ = 2 - (cos^{2} θ + sec^{2} θ)$ and squaring the given gives $cos^{2} θ + sec^{2} θ = k^{2} - 2$ at once, so the answer is $4 - k^{2}$ in two lines.

Common mistakes

Option A

Carried $k$ through instead of $k^{2}$ , forgetting that the given had to be squared before it could be used.

Option C

Computed $cos^{2} θ + sec^{2} θ - 2$ instead of $2 - (cos^{2} θ + sec^{2} θ)$ , flipping the sign of the whole answer.

Option D

Kept the $+ 2$ from $k^{2} = cos^{2} θ + 2 + sec^{2} θ$ instead of subtracting it.

Skills used

Pythagorean trigonometric identities Reciprocal and quotient relations between the ratios Square of a sum and of a difference Eliminate one variable from a linear pair

48Trigonometry • Trigonometric Identity • Complementary Angles

Very easyIdeal time: about 30 s

ABC is a triangle inscribed in a semicircle of diameter AB. What is $cos (A + B) + sin (A + B)$ equal to?

0

\frac{1}{4}

\frac{1}{2}

1

Show answer & solution

Option D

1

Approach

The angle in a semicircle is a right angle, which fixes $A + B$ before any trigonometry is done.

The angle at the centre is twice the angle at the circumference

Step-by-step solution

1
$∠ A C B = 9 0^{\circ}$
Why: $A B$ is a diameter and $C$ lies on the circle, so $∠ A C B$ is an angle in a semicircle -- a corollary of the angle-at-the-centre theorem.
The angle at the centre is twice the angle at the circumference
2
$A + B + C = 18 0^{\circ} \Rightarrow A + B = 9 0^{\circ}$
Why: The angle sum of the triangle converts the right angle at $C$ into the quantity the expression actually needs.
Angle sum of a triangle
3
$cos 9 0^{\circ} + sin 9 0^{\circ} = 0 + 1 = 1$
Why: Substituting the special-angle values finishes it; the answer does not depend on the shape of the triangle at all.
Exact trigonometric ratios of the standard angles

Common mistakes

Option A

Got $A + B = 9 0^{\circ}$ and $cos 9 0^{\circ} = 0$ , then stopped without adding $sin 9 0^{\circ} = 1$ .

Option B

No route produces $\frac{1}{4}$ .

Option C

No route produces $\frac{1}{2}$ ; it is the value of neither $cos 9 0^{\circ}$ nor $sin 9 0^{\circ}$ .

Skills used

The angle at the centre is twice the angle at the circumference Angle sum of a triangle Exact trigonometric ratios of the standard angles

49Trigonometry • Trigonometric Identity • Maximum and Minimum Values

EasyIdeal time: about 1 min

Consider the following statements:

1: $sin θ = x + \frac{1}{x}$ is possible for some real value of x.

2: $cos θ = x + \frac{1}{x}$ is possible for some real value of x.

Which of the above statements is/are correct?

1

onlyB

2

onlyCBoth

1

and

2

DNeither

1

nor

2

Show answer & solution

Option D

Neither

1

nor

2

Approach

Find the range of $x + \frac{1}{x}$ over real $x$ and compare it with the range of a sine or cosine.

A positive number plus its reciprocal is at least

2

Step-by-step solution

1
For $x > 0$ : $x + \frac{1}{x} \geq 2$ ; for $x < 0$ : $x + \frac{1}{x} \leq - 2$
Why: Both statements involve the same expression, so its range is the only thing worth computing. The bound follows from $x + \frac{1}{x} - 2 = \frac{( x - 1 ) ^{2}}{x}$ , and $x = 0$ is not allowed at all.
A positive number plus its reciprocal is at least $2$
2
So $x + \frac{1}{x} \geq 2$ for every real $x \neq = 0$
Why: The two cases together say the expression never lands strictly between $- 2$ and $2$ . That gap is exactly where the trigonometric ratios live.
3
$- 1 \leq sin θ \leq 1$ and $- 1 \leq cos θ \leq 1$
Why: Both ratios are confined to $[- 1, 1]$ , which is strictly inside $(- 2, 2)$ .
Sine and cosine lie between $- 1$ and $1$
4
Neither $sin θ = x + \frac{1}{x}$ nor $cos θ = x + \frac{1}{x}$ has a real solution
Why: The two ranges do not overlap, so both statements are false; the ratios are treated identically, which is why one cannot be correct without the other.

Shortcut

$x + \frac{1}{x} \geq 2 > 1 \geq ∣ sin θ ∣, ∣ cos θ ∣$ -- one line rules out both.

Common mistakes

Option A

No route treats the sine differently from the cosine; both have the same range.

Option B

No route treats the cosine differently from the sine.

Option C

Believed $x + \frac{1}{x}$ takes every real value as $x$ varies, so a value in $[- 1, 1]$ looked reachable.

Skills used

A positive number plus its reciprocal is at least

2

Sine and cosine lie between

- 1

and

1

50Geometry • Polygon • Interior and Exterior Angles

EasyIdeal time: about 50 s

What is the magnitude (in radian) of the interior angle of a regular pentagon?

\frac{π}{5}

\frac{2 π}{5}

\frac{3 π}{5}

\frac{4 π}{5}

Show answer & solution

Option C

\frac{3 π}{5}

Approach

Find the interior angle in degrees from the angle sum of a pentagon, then convert to radians.

Interior and exterior angle sums of a polygon

Step-by-step solution

1
Sum of interior angles $= (n - 2) \times 18 0^{\circ} = (5 - 2) \times 18 0^{\circ} = 54 0^{\circ}$
Why: A polygon's interior angles always total $(n - 2)$ straight angles, which is the only fact needed about the pentagon.
Interior and exterior angle sums of a polygon
2
Regular, so each interior angle $= \frac{54 0 ^{\circ}}{5} = 10 8^{\circ}$
Why: "Regular" is what allows the total to be divided equally among the five vertices.
3
$10 8^{\circ} = 108 \times \frac{π}{180} = \frac{3 π}{5}$ radian
Why: The options are in radians, so convert: $18 0^{\circ} = π$ radian, and $\frac{108}{180} = \frac{3}{5}$ .
Converting between degrees and radians

Shortcut

Interior angle $= 18 0^{\circ} - \frac{36 0 ^{\circ}}{5} = 10 8^{\circ}$ , and $\frac{108}{180} π = \frac{3 π}{5}$ .

Common mistakes

Option A

Halved the exterior angle $7 2^{\circ}$ to $3 6^{\circ} = \frac{π}{5}$ .

Option B

Gave the exterior angle $\frac{36 0 ^{\circ}}{5} = 7 2^{\circ} = \frac{2 π}{5}$ instead of the interior angle.

Option D

No route reaches $14 4^{\circ}$ for a pentagon; it is the interior angle of a regular decagon.

Skills used

Interior and exterior angle sums of a polygon Converting between degrees and radians

51Trigonometry • Trigonometric Identity • Degree and Radian

EasyIdeal time: about 55 s

The difference between two angles is $1 5^{\circ}$ and the sum of the angles in radian is $\frac{5 π}{12}$ . The bigger angle is k times the smaller angle. What is k equal to?

\frac{4}{3}

\frac{3}{2}

\frac{6}{5}

\frac{7}{6}

Show answer & solution

Option B

\frac{3}{2}

Approach

Put both givens into degrees, solve the two linear equations for the angles, then divide.

Converting between degrees and radians

Step-by-step solution

1
Sum $= \frac{5 π}{12}$ radian $= \frac{5}{12} \times 18 0^{\circ} = 7 5^{\circ}$
Why: One given is in radians and the other in degrees, so they must be brought to the same unit before they can be combined. Degrees are the easier choice because the difference is already $1 5^{\circ}$ .
Converting between degrees and radians
2
$A + B = 7 5^{\circ}$ and $A - B = 1 5^{\circ}$
Why: Two linear equations in the two angles, with $A$ the bigger one.
Turn a worded condition into a linear equation in one unknown
3
Adding: $2 A = 9 0^{\circ} \Rightarrow A = 4 5^{\circ}$ ; subtracting: $2 B = 6 0^{\circ} \Rightarrow B = 3 0^{\circ}$
Why: Adding and subtracting a sum-and-difference pair is faster than substitution, because each operation eliminates one unknown outright.
Add and subtract when the coefficients are reversed
4
$k = \frac{A}{B} = \frac{45}{30} = \frac{3}{2}$
Why: The question defines $k$ as bigger angle divided by smaller angle, so the order matters.
Unitary method: find one unit, then scale

Shortcut

Bigger angle $= \frac{sum + difference}{2} = \frac{75 + 15}{2} = 4 5^{\circ}$ , smaller $= \frac{75 - 15}{2} = 3 0^{\circ}$ , so $k = \frac{3}{2}$ .

Common mistakes

Option A

Converted $\frac{5 π}{12}$ to $10 5^{\circ}$ instead of $7 5^{\circ}$ , giving angles $6 0^{\circ}$ and $4 5^{\circ}$ and $k = \frac{4}{3}$ .

Option C

No route reaches $\frac{6}{5}$ ; it would need angles $9 0^{\circ}$ and $7 5^{\circ}$ , whose sum is not $7 5^{\circ}$ .

Option D

No route reaches $\frac{7}{6}$ .

Skills used

Converting between degrees and radians Turn a worded condition into a linear equation in one unknown Add and subtract when the coefficients are reversed Unitary method: find one unit, then scale

52Trigonometry • Trigonometric Equation

ModerateIdeal time: about 2 min

Consider the following statements:

1: The equation $2 sin^{2} θ - cos θ + 4 = 0$ is possible for all $θ$ .

2: $tan θ + cot θ$ cannot be less than $2$ , where $0 < θ < \frac{π}{2}$ .

Which of the above statements is/are correct?

1

onlyB

2

onlyCBoth

1

and

2

DNeither

1

nor

2

Show answer & solution

Option B

2

only

Approach

Turn statement 1 into a quadratic in $cos θ$ and check whether any root is a possible cosine; test statement 2 against the reciprocal bound.

Pythagorean trigonometric identities

Step-by-step solution

1
Statement 1: $2 (1 - cos^{2} θ) - cos θ + 4 = 0 \Rightarrow 2 cos^{2} θ + cos θ - 6 = 0$
Why: The equation mixes $sin^{2} θ$ and $cos θ$ ; replacing $sin^{2} θ$ by $1 - cos^{2} θ$ leaves a single ratio, and then it is just a quadratic.
Pythagorean trigonometric identities
2
$cos θ = \frac{- 1 \pm 1 + 48}{4} = \frac{- 1 \pm 7}{4}$ , so $cos θ = \frac{3}{2}$ or $cos θ = - 2$
Why: The discriminant $49$ is a perfect square, so both roots come out exactly -- no approximation is needed to judge them.
Quadratic formula
3
Both roots lie outside $[- 1, 1]$ , so no $θ$ satisfies the equation; statement 1 is false
Why: A cosine can never be $\frac{3}{2}$ or $- 2$ . The statement claims the equation holds for all $θ$ , and in fact it holds for none -- so it fails twice over.
Sine and cosine lie between $- 1$ and $1$
4
Statement 2: $tan θ + cot θ = t + \frac{1}{t}$ with $t = tan θ > 0$ on $0 < θ < \frac{π}{2}$
Why: $cot θ$ is the reciprocal of $tan θ$ , so the expression is a positive number plus its reciprocal -- and on the open first quadrant $tan θ$ is strictly positive.
Reciprocal and quotient relations between the ratios
5
$t + \frac{1}{t} \geq 2$ for $t > 0$ , so $tan θ + cot θ$ cannot be less than $2$ ; statement 2 is true
Why: Equality at $t = 1$ , i.e. $θ = 4 5^{\circ}$ , so the bound $2$ is actually attained and cannot be improved.
A positive number plus its reciprocal is at least $2$

Shortcut

For statement 1, note $2 sin^{2} θ \geq 0$ and $- cos θ \geq - 1$ , so the left side is at least $3$ and can never be $0$ . Statement 2 is the standard $t + \frac{1}{t} \geq 2$ .

Common mistakes

Option A

No route makes statement 1 hold for all $θ$ while statement 2 fails.

Option C

Verified statement 2 and then accepted statement 1 without testing whether its roots are legal cosines.

Option D

Judged statement 2 false too, believing $tan θ + cot θ$ can dip below $2$ somewhere in the first quadrant.

Skills used

Pythagorean trigonometric identities Quadratic formula Sine and cosine lie between

- 1

and

1

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

2

53Mensuration • Plane Figures • Circle

EasyIdeal time: about 1.5 min

A road curve is to be laid out on a circle. What radius should be used if the track is to change direction by $4 2^{\circ}$ in distance of $44$ m? (Assume $π = \frac{22}{7}$ )

60

66

75

80

Show answer & solution

Option A

60

Approach

The $44$ m of track is an arc of the circle and $4 2^{\circ}$ is its central angle, so use the arc-length formula and solve for $r$ .

Area and arc length of a circular sector

Step-by-step solution

1
Arc length $= \frac{θ}{36 0 ^{\circ}} \times 2 π r$ with $θ = 4 2^{\circ}$ and arc $= 44$ m
Why: "Change direction by $4 2^{\circ}$ in a distance of $44$ m" means the vehicle travels $44$ m along the curve while the tangent direction turns through $4 2^{\circ}$ -- and the tangent turns by exactly the central angle, so $44$ m is the arc of a $4 2^{\circ}$ sector.
Area and arc length of a circular sector
2
$44 = \frac{42}{360} \times 2 \times \frac{22}{7} \times r$
Why: Substitute the given $π = \frac{22}{7}$ ; the $7$ in the denominator is there to cancel the $42$ , which is why that angle was chosen.
Area and circumference of a circle
3
$\frac{42}{360} \times \frac{44}{7} = \frac{6}{360} \times 44 = \frac{44}{60}$ , so $44 = \frac{44}{60} r$
Why: Cancel $42$ against the $7$ first and the arithmetic stays in small numbers -- no long multiplication is needed.
Comparing two fractions
4
$r = 60$ m
Why: Both sides carry a factor $44$ , so it cancels outright.
Clear the denominators of an equation

Shortcut

$4 2^{\circ}$ is $\frac{7}{60}$ of a turn, so the whole circumference is $44 \times \frac{60}{7} = \frac{2640}{7}$ m; dividing by $2 π = \frac{44}{7}$ gives $r = 60$ m.

Common mistakes

Option B

No route reaches $66$ m.

Option C

No route reaches $75$ m.

Option D

No route reaches $80$ m.

Skills used

Area and arc length of a circular sector Area and circumference of a circle Comparing two fractions Clear the denominators of an equation

54Trigonometry • Trigonometric Identity • Maximum and Minimum Values

Very easyIdeal time: about 25 s

What is the maximum value of $3 sin θ - 4$ ?

- 4

- 1

0

1

Show answer & solution

Option B

- 1

Approach

The only variable part is $sin θ$ , so push it to its largest value and read off the result.

Sine and cosine lie between

- 1

and

1

Step-by-step solution

1
$- 1 \leq sin θ \leq 1$
Why: This is the only constraint in the question; the $3$ and the $- 4$ are fixed.
Sine and cosine lie between $- 1$ and $1$
2
$3$ is positive, so $3 sin θ$ is greatest when $sin θ$ is greatest, i.e. $sin θ = 1$
Why: Multiplying by a positive number preserves order, so the maximum of the expression sits at the maximum of $sin θ$ -- at $θ = 9 0^{\circ}$ .
Multiplying an inequality by a negative number reverses it
3
Maximum $= 3 (1) - 4 = - 1$
Why: Subtracting $4$ shifts the whole range down but does not move where the maximum occurs.

Common mistakes

Option A

Used $sin θ = 0$ instead of $sin θ = 1$ .

Option C

No route reaches $0$ ; it would need $sin θ = \frac{4}{3}$ .

Option D

Computed $3 - 4$ and then reported it as $+ 1$ , dropping the sign.

Skills used

Sine and cosine lie between

- 1

and

1

Multiplying an inequality by a negative number reverses it

55Trigonometry • Trigonometric Identity

ModerateIdeal time: about 1.5 min

If $sin θ + cos θ = 2$ , then what is $sin^{6} θ + cos^{6} θ + 6 sin^{2} θ cos^{2} θ$ equal to ?

\frac{1}{4}

\frac{3}{4}

1

\frac{7}{4}

Show answer & solution

Option D

\frac{7}{4}

Approach

Square the given to get $sin θ cos θ$ , then express $sin^{6} θ + cos^{6} θ$ in terms of that one product.

Cube of a sum and of a difference

Step-by-step solution

1
$(sin θ + cos θ)^{2} = 2 \Rightarrow 1 + 2 sin θ cos θ = 2 \Rightarrow sin θ cos θ = \frac{1}{2}$
Why: Squaring a sum of sine and cosine always produces $1$ plus twice the product, so one squaring converts the given into the product that the rest of the question needs.
Square of a sum and of a difference Pythagorean trigonometric identities
2
$sin^{2} θ cos^{2} θ = \frac{1}{4}$
Why: The target contains $sin^{2} θ cos^{2} θ$ , not $sin θ cos θ$ , so square the product found.
Power of a power, and the move to a common base
3
$sin^{6} θ + cos^{6} θ = (sin^{2} θ + cos^{2} θ)^{3} - 3 sin^{2} θ cos^{2} θ (sin^{2} θ + cos^{2} θ) = 1 - 3 sin^{2} θ cos^{2} θ$
Why: Treat $sin^{2} θ$ and $cos^{2} θ$ as the two quantities in $a^{3} + b^{3} = (a + b)^{3} - 3 ab (a + b)$ ; their sum is $1$ , which is what makes this collapse.
Cube of a sum and of a difference
4
$sin^{6} θ + cos^{6} θ = 1 - \frac{3}{4} = \frac{1}{4}$
Why: Substitute $sin^{2} θ cos^{2} θ = \frac{1}{4}$ .
5
$\frac{1}{4} + 6 (\frac{1}{4}) = \frac{1}{4} + \frac{3}{2} = \frac{7}{4}$
Why: The question asks for the sixth powers plus $6 sin^{2} θ cos^{2} θ$ , so the last term must be added, not ignored.
Comparing two fractions

Shortcut

$sin θ + cos θ = 2$ forces $θ = 4 5^{\circ}$ , so every term is a power of $\frac{1}{2}$ : $\frac{1}{8} + \frac{1}{8} + 6 \times \frac{1}{4} = \frac{7}{4}$ .

Common mistakes

Option A

Stopped at $sin^{6} θ + cos^{6} θ = \frac{1}{4}$ and never added the $6 sin^{2} θ cos^{2} θ$ term.

Option B

Took $sin^{2} θ cos^{2} θ = \frac{1}{8}$ , so the added term came out as $\frac{3}{4}$ and the sixth powers as $\frac{5}{8}$ .

Option C

Assumed the whole expression collapses to $1$ by the Pythagorean identity, as $sin^{2} θ + cos^{2} θ$ does.

Skills used

Cube of a sum and of a difference Square of a sum and of a difference Pythagorean trigonometric identities Power of a power, and the move to a common base Comparing two fractions

56Trigonometry • Trigonometric Identity • Maximum and Minimum Values

EasyIdeal time: about 50 s

What is the least value of $9 sin^{2} θ + 16 cos^{2} θ$ ?

0

9

16

25

Show answer & solution

Option B

9

Approach

Reduce to one trigonometric ratio with the Pythagorean identity, then the expression is linear in $sin^{2} θ$ .

Pythagorean trigonometric identities

Step-by-step solution

1
$9 sin^{2} θ + 16 cos^{2} θ = 9 sin^{2} θ + 16 (1 - sin^{2} θ)$
Why: Two different ratios cannot both be pushed to their extremes independently -- they are tied by $sin^{2} θ + cos^{2} θ = 1$ . Eliminating one is what makes the problem one-dimensional.
Pythagorean trigonometric identities
2
$= 16 - 7 sin^{2} θ$
Why: Collecting terms leaves a decreasing linear function of $sin^{2} θ$ , because the coefficient $- 7$ is negative.
Multiplying an inequality by a negative number reverses it
3
$0 \leq sin^{2} θ \leq 1$ , so the least value is at $sin^{2} θ = 1$ : $16 - 7 = 9$
Why: A negative coefficient means the smallest value of the expression comes from the largest value of $sin^{2} θ$ -- at $θ = 9 0^{\circ}$ .
Sine and cosine lie between $- 1$ and $1$

Shortcut

$a sin^{2} θ + b cos^{2} θ$ is a weighted average of $a$ and $b$ , so it lies between them: the least value is $min (9, 16) = 9$ .

Common mistakes

Option A

Treated $sin θ$ and $cos θ$ as independent and set both to $0$ , which the identity forbids.

Option C

Found the two end values $9$ and $16$ and quoted the greatest instead of the least.

Option D

Added the coefficients, $9 + 16 = 25$ , as if the expression were $9 + 16$ times something.

Skills used

Pythagorean trigonometric identities Multiplying an inequality by a negative number reverses it Sine and cosine lie between

- 1

and

1

57Trigonometry • Trigonometric Identity

HardIdeal time: about 2.5 min

If $cos 4 7^{\circ} + sin 4 7^{\circ} = k$ , then what is the value of $cos^{2} 4 7^{\circ} - sin^{2} 4 7^{\circ}$ ?

k 2 - k^{2}

- k 2 - k^{2}

k 1 - k^{2}

- k 1 - k^{2}

Show answer & solution

Option B

- k 2 - k^{2}

Approach

Factor the difference of squares, get $cos 4 7^{\circ} - sin 4 7^{\circ}$ by squaring, and settle its sign before taking the root.

Difference of two squares

Step-by-step solution

1
$cos^{2} 4 7^{\circ} - sin^{2} 4 7^{\circ} = (cos 4 7^{\circ} + sin 4 7^{\circ}) (cos 4 7^{\circ} - sin 4 7^{\circ}) = k (cos 4 7^{\circ} - sin 4 7^{\circ})$
Why: One factor is the given $k$ , so only the other factor remains to be found -- that is why the difference of squares is the first move.
Difference of two squares
2
$k^{2} = 1 + 2 sin 4 7^{\circ} cos 4 7^{\circ} \Rightarrow 2 sin 4 7^{\circ} cos 4 7^{\circ} = k^{2} - 1$
Why: Squaring the given produces the cross term, which is the only unknown left.
Square of a sum and of a difference Pythagorean trigonometric identities
3
$(cos 4 7^{\circ} - sin 4 7^{\circ})^{2} = 1 - 2 sin 4 7^{\circ} cos 4 7^{\circ} = 1 - (k^{2} - 1) = 2 - k^{2}$
Why: The same identity with a minus sign gives the square of the missing factor in terms of $k$ alone.
4
$4 7^{\circ} > 4 5^{\circ}$ , so $sin 4 7^{\circ} > cos 4 7^{\circ}$ and $cos 4 7^{\circ} - sin 4 7^{\circ} = - 2 - k^{2}$
Why: This is the step the question is really testing. Squaring loses the sign, and past $4 5^{\circ}$ the sine has overtaken the cosine, so the negative root is the correct one.
How the ratios move across the first quadrant denotes the non-negative root
5
$cos^{2} 4 7^{\circ} - sin^{2} 4 7^{\circ} = - k 2 - k^{2}$
Why: Multiply the two factors; $k$ itself is positive, so the answer is negative, as it must be for an angle above $4 5^{\circ}$ .

Common mistakes

Option A

Took the positive square root, missing that $sin 4 7^{\circ} > cos 4 7^{\circ}$ because $4 7^{\circ}$ is past $4 5^{\circ}$ .

Option C

Used $(cos θ - sin θ)^{2} = 1 - k^{2}$ , forgetting that the cross term is $k^{2} - 1$ rather than $k^{2}$ .

Option D

Same error in the bracket as option C, but with the sign of the root taken correctly.

Skills used

Difference of two squares Square of a sum and of a difference Pythagorean trigonometric identities How the ratios move across the first quadrant

denotes the non-negative root

58Trigonometry • Trigonometric Identity • Elimination of the Angle

HardIdeal time: about 2 min

If $cosec θ - sin θ = p^{3}$ and $sec θ - cos θ = q^{3}$ , then what is the value of $tan θ$ ?

\frac{p}{q}

\frac{q}{p}

pq

p^{2} q^{2}

Show answer & solution

Option B

\frac{q}{p}

Approach

Reduce each given to a single ratio, then form the products $p^{6} q^{3}$ and $p^{3} q^{6}$ , which come out as $cos^{3} θ$ and $sin^{3} θ$ .

Reciprocal and quotient relations between the ratios

Step-by-step solution

1
$p^{3} = cosec θ - sin θ = \frac{1 - sin ^{2} θ}{sin θ} = \frac{cos ^{2} θ}{sin θ}$
Why: Combine over one denominator and use $1 - sin^{2} θ = cos^{2} θ$ ; the reciprocal ratio disappears.
Reciprocal and quotient relations between the ratios Pythagorean trigonometric identities
2
$q^{3} = sec θ - cos θ = \frac{1 - cos ^{2} θ}{cos θ} = \frac{sin ^{2} θ}{cos θ}$
Why: The same move on the second given.
3
$p^{6} q^{3} = \frac{cos ^{4} θ}{sin ^{2} θ} \cdot \frac{sin ^{2} θ}{cos θ} = cos^{3} θ \Rightarrow p^{2} q = cos θ$
Why: The givens are cubes, so look for combinations that are perfect cubes on both sides. Squaring the first and multiplying by the second clears $sin θ$ entirely.
Power of a power, and the move to a common base
4
$p^{3} q^{6} = \frac{cos ^{2} θ}{sin θ} \cdot \frac{sin ^{4} θ}{cos ^{2} θ} = sin^{3} θ \Rightarrow p q^{2} = sin θ$
Why: The mirror-image combination clears $cos θ$ instead.
5
$tan θ = \frac{sin θ}{cos θ} = \frac{p q ^{2}}{p ^{2} q} = \frac{q}{p}$
Why: Dividing the two results cancels one $p$ and one $q$ ; the order matters, since $tan θ$ is sine over cosine.

Shortcut

Once $p^{3} = \frac{cos ^{2} θ}{sin θ}$ and $q^{3} = \frac{sin ^{2} θ}{cos θ}$ are in hand, divide them: $\frac{q ^{3}}{p ^{3}} = \frac{sin ^{3} θ}{cos ^{3} θ} = tan^{3} θ$ , so $tan θ = \frac{q}{p}$ immediately.

Common mistakes

Option A

Divided the wrong way round, computing $\frac{cos θ}{sin θ} = cot θ$ instead of $tan θ$ .

Option C

Multiplied $p^{2} q$ and $p q^{2}$ instead of dividing them, which gives $(sin θ cos θ)^{1/3}$ -type expressions rather than a tangent.

Option D

No route produces $p^{2} q^{2}$ ; it is a plausible-looking power of the givens.

Skills used

Reciprocal and quotient relations between the ratios Pythagorean trigonometric identities Power of a power, and the move to a common base

59Trigonometry • Trigonometric Equation

Very easyIdeal time: about 35 s

If $0 \leq α, β \leq 9 0^{\circ}$ such that $cos (α - β) = 1$ , then what is $sin α - sin β + cos α - cos β$ equal to ?

- 1

0

1

2

Show answer & solution

Option B

0

Approach

Use the range restriction to pin $α - β$ to a single value, then substitute.

Sine and cosine lie between

- 1

and

1

Step-by-step solution

1
$0 \leq α, β \leq 9 0^{\circ} \Rightarrow - 9 0^{\circ} \leq α - β \leq 9 0^{\circ}$
Why: The given bounds on the two angles bound their difference: $0 \leq β \leq 9 0^{\circ}$ becomes $- 9 0^{\circ} \leq - β \leq 0^{\circ}$ on multiplying by $- 1$ , which reverses the inequality, and adding the bounds on $α$ gives the range of $α - β$ .
Multiplying an inequality by a negative number reverses it
2
On that interval $cos (α - β) = 1$ only when $α - β = 0^{\circ}$
Why: The cosine reaches $1$ only at a whole number of turns, and $0^{\circ}$ is the only such value available in $[- 9 0^{\circ}, 9 0^{\circ}]$ . This is why the range was given.
Sine and cosine lie between $- 1$ and $1$
3
$α = β$
Why: The condition is a disguised way of saying the two angles are equal.
Eliminate one variable from a linear pair
4
$sin α - sin β + cos α - cos β = (sin α - sin α) + (cos α - cos α) = 0$
Why: With $β$ replaced by $α$ , each pair cancels, so the value is $0$ whatever $α$ happens to be.

Shortcut

$cos (α - β) = 1$ with both angles in $[0^{\circ}, 9 0^{\circ}]$ means $α = β$ , so every term cancels in pairs.

Common mistakes

Option A

No route reaches $- 1$ .

Option C

Read the expression as $sin α sin β + cos α cos β$ , which is $cos (α - β) = 1$ .

Option D

No route reaches $2$ ; it would need both differences to be $1$ at once.

Skills used

Sine and cosine lie between

- 1

and

1

Multiplying an inequality by a negative number reverses it Eliminate one variable from a linear pair

60Trigonometry • Trigonometric Identity • Complementary Angles

ModerateIdeal time: about 1.5 min

Consider the following statements :

1: The value of $cos 6 1^{\circ} + sin 2 9^{\circ}$ cannot exceed $1$ .

2: The value of $tan 2 3^{\circ} - cot 6 7^{\circ}$ is less than $0$ .

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

Both statements are pairs of complementary angles in disguise, so convert each to a single ratio and then compare with the stated bound.

Ratios of complementary angles

Step-by-step solution

1
$sin 2 9^{\circ} = cos (9 0^{\circ} - 2 9^{\circ}) = cos 6 1^{\circ}$
Why: $2 9^{\circ}$ and $6 1^{\circ}$ add to $9 0^{\circ}$ , so the cofunction identity turns the sum into a single term.
Ratios of complementary angles
2
$cos 6 1^{\circ} + sin 2 9^{\circ} = 2 cos 6 1^{\circ}$
Why: The expression is a fixed number, not a varying one -- so "cannot exceed $1$ " is simply a claim about its size.
3
$6 1^{\circ} > 6 0^{\circ}$ and $cos$ decreases on the first quadrant, so $cos 6 1^{\circ} < cos 6 0^{\circ} = \frac{1}{2}$ , hence $2 cos 6 1^{\circ} < 1$ ; statement 1 is correct
Why: Comparing with $6 0^{\circ}$ avoids needing a table: the special value $cos 6 0^{\circ} = \frac{1}{2}$ is exactly the threshold that $2 cos 6 1^{\circ}$ has to beat.
How the ratios move across the first quadrant Exact trigonometric ratios of the standard angles
4
$cot 6 7^{\circ} = tan (9 0^{\circ} - 6 7^{\circ}) = tan 2 3^{\circ}$ , so $tan 2 3^{\circ} - cot 6 7^{\circ} = 0$
Why: The same cofunction identity, now for tangent and cotangent: $2 3^{\circ}$ and $6 7^{\circ}$ are complementary.
5
$0$ is not less than $0$ , so statement 2 is incorrect; only statement 1 holds
Why: The difference is exactly zero, and the statement claims it is strictly negative -- so it fails, narrowly. This is the point of the question.

Shortcut

Both statements are complementary pairs: the first is $2 cos 6 1^{\circ}$ , which is below $1$ because $cos 6 1^{\circ} < \frac{1}{2}$ ; the second is $tan 2 3^{\circ} - tan 2 3^{\circ} = 0$ , which is not negative.

Common mistakes

Option B

Believed $2 cos 6 1^{\circ} > 1$ , and read $cot 6 7^{\circ}$ as $tan 6 7^{\circ}$ , making the second difference negative.

Option C

Read $cot 6 7^{\circ}$ as $tan 6 7^{\circ}$ , so $tan 2 3^{\circ} - tan 6 7^{\circ} < 0$ and statement 2 looked correct.

Option D

Judged statement 1 false by taking $cos 6 1^{\circ} > \frac{1}{2}$ , while correctly rejecting statement 2.

Skills used

Ratios of complementary angles How the ratios move across the first quadrant Exact trigonometric ratios of the standard angles