Elementary Mathematics 2019 (II)

Set C

Questions 21–40 of 100

21Arithmetic • Profit and Loss

EasyIdeal time: about 1 min

A trader sells two computers at the same price, making a profit of $30%$ on one and a loss of $30%$ on the other. What is the net loss or profit percentage on the transaction ?

6%

lossB

6%

gainC

9%

lossD

9%

gain

Show answer & solution

Option C

9%

loss

Approach

Equal selling prices force the two cost prices apart; choose a convenient common selling price so both cost prices come out whole.

Percentage change as a multiplier

Step-by-step solution

1
Take the common selling price as Rs. $91$ for each computer
Why: $91$ is divisible by both $1.3$ and $0.7$ , so both cost prices stay whole numbers — the only reason for picking it.
Percentage change as a multiplier
2
Profit computer: $SP = 1.3 \times CP \Rightarrow CP = \frac{91}{1.3} = Rs. 70$
Why: A $30%$ gain multiplies the cost price by $1.3$ , so dividing undoes it.
3
Loss computer: $SP = 0.7 \times CP \Rightarrow CP = \frac{91}{0.7} = Rs. 130$
Why: A $30%$ loss multiplies the cost price by $0.7$ . The loss-making machine cost nearly twice the other, which is why the deal cannot break even.
4
Total $CP = 70 + 130 = Rs. 200$ ; total $SP = 2 \times 91 = Rs. 182$
Why: Costs and revenues are added separately before any percentage is taken.
Gain percent is measured on the cost price
5
Loss $= 200 - 182 = Rs. 18$ ; loss percent $= \frac{18}{200} \times 100 = 9%$
Why: The percentage is always measured on the total cost price, never on the selling price.

Shortcut

With the same selling price and equal percentages up and down, there is always a loss of $\frac{x ^{2}}{100} %$ ; here $\frac{3 0 ^{2}}{100} = 9%$ loss.

Common mistakes

Option A

Nothing reaches $6%$ ; the loss fraction is $\frac{x ^{2}}{1 0 ^{4}}$ , which is $9%$ and not $6%$ at $x = 30$ .

Option B

Nothing reaches a $6%$ gain either.

Option D

Right magnitude, wrong direction: the dearer computer is the one sold at a loss, so the transaction must lose, not gain.

Skills used

Percentage change as a multiplier Gain percent is measured on the cost price

22Arithmetic • Ratio and Proportion

EasyIdeal time: about 1 min

The monthly incomes of $A$ and $B$ are in the ratio $4$ : $3$ . Each saves Rs. $600$ . If their expenditures are in the ratio $3$ : $2$ , then what is the monthly income of $A$ ?

ARs.

1800

BRs.

2000

CRs.

2400

DRs.

3600

Show answer & solution

Option C

Rs.

2400

Approach

Carry the income ratio as $4 x$ and $3 x$ , write each expenditure as income minus savings, and set the expenditure ratio to $3 : 2$ .

Introduce a common factor

k

for a ratio

Step-by-step solution

1
Incomes: $A = 4 x$ , $B = 3 x$
Why: A ratio fixes the two incomes only up to a common factor, so one unknown suffices.
Introduce a common factor $k$ for a ratio
2
Expenditures: $A = 4 x - 600$ , $B = 3 x - 600$
Why: Expenditure is income less savings, and both save the same Rs. $600$ .
Turn a worded condition into a linear equation in one unknown
3
$\frac{4 x - 600}{3 x - 600} = \frac{3}{2}$
Why: The second ratio is the equation that pins $x$ down.
4
$2 (4 x - 600) = 3 (3 x - 600) \Rightarrow 8 x - 1200 = 9 x - 1800 \Rightarrow x = 600$
Why: Cross-multiplying clears the fraction and leaves a linear equation.
Clear the denominators of an equation
5
$A$ 's income $= 4 x = Rs. 2400$
Why: The question asks for $A$ , not for the common factor $x$ .

Shortcut

$2 (4 x - 600) = 3 (3 x - 600)$ collapses to $x = 600$ in one line, so $A$ earns $4 \times 600 = Rs. 2400$ .

Common mistakes

Option A

Rs. $1800$ is $3 x$ , which is $B$ 's income — the right value for the wrong person.

Option B

Nothing reaches Rs. $2000$ ; it is not a multiple of $4$ times a common factor satisfying the second ratio.

Option D

Nothing reaches Rs. $3600$ ; it would need $x = 900$ , which makes the expenditure ratio $\frac{3000}{2100} = \frac{10}{7}$ , not $\frac{3}{2}$ .

Skills used

Introduce a common factor

k

for a ratio Turn a worded condition into a linear equation in one unknown Clear the denominators of an equation

23Arithmetic • Ratio and Proportion

EasyIdeal time: about 50 s

The train fare and bus fare between two stations is in the ratio $3$ : $4$ . If the train fare increases by $20%$ and bus fare increases by $30%$ , then what is the ratio between revised train fare and revised bus fare ?

\frac{9}{13}

\frac{17}{12}

\frac{32}{43}

\frac{19}{21}

Show answer & solution

Option A

\frac{9}{13}

Approach

Carry the fares as $3 k$ and $4 k$ , multiply each by its own percentage-increase factor, and reduce the new ratio.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
Train fare $= 3 k$ , bus fare $= 4 k$
Why: The ratio fixes the two fares only up to a common factor, and that factor will cancel at the end.
Introduce a common factor $k$ for a ratio
2
Revised train fare $= 3 k \times 1.2 = 3.6 k$
Why: A $20%$ rise multiplies by $1 + \frac{20}{100} = 1.2$ .
Percentage change as a multiplier
3
Revised bus fare $= 4 k \times 1.3 = 5.2 k$
Why: A $30%$ rise multiplies by $1.3$ . The two factors differ, which is why the ratio changes at all.
4
$\frac{3.6 k}{5.2 k} = \frac{36}{52} = \frac{9}{13}$
Why: $k$ cancels; multiplying numerator and denominator by $10$ and dividing by $4$ reduces the fraction.

Shortcut

$3 \times 12 : 4 \times 13 = 36 : 52 = 9 : 13$ — scale by the percentage factors as $12$ and $13$ and skip the decimals. The ratio must fall below $\frac{3}{4}$ , which already leaves only one option.

Common mistakes

Option B

Nothing reaches $\frac{17}{12}$ , which is greater than $1$ and so has the bus fare below the train fare.

Option C

Adding the percentages as absolute amounts — $3 + 0.2$ against $4 + 0.3$ — gives $\frac{3.2}{4.3} = \frac{32}{43}$ .

Option D

Nothing reaches $\frac{19}{21}$ ; it is larger than the original $\frac{3}{4}$ , which the larger bus-fare rise forbids.

Skills used

Introduce a common factor

k

for a ratio Percentage change as a multiplier

24Arithmetic • Number System • Divisibility

EasyIdeal time: about 1 min

When $N$ is divided by $17$ , the quotient is equal to $182$ . The difference between the quotient and the remainder is $175$ . What is the value of $N$ ?

2975

3094

3101

3269

Show answer & solution

Option C

3101

Approach

Write $N$ as divisor times quotient plus remainder, get the remainder from the stated difference, then add.

Turn a worded condition into a linear equation in one unknown

Step-by-step solution

1
$N = 17 \times 182 + r$ , where $r$ is the remainder and $0 \leq r < 17$
Why: Every division is of this shape, and the question gives the divisor and the quotient directly.
Turn a worded condition into a linear equation in one unknown
2
Quotient $-$ remainder $= 175 \Rightarrow 182 - r = 175 \Rightarrow r = 7$
Why: The stated difference is between the quotient $182$ and the remainder, so the remainder is $182 - 175$ .
3
$r = 7 < 17$ , so it is a legitimate remainder
Why: A remainder must be smaller than the divisor; this check is what rules out misreadings of the difference.
Enumerate a residue class within a bound
4
$N = 17 \times 182 + 7 = 3094 + 7 = 3101$
Why: $17 \times 182 = 17 \times 180 + 17 \times 2 = 3060 + 34 = 3094$ .

Shortcut

Remainder $= 182 - 175 = 7$ , so $N = 17 \times 182 + 7 = 3101$ . The remainder must be under $17$ , which immediately kills the option built on $+ 175$ .

Common mistakes

Option A

Taking $175$ as the quotient gives $17 \times 175 = 2975$ .

Option B

$17 \times 182 = 3094$ is the product alone, with the remainder never added.

Option D

Reading $175$ as the remainder gives $3094 + 175 = 3269$ — but a remainder on division by $17$ cannot exceed $16$ .

Skills used

Turn a worded condition into a linear equation in one unknown Enumerate a residue class within a bound

25Arithmetic • Time and Work

EasyIdeal time: about 40 s

A stock of food grains is enough for $240$ men for $48$ days. How long will the same stock last for $160$ men ?

72

daysB

64

daysC

60

daysD

54

days

Show answer & solution

Option A

72

days

Approach

The stock is a fixed amount of food, so men times days is constant: scale it from $240$ men to $160$ .

The work-equivalence chain rule

M_{1} D_{1} H_{1} = M_{2} D_{2} H_{2}

Step-by-step solution

1
Stock $= 240 \times 48 = 11520$ man-days of food
Why: Measuring the stock in man-days makes it a single number that does not depend on how many men eat it.
The work-equivalence chain rule $M_{1} D_{1} H_{1} = M_{2} D_{2} H_{2}$
2
$240 \times 48 = 160 \times D$
Why: The same stock, differently shared: fewer men means more days, so the two quantities vary inversely.
3
$D = \frac{11520}{160} = 72$ days
Why: Or directly: $48 \times \frac{240}{160} = 48 \times \frac{3}{2} = 72$ .

Shortcut

$48 \times \frac{240}{160} = 48 \times \frac{3}{2} = 72$ days.

Common mistakes

Option B

Adding a third of $48$ for the third of the men removed — $48 + 16 = 64$ — instead of multiplying by $\frac{3}{2}$ .

Option C

Nothing reaches $60$ days; it would need $160 \times 60 = 9600$ man-days, not $11520$ .

Option D

Nothing reaches $54$ days; that is $8640$ man-days.

Skills used

The work-equivalence chain rule

M_{1} D_{1} H_{1} = M_{2} D_{2} H_{2}

26Mensuration • Recasting and Melting

ModerateIdeal time: about 1.5 min

A hollow right circular cylindrical vessel of volume $V$ whose diameter is equal to its height, is completely filled with water. A heavy sphere of maximum possible volume is then completely immersed in the vessel. What volume of water remains in the vessel?

\frac{V}{2}

\frac{V}{3}

\frac{2 V}{3}

\frac{V}{4}

Show answer & solution

Option B

\frac{V}{3}

Approach

Let the radius be $r$ ; the condition "diameter equals height" makes the height $2 r$ , which is exactly the diameter of the largest sphere that fits.

Volume of a right circular cylinder

Step-by-step solution

1
Diameter $=$ height gives height $= 2 r$ , so $V = π r^{2} (2 r) = 2 π r^{3}$
Why: Writing the vessel's volume in terms of $r$ alone is what lets the answer come out as a multiple of $V$ .
Volume of a right circular cylinder
2
The largest sphere has radius $r$
Why: The sphere is limited by the narrower of the two dimensions: the width allows a diameter of $2 r$ and the height also allows $2 r$ , so a sphere of radius $r$ just fits.
Volume and surface area of a sphere
3
Sphere volume $= \frac{4}{3} π r^{3} = \frac{4}{3} π r^{3} \cdot \frac{V}{2 π r ^{3}} = \frac{2 V}{3}$
Why: Expressing the sphere as a fraction of $V$ removes $r$ and $π$ from the answer.
4
Water left $= V - \frac{2 V}{3} = \frac{V}{3}$
Why: The vessel was full, so immersing the sphere spills exactly its own volume of water; what is left fills the space the sphere does not occupy.
Recasting preserves volume

Shortcut

A sphere fills $\frac{2}{3}$ of its circumscribing cylinder, and this cylinder is exactly that, so $\frac{1}{3}$ of the water remains.

Common mistakes

Option A

Taking the sphere to fill half the cylinder leaves $\frac{V}{2}$ ; the true fraction is $\frac{2}{3}$ , not $\frac{1}{2}$ .

Option C

$\frac{2 V}{3}$ is the sphere's own volume, i.e. the water that spilled, not the water that stayed.

Option D

Nothing reaches $\frac{V}{4}$ .

Skills used

Volume of a right circular cylinder Volume and surface area of a sphere Recasting preserves volume

27Geometry • Line • Parallel Lines and Transversals

ModerateIdeal time: about 1.5 min

Three parallel lines $x, y$ and $z$ are cut by two transversals $m$ and $n$ . Transversal $m$ cuts the lines $x, y, z$ at $P, Q, R$ respectively; and Transversal $n$ cuts the lines $x, y, z$ at $L, M, N$ respectively. If $P Q = 3 cm$ , $QR = 9 cm$ and $M N = 10.5 cm$ , then what is the length of $L M$ ?

3 cm

3.5 cm

4 cm

4.5 cm

Show answer & solution

Option B

3.5 cm

Approach

Two transversals cutting three parallel lines cut them in the same ratio, so $P Q : QR = L M : M N$ .

Angles made by a transversal of two parallel lines

Step-by-step solution

1
Join $P$ to $N$ , meeting the middle line $y$ at $T$
Why: A single auxiliary line turns the configuration into two triangles, each with a side parallel to one of the given lines.
Angles made by a transversal of two parallel lines
2
In $△ P R N$ : $QT ∥ R N$ (both lie along $y$ and $z$ ), so $\frac{P Q}{P R} = \frac{P T}{P N} = \frac{3}{12} = \frac{1}{4}$
Why: A line parallel to one side cuts off a similar triangle, so it divides the other two sides in the same ratio.
Similar triangles have proportional sides
3
In $△ N P L$ : $T M ∥ P L$ , so $\frac{N M}{N L} = \frac{N T}{N P} = 1 - \frac{1}{4} = \frac{3}{4}$
Why: The same argument in the second triangle transfers the ratio to the other transversal.
4
$\frac{M N}{L N} = \frac{3}{4} \Rightarrow L N = \frac{4}{3} \times 10.5 = 14$ , so $L M = 14 - 10.5 = 3.5 cm$
Why: Equivalently $\frac{L M}{M N} = \frac{P Q}{QR} = \frac{3}{9} = \frac{1}{3}$ , giving $L M = \frac{10.5}{3} = 3.5$ cm directly.

Shortcut

$\frac{L M}{M N} = \frac{P Q}{QR} = \frac{3}{9} = \frac{1}{3}$ , so $L M = \frac{10.5}{3} = 3.5$ cm.

Common mistakes

Option A

Assuming the intercepts are equal across transversals, $L M = P Q = 3$ cm; equal intercepts need equally spaced parallels, which is not given.

Option C

Nothing reaches $4$ cm; the ratio would then be $4 : 10.5$ , not $1 : 3$ .

Option D

Nothing reaches $4.5$ cm.

Skills used

Angles made by a transversal of two parallel lines Similar triangles have proportional sides

28Mensuration • Plane Figures • Circle

EasyIdeal time: about 50 s

The area of a sector of a circle of radius $4 cm$ is $25.6 cm^{2}$ . What is the radian measure of the arc of the sector?

2.3

3.2

3.3

3.4

Show answer & solution

Option B

3.2

Approach

Use the radian form of the sector-area formula, $\frac{1}{2} r^{2} θ$ , and solve for $θ$ .

Area and arc length of a circular sector

Step-by-step solution

1
Sector area $= \frac{1}{2} r^{2} θ$ with $θ$ in radians
Why: The question asks for a radian measure, so the radian form of the formula avoids a conversion altogether.
Area and arc length of a circular sector
2
$25.6 = \frac{1}{2} (4)^{2} θ = 8 θ$
Why: Substituting $r = 4$ gives $\frac{1}{2} \times 16 = 8$ as the coefficient.
3
$θ = \frac{25.6}{8} = 3.2$
Why: A single division finishes it.

Shortcut

$θ = \frac{2 \times area}{r ^{2}} = \frac{51.2}{16} = 3.2$ .

Common mistakes

Option A

Transposing the digits of $3.2$ gives $2.3$ .

Option C

Nothing reaches $3.3$ ; it would make the area $26.4 cm^{2}$ .

Option D

Nothing reaches $3.4$ ; it would make the area $27.2 cm^{2}$ .

Skills used

Area and arc length of a circular sector

29Geometry • Triangle • Pythagoras Theorem

EasyIdeal time: about 35 s

Which one of the following is correct in respect of a right angled triangle?

AIts orthocentre lies inside the triangleBIts orthocentre lies outside the triangleCIts orthocentre lies on the triangleDIt has no orthocentre

Show answer & solution

Option C

Its orthocentre lies on the triangle

Approach

Locate the altitudes of a right-angled triangle; two of them are the legs themselves, so their intersection is forced.

The orthocentre and the orthocentric system

Step-by-step solution

1
In $△ A B C$ with $∠ B = 9 0^{\circ}$ , $A B ⊥ B C$
Why: The two legs are already perpendicular to one another, which is what makes them altitudes.
The orthocentre and the orthocentric system
2
$A B$ is the altitude from $A$ to $B C$ , and $C B$ is the altitude from $C$ to $A B$
Why: An altitude is the perpendicular from a vertex to the opposite side; here each leg already does that job.
3
These two altitudes meet at $B$ , so the orthocentre is $B$ — a vertex, hence a point on the triangle
Why: Two altitudes are enough to fix the orthocentre, since all three are concurrent.

Shortcut

In a right-angled triangle the two legs are themselves altitudes, so the orthocentre is the right-angle vertex — on the triangle.

Common mistakes

Option A

True of an acute triangle, not of a right-angled one; the vertex is a boundary point, not an interior one.

Option B

True of an obtuse triangle; the orthocentre leaves the triangle only once an angle exceeds $9 0^{\circ}$ .

Option D

Every triangle has an orthocentre, since the three altitudes are always concurrent.

Skills used

The orthocentre and the orthocentric system

30Geometry • Triangle • Angle Bisector

ModerateIdeal time: about 1.5 min

Let the bisector of the angle $B A C$ of a triangle $A B C$ meet $B C$ in $X$ . Which one of the following is correct?

A B < B X

A B > B X

A X = C X

DNone of the above

Show answer & solution

Option B

A B > B X

Approach

Work inside $△ A B X$ : compare the angle facing $A B$ with the angle facing $B X$ , then use "the greater side faces the greater angle".

Internal angle bisector of a triangle

Step-by-step solution

1
Let $∠ B A C = 2 α$ , so $∠ B A X = α$ ; write $∠ A B C = β$ and $∠ A C B = γ$ , with $2 α + β + γ = 18 0^{\circ}$
Why: The bisector halves the angle at $A$ , and naming the other two angles lets the triangle's angle sum be used.
Internal angle bisector of a triangle
2
In $△ A B X$ : $∠ A X B = 18 0^{\circ} - β - α$
Why: The angle sum of the smaller triangle $A B X$ , whose angles are $α$ at $A$ , $β$ at $B$ and $∠ A X B$ at $X$ .
Angle sum of a triangle
3
$∠ A X B - ∠ B A X = 18 0^{\circ} - β - 2 α = γ > 0$ , so $∠ A X B > ∠ B A X$
Why: Substituting the angle sum of $△ A B C$ turns the comparison into the single positive quantity $γ$ , which is positive for any genuine triangle.
4
In $△ A B X$ , $A B$ faces $∠ A X B$ and $B X$ faces $∠ B A X$ ; the greater angle faces the greater side, so $A B > B X$
Why: This holds for every triangle $A B C$ , whatever its shape, because $γ > 0$ always.
Base angles of an isosceles triangle

Shortcut

In $△ A B X$ , $∠ A X B = ∠ B A X + γ$ because $∠ A X B$ is an exterior angle of $△ A C X$ . So $∠ A X B$ is the larger, and the side facing it, $A B$ , is the longer.

Common mistakes

Option A

The inequality read the wrong way round, which happens if $∠ B A X$ is compared with $∠ B X A$ in the wrong order.

Option C

Treating the angle bisector as if it were the median or the perpendicular bisector of $B C$ ; the bisector divides $B C$ in the ratio $A B : A C$ , not into equal parts, and $A X = C X$ fails already for an equilateral triangle.

Option D

Chosen when none of the three is tested; B does hold for every triangle.

Skills used

Internal angle bisector of a triangle Angle sum of a triangle Base angles of an isosceles triangle

31Trigonometry • Trigonometric Identity

EasyIdeal time: about 55 s

What is the value of $lo g_{10} (cos θ) + lo g_{10} (sin θ) + lo g_{10} (tan θ) + lo g_{10} (cot θ) + lo g_{10} (sec θ) + lo g_{10} (cosec θ)$ ?

- 1

0

0.5

1

Show answer & solution

Option B

0

Approach

Add the six logarithms into one logarithm of a product; the six ratios pair off into three reciprocal pairs.

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

and

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

Step-by-step solution

1
Sum $= lo g_{10} (cos θ \cdot sin θ \cdot tan θ \cdot cot θ \cdot sec θ \cdot cosec θ)$
Why: The product law for logarithms turns a sum of six logarithms into one logarithm of the product.
$lo g (ab) = lo g a + lo g b$ and $lo g \frac{a}{b} = lo g a - lo g b$
2
$sin θ \cdot cosec θ = 1$ , $cos θ \cdot sec θ = 1$ , $tan θ \cdot cot θ = 1$
Why: Each ratio is paired with its own reciprocal, so each pair multiplies to $1$ wherever both are defined.
Reciprocal and quotient relations between the ratios
3
Product $= 1 \times 1 \times 1 = 1$
Why: The whole argument of the logarithm collapses, independently of $θ$ .
4
$lo g_{10} 1 = 0$
Why: Because $1 0^{0} = 1$ .
Logarithm to exponential form

Shortcut

The six ratios form three reciprocal pairs, so their product is $1$ and the sum of the logarithms is $0$ .

Common mistakes

Option A

Nothing reaches $- 1$ ; that would need the product to be $\frac{1}{10}$ .

Option C

Nothing reaches $0.5$ ; that would need the product to be $10$ .

Option D

Reporting the product itself, $1$ , instead of its logarithm.

Skills used

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

and

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

Reciprocal and quotient relations between the ratios Logarithm to exponential form

32Trigonometry • Trigonometric Identity

HardIdeal time: about 2.5 min

If $cos^{2} x + cos x = 1$ , then what is the value of $sin^{12} x + 3 sin^{10} x + 3 sin^{8} x + sin^{6} x$ ?

1

2

4

8

Show answer & solution

Option A

1

Approach

The given equation says $cos x = sin^{2} x$ ; the expression is a cube in disguise, so factorise it and substitute.

Pythagorean trigonometric identities

Step-by-step solution

1
$cos^{2} x + cos x = 1 \Rightarrow cos x = 1 - cos^{2} x = sin^{2} x$
Why: Rearranging and using $sin^{2} x + cos^{2} x = 1$ turns the condition into a direct substitution rule.
Pythagorean trigonometric identities
2
$sin^{12} x + 3 sin^{10} x + 3 sin^{8} x + sin^{6} x = sin^{6} x (sin^{6} x + 3 sin^{4} x + 3 sin^{2} x + 1)$
Why: Taking out $sin^{6} x$ leaves the coefficients $1, 3, 3, 1$ , which is the signature of a cube.
Cube of a sum and of a difference
3
$sin^{6} x + 3 sin^{4} x + 3 sin^{2} x + 1 = (sin^{2} x + 1)^{3}$
Why: With $a = sin^{2} x$ and $b = 1$ , $a^{3} + 3 a^{2} + 3 a + 1 = (a + 1)^{3}$ .
4
Expression $= (sin^{2} x)^{3} (sin^{2} x + 1)^{3} = [sin^{2} x (sin^{2} x + 1)]^{3}$
Why: Two cubes with the same exponent combine into the cube of the product.
Power of a power, and the move to a common base
5
$= [cos x (cos x + 1)]^{3} = [cos^{2} x + cos x]^{3} = 1^{3} = 1$
Why: Substituting $sin^{2} x = cos x$ reproduces exactly the left side of the given equation, which is $1$ .

Shortcut

The expression is $[sin^{2} x (1 + sin^{2} x)]^{3}$ and $sin^{2} x = cos x$ , so it is $(cos x + cos^{2} x)^{3} = 1^{3} = 1$ .

Common mistakes

Option B

Nothing reaches $2$ .

Option C

Nothing reaches $4$ .

Option D

Taking $sin^{2} x + 1 = 2$ (i.e. $sin^{2} x = 1$ ) and $sin^{6} x = 1$ gives $8$ ; but $sin^{2} x = cos x = 0.618$ , not $1$ .

Skills used

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

33Trigonometry • Trigonometric Identity

ModerateIdeal time: about 1.5 min

If $0 < θ < 9 0^{\circ}$ , $sin θ = \frac{3}{5}$ and $x = cot θ$ , then what is the value of $1 + 3 x + 9 x^{2} + 27 x^{3} + 81 x^{4} + 243 x^{5}$ ?

941

1000

1220

1365

Show answer & solution

Option D

1365

Approach

Get $cot θ$ from the $3$ - $4$ - $5$ triangle, then notice the polynomial is a sum of powers of $3 x$ .

Pythagorean trigonometric identities

Step-by-step solution

1
$sin θ = \frac{3}{5}$ and $0 < θ < 9 0^{\circ}$ , so $cos θ = 1 - \frac{9}{25} = \frac{4}{5}$
Why: The Pythagorean identity gives the cosine, and the restriction to the first quadrant makes it positive.
Pythagorean trigonometric identities
2
$x = cot θ = \frac{c o s θ}{s i n θ} = \frac{4/5}{3/5} = \frac{4}{3}$
Why: Cotangent is cosine over sine.
Reciprocal and quotient relations between the ratios
3
$3 x = 4$
Why: The cotangent was engineered so that $3 x$ is a whole number — that is the hint for how to group the polynomial.
4
$1 + 3 x + 9 x^{2} + 27 x^{3} + 81 x^{4} + 243 x^{5} = 1 + (3 x) + (3 x)^{2} + (3 x)^{3} + (3 x)^{4} + (3 x)^{5}$
Why: $9 x^{2} = (3 x)^{2}$ , $27 x^{3} = (3 x)^{3}$ and so on: the coefficients are powers of $3$ matched to the powers of $x$ .
Power of a power, and the move to a common base
5
$= 1 + 4 + 16 + 64 + 256 + 1024 = 1365$
Why: Substituting $3 x = 4$ reduces the whole expression to six powers of $4$ .

Shortcut

$cot θ = \frac{4}{3}$ , so $3 x = 4$ and the sum is $1 + 4 + 4^{2} + 4^{3} + 4^{4} + 4^{5} = \frac{4 ^{6} - 1}{3} = 1365$ .

Common mistakes

Option A

Nothing reaches $941$ .

Option B

Nothing reaches $1000$ ; it is a round-looking number with no route.

Option C

Mis-adding the six powers of $4$ — most easily $256 + 1024$ carried as $1280$ but $64$ dropped — gives figures near $1220$ .

Skills used

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

34Trigonometry • Heights and Distances

ModerateIdeal time: about 1.5 min

The angles of elevation of the tops of two pillars of heights $h$ and $2 h$ from a point $P$ on the line joining the feet of the two pillars are complementary. If the distances of the foot of the pillars from the point $P$ are $x$ and $y$ respectively, then which one of the following is correct?

2 h^{2} = x^{2} y

2 h^{2} = x y^{2}

2 h^{2} = x y

2 h^{2} = x^{2} y^{2}

Show answer & solution

Option C

2 h^{2} = x y

Approach

Write the tangent of each angle of elevation; complementary angles make the two tangents reciprocal, so their product is $1$ .

Angles of elevation and depression

Step-by-step solution

1
Pillar of height $h$ at distance $x$ : $tan α = \frac{h}{x}$
Why: The pillar, the ground and the line of sight form a right triangle, with the height opposite the angle of elevation.
Angles of elevation and depression
2
Pillar of height $2 h$ at distance $y$ : its elevation is $9 0^{\circ} - α$ , so $tan (9 0^{\circ} - α) = \frac{2 h}{y}$
Why: "Complementary" means the two angles of elevation add to $9 0^{\circ}$ .
Sine, cosine and tangent in a right triangle
3
$tan (9 0^{\circ} - α) = cot α = \frac{1}{t a n α}$
Why: The cofunction identity, with the reciprocal relation between tangent and cotangent.
Ratios of complementary angles
4
$tan α \cdot tan (9 0^{\circ} - α) = 1 \Rightarrow \frac{h}{x} \cdot \frac{2 h}{y} = 1$
Why: Multiplying the two tangents removes $α$ altogether, which is why the answer is a relation among $h$ , $x$ and $y$ only.
Reciprocal and quotient relations between the ratios
5
$\frac{2 h ^{2}}{x y} = 1 \Rightarrow 2 h^{2} = x y$
Why: Clearing the denominator gives the relation asked for.

Shortcut

Complementary elevations give $tan α tan (9 0^{\circ} - α) = 1$ , so $\frac{h}{x} \cdot \frac{2 h}{y} = 1$ and $2 h^{2} = x y$ . A dimension check alone rules out the other three.

Common mistakes

Option A

$2 h^{2} = x^{2} y$ does not balance dimensionally; it arises from squaring one tangent before multiplying.

Option B

$2 h^{2} = x y^{2}$ likewise fails the dimension check, from squaring the other tangent.

Option D

Multiplying the two right-hand sides without inverting one of them — $\frac{h}{x} \cdot \frac{2 h}{y}$ set against $\frac{1}{x y}$ — produces $2 h^{2} = x^{2} y^{2}$ .

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

35Trigonometry • Trigonometric Identity • Complementary Angles

EasyIdeal time: about 40 s

What is the value of $\frac{s i n 1 9 ^{\circ}}{c o s 7 1 ^{\circ}} + \frac{c o s 7 3 ^{\circ}}{s i n 1 7 ^{\circ}}$ ?

0

1

2

4

Show answer & solution

Option C

2

Approach

Each pair of angles in the two fractions is complementary, so the cofunction identity makes each fraction equal to $1$ .

Ratios of complementary angles

Step-by-step solution

1
$1 9^{\circ} + 7 1^{\circ} = 9 0^{\circ}$ , so $cos 7 1^{\circ} = cos (9 0^{\circ} - 1 9^{\circ}) = sin 1 9^{\circ}$
Why: Spotting the pair that adds to $9 0^{\circ}$ is what turns the first fraction into a ratio of equals.
Ratios of complementary angles
2
$\frac{s i n 1 9 ^{\circ}}{c o s 7 1 ^{\circ}} = \frac{s i n 1 9 ^{\circ}}{s i n 1 9 ^{\circ}} = 1$
Why: The numerator and denominator are the same number.
3
$7 3^{\circ} + 1 7^{\circ} = 9 0^{\circ}$ , so $sin 1 7^{\circ} = sin (9 0^{\circ} - 7 3^{\circ}) = cos 7 3^{\circ}$ , giving $\frac{c o s 7 3 ^{\circ}}{s i n 1 7 ^{\circ}} = 1$
Why: The same pairing in the second fraction.
4
Total $= 1 + 1 = 2$
Why: Add the two fractions.

Shortcut

Both fractions are a ratio of complementary cofunctions, hence $1$ each; the sum is $2$ .

Common mistakes

Option A

Treating the two fractions as cancelling against each other; they add, they do not subtract.

Option B

Stopping after the first fraction and reporting $1$ .

Option D

Doubling the answer, as if each fraction were worth $2$ .

Skills used

Ratios of complementary angles

36Geometry • Triangle • Similarity

ModerateIdeal time: about 1.5 min

The perimeter of a triangle is $22 cm$ . Through each vertex of the triangle, a straight line parallel to the opposite side is drawn. What is the perimeter of triangle formed by these lines?

33 cm

44 cm

66 cm

88 cm

Show answer & solution

Option B

44 cm

Approach

The three drawn lines make a larger triangle in which the original triangle is the medial triangle, so every side doubles.

Midpoint theorem

Step-by-step solution

1
Call the new triangle $△ QR S$ formed by the three drawn lines, with $△ A B C$ inside it
Why: Each drawn line is parallel to a side of $△ A B C$ and passes through the opposite vertex, so each vertex of $△ A B C$ lies on a side of the new triangle.
Midpoint theorem
2
$A$ , $B$ , $C$ are the mid-points of the sides of $△ QR S$
Why: The line through $A$ parallel to $B C$ , together with the lines through $B$ and $C$ , makes two parallelograms on either side of $A$ whose equal opposite sides put $A$ exactly halfway along its side of $△ QR S$ .
3
By the mid-point theorem each side of $△ A B C$ is half the parallel side of $△ QR S$
Why: The segment joining two mid-points is half the third side, so each side of the new triangle is twice the corresponding side of $△ A B C$ .
4
Perimeter of $△ QR S = 2 \times 22 = 44 cm$
Why: All three sides scale by the same factor $2$ , so the perimeter does too.
Similar triangles have proportional sides

Shortcut

The original triangle is the medial triangle of the new one, so lengths double: perimeter $= 2 \times 22 = 44$ cm.

Common mistakes

Option A

Nothing reaches $33$ cm; that would be a scale factor of $\frac{3}{2}$ .

Option C

Taking the scale factor as $3$ instead of $2$ gives $66$ cm.

Option D

Using the area ratio $2^{2} = 4$ as if it were the length ratio gives $88$ cm.

Skills used

Midpoint theorem Similar triangles have proportional sides

37Geometry • Quadrilateral • Trapezium

HardIdeal time: about 2.5 min

The sides $A D, B C$ of a trapezium $A B C D$ are parallel and the diagonals $A C$ and $B D$ meet at $O$ . If the area of triangle $A O B$ is $3 cm^{2}$ and the area of triangle $B D C$ is $8 cm^{2}$ , then what is the area of triangle $A O D$ ?

8 cm^{2}

5 cm^{2}

3.6 cm^{2}

1.8 cm^{2}

Show answer & solution

Option D

1.8 cm^{2}

Approach

The parallel sides make $△ A O D$ and $△ C O B$ similar with some ratio $k$ ; write all four corner areas in terms of $k$ and use the area of $△ B D C$ to find it.

Angles made by a transversal of two parallel lines

Step-by-step solution

1
$A D ∥ B C$ gives $∠ O A D = ∠ O C B$ and $∠ O D A = ∠ O B C$ , so $△ A O D \sim △ C O B$
Why: Alternate angles across the two parallel sides supply the AA similarity.
Angles made by a transversal of two parallel lines
2
Hence $\frac{A O}{O C} = \frac{D O}{O B} = \frac{A D}{C B} = k$
Why: Similar triangles put all corresponding lengths in one ratio, and that ratio is the only unknown in the figure.
Similar triangles have proportional sides
3
$△ A O D$ and $△ A O B$ share the apex $A$ and have bases $D O$ , $O B$ on one line, so $\frac{[ A O D ]}{[ A O B ]} = \frac{D O}{O B} = k \Rightarrow [A O D] = 3 k$
Why: Triangles with the same height compare by their bases alone.
Area of a triangle from base and height
4
$△ A O B$ and $△ B O C$ share the apex $B$ with bases $A O$ , $O C$ , so $\frac{[ A O B ]}{[ B O C ]} = \frac{A O}{O C} = k \Rightarrow [B O C] = \frac{3}{k}$
Why: The same base-comparison on the other diagonal.
5
Similarly $[D O C] = k \cdot [B O C] = 3$ , and $[B D C] = [B O C] + [D O C] = \frac{3}{k} + 3 = 8$
Why: $△ B D C$ is split by $O$ into the two corner triangles on the side $D C$ , which is what links the given $8 cm^{2}$ to $k$ .
6
$\frac{3}{k} = 5 \Rightarrow k = \frac{3}{5}$ , so $[A O D] = 3 k = \frac{9}{5} = 1.8 cm^{2}$
Why: With $k$ known the first expression gives the area asked for.

Shortcut

$[A O B] = [D O C] = 3$ , so $[B O C] = 8 - 3 = 5$ ; and $[A O D] \cdot [B O C] = [A O B]^{2}$ gives $[A O D] = \frac{9}{5} = 1.8 cm^{2}$ .

Common mistakes

Option A

Repeating the given $8 cm^{2}$ ; that is $[B D C]$ , which contains $△ B O C$ and $△ D O C$ , not $△ A O D$ .

Option B

Stopping at $[B O C] = 8 - 3 = 5 cm^{2}$ , which is the opposite corner, not the one asked for.

Option C

Taking $k = \frac{6}{5}$ instead of $\frac{3}{5}$ — from $\frac{3}{k} = \frac{5}{2}$ — gives $3 k = 3.6$ .

Skills used

Angles made by a transversal of two parallel lines Similar triangles have proportional sides Area of a triangle from base and height

38Geometry • Circle • Locus

ModerateIdeal time: about 1.5 min

A line segment $A B$ is the diameter of a circle with centre at $O$ having radius $6.5 cm$ . Point $P$ is in the plane of the circle such that $A P = x$ and $B P = y$ . In which one of the following cases the point $P$ does not lie on the circle?

x = 6.5 cm

and

y = 6.5 cm

x = 12 cm

and

y = 5 cm

x = 5 cm

and

y = 12 cm

x = 0 cm

and

y = 13 cm

Show answer & solution

Option A

x = 6.5 cm

and

y = 6.5 cm

Approach

$A B$ is a diameter, so $P$ lies on the circle exactly when $A P^{2} + B P^{2} = A B^{2}$ ; test each pair.

A circle is the locus of points at a fixed distance from a fixed point

Step-by-step solution

1
$A B = 2 \times 6.5 = 13 cm$
Why: The diameter is twice the radius.
A circle is the locus of points at a fixed distance from a fixed point
2
$P$ on the circle $\Rightarrow ∠ A P B = 9 0^{\circ}$ (angle in a semicircle) $\Rightarrow x^{2} + y^{2} = 169$
Why: The angle in a semicircle is a right angle, and the converse holds too, so $x^{2} + y^{2} = 169$ is exactly the condition for $P$ to be on the circle.
The angle at the centre is twice the angle at the circumference
3
A: $6. 5^{2} + 6. 5^{2} = 42.25 + 42.25 = 84.5 \neq = 169$
Why: This pair fails the condition, so this $P$ is not on the circle.
Pythagoras' theorem
4
B: $1 2^{2} + 5^{2} = 169$ ; C: $5^{2} + 1 2^{2} = 169$ ; D: $0^{2} + 1 3^{2} = 169$
Why: All three satisfy it. In D, $x = 0$ puts $P$ at $A$ itself, and $A$ is on the circle, with $B P = 13$ being the diameter.

Shortcut

$P$ is on the circle exactly when $x^{2} + y^{2} = 1 3^{2} = 169$ . Only $6. 5^{2} + 6. 5^{2} = 84.5$ fails.

Common mistakes

Option B

$1 2^{2} + 5^{2} = 169$ , so this $P$ does lie on the circle; it is the $5$ - $12$ - $13$ triangle.

Option C

The same point with $x$ and $y$ interchanged, also on the circle.

Option D

Rejecting $x = 0$ as impossible; it simply means $P$ coincides with $A$ , which is a point of the circle, and $y = 13$ is then the diameter.

Skills used

A circle is the locus of points at a fixed distance from a fixed point The angle at the centre is twice the angle at the circumference Pythagoras' theorem

39Geometry • Triangle • Similarity

EasyIdeal time: about 45 s

The perimeters of two similar triangles $A B C$ and $P QR$ are $75 cm$ and $50 cm$ respectively. If the length of one side of the triangle $P QR$ is $20 cm$ , then what is the length of corresponding side of the triangle $A B C$ ?

25 cm

30 cm

40 cm

45 cm

Show answer & solution

Option B

30 cm

Approach

In similar triangles the perimeters are in the same ratio as corresponding sides, so scale the given side by that ratio.

Step-by-step solution

1
$△ A B C \sim △ P QR$ , so $\frac{side of A B C}{corresponding side of P QR} = \frac{perimeter of A B C}{perimeter of P QR}$
Why: Similarity puts every corresponding length in one ratio, and the perimeter is just a sum of three such lengths, so it carries the same ratio.
Similar triangles have proportional sides
2
$\frac{75}{50} = \frac{3}{2}$
Why: The scale factor from $P QR$ to $A B C$ .
3
Required side $= 20 \times \frac{3}{2} = 30 cm$
Why: $A B C$ is the larger triangle, so the side must come out larger than $20$ cm.

Shortcut

Perimeters $75 : 50 = 3 : 2$ , so the side is $20 \times \frac{3}{2} = 30$ cm.

Common mistakes

Option A

Nothing reaches $25$ cm; the ratio would be $\frac{5}{4}$ , not $\frac{3}{2}$ .

Option C

Doubling instead of multiplying by $\frac{3}{2}$ gives $40$ cm.

Option D

Using the squared ratio $(\frac{3}{2})^{2} = \frac{9}{4}$ , which governs areas and not lengths, gives $45$ cm.

Skills used

40Geometry • Quadrilateral • Parallelogram

ModerateIdeal time: about 1.5 min

Let $P QR S$ be a parallelogram whose diagonals $P R$ and $QS$ intersect at $O$ . If triangle $QR S$ is an equilateral triangle having a side of length $10 cm$ , then what is the length of the diagonal $P R$ ?

53 cm

103 cm

153 cm

203 cm

Show answer & solution

Option B

103 cm

Approach

$O$ is the mid-point of both diagonals, so $R O$ is the median of $△ QR S$ to $QS$ ; in an equilateral triangle that median is the altitude.

The diagonals of a rectangle are equal and bisect each other

Step-by-step solution

1
In parallelogram $P QR S$ the diagonals bisect each other, so $O$ is the mid-point of $QS$ and of $P R$
Why: This is what lets a length in $△ QR S$ be converted into the full diagonal $P R$ .
The diagonals of a rectangle are equal and bisect each other
2
$R O$ joins $R$ to the mid-point of $QS$ , so it is the median of $△ QR S$ from $R$
Why: $QS$ is a side of $△ QR S$ and also a diagonal of the parallelogram — the two roles meet at $O$ .
3
$△ QR S$ is equilateral with side $10$ , so this median is its altitude: $R O = \frac{3}{2} \times 10 = 53 cm$
Why: In an equilateral triangle the median from any vertex is also the altitude, whose length is $\frac{3}{2}$ times the side.
Area and altitude of an equilateral triangle
4
$P R = 2 \times R O = 103 cm$
Why: $O$ bisects $P R$ , so the full diagonal is twice $R O$ .

Shortcut

$QR = R S = QS = 10$ makes $P QR S$ a rhombus of side $10$ with one diagonal $10$ ; the other is $2 1 0^{2} - 5^{2} = 103$ cm.

Common mistakes

Option A

$53$ cm is $R O$ , half the diagonal — one step short.

Option C

Nothing reaches $153$ cm; that would need $R O = 7.5 3$ .

Option D

Doubling a second time, as if $P R = 2 \times 103$ .

Skills used

The diagonals of a rectangle are equal and bisect each other Area and altitude of an equilateral triangle