Elementary Mathematics 2019 (II)

Set C

Questions 41–60 of 100

41Mensuration • Solids • Cuboid

EasyIdeal time: about 1 min

The areas of three adjacent faces of a cuboid are $x, y$ and $z$ . If $V$ is the volume of the cuboid, then which one of the following is correct?

V = x y z

V^{2} = x y z

V^{3} = x y z

V = (x y z)^{2}

Show answer & solution

Option B

V^{2} = x y z

Approach

Name the three edges, write each given face area as a product of two of them, then multiply all three areas together.

Volume and wall area of a cuboid

Step-by-step solution

1
$ab = x, b c = y, c a = z$
Why: Three faces meeting at one corner of a cuboid are the three different pairs of edges, so with edges $a, b, c$ the three adjacent face areas are exactly these products.
Volume and wall area of a cuboid
2
$x y z = (ab) (b c) (c a) = a^{2} b^{2} c^{2}$
Why: Multiplying the three areas uses every edge exactly twice, which is why the product is a perfect square rather than something involving $a, b, c$ singly.
Power of a power, and the move to a common base
3
$V = ab c \Rightarrow V^{2} = (ab c)^{2} = a^{2} b^{2} c^{2}$
Why: The volume of a cuboid is the product of its three edges, so $a^{2} b^{2} c^{2}$ is precisely $V^{2}$ .
4
$V^{2} = x y z$
Why: Equating the two expressions for $a^{2} b^{2} c^{2}$ gives the relation asked for.

Shortcut

Dimensions settle it in one line: an area is $(length)^{2}$ , so $x y z$ carries $(length)^{6}$ — the same as $V^{2}$ , not $V$ .

Common mistakes

Option A

Multiplied the three areas and read the result as $V$ itself, missing that each edge has been counted twice.

Option C

Reasoned that three faces must mean a third power, giving $V^{3} = x y z$ .

Option D

Got the relation $V^{2} = x y z$ the right way round in the head but wrote it out inverted as $V = (x y z)^{2}$ .

Skills used

Volume and wall area of a cuboid Power of a power, and the move to a common base

42Mensuration • Plane Figures • Triangle

EasyIdeal time: about 1 min

If $l$ is the length of the median of an equilateral triangle, then what is its area?

\frac{3 l ^{2}}{3}

\frac{3 l ^{2}}{2}

3 l^{2}

2 l^{2}

Show answer & solution

Option A

\frac{3 l ^{2}}{3}

Approach

In an equilateral triangle the median is the altitude, so recover the side from $l$ and substitute into the area formula.

Area and altitude of an equilateral triangle

Step-by-step solution

1
$l = \frac{3}{2} a$
Why: In an equilateral triangle the median to a side is also the perpendicular bisector of that side, so the median length is the altitude $\frac{3}{2} a$ .
Area and altitude of an equilateral triangle
2
$a = \frac{2 l}{3}, a^{2} = \frac{4 l ^{2}}{3}$
Why: The area formula needs $a^{2}$ , so square the rearranged side rather than carrying the surd through a longer expression.
$ab = a b$ and $a / b = a / b$
3
$Area = \frac{3}{4} a^{2} = \frac{3}{4} \cdot \frac{4 l ^{2}}{3} = \frac{3 l ^{2}}{3}$
Why: The $4$ in the denominator of the area formula cancels the $4$ coming from the squaring, leaving a single surd over $3$ .

Shortcut

Area $= \frac{1}{2} \times base \times median = \frac{1}{2} \cdot \frac{2 l}{3} \cdot l = \frac{l ^{2}}{3}$ , which is the same value with the surd left in the denominator.

Common mistakes

Option B

Used $l = \frac{a}{3}$ — the distance from the centre to a vertex rather than the altitude — giving $a = 3 l$ and area $\frac{3}{2} l^{2}$ .

Option C

Took the median as half the side, so $a = 2 l$ and area $= \frac{3}{4} (4 l^{2}) = 3 l^{2}$ .

Option D

No route reaches a surd-free answer: the area of an equilateral triangle always carries $3$ .

Skills used

Area and altitude of an equilateral triangle

ab = a b

and

a / b = a / b

43Mensuration • Plane Figures • Circle

ModerateIdeal time: about 1.5 min

A piece of wire is in the form of a sector of a circle of radius $20 cm$ , subtending an angle $15 0^{\circ}$ at the centre. If it is bent in the form of a circle, then what will be its radius?

\frac{19}{3} cm

7 cm

8 cm

DNone of the above

Show answer & solution

Option D

None of the above

Approach

Find the length of wire from the sector, then set that length equal to the circumference of the new circle.

Area and arc length of a circular sector

Step-by-step solution

1
$arc = \frac{150}{360} \times 2 π (20) = \frac{5}{12} \times 40 π = \frac{50 π}{3} cm$
Why: A $15 0^{\circ}$ sector is $\frac{5}{12}$ of the full circle, so its arc is $\frac{5}{12}$ of the circumference.
Area and arc length of a circular sector
2
$2 π r = \frac{50 π}{3}$
Why: Bending a wire changes its shape but not its length, so the circumference of the new circle is the length of wire available.
Area and circumference of a circle
3
$r = \frac{25}{3} = 8 \frac{1}{3} cm$
Why: The $π$ cancels on both sides, so no value of $π$ is needed and the answer is exact.
4
$\frac{25}{3} \neq = \frac{19}{3}, 7, 8$
Why: The derived radius matches none of the three numerical options, so the remaining option is the one that applies.

Shortcut

Scaling alone gets there: the new circle must have $\frac{5}{12}$ of the old circumference, hence $\frac{5}{12}$ of the old radius, $\frac{5}{12} \times 20 = \frac{25}{3}$ .

Common mistakes

Option A

$\frac{19}{3}$ corresponds to an arc of $\frac{38 π}{3}$ cm, which no reading of the given sector produces.

Option B

$7$ cm needs an arc of $44$ cm, i.e. a central angle of about $12 6^{\circ}$ rather than $15 0^{\circ}$ .

Option C

$8$ cm is the nearest whole number to the correct $8 \frac{1}{3}$ cm, and is what rounding the answer down lands on.

Skills used

Area and arc length of a circular sector Area and circumference of a circle

44Mensuration • Plane Figures • Triangle

EasyIdeal time: about 45 s

Suppose $P, Q$ and $R$ are the mid-points of sides of a triangle of area $128 cm^{2}$ . If a triangle $A B C$ is drawn by joining the mid-points of sides of triangle $P QR$ , then what is the area of triangle $A B C$ ?

4 cm^{2}

8 cm^{2}

16 cm^{2}

32 cm^{2}

Show answer & solution

Option B

8 cm^{2}

Approach

Apply the midpoint theorem twice: each medial triangle has one quarter of the area of the triangle it sits inside.

Midpoint theorem

Step-by-step solution

1
$[P QR] = \frac{1}{4} \times 128 = 32 cm^{2}$
Why: $P, Q, R$ are the midpoints of the sides of the original triangle, so $△ P QR$ is its medial triangle: similar with ratio $\frac{1}{2}$ , hence one quarter of the area.
Midpoint theorem
2
$[A B C] = \frac{1}{4} \times 32 = 8 cm^{2}$
Why: $A B C$ stands to $P QR$ exactly as $P QR$ stands to the original triangle, so the same quarter factor applies a second time.

Shortcut

Each midpoint construction divides the area by $4$ , so after two of them divide by $16$ : $128 \div 16 = 8$ .

Common mistakes

Option A

Ran the quarter factor three times ( $128 \to 32 \to 8 \to 4$ ), counting one medial triangle more than the question describes.

Option C

Took the medial triangle to have half the area rather than a quarter at one of the two stages, giving $128 \to 32 \to 16$ .

Option D

Stopped at $△ P QR$ , which is the triangle the construction starts from, not the one asked about.

Skills used

Midpoint theorem

45Geometry • Triangle • Similarity

ModerateIdeal time: about 1.5 min

Let two lines $p$ and $q$ be parallel. Consider two points $B$ and $C$ on the line $p$ and two points $D$ and $E$ on the line $q$ . The line through $B$ and $E$ intersects the line through $C$ and $D$ at $A$ in between the two lines $p$ and $q$ . If $A C : A D = 4 : 9$ , then what is the ratio of area of triangle $A B C$ to that of triangle $A D E$ ?

2 : 3

4 : 9

16 : 81

1 : 2

Show answer & solution

Option C

16 : 81

Approach

The two parallel lines and the crossing pair of lines make a pair of similar triangles; square the given side ratio to get the area ratio.

Step-by-step solution

1
$B, A, E$ collinear and $C, A, D$ collinear, so $∠ B A C = ∠ E A D$
Why: $A$ is the crossing point of lines $B E$ and $C D$ , so the angles at $A$ in the two triangles are vertically opposite and therefore equal.
Angles on a straight line and vertically opposite angles
2
$B C ∥ D E \Rightarrow ∠ A C B = ∠ A D E$
Why: $B C$ lies along $p$ and $D E$ along $q$ , which are parallel, and $C D$ is a transversal, so these alternate interior angles are equal.
Angles made by a transversal of two parallel lines
3
$△ A B C \sim △ A E D$
Why: Two pairs of equal angles is enough (AA), and writing the vertices in matching order is what makes $A C$ correspond to $A D$ .
Similar triangles have proportional sides
4
$\frac{[ A B C ]}{[ A D E ]} = (\frac{A C}{A D})^{2} = (\frac{4}{9})^{2} = \frac{16}{81}$
Why: Similar triangles have areas in the square of the ratio of corresponding sides, and $A C$ , $A D$ are corresponding sides.

Common mistakes

Option A

Square-rooted rather than squared, turning $4 : 9$ into $2 : 3$ .

Option B

Quoted the given side ratio as the area ratio, stopping before the squaring step.

Option D

No route produces $1 : 2$ from a side ratio of $4 : 9$ .

Skills used

46Mensuration • Plane Figures

EasyIdeal time: about 1.5 min

An equilateral triangle and a square are constructed using metallic wires of equal length. What is the ratio of area of triangle to that of square?

3 : 4

2 : 3

43 : 9

23 : 9

Show answer & solution

Option C

43 : 9

Approach

Call the common wire length $L$ , write each side as a fraction of $L$ , and divide the two areas.

Area and altitude of an equilateral triangle

Step-by-step solution

1
$triangle side = \frac{L}{3}, square side = \frac{L}{4}$
Why: Equal lengths of wire means equal perimeters, so each figure's side is $L$ divided by its number of sides.
Perimeter and area of a square
2
$[triangle] = \frac{3}{4} (\frac{L}{3})^{2} = \frac{3 L ^{2}}{36}$
Why: Substituting the side into the standard equilateral-triangle area keeps everything in terms of the one common quantity $L$ .
Area and altitude of an equilateral triangle
3
$[square] = (\frac{L}{4})^{2} = \frac{L ^{2}}{16}$
Why: The square's area is simply its side squared.
4
$\frac{[ triangle ]}{[ square ]} = \frac{3 L ^{2}}{36} \times \frac{16}{L ^{2}} = \frac{16 3}{36} = \frac{4 3}{9}$
Why: $L^{2}$ cancels, so the ratio is a pure number; dividing numerator and denominator by $4$ reduces it.
Introduce a common factor $k$ for a ratio

Common mistakes

Option A

Compared the two sides instead of the two areas, and then wrote the ratio the wrong way round — the sides are in the ratio $\frac{L}{3} : \frac{L}{4} = 4 : 3$ .

Option B

No route produces a surd-free $2 : 3$ ; the equilateral triangle's area always carries $3$ .

Option D

Lost a factor of $2$ while reducing $\frac{16 3}{36}$ , writing $\frac{2 3}{9}$ instead of $\frac{4 3}{9}$ .

Skills used

Area and altitude of an equilateral triangle Perimeter and area of a square Introduce a common factor

k

for a ratio

47Mensuration • Plane Figures • Quadrilateral

EasyIdeal time: about 1.5 min

All the four sides of a parallelogram are of equal length. The diagonals are in the ratio $1 : 2$ . If the sum of the lengths of the diagonals is $12 cm$ , then what is the area of the parallelogram?

9 cm^{2}

12 cm^{2}

16 cm^{2}

25 cm^{2}

Show answer & solution

Option C

16 cm^{2}

Approach

Equal sides make the parallelogram a rhombus, so its diagonals are perpendicular; split the sum $12$ in the ratio $1 : 2$ and add up the four right triangles.

The diagonals of a rectangle are equal and bisect each other

Step-by-step solution

1
$d_{1} + d_{2} = 12, d_{1} : d_{2} = 1 : 2 \Rightarrow d_{1} = 4, d_{2} = 8$
Why: Writing the diagonals as $k$ and $2 k$ gives $3 k = 12$ , so $k = 4$ — the ratio has three parts, not two.
Introduce a common factor $k$ for a ratio
2
$diagonals bisect each other at right angles$
Why: A parallelogram with all four sides equal is a rhombus, and in a rhombus the diagonals are perpendicular as well as bisecting each other.
The diagonals of a rectangle are equal and bisect each other
3
$Area = 4 \times \frac{1}{2} (\frac{d _{1}}{2}) (\frac{d _{2}}{2}) = \frac{1}{2} d_{1} d_{2}$
Why: The two diagonals cut the rhombus into four congruent right triangles with legs $\frac{d _{1}}{2}$ and $\frac{d _{2}}{2}$ ; adding them gives the familiar half-product rule.
Area of a triangle from base and height
4
$Area = \frac{1}{2} \times 4 \times 8 = 16 cm^{2}$
Why: Substituting the two diagonals finishes it.

Shortcut

For any quadrilateral whose diagonals cross at right angles the area is $\frac{1}{2} d_{1} d_{2}$ , so once $d_{1} = 4$ and $d_{2} = 8$ are read off, the area is one multiplication.

Common mistakes

Option A

Divided $12$ by $4$ instead of by the three parts of the ratio, getting diagonals $3$ and $6$ and area $\frac{1}{2} (3) (6) = 9$ .

Option B

Quoted the given sum of the diagonals, $12$ , as though it were the area.

Option D

No route reaches $25$ from diagonals summing to $12$ in the ratio $1 : 2$ .

Skills used

The diagonals of a rectangle are equal and bisect each other Introduce a common factor

k

for a ratio Area of a triangle from base and height

48Geometry • Triangle • Pythagoras Theorem

EasyIdeal time: about 1.5 min

$A B C$ is a triangle right angled at $B$ . If $A B = 5 cm$ and $B C = 10 cm$ , then what is the length of the perpendicular drawn from the vertex $B$ to the hypotenuse?

4 cm

25 cm

\frac{4}{5} cm

8 cm

Show answer & solution

Option B

25 cm

Approach

Compute the area from the two legs, then compute it again with the hypotenuse as base and the required perpendicular as height.

Area of a triangle from base and height

Step-by-step solution

1
$A C^{2} = A B^{2} + B C^{2} = 5^{2} + 1 0^{2} = 125 \Rightarrow A C = 55 cm$
Why: The right angle is at $B$ , so $A C$ is the hypotenuse and the perpendicular is dropped to it.
Pythagoras' theorem
2
$[A B C] = \frac{1}{2} \times A B \times B C = \frac{1}{2} \times 5 \times 10 = 25 cm^{2}$
Why: In a right triangle the two legs are already perpendicular, so they serve as base and height with no extra work.
Area of a triangle from base and height
3
$[A B C] = \frac{1}{2} \times A C \times h \Rightarrow 25 = \frac{1}{2} \times 55 \times h$
Why: Any side may be taken as the base provided the height is measured perpendicular to it — and $h$ , the perpendicular from $B$ to $A C$ , is exactly that height for base $A C$ .
4
$h = \frac{50}{5 5} = \frac{10}{5} = 25 cm$
Why: Rationalising $\frac{10}{5}$ by multiplying top and bottom by $5$ gives $\frac{10 5}{5} = 25$ .
Rationalise a denominator using the conjugate

Shortcut

The perpendicular from the right angle is $\frac{product of the legs}{hypotenuse} = \frac{5 \times 10}{5 5} = 25$ .

Common mistakes

Option A

Worked in decimals and rounded $4.47$ down to $4$ .

Option C

Mishandled the surd in $\frac{50}{5 5}$ , cancelling to $\frac{4}{5}$ instead of $\frac{10}{5}$ .

Option D

No route gives $8$ cm, which is in any case longer than the leg $A B = 5$ cm and so impossible for this perpendicular.

Skills used

Area of a triangle from base and height Pythagoras' theorem Rationalise a denominator using the conjugate

49Mensuration • Solids • Cylinder

EasyIdeal time: about 1 min

Two cylinders of equal volume have their heights in the ratio $2 : 3$ . What is the ratio of their radii?

3 : 1

3 : 2

2 : 3

3 : 2

Show answer & solution

Option B

3 : 2

Approach

Set the two volume expressions equal, substitute the heights as $2 k$ and $3 k$ , and solve for the ratio of the squared radii.

Volume of a right circular cylinder

Step-by-step solution

1
$h_{1} = 2 k, h_{2} = 3 k$
Why: A ratio fixes the heights only up to a common factor, and introducing $k$ lets the factor cancel later.
Introduce a common factor $k$ for a ratio
2
$π r_{1}^{2} (2 k) = π r_{2}^{2} (3 k)$
Why: Equal volumes, with the volume of a cylinder being base area times height.
Volume of a right circular cylinder
3
$\frac{r _{1}^{2}}{r _{2}^{2}} = \frac{3}{2}$
Why: $π$ and $k$ cancel, leaving the squared radii in the inverse ratio of the heights — the shorter cylinder must be the wider one.
4
$\frac{r _{1}}{r _{2}} = \frac{3}{2} = \frac{3}{2}$
Why: Taking the square root of both parts of the ratio, which is legitimate because both are positive.
$ab = a b$ and $a / b = a / b$

Common mistakes

Option A

Took the squared radii in the ratio $3 : 1$ rather than $3 : 2$ , dropping the second height.

Option C

Squared the height ratio before inverting it, getting $\frac{r _{1}}{r _{2}} = \frac{4}{3} = \frac{2}{3}$ .

Option D

Reached $\frac{r _{1}^{2}}{r _{2}^{2}} = \frac{3}{2}$ but rooted only the numerator, writing $3 : 2$ .

Skills used

Volume of a right circular cylinder Introduce a common factor

k

for a ratio

ab = a b

and

a / b = a / b

50Mensuration • Plane Figures • Quadrilateral

Very easyIdeal time: about 35 s

The length and breadth of a rectangle are increased by $20%$ and $10%$ respectively. What is the percentage increase in the area of the rectangle?

32%

30%

25%

15%

Show answer & solution

Option A

32%

Approach

Turn each percentage increase into a multiplier and multiply them, since area is length times breadth.

Percentage change as a multiplier

Step-by-step solution

1
$l \to 1.2 l, b \to 1.1 b$
Why: An increase of $p %$ multiplies a quantity by $1 + \frac{p}{100}$ , so $20%$ gives $1.2$ and $10%$ gives $1.1$ .
Percentage change as a multiplier
2
$new area = (1.2 l) (1.1 b) = 1.32 l b$
Why: The area of a rectangle is the product of its two dimensions, so the two multipliers multiply together.
Perimeter and area of a rectangle
3
$increase = 1.32 - 1 = 0.32 = 32%$
Why: A final multiplier of $1.32$ means the area is $132%$ of what it was, i.e. up by $32%$ .

Shortcut

For two changes of $a %$ and $b %$ on a product, the net change is $a + b + \frac{ab}{100} = 20 + 10 + 2 = 32%$ .

Common mistakes

Option B

Added the two percentages and stopped, missing the cross term $\frac{20 \times 10}{100} = 2$ .

Option C

No route gives $25%$ from increases of $20%$ and $10%$ .

Option D

Averaged the two percentages instead of combining them.

Skills used

Percentage change as a multiplier Perimeter and area of a rectangle

51Mensuration • Plane Figures • Triangle

ModerateIdeal time: about 1.5 min

If the length of the hypotenuse of a right angled triangle is $10$ cm, then what is the maximum area of such a right angled triangle?

100 cm^{2}

50 cm^{2}

25 cm^{2}

10 cm^{2}

Show answer & solution

Option C

25 cm^{2}

Approach

The two legs have a fixed sum of squares; the area is half their product, which is largest when the two legs are equal.

A product with a fixed sum is largest when the factors are equal

Step-by-step solution

1
$a^{2} + b^{2} = 1 0^{2} = 100$
Why: The legs $a$ and $b$ of a right triangle with hypotenuse $10$ satisfy Pythagoras, and this is the only constraint on them.
Pythagoras' theorem
2
$Area = \frac{1}{2} ab$
Why: In a right triangle the legs are perpendicular, so one is the base and the other the height.
Area of a triangle from base and height
3
$a^{2} b^{2} is largest when a^{2} = b^{2} = 50$
Why: The quantities $a^{2}$ and $b^{2}$ have the fixed sum $100$ , and a product with a fixed sum is largest when the two factors are equal — so maximise $a^{2} b^{2}$ rather than wrestle with $ab$ directly.
A product with a fixed sum is largest when the factors are equal
4
$ab = 50 \times 50 = 50, Area_{m a x} = \frac{1}{2} \times 50 = 25 cm^{2}$
Why: Equal legs means the triangle is right isosceles, with $a = b = 50 = 52$ , and half their product is the maximum area.

Shortcut

The right angle sits on the circle with the hypotenuse as diameter, so the height to the hypotenuse is at most the radius $5$ ; area $\leq \frac{1}{2} \times 10 \times 5 = 25$ .

Common mistakes

Option A

Squared the hypotenuse and quoted $100$ , which is four times the true maximum.

Option B

Took both legs equal to the hypotenuse itself, computing $\frac{1}{2} \times 10 \times 10 = 50$ .

Option D

No route reaches $10$ ; it is the hypotenuse's length rather than any area.

Skills used

A product with a fixed sum is largest when the factors are equal Pythagoras' theorem Area of a triangle from base and height

52Mensuration • Plane Figures • Composite and Shaded Regions

EasyIdeal time: about 1 min

A square is drawn such that its vertices are lying on a circle of radius $201$ mm. What is the ratio of area of circle to that of square?

11 : 7

7 : 11

20 : 19

19 : 20

Show answer & solution

Option A

11 : 7

Approach

The square's diagonal is a diameter of the circle, so write both areas in terms of $r$ and divide; $r$ cancels.

Side of a regular polygon inscribed in a circle

Step-by-step solution

1
$diagonal = 2 r \Rightarrow side a = \frac{2 r}{2} = r 2$
Why: All four vertices lie on the circle, so the centre of the square is the centre of the circle and each diagonal is a full diameter.
Side of a regular polygon inscribed in a circle
2
$[square] = a^{2} = 2 r^{2}$
Why: The side squared is the area; squaring $r 2$ removes the surd immediately.
Perimeter and area of a square
3
$\frac{[ circle ]}{[ square ]} = \frac{π r ^{2}}{2 r ^{2}} = \frac{π}{2}$
Why: $r^{2}$ cancels, so the radius $201$ mm never enters the answer — it is there only to look as though it matters.
Area and circumference of a circle
4
$\frac{π}{2} = \frac{22/7}{2} = \frac{11}{7}$
Why: The options are exact fractions, so $π$ must be taken as $\frac{22}{7}$ to land on one.

Shortcut

Area of the inscribed square is always $\frac{2}{π}$ of the circle, so the ratio circle : square is $π : 2 = 11 : 7$ — no computation with $201$ needed.

Common mistakes

Option B

Computed $\frac{π}{2}$ correctly but reported the ratio square : circle instead of circle : square.

Option C

$20 : 19$ has no route; it is built to look as though the $201$ mm radius is involved.

Option D

$19 : 20$ is below $1$ , which is impossible for a circle compared with a square inside it.

Skills used

Side of a regular polygon inscribed in a circle Perimeter and area of a square Area and circumference of a circle

53Mensuration • Solids • Cylinder

EasyIdeal time: about 55 s

A right circular cylinder has a diameter of $20$ cm and its curved surface area is $1000 cm^{2}$ . What is the volume of the cylinder?

4000 cm^{3}

4500 cm^{3}

5000 cm^{3}

5200 cm^{3}

Show answer & solution

Option C

5000 cm^{3}

Approach

Get the height from the curved surface area, then substitute into the volume — or better, notice that the volume is half the curved surface times the radius.

Volume of a right circular cylinder

Step-by-step solution

1
$r = \frac{20}{2} = 10 cm$
Why: The question gives the diameter; every cylinder formula takes the radius.
Volume of a right circular cylinder
2
$2 π r h = 1000 \Rightarrow 2 π (10) h = 1000 \Rightarrow π h = 50$
Why: The curved surface is the rectangle the wall unrolls into, of width $2 π r$ and height $h$ ; keeping $π h$ together avoids ever dividing by $π$ .
Curved and total surface of a cylinder
3
$V = π r^{2} h = r^{2} (π h) = 100 \times 50 = 5000 cm^{3}$
Why: Writing the volume as $r^{2}$ times the group $π h$ reuses the quantity already found, so no value of $π$ is ever needed and the answer is exact.

Shortcut

$V = π r^{2} h = \frac{r}{2} \times (2 π r h) = \frac{10}{2} \times 1000 = 5000$ cm $^{3}$ — the volume is half the radius times the curved surface area, so it is one multiplication.

Common mistakes

Option A

$4000$ has no route from a curved surface of $1000$ cm $^{2}$ and a radius of $10$ cm.

Option B

$4500$ has no route; it sits on the round ladder $4000, 4500, 5000, 5200$ to make the correct value look like a guess.

Option D

$5200$ has no route; rounding $h$ to $16$ cm would give about $5029$ cm $^{3}$ , not $5200$ .

Skills used

Volume of a right circular cylinder Curved and total surface of a cylinder

54Mensuration • Plane Figures • Circle

Very easyIdeal time: about 40 s

A piece of wire of length $33$ cm is bent into an arc of a circle of radius $14$ cm. What is the angle subtended by the arc at the centre of the circle?

7 5^{\circ}

9 0^{\circ}

13 5^{\circ}

15 0^{\circ}

Show answer & solution

Option C

13 5^{\circ}

Approach

The wire is the arc, so set the arc-length formula equal to $33$ cm and solve for the central angle.

Area and arc length of a circular sector

Step-by-step solution

1
$\frac{θ}{360} \times 2 π r = 33, r = 14$
Why: An arc takes the same fraction of the circumference as its central angle takes of $36 0^{\circ}$ .
Area and arc length of a circular sector
2
$2 π r = 2 \times \frac{22}{7} \times 14 = 88 cm$
Why: With $r = 14$ the $7$ in $\frac{22}{7}$ cancels exactly, which is why the question chose that radius — the whole circumference is a clean $88$ cm.
Area and circumference of a circle
3
$\frac{θ}{360} \times 88 = 33 \Rightarrow \frac{θ}{360} = \frac{33}{88} = \frac{3}{8}$
Why: Reducing $\frac{33}{88}$ to $\frac{3}{8}$ first keeps the final step to small numbers.
4
$θ = \frac{3}{8} \times 360 = 13 5^{\circ}$
Why: Three eighths of a full turn is the required angle.

Shortcut

The circumference is $88$ cm, so the $33$ cm arc is $\frac{3}{8}$ of the circle and the angle is $\frac{3}{8}$ of $36 0^{\circ} = 13 5^{\circ}$ .

Common mistakes

Option A

$7 5^{\circ}$ corresponds to an arc of about $18.3$ cm, not $33$ cm.

Option B

$9 0^{\circ}$ is the angle for a quarter of the circumference, i.e. an arc of $22$ cm.

Option D

$15 0^{\circ}$ corresponds to an arc of about $36.7$ cm.

Skills used

Area and arc length of a circular sector Area and circumference of a circle

55Mensuration • Plane Figures • Composite and Shaded Regions

ModerateIdeal time: about 2 min

What is the ratio of the area of a square inscribed in a semicircle of radius $r$ to the area of square inscribed in a circle of radius $r$ ?

1 : 2

2 : 5

2 : 3

3 : 5

Show answer & solution

Option B

2 : 5

Approach

Put each square in position, use Pythagoras from the centre to a top corner for the semicircle case, and divide the two areas.

Pythagoras' theorem

Step-by-step solution

1
$semicircle: side s, base on the diameter, centred$
Why: A square inscribed in a semicircle must stand on the diameter and be symmetric about the centre, otherwise one top corner leaves the arc before the other reaches it.
The perpendicular from the centre bisects a chord
2
$(\frac{s}{2})^{2} + s^{2} = r^{2} \Rightarrow \frac{5 s ^{2}}{4} = r^{2} \Rightarrow s^{2} = \frac{4 r ^{2}}{5}$
Why: The centre of the semicircle, the midpoint of the square's base and a top corner form a right triangle with legs $\frac{s}{2}$ and $s$ and hypotenuse the radius $r$ .
Pythagoras' theorem
3
$circle: diagonal = 2 r \Rightarrow side = r 2 \Rightarrow area = 2 r^{2}$
Why: In the full circle both diagonals of the inscribed square are diameters.
Side of a regular polygon inscribed in a circle
4
$\frac{4 r ^{2} /5}{2 r ^{2}} = \frac{4}{10} = \frac{2}{5}$
Why: $r^{2}$ cancels, so the ratio holds for every radius.
Perimeter and area of a square

Common mistakes

Option A

Took the square in the semicircle to have side $r$ — the length of the radius — giving area $r^{2}$ and the ratio $r^{2} : 2 r^{2} = 1 : 2$ .

Option C

$2 : 3$ has no route from $s^{2} = \frac{4 r ^{2}}{5}$ .

Option D

$3 : 5$ has no route; it shares the $5$ with the correct answer to make the near-miss look plausible.

Skills used

Pythagoras' theorem The perpendicular from the centre bisects a chord Side of a regular polygon inscribed in a circle Perimeter and area of a square

56Algebra • Polynomial • Factor and Remainder Theorem

ModerateIdeal time: about 1.5 min

The quotient when $x^{4} - x^{2} + 7 x + 5$ is divided by $(x + 2)$ is $a x^{3} + b x^{2} + c x + d$ . What are the values of $a$ , $b$ , $c$ and $d$ respectively?

1

- 2

3

1

- 1

2

3

1

1

- 2

- 3

- 1

- 1

2

- 3

- 1

Show answer & solution

Option A

1

- 2

3

1

Approach

Divide the polynomial by $(x + 2)$ , writing the missing $x^{3}$ term in explicitly so no column is skipped.

Factor and remainder theorem

Step-by-step solution

1
$x^{4} - x^{2} + 7 x + 5 = x^{4} + 0 \cdot x^{3} - x^{2} + 7 x + 5$
Why: The $x^{3}$ term is absent, and a division that quietly skips that column is the commonest way this question goes wrong.
Degree of a product of polynomials
2
$x^{4} + 0 x^{3} - x^{2} + 7 x + 5 = (x + 2) (a x^{3} + b x^{2} + c x + d) + R$
Why: Dividing a quartic by a linear factor leaves a cubic quotient and a constant remainder, which is the shape the question states.
3
$a = 1; 0 = b + 2 a \Rightarrow b = - 2; - 1 = c + 2 b \Rightarrow c = 3$
Why: Comparing coefficients of $x^{4}$ , $x^{3}$ and $x^{2}$ in turn: each coefficient of the product is the new term plus twice the previous one, which is exactly the synthetic-division pattern.
Factor and remainder theorem
4
$7 = d + 2 c \Rightarrow d = 7 - 6 = 1$
Why: The $x$ coefficient gives the last unknown of the quotient.
5
$a = 1, b = - 2, c = 3, d = 1$
Why: The four values asked for, in the order the question lists them.

Shortcut

Synthetic division by $- 2$ on the row $1, 0, - 1, 7, 5$ gives $1, - 2, 3, 1$ with remainder $3$ in four multiplications.

Common mistakes

Option B

Flipped the sign of the leading coefficient, writing $a = - 1$ and $b = 2$ as though dividing by $(2 - x)$ rather than $(x + 2)$ .

Option C

Carried the alternating signs on past where they stop, writing $c = - 3$ and $d = - 1$ .

Option D

Negated every coefficient of the correct quotient.

Skills used

Factor and remainder theorem Degree of a product of polynomials

57Mensuration • Plane Figures • Triangle

EasyIdeal time: about 1 min

The sides of a triangle are $30$ cm, $28$ cm and $16$ cm respectively. In order to determine its area, the logarithm of which of the quantities are required?

37

11

28

16

21

30

28

7

37

21

11

9

37

21

9

7

Show answer & solution

Option D

37

21

9

7

Approach

Heron's formula turns the area into a product of four numbers; taking logarithms turns that product into a sum, so the four numbers inside the root are exactly the ones whose logarithms are needed.

Heron's formula

Step-by-step solution

1
$s = \frac{30 + 28 + 16}{2} = \frac{74}{2} = 37$
Why: Heron's formula is the only route to the area here, since no height or angle is given; it starts from the semi-perimeter.
Heron's formula
2
$s - a = 37 - 30 = 7, s - b = 37 - 28 = 9, s - c = 37 - 16 = 21$
Why: These three differences, together with $s$ itself, are the four factors under the radical.
3
$Area = 37 \times 7 \times 9 \times 21$
Why: Substituting into $s (s - a) (s - b) (s - c)$ .
4
$lo g (Area) = \frac{1}{2} (lo g 37 + lo g 7 + lo g 9 + lo g 21)$
Why: The logarithm of a product is the sum of the logarithms, and the square root halves the total — so the quantities whose logarithms must be looked up are $37$ , $7$ , $9$ and $21$ .
$lo g (ab) = lo g a + lo g b$ and $lo g \frac{a}{b} = lo g a - lo g b$

Shortcut

Only $s$ and the three differences $s - a$ , $s - b$ , $s - c$ ever appear under Heron's radical, so compute $s = 37$ and subtract: $7, 9, 21$ .

Common mistakes

Option A

Kept the original side lengths $28$ and $16$ in the list, treating Heron's formula as though the sides themselves were multiplied.

Option B

Mixed the differences with the sides, listing $30$ and $28$ alongside $21$ and $7$ .

Option C

Got $21$ and $9$ right but wrote $11$ for $37 - 30$ instead of $7$ .

Skills used

Heron's formula

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

and

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

58Algebra • Logarithm • Characteristic and Mantissa

ModerateIdeal time: about 1.5 min

If $lo g_{10} 1995 = 3.3000$ , then what is the value of $(0.001995)^{\frac{1}{8}}$ ?

\frac{1}{1 0 ^{0.3475}}

\frac{1}{1 0 ^{0.3375}}

\frac{1}{1 0 ^{0.3275}}

\frac{1}{1 0 ^{0.3735}}

Show answer & solution

Option B

\frac{1}{1 0 ^{0.3375}}

Approach

Write $0.001995$ as $1995 \times 1 0^{- 6}$ , take logarithms, divide by $8$ , and read the result back as a power of $10$ .

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

Step-by-step solution

1
$0.001995 = 1995 \times 1 0^{- 6}$
Why: Shifting the decimal point six places turns the given decimal into the number whose logarithm is supplied.
Bar notation for a negative characteristic
2
$lo g_{10} (0.001995) = lo g_{10} 1995 - 6 = 3.3000 - 6 = - 2.7000$
Why: The logarithm of a quotient is the difference of the logarithms, and $lo g 1 0^{- 6} = - 6$ .
$lo g (ab) = lo g a + lo g b$ and $lo g \frac{a}{b} = lo g a - lo g b$
3
$lo g_{10} [(0.001995)^{1/8}] = \frac{1}{8} \times (- 2.7000) = - 0.3375$
Why: An exponent comes down in front of the logarithm, so taking an eighth root divides the logarithm by $8$ .
$lo g a^{n} = n lo g a$
4
$(0.001995)^{1/8} = 1 0^{- 0.3375} = \frac{1}{1 0 ^{0.3375}}$
Why: Converting back from logarithmic to exponential form, and a negative index is a reciprocal — which is the form the options are written in.
Negative exponent is a reciprocal

Common mistakes

Option A

Slipped in the division, writing $\frac{2.7}{8} = 0.3475$ .

Option C

Slipped the other way, writing $\frac{2.7}{8} = 0.3275$ .

Option D

Transposed the digits of the correct $0.3375$ into $0.3735$ .

Skills used

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

Bar notation for a negative characteristic

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

and

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

Negative exponent is a reciprocal

59Algebra • Expressions • Algebraic Identities

Very easyIdeal time: about 40 s

What is $(x - a) (x - b) (x - c)$ equal to?

x^{3} - (a + b + c) x^{2} + (b c + c a + ab) x - ab c

x^{3} + (a + b + c) x^{2} + (b c + c a + ab) x + ab c

x^{3} - (b c + c a + ab) x^{2} + (a + b + c) x - ab c

x^{3} + (b c + c a + ab) x^{2} - (a + b + c) x - ab c

Show answer & solution

Option A

x^{3} - (a + b + c) x^{2} + (b c + c a + ab) x - ab c

Approach

The expression is the monic cubic whose roots are $a$ , $b$ and $c$ , so its coefficients are the elementary symmetric sums with alternating signs.

Sum and product of the zeros of a cubic

Step-by-step solution

1
$(x - a) (x - b) = x^{2} - (a + b) x + ab$
Why: Multiplying the first two factors first keeps the work to one expansion at a time.
Square of a sum and of a difference
2
$[x^{2} - (a + b) x + ab] (x - c) = x^{3} - (a + b) x^{2} + ab x - c x^{2} + (a + b) c x - ab c$
Why: Distributing the third factor across the three terms of the quadratic.
3
$= x^{3} - (a + b + c) x^{2} + (ab + b c + c a) x - ab c$
Why: Collecting the two $x^{2}$ terms and the two $x$ terms: $ab + (a + b) c = ab + b c + c a$ .
Sum and product of the zeros of a cubic
4
$signs alternate: -, +, -$
Why: For a monic cubic with roots $a, b, c$ the coefficients are the sum, the sum of pairwise products and the product, taken with alternating signs — which also tells you the $x^{2}$ coefficient must carry $a + b + c$ , never $ab + b c + c a$ .

Shortcut

Set $x = 0$ : the left side is $- ab c$ , so the constant term must be $- ab c$ — that alone rules out two options, and checking the $x^{2}$ coefficient carries $a + b + c$ rules out the third.

Common mistakes

Option B

Expanded $(x + a) (x + b) (x + c)$ instead, so every sign came out positive.

Option C

Interchanged the coefficients of $x^{2}$ and $x$ , putting $b c + c a + ab$ where $a + b + c$ belongs.

Option D

Both interchanged the two middle coefficients and lost the signs.

Skills used

Sum and product of the zeros of a cubic Square of a sum and of a difference

60Algebra • Logarithm • Laws of Logarithms

HardIdeal time: about 2.5 min

Let $X Y Z$ be an equilateral triangle in which $X Y = 7$ cm. If $A$ denotes the area of the triangle, then what is the value of $lo g_{10} A^{4}$ ? (Given that $lo g_{10} 1050 = 3.0212$ and $lo g_{10} 35 = 1.5441$ )

5.3070

5.3700

5.5635

5.6535

Show answer & solution

Option A

5.3070

Approach

Write $A^{4}$ for the equilateral triangle of side $7$ , bring the fourth power down with the power rule, and assemble the answer from $lo g 3$ and $lo g 7 - lo g 2$ , which are what the two given logarithms supply.

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

Step-by-step solution

1
$A = \frac{3}{4} (7)^{2} = \frac{49 3}{4}$
Why: $X Y Z$ is equilateral with side $7$ , so its area comes straight from the standard formula.
Area and altitude of an equilateral triangle
2
$lo g A^{4} = 4 lo g A = 4 [\frac{1}{2} lo g 3 + 2 lo g 7 - 2 lo g 2] = 2 lo g 3 + 8 (lo g 7 - lo g 2)$
Why: The power rule brings down the $4$ , and splitting the product and quotient inside turns the area into logarithms of $3$ , $7$ and $2$ alone — grouping $lo g 7 - lo g 2$ is the key, because that difference is what the data can pin down.
$lo g a^{n} = n lo g a$
3
$lo g 3 = lo g \frac{1050}{35} - 1 = (3.0212 - 1.5441) - 1 = 1.4771 - 1 = 0.4771$
Why: $\frac{1050}{35} = 30$ , so the difference of the two given logarithms is $lo g 30 = 1 + lo g 3$ .
$lo g (ab) = lo g a + lo g b$ and $lo g \frac{a}{b} = lo g a - lo g b$
4
$lo g 35 = lo g \frac{7 \times 10}{2} = 1 + lo g 7 - lo g 2 \Rightarrow lo g 7 - lo g 2 = 0.5441$
Why: Writing $35$ as $\frac{70}{2}$ extracts exactly the grouped difference that step 2 needs; neither $lo g 7$ nor $lo g 2$ can be found on its own from the data, and neither is required.
5
$lo g A^{4} = 2 (0.4771) + 8 (0.5441) = 0.9542 + 4.3528 = 5.3070$
Why: Substituting the two assembled values finishes it.

Common mistakes

Option B

Transposed the digits of $5.3070$ into $5.3700$ .

Option C

$5.5635$ has no route from these two given logarithms.

Option D

Computed $lo g A^{2} = 2.6535$ along the way and then attached the wrong whole part, writing $5.6535$ instead of doubling it to $5.3070$ .

Skills used

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

Area and altitude of an equilateral triangle

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

and

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