Elementary Mathematics 2020 (II)

Set A

Questions 61–80 of 100

61Geometry • Triangle • Pythagoras Theorem

ModerateIdeal time: about 1.5 min

In a quadrilateral ABCD, $∠ B = 9 0^{\circ}$ and $A B^{2} + B C^{2} + C D^{2} - A D^{2} = 0$ , then what is $∠ A C D$ equal to ?

3 0^{\circ}

6 0^{\circ}

9 0^{\circ}

12 0^{\circ}

Show answer & solution

Option C

9 0^{\circ}

Approach

Use the right angle at $B$ to replace $A B^{2} + B C^{2}$ by $A C^{2}$ , then read the remaining relation as the converse of Pythagoras in $△ A C D$ .

Pythagoras' theorem

Step-by-step solution

1
In $△ A B C$ , $∠ B = 9 0^{\circ}$ , so $A B^{2} + B C^{2} = A C^{2}$
Why: The diagonal $A C$ splits the quadrilateral into two triangles, and the given right angle lives in the first of them. This is the only use the condition $∠ B = 9 0^{\circ}$ has.
Pythagoras' theorem
2
$A B^{2} + B C^{2} + C D^{2} - A D^{2} = 0 \Rightarrow A C^{2} + C D^{2} = A D^{2}$
Why: Substituting turns a four-term condition into a three-term one involving only the second triangle $A C D$ -- which is why the diagonal was drawn.
Eliminate one variable from a linear pair
3
In $△ A C D$ , $A C^{2} + C D^{2} = A D^{2}$ , so by the converse of Pythagoras the angle between $A C$ and $C D$ is a right angle
Why: The converse is what is needed here: the two sides whose squares add are $A C$ and $C D$ , so the right angle sits at their common vertex $C$ , with $A D$ as hypotenuse.
4
$∠ A C D = 9 0^{\circ}$
Why: That common vertex is exactly the angle asked for.

Shortcut

Read the condition as $A C^{2} + C D^{2} = A D^{2}$ after using $∠ B = 9 0^{\circ}$ ; the converse of Pythagoras puts the right angle at $C$ .

Common mistakes

Option A

No route reaches $3 0^{\circ}$ ; nothing in the data distinguishes a $30$ - $60$ split.

Option B

No route reaches $6 0^{\circ}$ .

Option D

No route reaches $12 0^{\circ}$ ; an obtuse angle at $C$ would need $A C^{2} + C D^{2} < A D^{2}$ strictly.

Skills used

Pythagoras' theorem Eliminate one variable from a linear pair

62Geometry • Triangle • Angle Bisector

EasyIdeal time: about 50 s

In a $△ A B C$ , $A C = 12$ cm, $A B = 16$ cm and AD is the bisector of $∠ A$ . If $B D = 4$ cm, then what is DC equal to ?

2

cmB

3

cmC

4

cmD

5

Show answer & solution

Option B

3

Approach

Apply the internal angle bisector theorem: the bisector from $A$ cuts $B C$ in the ratio of the two sides containing $∠ A$ .

Internal angle bisector of a triangle

Step-by-step solution

1
$A D$ bisects $∠ A$ , so $\frac{B D}{D C} = \frac{A B}{A C}$
Why: The bisector theorem is the only result that links a bisector to the segments it makes on the opposite side. The side adjacent to $B$ goes over the segment adjacent to $B$ -- getting this correspondence right is the whole question.
Internal angle bisector of a triangle
2
$\frac{4}{D C} = \frac{16}{12} = \frac{4}{3}$
Why: Substitute $A B = 16$ , $A C = 12$ , $B D = 4$ , and reduce the ratio before cross-multiplying to keep the numbers small.
Unitary method: find one unit, then scale
3
$D C = 4 \times \frac{3}{4} = 3$ cm
Why: Cross-multiplying gives $4 \times 12 = 16 \times D C$ , i.e. $D C = 3$ .
Clear the denominators of an equation

Shortcut

$B D : D C = A B : A C = 4 : 3$ , and $B D = 4$ , so $D C = 3$ cm.

Common mistakes

Option A

No route reaches $2$ cm.

Option C

Assumed the bisector also bisects the opposite side, so $D C = B D = 4$ cm.

Option D

Set up the ratio the wrong way round, $\frac{B D}{D C} = \frac{A C}{A B}$ , giving $D C = \frac{16}{3} = 5.33$ cm and rounding to the nearest option.

Skills used

Internal angle bisector of a triangle Unitary method: find one unit, then scale Clear the denominators of an equation

63Geometry • Circle • Arcs and Angles

HardIdeal time: about 2.5 min

ABCD is a cyclic quadrilateral. The bisectors of the angles A, B, C and D cut the circle at P, Q, R and S respectively. What is $∠ P QR + ∠ R S P$ equal to ?

9 0^{\circ}

13 5^{\circ}

18 0^{\circ}

27 0^{\circ}

Show answer & solution

Option C

18 0^{\circ}

Approach

Show that $P$ , $Q$ , $R$ , $S$ lie on the same circle in that cyclic order, so $∠ P QR$ and $∠ R S P$ are opposite angles of a cyclic quadrilateral.

Opposite angles of a cyclic quadrilateral are supplementary

Step-by-step solution

1
Let the arcs cut off by the sides be $A B = 2 a$ , $B C = 2 b$ , $C D = 2 c$ , $D A = 2 d$ , with $a + b + c + d = 18 0^{\circ}$
Why: Naming the four arcs is the cleanest way to track where each bisector lands; the arcs total $36 0^{\circ}$ , which is where $a + b + c + d = 18 0^{\circ}$ comes from.
The angle at the centre is twice the angle at the circumference
2
The bisector of $∠ A$ meets the circle at the mid-point $P$ of arc $B C D$ ; similarly $Q$ bisects arc $C D A$ , $R$ bisects arc $D A B$ and $S$ bisects arc $A B C$
Why: Equal inscribed angles stand on equal arcs. Since $A P$ splits $∠ B A D$ into two equal angles, it splits the opposite arc $B C D$ into two equal arcs -- so each bisector lands at an arc mid-point.
Internal angle bisector of a triangle
3
Measuring arcs from $A$ : $S$ at $a + b$ , $P$ at $2 a + b + c$ , $Q$ at $2 a + 2 b + c + d$ , $R$ at $3 a + 2 b + 2 c + d$
Why: Writing down the four positions settles the order. The gaps are $a + c$ , $b + d$ , $a + c$ and $b + d$ , all positive, so the four points occur in the order $S, P, Q, R$ round the circle -- that is, $P QR S$ is a quadrilateral with its vertices in order.
4
$P QR S$ is a cyclic quadrilateral, and $∠ P QR$ , $∠ R S P$ are its angles at the opposite vertices $Q$ and $S$
Why: This is the point of the question: nothing about $A B C D$ beyond its being cyclic is needed, because the four bisector feet are themselves concyclic by construction.
Opposite angles of a cyclic quadrilateral are supplementary
5
$∠ P QR + ∠ R S P = 18 0^{\circ}$
Why: Opposite angles of a cyclic quadrilateral are supplementary.

Shortcut

$P$ , $Q$ , $R$ , $S$ all lie on the given circle, and the bisectors from $A, B, C, D$ land in that cyclic order, so $P QR S$ is a cyclic quadrilateral and its opposite angles at $Q$ and $S$ sum to $18 0^{\circ}$ -- no arcs need be computed.

Common mistakes

Option A

No route reaches $9 0^{\circ}$ ; it is the value each angle happens to take in the square-like case, not the sum.

Option B

No route reaches $13 5^{\circ}$ .

Option D

No route reaches $27 0^{\circ}$ ; the four angles of $P QR S$ total $36 0^{\circ}$ , so no opposite pair can exceed that.

Skills used

Opposite angles of a cyclic quadrilateral are supplementary The angle at the centre is twice the angle at the circumference Internal angle bisector of a triangle

64Geometry • Triangle • Pythagoras Theorem

ModerateIdeal time: about 1.5 min

ABC is an equilateral triangle. The side BC is trisected at D such that $B C = 3 B D$ . What is the ratio of $A D^{2}$ to $A B^{2}$ ?

7 : 9

1 : 3

5 : 7

1 : 2

Show answer & solution

Option A

7 : 9

Approach

Drop the altitude from $A$ to $B C$ , locate $D$ relative to its foot, and use Pythagoras twice.

Pythagoras' theorem

Step-by-step solution

1
Let $A B = B C = C A = a$ . Then $B D = \frac{a}{3}$
Why: $B C = 3 B D$ is the given, so $D$ is the trisection point nearer $B$ .
Unitary method: find one unit, then scale
2
Let $M$ be the foot of the altitude from $A$ . Then $B M = \frac{a}{2}$ and $A M^{2} = a^{2} - \frac{a ^{2}}{4} = \frac{3 a ^{2}}{4}$
Why: In an equilateral triangle the altitude from a vertex bisects the opposite side (the two halves give congruent right triangles), so $B M$ is half of $B C$ and $A M$ follows from Pythagoras in $△ A B M$ .
Congruence criteria for triangles Pythagoras' theorem
3
$D M = B M - B D = \frac{a}{2} - \frac{a}{3} = \frac{a}{6}$
Why: $D$ and $M$ both lie on $B C$ on the same side of $B$ , so the gap between them is a subtraction. Working with $D M$ rather than $B D$ is what lets the second Pythagoras be applied to a right triangle.
4
$A D^{2} = A M^{2} + D M^{2} = \frac{3 a ^{2}}{4} + \frac{a ^{2}}{36} = \frac{27 a ^{2} + a ^{2}}{36} = \frac{28 a ^{2}}{36} = \frac{7 a ^{2}}{9}$
Why: $△ A D M$ is right-angled at $M$ because $A M$ is the altitude, so Pythagoras applies again.
5
$\frac{A D ^{2}}{A B ^{2}} = \frac{7 a ^{2} /9}{a ^{2}} = \frac{7}{9}$ , i.e. $7 : 9$
Why: The side $a$ cancels, as it must -- the ratio cannot depend on the size of the triangle.
Similar triangles have proportional sides

Shortcut

With $a = 3$ : $A M = \frac{3 3}{2}$ , $D M = \frac{1}{2}$ , so $A D^{2} = \frac{27}{4} + \frac{1}{4} = 7$ while $A B^{2} = 9$ .

Common mistakes

Option B

No route gives $A D^{2} = \frac{a ^{2}}{3}$ .

Option C

No route reaches $5 : 7$ .

Option D

No route reaches $1 : 2$ ; it would need $A D$ to be shorter than the altitude.

Skills used

Pythagoras' theorem Unitary method: find one unit, then scale Congruence criteria for triangles Similar triangles have proportional sides

65Geometry • Quadrilateral • Trapezium

EasyIdeal time: about 1 min

Consider the following statements :

1: The diagonals of a trapezium divide each other proportionally.

2: Any line drawn parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.

Which of the above statements is/are correct ?

A1 onlyB2 onlyCBoth 1 and 2DNeither 1 nor 2

Show answer & solution

Option C

Both 1 and 2

Approach

Both statements are the basic proportionality theorem applied inside a trapezium; check each against it.

Parallel lines cut transversals proportionally

Step-by-step solution

1
In trapezium $A B C D$ with $A B ∥ C D$ , the diagonals meet at $O$ ; then $△ A O B \sim △ C O D$
Why: $A B ∥ C D$ makes the alternate angles at $A$ and $C$ , and at $B$ and $D$ , equal, so the two triangles are similar by AA.
Similar triangles have proportional sides
2
$\frac{A O}{O C} = \frac{B O}{O D}$ , so statement 1 is correct
Why: Similar triangles have proportional sides, and these are exactly the four segments into which the diagonals cut each other.
3
A line $P Q ∥ A B ∥ C D$ with $P$ on $A D$ and $Q$ on $B C$ cuts three parallel lines with the two non-parallel sides as transversals, so $\frac{A P}{P D} = \frac{B Q}{QC}$ ; statement 2 is correct
Why: This is the intercept theorem in its three-parallel-lines form. The statement says "any" such line, and the theorem holds for every one of them, so the generality is not a problem.
Parallel lines cut transversals proportionally
4
Both statements hold
Why: Nothing in either statement needs the trapezium to be isosceles or the line to pass through a special point.

Shortcut

Both are the basic proportionality theorem: the diagonals give $△ A O B \sim △ C O D$ , and a line parallel to the parallel sides cuts the legs proportionally.

Common mistakes

Option A

Accepted the diagonal result but doubted statement 2, reading "any line" as needing to pass through the intersection of the diagonals.

Option B

Rejected statement 1, confusing a trapezium with a general quadrilateral whose diagonals need not divide proportionally.

Option D

No route rejects both.

Skills used

Parallel lines cut transversals proportionally Similar triangles have proportional sides

66Mensuration • Solids • Cone

ModerateIdeal time: about 2 min

If H, C and V are respectively the height, curved surface area and volume of a cone, then what is $3 π V H^{3} + 9 V^{2}$ equal to ?

C^{2} H^{2}

2 C^{2} H^{2}

5 C^{2} H^{2}

7 C^{2} H^{2}

Show answer & solution

Option A

C^{2} H^{2}

Approach

Write $V$ and $C$ in terms of $r$ and $H$ , substitute into the expression, and factor out $π^{2} r^{2} H^{2}$ .

Volume of a right circular cone

Step-by-step solution

1
$V = \frac{1}{3} π r^{2} H$ and $C = π r l$ with $l^{2} = r^{2} + H^{2}$
Why: Only the radius is not named in the question, so everything must be expressed in $r$ and $H$ and the $r$ must disappear by the end.
Volume of a right circular cone Slant height and curved surface of a cone
2
$3 π V H^{3} = 3 π \cdot \frac{1}{3} π r^{2} H \cdot H^{3} = π^{2} r^{2} H^{4}$
Why: The factor $3$ and the $\frac{1}{3}$ in the volume cancel, which is a hint that the expression was built from this formula.
3
$9 V^{2} = 9 \cdot \frac{1}{9} π^{2} r^{4} H^{2} = π^{2} r^{4} H^{2}$
Why: Likewise the $9$ cancels the $\frac{1}{9}$ from squaring $\frac{1}{3}$ .
Power of a power, and the move to a common base
4
Sum $= π^{2} r^{2} H^{2} (H^{2} + r^{2}) = π^{2} r^{2} H^{2} l^{2}$
Why: Taking out the common factor $π^{2} r^{2} H^{2}$ leaves exactly $r^{2} + H^{2}$ , which Pythagoras in the cone's axial section turns into $l^{2}$ . This is the step the question is built around.
Pythagoras' theorem
5
$π^{2} r^{2} l^{2} H^{2} = (π r l)^{2} H^{2} = C^{2} H^{2}$
Why: Regrouping the squares recovers the curved surface area, so the $r$ has gone and only the named quantities remain.

Shortcut

Test $r = 3$ , $H = 4$ , $l = 5$ : the expression comes to $3600 π^{2}$ and $C^{2} H^{2} = 3600 π^{2}$ , so the factor is $1$ .

Common mistakes

Option B

No route produces a factor $2$ ; a guessed coefficient.

Option C

No route produces a factor $5$ .

Option D

No route produces a factor $7$ .

Skills used

Volume of a right circular cone Slant height and curved surface of a cone Power of a power, and the move to a common base Pythagoras' theorem

67Mensuration • Recasting and Melting

EasyIdeal time: about 1 min

How many solid lead balls each of diameter $2$ mm can be made from a solid lead ball of radius $8$ cm ?

512

1024

256000

512000

Show answer & solution

Option D

512000

Approach

Melting conserves volume, so the count is the ratio of volumes -- which for spheres is the cube of the ratio of radii.

Recasting preserves volume

Step-by-step solution

1
Big ball: $R = 8$ cm. Small ball: diameter $2$ mm, so $r = 1$ mm $= 0.1$ cm
Why: The two measurements are in different units and the radius is given for one but the diameter for the other -- both must be fixed before anything is cubed, because the error is then cubed too.
Volume and surface area of a sphere
2
Number $= \frac{\frac{4}{3} π R ^{3}}{\frac{4}{3} π r ^{3}} = (\frac{R}{r})^{3}$
Why: Recasting conserves volume, so the number of small balls is the volume ratio; the common factor $\frac{4}{3} π$ cancels, leaving only the cube of the radius ratio.
Recasting preserves volume
3
$\frac{R}{r} = \frac{8}{0.1} = 80$
Why: Working in centimetres throughout; equivalently $\frac{80 mm}{1 mm} = 80$ .
Unitary method: find one unit, then scale
4
$8 0^{3} = 512000$
Why: $8^{3} = 512$ and the three factors of $10$ contribute $1000$ .
Power of a power, and the move to a common base

Shortcut

Work in millimetres: $R = 80$ mm, $r = 1$ mm, so the count is $8 0^{3} = 512000$ .

Common mistakes

Option A

Ignored the units and used $\frac{8}{1} = 8$ , giving $8^{3} = 512$ .

Option B

Doubled $512$ , treating the given diameter as though it still needed halving after the ratio was formed.

Option C

Halved $512000$ , as if the small ball's diameter of $2$ mm meant two balls per unit volume.

Skills used

Recasting preserves volume Volume and surface area of a sphere Unitary method: find one unit, then scale Power of a power, and the move to a common base

68Mensuration • Plane Figures • Triangle

EasyIdeal time: about 1 min

The two sides of a triangle are $40$ cm and $41$ cm. If the perimeter of the triangle is $90$ cm, what is its area ?

90

^{2}

135

^{2}

150

^{2}

180

^{2}

Show answer & solution

Option D

180

^{2}

Approach

Get the third side from the perimeter, then use Heron's formula -- and notice on the way that the triangle is right-angled.

Pythagoras' theorem

Step-by-step solution

1
Third side $= 90 - (40 + 41) = 9$ cm
Why: The perimeter is the only given that has not been used, and it fixes the missing side.
Turn a worded condition into a linear equation in one unknown
2
$9^{2} + 4 0^{2} = 81 + 1600 = 1681 = 4 1^{2}$
Why: Before reaching for Heron, test the three sides against Pythagoras -- CDS numbers like $9, 40, 41$ are usually a Pythagorean triple, and if they are, the area is a one-line computation.
Pythagoras' theorem
3
The triangle is right-angled, with legs $9$ and $40$ and hypotenuse $41$
Why: The converse of Pythagoras makes the two shorter sides perpendicular, so they are the base and the height.
4
Area $= \frac{1}{2} \times 9 \times 40 = 180$ cm $^{2}$
Why: Half base times height, with no square roots at all.
Area of a triangle from base and height

Shortcut

$9, 40, 41$ is a Pythagorean triple, so the area is $\frac{1}{2} \times 9 \times 40 = 180$ cm $^{2}$ .

Common mistakes

Option A

No route reaches $90$ cm $^{2}$ ; it is the perimeter's value carried across.

Option B

No route reaches $135$ cm $^{2}$ .

Option C

No route reaches $150$ cm $^{2}$ .

Skills used

Pythagoras' theorem Turn a worded condition into a linear equation in one unknown Area of a triangle from base and height

69Mensuration • Plane Figures • Quadrilateral

HardIdeal time: about 2.5 min

The diagonals of a rhombus differ by $2$ units and its perimeter exceeds the sum of the diagonals by $6$ units. What is the area of the rhombus ?

48

square unitsB

36

square unitsC

24

square unitsD

12

square units

Show answer & solution

Option C

24

square units

Approach

Turn both conditions into equations in the diagonals, using the fact that half-diagonals and a side form a right triangle.

Area from perpendicular diagonals

Step-by-step solution

1
Let the diagonals be $d_{1}$ and $d_{2}$ with $d_{1} - d_{2} = 2$
Why: Name the two quantities the answer depends on; the area of a rhombus needs nothing else.
Turn a worded condition into a linear equation in one unknown
2
Side $= (\frac{d _{1}}{2})^{2} + (\frac{d _{2}}{2})^{2} = \frac{1}{2} d_{1}^{2} + d_{2}^{2}$ , so perimeter $= 2 d_{1}^{2} + d_{2}^{2}$
Why: The diagonals of a rhombus bisect each other at right angles, so each side is the hypotenuse of a right triangle with legs the two half-diagonals.
Area from perpendicular diagonals Pythagoras' theorem
3
$2 d_{1}^{2} + d_{2}^{2} = d_{1} + d_{2} + 6$ , and with $d_{2} = b$ , $d_{1} = b + 2$ : $2 2 b^{2} + 4 b + 4 = 2 b + 8$
Why: "Perimeter exceeds the sum of the diagonals by $6$ " is the second equation. Substituting $d_{1} = b + 2$ reduces everything to one unknown.
Eliminate one variable from a linear pair
4
$2 b^{2} + 4 b + 4 = b + 4 \Rightarrow 2 b^{2} + 4 b + 4 = b^{2} + 8 b + 16 \Rightarrow b^{2} - 4 b - 12 = 0$
Why: Divide by $2$ first so the squaring stays in small numbers, then square both sides to clear the radical.
Square both sides to clear a radical
5
$(b - 6) (b + 2) = 0 \Rightarrow b = 6$ (a length cannot be negative), so $d_{2} = 6$ , $d_{1} = 8$ , and area $= \frac{1}{2} \times 6 \times 8 = 24$ square units
Why: Of the two roots only the positive one is a diagonal; the area of a rhombus is half the product of its diagonals.
Factorise a quadratic by splitting the middle term

Shortcut

The side is a whole number here, so look for a Pythagorean triple whose legs are half-diagonals differing by $1$ : $3, 4, 5$ gives diagonals $6$ and $8$ , perimeter $20$ , and $20 - 14 = 6$ . Area $= 24$ .

Common mistakes

Option A

Used $d_{1} d_{2} = 48$ as the area, forgetting the factor $\frac{1}{2}$ .

Option B

No route reaches $36$ square units.

Option D

Halved twice, taking the area as $\frac{1}{2} \times \frac{1}{2} d_{1} d_{2}$ -- the area of one of the four right triangles times two.

Skills used

Area from perpendicular diagonals Turn a worded condition into a linear equation in one unknown Pythagoras' theorem Eliminate one variable from a linear pair Square both sides to clear a radical Factorise a quadratic by splitting the middle term

70Mensuration • Plane Figures • Triangle

EasyIdeal time: about 1 min

What is the area of a right-angled triangle, if the radius of the circumcircle is $5$ cm and altitude drawn to the hypotenuse is $4$ cm ?

20

^{2}

18

^{2}

16

^{2}

10

^{2}

Show answer & solution

Option A

20

^{2}

Approach

In a right triangle the hypotenuse is a diameter of the circumcircle, so the circumradius gives the hypotenuse; take it as the base and the given altitude as the height.

The circumcentre is the point equidistant from the vertices

Step-by-step solution

1
The circumcentre of a right triangle is the mid-point of the hypotenuse, so the hypotenuse $= 2 R = 10$ cm
Why: The right angle stands in a semicircle, so the hypotenuse must be a diameter. This converts the circumradius into a length of the triangle, which is the only way the given $R = 5$ can be used.
The circumcentre is the point equidistant from the vertices The angle at the centre is twice the angle at the circumference
2
The altitude to the hypotenuse is $4$ cm, and it is perpendicular to it by definition
Why: Area needs a base and the height to that base; the question supplies exactly this pair, so no side lengths are required.
Area of a triangle from base and height
3
Area $= \frac{1}{2} \times 10 \times 4 = 20$ cm $^{2}$
Why: Half base times height with the hypotenuse as base.

Shortcut

Hypotenuse $= 2 R = 10$ cm, so area $= \frac{1}{2} \times 10 \times 4 = 20$ cm $^{2}$ .

Common mistakes

Option B

No route reaches $18$ cm $^{2}$ .

Option C

No route reaches $16$ cm $^{2}$ .

Option D

Used the radius $5$ cm as the base instead of the diameter $10$ cm.

Skills used

The circumcentre is the point equidistant from the vertices The angle at the centre is twice the angle at the circumference Area of a triangle from base and height

71Geometry • Triangle • Angles of a Triangle

EasyIdeal time: about 1 min

In a triangle, values of all the angles are integers (in degree measure). Which one of the following cannot be the proportion of their measures ?

1 : 2 : 3

3 : 4 : 5

5 : 6 : 7

6 : 7 : 8

Show answer & solution

Option D

6 : 7 : 8

Approach

Write the three angles as a common factor $k$ times the parts of the proportion; the proportion is possible only if $k = \frac{180}{sum of the parts}$ leaves every angle a whole number.

Angle sum of a triangle

Step-by-step solution

1
$a : b : c = p : q : r \Rightarrow a = p k, b = q k, c = r k$
Why: A proportion fixes the three angles only up to one common factor $k$ .
Introduce a common factor $k$ for a ratio
2
$p k + q k + r k = 18 0^{\circ} \Rightarrow k = \frac{18 0 ^{\circ}}{p + q + r}$
Why: The three interior angles of a triangle add to $18 0^{\circ}$ , and that single equation pins down $k$ . So the angles are integers exactly when $p + q + r$ divides $180$ suitably.
Angle sum of a triangle
3
$1 + 2 + 3 = 6, k = 3 0^{\circ}$ ; $3 + 4 + 5 = 12, k = 1 5^{\circ}$ ; $5 + 6 + 7 = 18, k = 1 0^{\circ}$
Why: Each of these part-sums divides $180$ exactly, so $k$ is itself a whole number of degrees and all three angles come out as integers: $30, 60, 90$ ; $45, 60, 75$ ; $50, 60, 70$ .
4
$6 + 7 + 8 = 21$ , and $\frac{180}{21} = \frac{60}{7}$
Why: $21$ does not divide $180$ , so $k$ is a fraction and the angles would be $\frac{36 0 ^{\circ}}{7}$ , $\frac{42 0 ^{\circ}}{7}$ , $\frac{48 0 ^{\circ}}{7}$ .
5
Only $\frac{420}{7} = 60$ is an integer; $\frac{360}{7}$ and $\frac{480}{7}$ are not
Why: Two of the three angles fail to be whole numbers of degrees, so no triangle with integer angles has its angles in the proportion $6 : 7 : 8$ .

Shortcut

Only the sum of the parts matters: $6$ , $12$ and $18$ all divide $180$ , while $21$ does not.

Common mistakes

Option A

No error reaches this: $1 : 2 : 3$ gives the $3 0^{\circ}$ - $6 0^{\circ}$ - $9 0^{\circ}$ triangle, so it is a possible proportion.

Option B

Read $3 : 4 : 5$ as the familiar side ratio of a right triangle and ruled it out as an angle proportion, without testing whether $12$ divides $180$ - it does, giving $4 5^{\circ}, 6 0^{\circ}, 7 5^{\circ}$ .

Option C

Slipped in dividing $180$ by $18$ and judged the quotient non-integral; in fact $k = 1 0^{\circ}$ and the angles are $5 0^{\circ}, 6 0^{\circ}, 7 0^{\circ}$ .

Skills used

Angle sum of a triangle Introduce a common factor

k

for a ratio

72Mensuration • Plane Figures • Quadrilateral

EasyIdeal time: about 40 s

The length of a rectangle is increased by $10%$ and breadth is decreased by $10%$ . Then the area of the new rectangle is

Aneither increased nor decreasedBincreased by

1%

Cdecreased by

1%

Ddecreased by

10%

Show answer & solution

Option C

decreased by

1%

Approach

Turn each percentage change into a multiplier and multiply the two, since area is the product of the two dimensions.

Percentage change as a multiplier

Step-by-step solution

1
$l \to (1 + \frac{10}{100}) l = 1.1 l$ and $b \to (1 - \frac{10}{100}) b = 0.9 b$
Why: A rise of $10%$ multiplies a quantity by $1.1$ and a fall of $10%$ multiplies it by $0.9$ ; percentage changes act as factors, not as additions.
Percentage change as a multiplier
2
New area $= (1.1 l) (0.9 b) = 0.99 l b$
Why: The area of a rectangle is $l b$ , so the two multipliers simply multiply together.
Perimeter and area of a rectangle
3
$0.99 l b = l b - 0.01 l b$
Why: The new area is $99%$ of the old, i.e. $1%$ smaller - a decrease, not a wash.

Shortcut

A rise of $p %$ followed by a fall of $p %$ always loses $\frac{p ^{2}}{100} %$ ; here $\frac{1 0 ^{2}}{100} = 1%$ .

Common mistakes

Option A

Added the percentages, $+ 10%$ and $- 10%$ , and concluded they cancel; percentage changes compose by multiplying their factors, and $1.1 \times 0.9 = 0.99 \neq = 1$ .

Option B

Got the $1%$ but the wrong direction, reading $0.99$ as a gain instead of a loss.

Option D

Applied only the breadth's $10%$ fall and ignored the length's rise.

Skills used

Percentage change as a multiplier Perimeter and area of a rectangle

73Mensuration • Solids • Sphere

EasyIdeal time: about 45 s

The surface areas of two spheres are in the ratio $1 : 4$ . What is the ratio of their volumes ?

1 : 16

1 : 12

1 : 10

1 : 8

Show answer & solution

Option D

1 : 8

Approach

Get the ratio of the radii from the surface areas, then cube it, since volume varies as the cube of the radius.

Volume and surface area of a sphere

Step-by-step solution

1
$4 π r_{1}^{2} : 4 π r_{2}^{2} = 1 : 4 \Rightarrow r_{1}^{2} : r_{2}^{2} = 1 : 4$
Why: The surface area of a sphere is $4 π r^{2}$ , and the constant $4 π$ cancels from both parts of a ratio.
Volume and surface area of a sphere
2
$r_{1} : r_{2} = 1 : 2$
Why: Surface area varies as the square of the radius, so the radii are in the square root of the given ratio; a radius is positive, so only the positive root is taken.
Direct variation as a power: $y \propto x^{n}$
3
$V_{1} : V_{2} = \frac{4}{3} π r_{1}^{3} : \frac{4}{3} π r_{2}^{3} = r_{1}^{3} : r_{2}^{3} = 1^{3} : 2^{3} = 1 : 8$
Why: Volume is $\frac{4}{3} π r^{3}$ , so it varies as the cube of the radius and the constant again cancels.

Shortcut

Surface ratio $1 : 4$ means a linear ratio $1 : 2$ ; the volume ratio is then $1 : 2^{3}$ , read off without writing either formula.

Common mistakes

Option A

Squared the surface-area ratio instead of passing through the radii, giving $1 : 4^{2}$ .

Option B

No route reaches $1 : 12$ ; it is not a power of the ratio $1 : 2$ .

Option C

No route reaches $1 : 10$ ; it is not a power of the ratio $1 : 2$ .

Skills used

Volume and surface area of a sphere Direct variation as a power:

y \propto x^{n}

74Mensuration • Solids • Cuboid

ModerateIdeal time: about 1.5 min

The length, breadth and height of a brick are $20$ cm, $15$ cm and $10$ cm respectively. The number of bricks required to construct a wall with dimensions $45$ m length, $0.15$ m breadth and $3$ m height is

12450

11250

6750

DNone of the above

Show answer & solution

Option C

6750

Approach

Put every length in one unit, then divide the volume of the wall by the volume of one brick.

Volume and wall area of a cuboid

Step-by-step solution

1
Volume of one brick $= 20 \times 15 \times 10 = 3000 cm^{3}$
Why: A brick is a cuboid, so its volume is the product of its three edges.
Volume and wall area of a cuboid
2
$45 m = 4500 cm$ , $0.15 m = 15 cm$ , $3 m = 300 cm$
Why: The brick is given in centimetres and the wall in metres; mixing the two units is the one slip this sum is built to punish.
3
Volume of the wall $= 4500 \times 15 \times 300 = 20250000 cm^{3}$
Why: The wall is also a cuboid, so the same product applies.
4
$N = \frac{20 250 000}{3000} = 6750$
Why: The bricks fill the wall with no waste, so the number required is the wall's volume divided by one brick's volume.
Recasting preserves volume

Shortcut

Work in metres throughout: brick $0.2 \times 0.15 \times 0.1 = 0.003 m^{3}$ , wall $45 \times 0.15 \times 3 = 20.25 m^{3}$ , and $20.25 \div 0.003 = 6750$ - far shorter digits.

Common mistakes

Option A

No route reaches $12450$ ; it is not a quotient of the wall volume by any plausible brick volume.

Option B

No route reaches $11250$ from the stated dimensions; it would need a wall volume of $33.75 m^{3}$ , which none of the givens produces.

Option D

Concluded that no printed option matched after a slip in the long division or in the metre-to-centimetre conversion; the correct quotient is $6750$ , which is printed.

Skills used

Volume and wall area of a cuboid Recasting preserves volume

75Geometry • Polygon • Interior and Exterior Angles

Very easyIdeal time: about 35 s

If the sum of all interior angles of a regular polygon is twice the sum of all its exterior angles, then the polygon is

AHexagonBOctagonCNonagonDDecagon

Show answer & solution

Option A

Hexagon

Approach

Set the interior-angle sum of an $n$ -gon against twice $36 0^{\circ}$ and solve the resulting linear equation for $n$ .

Interior and exterior angle sums of a polygon

Step-by-step solution

1
$(n - 2) \times 18 0^{\circ} = 2 \times 36 0^{\circ}$
Why: The interior angles of an $n$ -sided polygon add to $(n - 2) \times 18 0^{\circ}$ , while the exterior angles add to $36 0^{\circ}$ whatever $n$ is - so 'twice the sum of the exterior angles' is simply $72 0^{\circ}$ .
Interior and exterior angle sums of a polygon
2
$(n - 2) \times 18 0^{\circ} = 72 0^{\circ} \Rightarrow n - 2 = 4$
Why: Divide both sides by $18 0^{\circ}$ .
3
$n = 6$
Why: A polygon with six sides is a hexagon.

Shortcut

The exterior sum is always $36 0^{\circ}$ , so the condition reads 'interior sum $= 72 0^{\circ}$ ', and $720 \div 180 = 4$ triangles means $6$ sides.

Common mistakes

Option B

Compared the interior sum with three times the exterior sum, solving $(n - 2) 18 0^{\circ} = 108 0^{\circ}$ to get $n = 8$ .

Option C

No route reaches nine sides: a nonagon's interior sum is $126 0^{\circ}$ , which is not a whole multiple of $36 0^{\circ}$ .

Option D

Compared the interior sum with four times the exterior sum, solving $(n - 2) 18 0^{\circ} = 144 0^{\circ}$ to get $n = 10$ .

Skills used

Interior and exterior angle sums of a polygon

76Mensuration • Plane Figures • Circle

EasyIdeal time: about 1 min

A bicycle wheel makes $5000$ revolutions in moving $11$ km. What is the radius of the wheel ? (Assume $π = \frac{22}{7}$ )

17.5

cmB

35

cmC

70

cmD

140

Show answer & solution

Option B

35

Approach

One revolution advances the wheel by one circumference; get the circumference from the distance and the count, then solve $2 π r =$ circumference.

Distance covered by a rolling wheel

Step-by-step solution

1
$11 km = 11 \times 1000 \times 100 = 1100000 cm$
Why: The options are in centimetres, so the journey must be expressed in centimetres before it is divided.
2
Circumference $= \frac{1 100 000}{5000} = 220 cm$
Why: A wheel rolling without slipping covers exactly one circumference per revolution, so distance $= N \times$ circumference.
Distance covered by a rolling wheel
3
$2 \times \frac{22}{7} \times r = 220 \Rightarrow \frac{44 r}{7} = 220$
Why: The circumference of a circle is $2 π r$ ; substitute the given $π = \frac{22}{7}$ .
Area and circumference of a circle
4
$r = 220 \times \frac{7}{44} = 35 cm$
Why: $44$ divides $220$ five times, so $r = 5 \times 7$ .

Common mistakes

Option A

Took the $220$ cm as $2 π d$ rather than $2 π r$ , so the $35$ obtained was treated as the diameter and halved once more.

Option C

Dropped the factor $2$ and solved $π r = 220$ , giving $r = 70$ - which is in fact the diameter.

Option D

Solved $π r = 220$ for $70$ and then doubled it, confusing radius with diameter in the opposite direction.

Skills used

Distance covered by a rolling wheel Area and circumference of a circle

77Mensuration • Solids • Cone

ModerateIdeal time: about 1.5 min

The volumes of two cones are in the ratio $1 : 4$ and their diameters are in the ratio $4 : 5$ . What is the ratio of their heights ?

25 : 64

16 : 25

9 : 16

5 : 9

Show answer & solution

Option A

25 : 64

Approach

Write both volumes with the cone formula, divide one by the other, and rearrange so the height ratio is left in terms of the two given ratios.

Volume of a right circular cone

Step-by-step solution

1
$d_{1} : d_{2} = 4 : 5 \Rightarrow r_{1} : r_{2} = 4 : 5$
Why: A radius is half its diameter, and halving both parts of a ratio leaves the ratio unchanged.
Introduce a common factor $k$ for a ratio
2
$\frac{V _{1}}{V _{2}} = \frac{\frac{1}{3} π r _{1}^{2} h _{1}}{\frac{1}{3} π r _{2}^{2} h _{2}} = \frac{r _{1}^{2} h _{1}}{r _{2}^{2} h _{2}}$
Why: The volume of a cone is $\frac{1}{3} π r^{2} h$ , and the constant $\frac{1}{3} π$ cancels in the quotient.
Volume of a right circular cone
3
$\frac{h _{1}}{h _{2}} = \frac{V _{1}}{V _{2}} \times (\frac{r _{2}}{r _{1}})^{2} = \frac{1}{4} \times \frac{25}{16}$
Why: Rearrange for the one ratio asked for; the radius ratio enters inverted and squared because $r^{2}$ sits in the denominator of $\frac{h _{1}}{h _{2}}$ .
4
$\frac{h _{1}}{h _{2}} = \frac{25}{64}$ , i.e. $25 : 64$
Why: Multiply the two fractions: $1 \times 25 = 25$ and $4 \times 16 = 64$ .

Shortcut

$h \propto \frac{V}{r ^{2}}$ , so the height ratio is the volume ratio divided by the square of the radius ratio: $\frac{1}{4} \div \frac{16}{25}$ .

Common mistakes

Option B

Took the heights to be in the square of the diameter ratio, $(4 : 5)^{2}$ , ignoring the volume ratio altogether.

Option C

No route reaches $9 : 16$ from a volume ratio $1 : 4$ and a radius ratio $4 : 5$ .

Option D

No route reaches $5 : 9$ from the two given ratios.

Skills used

Volume of a right circular cone Introduce a common factor

k

for a ratio

78Geometry • Triangle • Angles of a Triangle

EasyIdeal time: about 55 s

In a triangle ABC, if $2∠ A = 3∠ B = 6∠ C$ , then what is $∠ A + ∠ C$ equal to ?

9 0^{\circ}

12 0^{\circ}

13 5^{\circ}

15 0^{\circ}

Show answer & solution

Option B

12 0^{\circ}

Approach

Name the common value of the three equal products, turn it into the proportion of the angles, then use the angle sum.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
$2∠ A = 3∠ B = 6∠ C = t$
Why: Naming the shared value $t$ lets each of the three angles be written separately.
2
$∠ A = \frac{t}{2}, ∠ B = \frac{t}{3}, ∠ C = \frac{t}{6}$ , so $∠ A : ∠ B : ∠ C = 3 : 2 : 1$
Why: Multiplying all three by $6$ clears the denominators and leaves the proportion; note it is the reciprocals of $2, 3, 6$ , not $2 : 3 : 6$ .
Introduce a common factor $k$ for a ratio
3
$3 k + 2 k + k = 18 0^{\circ} \Rightarrow 6 k = 18 0^{\circ} \Rightarrow k = 3 0^{\circ}$
Why: The three interior angles of a triangle add to $18 0^{\circ}$ .
Angle sum of a triangle
4
$∠ A = 9 0^{\circ}, ∠ B = 6 0^{\circ}, ∠ C = 3 0^{\circ}$ , so $∠ A + ∠ C = 12 0^{\circ}$
Why: The question asks for the sum $∠ A + ∠ C$ , not for either angle on its own.

Shortcut

$∠ A + ∠ C = 18 0^{\circ} - ∠ B$ , so only $∠ B = 6 0^{\circ}$ is needed: $18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ .

Common mistakes

Option A

Stopped at $∠ A = 9 0^{\circ}$ instead of adding $∠ C$ .

Option C

No route reaches $13 5^{\circ}$ : no pair of these angles sums to it, and reading the proportion as $2 : 3 : 6$ gives $∠ A + ∠ C \approx 13 1^{\circ}$ , not $13 5^{\circ}$ .

Option D

Added the wrong pair, $∠ A + ∠ B = 9 0^{\circ} + 6 0^{\circ} = 15 0^{\circ}$ .

Skills used

Introduce a common factor

k

for a ratio Angle sum of a triangle

79Mensuration • Plane Figures

EasyIdeal time: about 1 min

If the perimeter of a circle and a square are equal, then what is the ratio of the area of the circle to that of the square ?

1 : π

2 : π

3 : π

4 : π

Show answer & solution

Option D

4 : π

Approach

Call the common perimeter $P$ , express each area in terms of $P$ alone, and divide.

Area and circumference of a circle

Step-by-step solution

1
$2 π r = P \Rightarrow r = \frac{P}{2 π}$ , so the circle's area is $π r^{2} = π \cdot \frac{P ^{2}}{4 π ^{2}} = \frac{P ^{2}}{4 π}$
Why: Eliminating $r$ between circumference and area is exactly what a fixed-perimeter comparison needs; nothing else about the circle is given.
Area and circumference of a circle
2
$4 a = P \Rightarrow a = \frac{P}{4}$ , so the square's area is $a^{2} = \frac{P ^{2}}{16}$
Why: The same elimination for the square: perimeter $4 a$ , area $a^{2}$ .
Perimeter and area of a square
3
$\frac{P ^{2}}{4 π} : \frac{P ^{2}}{16} = \frac{1}{4 π} : \frac{1}{16}$
Why: $P^{2}$ is common to both parts of the ratio and cancels, which is why the actual perimeter never has to be known.
4
$\frac{1}{4 π} : \frac{1}{16} = 16 : 4 π = 4 : π$
Why: Multiply both parts by $16 \times 4 π$ to clear the fractions, then divide both by $4$ .

Shortcut

Use area $= \frac{P ^{2}}{4 π}$ for the circle and $\frac{P ^{2}}{16}$ for the square directly; the comparison is then one division.

Common mistakes

Option A

No route reaches $1 : π$ ; it would need the circle to have the smaller area, which cannot happen at equal perimeter.

Option B

No route reaches $2 : π$ , which is also below $1$ and so has the inequality the wrong way round.

Option C

No route reaches $3 : π$ ; it is below $1$ as well.

Skills used

Area and circumference of a circle Perimeter and area of a square

80Geometry • Triangle • Pythagoras Theorem

ModerateIdeal time: about 1.5 min

The lengths of the sides of a right-angled triangle are consecutive even integers (in cm). What is the product of these integers ?

60

120

360

480

Show answer & solution

Option D

480

Approach

Write the three consecutive even sides algebraically, apply Pythagoras to get a quadratic, solve it, then multiply the three sides.

Pythagoras' theorem

Step-by-step solution

1
Sides $2 n, 2 n + 2, 2 n + 4$ , with $2 n + 4$ the hypotenuse
Why: Consecutive even integers differ by $2$ , and in a right triangle the longest side faces the right angle.
Write an odd number as $2 m + 1$
2
$(2 n)^{2} + (2 n + 2)^{2} = (2 n + 4)^{2}$
Why: Pythagoras' theorem: the squares on the two legs add to the square on the hypotenuse.
Pythagoras' theorem
3
$4 n^{2} + (4 n^{2} + 8 n + 4) = 4 n^{2} + 16 n + 16$
Why: Expand both brackets with $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ .
Square of a sum and of a difference
4
$4 n^{2} - 8 n - 12 = 0 \Rightarrow n^{2} - 2 n - 3 = 0 \Rightarrow (n - 3) (n + 1) = 0$
Why: Collect everything on one side, divide through by $4$ , and split the middle term using $- 3 \times 1 = - 3$ with $- 3 + 1 = - 2$ .
Factorise a quadratic by splitting the middle term
5
$n = 3$ or $n = - 1$ ; take $n = 3$ , giving sides $6, 8, 10$
Why: A product is zero only if a factor is, and a side length must be positive, so $n = - 1$ is rejected.
A product is zero only when a factor is zero
6
$6 \times 8 \times 10 = 480$
Why: The question asks for the product of the three integers, not for the area or the perimeter.

Shortcut

$6$ - $8$ - $10$ is the only right triangle with consecutive even sides - it is the $3$ - $4$ - $5$ triple doubled - so the product is $480$ on sight.

Common mistakes

Option A

Used consecutive integers $3, 4, 5$ instead of consecutive even integers, giving the product $60$ .

Option B

No route reaches $120$ ; it is neither $6 \times 8 \times 10$ nor the product of any right triangle with consecutive even sides.

Option C

No route reaches $360$ from the sides $6, 8, 10$ .

Skills used

Pythagoras' theorem Write an odd number as

2 m + 1

Square of a sum and of a difference Factorise a quadratic by splitting the middle term A product is zero only when a factor is zero