Elementary Mathematics 2016 (II)

Set A

Questions 41–60 of 100

41Mensuration • Solids • Cuboid

EasyIdeal time: about 50 s

The area of four walls of a room is $120$ m $^{2}$ . The length of the room is twice its breadth. If the height of the room is $4$ m, what is area of the floor?

40

^{2}

50

^{2}

60

^{2}

80

^{2}

Show answer & solution

Option B

50

^{2}

Approach

Turn the four-wall area into one equation in the breadth, solve it, then multiply length by breadth.

Volume and wall area of a cuboid

Step-by-step solution

1
$2 (l + b) h = 120$
Why: The four walls of a cuboidal room are its lateral surface, whose area is $2 (l + b) h$ ; the floor and ceiling are excluded.
Volume and wall area of a cuboid
2
$l = 2 b, h = 4 \Rightarrow 2 (2 b + b) (4) = 120$
Why: Substituting the two given relations leaves a single unknown $b$ .
Turn a worded condition into a linear equation in one unknown
3
$24 b = 120 \Rightarrow b = 5, l = 2 b = 10$
Why: Solve the linear equation, then read the length back off $l = 2 b$ .
4
$floor area = l \times b = 10 \times 5 = 50$ m $^{2}$
Why: The floor is the $l \times b$ rectangle; the height has done its work and plays no further part.

Shortcut

The wall area gives $l + b = \frac{120}{2 \times 4} = 15$ in one line, and $l = 2 b$ splits $15$ into $10$ and $5$ .

Common mistakes

Option A

Multiplied length by height, $10 \times 4$ , instead of length by breadth.

Option C

Halved the wall area, treating the floor as one opposite pair of walls.

Option D

No plausible route reaches $80$ m $^{2}$ .

Skills used

Volume and wall area of a cuboid Turn a worded condition into a linear equation in one unknown

42Geometry • Circle • Inscribed Figures

ModerateIdeal time: about 1.5 min

If a square of side $x$ and an equilateral triangle of side $y$ are inscribed in a circle, then what is the ratio of $x$ to $y$ ?

\frac{2}{3}

\frac{3}{2}

\frac{3}{2}

\frac{2}{3}

Show answer & solution

Option A

\frac{2}{3}

Approach

Express both sides in terms of the shared circumradius $R$ , then divide so that $R$ cancels.

Side of a regular polygon inscribed in a circle

Step-by-step solution

1
$x 2 = 2 R \Rightarrow x = R 2$
Why: The diagonal of an inscribed square passes through the centre, so it is a diameter; by Pythagoras that diagonal is $x 2$ .
Pythagoras' theorem Side of a regular polygon inscribed in a circle
2
$(\frac{y}{2})^{2} + (\frac{R}{2})^{2} = R^{2} \Rightarrow y = R 3$
Why: Each side of the inscribed equilateral triangle subtends $12 0^{\circ}$ at the centre, so the perpendicular from $O$ to that side has length $\frac{R}{2}$ and meets it at its midpoint.
3
$\frac{x}{y} = \frac{R 2}{R 3} = \frac{2}{3}$
Why: $R$ cancels, so the ratio is the same whatever circle is drawn.

Shortcut

Recall the inscribed-side table: square $= R 2$ , equilateral triangle $= R 3$ , hexagon $= R$ . The answer is $2 : 3$ on sight.

Common mistakes

Option B

Formed $\frac{y}{x}$ instead of $\frac{x}{y}$ .

Option C

No plausible route reaches $\frac{3}{2}$ .

Option D

Reached $\frac{2}{3}$ and then struck the radical off the denominator, leaving $\frac{2}{3}$ .

Skills used

Side of a regular polygon inscribed in a circle Pythagoras' theorem

43Mensuration • Solids • Cylinder

ModerateIdeal time: about 1.5 min

The ratio of the curved surface area to the total surface area of a right circular cylinder is $1 : 2$ . If the total surface area is $616$ cm $^{2}$ , what is the volume of the cylinder?

539

^{3}

616

^{3}

1078

^{3}

1232

^{3}

Show answer & solution

Option C

1078

^{3}

Approach

Read $1 : 2$ as 'the curved surface is half the total', which forces $h = r$ , then get $r$ from the total surface area.

Curved and total surface of a cylinder

Step-by-step solution

1
$CSA = 2 π r h, TSA = 2 π r h + 2 π r^{2}$
Why: A solid cylinder's total surface is its curved surface plus the two circular ends.
Curved and total surface of a cylinder
2
$CSA = k, TSA = 2 k \Rightarrow 2 π r h = 2 π r^{2} \Rightarrow h = r$
Why: A part-to-whole ratio of $1 : 2$ means the curved surface and the two ends together are equal halves of the total.
Introduce a common factor $k$ for a ratio
3
$2 π r^{2} = \frac{616}{2} = 308 \Rightarrow r^{2} = \frac{308 \times 7}{2 \times 22} = 49 \Rightarrow r = 7$ cm
Why: The two ends account for half of $616$ ; solve for $r$ with $π = \frac{22}{7}$ .
4
$V = π r^{2} h = \frac{22}{7} \times 49 \times 7 = 1078$ cm $^{3}$
Why: With $h = r = 7$ the volume follows directly.
Volume of a right circular cylinder

Shortcut

$CSA : TSA = 1 : 2$ is just $h = r$ ; from there $TSA = 4 π r^{2} = 616$ gives $r = 7$ at once.

Common mistakes

Option A

Halved the volume, $\frac{1078}{2}$ , perhaps carrying the $1 : 2$ across into the final answer.

Option B

Copied the given total surface area $616$ as though it were the volume.

Option D

No plausible route reaches $1232$ cm $^{3}$ .

Skills used

Curved and total surface of a cylinder Volume of a right circular cylinder Introduce a common factor

k

for a ratio

44Mensuration • Recasting and Melting

ModerateIdeal time: about 1.5 min

A cubic metre of copper weighing $9000$ kg is rolled into a square bar $9$ m long. An exact cube is cut off from the bar. How much does the cube weigh?

1000

kgB

\frac{1000}{3}

kgC

300

kgD

\frac{500}{3}

Show answer & solution

Option B

\frac{1000}{3}

Approach

Rolling conserves volume, so the bar still holds $1$ m $^{3}$ ; that fixes its square cross-section and hence the cube's edge.

Recasting preserves volume

Step-by-step solution

1
$volume of bar = 1$ m $^{3}$ , length $= 9$ m
Why: Rolling reshapes the copper without adding or losing any of it.
Recasting preserves volume
2
$cross-section = \frac{1}{9}$ m $^{2} \Rightarrow edge = \frac{1}{9} = \frac{1}{3}$ m
Why: Volume $=$ cross-section $\times$ length, and the cross-section is a square, so its side is the square root of its area.
Volume and wall area of a cuboid
3
$cube volume = (\frac{1}{3})^{3} = \frac{1}{27}$ m $^{3}$
Why: An exact cube cut from the bar can only have edge equal to the bar's thickness, $\frac{1}{3}$ m.
4
$weight = 9000 \times \frac{1}{27} = \frac{1000}{3}$ kg
Why: One cubic metre weighs $9000$ kg, so scale that rate down to $\frac{1}{27}$ m $^{3}$ .
Unitary method: find one unit, then scale

Shortcut

The bar is $9$ m long and holds $1$ m $^{3}$ , so it is $\frac{1}{3}$ m square; $27$ such cubes make the original metre, so each weighs $\frac{9000}{27}$ .

Common mistakes

Option A

Took the cube's volume as $\frac{1}{9}$ m $^{3}$ , reusing the cross-sectional area as a volume.

Option C

Rounded $\frac{1000}{3} = 333. 3$ down to $300$ .

Option D

No plausible route reaches $\frac{500}{3}$ kg.

Skills used

Recasting preserves volume Volume and wall area of a cuboid Unitary method: find one unit, then scale

45Mensuration • Solids • Cone

ModerateIdeal time: about 1.5 min

Into a conical tent of radius $8.4$ m and vertical height $3.5$ m, how many full bags of wheat can be emptied, if space required for the wheat in each bag is $1.96$ m $^{3}$ ?

264

201

132

105

Show answer & solution

Option C

132

Approach

Compute the cone's volume, then divide by the space one bag needs.

Volume of a right circular cone

Step-by-step solution

1
$V = \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times (8.4)^{2} \times 3.5$
Why: The tent is a right circular cone of radius $8.4$ m and vertical height $3.5$ m.
Volume of a right circular cone
2
$\frac{22}{7} \times 70.56 = 221.76$ , so $V = \frac{221.76 \times 3.5}{3} = 258.72$ m $^{3}$
Why: Cancel the $7$ against $8. 4^{2} = 70.56$ before multiplying, to keep the decimals short.
3
$\frac{258.72}{1.96} = 132$
Why: Each bag needs $1.96$ m $^{3}$ , so the number of bags is the volume divided by that rate; it comes out exactly whole, so $132$ full bags fit.
Unitary method: find one unit, then scale

Shortcut

$\frac{1}{3} π r^{2} h \div 1.96$ collapses to $\frac{22 \times 8.4 \times 8.4 \times 3.5}{7 \times 3 \times 1.96} = 132$ with everything cancelled before any multiplication.

Common mistakes

Option A

Used $\frac{2}{3} π r^{2} h$ , doubling the cone's volume.

Option B

No plausible route reaches $201$ .

Option D

No plausible route reaches $105$ .

Skills used

Volume of a right circular cone Unitary method: find one unit, then scale

46Trigonometry • Triangle

HardIdeal time: about 2.5 min

Let $A B C$ be a right angled triangle with $B C = 5$ cm and $A C = 12$ cm. Let $D$ be a point on the hypotenuse $A B$ such that $∠ B C D = 3 0^{\circ}$ . What is length of $C D$ ?

\frac{60}{13}

cmB

\frac{17}{2}

cmC

\frac{120}{5 + 12 2}

cmD

\frac{120}{5 + 12 3}

Show answer & solution

Option D

\frac{120}{5 + 12 3}

Approach

Split $△ A B C$ along $C D$ into two triangles and equate the sum of their areas to the whole.

Area of a triangle from two sides and the included angle

Step-by-step solution

1
$∠ A C B = 9 0^{\circ}$ , $∠ B C D = 3 0^{\circ} \Rightarrow ∠ A C D = 6 0^{\circ}$
Why: $A B$ is named as the hypotenuse, so the right angle sits at $C$ ; $C D$ therefore cuts that right angle into $3 0^{\circ}$ and $6 0^{\circ}$ .
2
$[A B C] = [A C D] + [B C D]$
Why: $D$ lies on $A B$ , so $C D$ divides the triangle into two pieces whose areas add to the whole.
3
$\frac{1}{2} (12) (5) = \frac{1}{2} (12) (C D) sin 6 0^{\circ} + \frac{1}{2} (5) (C D) sin 3 0^{\circ}$
Why: Each piece has two known sides enclosing a known angle at $C$ , so use $area = \frac{1}{2} ab sin C$ . $C D$ is the one unknown and appears linearly in both pieces.
Area of a triangle from two sides and the included angle Exact trigonometric ratios of the standard angles
4
$60 = C D (12 \cdot \frac{3}{2} + 5 \cdot \frac{1}{2}) = C D \cdot \frac{12 3 + 5}{2}$
Why: Multiply through by $2$ and substitute the exact values $sin 6 0^{\circ} = \frac{3}{2}$ , $sin 3 0^{\circ} = \frac{1}{2}$ .
5
$C D = \frac{120}{5 + 12 3}$ cm
Why: Divide; the printed option is left unrationalised in the same form.

Shortcut

For a cevian from the right angle splitting it into $α$ and $9 0^{\circ} - α$ , the area split gives $C D = \frac{ab}{a c o s α + b s i n α}$ directly, with $a = 12$ , $b = 5$ , $α = 6 0^{\circ}$ .

Common mistakes

Option A

Gave $\frac{60}{13}$ , the altitude from $C$ to $A B$ — the answer only if $C D$ were perpendicular to $A B$ , which $∠ B C D = 3 0^{\circ}$ forbids.

Option B

No plausible route reaches $\frac{17}{2}$ cm.

Option C

Used $sin 6 0^{\circ} = \frac{2}{2}$ , the value belonging to $4 5^{\circ}$ , turning $123$ into $122$ .

Skills used

Area of a triangle from two sides and the included angle Exact trigonometric ratios of the standard angles

47Mensuration • Plane Figures • Circle

ModerateIdeal time: about 1.5 min

Two circles touch externally and sum of their areas is $130 π$ cm $^{2}$ and the distance between their centres is $14$ cm. What is the difference in the radii of the circles?

5

cmB

6

cmC

7

cmD

8

Show answer & solution

Option D

8

Approach

Touching externally gives the sum of the radii, the area gives the sum of their squares, and one identity turns those into the difference.

Square of a sum and of a difference

Step-by-step solution

1
$r_{1} + r_{2} = 14$
Why: Two circles that touch externally have their centres exactly $r_{1} + r_{2}$ apart.
Distance between the centres of touching circles
2
$π r_{1}^{2} + π r_{2}^{2} = 130 π \Rightarrow r_{1}^{2} + r_{2}^{2} = 130$
Why: Divide the given area sum by $π$ ; no numerical value of $π$ is needed.
Area and circumference of a circle
3
$(r_{1} - r_{2})^{2} = 2 (r_{1}^{2} + r_{2}^{2}) - (r_{1} + r_{2})^{2} = 260 - 196 = 64$
Why: Expanding both squares shows $(a + b)^{2} + (a - b)^{2} = 2 (a^{2} + b^{2})$ , so the difference follows without ever finding the radii separately.
Square of a sum and of a difference
4
$r_{1} - r_{2} = 8$ cm
Why: A difference of radii is asked for as a length, so the non-negative root is the one wanted.
denotes the non-negative root

Shortcut

$(r_{1} - r_{2})^{2} = 2 \times 130 - 1 4^{2} = 64$ is a single line from the two givens.

Common mistakes

Option A

No plausible route reaches $5$ cm.

Option B

No plausible route reaches $6$ cm.

Option C

Assumed the two circles are equal and reported the common radius $\frac{14}{2} = 7$ .

Skills used

Square of a sum and of a difference Area and circumference of a circle Distance between the centres of touching circles

denotes the non-negative root

48Geometry • Circle • Chords

ModerateIdeal time: about 1.5 min

In a circle of radius $3$ units, a diameter $A B$ intersects a chord of length $2$ units perpendicularly at $P$ . If $A P > B P$ , then what is the ratio of $A P$ to $B P$ ?

3 + 10 : 3 - 10

3 + 8 : 3 - 8

3 + 3 : 3 - 3

3 : 3

Show answer & solution

Option B

3 + 8 : 3 - 8

Approach

The diameter bisects the perpendicular chord, so Pythagoras gives the distance $O P$ , and $A P$ , $B P$ are $r \pm O P$ .

The perpendicular from the centre bisects a chord

Step-by-step solution

1
$M P = \frac{2}{2} = 1$ , where $M$ is an end of the chord
Why: A diameter perpendicular to a chord bisects it, so each half of the chord is $1$ unit.
The perpendicular from the centre bisects a chord
2
$O P^{2} + 1^{2} = 3^{2} \Rightarrow O P = 8$
Why: $O$ , $P$ and the chord's endpoint form a right triangle whose hypotenuse is the radius $3$ .
Pythagoras' theorem
3
$A P = 3 + 8, B P = 3 - 8$
Why: $P$ sits on the diameter $8$ from the centre; $A P > B P$ places $A$ on the far side, so $A P$ takes the plus sign.
4
$A P : B P = 3 + 8 : 3 - 8$
Why: Both are already in the printed form, with $8$ left unsimplified.

Shortcut

Intersecting chords give $A P \cdot B P = 1^{2} = 1$ and $A P + B P = 6$ , so $A P$ and $B P$ are the roots of $t^{2} - 6 t + 1 = 0$ , namely $3 \pm 22$ .

Common mistakes

Option A

Added inside Pythagoras, taking $O P = 9 + 1 = 10$ instead of $9 - 1$ .

Option C

No plausible route reaches $3$ .

Option D

No plausible route reaches $3 : 3$ .

Skills used

The perpendicular from the centre bisects a chord Pythagoras' theorem

49Mensuration • Plane Figures • Circle

EasyIdeal time: about 50 s

What is the number of rounds that a wheel of diameter $\frac{5}{11}$ m will make in traversing $7$ km?

3300

3500

4400

4900

Show answer & solution

Option D

4900

Approach

Find the distance covered in one round, put the journey in the same unit, and divide.

Distance covered by a rolling wheel

Step-by-step solution

1
$C = π d = \frac{22}{7} \times \frac{5}{11} = \frac{10}{7}$ m
Why: One round of a wheel covers its circumference; $\frac{22}{11} = 2$ cancels immediately.
Area and circumference of a circle Distance covered by a rolling wheel
2
$7 km = 7000$ m
Why: The circumference is in metres, so the journey must be too.
3
$\frac{7000}{\frac{10}{7}} = 7000 \times \frac{7}{10} = 4900$
Why: One round covers $\frac{10}{7}$ m, so the number of rounds is the total distance divided by that per-round distance.
Unitary method: find one unit, then scale

Shortcut

$\frac{22}{7} \times \frac{5}{11} = \frac{10}{7}$ , so rounds $= 7000 \times \frac{7}{10} = 4900$ with no long division.

Common mistakes

Option A

No plausible route reaches $3300$ .

Option B

No plausible route reaches $3500$ .

Option C

No plausible route reaches $4400$ .

Skills used

Distance covered by a rolling wheel Area and circumference of a circle Unitary method: find one unit, then scale

50Mensuration • Combination of Solids

ModerateIdeal time: about 2 min

A building is in the form of a cylinder surmounted by a hemispherical dome on the diameter of the cylinder. The height of the building is three times the radius of the base of the cylinder. The building contains $67 \frac{1}{21}$ m $^{3}$ of air. What is the height of the building?

6

4

3

2

Show answer & solution

Option A

6

Approach

Write every length in terms of the base radius, add the cylinder and hemisphere volumes, and solve for the radius.

Volume and surface of a composite solid

Step-by-step solution

1
$radius = k, total height = 3 k \Rightarrow cylinder height = 3 k - k = 2 k$
Why: The dome sits on the cylinder's diameter, so the hemisphere has the same radius $k$ and adds exactly $k$ to the height; height $:$ radius $= 3 : 1$ fixes both up to the common factor $k$ .
Introduce a common factor $k$ for a ratio
2
$V = π k^{2} (2 k) + \frac{2}{3} π k^{3} = \frac{8}{3} π k^{3}$
Why: Cylinder volume plus hemisphere volume; the air inside the building is the sum of the two.
Volume and surface of a composite solid Volume of a right circular cylinder Volume and surface area of a sphere
3
$67 \frac{1}{21} = \frac{1408}{21}$ , so $\frac{8}{3} \times \frac{22}{7} k^{3} = \frac{176}{21} k^{3} = \frac{1408}{21}$
Why: Convert the mixed number first — the denominator $21 = 3 \times 7$ is a strong hint that $π = \frac{22}{7}$ is intended.
4
$k^{3} = \frac{1408}{176} = 8 \Rightarrow k = 2$
Why: The $21$ s cancel, leaving an exact cube.
5
$height = 3 k = 6$ m
Why: The question asks for the height of the building, which is $3$ times the radius.

Shortcut

$\frac{8}{3} π k^{3} = \frac{1408}{21}$ reduces to $k^{3} = 8$ ; spotting $k = 2$ then gives the height as $6$ without further work.

Common mistakes

Option B

Reported the cylinder's height $2 k = 4$ m rather than the building's.

Option C

No plausible route reaches $3$ m.

Option D

Stopped at the radius, $k = 2$ m.

Skills used

Volume and surface of a composite solid Volume of a right circular cylinder Volume and surface area of a sphere Introduce a common factor

k

for a ratio

51Mensuration • Solids • Cylinder

ModerateIdeal time: about 2 min

The radius of the base and the height of a solid right circular cylinder are in the ratio $2 : 3$ and its volume is $1617$ cm $^{3}$ . What is the total surface area of the cylinder?

462

^{2}

616

^{2}

770

^{2}

786

^{2}

Show answer & solution

Option C

770

^{2}

Approach

Introduce $k$ for the $2 : 3$ ratio, pin $k$ down with the volume, then apply the total-surface formula.

Curved and total surface of a cylinder

Step-by-step solution

1
$r = 2 k, h = 3 k$
Why: A ratio fixes the radius and height only up to a common factor.
Introduce a common factor $k$ for a ratio
2
$π r^{2} h = \frac{22}{7} (4 k^{2}) (3 k) = \frac{264}{7} k^{3} = 1617$
Why: Substitute into the cylinder volume; everything is now in $k$ .
Volume of a right circular cylinder
3
$k^{3} = \frac{1617 \times 7}{264} = \frac{343}{8} \Rightarrow k = \frac{7}{2}$
Why: $343 = 7^{3}$ and $8 = 2^{3}$ , so the cube root is exact and no decimals are needed.
4
$r = 2 k = 7$ cm, $h = 3 k = \frac{21}{2}$ cm
Why: Read both dimensions back from $k$ .
5
$TSA = 2 π r (r + h) = 2 \times \frac{22}{7} \times 7 \times \frac{35}{2} = 770$ cm $^{2}$
Why: Total surface is the curved surface plus both circular ends, which factorises as $2 π r (r + h)$ .
Curved and total surface of a cylinder

Shortcut

$1617 = 3 \times 7^{2} \times 11$ ; matching it against $\frac{264}{7} k^{3}$ gives $k^{3} = \frac{343}{8}$ by inspection, so $r = 7$ and $h = 10.5$ .

Common mistakes

Option A

Stopped at the curved surface $2 π r h = 462$ cm $^{2}$ .

Option B

Added only one circular end: $2 π r h + π r^{2} = 462 + 154 = 616$ cm $^{2}$ .

Option D

No plausible route reaches $786$ cm $^{2}$ .

Skills used

Curved and total surface of a cylinder Volume of a right circular cylinder Introduce a common factor

k

for a ratio

52Mensuration • Solids • Sphere

EasyIdeal time: about 55 s

The cost of painting a spherical vessel of diameter $14$ cm is Rs. $8008$ . What is the cost of painting per square centimetre?

ARs.

8

BRs.

9

CRs.

13

DRs.

14

Show answer & solution

Option C

Rs.

13

Approach

Find the sphere's surface area, then share the total cost over it.

Volume and surface area of a sphere

Step-by-step solution

1
$r = \frac{14}{2} = 7$ cm
Why: The diameter is given, but the surface formula takes the radius.
2
$S = 4 π r^{2} = 4 \times \frac{22}{7} \times 49 = 616$ cm $^{2}$
Why: The whole outer surface of a sphere is $4 π r^{2}$ ; the $7$ cancels cleanly.
Volume and surface area of a sphere
3
$\frac{8008}{616} = 13$
Why: Cost per square centimetre is the total cost divided by the area painted.
Unitary method: find one unit, then scale

Shortcut

$4 π r^{2}$ with $r = 7$ is $616$ , a value worth knowing; $8008 = 616 \times 13$ then falls out because $8008 = 8 \times 7 \times 11 \times 13$ .

Common mistakes

Option A

No plausible route reaches Rs. $8$ .

Option B

No plausible route reaches Rs. $9$ .

Option D

No plausible route reaches Rs. $14$ .

Skills used

Volume and surface area of a sphere Unitary method: find one unit, then scale

53Geometry • Quadrilateral • Tangential

ModerateIdeal time: about 1 min

If a quadrilateral has an inscribed circle, then the sum of a pair of opposite sides equals

AHalf the sum of the diagonalsBSum of the other pair of opposite sidesCSum of two adjacent sidesDNone of the above

Show answer & solution

Option B

Sum of the other pair of opposite sides

Approach

Test the four claims on concrete tangential quadrilaterals: a square first, which is not discriminating enough, then a deliberately lopsided one built from tangent lengths.

Pitot's theorem for a tangential quadrilateral

Step-by-step solution

1
$Square of side 4 : A B + C D = 8, B C + A D = 8, \frac{1}{2} (A C + B D) = 42 \approx 5.66$
Why: A square has an inscribed circle, so it is a legal test case; it already kills option A, since $5.66 \neq = 8$ .
2
$Square: A B + B C = 8 too, so options B and C both survive this case$
Why: A symmetric example cannot separate "opposite pair" from "adjacent pair" — the test has to be tightened with an asymmetric one.
3
$Tangent lengths p = 1, q = 2, r = 3, s = 4 : sides 3, 5, 7, 5$
Why: Any four positive tangent lengths build a quadrilateral circumscribing a circle, because the two tangents from each vertex are equal.
The two tangents from an external point
4
$A B + C D = 3 + 7 = 10, B C + A D = 5 + 5 = 10, A B + B C = 3 + 5 = 8$
Why: Here the opposite pairs still agree while the adjacent pair does not, so C fails and B survives.
5
$(p + q) + (r + s) = (q + r) + (s + p) = p + q + r + s$
Why: In general both opposite-side sums equal the total of the four tangent lengths, which is why B holds for every tangential quadrilateral and not just these two.
Pitot's theorem for a tangential quadrilateral

Shortcut

Pitot: in any quadrilateral with an inscribed circle, $A B + C D = B C + A D$ .

Common mistakes

Option A

Confused the tangential-quadrilateral property with a diagonal relation of Ptolemy type.

Option C

Generalised from a square, where adjacent sides happen to add to the same total.

Option D

Chosen when the Pitot property is not recognised at all.

Skills used

Pitot's theorem for a tangential quadrilateral The two tangents from an external point

54Mensuration • Plane Figures • Circle

EasyIdeal time: about 55 s

A circle and a square have the same perimeter. Which one of the following is correct?

ATheir areas are equalBThe area of the circle is largerCThe area of the square is

\frac{π}{2}

times area of circleDThe area of the square is

π

times area of circle

Show answer & solution

Option B

The area of the circle is larger

Approach

Write both areas in terms of the shared perimeter $P$ and compare the two coefficients.

Area and circumference of a circle

Step-by-step solution

1
$2 π r = P \Rightarrow r = \frac{P}{2 π}$ , so $area of circle = π r^{2} = \frac{P ^{2}}{4 π}$
Why: Fix the common perimeter as $P$ and express the circle's radius, hence its area, through it.
Area and circumference of a circle
2
$4 a = P \Rightarrow a = \frac{P}{4}$ , so $area of square = \frac{P ^{2}}{16}$
Why: Do the same for the square, whose perimeter is four times its side.
Perimeter and area of a square
3
$\frac{square}{circle} = \frac{P ^{2} /16}{P ^{2} / ( 4 π )} = \frac{4 π}{16} = \frac{π}{4} \approx 0.785$
Why: $P^{2}$ cancels, so the comparison does not depend on the size of either figure.
4
$\frac{π}{4} < 1$ , so the circle has the larger area.
Why: Dividing $P^{2}$ by the smaller number $4 π \approx 12.57$ rather than by $16$ gives the bigger area.

Shortcut

Of all plane figures with a given perimeter the circle encloses the greatest area; the square is only $\frac{π}{4} \approx 79%$ of it.

Common mistakes

Option A

Assumed equal perimeters must force equal areas.

Option C

Reached the factor as $\frac{π}{2}$ instead of $\frac{π}{4}$ , halving rather than quartering the perimeter somewhere in the algebra.

Option D

Read the factor straight off the $π$ in $π r^{2}$ without carrying the $\frac{1}{4}$ .

Skills used

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

55Mensuration • Solids • Frustum

ModerateIdeal time: about 1.5 min

A drinking glass of height $24$ cm is in the shape of frustum of a cone and diameters of its bottom and top circular ends are $4$ cm and $18$ cm respectively. If we take capacity of the glass as $π x$ cm $^{3}$ , then what is the value of $x$ ?

824

1236

1628

2472

Show answer & solution

Option A

824

Approach

Halve both diameters and substitute into the frustum volume formula.

Volume of a frustum of a cone

Step-by-step solution

1
$R = \frac{18}{2} = 9$ cm, $r = \frac{4}{2} = 2$ cm, $h = 24$ cm
Why: The formula takes radii; the question gives diameters.
2
$V = \frac{π h}{3} (R^{2} + R r + r^{2}) = \frac{π ( 24 )}{3} (81 + 18 + 4)$
Why: A frustum's volume is the cone formula with $r^{2}$ replaced by $R^{2} + R r + r^{2}$ .
Volume of a frustum of a cone
3
$V = 8 π \times 103 = 824 π$ , so $x = 824$
Why: The capacity is written as $π x$ , so $x$ is everything except the $π$ .

Shortcut

$\frac{h}{3} = 8$ and $R^{2} + R r + r^{2} = 103$ , so $x = 8 \times 103 = 824$ in two multiplications.

Common mistakes

Option B

Divided by $2$ instead of $3$ , giving $12 \times 103$ .

Option C

No plausible route reaches $1628$ .

Option D

Dropped the $\frac{1}{3}$ , using $π h (R^{2} + R r + r^{2}) = 24 \times 103$ .

Skills used

Volume of a frustum of a cone

56Geometry • Triangle • Similarity

ModerateIdeal time: about 1 min

In an equilateral triangle another equilateral triangle is drawn inside joining the mid-points of the sides of given equilateral triangle and the process is continued up to $7$ times. What is the ratio of area of fourth triangle to that of seventh triangle?

256 : 1

128 : 1

64 : 1

16 : 1

Show answer & solution

Option C

64 : 1

Approach

Each midpoint triangle has half the side and so a quarter of the area; count the halvings between the fourth and the seventh.

Step-by-step solution

1
$side_{n + 1} = \frac{1}{2} side_{n}$
Why: The segment joining the midpoints of two sides of a triangle is half the third side, so the inner triangle is equilateral with half the side.
Midpoint theorem
2
$\frac{A _{n + 1}}{A _{n}} = (\frac{1}{2})^{2} = \frac{1}{4}$
Why: The triangles are similar, and similar figures have areas in the square of the side ratio.
Similar triangles have proportional sides
3
$\frac{A _{4}}{A _{7}} = 4^{7 - 4} = 4^{3} = 64$
Why: Three halvings separate the fourth triangle from the seventh; the answer depends only on that gap of three, not on which triangle is called the first.
4
$A_{4} : A_{7} = 64 : 1$
Why: State it as the ratio the question asks for.

Shortcut

Only the gap matters: each step divides the area by $4$ , and the gap is $3$ , so the ratio is $4^{3} : 1$ .

Common mistakes

Option A

Counted four halvings instead of three, giving $4^{4}$ .

Option B

No plausible route reaches $128 : 1$ ; it is $2^{7}$ , unrelated to the gap of three.

Option D

Counted two halvings instead of three, giving $4^{2}$ .

Skills used

57Mensuration • Solids • Cylinder

EasyIdeal time: about 1 min

Rain water from a roof $22$ m $\times$ $20$ m drains into a cylindrical vessel having diameter of base $2$ m and height $3.5$ m. If the vessel is just full, what is the rainfall?

3.5

cmB

3

cmC

2.5

cmD

2

Show answer & solution

Option C

2.5

Approach

The rain that fell on the roof is exactly the water now in the vessel, so divide that volume by the roof's area.

Volume of a right circular cylinder

Step-by-step solution

1
$roof area = 22 \times 20 = 440$ m $^{2}$
Why: Rainfall is a depth, measured over the area the rain fell on.
2
$r = 1$ m, $h = 3.5$ m $\Rightarrow V = π r^{2} h = \frac{22}{7} \times 1 \times 3.5 = 11$ m $^{3}$
Why: The vessel is just full, so it holds all the rain; halve the diameter of $2$ m to get the radius.
Volume of a right circular cylinder
3
$depth = \frac{11}{440} = 0.025$ m
Why: Spreading the same volume over the roof at a uniform depth gives volume $=$ area $\times$ depth.
Volume and wall area of a cuboid
4
$0.025 m = 2.5$ cm
Why: The options are in centimetres, and $1$ m $= 100$ cm.

Shortcut

$\frac{π r ^{2} h}{roof area} = \frac{11}{440} = \frac{1}{40}$ m, and $\frac{100}{40} = 2.5$ cm.

Common mistakes

Option A

Quoted the vessel's height $3.5$ cm as the rainfall.

Option B

No plausible route reaches $3$ cm.

Option D

No plausible route reaches $2$ cm.

Skills used

Volume of a right circular cylinder Volume and wall area of a cuboid

58Mensuration • Solids • Frustum

ModerateIdeal time: about 1 min

The height of a cone is $60$ cm. A small cone is cut off at the top by a plane parallel to the base and its volume is $\frac{1}{64}$ the volume of original cone. What is the height from the base at which the section is made?

15

cmB

20

cmC

30

cmD

45

Show answer & solution

Option D

45

Approach

The cut-off cone is similar to the whole, so the volume ratio is the cube of the height ratio; then measure from the base rather than the apex.

Step-by-step solution

1
$(\frac{h}{60})^{3} = \frac{1}{64}$
Why: A plane parallel to the base cuts off a cone similar to the original, and similar solids have volumes in the cube of their length ratio.
Similar triangles have proportional sides
2
$\frac{h}{60} = \frac{1}{4} \Rightarrow h = 15$ cm
Why: Take the cube root of $\frac{1}{64}$ .
3
$height above the base = 60 - 15 = 45$ cm
Why: $h$ is measured down from the apex; the question asks how high the section is above the base.

Shortcut

$\frac{1}{64} = (\frac{1}{4})^{3}$ , so the cut is a quarter of the way down from the apex, i.e. three quarters of the way up: $\frac{3}{4} \times 60 = 45$ cm.

Common mistakes

Option A

Gave the small cone's height $15$ cm, which is measured from the apex, not from the base.

Option B

No plausible route reaches $20$ cm.

Option C

No plausible route reaches $30$ cm.

Skills used

59Geometry • Circle • Tangents

EasyIdeal time: about 55 s

If two tangents inclined at an angle $6 0^{\circ}$ are drawn to a circle of radius $3$ cm, then what is the length of each tangent?

33

cmB

3

cmC

6

cmD

22

Show answer & solution

Option A

33

Approach

The line from the external point to the centre bisects the angle between the tangents; solve the resulting right triangle.

The two tangents from an external point

Step-by-step solution

1
$∠ A P B = 6 0^{\circ} \Rightarrow ∠ A P O = 3 0^{\circ}$
Why: The two tangents from $P$ are equal, so $O P$ is the axis of symmetry of the figure and bisects the angle between them.
The two tangents from an external point
2
$O A ⊥ P A$ with $O A = 3$ cm
Why: A radius drawn to the point of contact is perpendicular to the tangent there, so $△ O A P$ is right-angled at $A$ .
3
$tan 3 0^{\circ} = \frac{O A}{P A} \Rightarrow P A = \frac{3}{t a n 3 0 ^{\circ}} = 33$ cm
Why: The radius is opposite the $3 0^{\circ}$ angle and the tangent length is adjacent to it.
Exact trigonometric ratios of the standard angles

Shortcut

The tangent length from a point where the tangents meet at $2 θ$ is $r cot θ$ ; here $r cot 3 0^{\circ} = 33$ .

Common mistakes

Option B

Used the full $6 0^{\circ}$ rather than the half-angle: $\frac{3}{t a n 6 0 ^{\circ}} = 3$ .

Option C

Found $O P = 6$ cm, the distance from the point to the centre, instead of the tangent length.

Option D

No plausible route reaches $22$ cm.

Skills used

The two tangents from an external point Exact trigonometric ratios of the standard angles

60Mensuration • Solids • Sphere

Very easyIdeal time: about 25 s

What is the volume of a sphere of radius $3$ cm?

36 π

^{3}

18 π

^{3}

9 π

^{3}

6 π

^{3}

Show answer & solution

Option A

36 π

^{3}

Approach

Substitute $r = 3$ into the sphere volume formula.

Volume and surface area of a sphere

Step-by-step solution

1
$V = \frac{4}{3} π r^{3}$
Why: The volume of a sphere of radius $r$ ; the options are left in terms of $π$ , so no numerical value of $π$ is needed.
Volume and surface area of a sphere
2
$V = \frac{4}{3} π (3)^{3} = \frac{4}{3} π \times 27 = 36 π$ cm $^{3}$
Why: $\frac{27}{3} = 9$ and $4 \times 9 = 36$ .

Shortcut

For $r = 3$ the $\frac{1}{3}$ cancels the cube's $3$ , leaving $4 \times 9 π = 36 π$ in one line.

Common mistakes

Option B

Used the hemisphere formula $\frac{2}{3} π r^{3}$ .

Option C

Computed $π r^{2} = 9 π$ , the area of a great circle rather than a volume.

Option D

Computed $2 π r = 6 π$ , a circumference rather than a volume.

Skills used

Volume and surface area of a sphere