Elementary Mathematics 2020 (II)

Set A

Questions 81–100 of 100

81Geometry • Circle • Inscribed Figures

EasyIdeal time: about 1 min

A circle is inscribed in a triangle ABC. It touches the sides AB and AC at M and N respectively. If O is the centre of the circle and $∠ A = 7 0^{\circ}$ , then what is $∠ M O N$ equal to?

9 0^{\circ}

10 0^{\circ}

11 0^{\circ}

12 0^{\circ}

Show answer & solution

Option C

11 0^{\circ}

Approach

Use the two right angles at the points of contact and the angle sum of the quadrilateral $A M O N$ .

The two tangents from an external point

Step-by-step solution

1
$A B$ touches the circle at $M$ and $A C$ touches it at $N$ , so $O M ⊥ A B$ and $O N ⊥ A C$ , giving $∠ O M A = ∠ O N A = 9 0^{\circ}$
Why: A tangent is perpendicular to the radius drawn to its point of contact - the inscribed circle touches each side, so each side is a tangent.
The two tangents from an external point
2
$A M O N$ is a quadrilateral, so $∠ M A N + ∠ O M A + ∠ M O N + ∠ O N A = 36 0^{\circ}$
Why: The interior angles of a quadrilateral add to $(4 - 2) \times 18 0^{\circ} = 36 0^{\circ}$ .
Interior and exterior angle sums of a polygon
3
$7 0^{\circ} + 9 0^{\circ} + ∠ M O N + 9 0^{\circ} = 36 0^{\circ}$
Why: Substitute $∠ M A N = ∠ A = 7 0^{\circ}$ and the two right angles from step 1.
4
$∠ M O N = 36 0^{\circ} - 25 0^{\circ} = 11 0^{\circ}$
Why: Subtract; the two right angles use up $18 0^{\circ}$ , so $∠ M O N$ is simply the supplement of $∠ A$ .

Shortcut

For a circle touching both arms of an angle, the angle between the two radii to the points of contact is the supplement of that angle: $18 0^{\circ} - 7 0^{\circ} = 11 0^{\circ}$ .

Common mistakes

Option A

Assumed the two radii to the points of contact meet at a right angle, which happens only when $∠ A = 9 0^{\circ}$ .

Option B

Slipped in the subtraction $360 - 250$ , taking it as $100$ rather than $110$ .

Option D

No route reaches $12 0^{\circ}$ ; it would require $∠ A = 6 0^{\circ}$ .

Skills used

The two tangents from an external point Interior and exterior angle sums of a polygon

82Geometry • Triangle • Pythagoras Theorem

EasyIdeal time: about 55 s

The sum of the squares of sides of a right-angled triangle is $8, 450$ square units. What is the length of its hypotenuse?

50

unitsB

55

unitsC

60

unitsD

65

units

Show answer & solution

Option D

65

units

Approach

Use Pythagoras to replace the squares on the two legs by the square on the hypotenuse, so the stated total is twice $c^{2}$ .

Pythagoras' theorem

Step-by-step solution

1
Legs $a$ , $b$ and hypotenuse $c$ , with $a^{2} + b^{2} = c^{2}$
Why: Pythagoras' theorem for the right angle enclosed by the two legs.
Pythagoras' theorem
2
$a^{2} + b^{2} + c^{2} = 8450$
Why: 'The sum of the squares of sides' counts all three sides, the hypotenuse included - that is the reading the numbers bear out.
3
$c^{2} + c^{2} = 8450 \Rightarrow 2 c^{2} = 8450 \Rightarrow c^{2} = 4225$
Why: Substitute $a^{2} + b^{2} = c^{2}$ from step 1, which collapses the three squares into two.
4
$c = 4225 = 65$
Why: $6 5^{2} = 4225$ , and a length takes the non-negative square root.
denotes the non-negative root

Shortcut

In any right triangle the sum of the squares on all three sides is $2 c^{2}$ , so halve the given total and take the square root.

Common mistakes

Option A

No route reaches $50$ : $2 \times 5 0^{2} = 5000$ , not $8450$ .

Option B

No route reaches $55$ : $2 \times 5 5^{2} = 6050$ , not $8450$ .

Option C

No route reaches $60$ : $2 \times 6 0^{2} = 7200$ , not $8450$ .

Skills used

Pythagoras' theorem

denotes the non-negative root

83Mensuration • Plane Figures

EasyIdeal time: about 40 s

A triangle and a parallelogram have equal areas and equal bases. If the altitude of the triangle is k times the altitude of the parallelogram, then what is the value of k?

4

2

1

\frac{1}{2}

Show answer & solution

Option B

2

Approach

Write both areas on the common base, set them equal, and cancel the base.

Area of a triangle from base and height

Step-by-step solution

1
Triangle: area $= \frac{1}{2} b h_{t}$
Why: A triangle on base $b$ with perpendicular height $h_{t}$ to that base has area $\frac{1}{2} b h_{t}$ .
Area of a triangle from base and height
2
Parallelogram: area $= \frac{1}{2} (b + b) h_{p} = b h_{p}$
Why: A parallelogram is a trapezium whose two parallel sides are both equal to $b$ , so the trapezium formula $\frac{1}{2} (a + b) h$ reduces to base times height.
Area of a trapezium
3
$\frac{1}{2} b h_{t} = b h_{p}$
Why: The two areas are equal and the bases are equal, so the one letter $b$ serves for both.
4
$h_{t} = 2 h_{p}$ , so $k = 2$
Why: Cancel the non-zero base $b$ and multiply through by $2$ ; $k$ was defined by $h_{t} = k h_{p}$ .

Shortcut

A triangle carries the factor $\frac{1}{2}$ that a parallelogram does not, so at equal area and equal base the triangle must be twice as tall.

Common mistakes

Option A

Doubled twice, or used $\frac{1}{4} bh$ for the triangle's area in place of $\frac{1}{2} bh$ .

Option C

Treated both areas as base times height, dropping the triangle's factor $\frac{1}{2}$ , so the two altitudes came out equal.

Option D

Solved for the parallelogram's altitude as a multiple of the triangle's, giving the reciprocal $\frac{1}{2}$ instead of $k$ .

Skills used

Area of a triangle from base and height Area of a trapezium

84Mensuration • Plane Figures • Quadrilateral

EasyIdeal time: about 45 s

Areas of two squares are in the ratio $m^{2} : n^{4}$ . What is the ratio of their perimeters?

m : n

n : m

m : n^{2}

m^{2} : n

Show answer & solution

Option C

m : n^{2}

Approach

Pass from areas to sides by taking square roots, then note that the perimeter is proportional to the side.

Perimeter and area of a square

Step-by-step solution

1
$a_{1}^{2} : a_{2}^{2} = m^{2} : n^{4}$
Why: A square of side $a$ has area $a^{2}$ , so the given area ratio is the ratio of the squares of the sides.
Perimeter and area of a square
2
$a_{1} : a_{2} = m^{2} : n^{4} = m : n^{2}$
Why: Area varies as the square of the side, so the sides are in the square root of the area ratio; $n^{4} = n^{2}$ , not $n$ .
Direct variation as a power: $y \propto x^{n}$
3
$4 a_{1} : 4 a_{2} = a_{1} : a_{2} = m : n^{2}$
Why: The perimeter of a square is $4 a$ , and the common factor $4$ cancels from both parts of the ratio - so perimeters are in the same ratio as sides.

Shortcut

Perimeter varies as the first power of the side and area as the second, so the perimeter ratio is always the square root of the area ratio.

Common mistakes

Option A

Rooted $n^{4}$ down to $n$ instead of $n^{2}$ , halving the exponent twice.

Option B

Wrote the ratio the other way round after rooting, reporting the second square's side first.

Option D

Square-rooted only the second term and left $m^{2}$ untouched.

Skills used

Perimeter and area of a square Direct variation as a power:

y \propto x^{n}

85Geometry • Triangle • Medians and Centroid

ModerateIdeal time: about 1.5 min

AD is the median of the triangle ABC. If P is any point on AD, then which one of the following is correct?

AArea of triangle PAB is greater than the area of triangle PACBArea of triangle PAB is equal to area of triangle PACCArea of triangle PAB is one-fourth of the area of triangle PACDArea of triangle PAB is half of the area of triangle PAC

Show answer & solution

Option B

Area of triangle PAB is equal to area of triangle PAC

Approach

Writing $[X Y Z]$ for the area of $△ X Y Z$ , compare the two triangles by subtracting equal pieces from the two equal halves that the median cuts.

A median bisects the area of a triangle

Step-by-step solution

1
$A D$ is a median, so $D$ is the midpoint of $B C$ and $B D = D C$
Why: That is what a median is: the segment from a vertex to the midpoint of the opposite side.
2
$[A B D] = [A C D]$
Why: Equal bases $B D = D C$ with the same height, namely the perpendicular from $A$ to the line $B C$ ; so a median always bisects the area of a triangle.
Area of a triangle from base and height A median bisects the area of a triangle
3
$[P B D] = [P C D]$
Why: The same two equal bases, now with apex $P$ . Since $P$ lies on $A D$ , its perpendicular distance to $B C$ is one number serving both triangles.
4
$[P A B] = [A B D] - [P B D]$ and $[P A C] = [A C D] - [P C D]$
Why: On each side of the median, the triangle asked about is the half-triangle less the small triangle cut off below $P$ .
5
$[P A B] = [A C D] - [P C D] = [P A C]$
Why: Substitute the two equalities from steps 2 and 3: equal quantities less equal quantities are equal, whatever $P$ is.

Shortcut

$A D$ bisects the area of $△ A B C$ and also of $△ P B C$ ; subtracting the second from the first leaves the two halves still equal.

Common mistakes

Option A

Assumed that moving $P$ along $A D$ tilts the two triangles unequally; in fact both keep the same pair of equal bases and a common height.

Option C

No route reaches one-fourth: the two areas are equal for every position of $P$ on $A D$ .

Option D

Imported the centroid's $2 : 1$ division of the median and turned it into an area ratio, which it is not.

Skills used

A median bisects the area of a triangle Area of a triangle from base and height

86Mensuration • Plane Figures • Circle

ModerateIdeal time: about 1.5 min

What is the area of a segment of a circle of radius r subtending an angle $θ$ at the centre?

\frac{1}{2} r^{2} θ

\frac{1}{2} r^{2} (θ - 2 sin \frac{θ}{2} cos \frac{θ}{2})

\frac{1}{2} r^{2} (θ - sin \frac{θ}{2} cos \frac{θ}{2})

\frac{1}{2} r^{2} sin \frac{θ}{2} cos \frac{θ}{2}

Show answer & solution

Option B

\frac{1}{2} r^{2} (θ - 2 sin \frac{θ}{2} cos \frac{θ}{2})

Approach

A segment is the sector less the triangle on the two bounding radii; compute both and then rewrite $sin θ$ in the half-angle form the options use.

Area and arc length of a circular sector

Step-by-step solution

1
Sector area $= \frac{θ}{2 π} \times π r^{2} = \frac{1}{2} r^{2} θ$
Why: The sector is the fraction $\frac{θ}{2 π}$ of the whole circle. The $\frac{1}{2} r^{2} θ$ form requires $θ$ in radians, which is how the printed options are written.
Area and arc length of a circular sector Converting between degrees and radians
2
Triangle $O A B$ has two sides $r$ enclosing the angle $θ$ , so its area is $\frac{1}{2} r \cdot r sin θ = \frac{1}{2} r^{2} sin θ$
Why: Area from two sides and the included angle; both sides here are radii, so both are $r$ .
Area of a triangle from two sides and the included angle
3
Segment $=$ sector $-$ triangle $= \frac{1}{2} r^{2} θ - \frac{1}{2} r^{2} sin θ = \frac{1}{2} r^{2} (θ - sin θ)$
Why: The chord $A B$ splits the sector into the triangle $O A B$ and the segment beyond it, so removing the triangle leaves the segment.
Cut a compound plane figure into standard pieces
4
$sin θ = 2 sin \frac{θ}{2} cos \frac{θ}{2}$ , so the area is $\frac{1}{2} r^{2} (θ - 2 sin \frac{θ}{2} cos \frac{θ}{2})$
Why: The options carry the half-angle form, so the double-angle identity is needed to match; the factor $2$ in front of the product is what separates the right option from the near-miss.
$sin 2 A = 2 sin A cos A$

Shortcut

Segment $= \frac{1}{2} r^{2} (θ - sin θ)$ is the form worth remembering; only then look for the printed equivalent.

Common mistakes

Option A

Gave the sector and stopped, without subtracting the triangle $O A B$ .

Option C

Dropped the factor $2$ when rewriting $sin θ$ as $2 sin \frac{θ}{2} cos \frac{θ}{2}$ , so only half the triangle was removed.

Option D

Gave half the triangle's area instead of the segment, subtracting nothing and keeping the wrong half of the difference.

Skills used

Area and arc length of a circular sector Converting between degrees and radians Area of a triangle from two sides and the included angle Cut a compound plane figure into standard pieces

sin 2 A = 2 sin A cos A

87Geometry • Triangle • Pythagoras Theorem

ModerateIdeal time: about 1.5 min

ABC is a triangle right-angled at C. Let P be any point on AC and Q be any point on BC. Which of the following statements is/are correct?

1: $A Q^{2} + B P^{2} = A B^{2} + P Q^{2}$

2: $A B = 2 P Q$

Select the correct answer using the code given below:

A1 onlyB2 onlyCBoth 1 and 2DNeither 1 nor 2

Show answer & solution

Option A

1 only

Approach

Every one of $A Q$ , $B P$ , $A B$ , $P Q$ is the hypotenuse of a right triangle at $C$ , so Pythagoras turns statement 1 into an identity; statement 2 is then tested on one configuration.

Pythagoras' theorem

Step-by-step solution

1
$∠ A C B = 9 0^{\circ}$ with $P$ on $C A$ and $Q$ on $C B$ , so $∠ A C Q = ∠ P C B = ∠ P C Q = 9 0^{\circ}$
Why: $P$ and $Q$ sit on the two arms of the right angle at $C$ , so any pair taken one from each arm encloses that same right angle.
2
$A Q^{2} = A C^{2} + C Q^{2}$ and $B P^{2} = B C^{2} + C P^{2}$
Why: Pythagoras in $△ A C Q$ and in $△ B C P$ .
Pythagoras' theorem
3
$A Q^{2} + B P^{2} = (A C^{2} + B C^{2}) + (C Q^{2} + C P^{2})$
Why: Add the two and regroup the four squares into the pairs belonging to $A B$ and to $P Q$ - this regrouping is the whole idea.
4
$A C^{2} + B C^{2} = A B^{2}$ and $C P^{2} + C Q^{2} = P Q^{2}$ , so $A Q^{2} + B P^{2} = A B^{2} + P Q^{2}$
Why: Pythagoras again, in $△ A B C$ and in $△ P C Q$ . Statement 1 therefore holds for every position of $P$ and $Q$ .
5
Take $A C = B C = 2$ and $C P = C Q = \frac{1}{2}$ : then $P Q = \frac{1}{4} + \frac{1}{4} = \frac{2}{2}$ and $A B = 22$ , so $2 P Q = 2 \neq = 22 = A B$
Why: One counterexample refutes a universal claim. $A B = 2 P Q$ does hold when $P$ and $Q$ are the midpoints of $C A$ and $C B$ , by the midpoint theorem, but 'any point' allows every other position too, so statement 2 is not correct.
Midpoint theorem
6
Statement 1 only
Why: Statement 1 is an identity; statement 2 holds in one special case and fails otherwise.

Shortcut

Statement 2 dies instantly at $P = A$ , $Q = B$ , where $P Q = A B$ ; statement 1 is the four-squares regrouping.

Common mistakes

Option B

Accepted statement 2 from the midpoint case and rejected statement 1 without regrouping the four squares.

Option C

Tested only $P$ and $Q$ at the midpoints, where $A B = 2 P Q$ happens to hold, and took that as general.

Option D

Rejected statement 1 as well, not seeing that $A C^{2} + B C^{2}$ and $C P^{2} + C Q^{2}$ reassemble into $A B^{2}$ and $P Q^{2}$ .

Skills used

Pythagoras' theorem Midpoint theorem

88Mensuration • Plane Figures • Composite and Shaded Regions

ModerateIdeal time: about 1.5 min

Four circular coins of equal radius are placed with their centres coinciding with four vertices of a square. Each coin touches two other coins. If the uncovered area of the square is $42 cm^{2}$ , then what is the radius of each coin? (Assume $π = \frac{22}{7}$ )

5 cm

7 cm

10 cm

14 cm

Show answer & solution

Option B

7 cm

Approach

Touching coins fix the square's side at $2 r$ ; inside the square each coin contributes a quarter circle, and the four quarters make exactly one full circle.

Cut a compound plane figure into standard pieces

Step-by-step solution

1
Adjacent coins touch externally, so the distance between their centres is $r + r = 2 r$ ; those centres are adjacent vertices of the square, so the side is $2 r$
Why: Two circles touching externally have their centres the sum of the radii apart, and the coins' centres are the square's four vertices.
Distance between the centres of touching circles
2
Area of the square $= (2 r)^{2} = 4 r^{2}$
Why: Side squared.
Perimeter and area of a square
3
Covered area $= 4 \times \frac{90}{360} \times π r^{2} = π r^{2}$
Why: Each vertex of a square is a $9 0^{\circ}$ corner, so the part of each coin lying inside the square is a quarter of that coin; four quarters amount to one whole circle. The quarters do not overlap, since the coins only touch.
Area and arc length of a circular sector Cut a compound plane figure into standard pieces
4
$4 r^{2} - π r^{2} = 42 \Rightarrow r^{2} (4 - \frac{22}{7}) = 42$
Why: The uncovered area is the square less the covered part; substitute $π = \frac{22}{7}$ .
5
$r^{2} \times \frac{6}{7} = 42 \Rightarrow r^{2} = 49 \Rightarrow r = 7 cm$
Why: $4 - \frac{22}{7} = \frac{28 - 22}{7} = \frac{6}{7}$ , and a radius takes the positive root.
denotes the non-negative root

Shortcut

Uncovered area $= r^{2} (4 - π) = \frac{6}{7} r^{2}$ , so $r^{2} = 42 \times \frac{7}{6} = 49$ .

Common mistakes

Option A

No route reaches $5$ cm: it would leave an uncovered area of $\frac{6}{7} \times 25 \approx 21.4 cm^{2}$ .

Option C

No route reaches $10$ cm: the uncovered area would be $\frac{600}{7} \approx 85.7 cm^{2}$ .

Option D

Answered with the diameter $2 r = 14$ cm, or equivalently read the square's side as the radius.

Skills used

Cut a compound plane figure into standard pieces Distance between the centres of touching circles Perimeter and area of a square Area and arc length of a circular sector

denotes the non-negative root

89Mensuration • Solids • Frustum

ModerateIdeal time: about 1.5 min

The radii of the flat circular faces of a bucket are x and $2 x$ . If the height of the bucket is $3 x$ , what is the capacity of the bucket? (Assume $π = \frac{22}{7}$ )

11 x^{3}

22 x^{3}

44 x^{3}

55 x^{3}

Show answer & solution

Option B

22 x^{3}

Approach

A bucket with two unequal flat circular faces is a frustum of a cone; substitute the two radii and the height into the frustum volume formula.

Volume of a frustum of a cone

Step-by-step solution

1
$R = 2 x$ , $r = x$ , $h = 3 x$
Why: The two flat circular faces are the parallel ends of the frustum, the larger one being the rim; the stated height is the perpendicular distance between them.
2
$V = \frac{π h}{3} (R^{2} + R r + r^{2})$
Why: Volume of a frustum of a cone - note the cross term $R r$ , which a cylinder or cone formula does not have.
Volume of a frustum of a cone
3
$V = \frac{π ( 3 x )}{3} (4 x^{2} + 2 x^{2} + x^{2}) = π x \times 7 x^{2} = 7 π x^{3}$
Why: The $3$ of the height cancels the $3$ in the denominator, and the bracket totals $7 x^{2}$ .
4
$V = 7 \times \frac{22}{7} x^{3} = 22 x^{3}$
Why: With $π = \frac{22}{7}$ the $7$ cancels exactly, which is why the question fixes $π$ at that value.

Common mistakes

Option A

Halved the result, typically by using $\frac{π h}{2}$ or by dropping one of the three terms in the bracket.

Option C

Doubled the result, for instance by forgetting the $\frac{1}{3}$ only partially or by taking the bracket as $14 x^{2}$ .

Option D

No route reaches $55 x^{3}$ from radii $x$ and $2 x$ with height $3 x$ .

Skills used

Volume of a frustum of a cone

90Mensuration • Solids • Cuboid

EasyIdeal time: about 1 min

If p, q, r, s and t represent length, breadth, height, surface area and volume of a cuboid respectively, then what is $\frac{1}{p} + \frac{1}{q} + \frac{1}{r}$ equal to?

\frac{s}{t}

\frac{2 t}{s}

\frac{s}{2 t}

\frac{2 s}{t}

Show answer & solution

Option C

\frac{s}{2 t}

Approach

Add the three reciprocals over the common denominator $pq r$ and recognise the numerator as half the surface area.

Volume and wall area of a cuboid

Step-by-step solution

1
$s = 2 (pq + q r + r p)$ and $t = pq r$
Why: Total surface area and volume of a cuboid with edges $p$ , $q$ , $r$ .
Volume and wall area of a cuboid
2
$\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{q r + r p + pq}{pq r}$
Why: Put the three fractions over the common denominator $pq r$ ; each numerator is the product of the other two edges.
3
$pq + q r + r p = \frac{s}{2}$
Why: Read step 1 backwards: the bracket in the surface-area formula is exactly half of $s$ . This is the step the three wrong options all miss.
4
$\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{s /2}{t} = \frac{s}{2 t}$
Why: Substitute both results and tidy the double fraction.

Shortcut

Note that $\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{pq + q r + r p}{pq r}$ , and that the numerator is half the surface area while the denominator is the volume.

Common mistakes

Option A

Forgot the factor $2$ in the surface-area formula, taking $s = pq + q r + r p$ .

Option B

Inverted the result, reporting the reciprocal of the correct expression.

Option D

Multiplied by $2$ instead of dividing, putting the $2$ on the wrong side of the fraction.

Skills used

Volume and wall area of a cuboid

91Statistics • Median

HardIdeal time: about 2 min

Fifteen candidates appeared in an examination. The marks of the candidates who passed in the examination are $9, 6, 7, 8, 8, 9, 6, 5, 4$ and $7$ . What is the median of marks of all the fifteen candidates?

6

6.5

7

7.5

Show answer & solution

Option A

6

Approach

The five candidates who failed must have scored below every passing mark, so they fill the bottom five places; the median of fifteen is then the eighth value of the combined list.

Median of a data set

Step-by-step solution

1
$n = 15$ is odd, so the median is the $\frac{15 + 1}{2} = 8$ th value when all fifteen marks are put in increasing order
Why: For an odd number of observations the median is the single middle one.
Median of a data set
2
The ten passing marks in order: $4, 5, 6, 6, 7, 7, 8, 8, 9, 9$
Why: Ordering the data is the step this sum turns on, and the marks are printed out of order.
3
$15 - 10 = 5$ candidates failed, and each of their marks is below the pass mark, hence below $4$
Why: Only the passers' marks are listed. The five failing marks are unknown, but every one of them is smaller than the lowest mark that passed.
4
Full ordered list: five unknown marks, all below $4$ , then $4, 5, 6, 6, 7, 7, 8, 8, 9, 9$
Why: The five failures occupy places $1$ to $5$ whatever their values are, so those values never have to be known - which is why the question is answerable at all.
5
Places $6, 7, 8$ hold $4, 5, 6$ , so the $8$ th value is $6$
Why: Count up from place $6$ ; the median of all fifteen candidates is therefore $6$ .

Shortcut

Inserting five marks below everything pushes the middle five places down: the $8$ th of the fifteen is the $8 - 5 = 3$ rd of the listed ten, which is $6$ .

Common mistakes

Option B

Located the middle of fifteen one place too high and averaged the $9$ th and $10$ th values, $6$ and $7$ .

Option C

Took the median of the ten listed passing marks only, $\frac{7 + 7}{2} = 7$ , ignoring the five who failed.

Option D

No route reaches $7.5$ ; it is the mean of the $11$ th and $12$ th values and corresponds to no reading of 'median of all fifteen'.

Skills used

Median of a data set

92Statistics • Median

Very easyIdeal time: about 35 s

If the yield (in gm) of barley from $7$ plots of size one square yard each, were found to be $180, 191, 175, 111, 154, 141$ and $176$ , then what is the median yield?

111 gm

154 gm

175 gm

176 gm

Show answer & solution

Option C

175 gm

Approach

Order the seven yields and read off the middle one.

Median of a data set

Step-by-step solution

1
$111, 141, 154, 175, 176, 180, 191$
Why: The median is defined on ordered data, and the yields are printed in the order the plots were harvested, not in increasing order.
Median of a data set
2
$n = 7$ is odd, so the median is the $\frac{7 + 1}{2} = 4$ th value
Why: With an odd count there is a single middle observation.
3
$4$ th value $= 175$ gm
Why: Three yields lie below it and three above, which is what being the median means.

Common mistakes

Option A

Did not order the data and took the $4$ th yield as printed, which is $111$ - also the smallest of the seven.

Option B

Took the $3$ rd value of the ordered list instead of the $4$ th.

Option D

Took the $5$ th value of the ordered list instead of the $4$ th.

Skills used

Median of a data set

93Statistics • Mode

Very easyIdeal time: about 25 s

Which one of the following measures of central tendency will be used to determine the average size of the shoe sold in the shop?

AArithmetic meanBGeometric meanCMedianDMode

Show answer & solution

Option D

Mode

Approach

Decide which average answers the question a shoe shop is actually asking, then name it.

Mode: the most frequent value, and it need not be unique

Step-by-step solution

1
The variable is shoe size, and what the shop needs is the size it sells most of
Why: 'Average size sold' in a stock-keeping sense means the size in greatest demand; that is the only reading on which the answer is of any use to the shop.
2
An arithmetic mean of sizes, e.g. $\frac{7 + 8 + 11}{3} = \frac{26}{3}$ , need not be a size that exists
Why: Shoe sizes come in fixed steps, so a mean - arithmetic or geometric - can land between two stocked sizes and tell the shop nothing about what to order.
Mean of a frequency distribution
3
The value occurring with the greatest frequency is the mode
Why: The mode is the most frequent value and is always one of the values actually present, so it names a size that can be stocked. It is also the one average that works on a variable treated as a set of discrete categories.
Mode: the most frequent value, and it need not be unique

Shortcut

'Most commonly sold' is the mode by definition; the other three averages are arithmetic combinations and need not land on a real size.

Common mistakes

Option A

Averaged the sizes arithmetically, which gives a number nobody wears and is distorted by a few very large or very small feet.

Option B

The geometric mean is for ratios and growth factors; it has no meaning for shoe sizes at all.

Option C

Gave the middle size when the sizes are ranked, which is not the size sold most often.

Skills used

Mode: the most frequent value, and it need not be unique Mean of a frequency distribution

94Statistics • Frequency Distribution

Very easyIdeal time: about 25 s

When the class intervals have equal width, the height of a rectangle in a histogram represents

AWidth of the classBLower class limitCUpper class limitDFrequency of the class

Show answer & solution

Option D

Frequency of the class

Approach

Start from what a histogram's bar is drawn to represent in general, then specialise to equal class widths.

Frequency density

Step-by-step solution

1
In a histogram the AREA of each bar represents the frequency of its class, so the height is $\frac{frequency}{class width}$ , the frequency density
Why: That is what lets classes of different widths appear on one diagram without distorting the picture - the histogram is an area diagram, not a bar chart.
Frequency density
2
If every width equals $w$ , then height $= \frac{f}{w}$ , which is proportional to $f$
Why: A common divisor $w$ only rescales the vertical axis; the heights then rank and compare exactly as the frequencies do, and the axis is conventionally labelled with the frequencies themselves.
3
The height therefore represents the frequency of the class
Why: The width of a bar is the class interval and the two ends of that interval are the class limits, so none of those is what the height shows.

Shortcut

Area represents frequency; with equal widths the height does too.

Common mistakes

Option A

Confused the bar's height with its width; the width is the class interval, which is what is equal here and so carries no information at all.

Option B

The lower class limit fixes where the bar starts along the horizontal axis, not how tall it is.

Option C

The upper class limit fixes where the bar ends along the horizontal axis, not how tall it is.

Skills used

Frequency density

95Statistics • Mean

ModerateIdeal time: about 1.5 min

The ages of $7$ family members are $2, 5, 12, 18, 38, 40$ and $60$ years respectively. After $5$ years a new member aged x years is added. If the mean age of the family now goes up by $1.5$ years, then what is the value of x?

1

2

3

4

Show answer & solution

Option B

2

Approach

Age all seven members by five years to get the present total, then set the new eight-member mean $1.5$ years above the family's original mean.

Mean of a frequency distribution

Step-by-step solution

1
$2 + 5 + 12 + 18 + 38 + 40 + 60 = 175$ , so the original mean is $\frac{175}{7} = 25$ years
Why: Mean is the total divided by the count; this is the figure the increase is measured from.
Mean of a frequency distribution
2
After five years every age rises by $5$ , so the total becomes $175 + 7 \times 5 = 210$
Why: Adding a constant to each of seven observations adds it seven times to the total.
3
With the new member there are $8$ members and the total is $210 + x$
Why: The newcomer is $x$ years old at that time, five years from now.
4
$\frac{210 + x}{8} = 25 + 1.5 = 26.5$
Why: The mean age of the family is to stand $1.5$ years above its original value of $25$ .
5
$210 + x = 212 \Rightarrow x = 2$
Why: Multiply out and subtract.

Common mistakes

Option A

No route reaches $1$ : it would make the new mean $\frac{211}{8} = 26.375$ , a rise of $1.375$ years.

Option C

No route reaches $3$ : it would make the new mean $\frac{213}{8} = 26.625$ , a rise of $1.625$ years.

Option D

No route reaches $4$ : it would make the new mean $\frac{214}{8} = 26.75$ , a rise of $1.75$ years.

Skills used

Mean of a frequency distribution

96Statistics • Mean • Weighted Mean

EasyIdeal time: about 55 s

The mean weight of $100$ students in a class is $46 kg$ . The mean weight of boys is $50 kg$ and that of girls is $40 kg$ . The number of boys exceeds the number of girls by

10

15

20

25

Show answer & solution

Option C

20

Approach

Let the numbers of boys and girls be $b$ and $g$ , write the class's total weight in two ways, and solve the pair.

Mean of a frequency distribution

Step-by-step solution

1
$b + g = 100$
Why: The class has $100$ students in all, boys and girls together.
2
$50 b + 40 g = 46 \times 100 = 4600$
Why: Total weight equals mean times count. Computed from the whole class it is $4600$ kg; computed group by group it is $50 b + 40 g$ , and the two must agree.
Mean of a frequency distribution
3
$50 b + 40 (100 - b) = 4600 \Rightarrow 10 b + 4000 = 4600 \Rightarrow b = 60$
Why: Substitute $g = 100 - b$ to eliminate $g$ , leaving one linear equation.
Eliminate one variable from a linear pair
4
$g = 100 - 60 = 40$ , so $b - g = 20$
Why: The question asks by how much the boys exceed the girls, not how many boys there are.

Shortcut

Alligation: the class mean $46$ is $6$ above the girls' $40$ and $4$ below the boys' $50$ , so boys : girls $= 6 : 4 = 3 : 2$ , i.e. $60$ and $40$ , a difference of $20$ .

Common mistakes

Option A

Answered with the difference of the two mean weights, $50 - 40 = 10$ kg, rather than the difference in the numbers of students.

Option B

No route reaches $15$ : it would need $b = 57.5$ , which is not a whole number of students.

Option D

No route reaches $25$ : it would need $b = 62.5$ , again not a whole number.

Skills used

Mean of a frequency distribution Eliminate one variable from a linear pair

97Statistics • Mean

EasyIdeal time: about 40 s

What is the algebraic sum of the deviations from the mean of a set of values $25, 65, 73, 75, 83, 76, 17, 15, 7, 14$ ?

- 1

0

1

2

Show answer & solution

Option B

0

Approach

The deviations about the mean always cancel; prove it in general and the particular ten numbers never need adding.

Deviations about the mean add to zero

Step-by-step solution

1
$\sum_{i = 1}^{n} (x_{i} - x ˉ) = \sum_{i = 1}^{n} x_{i} - n x ˉ$
Why: Separate the sum of the values from the $n$ copies of the mean that are subtracted, one per observation.
2
$x ˉ = \frac{\sum x _{i}}{n}$ , so $n x ˉ = \sum x_{i}$
Why: That is the definition of the mean, rearranged.
Mean of a frequency distribution
3
$\sum (x_{i} - x ˉ) = \sum x_{i} - \sum x_{i} = 0$
Why: The two totals are the same number, so they cancel. The algebraic sum of the deviations about the mean is zero for every data set, whatever the values are.
Deviations about the mean add to zero

Shortcut

The answer is $0$ for any data set whatever; no addition is needed and the ten values can be ignored.

Common mistakes

Option A

Slipped in the long addition of the ten values, got a mean slightly off $45$ , and was left with a small negative residue.

Option C

The same slip with the opposite sign, leaving a small positive residue.

Option D

No route reaches $2$ ; a correct mean forces the deviations to cancel exactly.

Skills used

Deviations about the mean add to zero Mean of a frequency distribution

98Statistics • Mean

EasyIdeal time: about 50 s

The mean of five observations x, x + 2, x + 4, x + 6, x + 8 is m. What is the mean of the first three observations?

m

m - 1

m - 2

m - 3

Show answer & solution

Option C

m - 2

Approach

The five observations are in arithmetic progression, so each mean is the middle term of its own run.

An evenly spaced run averages to its middle term

Step-by-step solution

1
$x, x + 2, x + 4, x + 6, x + 8$ is an AP with common difference $2$
Why: Each term exceeds the previous one by $2$ , which is what makes the middle-term shortcut available.
2
Mean of the five $=$ the third term $= x + 4$ , so $x + 4 = m$
Why: For an odd number of terms in AP the mean is exactly the middle term.
An evenly spaced run averages to its middle term
3
The first three are $x, x + 2, x + 4$ , themselves an AP, so their mean is the middle one, $x + 2$
Why: Three terms is again an odd count with a constant step, so the same rule applies to the sub-run.
4
$x + 2 = (x + 4) - 2 = m - 2$
Why: Substitute $x + 4 = m$ from step 2, which removes $x$ without ever solving for it.

Shortcut

Dropping the two largest terms moves the centre of the run back by one place, from $x + 4$ to $x + 2$ , so the mean falls by the common difference $2$ .

Common mistakes

Option A

Assumed that the mean of a sub-run equals the mean of the whole run, which holds only when the sub-run is centred on the same term.

Option B

Took the middle of the first three to be $x + 3$ , shifting the centre by one step instead of two.

Option D

No route reaches $m - 3$ : the centre moves from $x + 4$ to $x + 2$ , a fall of $2$ , not $3$ .

Skills used

An evenly spaced run averages to its middle term

99Statistics • Median

EasyIdeal time: about 50 s

What is the median of $2, 4, 6$ , ..., $100$ ?

48

49

50

51

Show answer & solution

Option D

51

Approach

Count the terms, find they are even in number, and average the two middle ones.

Median of a data set

Step-by-step solution

1
The list is $2, 4, 6, \dots, 100$ , whose $k$ th term is $2 k$ ; $2 k = 100$ at $k = 50$ , so there are $50$ terms
Why: Counting the terms is the step the answer turns on, and dividing the last term by the common difference does it here.
2
$n = 50$ is even, so the median is the mean of the $25$ th and $26$ th terms
Why: With an even count there is no single middle value, so the median is the average of the two central ones.
Median of a data set
3
$25$ th term $= 2 \times 25 = 50$ and $26$ th term $= 2 \times 26 = 52$
Why: Read both off the $k$ th-term rule $2 k$ .
4
Median $= \frac{50 + 52}{2} = 51$
Why: Average the two central terms. The list is an AP, so its median also equals its mean, $\frac{2 + 100}{2} = 51$ - the two routes must agree, and they do.
An evenly spaced run averages to its middle term

Shortcut

For any AP the median equals the mean, so $\frac{first + last}{2} = \frac{2 + 100}{2} = 51$ in one line.

Common mistakes

Option A

No route reaches $48$ ; it is the $24$ th term and sits two places off the centre.

Option B

Averaged the $24$ th and $25$ th terms, $48$ and $50$ , locating the centre one place too low.

Option C

Took the $25$ th term alone as the median, treating a set of $50$ as though it had a single middle member.

Skills used

Median of a data set An evenly spaced run averages to its middle term

100Statistics • Mean

EasyIdeal time: about 55 s

The harmonic mean and the geometric mean of two numbers are $10$ and $12$ respectively. What is their arithmetic mean?

\frac{25}{3}

120

11

14.4

Show answer & solution

Option D

14.4

Approach

Turn each given mean into an equation in the product and the sum of the two numbers; the arithmetic mean is then half the sum.

Geometric mean: the

n

th root of the product

Step-by-step solution

1
Let the numbers be $a$ and $b$ . Geometric mean: $ab = 12 \Rightarrow ab = 144$
Why: The geometric mean of two numbers is the square root of their product, so squaring turns the statement into a product equation.
Geometric mean: the $n$ th root of the product
2
Harmonic mean: $\frac{2 ab}{a + b} = 10 \Rightarrow a + b = \frac{2 \times 144}{10} = 28.8$
Why: The harmonic mean of two numbers is the reciprocal of the average of their reciprocals, which simplifies to $\frac{2 ab}{a + b}$ ; with $ab$ already known it delivers the sum.
Harmonic mean: $n$ over the sum of the reciprocals
3
$A = \frac{a + b}{2} = \frac{28.8}{2} = 14.4$
Why: The arithmetic mean of two numbers is half their sum, and the sum is now known without finding $a$ and $b$ separately.
Mean of a frequency distribution

Shortcut

For two numbers $G^{2} = A \times H$ , so $A = \frac{G ^{2}}{H} = \frac{144}{10} = 14.4$ in one line.

Common mistakes

Option A

Divided the wrong way round, computing $\frac{H ^{2}}{G} = \frac{100}{12} = \frac{25}{3}$ instead of $\frac{G ^{2}}{H}$ .

Option B

Took the geometric mean of the two given means, $10 \times 12 = 120$ , rather than using the relation between them.

Option C

Averaged the two given means, $\frac{10 + 12}{2} = 11$ , which is not what the relation $G^{2} = A H$ says.

Skills used

Geometric mean: the

n

th root of the product Harmonic mean:

n

over the sum of the reciprocals Mean of a frequency distribution