Elementary Mathematics 2017 (II)

Set A

Questions 81–100 of 100

81Mensuration • Plane Figures • Circle

EasyIdeal time: about 45 s

An arc of a circle subtends an angle $π$ at the centre. If the length of the arc is $22$ cm, then what is the radius of the circle? (Take $π = \frac{22}{7}$ )

5

cmB

7

cmC

9

cmD

11

Show answer & solution

Option B

7

Approach

An angle of $π$ radians is a straight angle, so the arc is a semicircle; apply arc $= r θ$ .

Converting between degrees and radians

Step-by-step solution

1
$θ = π$ radians $= 18 0^{\circ}$ , so the arc is half the circle.
Why: A straight angle is $π$ radians; the symbol $π$ here is an angle in radian measure, not a stray constant.
Converting between degrees and radians
2
Arc $= r θ \Rightarrow 22 = π r$
Why: With the central angle in radians the arc length is simply $r θ$ ; equivalently a semicircular arc is half of $2 π r$ .
Area and arc length of a circular sector
3
$r = \frac{22}{π} = 22 \times \frac{7}{22} = 7$ cm
Why: Substituting $π = \frac{22}{7}$ as instructed makes the $22$ s cancel.

Shortcut

Semicircular arc $= π r = 22$ , so $r = 7$ cm.

Common mistakes

Option A

$5$ cm: no route reaches it.

Option C

$9$ cm: no route reaches it.

Option D

$11$ cm is half the arc length, as though the arc were a diameter.

Skills used

Converting between degrees and radians Area and arc length of a circular sector

82Geometry • Triangle • Similarity

ModerateIdeal time: about 1.5 min

One-fifth of the area of a triangle ABC is cut off by a line DE drawn parallel to BC such that D is on AB and E is on AC. If BC = $10$ cm, then what is DE equal to?

5

cmB

25

cmC

35

cmD

45

Show answer & solution

Option B

25

Approach

The piece cut off at the apex is similar to the whole triangle, so the area ratio is the square of $\frac{D E}{B C}$ .

Step-by-step solution

1
$D E ∥ B C$ with $D$ on $A B$ and $E$ on $A C$ , so $△ A D E \sim △ A B C$ .
Why: A line parallel to one side cuts off a triangle similar to the original: the angle at $A$ is common and the other two correspond across the parallels.
Similar triangles have proportional sides
2
$\frac{[ △ A D E ]}{[ △ A B C ]} = (\frac{D E}{B C})^{2} = \frac{1}{5}$
Why: The triangle cut off carries one-fifth of the area, and similar figures have areas in the square of their linear ratio.
Direct variation as a power: $y \propto x^{n}$
3
$D E^{2} = \frac{B C ^{2}}{5} = \frac{100}{5} = 20 \Rightarrow D E = 20 = 25$ cm
Why: Take the positive square root and simplify the surd by pulling out the perfect square $4$ .
$ab = a b$ and $a / b = a / b$

Shortcut

$D E = B C \frac{1}{5} = \frac{10}{5} = 25$ cm.

Common mistakes

Option A

$5$ cm: no route reaches it.

Option C

$35$ cm: no route reaches it.

Option D

$45$ cm reads the trapezium $D B C E$ as the fifth that is cut off, leaving $△ A D E$ with $\frac{4}{5}$ of the area and $D E = 10 \frac{4}{5}$ .

Skills used

y \propto x^{n}

ab = a b

and

a / b = a / b

83Geometry • Line

EasyIdeal time: about 1 min

There are $8$ lines in a plane, no two of which are parallel. What is the maximum number of points at which they can intersect?

15

21

28

DNone of the above

Show answer & solution

Option C

28

Approach

Enumerate line by line: each new line can cross every line already drawn at a fresh point, and sum the running total.

Sums of the first

n

natural numbers, their squares and their cubes

Step-by-step solution

1
Line $1$ : $0$ new intersection points (nothing to cross yet).
Why: Baseline of the enumeration.
2
Line $2$ adds $1$ point, line $3$ adds $2$ , \ldots, line $8$ adds $7$ — each new line meets every earlier line exactly once since no two are parallel.
Why: Maximality means every possible crossing actually happens and none coincide.
3
Total $= 0 + 1 + 2 + \dots + 7 = \frac{7 \times 8}{2} = 28$ , matching option C.
Why: Sum of the first $7$ natural numbers.
Sums of the first $n$ natural numbers, their squares and their cubes

Shortcut

Equivalently, every pair of the $8$ lines contributes exactly one crossing: $(2 8) = 28$ .

Common mistakes

Option A

Matches $2 \times 8 - 1 = 15$ , as if intersections only chain along one path instead of counting every pair.

Option B

Matches $(2 7) = 21$ , i.e. using $7$ lines instead of $8$ .

Option D

Would only be correct if the true maximum weren't among the printed options, but $28$ is option C.

Skills used

Sums of the first

n

natural numbers, their squares and their cubes

84Geometry • Polygon • Interior and Exterior Angles

EasyIdeal time: about 1 min

A closed polygon has six sides and one of its angles is $3 0^{\circ}$ greater than each of the other five equal angles. What is the value of one of the equal angles?

5 5^{\circ}

11 5^{\circ}

15 0^{\circ}

17 5^{\circ}

Show answer & solution

Option B

11 5^{\circ}

Approach

Use the interior angle sum of a hexagon and write all six angles in terms of the repeated one.

Interior and exterior angle sums of a polygon

Step-by-step solution

1
Interior angle sum $= (n - 2) \times 18 0^{\circ} = (6 - 2) \times 18 0^{\circ} = 72 0^{\circ}$
Why: A polygon of $n$ sides splits from one vertex into $n - 2$ triangles, each contributing $18 0^{\circ}$ .
Angle sum of a triangle Interior and exterior angle sums of a polygon
2
Let each of the five equal angles be $x$ ; the sixth is then $x + 3 0^{\circ}$ .
Why: ' $3 0^{\circ}$ greater than each of the other five equal angles' names one odd angle and five equal ones.
Turn a worded condition into a linear equation in one unknown
3
$5 x + (x + 3 0^{\circ}) = 72 0^{\circ} \Rightarrow 6 x = 69 0^{\circ} \Rightarrow x = 11 5^{\circ}$
Why: The six interior angles must add to the sum found in step 1.

Shortcut

$6 x + 3 0^{\circ} = 72 0^{\circ}$ , so $x = 11 5^{\circ}$ .

Common mistakes

Option A

$5 5^{\circ}$ solves $6 x + 3 0^{\circ} = 36 0^{\circ}$ , using the EXTERIOR angle sum $36 0^{\circ}$ in place of the interior one.

Option C

$15 0^{\circ}$ takes the average angle $\frac{72 0 ^{\circ}}{6} = 12 0^{\circ}$ and adds the $3 0^{\circ}$ , so it answers with the odd angle rather than one of the equal ones.

Option D

$17 5^{\circ}$ solves $6 x + 3 0^{\circ} = 108 0^{\circ}$ , using $n \times 18 0^{\circ}$ instead of $(n - 2) \times 18 0^{\circ}$ .

Skills used

Interior and exterior angle sums of a polygon Angle sum of a triangle Turn a worded condition into a linear equation in one unknown

85Geometry • Triangle • Triangle Centres

ModerateIdeal time: about 1.5 min

Consider the following statements:

1: The point of intersection of the perpendicular bisectors of the sides of a triangle may lie outside the triangle.

2: The point of intersection of the perpendiculars drawn from the vertices to the opposite sides of a triangle may lie on two sides.

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

Name the centre each statement describes, then produce a triangle in which the claimed position actually occurs.

The circumcentre is the point equidistant from the vertices

Step-by-step solution

1
Statement 1 describes the circumcentre, the common point of the perpendicular bisectors of the sides.
Why: That is the defining construction of the circumcentre.
The circumcentre is the point equidistant from the vertices
2
Take $A = (0, 0)$ , $B = (4, 0)$ , $C = (5, 1)$ : the perpendicular bisectors meet at $(2, 3)$ , while every point of the triangle has $0 \leq y \leq 1$ . The circumcentre is outside. TRUE.
Why: The triangle is obtuse at $B$ (since $A C^{2} = 26 > A B^{2} + B C^{2} = 18$ ), and the circumcentre of an obtuse triangle always falls outside it.
Pythagoras' theorem
3
Statement 2 describes the orthocentre, the common point of the altitudes.
Why: 'Perpendiculars drawn from the vertices to the opposite sides' are precisely the altitudes.
The orthocentre and the orthocentric system
4
In a right triangle two of the altitudes are the legs themselves, so they meet at the right-angle vertex - a point lying on both of those sides. TRUE.
Why: For $A = (0, 0)$ , $B = (3, 0)$ , $C = (0, 4)$ the altitude from $B$ is $B A$ and the altitude from $C$ is $C A$ ; they meet at $A$ , which lies on side $A B$ and on side $A C$ .
5
Both statements are correct.
Why: Each has been realised by an explicit triangle, and 'may lie' asks only for possibility.

Common mistakes

Option A

1 only: missed the right-angled case, where the orthocentre sits at a vertex and so lies on two sides at once.

Option B

2 only: assumed the circumcentre is always inside, which fails for every obtuse triangle.

Option D

Neither: both of the above oversights together.

Skills used

The circumcentre is the point equidistant from the vertices Pythagoras' theorem The orthocentre and the orthocentric system

86Statistics • Data Interpretation

EasyIdeal time: about 1.5 min

For the next five (05) items that follow :

In a University there are $1200$ students studying four different subjects, Mathematics, Statistics, Physics and Chemistry. $20%$ of the total number of students are studying Mathematics, one-fourth of the total number of students are studying Physics, $320$ students are studying Statistics and remaining students are studying Chemistry. Three-fifth of the total number of students studying Chemistry are girls. $150$ boys are studying Mathematics. $60%$ of students studying Physics are boys. $250$ girls are studying Statistics.

What is the total number of boys studying Statistics and Physics?

180

240

250

310

Show answer & solution

Option C

250

Approach

Build the subject totals from the stem, split each into boys and girls, then add the two entries asked for.

x %

of a quantity

Step-by-step solution

1
Mathematics $= 20%$ of $1200 = 240$ ; Physics $= \frac{1}{4} \times 1200 = 300$ ; Statistics $= 320$ ; Chemistry $= 1200 - 240 - 300 - 320 = 340$
Why: Three of the four subject totals are given outright or as a share of $1200$ ; Chemistry is whatever is left.
$x %$ of a quantity
2
Statistics boys $= 320 - 250 = 70$
Why: $250$ of the $320$ Statistics students are girls, so the rest are boys.
3
Physics boys $= 60%$ of $300 = 180$
Why: The stem gives the boys' share of Physics as a percentage of the Physics enrolment, not of the whole university.
4
$70 + 180 = 250$
Why: The question asks for the boys in the two subjects taken together.

Common mistakes

Option A

$180$ is the Physics boys alone, with Statistics left out.

Option B

$240$ is the Mathematics enrolment - a figure lifted from the wrong row of the table.

Option D

$310$ : no combination of the table's entries produces it.

Skills used

x %

of a quantity

87Statistics • Data Interpretation

EasyIdeal time: about 1.5 min

Same directions as Q86

The number of girls studying Statistics is what percent (approximate) of the total number of students studying Chemistry?

58.8

73.5

78.7

80.6

Show answer & solution

Option B

73.5

Approach

Take the Statistics girls from the stem, find the Chemistry enrolment as the remainder, and express the first as a percentage of the second.

x %

of a quantity

Step-by-step solution

1
Mathematics $= 20%$ of $1200 = 240$ ; Physics $= \frac{1}{4} \times 1200 = 300$ ; Statistics $= 320$ ; Chemistry $= 1200 - 240 - 300 - 320 = 340$
Why: Three of the four subject totals are given outright or as a share of $1200$ ; Chemistry is whatever is left.
$x %$ of a quantity
2
Girls studying Statistics $= 250$
Why: Given directly in the stem; the denominator, not the numerator, is the part that must be computed.
3
$\frac{250}{340} \times 100 = \frac{25000}{340} = 73.529 \dots \approx 73.5$
Why: 'What percent of' puts the Chemistry total in the denominator, and the question asks for an approximate value.

Common mistakes

Option A

$58.8$ : no pairing of the table's figures gives it.

Option C

$78.7$ : no pairing of the table's figures gives it.

Option D

$80.6$ : no pairing of the table's figures gives it.

Skills used

x %

of a quantity

88Statistics • Data Interpretation

ModerateIdeal time: about 2 min

Same directions as Q86

In which subjects is the difference between the number of boys and girls equal?

AMathematics and ChemistryBStatistics and ChemistryCMathematics and PhysicsDMathematics and Statistics

Show answer & solution

Option C

Mathematics and Physics

Approach

Split every subject into boys and girls, take the four differences, and look for the pair that matches.

x %

of a quantity

Step-by-step solution

1
Mathematics $= 20%$ of $1200 = 240$ ; Physics $= \frac{1}{4} \times 1200 = 300$ ; Statistics $= 320$ ; Chemistry $= 1200 - 240 - 300 - 320 = 340$
Why: Three of the four subject totals are given outright or as a share of $1200$ ; Chemistry is whatever is left.
$x %$ of a quantity
2
Mathematics: boys $150$ (given), girls $240 - 150 = 90$ , difference $60$
Why: Each subject's girls are its total less its boys.
3
Statistics: girls $250$ (given), boys $320 - 250 = 70$ , difference $180$
Why: Here it is the girls that are given and the boys that must be deduced.
4
Physics: boys $60%$ of $300 = 180$ , girls $300 - 180 = 120$ , difference $60$
Why: The $60%$ applies to the Physics enrolment of $300$ .
5
Chemistry: girls $\frac{3}{5} \times 340 = 204$ , boys $340 - 204 = 136$ , difference $68$
Why: Three-fifths of the Chemistry students are girls, and Chemistry is the $340$ left over.
6
Differences: $60$ , $180$ , $60$ , $68$ . The equal pair is Mathematics and Physics.
Why: Only two of the four differences coincide, so the answer is forced.

Common mistakes

Option A

Mathematics and Chemistry: $60$ against $68$ - a near miss that survives only if $\frac{3}{5} \times 340$ is slurred.

Option B

Statistics and Chemistry: $180$ against $68$ , nowhere near equal.

Option D

Mathematics and Statistics: $60$ against $180$ .

Skills used

x %

of a quantity

89Statistics • Data Interpretation

EasyIdeal time: about 1 min

Same directions as Q86

What is the difference between the number of boys studying Mathematics and the number of girls studying Physics?

20

30

60

80

Show answer & solution

Option B

30

Approach

Read off the Mathematics boys from the stem, compute the Physics girls, and subtract.

x %

of a quantity

Step-by-step solution

1
Mathematics $= 20%$ of $1200 = 240$ ; Physics $= \frac{1}{4} \times 1200 = 300$ ; Statistics $= 320$ ; Chemistry $= 1200 - 240 - 300 - 320 = 340$
Why: Three of the four subject totals are given outright or as a share of $1200$ ; Chemistry is whatever is left.
$x %$ of a quantity
2
Boys studying Mathematics $= 150$ (given)
Why: The stem states this outright.
3
Physics boys $= 60%$ of $300 = 180$ , so Physics girls $= 300 - 180 = 120$
Why: The girls are the remainder of the Physics enrolment once its boys are removed.
4
$150 - 120 = 30$
Why: The question asks for the difference between these two cells, one from each subject.

Common mistakes

Option A

$20$ : no pair of table entries differs by it.

Option C

$60$ is the Mathematics boys-minus-girls gap ( $150 - 90$ ), not the cross-subject difference asked for.

Option D

$80$ : no pair of table entries differs by it.

Skills used

x %

of a quantity

90Statistics • Data Interpretation

ModerateIdeal time: about 2 min

Same directions as Q86

What is the ratio of the total number of boys to the total number of girls?

67 : 83

17 : 26

27 : 19

189 : 179

Show answer & solution

Option A

67 : 83

Approach

Fill in all eight cells of the table, total each sex, and reduce the ratio.

x %

of a quantity

Step-by-step solution

1
Mathematics $= 20%$ of $1200 = 240$ ; Physics $= \frac{1}{4} \times 1200 = 300$ ; Statistics $= 320$ ; Chemistry $= 1200 - 240 - 300 - 320 = 340$
Why: Three of the four subject totals are given outright or as a share of $1200$ ; Chemistry is whatever is left.
$x %$ of a quantity
2
Boys: Mathematics $150$ , Statistics $320 - 250 = 70$ , Physics $60%$ of $300 = 180$ , Chemistry $\frac{2}{5} \times 340 = 136$
Why: Each subject's boys come from a different clue - a count, a complement, a percentage and a fraction.
3
Total boys $= 150 + 70 + 180 + 136 = 536$
Why: Sum the four cells of the boys' row.
4
Total girls $= 1200 - 536 = 664$
Why: Every student is a boy or a girl, so the girls are the complement of the boys in the total of $1200$ - quicker than adding $90 + 250 + 120 + 204$ .
5
$536 : 664 = \frac{536}{8} : \frac{664}{8} = 67 : 83$
Why: Divide both terms by their HCF $8$ to reach lowest terms, the form the options are printed in.
Prime factorisation

Shortcut

Boys $= 150 + 70 + 180 + 136 = 536$ , girls $= 1200 - 536 = 664$ , and dividing by $8$ gives $67 : 83$ .

Common mistakes

Option B

$17 : 26$ : no grouping of the table's totals reduces to it.

Option C

$27 : 19$ : no grouping of the table's totals reduces to it, and it has boys outnumbering girls, which is the wrong way round.

Option D

$189 : 179$ : no grouping of the table's totals reduces to it.

Skills used

x %

of a quantity Prime factorisation

91Statistics • Frequency Distribution • Frequency Density

Very easyIdeal time: about 25 s

Frequency density of a class is computed by the ratio

AClass frequency to the class widthBClass frequency to total frequencyCClass frequency to total number of classesDCumulative frequency up to that class to total frequency

Show answer & solution

Option A

Class frequency to the class width

Approach

Recall what frequency density is for: making histogram bars of unequal width comparable.

Frequency density

Step-by-step solution

1
Frequency density $= \frac{class frequency}{class width}$
Why: It is defined as the frequency per unit of the variable, which is what a histogram plots as the height of a bar.
Frequency density
2
$height \times width = \frac{f}{w} \times w = f$
Why: This is the property the definition exists to secure - the AREA of each bar equals that class's frequency - and it fails for any other ratio.

Common mistakes

Option B

Class frequency to total frequency is the RELATIVE frequency, a proportion of the whole rather than a rate per unit width.

Option C

Class frequency to the number of classes is not a standard measure at all.

Option D

Cumulative frequency to total frequency is the relative cumulative frequency, the quantity read off an ogive.

Skills used

Frequency density

92Statistics • Mean

EasyIdeal time: about 1 min

A small company pays each of its $5$ category 'C' workers Rs. $20, 000$ , each of its $3$ category 'B' workers Rs. $25, 000$ and a category 'A' worker Rs. $65, 000$ . The number of workers earning less than the mean salary is

8

5

4

3

Show answer & solution

Option A

8

Approach

Total the payroll, divide by the head count for the mean, then count the individual salaries that fall below it.

Mean of a frequency distribution

Step-by-step solution

1
Total $= 5 \times 20000 + 3 \times 25000 + 1 \times 65000 = 100000 + 75000 + 65000 = 240000$
Why: The salaries come with frequencies $5$ , $3$ and $1$ , so the total is $\sum f_{i} x_{i}$ .
Mean of a frequency distribution
2
Head count $= 5 + 3 + 1 = 9$
Why: $\sum f_{i}$ counts workers, not salary grades.
3
$x ˉ = \frac{240000}{9} = 26666. 6$
Why: The mean is the total divided by the number of workers.
4
$20000 < x ˉ$ and $25000 < x ˉ$ , so $5 + 3 = 8$ workers earn less than the mean.
Why: Only the single category 'A' salary of $65000$ exceeds the mean; one large value drags the mean above all eight of the others.

Shortcut

The mean is about $26667$ , above both $20000$ and $25000$ , so every worker except the single $65000$ earner is below it: $9 - 1 = 8$ .

Common mistakes

Option B

$5$ counts only the category 'C' workers, overlooking that $25000$ is also below $26666. 6$ .

Option C

$4$ : no route reaches it.

Option D

$3$ counts only the category 'B' workers.

Skills used

Mean of a frequency distribution

93Arithmetic • Time, Speed and Distance • Average Speed

EasyIdeal time: about 45 s

A man travelled $12$ km at a speed of $4$ km/hr and further $10$ km at a speed of $5$ km/hr. What was his average speed?

4.4

km/hrB

4.5

km/hrC

5.0

km/hrD

2.5

km/hr

Show answer & solution

Option A

4.4

km/hr

Approach

Average speed is total distance over total time - never the mean of the two speeds.

Speed, distance and time

Step-by-step solution

1
$t_{1} = \frac{12}{4} = 3$ hr and $t_{2} = \frac{10}{5} = 2$ hr
Why: Each leg's time comes from its own distance and speed.
Speed, distance and time
2
Total distance $= 12 + 10 = 22$ km; total time $= 3 + 2 = 5$ hr
Why: For a journey in stages the distances add and the times add.
3
Average speed $= \frac{22}{5} = 4.4$ km/hr
Why: The average speed is the single constant speed that would cover the whole distance in the whole time.

Common mistakes

Option B

$4.5$ km/hr is $\frac{4 + 5}{2}$ , the plain mean of the speeds - correct only if equal TIMES were spent at each, which they were not.

Option C

$5.0$ km/hr is the faster leg's speed alone.

Option D

$2.5$ km/hr: no route reaches it.

Skills used

Speed, distance and time

94Statistics • Data Interpretation • Pie Chart

ModerateIdeal time: about 1.5 min

The pie diagrams on the monthly expenditure of two families A and B are drawn with radii of two circles taken in the ratio $16 : 9$ to compare their expenditures. Which one of the following is the appropriate data used for the above mentioned pie diagrams?

ARs.

16, 000

and Rs.

9, 000

BRs.

8, 000

and Rs.

4, 500

CRs.

25, 600

and Rs.

8, 100

DRs.

4, 000

and Rs.

3, 000

Show answer & solution

Option C

Rs.

25, 600

and Rs.

8, 100

Approach

A pie diagram represents its total by the AREA of the circle, so the two totals must be in the ratio of the squares of the radii.

Area and circumference of a circle

Step-by-step solution

1
$\frac{Total _{A}}{Total _{B}} = \frac{π r _{A}^{2}}{π r _{B}^{2}} = (\frac{r _{A}}{r _{B}})^{2} = (\frac{16}{9})^{2} = \frac{256}{81}$
Why: The whole circle stands for the whole expenditure, and what a circle displays is its area, which varies as the square of the radius.
Area and circumference of a circle Direct variation as a power: $y \propto x^{n}$
2
Required expenditure ratio $= 256 : 81$
Why: This is the ratio the data must have for radii of $16 : 9$ to be the appropriate choice.
3
$25600 : 8100 = 256 : 81$ after dividing both terms by $100$
Why: Only this pair of amounts reduces to the square of the radius ratio.

Shortcut

Squares of $16$ and $9$ are $256$ and $81$ ; only Rs. $25, 600$ and Rs. $8, 100$ are in that ratio.

Common mistakes

Option A

Rs. $16, 000$ and Rs. $9, 000$ are in the ratio of the RADII themselves, as if the radius rather than the area carried the quantity.

Option B

Rs. $8, 000$ and Rs. $4, 500$ is again $16 : 9$ , the same error with smaller figures.

Option D

Rs. $4, 000$ and Rs. $3, 000$ is $4 : 3$ , the ratio of the square ROOTS of $16 : 9$ - the squaring went the wrong way.

Skills used

Area and circumference of a circle Direct variation as a power:

y \propto x^{n}

95Statistics • Mode

EasyIdeal time: about 1 min

Consider the following statements:

Statement I: The value of a random variable having the highest frequency is mode.

Statement II: Mode is unique.

Which one of the following is correct in respect of the above statements?

ABoth Statement I and Statement II are true and Statement II is the correct explanation of Statement IBBoth Statement I and Statement II are true but Statement II is not the correct explanation of Statement ICStatement I is true but Statement II is falseDStatement I is false but Statement II is true

Show answer & solution

Option C

Statement I is true but Statement II is false

Approach

Check each statement against the definition of the mode, and only then ask whether the second explains the first.

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

Step-by-step solution

1
Statement I: the mode is the value of the variable occurring with the greatest frequency. TRUE.
Why: That is the definition of the mode, and 'random variable' here is just the variable being observed.
Mode: the most frequent value, and it need not be unique
2
Statement II: in $1, 1, 2, 2, 3$ both $1$ and $2$ occur twice, so the data are bimodal and the mode is not unique. FALSE.
Why: A data set may have two or more modes, or none at all when every value occurs equally often; uniqueness is exactly the property the mode lacks and the mean possesses.
3
Statement I is true and Statement II is false, so no question of explanation arises.
Why: The four options pair the two verdicts; only one pairs 'true' with 'false' in this order.

Common mistakes

Option A

Both true, II explaining I: accepts uniqueness, which any bimodal data set disproves.

Option B

Both true, II not explaining I: the same error about uniqueness, differing only on the explanation.

Option D

I false, II true: rejects the standard definition of the mode while keeping the false claim.

Skills used

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

96Statistics • Data Interpretation • Pie Chart

ModerateIdeal time: about 1.5 min

Which one of the following is not correct? The proportion of various items in a pie diagram is proportional to the

AAreas of slicesBAngles of slicesCLengths of the curved arcs of the slicesDPerimeters of the slices

Show answer & solution

Option D

Perimeters of the slices

Approach

Write down what each listed measurement of a slice actually equals, and find the one that is not a constant multiple of the share.

A pie chart sector's angle is proportional to its share

Step-by-step solution

1
Share of a component $= \frac{θ}{36 0 ^{\circ}}$ , so the central angle is proportional to the component.
Why: This is how a pie diagram is drawn in the first place: the angle is allotted in proportion.
A pie chart sector's angle is proportional to its share
2
Sector area $= \frac{θ}{36 0 ^{\circ}} π r^{2}$ and arc length $= \frac{θ}{36 0 ^{\circ}} 2 π r$ ; with $r$ the same for every slice, both are constant multiples of $θ$ .
Why: Anything of the form (constant) $\times θ$ is proportional to $θ$ , hence to the share.
Area and arc length of a circular sector
3
Perimeter of a slice $= 2 r + \frac{θ}{36 0 ^{\circ}} 2 π r$ , which is a multiple of $θ$ PLUS a fixed $2 r$ .
Why: The two straight radii add the same $2 r$ to every slice no matter how small it is, and a quantity of the form $a θ + b$ with $b \neq = 0$ is not proportional to $θ$ .
4
So the perimeters are NOT proportional to the components: option D is the incorrect statement.
Why: The question asks which of the four is not correct, so the odd one out is the answer.

Common mistakes

Option A

Areas of slices ARE proportional; the question asks for the statement that is not correct.

Option B

Angles of slices ARE proportional, by the construction of the diagram.

Option C

Arc lengths ARE proportional, being $r θ$ with $r$ fixed.

Skills used

A pie chart sector's angle is proportional to its share Area and arc length of a circular sector

97Statistics • Mean • Geometric Mean

EasyIdeal time: about 55 s

The geometric mean of x and y is $6$ and the geometric mean of x, y and z is also $6$ . Then the value of z is

12

6

6

36

Show answer & solution

Option C

6

Approach

Turn each geometric mean into a product equation and divide one by the other.

Geometric mean: the

n

th root of the product

Step-by-step solution

1
$x y = 6 \Rightarrow x y = 36$
Why: The geometric mean of two numbers is the square root of their product, so squaring clears the root.
Geometric mean: the $n$ th root of the product
2
$(x y z)^{1/3} = 6 \Rightarrow x y z = 6^{3} = 216$
Why: The geometric mean of $n$ numbers is the $n$ th root of their product; cubing clears the cube root.
3
$36 z = 216 \Rightarrow z = \frac{216}{36} = 6$
Why: Substitute $x y = 36$ into the second product; neither $x$ nor $y$ needs to be found separately.

Shortcut

Adjoining a value equal to the geometric mean cannot change it, so $z = 6$ .

Common mistakes

Option A

$12$ adds the two given means, $6 + 6$ , instead of dividing the products.

Option B

$6$ : no route reaches it.

Option D

$36$ takes a cube root of $6$ rather than solving $36 z = 216$ .

Skills used

Geometric mean: the

n

th root of the product

98Statistics • Data Classification

Very easyIdeal time: about 30 s

The total number of live births in a specific locality during different months of a specific year was obtained from the office of the Birth Registrar. This set of data may be called

APrimary dataBSecondary dataCRecorded dataDCountable data

Show answer & solution

Option B

Secondary data

Approach

Ask who collected the figures and why: the user here is borrowing records compiled by somebody else for another purpose.

Primary and secondary data

Step-by-step solution

1
The birth counts were compiled by the Birth Registrar's office, as part of its own registration work.
Why: Primary data are those the investigator collects first-hand for the enquiry in hand; these were not.
Primary and secondary data
2
The investigator has obtained them from that existing record, so for this investigator they are secondary data.
Why: The classification is relative to the user, not absolute: the same figures are primary to the Registrar who recorded them and secondary to anyone who takes them from the register.

Common mistakes

Option A

Primary data would require the investigator to have collected the births first-hand, rather than obtaining them from the Registrar's office.

Option C

'Recorded data' is not a term in the primary/secondary classification.

Option D

'Countable data' describes the variable - births are discrete - and says nothing about the source.

Skills used

Primary and secondary data

99Statistics • Mean

EasyIdeal time: about 55 s

The heights (in cm) of $5$ students are $150, 165, 161, 144$ and $155$ . What are the values of mean and median (in cm) respectively?

165

and

161

155

and

155

160

and

155

155

and

161

Show answer & solution

Option B

155

and

155

Approach

Add and divide for the mean; sort the list and take the middle value for the median.

Mean of a frequency distribution

Step-by-step solution

1
$150 + 165 + 161 + 144 + 155 = 775$
Why: The mean needs the total of all five heights.
Mean of a frequency distribution
2
Mean $= \frac{775}{5} = 155$ cm
Why: Divide the total by the number of observations.
3
Sorted: $144, 150, 155, 161, 165$
Why: The median is defined on ORDERED data, and the heights as given are not in order.
Median of a data set
4
$n = 5$ is odd, so the median is the $\frac{5 + 1}{2} = 3$ rd value, namely $155$ cm.
Why: For an odd number of observations the median is a single middle value, not an average of two.

Common mistakes

Option A

$165$ and $161$ are simply the largest and second largest heights, not averages at all.

Option C

$160$ and $155$ follows a mis-addition of the five heights (a total of $800$ rather than $775$ ).

Option D

$155$ and $161$ takes the median from the UNSORTED list, where $161$ happens to sit third.

Skills used

Mean of a frequency distribution Median of a data set

100Statistics • Mean • Weighted Mean

EasyIdeal time: about 1 min

The average height of $22$ students of a class is $140$ cm and the average height of $28$ students of another class is $152$ cm. What is the average height of students of both the classes?

144.32

cmB

145.52

cmC

146.72

cmD

147.92

Show answer & solution

Option C

146.72

Approach

Pool the two classes: combine the totals, not the two means.

Mean of a frequency distribution

Step-by-step solution

1
Total height of the first class $= 22 \times 140 = 3080$ cm
Why: A mean times its head count recovers the group total.
Mean of a frequency distribution
2
Total height of the second class $= 28 \times 152 = 4256$ cm
Why: Same move for the second class.
3
Combined mean $= \frac{3080 + 4256}{22 + 28} = \frac{7336}{50} = 146.72$ cm
Why: The mean of a pooled group is the pooled total over the pooled count - a weighted mean, not the average of the two means.

Shortcut

$140 + \frac{28}{50} \times 12 = 140 + 6.72 = 146.72$ cm.

Common mistakes

Option A

$144.32$ : no weighting of $140$ and $152$ produces it.

Option B

$145.52$ : no weighting of $140$ and $152$ produces it.

Option D

$147.92$ : no weighting of $140$ and $152$ produces it.

Skills used

Mean of a frequency distribution