Elementary Mathematics 2019 (I)

Set A

Questions 61–80 of 100

61Statistics • Mean

ModerateIdeal time: about 2 min

Consider the following grouped frequency distribution:

x	f
0 - 10	8
10 - 20	12
20 - 30	10
30 - 40	p
40 - 50	9

If the mean of the above data is $25.2$ , then what is the value of $p$ ?

9

10

11

12

Show answer & solution

Option C

11

Approach

Write the mean using class midpoints, set it equal to $25.2$ and solve the resulting linear equation for $p$ .

Mean of a frequency distribution

Step-by-step solution

1
Midpoints $5, 15, 25, 35, 45$ with frequencies $8, 12, 10, p, 9$
Why: For grouped data the mean uses the midpoint of each class, never its limits.
Mean of a frequency distribution
2
$\sum f = 8 + 12 + 10 + p + 9 = 39 + p$
Why: The divisor is the total frequency, which still carries the unknown — not the number of classes.
3
$\sum f x = 40 + 180 + 250 + 35 p + 405 = 875 + 35 p$
Why: Multiply each midpoint by its frequency and add, keeping the $p$ term symbolic.
4
$\frac{875 + 35 p}{39 + p} = 25.2 \Rightarrow 875 + 35 p = 982.8 + 25.2 p$
Why: Cross-multiply to clear the fraction; $25.2 \times 39 = 982.8$ .
Clear the denominators of an equation
5
$9.8 p = 107.8 \Rightarrow p = 11$
Why: Collect the $p$ terms and divide: $\frac{107.8}{9.8} = 11$ .

Common mistakes

Option A

No route reaches $9$ ; it is the last printed frequency, repeated.

Option B

Correct equation, but $\frac{107.8}{9.8}$ was mis-divided.

Option D

Correct equation, but $\frac{107.8}{9.8}$ was mis-divided in the other direction.

Skills used

Mean of a frequency distribution Clear the denominators of an equation

62Statistics • Median

ModerateIdeal time: about 1.5 min

Consider the following frequency distribution:

x	f
8	6
5	4
6	5
10	8
9	9
4	6
7	4

What is the median for the distribution?

6

7

8

9

Show answer & solution

Option C

8

Approach

Order the values, build the cumulative frequencies, and read off the value sitting at the two middle positions.

Median of a data set

Step-by-step solution

1
$\sum f = 6 + 4 + 5 + 8 + 9 + 6 + 4 = 42$
Why: The median's position depends on how many observations there are, not on how many distinct values the table lists.
Cumulative frequency
2
Ordered values with cumulative frequencies: $4 (6)$ , $5 (10)$ , $6 (15)$ , $7 (19)$ , $8 (25)$ , $9 (34)$ , $10 (42)$
Why: The table is printed out of order; ordering it is the step most often skipped and the one the wrong options punish.
3
$n = 42$ is even, so the median is the mean of the $21$ st and $22$ nd observations
Why: With an even count there is no single middle observation.
Median of a data set
4
The cumulative frequency jumps from $19$ to $25$ at $x = 8$ , so positions $20$ to $25$ all carry the value $8$ ; both the $21$ st and the $22$ nd are $8$ , and the median is $8$
Why: Locating the first cumulative total that reaches the middle position is how a median is read off a frequency table.

Common mistakes

Option A

Took the median of the frequency column instead of the data: the frequencies in order are $4, 4, 5, 6, 6, 8, 9$ , whose middle value is $6$ .

Option B

Ignored the frequencies and took the median of the seven distinct values $4, 5, 6, 7, 8, 9, 10$ , which is $7$ .

Option D

Gave the mode: $9$ carries the largest frequency, but the question asks for the median.

Skills used

Median of a data set Cumulative frequency

63Statistics • Mean

EasyIdeal time: about 1 min

The average of $50$ consecutive natural numbers is $x$ . What will be the new average when the next four natural numbers are also included?

x + 1

x + 2

x + 4

x + (x /54)

Show answer & solution

Option B

x + 2

Approach

An evenly spaced run averages to the midpoint of its first and last terms, so write both averages that way and subtract.

An evenly spaced run averages to its middle term

Step-by-step solution

1
Let the numbers be $a, a + 1, \dots, a + 49$
Why: Consecutive natural numbers form an arithmetic progression with common difference $1$ .
An evenly spaced run averages to its middle term
2
$x = \frac{a + ( a + 49 )}{2} = a + 24.5$
Why: For an evenly spaced run the mean is the average of the first and last terms.
3
Adding the next four gives $54$ numbers $a, \dots, a + 53$ , with mean $\frac{a + ( a + 53 )}{2} = a + 26.5$
Why: The longer run is still evenly spaced, so the same rule applies to it unchanged.
4
New mean $-$ old mean $= 26.5 - 24.5 = 2$ , so the new average is $x + 2$
Why: Both means are offsets from the same $a$ , so $a$ cancels and the answer does not depend on where the run starts.

Shortcut

Extending an evenly spaced run by four terms at the top moves its midpoint up by $\frac{4}{2} = 2$ steps, i.e. by $2$ .

Common mistakes

Option A

Took the shift as $1$ , as though the mean moved up one place rather than by half the four terms added.

Option C

Added the number of extra terms, $4$ , straight on to the mean.

Option D

Manipulated the totals — new mean as $\frac{50 x + something}{54}$ — without using the even spacing, which is what makes the answer independent of $x$ .

Skills used

An evenly spaced run averages to its middle term

64Statistics • Mean

EasyIdeal time: about 55 s

Consider two-digit numbers which remain the same when the digits interchange their positions. What is the average of such two-digit numbers?

33

44

55

66

Show answer & solution

Option C

55

Approach

Find which two-digit numbers survive reversing, then average the list.

Place-value expansion of a numeral

Step-by-step solution

1
A two-digit number with digits $a$ and $b$ is $10 a + b$ ; reversed it is $10 b + a$
Why: Place-value expansion turns a condition about digits into an ordinary equation.
Place-value expansion of a numeral
2
$10 a + b = 10 b + a \Rightarrow 9 a = 9 b \Rightarrow a = b$
Why: The numeral is unchanged by interchanging its digits exactly when the two digits are equal.
3
With $a = b$ and $a \neq = 0$ the numbers are $11, 22, 33, \dots, 99$ — nine of them
Why: The leading digit of a two-digit number cannot be $0$ , so $00$ is not in the list.
4
They are evenly spaced by $11$ , so the mean is $\frac{11 + 99}{2} = 55$
Why: An evenly spaced run averages to the midpoint of its ends, which here is also the middle term.
An evenly spaced run averages to its middle term

Common mistakes

Option A

Averaged a truncated list, $11$ to $55$ , whose midpoint is $33$ .

Option B

Stopped the list at $77$ , giving the midpoint $\frac{11 + 77}{2} = 44$ .

Option D

Started the list at $33$ , giving the midpoint $\frac{33 + 99}{2} = 66$ .

Skills used

Place-value expansion of a numeral An evenly spaced run averages to its middle term

65Statistics • Data Classification

EasyIdeal time: about 40 s

Diagrammatic representation of data includes which of the following?

1: Bar diagram

2: Pie-diagram

3: Pictogram

Select the correct answer using the code given below:

A1 and 2 onlyB2 and 3 onlyC1 and 3 onlyD1, 2 and 3

Show answer & solution

Option D

1, 2 and 3

Approach

Separate diagrams from graphs: all three devices named are diagrams.

Diagrammatic and graphic presentation of data

Step-by-step solution

1
A bar diagram, a pie-diagram and a pictogram each show the figures as a picture — by bar lengths, by sector angles, and by repeated symbols respectively
Why: Diagrammatic presentation means the data are carried by a picture whose sizes stand for the figures, rather than plotted against a measured scale.
Diagrammatic and graphic presentation of data
2
Graphic presentation is the separate category — histogram, frequency polygon, ogive — and none of those is offered here
Why: Checking what is absent confirms the reading: the question is testing the diagram-versus-graph split, and every listed item falls on the diagram side.
3
So $1$ , $2$ and $3$ all belong
Why: Nothing has to be excluded, which is what makes the all-three code correct.

Common mistakes

Option A

Excluded the pictogram, treating a picture made of repeated symbols as something other than a diagram.

Option B

Excluded the bar diagram, on the view that bars belong with histograms and so count as a graph.

Option C

Excluded the pie-diagram.

Skills used

Diagrammatic and graphic presentation of data

66Statistics • Data Classification

EasyIdeal time: about 35 s

The data collected from which one of the following methods is not a primary data?

ABy direct personal interviewsBBy indirect personal interviewsCBy schedules sent through enumeratorsDFrom published thesis

Show answer & solution

Option D

From published thesis

Approach

Primary data are collected first-hand for the enquiry in hand; look for the option where the figures are read off somebody else's record.

Primary and secondary data

Step-by-step solution

1
Direct personal interviews, indirect personal interviews and schedules sent through enumerators are all ways the investigator gathers figures for this enquiry
Why: In all three the data are being created for the investigation in hand, whoever physically puts the questions — an agent or an enumerator is still collecting on the investigator's behalf.
Primary and secondary data
2
A published thesis is an existing record compiled by somebody else for their own purpose, so figures taken from it are secondary
Why: The classification turns on who collected the data and why, not on how recent or how reliable they are.
3
So the published thesis is the method that does not yield primary data
Why: The question asks which one is not primary, so the single secondary source is the answer.

Common mistakes

Option A

Answered the reverse question and picked the most obviously first-hand method.

Option B

Read 'indirect' as meaning second-hand; it means the investigator questions witnesses or third parties rather than the informant, but the collection is still the investigator's own.

Option C

Read 'through enumerators' as somebody else's data, when enumerators are the investigator's own field staff.

Skills used

Primary and secondary data

67Statistics • Data Interpretation • Bar Chart

ModerateIdeal time: about 1.5 min

The monthly expenditure of a person is Rs. $6, 000$ . The distribution of expenditure on various items is as follows:

Item of expenditure	Amount (in Rs.)
1. Food	2,000
2. Clothing	660
3. Fuel and rent	1,200
4. Education	480
5. Miscellaneous	1,660

If the above data is represented by a percentage bar diagram of height $15 cm$ , then what are the lengths of the two segments of the bar diagram corresponding to education and miscellaneous respectively?

1.25 cm

and

5 cm

1.2 cm

and

4.15 cm

1.2 cm

and

3.5 cm

4.15 cm

and

6 cm

Show answer & solution

Option B

1.2 cm

and

4.15 cm

Approach

In a percentage bar diagram the whole bar stands for the whole expenditure, so each segment's length is that item's share of the $15 cm$ .

Unitary method: find one unit, then scale

Step-by-step solution

1
Education's share $= \frac{480}{6000} = 0.08$ , i.e. $8%$
Why: A percentage bar diagram divides one bar in proportion to the components, so each component's share of the total is the first thing needed.
$x %$ of a quantity
2
Length for education $= 0.08 \times 15 = 1.2 cm$
Why: The full $15 cm$ represents $100%$ , so a share of $8%$ takes up $8%$ of the height.
Unitary method: find one unit, then scale
3
Miscellaneous: $\frac{1660}{6000} \times 15 = \frac{1660}{400} = 4.15 cm$
Why: The same calculation; cancelling $\frac{15}{6000}$ to $\frac{1}{400}$ first keeps the arithmetic to one short division.
4
Education $1.2 cm$ and miscellaneous $4.15 cm$ , in that order
Why: The question names education first, so the pair must be reported that way round.

Shortcut

Every rupee of expenditure is $\frac{15}{6000} = 0.0025 cm$ of bar, so multiply each amount by $0.0025$ .

Common mistakes

Option A

Read the Food row for miscellaneous, giving $\frac{2000}{6000} \times 15 = 5 cm$ , and rounded education's $480$ to $500$ to get $1.25 cm$ .

Option C

Education right, but miscellaneous mis-divided: $3.5 cm$ corresponds to Rs. $1, 400$ , not Rs. $1, 660$ .

Option D

Gave the two lengths the other way round, and used $6 cm$ for education.

Skills used

Unitary method: find one unit, then scale

x %

of a quantity

68Statistics • Mean • Weighted Mean

ModerateIdeal time: about 1.5 min

If the mean of $m$ observations out of $n$ observations is $n$ and the mean of remaining observations is $m$ , then what is the mean of all $n$ observations?

2 m - \frac{m ^{2}}{n}

2 m + \frac{m ^{2}}{n}

m - \frac{m ^{2}}{n}

m + \frac{m ^{2}}{n}

Show answer & solution

Option A

2 m - \frac{m ^{2}}{n}

Approach

Add the two group totals and divide by $n$ — means never add, totals always do.

Mean of a frequency distribution

Step-by-step solution

1
First group: $m$ observations with mean $n$ , so its total is $mn$
Why: Total $=$ mean $\times$ number of observations, which is the only way two groups can be combined.
Mean of a frequency distribution
2
Second group: $n - m$ observations with mean $m$ , so its total is $m (n - m)$
Why: The remaining count is $n - m$ , not $m$ — this is where the question tries to trip you.
3
Grand total $= mn + m (n - m) = mn + mn - m^{2} = 2 mn - m^{2}$
Why: Two means can only be averaged directly when the groups are the same size, so go through the totals.
4
Mean of all $n$ observations $= \frac{2 mn - m ^{2}}{n} = 2 m - \frac{m ^{2}}{n}$
Why: Divide by the full count $n$ and split the fraction into the two printed terms.

Common mistakes

Option B

Right magnitude but the sign of the $\frac{m ^{2}}{n}$ term flipped when $mn + mn - m^{2}$ was collected.

Option C

Divided only the second group's total $m (n - m)$ by $n$ , leaving out the first group's $mn$ altogether.

Option D

Took the remaining group to have $m$ observations rather than $n - m$ , giving a grand total of $mn + m^{2}$ .

Skills used

Mean of a frequency distribution

69Statistics • Median

EasyIdeal time: about 45 s

Which one of the following pairs is correctly matched?

AMedian - Graphical locationBMean - Graphical locationCGeometric mean - OgiveDMode - Ogive

Show answer & solution

Option A

Median - Graphical location

Approach

Ask of each average whether it can be located on a graph, and if so on which graph.

Locating the median graphically from an ogive

Step-by-step solution

1
The median can be located graphically: draw the 'less than' and 'more than' cumulative frequency curves and drop a perpendicular from their intersection to the $x$ -axis
Why: The median is a positional average, and a cumulative frequency curve is precisely a plot of position against value — which is why a graph can find it.
Locating the median graphically from an ogive Cumulative frequency Median of a data set
2
The mean is computed from every observation, so no point on any graph marks it
Why: That rules out option B: a graphical location is not a property the mean has.
Mean of a frequency distribution
3
The mode does have a graphical location, but from the histogram's modal class, not from an ogive; and the geometric mean has no graphical construction at all
Why: Both remaining options pair an average with the wrong graph, which is the standard way this matching question is set.
Mode: the most frequent value, and it need not be unique Geometric mean: the $n$ th root of the product
4
Only 'Median $-$ Graphical location' is correctly matched
Why: It is the one pair where the average and the method genuinely belong together.

Common mistakes

Option B

Confused the mean with the median; it is the median that the cumulative frequency curves locate.

Option C

No graphical construction gives the geometric mean, and an ogive gives the median.

Option D

The mode is found graphically, but from the histogram, not from an ogive.

Skills used

Locating the median graphically from an ogive Cumulative frequency Median of a data set Mean of a frequency distribution Mode: the most frequent value, and it need not be unique Geometric mean: the

n

th root of the product

70Statistics • Frequency Distribution

ModerateIdeal time: about 1 min

The following pairs relate to frequency distribution of a discrete variable and its frequency polygon. Which one of the following pairs is not correctly matched?

ABase line of the polygon - X-axisBOrdinates of the vertices of the polygon - Class frequenciesCAbscissa of the vertices of the polygon - Class marks of the frequency distributionDArea of the polygon - Total frequency of the distribution

Show answer & solution

Option D

Area of the polygon - Total frequency of the distribution

Approach

Check each pair against how a frequency polygon is actually constructed; the one that fails is the pair about area.

Frequency polygon and the frequency curve it tends to

Step-by-step solution

1
The polygon is closed by bringing both ends down to the horizontal axis, so its base line is the $X$ -axis — pair A is correctly matched
Why: The horizontal axis carries the variable, and the polygon rests on it once the ends are brought down to zero frequency.
Frequency polygon and the frequency curve it tends to
2
Each vertex is plotted at a height equal to that class's frequency — pair B is correctly matched
Why: A frequency polygon plots the frequency itself; it is the ogive that plots the cumulative frequency.
3
Each vertex sits horizontally at the class mark, i.e. the mid-value — pair C is correctly matched
Why: Plotting against mid-points rather than class limits is exactly what distinguishes a frequency polygon from an ogive.
4
The area enclosed by the polygon equals the area of the corresponding histogram, which is $\sum f \times (class width)$ — equal to the total frequency only when the width happens to be $1$ . So pair D is not correctly matched
Why: An area here has the dimensions of variable $\times$ frequency, so it cannot in general be a bare frequency count; that mismatch of dimensions is the quickest way to see which pair is wrong.

Common mistakes

Option A

No route makes the base line anything but the $X$ -axis.

Option B

Thought the vertices carry cumulative frequencies, confusing the polygon with an ogive.

Option C

Objected to the phrase 'class marks' on the ground that a discrete variable has no class intervals; for a discrete distribution the values are the class marks, so the pair stands.

Skills used

Frequency polygon and the frequency curve it tends to

71Mensuration • Plane Figures • Quadrilateral

EasyIdeal time: about 1 min

In a rectangle, length is three times its breadth. If the length and the breadth of the rectangle are increased by $30%$ and $10%$ respectively, then its perimeter increases by

\frac{40}{3} %

20%

25%

27%

Show answer & solution

Option C

25%

Approach

Put the breadth at $b$ , write the old and the new perimeter, and compare them.

Percentage change as a multiplier

Step-by-step solution

1
breadth $= b$ , length $= 3 b$ , so $P = 2 (3 b + b) = 8 b$
Why: A ratio fixes the two sides only up to a common factor, so one letter carries the whole rectangle and it will cancel at the end.
Introduce a common factor $k$ for a ratio Perimeter and area of a rectangle
2
new length $= 1.3 \times 3 b = 3.9 b$ , new breadth $= 1.1 b$
Why: An increase of $p %$ multiplies a quantity by $(1 + \frac{p}{100})$ , so the two sides take different multipliers.
Percentage change as a multiplier
3
$P^{'} = 2 (3.9 b + 1.1 b) = 2 (5 b) = 10 b$
Why: The new perimeter comes from the same formula applied to the new sides.
4
increase $= \frac{10 b - 8 b}{8 b} = \frac{1}{4} = 25%$
Why: A percentage rise is always measured against the original value, here the original perimeter $8 b$ .

Shortcut

The perimeter depends only on $l + b$ , so the percentage rise in the perimeter is the percentage rise in $l + b$ . Length is $\frac{3}{4}$ of $l + b$ and breadth $\frac{1}{4}$ , so the answer is the weighted average $30 \times \frac{3}{4} + 10 \times \frac{1}{4} = 25$ .

Common mistakes

Option A

Divided the total of the two percentages by $3$ , the ratio of the sides, instead of weighting them by it: $\frac{30 + 10}{3} = \frac{40}{3}$ .

Option B

Took the plain average of $30%$ and $10%$ , which would be right only if length and breadth contributed equally to the perimeter.

Option D

No route reaches $27%$ .

Skills used

Introduce a common factor

k

for a ratio Perimeter and area of a rectangle Percentage change as a multiplier

72Mensuration • Plane Figures • Triangle

EasyIdeal time: about 45 s

What is the percentage decrease in the area of a triangle if its each side is halved?

75%

50%

25%

DNo change

Show answer & solution

Option A

75%

Approach

Halving every side gives a similar triangle, and areas of similar figures go as the square of the length ratio.

Step-by-step solution

1
the new triangle has all three sides half the old ones, so it is similar to it with ratio $k = \frac{1}{2}$
Why: Three pairs of sides in a constant ratio is SSS similarity; no angle information is needed.
Similar triangles have proportional sides
2
$\frac{new area}{old area} = k^{2} = (\frac{1}{2})^{2} = \frac{1}{4}$
Why: Area is a two-dimensional measure, so it scales by $k^{2}$ while lengths scale by $k$ .
3
decrease $= 1 - \frac{1}{4} = \frac{3}{4} = 75%$
Why: The question asks what was lost, not what is left.
Percentage change as a multiplier

Common mistakes

Option B

Assumed the area falls in the same ratio as the sides, so halving the sides halves the area.

Option C

Found that the new area is $25%$ of the old and reported that figure as the decrease.

Option D

No route reaches 'no change'.

Skills used

73Mensuration • Solids • Sphere

EasyIdeal time: about 1 min

The volume of a spherical balloon is increased by $700%$ . What is the percentage increase in its surface area?

300%

400%

450%

500%

Show answer & solution

Option A

300%

Approach

Turn the $700%$ rise into a volume factor of $8$ , take the cube root for the radius factor, then square it for the surface.

Percentage change as a multiplier

Step-by-step solution

1
$V^{'} = V (1 + \frac{700}{100}) = 8 V$
Why: 'Increased by $700%$ ' means the original plus seven more copies of it, so the factor is $8$ and not $7$ .
Percentage change as a multiplier
2
$\frac{4}{3} π r^{'3} = 8 \times \frac{4}{3} π r^{3} \Rightarrow r^{'3} = 8 r^{3} \Rightarrow r^{'} = 2 r$
Why: The constants cancel, so the volume factor passes straight onto the cube of the radius.
Volume and surface area of a sphere
3
$\frac{4 π r ^{'2}}{4 π r ^{2}} = (\frac{r ^{'}}{r})^{2} = 2^{2} = 4$
Why: Surface area goes as the square of the radius, so the radius factor of $2$ becomes a surface factor of $4$ .
4
increase $= 4 - 1 = 3 = 300%$
Why: A factor of $4$ is a rise of three times the original, not of four.

Shortcut

Volume scales as the cube of the radius and surface as the square, so the surface factor is the volume factor to the power $\frac{2}{3}$ : $8^{2/3} = 4$ , a rise of $300%$ .

Common mistakes

Option B

Reported the surface factor of $4$ as the increase; being four times as large is a rise of $300%$ .

Option C

No route reaches $450%$ .

Option D

No route reaches $500%$ .

Skills used

Percentage change as a multiplier Volume and surface area of a sphere

74Geometry • Circle • Chords

EasyIdeal time: about 1 min

If the lengths of two parallel chords in a circle of radius $10 cm$ are $12 cm$ and $16 cm$ , then what is the distance between these two chords?

1 cm

7 cm

2 cm

14 cm

3 cm

21 cm

4 cm

28 cm

Show answer & solution

Option B

2 cm

14 cm

Approach

Use $d^{2} + m^{2} = r^{2}$ on each chord to get its distance from the centre, then add or subtract those distances according to which side of the centre the chords lie.

The perpendicular from the centre bisects a chord

Step-by-step solution

1
for the $12$ cm chord, $m_{1} = 6$ and $d_{1} = 1 0^{2} - 6^{2} = 64 = 8$ cm
Why: The perpendicular from the centre bisects the chord, so half the chord, the distance from the centre and the radius form a right triangle.
The perpendicular from the centre bisects a chord Pythagoras' theorem
2
for the $16$ cm chord, $m_{2} = 8$ and $d_{2} = 1 0^{2} - 8^{2} = 36 = 6$ cm
Why: The same right triangle, with the longer chord coming out nearer the centre as it must.
3
chords on opposite sides of the centre: distance $= d_{1} + d_{2} = 8 + 6 = 14$ cm
Why: Both perpendiculars lie along one line through the centre, and with the centre between the chords the two distances add.
4
chords on the same side of the centre: distance $= d_{1} - d_{2} = 8 - 6 = 2$ cm
Why: With both chords on one side, the nearer one sits between the centre and the farther one, so the distances subtract.
5
both arrangements are drawable, so the distance is $2$ cm or $14$ cm
Why: Nothing in the question says which side of the centre each chord is on, so neither case may be dropped.

Common mistakes

Option A

Halved both distances to $4$ and $3$ , as though the perpendicular from the centre bisected the distance as well as the chord.

Option C

No route reaches distances of $12$ and $9$ .

Option D

Doubled both distances to $16$ and $12$ , treating the computed $d$ as a half-distance.

Skills used

The perpendicular from the centre bisects a chord Pythagoras' theorem

75Mensuration • Plane Figures • Quadrilateral

HardIdeal time: about 2.5 min

Considering two opposite vertices of a square of side ' $a$ ' as centres, two circular arcs are drawn within the square joining the other two vertices, thus forming two sectors. What is the common area in these two sectors?

a^{2} (π + \frac{1}{2})

a^{2} (π - \frac{1}{2})

a^{2} (\frac{π}{2} - 1)

a^{2} (\frac{π}{2} + 1)

Show answer & solution

Option C

a^{2} (\frac{π}{2} - 1)

Approach

The two quarter-circles together cover the square exactly, so the overlap is (sum of the two sectors) minus (area of the square).

Inclusion-exclusion for two sets

Step-by-step solution

1
name the square $A B C D$ of side $a$ with the two centres at the opposite vertices $A$ and $C$ ; each arc joins $B$ to $D$ , so each sector is a quarter circle of radius $a$
Why: $B$ and $D$ are both at distance $a$ from $A$ , so the arc through them centred at $A$ has radius $a$ , and the angle at $A$ is the square's right angle.
Perimeter and area of a square
2
area of each sector $= \frac{90}{360} \times π a^{2} = \frac{π a ^{2}}{4}$
Why: A sector takes the same fraction of the circle as its angle does of a full turn, and $9 0^{\circ}$ is a quarter.
Area and arc length of a circular sector
3
both arcs run from $B$ to $D$ , and the arc centred at $C$ lies nearer $A$ than the arc centred at $A$ does, so the two arcs cut the square into exactly three pieces: sector $A$ only, the common part, and sector $C$ only
Why: There is no fourth piece, so no point of the square escapes both sectors — that is what makes the two sector areas comparable with the square's area.
Cut a compound plane figure into standard pieces
4
$\frac{π a ^{2}}{4} + \frac{π a ^{2}}{4} - X = a^{2} \Rightarrow X = \frac{π a ^{2}}{2} - a^{2} = a^{2} (\frac{π}{2} - 1)$
Why: Adding the two sectors counts the common part twice while covering the square once, so subtracting the square's area leaves exactly that common part.
Inclusion-exclusion for two sets

Common mistakes

Option A

No route reaches $a^{2} (π + \frac{1}{2})$ , which is larger than the square itself.

Option B

No route reaches $a^{2} (π - \frac{1}{2})$ , which is still larger than the square.

Option D

Added the square's area instead of subtracting it when undoing the double count.

Skills used

Perimeter and area of a square Area and arc length of a circular sector Cut a compound plane figure into standard pieces Inclusion-exclusion for two sets

76Geometry • Polygon

ModerateIdeal time: about 1.5 min

The corners of a square of side 'a' are cut away so as to form a regular octagon. What is the side of the octagon?

a (2 - 1)

a (3 - 1)

\frac{a}{2 + 2}

\frac{a}{3}

Show answer & solution

Option A

a (2 - 1)

Approach

Call the leg cut from each corner $x$ ; the octagon is regular exactly when the hypotenuse of the cut triangle equals what is left of the square's side.

Pythagoras' theorem

Step-by-step solution

1
each corner is cut off as an isosceles right triangle with legs $x$ , so its hypotenuse — one side of the octagon — is $x^{2} + x^{2} = x 2$
Why: The corner of the square is a right angle, and the two legs must be equal or the octagon's angles will not all match.
Pythagoras' theorem
2
each side of the square loses a leg at both ends, so the piece that survives — also a side of the octagon — is $a - 2 x$
Why: Two of the four cut corners sit on every side of the square.
Perimeter and area of a square
3
regular means all eight sides are equal: $a - 2 x = x 2 \Rightarrow a = x (2 + 2) \Rightarrow x = \frac{a}{2 + 2}$
Why: The eight sides alternate between hypotenuses and leftovers, so one equation between those two lengths is the whole regularity condition.
Turn a worded condition into a linear equation in one unknown
4
side $= x 2 = \frac{a 2}{2 + 2} = \frac{a 2 ( 2 - 2 )}{( 2 + 2 ) ( 2 - 2 )} = \frac{a ( 2 2 - 2 )}{2} = a (2 - 1)$
Why: Multiplying by the conjugate clears the surd from the denominator and lands the answer in the form the options use.
Rationalise a denominator using the conjugate

Shortcut

Work with the octagon's side $s$ from the start: the square's side is $s$ plus the two legs either side of it, each $\frac{s}{2}$ , so $a = s (1 + 2)$ and $s = \frac{a}{1 + 2} = a (2 - 1)$ .

Common mistakes

Option B

Used $3$ for the hypotenuse of the cut corner, as though it were half an equilateral triangle rather than half a square.

Option C

Stopped at $x$ , the leg cut off the corner, instead of the octagon's side $x 2$ .

Option D

Split the square's side into three equal parts, which would need each cut leg to equal the octagon's side — it equals the side divided by $2$ .

Skills used

Pythagoras' theorem Perimeter and area of a square Turn a worded condition into a linear equation in one unknown Rationalise a denominator using the conjugate

77Geometry • Triangle • Pythagoras Theorem

ModerateIdeal time: about 1.5 min

Three consecutive integers form the lengths of a right-angled triangle. How many sets of such three consecutive integers is/are possible?

AOnly oneBOnly twoCOnly threeDInfinitely many

Show answer & solution

Option A

Only one

Approach

Write the three consecutive integers as $n - 1$ , $n$ , $n + 1$ , impose Pythagoras on the largest, and count the usable roots.

Pythagoras' theorem

Step-by-step solution

1
let the sides be $n - 1$ , $n$ , $n + 1$ ; the largest, $n + 1$ , must be the hypotenuse
Why: The hypotenuse faces the right angle and is always the longest side, so there is only one way to assign the three lengths.
Pythagoras' theorem
2
$(n - 1)^{2} + n^{2} = (n + 1)^{2}$
Why: Centring the run at $n$ rather than starting it at $n$ keeps the algebra symmetric and makes the squares cancel.
3
$n^{2} - 2 n + 1 + n^{2} = n^{2} + 2 n + 1 \Rightarrow n^{2} - 4 n = 0 \Rightarrow n (n - 4) = 0$
Why: Expanding both sides lets $n^{2}$ and the $1$ cancel, leaving a quadratic that factorises with no constant term.
Square of a sum and of a difference
4
$n = 0$ or $n = 4$ ; $n = 0$ gives the sides $- 1, 0, 1$ , which are not lengths, so only $n = 4$ survives
Why: A product is zero only when a factor is zero, and each root then has to be tested against the geometry.
A product is zero only when a factor is zero
5
$n = 4$ gives $3, 4, 5$ and $3^{2} + 4^{2} = 5^{2}$ : exactly one such set exists
Why: The algebra produced every candidate, so once $n = 0$ is rejected there is nothing further to count.

Common mistakes

Option B

Kept the root $n = 0$ and counted $- 1, 0, 1$ (or $0, 1, 2$ ) as a second set, although neither is a triangle.

Option C

No route produces three such sets.

Option D

Confused this with the fact that there are infinitely many Pythagorean triples; only one of them is three consecutive integers.

Skills used

Pythagoras' theorem Square of a sum and of a difference A product is zero only when a factor is zero

78Mensuration • Plane Figures • Circle

EasyIdeal time: about 1 min

Two circles are drawn with the same centre. The circumference of the smaller circle is $44$ cm and that of the bigger circle is double the smaller one. What is the area between these two circles?

154

square cmB

308

square cmC

462

square cmD

616

square cm

Show answer & solution

Option C

462

square cm

Approach

Recover both radii from the circumferences, then subtract the two areas.

Area and circumference of a circle

Step-by-step solution

1
$2 π r = 44 \Rightarrow r = \frac{44 \times 7}{2 \times 22} = 7$ cm
Why: With $π = \frac{22}{7}$ the $44$ divides cleanly, which is why the question chose it.
Area and circumference of a circle
2
bigger circumference $= 2 \times 44 = 88$ cm, so $2 π R = 88 \Rightarrow R = 14$ cm
Why: Circumference is proportional to radius, so doubling one doubles the other.
3
ring area $= π R^{2} - π r^{2} = π (1 4^{2} - 7^{2}) = \frac{22}{7} \times 147 = 462$ square cm
Why: The region between concentric circles is the big disc with the small disc removed; factorising $196 - 49$ as $147 = 3 \times 49$ keeps the arithmetic to one multiplication.
Difference of two squares

Shortcut

Doubling the radius quadruples the area, so the ring is three times the inner circle: $3 \times \frac{22}{7} \times 49 = 3 \times 154 = 462$ .

Common mistakes

Option A

Gave the smaller circle's area, $\frac{22}{7} \times 49 = 154$ , and stopped.

Option B

Doubled the smaller circle's area because the circumference doubled; area goes up fourfold, not twofold.

Option D

Gave the bigger circle's area, $\frac{22}{7} \times 196 = 616$ , without removing the smaller one.

Skills used

Area and circumference of a circle Difference of two squares

79Mensuration • Plane Figures • Quadrilateral

ModerateIdeal time: about 1.5 min

A rectangular red carpet of size $6$ ft $\times$ $12$ ft has a dark red border $6$ inches wide. What is the area of the dark red border?

9

square feetB

15

square feetC

17

square feetD

18

square feet

Show answer & solution

Option C

17

square feet

Approach

Convert the border width to feet, then subtract the inner rectangle from the whole carpet.

Cut a compound plane figure into standard pieces

Step-by-step solution

1
$6$ inches $= \frac{6}{12} = 0.5$ ft
Why: The carpet is measured in feet, so the border width has to be converted before anything is subtracted.
2
carpet area $= 6 \times 12 = 72$ square feet
Why: The border is part of the carpet, so the whole rectangle is the starting total.
Perimeter and area of a rectangle
3
inner rectangle $= (6 - 2 \times 0.5) \times (12 - 2 \times 0.5) = 5 \times 11 = 55$ square feet
Why: Each dimension loses the border width twice, once at each end — this is the step the question is built around.
4
border $= 72 - 55 = 17$ square feet
Why: What is left after removing the plain middle is exactly the border.
Cut a compound plane figure into standard pieces

Common mistakes

Option A

Counted the border along one length and one breadth only: $6 \times 0.5 + 12 \times 0.5 = 9$ .

Option B

No route reaches $15$ square feet.

Option D

Took perimeter $\times$ width $= 36 \times 0.5 = 18$ , which counts each of the four corner squares twice; removing $4 \times 0. 5^{2} = 1$ gives $17$ .

Skills used

Perimeter and area of a rectangle Cut a compound plane figure into standard pieces

80Mensuration • Plane Figures • Triangle

EasyIdeal time: about 55 s

The perimeter of a right-angled triangle is k times the shortest side. If the ratio of the other side to hypotenuse is $4$ : $5$ , then what is the value of k?

2

3

4

5

Show answer & solution

Option C

4

Approach

The $4 : 5$ ratio makes this a $3$ - $4$ - $5$ triangle; carry a common factor $t$ and divide the perimeter by the shortest side.

Pythagoras' theorem

Step-by-step solution

1
let the other side be $4 t$ and the hypotenuse $5 t$
Why: A ratio fixes the two lengths only up to a common factor, and the factor will cancel in $k$ .
Introduce a common factor $k$ for a ratio
2
shortest side $= (5 t)^{2} - (4 t)^{2} = 9 t^{2} = 3 t$
Why: The $4 : 5$ pair is a leg and the hypotenuse, so the remaining leg comes from Pythagoras; $3 t < 4 t$ confirms it really is the shortest side.
Pythagoras' theorem
3
perimeter $= 3 t + 4 t + 5 t = 12 t$
Why: All three sides are now in the same unit $t$ .
4
$12 t = k \times 3 t \Rightarrow k = 4$
Why: $t$ cancels, so $k$ is a property of the shape alone and does not depend on the triangle's size.

Common mistakes

Option A

No route reaches $2$ ; in fact no triangle can have $k = 2$ , because the two shorter sides always exceed the longest.

Option B

Treated the $4$ in the ratio as the shortest side, giving $k = \frac{12}{4} = 3$ ; the side in the $4 : 5$ ratio is the longer leg.

Option D

No route reaches $5$ .

Skills used

Introduce a common factor

k

for a ratio Pythagoras' theorem