Elementary Mathematics 2016 (II)

Set A

Questions 81–100 of 100

81Trigonometry • Trigonometric Equation

EasyIdeal time: about 50 s

If $tan^{2} x + \frac{1}{t a n ^{2} x} = 2$ and $0^{\circ} < x < 9 0^{\circ}$ , then what is the value of $x$ ?

1 5^{\circ}

3 0^{\circ}

4 5^{\circ}

6 0^{\circ}

Show answer & solution

Option C

4 5^{\circ}

Approach

Treat $tan^{2} x$ as a single unknown; $t + \frac{1}{t} = 2$ has the one solution $t = 1$ .

Exact trigonometric ratios of the standard angles

Step-by-step solution

1
$t = tan^{2} x, t + \frac{1}{t} = 2$
Why: The equation involves $tan^{2} x$ only through itself and its reciprocal, so one letter suffices.
2
$t^{2} - 2 t + 1 = 0 \Rightarrow (t - 1)^{2} = 0 \Rightarrow t = 1$
Why: Multiplying by $t$ - non-zero, since $\frac{1}{t}$ was written down - gives a perfect square with a single repeated root.
Square of a sum and of a difference A product is zero only when a factor is zero
3
$tan^{2} x = 1 \Rightarrow tan x = 1 for 0^{\circ} < x < 9 0^{\circ}$
Why: The tangent is positive in the first quadrant, so the root $tan x = - 1$ is rejected by the given range.
4
$x = 4 5^{\circ}$
Why: $tan 4 5^{\circ} = 1$ , and it is the only such angle in the range.
Exact trigonometric ratios of the standard angles

Shortcut

For positive $t$ , $t + \frac{1}{t} \geq 2$ with equality only at $t = 1$ ; so $tan^{2} x = 1$ and $x = 4 5^{\circ}$ .

Common mistakes

Option A

No standard angle gives $tan^{2} x + cot^{2} x = 2$ other than $4 5^{\circ}$ ; $1 5^{\circ}$ gives roughly $14.4$ .

Option B

$tan^{2} 3 0^{\circ} = \frac{1}{3}$ gives $\frac{1}{3} + 3 = \frac{10}{3}$ , not $2$ .

Option D

$tan^{2} 6 0^{\circ} = 3$ gives $3 + \frac{1}{3} = \frac{10}{3}$ , not $2$ ; also reached by mis-solving $(t - 1)^{2} = 0$ as $t = 3$ .

Skills used

Exact trigonometric ratios of the standard angles Square of a sum and of a difference A product is zero only when a factor is zero

82Trigonometry • Heights and Distances

Very easyIdeal time: about 25 s

If the length of the shadow of a tower is equal to its height, then what is the Sun's altitude at that time?

1 5^{\circ}

3 0^{\circ}

4 5^{\circ}

6 0^{\circ}

Show answer & solution

Option C

4 5^{\circ}

Approach

The tower and its shadow are the two legs of a right triangle whose acute angle at the tip of the shadow is the Sun's altitude.

Angles of elevation and depression

Step-by-step solution

1
$tan α = \frac{height of tower}{length of shadow}$
Why: The Sun's altitude is the angle of elevation of the Sun measured at the far end of the shadow, and the tower is the side opposite it.
Angles of elevation and depression Sine, cosine and tangent in a right triangle
2
$shadow = height \Rightarrow tan α = 1$
Why: Equal legs make the ratio exactly $1$ .
3
$α = 4 5^{\circ}$
Why: $tan 4 5^{\circ} = 1$ , and the altitude of the Sun is an acute angle, so this is the only solution.
Exact trigonometric ratios of the standard angles

Shortcut

Shadow equal to height means an isosceles right triangle, so the altitude is $4 5^{\circ}$ .

Common mistakes

Option A

$tan 1 5^{\circ} \approx 0.27$ would make the shadow almost four times the height.

Option B

$tan 3 0^{\circ} = \frac{1}{3}$ makes the shadow $3$ times the height, not equal to it.

Option D

$tan 6 0^{\circ} = 3$ makes the shadow one third of $3$ times the height.

Skills used

Angles of elevation and depression Sine, cosine and tangent in a right triangle Exact trigonometric ratios of the standard angles

83Trigonometry • Trigonometric Identity • Complementary Angles

ModerateIdeal time: about 2 min

Consider the following:

1: $\frac{c o s 7 5 ^{\circ}}{s i n 1 5 ^{\circ}} + \frac{s i n 1 2 ^{\circ}}{c o s 7 8 ^{\circ}} - \frac{c o s 1 8 ^{\circ}}{s i n 7 2 ^{\circ}} = 1$

2: $\frac{c o s 3 5 ^{\circ}}{s i n 5 5 ^{\circ}} - \frac{s i n 1 1 ^{\circ}}{c o s 7 9 ^{\circ}} + cos 2 8^{\circ} cosec 6 2^{\circ} = 1$

3: $\frac{s i n 8 0 ^{\circ}}{c o s 1 0 ^{\circ}} - sin 5 9^{\circ} sec 3 1^{\circ} = 0$

Which of the above are correct?

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

Show answer & solution

Option D

1, 2 and 3

Approach

Every ratio in the three statements pairs an angle with its complement, so each fraction or product reduces to $1$ once $sin (9 0^{\circ} - θ) = cos θ$ is applied.

Ratios of complementary angles

Step-by-step solution

1
$cos 7 5^{\circ} = sin 1 5^{\circ}, sin 1 2^{\circ} = cos 7 8^{\circ}, cos 1 8^{\circ} = sin 7 2^{\circ}$
Why: Each pair of angles adds to $9 0^{\circ}$ , so each ratio equals the co-ratio of its complement and every fraction in statement 1 is $1$ .
Ratios of complementary angles
2
$Statement 1 : 1 + 1 - 1 = 1 - correct$
Why: The three fractions are each $1$ , and the statement claims the value $1$ .
3
$cos 3 5^{\circ} = sin 5 5^{\circ}, sin 1 1^{\circ} = cos 7 9^{\circ}, cos 2 8^{\circ} = sin 6 2^{\circ} \Rightarrow cos 2 8^{\circ} cosec 6 2^{\circ} = \frac{s i n 6 2 ^{\circ}}{s i n 6 2 ^{\circ}} = 1$
Why: The same complement rule; the third term is a product rather than a fraction, but $cosec 6 2^{\circ} = \frac{1}{s i n 6 2 ^{\circ}}$ makes it one.
Reciprocal and quotient relations between the ratios
4
$Statement 2 : 1 - 1 + 1 = 1 - correct$
Why: Again each term is $1$ , and the claimed value is $1$ .
5
$sin 8 0^{\circ} = cos 1 0^{\circ}, sin 5 9^{\circ} = cos 3 1^{\circ} \Rightarrow sin 5 9^{\circ} sec 3 1^{\circ} = \frac{c o s 3 1 ^{\circ}}{c o s 3 1 ^{\circ}} = 1$
Why: Both terms of statement 3 are $1$ by the same rule.
6
$Statement 3 : 1 - 1 = 0 - correct$
Why: The statement claims $0$ , which is what the two unit terms give.

Shortcut

Check only that each pair of angles sums to $9 0^{\circ}$ ; every term is then $1$ , and the three arithmetic claims $1 + 1 - 1 = 1$ , $1 - 1 + 1 = 1$ , $1 - 1 = 0$ are all true.

Common mistakes

Option A

Judged statement 3 false, not seeing that $sin 5 9^{\circ} sec 3 1^{\circ}$ is also a complementary pair worth $1$ .

Option B

Judged statement 1 false, most easily by mishandling the three-term combination $1 + 1 - 1$ .

Option C

Judged statement 2 false, missing that $cos 2 8^{\circ} cosec 6 2^{\circ} = 1$ because $2 8^{\circ}$ and $6 2^{\circ}$ are complementary.

Skills used

Ratios of complementary angles Reciprocal and quotient relations between the ratios

84Trigonometry • Heights and Distances

EasyIdeal time: about 1 min

A pole stands vertically inside a triangular park $A B C$ . If the angle of elevation of the top of the pole from each corner of the park is same, then in the triangle $A B C$ , the foot of the pole is at the

ACentroidBCircumcentreCIncentreDOrthocentre

Show answer & solution

Option B

Circumcentre

Approach

Equal angles of elevation from the three corners force the three horizontal distances to be equal, which identifies the foot as the point equidistant from the vertices.

The circumcentre is the point equidistant from the vertices

Step-by-step solution

1
$tan θ = \frac{h}{P A} = \frac{h}{P B} = \frac{h}{P C}$
Why: The pole has a single height $h$ , and the angle of elevation of its top is the same $θ$ from each of $A$ , $B$ and $C$ .
Sine, cosine and tangent in a right triangle
2
$P A = P B = P C$
Why: Equal ratios with the same numerator $h$ force the three denominators to be equal.
3
$P is equidistant from A, B and C \Rightarrow P is the circumcentre of △ A B C$
Why: The circumcentre is defined as the point equidistant from all three vertices - the centre of the circle through them - and it is the only point in the plane with that property.
The circumcentre is the point equidistant from the vertices

Shortcut

Same height, same angle of elevation, therefore same horizontal distance to each vertex: the circumcentre.

Common mistakes

Option A

The centroid divides each median in the ratio $2 : 1$ and is equidistant from the vertices only in an equilateral triangle.

Option C

The incentre is equidistant from the three SIDES, not from the three vertices - the two properties are easy to interchange.

Option D

The orthocentre is the common point of the altitudes and carries no equidistance property at all.

Skills used

The circumcentre is the point equidistant from the vertices Sine, cosine and tangent in a right triangle

85Trigonometry • Trigonometric Identity • Complementary Angles

EasyIdeal time: about 55 s

What is the value of $tan 1^{\circ} tan 2^{\circ} tan 3^{\circ} tan 4^{\circ} \dots tan 8 9^{\circ}$ ?

0

1

2

3

Show answer & solution

Option B

1

Approach

Pair each tangent with the tangent of the complementary angle; each pair multiplies to $1$ , and the unpaired middle factor is $tan 4 5^{\circ} = 1$ .

Ratios of complementary angles

Step-by-step solution

1
$tan θ \cdot tan (9 0^{\circ} - θ) = tan θ cot θ = 1$
Why: The tangent of the complement is the cotangent, and any ratio times its reciprocal is $1$ .
Ratios of complementary angles Reciprocal and quotient relations between the ratios
2
$tan 1^{\circ} tan 8 9^{\circ} = tan 2^{\circ} tan 8 8^{\circ} = \dots = tan 4 4^{\circ} tan 4 6^{\circ} = 1$
Why: Of the $89$ factors, the $88$ other than $tan 4 5^{\circ}$ fall into exactly $44$ such complementary pairs.
3
$tan 4 5^{\circ} = 1$
Why: The middle factor has no partner - $4 5^{\circ}$ is its own complement - and its value is $1$ anyway.
Exact trigonometric ratios of the standard angles
4
$tan 1^{\circ} tan 2^{\circ} \dots tan 8 9^{\circ} = 1^{44} \times 1 = 1$
Why: Every group contributes a factor of $1$ .

Shortcut

Pair $θ$ with $9 0^{\circ} - θ$ : the product collapses to $tan 4 5^{\circ} = 1$ .

Common mistakes

Option A

Assumed some factor vanishes; no tangent between $1^{\circ}$ and $8 9^{\circ}$ is zero, and $tan 0^{\circ}$ is not in the product.

Option C

No route reaches $2$ ; the product is a run of reciprocal pairs.

Option D

$3$ is $tan 6 0^{\circ}$ , the one factor a candidate is likely to evaluate on its own.

Skills used

Ratios of complementary angles Reciprocal and quotient relations between the ratios Exact trigonometric ratios of the standard angles

86Statistics • Mean • Weighted Mean

ModerateIdeal time: about 2 min

The average score of class $X$ is $83$ . The average score of class $Y$ is $76$ . The average score of class $Z$ is $85$ . The average score of class $X$ and $Y$ is $79$ and average score of class $Y$ and $Z$ is $81$ . What is the average score of $X$ , $Y$ and $Z$ ?

81.5

80.5

79.0

78.0

Show answer & solution

Option A

81.5

Approach

Each pairwise average fixes a ratio between two class sizes; chain the two ratios and take the weighted mean of all three classes.

Mean of a frequency distribution

Step-by-step solution

1
$\frac{83 x + 76 y}{x + y} = 79 \Rightarrow 83 x + 76 y = 79 x + 79 y \Rightarrow 4 x = 3 y$
Why: A combined average is the weighted mean of the parts; clearing the denominator leaves a relation between the two class sizes $x$ and $y$ .
Mean of a frequency distribution
2
$\frac{76 y + 85 z}{y + z} = 81 \Rightarrow 76 y + 85 z = 81 y + 81 z \Rightarrow 4 z = 5 y$
Why: The same move on the second given pair.
3
$y = \frac{4 x}{3}, z = \frac{5 y}{4} = \frac{5 x}{3} \Rightarrow x : y : z = 3 : 4 : 5$
Why: Both relations are anchored on $y$ , so the three sizes lie in a single ratio chain.
Introduce a common factor $k$ for a ratio
4
$x = 3 k, y = 4 k, z = 5 k - take k = 1$
Why: Only the ratio matters: the weighted mean is homogeneous in the sizes, so $k$ cancels and any convenient multiple serves.
5
$x ˉ = \frac{83 ( 3 ) + 76 ( 4 ) + 85 ( 5 )}{3 + 4 + 5} = \frac{249 + 304 + 425}{12} = \frac{978}{12} = 81.5$
Why: The weighted mean of all three classes together.

Shortcut

The deviations from $79$ give $4 x = 3 y$ and from $81$ give $4 z = 5 y$ at a glance, so $x : y : z = 3 : 4 : 5$ and the answer is $\frac{978}{12}$ .

Common mistakes

Option B

Averaged the class means of $Y$ and $Z$ directly, $\frac{76 + 85}{2}$ , ignoring the class sizes.

Option C

Quoted the given average of $X$ and $Y$ instead of computing the average of all three.

Option D

No route reaches $78$ ; the answer must exceed the smallest class mean by more than that.

Skills used

Mean of a frequency distribution Introduce a common factor

k

for a ratio

87Statistics • Median

EasyIdeal time: about 1.5 min

For $x > 0$ , if a variable takes discrete values $x + 4$ , $x - 3.5$ , $x - 2.5$ , $x - 3$ , $x - 2$ , $x + 0.5$ , $x - 0.5$ , $x + 5$ , then what is the median?

x - 1.25

x - 0.5

x + 0.5

x + 1.25

Show answer & solution

Option A

x - 1.25

Approach

All eight values are $x$ plus a constant, so order the constants and average the fourth and fifth.

Median of a data set

Step-by-step solution

1
$Offsets from x : + 4, - 3.5, - 2.5, - 3, - 2, + 0.5, - 0.5, + 5$
Why: Since $x$ is common to every value, the order of the values is the order of the offsets, whatever $x > 0$ is.
2
$In order : - 3.5, - 3, - 2.5, - 2, - 0.5, + 0.5, + 4, + 5$
Why: The median requires the data arranged in order - the step most often skipped.
Median of a data set
3
$n = 8 is even \Rightarrow median = \frac{4 th + 5 th}{2}$
Why: For an even count the median is the mean of the two middle observations, not a single value.
4
$median = \frac{( x - 2 ) + ( x - 0.5 )}{2} = \frac{2 x - 2.5}{2} = x - 1.25$
Why: Averaging the fourth and fifth ordered values.

Shortcut

Order the offsets, take the middle two, $- 2$ and $- 0.5$ , and average them: $- 1.25$ .

Common mistakes

Option B

Took the fifth ordered value alone as the median, forgetting that an even count needs two.

Option C

Took the sixth ordered value, counting the middle position from the wrong end.

Option D

Averaged the correct pair but dropped the minus signs on the offsets.

Skills used

Median of a data set

88Statistics • Median

EasyIdeal time: about 1 min

The median of set of $9$ distinct observations is $20.5$ . If each of the largest $4$ observations of the set is increased by $2$ , then the median of the new set

Ais increased by

2

Bis decreased by

2

Cis two times the original medianDremains the same as that of original set

Show answer & solution

Option D

remains the same as that of original set

Approach

With nine observations the median is the fifth in order, and the change touches only positions six to nine.

Median of a data set

Step-by-step solution

1
$n = 9 is odd \Rightarrow median = (\frac{9 + 1}{2}) th = 5 th observation = 20.5$
Why: For an odd count the median is a single ordered value, here the fifth.
Median of a data set
2
$The largest 4 observations occupy positions 6, 7, 8, 9$
Why: The observations are distinct, so the ordering is strict and 'the largest four' is unambiguous.
3
$Adding 2 to each of them leaves them the largest four$
Why: Increasing values that are already above the fifth cannot push any of them below it, so the ordered positions are unchanged.
4
$New median = new 5 th observation = 20.5$
Why: Nothing at or below position five was altered, so the median is untouched.

Shortcut

The median of nine values depends only on the fifth; the top four are not the fifth.

Common mistakes

Option A

Assumed every observation was increased by $2$ , which would indeed shift the median by $2$ .

Option B

Read 'increased' as 'decreased' and then applied the shift to the median as well.

Option C

No route doubles the median; nothing in the change is multiplicative.

Skills used

Median of a data set

89Statistics • Mean

EasyIdeal time: about 1 min

A cricketer has a certain average of $10$ innings. In the eleventh inning he scored $108$ runs, thereby increasing his average by $6$ runs. What is his new average?

42

47

48

60

Show answer & solution

Option C

48

Approach

Name the old average, write the eleven-innings total two ways, and solve; then answer the question actually asked.

Mean of a frequency distribution

Step-by-step solution

1
$Let the average of the first 10 innings be a, so the total is 10 a$
Why: Naming the old average makes both the old and the new total expressible in one unknown.
Turn a worded condition into a linear equation in one unknown
2
$\frac{10 a + 108}{11} = a + 6$
Why: After the eleventh innings the total is $10 a + 108$ over $11$ innings, and the new average is $6$ more than the old.
Mean of a frequency distribution
3
$10 a + 108 = 11 a + 66 \Rightarrow a = 42$
Why: Clearing the denominator and collecting terms.
4
$New average = a + 6 = 48$
Why: The question asks for the NEW average, not the old one.

Shortcut

The $108$ runs cover the new average plus $6$ extra runs for each of the ten earlier innings: new average $= 108 - 10 \times 6 = 48$ .

Common mistakes

Option A

Reported the old average $42$ - the correct intermediate value, one step short of the question.

Option B

No route reaches $47$ ; it is a near neighbour of the key placed to punish rough estimation.

Option D

No route reaches $60$ ; $108 - 48$ is the extra the innings carried, not an average.

Skills used

Mean of a frequency distribution Turn a worded condition into a linear equation in one unknown

90Statistics • Mean

EasyIdeal time: about 55 s

The mean of $20$ observations is $17$ . On checking it was found that the two observations were wrongly copied as $3$ and $6$ . If wrong observations are replaced by correct values $8$ and $9$ , then what is the correct mean?

17.4

16.6

15.8

14.2

Show answer & solution

Option A

17.4

Approach

Recover the total from the wrong mean, correct it by the net change in the two entries, and divide again.

Mean of a frequency distribution

Step-by-step solution

1
$\sum x = 20 \times 17 = 340$
Why: The mean times the number of observations recovers the total that was used.
Mean of a frequency distribution
2
$Net change = (8 - 3) + (9 - 6) = 5 + 3 = 8$
Why: Each wrong entry is replaced rather than added, so only the difference it makes to the total counts.
3
$Correct total = 340 + 8 = 348$
Why: The count of observations is unchanged at $20$ ; only the total moves.
4
$Correct mean = \frac{348}{20} = 17.4$
Why: Dividing the corrected total by the same $20$ observations.

Shortcut

The mean rises by $\frac{8}{20} = 0.4$ , so the correct mean is $17 + 0.4 = 17.4$ .

Common mistakes

Option B

Subtracted the net change instead of adding it, giving $\frac{332}{20}$ .

Option C

No route reaches $15.8$ ; replacing small values by larger ones can only raise the mean.

Option D

No route reaches $14.2$ , and like C it moves the mean the wrong way.

Skills used

Mean of a frequency distribution

91Statistics • Data Classification

EasyIdeal time: about 40 s

Number of credit cards held by an individual can be treated as

AQualitative dataBDiscrete dataCCategorical dataDNone of the above

Show answer & solution

Option B

Discrete data

Approach

Classify by what the variable's values are: a count is numerical, and it can only take isolated whole-number values.

Qualitative, discrete and continuous data

Step-by-step solution

1
$The variable recorded is the NUMBER of cards : 0, 1, 2, 3, \dots$
Why: What is measured is a count, not an attribute; the cards themselves may be categorised by type, but that is not what the question records.
2
$Numerical \Rightarrow neither qualitative nor categorical$
Why: Qualitative (equivalently categorical) data record an attribute such as colour, brand or bank, which cannot be added or averaged; a count can.
Qualitative, discrete and continuous data
3
$Isolated whole-number values \Rightarrow discrete, not continuous$
Why: A discrete variable takes values with gaps between them; nobody holds $2.4$ credit cards, whereas a continuous variable such as height takes every value in a range.

Shortcut

Counts are always discrete quantitative data.

Common mistakes

Option A

Confused the objects being counted (credit cards, which do have categories) with the variable being recorded, which is how many.

Option C

The same confusion under the other name; categorical and qualitative are synonyms here, so this option and A cannot be told apart.

Option D

Chosen when 'discrete' is not recognised as a standard data type.

Skills used

Qualitative, discrete and continuous data

92Statistics • Mean • Weighted Mean

EasyIdeal time: about 1 min

In a class of $100$ students, there are $70$ boys whose average marks in a subject are $75$ . If the average marks of the complete class is $72$ , then what is the average marks of the girls?

64

65

68

74

Show answer & solution

Option B

65

Approach

Recover the total marks of the class and of the boys; the difference, over the number of girls, is the girls' average.

Mean of a frequency distribution

Step-by-step solution

1
$Total marks of the class = 100 \times 72 = 7200$
Why: Mean times count gives the total.
Mean of a frequency distribution
2
$Total marks of the boys = 70 \times 75 = 5250$
Why: The same move applied to the boys alone.
3
$Number of girls = 100 - 70 = 30, their total = 7200 - 5250 = 1950$
Why: The class splits into boys and girls with nothing left over, so both the counts and the totals subtract.
4
$Girls’ average = \frac{1950}{30} = 65$
Why: Their total over their number.

Shortcut

The boys are $3$ marks above the class mean, a surplus of $70 \times 3 = 210$ ; the $30$ girls must be $\frac{210}{30} = 7$ below it, so $72 - 7 = 65$ .

Common mistakes

Option A

An arithmetic slip in $7200 - 5250$ or in the final division by $30$ .

Option C

No route reaches $68$ ; it sits between the class mean and the key as a plausible-looking guess.

Option D

Above the class average, which is impossible when the boys are already above it and outnumber the girls.

Skills used

Mean of a frequency distribution

93Statistics • Data Interpretation • Pie Chart

Very easyIdeal time: about 30 s

A pie chart depicts the classification of total funds of an organization according to different sources of funds. A particular sector of pie chart for corporate tax has $10 8^{\circ}$ angle at the centre. What is the percentage of income from corporate tax to total funds?

20%

25%

30%

35%

Show answer & solution

Option C

30%

Approach

A pie chart shares $36 0^{\circ}$ out in proportion to the parts, so a sector's share of the angle is its share of the total funds.

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

Step-by-step solution

1
$p % of 36 0^{\circ} = 10 8^{\circ}$
Why: The whole circle represents the whole fund, so the fraction of the circle a sector occupies is the fraction of the fund it represents.
A pie chart sector's angle is proportional to its share
2
$\frac{p}{100} \times 360 = 108$
Why: Writing 'p percent of $360$ ' as an ordinary product.
$x %$ of a quantity
3
$p = \frac{108 \times 100}{360} = 30$
Why: Solving for the number of percent.

Shortcut

$\frac{108}{360} = \frac{3}{10}$ , so $30%$ .

Common mistakes

Option A

$20%$ of the circle is $7 2^{\circ}$ , not $10 8^{\circ}$ .

Option B

Read the sector as a quarter of the circle, which would be $9 0^{\circ}$ .

Option D

$35%$ of the circle is $12 6^{\circ}$ , not $10 8^{\circ}$ .

Skills used

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

x %

of a quantity

94Statistics • Frequency Distribution • Class Interval

ModerateIdeal time: about 1.5 min

Consider the following statements:

1: The classes of type $15 - 19$ , $20 - 24$ , $25 - 29$ etc. are exclusive classes.

2: The classes of type $15 - 20$ , $20 - 25$ , $25 - 30$ etc. are inclusive classes.

Which of the above statements is/are correct?

A1 onlyB2 onlyCBoth 1 and 2DNeither 1 nor 2

Show answer & solution

Option D

Neither 1 nor 2

Approach

Test each family of classes against the definitions: inclusive classes keep a gap between one upper limit and the next lower limit, exclusive classes share the boundary.

Inclusive and exclusive class intervals

Step-by-step solution

1
$15 - 19, 20 - 24, 25 - 29$
Why: The upper limit $19$ and the next lower limit $20$ are different, so no value is claimed by two classes and both limits belong to their own class.
2
$These are INCLUSIVE classes, so statement 1 is not correct$
Why: An inclusive class includes both of its stated limits; the gap it leaves has to be closed by the class-boundary correction ( $14.5 - 19.5$ and so on) before a histogram or a median is computed.
Inclusive and exclusive class intervals
3
$15 - 20, 20 - 25, 25 - 30$
Why: Here $20$ is the upper limit of one class and the lower limit of the next, so it cannot belong to both.
4
$These are EXCLUSIVE classes, so statement 2 is not correct$
Why: In an exclusive class the upper limit is excluded and counted in the following class, which is exactly what makes a shared boundary unambiguous.
5
$Neither statement 1 nor statement 2 is correct$
Why: Each statement attaches the other family's name to its family.

Shortcut

The two statements are exact mirror images with the labels swapped; since the labels are swapped, both are wrong together.

Common mistakes

Option A

Judged statement 1 correct, reading the gap between $19$ and $20$ as an exclusion of values.

Option B

Judged statement 2 correct, reading the shared boundary $20$ as an inclusion.

Option C

Interchanged the two definitions wholesale, which makes both statements read as correct.

Skills used

Inclusive and exclusive class intervals

95Statistics • Frequency Distribution • Frequency Density

EasyIdeal time: about 45 s

Suppose the class interval $10 - 15$ has frequency $30$ , then what is the frequency density of this class interval?

2

3

5

6

Show answer & solution

Option D

6

Approach

Frequency density is frequency per unit of class width, so divide the frequency by the width of the interval.

Frequency density

Step-by-step solution

1
$Class width = 15 - 10 = 5$
Why: The width is the difference between the two class limits.
2
$Frequency density = \frac{frequency}{class width}$
Why: This is what makes histogram bars of unequal width comparable: the AREA of a bar, not its height, must represent the frequency.
Frequency density
3
$Frequency density = \frac{30}{5} = 6$
Why: Substituting the frequency and the width.

Shortcut

$\frac{30}{5} = 6$ .

Common mistakes

Option A

Divided by the upper class limit $15$ instead of by the width.

Option B

Divided by the lower class limit $10$ instead of by the width.

Option C

Reported the class width itself rather than the density.

Skills used

Frequency density

96Arithmetic • Number System • Divisibility

HardIdeal time: about 3 min

Let $S$ be a set of first ten natural numbers. What is the possible number of pairs $(a, b)$ where $a, b \in S$ and $a \neq = b$ such that the product $ab$ $(> 12)$ leaves remainder $4$ when divided by $12$ ?

4

6

8

10

Show answer & solution

Option C

8

Approach

List the products that can leave remainder $4$ on division by $12$ within the reachable range, factor each into two distinct members of $S$ , then count the ordered pairs.

Enumerate a residue class within a bound

Step-by-step solution

1
$S = {1, 2, \dots, 10}, ab \leq 9 \times 10 = 90$
Why: The largest product of two distinct members of $S$ bounds the search, which makes an exhaustive check possible.
2
$ab \equiv 4 (mod 12) and ab > 12 \Rightarrow ab \in {16, 28, 40, 52, 64, 76, 88}$
Why: These are the numbers of the form $12 k + 4$ that exceed $12$ and do not exceed $90$ ; the condition $ab > 12$ is what removes $ab = 4$ itself.
Enumerate a residue class within a bound
3
$16 = 2 \times 8; 28 = 4 \times 7; 40 = 4 \times 10 = 5 \times 8; 52 = 4 \times 13; 64 = 8 \times 8; 76 = 4 \times 19; 88 = 8 \times 11$
Why: Factorising each candidate and keeping only factorisations into two DISTINCT members of $S$ : $52$ , $76$ and $88$ need a factor above $10$ , and $64$ only repeats $8$ .
Prime factorisation
4
$Unordered pairs : {2, 8}, {4, 7}, {4, 10}, {5, 8} - four of them$
Why: These are all the two-element subsets of $S$ whose product meets both conditions.
5
$(a, b) is an ordered pair, so each subset yields two : 4 \times 2 = 8$
Why: $(2, 8)$ and $(8, 2)$ are different ordered pairs and each satisfies every stated condition, so both are counted.

Shortcut

Only $16$ , $28$ and $40$ are reachable, giving the four subsets ${2, 8}, {4, 7}, {4, 10}, {5, 8}$ and hence $8$ ordered pairs.

Common mistakes

Option A

Counted unordered pairs ${a, b}$ rather than the ordered pairs $(a, b)$ the question names; this is the reading under which $4$ is correct.

Option B

Missed one of the four subsets - most easily ${4, 10}$ , since $40$ has two admissible factorisations - and doubled the remaining three.

Option D

Counted $64 = 8 \times 8$ as a fifth case despite $a \neq = b$ , giving five subsets and ten ordered pairs.

Skills used

Enumerate a residue class within a bound Prime factorisation

97Arithmetic • Number System • Divisibility

ModerateIdeal time: about 1.5 min

What is the remainder when $1 3^{5} + 1 4^{5} + 1 5^{5} + 1 6^{5}$ is divided by $29$ ?

8

5

3

0

Show answer & solution

Option D

0

Approach

Pair the four terms so that each pair has bases adding to $29$ ; an odd exponent then makes each pair a multiple of $29$ .

a + b

divides

a^{n} + b^{n}

for odd

n

Step-by-step solution

1
$13 + 16 = 29 and 14 + 15 = 29$
Why: The modulus is itself the sum of the outer pair of bases and of the inner pair - the structure the question is built on, and the reason the four powers were chosen.
2
$a^{n} + b^{n} = (a + b) (a^{n - 1} - a^{n - 2} b + \dots + b^{n - 1}) for odd n \Rightarrow (a + b) ∣ (a^{n} + b^{n})$
Why: For an odd exponent the alternating factorisation closes up; $n = 3$ is the familiar sum-of-cubes case, and $n = 5$ behaves the same way.
$a + b$ divides $a^{n} + b^{n}$ for odd $n$
3
$29 ∣ (1 3^{5} + 1 6^{5}) and 29 ∣ (1 4^{5} + 1 5^{5})$
Why: Each pair has base-sum exactly $29$ , and the exponent $5$ is odd.
4
$1 3^{5} + 1 4^{5} + 1 5^{5} + 1 6^{5} = (1 3^{5} + 1 6^{5}) + (1 4^{5} + 1 5^{5}) \equiv 0 (mod 29)$
Why: A sum of two multiples of $29$ is a multiple of $29$ , so the remainder is $0$ .

Shortcut

Working modulo $29$ , $16 \equiv - 13$ and $15 \equiv - 14$ , so $1 6^{5} \equiv - 1 3^{5}$ and $1 5^{5} \equiv - 1 4^{5}$ ; the four terms cancel in pairs.

Common mistakes

Option A

No route reaches $8$ ; the sum is exactly divisible, so any non-zero remainder comes from an arithmetic error rather than a method.

Option B

No route reaches $5$ ; same reason as A.

Option C

No route reaches $3$ ; the bases sum to $58 = 2 \times 29$ , which tempts a guess of a small non-zero remainder.

Skills used

a + b

divides

a^{n} + b^{n}

for odd

n

98Algebra • Sequence and Series • Summation

EasyIdeal time: about 1 min

What is the difference between the sum of the cubes and that of squares of first ten natural numbers?

2280

2640

3820

4130

Show answer & solution

Option B

2640

Approach

Apply the two standard summation formulas at $n = 10$ and subtract in the order the question states.

Sums of the first

n

natural numbers, their squares and their cubes

Step-by-step solution

1
$\sum_{r = 1}^{n} r^{3} = [\frac{n ( n + 1 )}{2}]^{2} \Rightarrow \sum_{r = 1}^{10} r^{3} = (\frac{10 \times 11}{2})^{2} = 5 5^{2} = 3025$
Why: The sum of the first $n$ cubes is the square of the sum of the first $n$ natural numbers.
Sums of the first $n$ natural numbers, their squares and their cubes
2
$\sum_{r = 1}^{n} r^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6} \Rightarrow \sum_{r = 1}^{10} r^{2} = \frac{10 \times 11 \times 21}{6} = 385$
Why: The standard sum-of-squares formula, evaluated at $n = 10$ .
3
$3025 - 385 = 2640$
Why: The question asks for the sum of the cubes minus the sum of the squares, in that order, so the answer is positive.

Shortcut

$5 5^{2} - 385 = 3025 - 385 = 2640$ .

Common mistakes

Option A

No route reaches $2280$ ; it is a near neighbour that punishes a mis-evaluated sum-of-squares.

Option C

No route reaches $3820$ ; it exceeds the cube sum minus any plausible square sum.

Option D

No route reaches $4130$ ; it exceeds $3025$ and so cannot be a difference of these two sums.

Skills used

Sums of the first

n

natural numbers, their squares and their cubes

99Arithmetic • Time, Speed and Distance • Average Speed

EasyIdeal time: about 1 min

A motorist travels to a place $150$ km away at an average speed of $50$ km/hour and returns at $30$ km/hour. What is the average speed for the whole journey?

35

km/hourB

37

km/hourC

37.5

km/hourD

40

km/hour

Show answer & solution

Option C

37.5

km/hour

Approach

Average speed is total distance over total time, which is not the arithmetic mean of the two speeds.

Speed, distance and time

Step-by-step solution

1
$t_{1} = \frac{150}{50} = 3 hours, t_{2} = \frac{150}{30} = 5 hours$
Why: Each leg covers the same $150$ km, so each time follows directly from its speed.
Speed, distance and time
2
$Total distance = 150 + 150 = 300 km, Total time = 3 + 5 = 8 hours$
Why: For a journey made in stages the distances add and the times add.
3
$Average speed = \frac{300}{8} = 37.5 km/hour$
Why: The definition of average speed over the whole journey.

Shortcut

Equal distances at two speeds give the harmonic mean: $\frac{2 \times 50 \times 30}{50 + 30} = \frac{3000}{80} = 37.5$ km/hour.

Common mistakes

Option A

No route reaches $35$ ; it is below the harmonic mean and would need more than $8$ hours.

Option B

Rounded $37.5$ down to a whole number.

Option D

Took the arithmetic mean of the two speeds, $\frac{50 + 30}{2}$ - the error this question exists to punish.

Skills used

Speed, distance and time

100Algebra • Expressions • Simplification

EasyIdeal time: about 1 min

If $x + \frac{1}{1 + \frac{1}{2 + \frac{1}{3}}} = 2$ , then what is $x$ equal to?

\frac{7}{10}

\frac{13}{10}

\frac{11}{10}

\frac{17}{10}

Show answer & solution

Option B

\frac{13}{10}

Approach

Evaluate the nested fraction from the innermost level outwards, then solve the one-step equation that remains.

Evaluate a nested fraction from the inside out

Step-by-step solution

1
$2 + \frac{1}{3} = \frac{7}{3}$
Why: The innermost level must be resolved before the level enclosing it can be.
Evaluate a nested fraction from the inside out
2
$1 + \frac{1}{\frac{7}{3}} = 1 + \frac{3}{7} = \frac{10}{7}$
Why: Dividing by a fraction is multiplying by its reciprocal, which is what unwinds each level.
3
$\frac{1}{\frac{10}{7}} = \frac{7}{10}$
Why: The outermost reciprocal, giving the constant that sits beside $x$ .
4
$x + \frac{7}{10} = 2 \Rightarrow x = 2 - \frac{7}{10} = \frac{20 - 7}{10} = \frac{13}{10}$
Why: A single subtraction now answers the question.

Shortcut

The nested fraction is $\frac{7}{10} = 0.7$ , so $x = 2 - 0.7 = 1.3$ .

Common mistakes

Option A

Reported the value of the nested fraction, $\frac{7}{10}$ , instead of $x$ .

Option C

Inverted $\frac{10}{7}$ as $\frac{9}{10}$ , giving $2 - \frac{9}{10}$ .

Option D

Took the nested fraction as $\frac{3}{10}$ , one inversion short, giving $2 - \frac{3}{10}$ .

Skills used

Evaluate a nested fraction from the inside out