Elementary Mathematics 2018 (I)

Set A

Questions 41–60 of 100

41Arithmetic • Number System • Unit Digit

EasyIdeal time: about 45 s

All odd prime numbers upto $110$ are multiplied together. What is the unit digit in this product ?

0

3

5

DNone of the above

Show answer & solution

Option C

5

Approach

Decide the parity of the product and whether $5$ divides it; those two facts alone fix the unit digit.

Write an odd number as

2 m + 1

Step-by-step solution

1
The odd primes up to $110$ are $3, 5, 7, 11, \dots, 109$ ; the prime $2$ is excluded because it is even
Why: Only odd primes are multiplied together, so the one even prime is not a factor.
Prime factorisation
2
Every factor is odd, so the product is odd
Why: A product of odd numbers is odd, since no factor of $2$ is ever introduced.
Write an odd number as $2 m + 1$
3
$5$ is one of the factors, so the product is a multiple of $5$
Why: The list of odd primes contains $5$ .
4
A multiple of $5$ ends in $0$ or $5$ , and $0$ would make it even, so the unit digit is $5$
Why: An odd multiple of $5$ can only end in $5$ .

Shortcut

The product is odd and contains the factor $5$ , so it ends in $5$ .

Common mistakes

Option A

Slipped the prime $2$ into the list, which would make the product a multiple of $10$ and so end in $0$ .

Option B

No route reaches a unit digit of $3$ .

Option D

'None of the above' is wrong because $5$ is printed as option C.

Skills used

Write an odd number as

2 m + 1

Prime factorisation

42Arithmetic • Mixtures and Alligation

EasyIdeal time: about 1 min

An alloy A contains two elements, copper and tin in the ratio $2 : 3$ , whereas an alloy B contains the same elements in the ratio $3 : 4$ . If $20$ kg of alloy A, $28$ kg of alloy B and some more pure copper are mixed to form a third alloy C which now contains copper and tin in the ratio $6 : 7$ , then what is the quantity of pure copper mixed in the alloy C ?

3

kgB

4

kgC

5

kgD

7

Show answer & solution

Option B

4

Approach

Split each alloy into copper and tin, note that the tin total is fixed, and impose the final ratio on the copper.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
Alloy A: $20$ kg as $2 k : 3 k$ with $5 k = 20$ gives $8$ kg copper and $12$ kg tin
Why: A ratio fixes the two parts only up to a common factor, which the total determines.
Introduce a common factor $k$ for a ratio
2
Alloy B: $28$ kg as $3 k : 4 k$ with $7 k = 28$ gives $12$ kg copper and $16$ kg tin
Why: The same move on the second alloy.
3
With $x$ kg of pure copper added: copper $= 8 + 12 + x = 20 + x$ , tin $= 12 + 16 = 28$
Why: Pure copper adds to the copper only, so the tin total is already final.
Turn a worded condition into a linear equation in one unknown
4
$\frac{20 + x}{28} = \frac{6}{7} \Rightarrow 7 (20 + x) = 168 \Rightarrow 20 + x = 24 \Rightarrow x = 4$
Why: Alloy C holds copper and tin in the ratio $6 : 7$ .
Clear the denominators of an equation

Shortcut

The tin is fixed at $12 + 16 = 28$ kg, so a ratio of $6 : 7$ forces the copper to be $24$ kg; since $20$ kg is already there, $4$ kg is added.

Common mistakes

Option A

No route reaches $3$ kg.

Option C

No route reaches $5$ kg.

Option D

No route reaches $7$ kg; the $7$ is the tin part of the final ratio.

Skills used

Introduce a common factor

k

for a ratio Turn a worded condition into a linear equation in one unknown Clear the denominators of an equation

43Algebra • Polynomial • Factor and Remainder Theorem

EasyIdeal time: about 1 min

A quadratic polynomial $a x^{2} + b x + c$ is such that when it is divided by $x$ , $(x - 1)$ and $(x + 1)$ , the remainders are $3$ , $6$ and $4$ respectively. What is the value of $(a + b)$ ?

3

2

1

- 1

Show answer & solution

Option A

3

Approach

The remainder on dividing by a linear factor is the polynomial evaluated at that factor's root, so read off $c$ and then $a + b$ .

Factor and remainder theorem

Step-by-step solution

1
$p (x) = a x^{2} + b x + c$ ; division by $x$ leaves $p (0) = c = 3$
Why: The remainder on division by $x - 0$ is $p (0)$ , which is the constant term.
Factor and remainder theorem
2
Division by $x - 1$ leaves $p (1) = a + b + c = 6$
Why: The remainder on division by $x - 1$ is $p (1)$ .
3
$a + b = 6 - c = 6 - 3 = 3$
Why: Subtracting the known constant term isolates exactly what is asked.

Shortcut

$a + b = p (1) - p (0) = 6 - 3 = 3$ ; the third remainder is never needed.

Common mistakes

Option B

Solved the full system and quoted $a = 2$ rather than $a + b$ .

Option C

Quoted $b = 1$ , or equivalently $a - b$ , instead of $a + b$ .

Option D

Computed $b - a = - 1$ , the required sum with a sign slip on one coefficient.

Skills used

Factor and remainder theorem

44Statistics • Mean

Very easyIdeal time: about 30 s

If the average of $9$ consecutive positive integers is $55$ , then what is the largest integer ?

57

58

59

60

Show answer & solution

Option C

59

Approach

An odd-length run of consecutive integers averages to its middle term, so locate the middle and step to the end.

An evenly spaced run averages to its middle term

Step-by-step solution

1
With $9$ consecutive integers the mean is the $5$ th term, so the middle term is $55$
Why: In any evenly spaced run the mean equals the middle term.
An evenly spaced run averages to its middle term
2
The largest term is four steps past the middle: $55 + 4 = 59$
Why: Nine terms leave four on each side of the middle one.

Common mistakes

Option A

No route reaches $57$ .

Option B

No route reaches $58$ .

Option D

Counted five steps up from the middle term instead of four.

Skills used

An evenly spaced run averages to its middle term

45Statistics • Mean • Weighted Mean

EasyIdeal time: about 55 s

The average of the ages of $15$ students in a class is $19$ years. When $5$ new students are admitted to the class, the average age of the class becomes $18.5$ years. What is the average age of the $5$ newly admitted students ?

17

yearsB

17.5

yearsC

18

yearsD

18.5

years

Show answer & solution

Option A

17

years

Approach

Work in totals: the new total minus the old total is exactly what the five new students bring.

Mean of a frequency distribution

Step-by-step solution

1
$15 \times 19 = 285$ years
Why: A total is the mean times the number of observations.
Mean of a frequency distribution
2
$20 \times 18.5 = 370$ years
Why: After the admissions there are $20$ students at the new mean.
3
$370 - 285 = 85$ years belong to the $5$ new students
Why: The original students' ages are unchanged, so the totals simply subtract.
4
$\frac{85}{5} = 17$
Why: Their mean is their total over their count.

Shortcut

$\frac{20 \times 18.5 - 15 \times 19}{5} = \frac{370 - 285}{5} = 17$ .

Common mistakes

Option B

Spread the half-year fall over the original $15$ students rather than all $20$ : $19 - \frac{7.5}{5} = 17.5$ .

Option C

No route reaches $18$ years.

Option D

Quoted the new class average instead of the newcomers' average.

Skills used

Mean of a frequency distribution

46Arithmetic • Time, Speed and Distance • Boats and Streams

EasyIdeal time: about 1 min

A man can row at a speed of $x$ km/hr in still water. If in a stream which is flowing at a speed of $y$ km/hr it takes him $z$ hours to row to a place and back, then what is the distance between the two places ?

\frac{z ( x ^{2} - y ^{2} )}{2 y}

\frac{z ( x ^{2} - y ^{2} )}{2 x}

\frac{( x ^{2} - y ^{2} )}{2 z x}

\frac{z ( x ^{2} - y ^{2} )}{x}

Show answer & solution

Option B

\frac{z ( x ^{2} - y ^{2} )}{2 x}

Approach

Add the downstream and upstream times, set the sum to $z$ , and solve for the one-way distance.

Still-water speed and current from the two legs

Step-by-step solution

1
$downstream speed = x + y$ and $upstream speed = x - y$
Why: The current adds to the rowing speed one way and subtracts the other.
Still-water speed and current from the two legs
2
$\frac{d}{x + y} + \frac{d}{x - y} = z$
Why: Both legs cover the same distance $d$ and together take $z$ hours.
Speed, distance and time
3
$d (\frac{( x - y ) + ( x + y )}{( x + y ) ( x - y )}) = z$ , and $(x + y) (x - y) = x^{2} - y^{2}$
Why: Put the two fractions over their common denominator, which is a difference of squares.
Difference of two squares Clear the denominators of an equation
4
$\frac{2 x d}{x ^{2} - y ^{2}} = z \Rightarrow d = \frac{z ( x ^{2} - y ^{2} )}{2 x}$
Why: The two numerators add to $2 x$ , and the rest is rearrangement.

Common mistakes

Option A

Put $y$ in the denominator, as if the current rather than the rowing speed governed the round-trip time; the expression even fails when $y = 0$ .

Option C

Divided by $z$ instead of multiplying by it, so the distance shrinks as the journey takes longer.

Option D

Lost the factor $2$ coming from $(x - y) + (x + y) = 2 x$ .

Skills used

Still-water speed and current from the two legs Speed, distance and time Difference of two squares Clear the denominators of an equation

47Arithmetic • Time and Work • Pipes and Cisterns

ModerateIdeal time: about 2 min

A water tank has been fitted with two taps P and Q and a drain pipe R. Taps P and Q fill at the rate of $12$ litres per minute and $10$ litres per minute respectively.

Consider the following statements S1, S2 and S3 :

S1: Pipe R drains out at the rate of $6$ litres per minute.

S2: If both the taps and the drain pipe are opened simultaneously, then the tank is filled in $5$ hours $45$ minutes.

S3: Pipe R drains out (fully) the filled tank in $15$ hours $20$ minutes.

To know what is the capacity of the tank, which one of the following is correct ?

AS2 is only sufficientBS1, S2 and S3 are necessaryCAny two out of S1, S2 and S3 are sufficientDNone of the above

Show answer & solution

Option C

Any two out of S1, S2 and S3 are sufficient

Approach

The capacity follows as soon as the drain rate and one timing are both known, so test what each pair of statements supplies.

Take total work as the LCM of the times

Step-by-step solution

1
Taps $P$ and $Q$ together fill at $12 + 10 = 22$ litres per minute; let the drain rate be $r$ litres per minute and the capacity $C$ litres
Why: Rates add for the taps and the drain opposes them, so the unknowns are just $r$ and $C$ .
Take total work as the LCM of the times
2
S1 and S2: net rate $22 - 6 = 16$ and time $5$ h $45$ min $= 345$ min give $C = 16 \times 345 = 5520$
Why: Capacity is the net filling rate times the filling time.
3
S1 and S3: the drain alone empties the tank, so $C = 6 \times 920 = 5520$ , since $15$ h $20$ min $= 920$ min
Why: Draining the full tank at a known rate for a known time measures the capacity directly.
4
S2 and S3: $(22 - r) \times 345 = C = r \times 920$ , so $7590 = 1265 r$ , $r = 6$ and $C = 5520$
Why: Both statements describe the same capacity, which eliminates $C$ and leaves one equation in $r$ .
Turn a worded condition into a linear equation in one unknown
5
No single statement is enough: S1 alone gives no timing, and S2 or S3 alone leaves both $r$ and $C$ unknown
Why: One equation cannot settle two unknowns, so at least two statements are needed.

Common mistakes

Option A

Treated S2 as self-contained, forgetting that without S1 or S3 the drain rate is unknown.

Option B

Assumed every statement is needed; in fact each pair already fixes the capacity, so the third is redundant.

Option D

'None of the above' is wrong because option C does hold.

Skills used

Take total work as the LCM of the times Turn a worded condition into a linear equation in one unknown

48Arithmetic • Time, Speed and Distance • Average Speed

EasyIdeal time: about 55 s

A car has an average speed of $60$ km per hour while going from Delhi to Agra and has an average speed of $y$ km per hour while returning to Delhi from Agra (by travelling the same distance). If the average speed of the car for the whole journey is $48$ km per hour, then what is the value of $y$ ?

30

km per hourB

35

km per hourC

40

km per hourD

45

km per hour

Show answer & solution

Option C

40

km per hour

Approach

Over equal distances the whole-journey average is the harmonic mean of the two speeds; set it to $48$ and solve.

Average speed over two equal distances

Step-by-step solution

1
$\frac{2 \times 60 \times y}{60 + y} = 48$
Why: The two legs are the same distance, so the average speed is $\frac{2 uv}{u + v}$ , not the arithmetic mean.
Average speed over two equal distances
2
$120 y = 48 (60 + y) = 2880 + 48 y$
Why: Clear the denominator.
Clear the denominators of an equation
3
$72 y = 2880 \Rightarrow y = 40$
Why: Collect the terms in $y$ and divide.

Common mistakes

Option A

No route reaches $30$ km per hour.

Option B

No route reaches $35$ km per hour; the arithmetic-mean error gives $36$ , which is not printed.

Option D

No route reaches $45$ km per hour.

Skills used

Average speed over two equal distances Clear the denominators of an equation

49Arithmetic • Profit and Loss

EasyIdeal time: about 55 s

An article is sold at a profit of $32%$ . If the cost price is increased by $20%$ and the sale price remains the same, then the profit percentage becomes

10%

12%

15%

20%

Show answer & solution

Option A

10%

Approach

Take the cost as $100$ , fix the selling price it implies, then raise the cost and measure the gain on the new cost.

Gain percent is measured on the cost price

Step-by-step solution

1
$C = 100 \Rightarrow S = 132$
Why: A gain of $32%$ is measured on the cost price, so the selling price is $132$ .
Gain percent is measured on the cost price
2
$C^{'} = 100 \times (1 + \frac{20}{100}) = 120$ , with $S$ still $132$
Why: The cost rises by a fifth while the sale price is held fixed.
Percentage change as a multiplier
3
$gain % = \frac{132 - 120}{120} \times 100 = \frac{12}{120} \times 100 = 10%$
Why: The gain percent is always taken on the cost price actually in force.

Shortcut

The sale price is unchanged, so the new factor is $\frac{1.32}{1.20} = 1.10$ , a gain of $10%$ .

Common mistakes

Option B

Divided the gain of $12$ by the old cost $100$ instead of by the new cost $120$ .

Option C

No route reaches $15%$ .

Option D

Quoted the $20%$ rise in cost as though it were the new profit percentage.

Skills used

Gain percent is measured on the cost price Percentage change as a multiplier

50Arithmetic • Profit and Loss • Partnership

ModerateIdeal time: about 2 min

A, B, C, D and E start a partnership firm. Capital contributed by A is three times that contributed by D. E contributes half of A's contribution, B contributes one-third of E's contribution and C contributes two-third of A's contribution. If the difference between the combined shares of A, D and E and the combined shares of B and C in the total profit of the firm is Rs. $13, 500$ , what is the combined share of B, C and E ? (The shares are supposed to be proportional to the contributions)

ARs.

13, 500

BRs.

18, 000

CRs.

19, 750

DRs.

20, 250

Show answer & solution

Option B

Rs.

18, 000

Approach

Put all five contributions in units of the smallest, read the profit shares off that ratio, and let the stated difference fix the value of one unit.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
Take $A = 6$ units; then $D = \frac{A}{3} = 2$ , $E = \frac{A}{2} = 3$ , $B = \frac{E}{3} = 1$ and $C = \frac{2 A}{3} = 4$
Why: Choosing $A = 6$ clears all three fractions at once, so every contribution is a whole number of units.
Introduce a common factor $k$ for a ratio
2
Shares are in the ratio $A : B : C : D : E = 6 : 1 : 4 : 2 : 3$
Why: Profit is shared in the ratio of capital times time, and every partner is in for the same period.
Profit is shared in the ratio of capital $\times$ time
3
$(A + D + E) - (B + C) = (6 + 2 + 3) - (1 + 4) = 6$ units $= 13500$ , so one unit is $2250$
Why: The stated difference is what calibrates the unit in rupees.
Turn a worded condition into a linear equation in one unknown
4
$B + C + E = 1 + 4 + 3 = 8$ units $= 8 \times 2250 = 18000$
Why: The combined share asked for is eight of those units.

Shortcut

The difference is $6$ units and the group asked for is $8$ units, so the answer is $\frac{8}{6} \times 13500 = 18000$ without ever finding a single share.

Common mistakes

Option A

Quoted the given difference of Rs. $13, 500$ as the answer.

Option C

No route reaches Rs. $19, 750$ ; it is not even a whole number of units.

Option D

No route reaches Rs. $20, 250$ , which would be nine units rather than eight.

Skills used

Introduce a common factor

k

for a ratio Profit is shared in the ratio of capital

\times

time Turn a worded condition into a linear equation in one unknown

51Statistics • Data Interpretation • Pie Chart

EasyIdeal time: about 45 s

A Pie Chart is drawn for the following data :

Sector	Percentage
Agriculture and Rural Development	12.9
Irrigation	12.5
Energy	27.2
Industry and Minerals	15.4
Transport and Communication	15.9
Social Services	16.1

What is the angle (approximately) subtended by the Social Services Sector at the centre of the circle ?

4 5^{\circ}

4 6^{\circ}

5 8^{\circ}

9 8^{\circ}

Show answer & solution

Option C

5 8^{\circ}

Approach

A sector's central angle is its share of the total multiplied by $36 0^{\circ}$ .

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

Step-by-step solution

1
Social Services accounts for $16.1$ out of $100$
Why: Read the Social Services row of the table; the six percentages already total $100$ .
A pie chart sector's angle is proportional to its share
2
$θ = \frac{16.1}{100} \times 36 0^{\circ} = 16.1 \times 3. 6^{\circ}$
Why: One percent of a full circle is exactly $3. 6^{\circ}$ .
3
$16.1 \times 3.6 = 57.96$ , so $θ \approx 5 8^{\circ}$
Why: The question asks for the angle only approximately.

Common mistakes

Option A

Used Irrigation's $12.5%$ : $12.5 \times 3.6 = 45$ .

Option B

Used Agriculture and Rural Development's $12.9%$ : $12.9 \times 3.6 \approx 46$ .

Option D

Used Energy's $27.2%$ : $27.2 \times 3.6 \approx 98$ .

Skills used

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

52Statistics • Mean

EasyIdeal time: about 50 s

The arithmetic mean of two numbers is $10$ and their geometric mean is $8$ . What are the two numbers ?

15, 5

12, 8

16, 4

18, 2

Show answer & solution

Option C

16, 4

Approach

Turn the two means into the sum and the product of the numbers, then solve the quadratic those determine.

Geometric mean: the

n

th root of the product

Step-by-step solution

1
$\frac{a + b}{2} = 10 \Rightarrow a + b = 20$
Why: The arithmetic mean of two numbers is half their sum.
Mean of a frequency distribution
2
$ab = 8 \Rightarrow ab = 64$
Why: The geometric mean of two numbers is the square root of their product.
Geometric mean: the $n$ th root of the product
3
$a$ and $b$ are the roots of $t^{2} - 20 t + 64 = 0$
Why: A monic quadratic is determined by the sum and the product of its roots.
Sum and product of roots (Vieta)
4
$t = \frac{20 \pm 400 - 256}{2} = \frac{20 \pm 12}{2} = 16 or 4$
Why: Solve by the quadratic formula; the discriminant $144$ is a perfect square.
Quadratic formula

Shortcut

$t^{2} - 20 t + 64$ factorises by inspection as $(t - 16) (t - 4)$ .

Common mistakes

Option A

Checked only the arithmetic mean: $15 + 5 = 20$ , but $75 \approx 8.7$ , not $8$ .

Option B

$12 + 8 = 20$ , but $96 \approx 9.8$ .

Option D

$18 + 2 = 20$ , but $36 = 6$ .

Skills used

Geometric mean: the

n

th root of the product Mean of a frequency distribution Sum and product of roots (Vieta)Quadratic formula

53Statistics • Mean • Weighted Mean

EasyIdeal time: about 1 min

The arithmetic mean of $11$ observations is $11$ . The arithmetic mean of the first $6$ observations is $10.5$ and the arithmetic mean of the last $6$ observations is $11.5$ . What is the sixth observation ?

10.0

10.5

11.0

11.5

Show answer & solution

Option C

11.0

Approach

Compare the total of all eleven observations with the totals of the two overlapping blocks of six.

Mean of a frequency distribution

Step-by-step solution

1
$total of all 11 = 11 \times 11 = 121$
Why: A total is the mean times the number of observations.
Mean of a frequency distribution
2
$first 6 = 6 \times 10.5 = 63$ and $last 6 = 6 \times 11.5 = 69$
Why: The same rule applied to each block of six.
3
$63 + 69 = 132$ , which counts the $6$ th observation twice
Why: Among eleven observations the first six and the last six overlap in exactly the $6$ th.
4
$6th observation = 132 - 121 = 11$
Why: The excess over the true total is precisely the double-counted value.

Shortcut

The two blocks overlap in one observation, so it equals $63 + 69 - 121 = 11$ .

Common mistakes

Option A

No route reaches $10.0$ .

Option B

Quoted the mean of the first six observations.

Option D

Quoted the mean of the last six observations.

Skills used

Mean of a frequency distribution

54Trigonometry • Trigonometric Identity

EasyIdeal time: about 50 s

What is $sin^{4} θ - cos^{4} θ$ equal to for any real number $θ$ ?

1

1 - 2 sin^{2} θ

2 cos^{2} θ + 1

1 - 2 cos^{2} θ

Show answer & solution

Option D

1 - 2 cos^{2} θ

Approach

Factorise as a difference of two squares, collapse the first bracket with $sin^{2} + cos^{2} = 1$ , and then write the result in terms of $cos θ$ .

Difference of two squares

Step-by-step solution

1
$sin^{4} θ - cos^{4} θ = (sin^{2} θ + cos^{2} θ) (sin^{2} θ - cos^{2} θ)$
Why: $a^{2} - b^{2} = (a + b) (a - b)$ with $a = sin^{2} θ$ and $b = cos^{2} θ$ .
Difference of two squares
2
$= 1 \times (sin^{2} θ - cos^{2} θ) = sin^{2} θ - cos^{2} θ$
Why: The first bracket is the Pythagorean identity and equals $1$ .
Pythagorean trigonometric identities
3
$= (1 - cos^{2} θ) - cos^{2} θ = 1 - 2 cos^{2} θ$
Why: Replacing $sin^{2} θ$ by $1 - cos^{2} θ$ puts the answer in the single-ratio form the options use.

Shortcut

$sin^{2} θ - cos^{2} θ = - cos 2 θ$ , and $- cos 2 θ = 1 - 2 cos^{2} θ$ .

Common mistakes

Option A

Mistook the expression for the Pythagorean identity itself and answered $1$ .

Option B

$1 - 2 sin^{2} θ$ is the negative of the correct value, which is what $cos^{4} θ - sin^{4} θ$ gives.

Option C

No route reaches $2 cos^{2} θ + 1$ , which exceeds $1$ for most $θ$ while the expression never does.

Skills used

Difference of two squares Pythagorean trigonometric identities

55Trigonometry • Trigonometric Identity • Complementary Angles

EasyIdeal time: about 45 s

What is $cot 1^{\circ} cot 2 3^{\circ} cot 4 5^{\circ} cot 6 7^{\circ} cot 8 9^{\circ}$ equal to ?

0

1

\frac{1}{2}

\frac{1}{3}

Show answer & solution

Option B

1

Approach

Pair the angles that add to $9 0^{\circ}$ ; each pair turns into a ratio times its own reciprocal.

Ratios of complementary angles

Step-by-step solution

1
$1^{\circ} + 8 9^{\circ} = 9 0^{\circ}$ and $2 3^{\circ} + 6 7^{\circ} = 9 0^{\circ}$ , so $cot 8 9^{\circ} = tan 1^{\circ}$ and $cot 6 7^{\circ} = tan 2 3^{\circ}$
Why: $cot (9 0^{\circ} - θ) = tan θ$ .
Ratios of complementary angles
2
$cot 1^{\circ} cot 8 9^{\circ} = cot 1^{\circ} tan 1^{\circ} = 1$ and $cot 2 3^{\circ} cot 6 7^{\circ} = cot 2 3^{\circ} tan 2 3^{\circ} = 1$
Why: Tangent and cotangent of the same angle are reciprocals.
Reciprocal and quotient relations between the ratios
3
$cot 4 5^{\circ} = 1$
Why: At $4 5^{\circ}$ the two legs of the right triangle are equal.
Exact trigonometric ratios of the standard angles
4
$product = 1 \times 1 \times 1 = 1$
Why: Every factor has been paired off, and the middle one is self-complementary.

Common mistakes

Option A

No route reaches $0$ ; none of the five cotangents is zero, since that needs an angle of $9 0^{\circ}$ .

Option C

No route reaches $\frac{1}{2}$ .

Option D

No route reaches $\frac{1}{3}$ .

Skills used

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

56Arithmetic • Clocks • Angle Between Hands

Very easyIdeal time: about 30 s

What angle does the hour hand of a clock describe in $10$ minutes of time ?

1^{\circ}

5^{\circ}

6^{\circ}

1 0^{\circ}

Show answer & solution

Option B

5^{\circ}

Approach

Convert the hour hand's full revolution into degrees per minute and multiply by ten.

Angle between the hands of a clock

Step-by-step solution

1
The hour hand turns $36 0^{\circ}$ in $12$ hours, so $\frac{360}{12} = 3 0^{\circ}$ per hour
Why: Twelve hours is one complete revolution of the hour hand.
Angle between the hands of a clock
2
$3 0^{\circ}$ per hour is $\frac{30}{60} = \frac{1}{2}$ degree per minute
Why: There are $60$ minutes in an hour, and the hand moves uniformly.
3
$10 \times \frac{1}{2} = 5^{\circ}$
Why: Ten minutes at half a degree a minute.

Shortcut

The hour hand moves $\frac{1}{2}$ degree a minute, so $10$ minutes is $5^{\circ}$ .

Common mistakes

Option A

No route reaches $1^{\circ}$ .

Option C

Used the minute hand's rate of $6^{\circ}$ per minute, or read the question as being about the minute hand.

Option D

Read the $10$ minutes off as $10$ degrees.

Skills used

Angle between the hands of a clock

57Trigonometry • Trigonometric Identity

ModerateIdeal time: about 2 min

Consider the following statements :

1: $(sec^{2} θ - 1) (1 - cosec^{2} θ) = 1$

2: $sin θ (1 + cos θ)^{- 1} + (1 + cos θ) (sin θ)^{- 1} = 2 cosec θ$

Which of the above is/are correct ?

1

onlyB

2

onlyCBoth

1

and

2

DNeither

1

nor

2

Show answer & solution

Option B

2

only

Approach

Reduce each statement to a closed form and compare it with what the statement claims.

Pythagorean trigonometric identities

Step-by-step solution

1
$sec^{2} θ - 1 = tan^{2} θ$ and $1 - cosec^{2} θ = - cot^{2} θ$
Why: The Pythagorean identities, the second one rearranged so that the sign is visible.
Pythagorean trigonometric identities
2
$tan^{2} θ \times (- cot^{2} θ) = - 1 \neq = 1$ , so S1 is false
Why: Tangent and cotangent are reciprocals, so their squares multiply to $1$ , but the minus sign survives.
Reciprocal and quotient relations between the ratios
3
$\frac{s i n θ}{1 + c o s θ} + \frac{1 + c o s θ}{s i n θ} = \frac{s i n ^{2} θ + ( 1 + c o s θ ) ^{2}}{s i n θ ( 1 + c o s θ )}$
Why: $(1 + cos θ)^{- 1}$ and $(sin θ)^{- 1}$ are reciprocals, so S2 is a sum of two fractions; put them over a common denominator.
Negative exponent is a reciprocal Clear the denominators of an equation
4
$sin^{2} θ + 1 + 2 cos θ + cos^{2} θ = 2 + 2 cos θ = 2 (1 + cos θ)$
Why: Expand the square, then use $sin^{2} θ + cos^{2} θ = 1$ .
Square of a sum and of a difference
5
$\frac{2 ( 1 + c o s θ )}{s i n θ ( 1 + c o s θ )} = \frac{2}{s i n θ} = 2 cosec θ$ , so S2 is true
Why: The factor $1 + cos θ$ cancels, leaving exactly what S2 claims.

Common mistakes

Option A

Read $1 - cosec^{2} θ$ as $+ cot^{2} θ$ , making S1 come out as $1$ , and rejected S2 as well.

Option C

Accepted S1 because $tan^{2} θ cot^{2} θ = 1$ , missing that the bracket is $- cot^{2} θ$ .

Option D

Rejected S2, usually by expanding $(1 + cos θ)^{2}$ as $1 + cos^{2} θ$ .

Skills used

Pythagorean trigonometric identities Reciprocal and quotient relations between the ratios Negative exponent is a reciprocal Clear the denominators of an equation Square of a sum and of a difference

58Trigonometry • Heights and Distances

HardIdeal time: about 2 min

Each side of a square subtends an angle of $6 0^{\circ}$ at the tip of a tower of height $h$ metres standing at the centre of the square. If $l$ is the length of each side of the square, then what is $h^{2}$ equal to ?

2 l^{2}

\frac{l ^{2}}{2}

\frac{3 l ^{2}}{2}

\frac{2 l ^{2}}{3}

Show answer & solution

Option B

\frac{l ^{2}}{2}

Approach

By symmetry each side of the square and the tip form an isosceles triangle, so a $6 0^{\circ}$ apex makes it equilateral; Pythagoras from the centre then gives $h$ .

Base angles of an isosceles triangle

Step-by-step solution

1
Let $O$ be the centre of square $A B C D$ and $P$ the tip of the tower, so $O P = h$ is vertical and $∠ P O A = 9 0^{\circ}$
Why: The tower stands at the centre, so $O P$ is perpendicular to the plane of the square.
Angles of elevation and depression
2
$O A = O B = O C = O D$ , so $P A = P B = P C = P D$
Why: Equal horizontal legs with the common vertical leg $O P$ give equal slant distances.
Pythagoras' theorem
3
In $△ P A B$ , $P A = P B$ and $∠ A P B = 6 0^{\circ}$ , so the base angles are each $\frac{18 0 ^{\circ} - 6 0 ^{\circ}}{2} = 6 0^{\circ}$ and the triangle is equilateral: $P A = A B = l$
Why: An isosceles triangle with a $6 0^{\circ}$ apex is equilateral, which is what converts the given angle into a length.
Base angles of an isosceles triangle Angle sum of a triangle
4
$O A = \frac{l 2}{2} = \frac{l}{2}$ , since the diagonal $l 2$ is a diameter of the square's circumcircle
Why: The distance from the centre of a square to a vertex is half its diagonal.
Side of a regular polygon inscribed in a circle
5
$h^{2} = P A^{2} - O A^{2} = l^{2} - \frac{l ^{2}}{2} = \frac{l ^{2}}{2}$
Why: Pythagoras in the right triangle $P O A$ .

Common mistakes

Option A

No route reaches $2 l^{2}$ ; it would need the tower to be taller than the distance to a vertex.

Option C

Added instead of subtracting, taking $h^{2} = l^{2} + \frac{l ^{2}}{2}$ .

Option D

No route reaches $\frac{2 l ^{2}}{3}$ .

Skills used

Base angles of an isosceles triangle Angles of elevation and depression Pythagoras' theorem Angle sum of a triangle Side of a regular polygon inscribed in a circle

59Trigonometry • Heights and Distances

ModerateIdeal time: about 2 min

From a height of $h$ units, a man observes the angle of elevation as $α$ and angle of depression as $β$ of the top and the bottom respectively of a tower of height $H$ $(> 4 h)$ . To what further height should he climb so that the values of angle of elevation and angle of depression get interchanged for the top and bottom of the tower ?

H - h

unitsB

H - 2 h

unitsC

H - 3 h

unitsD

H - 4 h

units

Show answer & solution

Option B

H - 2 h

units

Approach

Both angles are read from the same horizontal distance, so write the two tangents at eye height $h$ and demand that they swap at the new height.

Angles of elevation and depression

Step-by-step solution

1
Let $d$ be the horizontal distance to the tower. At height $h$ : $tan β = \frac{h}{d}$ and $tan α = \frac{H - h}{d}$
Why: The depression sights the foot, which is $h$ below the eye; the elevation sights the top, $H - h$ above it.
Angles of elevation and depression Sine, cosine and tangent in a right triangle
2
At a new height $h^{'}$ the two tangents are $\frac{h ^{'}}{d}$ for the depression and $\frac{H - h ^{'}}{d}$ for the elevation
Why: Climbing changes only the height of the eye, never the horizontal distance $d$ .
3
Interchanged values mean the new depression is $α$ : $\frac{h ^{'}}{d} = tan α = \frac{H - h}{d} \Rightarrow h^{'} = H - h$
Why: The angle of depression must take over the old angle of elevation.
Turn a worded condition into a linear equation in one unknown
4
The other half of the condition agrees: $\frac{H - h ^{'}}{d} = tan β = \frac{h}{d}$ also gives $h^{'} = H - h$
Why: Both requirements give the same height, so the interchange really is possible.
5
$further climb = h^{'} - h = (H - h) - h = H - 2 h$
Why: The question asks for the extra height climbed, not the new height reached.

Common mistakes

Option A

Quoted the new height $H - h$ instead of the additional climb.

Option C

No route reaches $H - 3 h$ .

Option D

No route reaches $H - 4 h$ ; it only echoes the condition $H > 4 h$ printed in the stem.

Skills used

Angles of elevation and depression Sine, cosine and tangent in a right triangle Turn a worded condition into a linear equation in one unknown

60Trigonometry • Trigonometric Identity

EasyIdeal time: about 1 min

If $sec x cosec x = 2$ , then what is $tan^{n} x + cot^{n} x$ equal to ?

2

2^{n + 1}

2^{n}

2^{n - 1}

Show answer & solution

Option A

2

Approach

Turn the condition into a statement about $sin x cos x$ ; it forces $tan x = cot x = 1$ , after which the exponent is irrelevant.

Reciprocal and quotient relations between the ratios

Step-by-step solution

1
$sec x cosec x = \frac{1}{c o s x s i n x} = 2 \Rightarrow sin x cos x = \frac{1}{2}$
Why: Secant and cosecant are the reciprocals of cosine and sine.
Reciprocal and quotient relations between the ratios
2
$(sin x - cos x)^{2} = sin^{2} x + cos^{2} x - 2 sin x cos x = 1 - 1 = 0$ , so $sin x = cos x$
Why: Expanding the square and using the Pythagorean identity shows the two ratios are equal.
Square of a sum and of a difference Pythagorean trigonometric identities
3
$tan x = \frac{s i n x}{c o s x} = 1$ and $cot x = \frac{1}{t a n x} = 1$
Why: Equal sine and cosine make the tangent $1$ , and the cotangent is its reciprocal.
4
$tan^{n} x + cot^{n} x = 1^{n} + 1^{n} = 2$ for every $n$
Why: Both powers are $1$ whatever the exponent is, so $n$ never enters the answer.

Shortcut

$sec x cosec x = \frac{2}{s i n 2 x}$ , so the condition is $sin 2 x = 1$ , i.e. $x = 4 5^{\circ}$ , and the sum is $1 + 1 = 2$ .

Common mistakes

Option B

No route reaches $2^{n + 1}$ .

Option C

Raised the whole of $tan x + cot x = 2$ to the $n$ th power, as if $tan^{n} x + cot^{n} x = (tan x + cot x)^{n}$ .

Option D

No route reaches $2^{n - 1}$ , though it does agree with the correct answer at $n = 2$ .

Skills used

Reciprocal and quotient relations between the ratios Square of a sum and of a difference Pythagorean trigonometric identities