Elementary Mathematics 2017 (II)

Set A

Questions 41–60 of 100

41Arithmetic • Ratio and Proportion • Division into Parts

EasyIdeal time: about 1 min

If $78$ is divided into $3$ parts which are proportional to $1, \frac{1}{3}, \frac{1}{6}$ , then the middle part is

\frac{28}{3}

13

\frac{52}{3}

\frac{55}{3}

Show answer & solution

Option C

\frac{52}{3}

Approach

Clear the fractions to turn $1 : \frac{1}{3} : \frac{1}{6}$ into a whole-number ratio, introduce a common factor, and read off the middle part.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
$1 : \frac{1}{3} : \frac{1}{6} = 6 : 2 : 1$
Why: Multiplying every term of a ratio by the same number (here $6$ , the LCM of the denominators) leaves the ratio unchanged.
Introduce a common factor $k$ for a ratio
2
Parts $= 6 k, 2 k, k$ with $6 k + 2 k + k = 78$
Why: A ratio fixes the three parts only up to a common factor $k$ , which the given total then pins down.
3
$9 k = 78 \Rightarrow k = \frac{26}{3}$
Why: Solve for the common factor.
4
Middle part $= 2 k = \frac{52}{3}$
Why: The parts are $52, \frac{52}{3}, \frac{26}{3}$ , so the middle one in both listing order and size is $2 k$ .

Shortcut

The middle part is $\frac{2}{9}$ of $78$ , which is $\frac{156}{9} = \frac{52}{3}$ .

Common mistakes

Option A

No natural route reaches $\frac{28}{3}$ .

Option B

Divided $78$ by $6$ , using the first term of the cleared ratio as if it were the number of parts.

Option D

Slip in the final multiplication or in solving for $k$ , landing one unit above $\frac{52}{3}$ .

Skills used

Introduce a common factor

k

for a ratio

42Arithmetic • Ratio and Proportion

EasyIdeal time: about 1 min

There are $350$ boys in the first three standards. The ratio of the number of boys in first and second standards is $2 : 3$ , while that of boys in second and third standards is $4 : 5$ . What is the total number of boys in first and third standards ?

302

280

242

230

Show answer & solution

Option D

230

Approach

Chain the two ratios through the common second standard, then scale to the total of $350$ .

Introduce a common factor

k

for a ratio

Step-by-step solution

1
$I : II = 2 : 3 = 8 : 12$ and $II : III = 4 : 5 = 12 : 15$
Why: Both ratios are rescaled so that the shared term, standard II, carries the same number $12$ in each.
Introduce a common factor $k$ for a ratio
2
$I : II : III = 8 : 12 : 15$
Why: With the common term matched, the two ratios join into one three-term ratio.
3
$8 k + 12 k + 15 k = 350 \Rightarrow 35 k = 350 \Rightarrow k = 10$
Why: The total fixes the common factor.
4
$I + III = 8 k + 15 k = 23 k = 230$
Why: The question asks only for the first and third standards.

Shortcut

The first and third together are $\frac{23}{35}$ of $350$ , i.e. $23 \times 10 = 230$ .

Common mistakes

Option A

No natural route reaches $302$ .

Option B

Mis-scaled the chained ratio, so the parts no longer match the two given pairwise ratios.

Option C

No natural route reaches $242$ .

Skills used

Introduce a common factor

k

for a ratio

43Arithmetic • Interest • Difference of SI and CI

EasyIdeal time: about 45 s

The difference between the compound interest (compounded annually) and simple interest on a sum of money deposited for $2$ years at $5%$ per annum is Rs. $15$ . What is the sum of money deposited ?

ARs.

6, 000

BRs.

4, 800

CRs.

3, 600

DRs.

2, 400

Show answer & solution

Option A

Rs.

6, 000

Approach

Apply the two-year formula $CI - SI = P (\frac{r}{100})^{2}$ and solve for $P$ .

Two-year difference between compound and simple interest

Step-by-step solution

1
$CI - SI = P (\frac{r}{100})^{2}$ for $2$ years
Why: Over two years the whole excess of compound over simple interest is one year's interest on the first year's interest.
Two-year difference between compound and simple interest
2
$15 = P (\frac{5}{100})^{2} = \frac{P}{400}$
Why: Substitute $r = 5$ and the given difference.
3
$P = 15 \times 400 = 6000$
Why: Solve for the principal.

Shortcut

The difference is $(\frac{5}{100})^{2} = \frac{1}{400}$ of the principal, so the principal is $400$ times Rs. $15$ .

Common mistakes

Option B

Slip in dividing or in squaring $\frac{5}{100}$ , giving a principal below the true one.

Option C

No natural route reaches Rs. $3, 600$ .

Option D

No natural route reaches Rs. $2, 400$ .

Skills used

Two-year difference between compound and simple interest

44Arithmetic • Percentage • Price and Consumption

EasyIdeal time: about 1 min

When prices rise by $12%$ , if the expenditure is to be the same, what is the percentage of consumption to be reduced ?

16 \frac{2}{3} %

10 \frac{2}{7} %

16 \frac{3}{5} %

10 \frac{5}{7} %

Show answer & solution

Option D

10 \frac{5}{7} %

Approach

Hold expenditure fixed, so consumption must be divided by the price multiplier; the reduction is what is left over.

Consumption cut that absorbs a price rise

Step-by-step solution

1
New price $= 1 \cdot 12 \times$ old price
Why: A rise of $12%$ multiplies the price by $1 + \frac{12}{100}$ .
Percentage change as a multiplier
2
Expenditure $=$ price $\times$ consumption is to be unchanged, so new consumption $= \frac{1}{1 \cdot 12} \times$ old consumption
Why: A change is reversed by dividing by its multiplier, not by applying the opposite percentage.
3
Reduction $= 1 - \frac{1}{1 \cdot 12} = \frac{0 \cdot 12}{1 \cdot 12} = \frac{12}{112} = \frac{3}{28}$
Why: The fall is measured against the ORIGINAL consumption, so the denominator is the new price level $112$ .
Consumption cut that absorbs a price rise
4
$\frac{3}{28} \times 100 = \frac{300}{28} = \frac{75}{7} = 10 \frac{5}{7} %$
Why: Convert the fraction to a percentage.

Shortcut

The reduction is always $\frac{p}{100 + p}$ of the consumption: $\frac{12}{112} = \frac{3}{28} = 10 \frac{5}{7} %$ .

Common mistakes

Option A

$16 \frac{2}{3} %$ is $\frac{20}{120}$ , the answer to the same question with a $20%$ rise rather than a $12%$ one.

Option B

Reached $\frac{300}{28}$ but botched the conversion to a mixed number, writing $10 \frac{2}{7}$ for $10 \frac{5}{7}$ .

Option C

No natural route reaches $16 \frac{3}{5} %$ .

Skills used

Consumption cut that absorbs a price rise Percentage change as a multiplier

45Arithmetic • Time, Speed and Distance • Boats and Streams

ModerateIdeal time: about 1.5 min

A man rows down a river $18$ km in $4$ hours with the stream and returns in $10$ hours. Consider the following statements :

1: The speed of the man against the stream is $1 \cdot 8$ km/hr.

2: The speed of the man in still water is $3 \cdot 15$ km/hr.

3: The speed of the stream is $1 \cdot 35$ km/hr.

Which of the above statements are correct ?

1

and

2

onlyB

2

and

3

onlyC

1

and

3

onlyD

1

2

and

3

Show answer & solution

Option D

1

2

and

3

Approach

Get the downstream and upstream speeds from the two legs, then take their half-sum and half-difference.

Still-water speed and current from the two legs

Step-by-step solution

1
Downstream speed $= \frac{18}{4} = 4 \cdot 5$ km/hr
Why: Speed is distance over time, and the $18$ km with the stream took $4$ hours.
Speed, distance and time
2
Upstream speed $= \frac{18}{10} = 1 \cdot 8$ km/hr
Why: The return covers the same $18$ km in $10$ hours, so statement $1$ is correct.
3
Still water $= \frac{4 \cdot 5 + 1 \cdot 8}{2} = 3 \cdot 15$ km/hr
Why: The man's own speed is the average of the two, since the current helps by as much as it hinders. Statement $2$ is correct.
Still-water speed and current from the two legs
4
Stream $= \frac{4 \cdot 5 - 1 \cdot 8}{2} = 1 \cdot 35$ km/hr
Why: The current is half the gap between the two speeds. Statement $3$ is correct.

Shortcut

$4 \cdot 5$ and $1 \cdot 8$ have half-sum $3 \cdot 15$ and half-difference $1 \cdot 35$ , matching statements $2$ and $3$ on sight once statement $1$ is read off.

Common mistakes

Option A

Rejected statement $3$ by taking the stream's speed as the whole difference $4 \cdot 5 - 1 \cdot 8 = 2 \cdot 7$ instead of half of it.

Option B

Rejected statement $1$ by reading 'speed against the stream' as the speed of the stream itself rather than the upstream speed of the man.

Option C

Rejected statement $2$ by swapping the two formulas, computing the still-water speed as the half-difference $1 \cdot 35$ .

Skills used

Still-water speed and current from the two legs Speed, distance and time

46Trigonometry • Triangle

EasyIdeal time: about 50 s

If a triangle has sides $5$ , $13$ and $12$ units and $θ$ is the acute angle of the triangle, then what is the value of $(sin θ + cos θ)$ ?

\frac{5}{13}

\frac{7}{13}

\frac{12}{13}

\frac{17}{13}

Show answer & solution

Option D

\frac{17}{13}

Approach

Check that $5, 12, 13$ is a right triangle, then add the sine and cosine of either acute angle.

Sine, cosine and tangent in a right triangle

Step-by-step solution

1
$5^{2} + 1 2^{2} = 25 + 144 = 169 = 1 3^{2}$
Why: The converse of Pythagoras' theorem makes the angle opposite $13$ a right angle, so $13$ is the hypotenuse.
Pythagoras' theorem
2
For the acute angle $θ$ opposite the side $5$ : $sin θ = \frac{5}{13}$ , $cos θ = \frac{12}{13}$
Why: Sine is opposite over hypotenuse and cosine is adjacent over hypotenuse.
Sine, cosine and tangent in a right triangle
3
$sin θ + cos θ = \frac{5}{13} + \frac{12}{13} = \frac{17}{13}$
Why: Add the two ratios over their common denominator.

Shortcut

For either acute angle the two legs $5$ and $12$ simply swap roles, so the sum is always $\frac{5 + 12}{13} = \frac{17}{13}$ and the choice of $θ$ never has to be made.

Common mistakes

Option A

Gave $sin θ$ alone and stopped.

Option B

Subtracted instead of adding, giving $cos θ - sin θ = \frac{12 - 5}{13}$ .

Option C

Gave $cos θ$ alone and stopped.

Skills used

Sine, cosine and tangent in a right triangle Pythagoras' theorem

47Trigonometry • Trigonometric Identity • Maximum and Minimum Values

ModerateIdeal time: about 1 min

If $0 < x < \frac{π}{2}$ , then $(sin x + cosec x)$ is

> 2

< 2

\geq 2

\leq 2

Show answer & solution

Option A

> 2

Approach

Write the expression as $t + \frac{1}{t}$ with $t = sin x$ , and use that $t$ is strictly between $0$ and $1$ on the open interval.

A positive number plus its reciprocal is at least

2

Step-by-step solution

1
$sin x + cosec x = t + \frac{1}{t}$ where $t = sin x$
Why: Cosec is the reciprocal of sine, so the expression depends on $sin x$ alone.
Reciprocal and quotient relations between the ratios
2
$0 < x < \frac{π}{2} \Rightarrow 0 < t < 1$
Why: On the OPEN first-quadrant interval the sine rises strictly from $0$ towards $1$ without ever reaching $1$ .
How the ratios move across the first quadrant Sine and cosine lie between $- 1$ and $1$
3
$t + \frac{1}{t} - 2 = \frac{( 1 - t ) ^{2}}{t}$
Why: The two terms have fixed product $1$ , so their sum is least when they are equal; the square makes the excess over $2$ explicit.
A positive number plus its reciprocal is at least $2$
4
$t \neq = 1 \Rightarrow (1 - t)^{2} > 0$ and $t > 0$ , so $t + \frac{1}{t} > 2$
Why: Equality would need $t = 1$ , i.e. $x = \frac{π}{2}$ , which the open interval excludes.

Shortcut

$t + \frac{1}{t} \geq 2$ always, with equality only at $t = 1$ ; since $sin x$ never reaches $1$ here, the inequality is strict.

Common mistakes

Option B

Reasoned from $sin x \leq 1$ that the sum stays below $2$ , ignoring that $cosec x \geq 1$ pushes the other way.

Option C

Gave the inequality for the CLOSED interval, missing that $x = \frac{π}{2}$ is excluded so equality is never attained.

Option D

Inverted the direction of the AM-GM bound, taking $2$ as a maximum rather than a minimum.

Skills used

A positive number plus its reciprocal is at least

2

Reciprocal and quotient relations between the ratios How the ratios move across the first quadrant Sine and cosine lie between

- 1

and

1

48Trigonometry • Trigonometric Identity

EasyIdeal time: about 1 min

If $sin θ = \frac{m ^{2} - n ^{2}}{m ^{2} + n ^{2}}$ and $0 < θ < \frac{π}{2}$ , then what is the value of $cos θ$ ?

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

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

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

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

Show answer & solution

Option A

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

Approach

Use $cos^{2} θ = 1 - sin^{2} θ$ , simplify the numerator as a difference of squares, and take the positive root.

Pythagorean trigonometric identities

Step-by-step solution

1
$cos^{2} θ = 1 - (\frac{m ^{2} - n ^{2}}{m ^{2} + n ^{2}})^{2} = \frac{( m ^{2} + n ^{2} ) ^{2} - ( m ^{2} - n ^{2} ) ^{2}}{( m ^{2} + n ^{2} ) ^{2}}$
Why: The Pythagorean identity converts the given sine into the cosine, over the common denominator $(m^{2} + n^{2})^{2}$ .
Pythagorean trigonometric identities
2
$(m^{2} + n^{2})^{2} - (m^{2} - n^{2})^{2} = [(m^{2} + n^{2}) + (m^{2} - n^{2})] [(m^{2} + n^{2}) - (m^{2} - n^{2})] = 2 m^{2} \cdot 2 n^{2}$
Why: Factorising as a difference of two squares is far quicker than expanding both brackets.
Difference of two squares
3
$cos^{2} θ = \frac{4 m ^{2} n ^{2}}{( m ^{2} + n ^{2} ) ^{2}}$
Why: Collect the numerator.
4
$cos θ = \frac{2 mn}{m ^{2} + n ^{2}}$
Why: $0 < θ < \frac{π}{2}$ puts $θ$ in the first quadrant, where the cosine is positive, so the positive square root is the one to take.
denotes the non-negative root

Shortcut

Recognise the Pythagorean triple $(m^{2} - n^{2}, 2 mn, m^{2} + n^{2})$ : with the given sine, the adjacent side is $2 mn$ and the hypotenuse $m^{2} + n^{2}$ .

Common mistakes

Option B

Divided by $m^{2} - n^{2}$ instead of $m^{2} + n^{2}$ , which gives $cot θ$ rather than $cos θ$ .

Option C

Inverted the correct answer, giving $sec θ$ instead of $cos θ$ .

Option D

Gave $tan θ = \frac{sin θ}{cos θ}$ instead of the cosine that was asked for.

Skills used

Pythagorean trigonometric identities Difference of two squares

denotes the non-negative root

49Trigonometry • Triangle

EasyIdeal time: about 55 s

If angle A of triangle ABC is $3 0^{\circ}$ and the circum-radius of the triangle is $10$ cm, then what is the length of side BC ?

5

cmB

10

cmC

53

cmD

103

Show answer & solution

Option B

10

Approach

Put the circumcentre in: the angle BC subtends at the centre is twice angle $A$ , and the two radii to $B$ and $C$ then fix $B C$ .

The angle at the centre is twice the angle at the circumference

Step-by-step solution

1
Let $O$ be the circumcentre, so $O B = O C = R = 10$ cm
Why: The circumcentre is by definition equidistant from all three vertices, the common distance being the circumradius.
The circumcentre is the point equidistant from the vertices
2
$∠ B O C = 2∠ B A C = 2 \times 3 0^{\circ} = 6 0^{\circ}$
Why: The arc $B C$ subtends at the centre twice the angle it subtends at the remaining point $A$ of the circle.
The angle at the centre is twice the angle at the circumference
3
In $△ O B C$ : $O B = O C$ , so the base angles are equal and each is $\frac{18 0 ^{\circ} - 6 0 ^{\circ}}{2} = 6 0^{\circ}$
Why: An isosceles triangle has equal base angles, and the three angles of a triangle sum to $18 0^{\circ}$ .
Base angles of an isosceles triangle Angle sum of a triangle
4
All three angles are $6 0^{\circ}$ , so $△ O B C$ is equilateral and $B C = O B = 10$ cm
Why: Equal angles face equal sides, so the base equals the two radii.

Shortcut

$\frac{a}{sin A} = 2 R$ gives $B C = 2 R sin A = 2 \times 10 \times sin 3 0^{\circ} = 10$ cm in one line.

Common mistakes

Option A

Used $R sin A$ instead of $2 R sin A$ , dropping the factor $2$ in the extended sine rule.

Option C

Used $2 R cos A$ and then halved, or equivalently mixed the sine and cosine of $3 0^{\circ}$ with the missing factor $2$ .

Option D

Used $2 R cos A = 20 cos 3 0^{\circ}$ , taking the cosine of the angle in place of its sine.

Skills used

The angle at the centre is twice the angle at the circumference The circumcentre is the point equidistant from the vertices Base angles of an isosceles triangle Angle sum of a triangle

50Trigonometry • Trigonometric Identity

ModerateIdeal time: about 1.5 min

If $A = \frac{s i n 4 5 ^{\circ} - s i n 3 0 ^{\circ}}{c o s 4 5 ^{\circ} + c o s 6 0 ^{\circ}}$ and $B = \frac{s e c 4 5 ^{\circ} - t a n 4 5 ^{\circ}}{cosec 4 5 ^{\circ} + c o t 4 5 ^{\circ}}$ , then which one of the following is correct ?

A = B

A > B > 0

A < B

B < A < 0

Show answer & solution

Option A

A = B

Approach

Substitute the standard-angle values into both expressions and simplify each to a single surd fraction.

Exact trigonometric ratios of the standard angles

Step-by-step solution

1
$A = \frac{\frac{1}{2} - \frac{1}{2}}{\frac{1}{2} + \frac{1}{2}}$
Why: Substitute $sin 4 5^{\circ} = cos 4 5^{\circ} = \frac{1}{2}$ , $sin 3 0^{\circ} = cos 6 0^{\circ} = \frac{1}{2}$ .
Exact trigonometric ratios of the standard angles
2
$A = \frac{\frac{2}{2} - 1}{\frac{2}{2} + 1} = \frac{2 - 1}{2 + 1}$
Why: Multiplying numerator and denominator by $2$ clears the halves, and $\frac{2}{2} = 2$ .
3
$B = \frac{2 - 1}{2 + 1}$
Why: $sec 4 5^{\circ} = cosec 4 5^{\circ} = 2$ and $tan 4 5^{\circ} = cot 4 5^{\circ} = 1$ , so $B$ is that same fraction directly.
Reciprocal and quotient relations between the ratios
4
$A = B$
Why: The two expressions reduce to the identical surd fraction.

Shortcut

Double $A$ 's numerator and denominator: $\frac{2 - 1}{2 + 1}$ , which is exactly what $B$ reads as on substitution.

Common mistakes

Option B

Rationalised only one of the two and compared the resulting decimals carelessly, concluding $A$ is the larger.

Option C

Mis-simplified $A$ , for instance leaving $\frac{2}{2}$ as $2$ , so that $A$ came out below $B$ .

Option D

Sign slip making both expressions negative; in fact both numerators are positive because $\frac{1}{2} > \frac{1}{2}$ and $2 > 1$ .

Skills used

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

51Arithmetic • Clocks • Angle Between Hands

ModerateIdeal time: about 1.5 min

If $θ$ measured in radians is the angle between the hour hand and the minute hand of a clock when the time is $4 : 36$ pm, then which one of the following is correct ?

\frac{3 π}{5} < θ < \frac{4 π}{5}

\frac{2 π}{5} < θ < \frac{3 π}{5}

\frac{π}{5} \leq θ \leq \frac{2 π}{5}

\frac{7 π}{15} \leq θ \leq \frac{8 π}{15}

Show answer & solution

Option B

\frac{2 π}{5} < θ < \frac{3 π}{5}

Approach

Place both hands in degrees measured from $12$ , take the difference, then convert to radians and locate it among the option ranges.

Angle between the hands of a clock

Step-by-step solution

1
Minute hand: $36 \times 6^{\circ} = 21 6^{\circ}$
Why: The minute hand sweeps $36 0^{\circ}$ in $60$ minutes, i.e. $6^{\circ}$ per minute.
Angle between the hands of a clock
2
Hour hand: $4 \times 3 0^{\circ} + 36 \times 0 \cdot 5^{\circ} = 12 0^{\circ} + 1 8^{\circ} = 13 8^{\circ}$
Why: The hour hand moves $3 0^{\circ}$ per hour and, crucially, a further $0 \cdot 5^{\circ}$ per minute, so at $4 : 36$ it has already left the $4$ .
3
$θ = 21 6^{\circ} - 13 8^{\circ} = 7 8^{\circ}$
Why: The angle between the hands is the difference of their positions, and $7 8^{\circ} < 18 0^{\circ}$ so it is already the smaller angle.
4
$7 8^{\circ} = \frac{π \times 78}{180} = \frac{13 π}{30}$ radians
Why: Multiply by $\frac{π}{180}$ to convert degrees to radians.
Converting between degrees and radians
5
$\frac{2 π}{5} = \frac{12 π}{30} < \frac{13 π}{30} < \frac{18 π}{30} = \frac{3 π}{5}$
Why: Putting the option bounds over the common denominator $30$ decides the containment at a glance.

Shortcut

The hands are $\frac{11}{2} M - 30 H = \frac{11}{2} \times 36 - 120 = 7 8^{\circ}$ apart, and $7 8^{\circ}$ sits between $7 2^{\circ}$ and $10 8^{\circ}$ .

Common mistakes

Option A

No natural route reaches the range $10 8^{\circ}$ to $14 4^{\circ}$ .

Option C

Slip in placing the hands that leaves the gap below $7 2^{\circ}$ , for instance reading the time as $4 : 35$ and dropping the hour hand's drift.

Option D

Left the hour hand on the $4$ at $12 0^{\circ}$ , giving $21 6^{\circ} - 12 0^{\circ} = 9 6^{\circ}$ , which falls inside the closed range $[8 4^{\circ}, 9 6^{\circ}]$ .

Skills used

Angle between the hands of a clock Converting between degrees and radians

52Trigonometry • Trigonometric Identity • Maximum and Minimum Values

HardIdeal time: about 2.5 min

Consider the following statements :

1: If $4 5^{\circ} < θ < 6 0^{\circ}$ , then $sec^{2} θ + cosec^{2} θ = α^{2}$ for some real number $α > 1$ .

2: If $0^{\circ} < θ < 4 5^{\circ}$ , then $\frac{1 + c o s θ}{1 - c o s θ} = x^{2}$ for some real number $x > 2$ .

3: If $0^{\circ} < θ < 4 5^{\circ}$ , then $\frac{c o s θ}{1 - t a n θ} + \frac{s i n θ}{1 - c o t θ} \geq 2$ .

What is the number of true statements ?

AZeroBOneCTwoDThree

Show answer & solution

Option C

Two

Approach

Rewrite every ratio in $sin θ, cos θ$ , evaluate each statement's expression as a function of $θ$ across its stated open interval, and check whether it can actually satisfy the claimed bound — a genuine counterexample or a clean bounding identity settles each one independently.

Sine and cosine lie between

- 1

and

1

Step-by-step solution

1
$sec^{2} θ + cosec^{2} θ = \frac{1}{sin ^{2} θ cos ^{2} θ} = \frac{4}{sin ^{2} 2 θ}$
Why: Rewrite statement 1 in sine and cosine and combine over a common denominator.
Reciprocal and quotient relations between the ratios
2
For $4 5^{\circ} < θ < 6 0^{\circ}$ , $2 θ \in (9 0^{\circ}, 12 0^{\circ})$ , so $sin 2 θ \in (\frac{3}{2}, 1)$ , giving $sec^{2} θ + cosec^{2} θ \in (4, \frac{16}{3})$
Why: Sine is decreasing past $9 0^{\circ}$ , so it shrinks from $1$ towards $sin 12 0^{\circ}$ , pushing the reciprocal square up from $4$ .
Sine and cosine lie between $- 1$ and $1$ How the ratios move across the first quadrant
3
Any value in $(4, \frac{16}{3})$ is $α^{2}$ for $α = value > 2 > 1$ , so statement 1 is true
Why: The claim only needs some real $α > 1$ to exist, and it does.
4
$\frac{1 + cos θ}{1 - cos θ} = cot^{2} \frac{θ}{2}$ ; for $0^{\circ} < θ < 4 5^{\circ}$ , $\frac{θ}{2} \in (0^{\circ}, 22. 5^{\circ})$ so $cot \frac{θ}{2} > cot 22. 5^{\circ} = 1 + 2 \approx 2.414$
Why: Cotangent decreases as its angle grows, so shrinking the half-angle below $22. 5^{\circ}$ only makes it bigger.
5
Hence $x = cot \frac{θ}{2} > 2.414 > 2$ throughout the interval, so statement 2 is true
Why: Direct substitution of the boundary value bounds the whole range.
6
$\frac{cos θ}{1 - tan θ} + \frac{sin θ}{1 - cot θ} = \frac{cos ^{2} θ}{cos θ - sin θ} - \frac{sin ^{2} θ}{cos θ - sin θ} = \frac{cos ^{2} θ - sin ^{2} θ}{cos θ - sin θ} = cos θ + sin θ$
Why: Clear both fractions to the same denominator $cos θ - sin θ$ (non-zero since $θ < 4 5^{\circ}$ ) and factor as a difference of squares.
Difference of two squares
7
For $0^{\circ} < θ < 4 5^{\circ}$ , $sin θ + cos θ = 2 sin (θ + 4 5^{\circ}) \in (1, 2)$ , which never reaches $2$
Why: Sine and cosine are each at most $1$ , so their sum caps out at $2 \approx 1.414$ — the counterexample to statement 3 is every $θ$ in the interval, not a special one.
8
Statements 1 and 2 hold, statement 3 fails $\Rightarrow$ exactly two true statements

Shortcut

Statement 3 alone can be killed on sight: it simplifies to $sin θ + cos θ$ , and no sum of two numbers each below $1$ reaches $2$ — no computation needed.

Common mistakes

Option A

Assumed some sign or direction reverses across the open interval and concluded no statement survives.

Option B

Miscalculated $\frac{1 + cos θ}{1 - cos θ}$ near $θ = 4 5^{\circ}$ , getting a value under $4$ so $x < 2$ , and counted only statement 1 as true.

Option D

Mistook $sin θ + cos θ$ for the familiar $x + \frac{1}{x} \geq 2$ pattern (as with $sec θ + cosec θ$ ) and accepted statement 3 as always true, counting all three.

Skills used

Reciprocal and quotient relations between the ratios Sine and cosine lie between

- 1

and

1

Difference of two squares How the ratios move across the first quadrant

53Trigonometry • Heights and Distances

HardIdeal time: about 2.5 min

Let AB represent a building of height h metre with A being its top, B being its bottom. Let A'B' represent a tower of height (h + x) metre (x > 0) with A' being its top and B' being its bottom. Let BB' = d metre. Let the angle of elevation of A' as seen from A be $4 5^{\circ}$ .

Consider the following statements :

Statement I: $h + x > d$

Statement II: The angle of depression of B as seen from A' is less than $4 5^{\circ}$ .

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

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

Show answer & solution

Option C

Statement I is true but Statement II is false

Approach

Turn the $4 5^{\circ}$ elevation into the relation $x = d$ , then test each statement against it.

Angles of elevation and depression

Step-by-step solution

1
From $A$ the horizontal run to the tower is $d$ and the rise to $A^{'}$ is $(h + x) - h = x$
Why: $A$ and $B^{'}$ are at horizontal distance $B B^{'} = d$ , and $A$ is already at height $h$ , so only the excess height $x$ is seen above the horizontal.
Angles of elevation and depression
2
$tan 4 5^{\circ} = \frac{x}{d} = 1 \Rightarrow x = d$
Why: The angle of elevation of $A^{'}$ from $A$ is given as $4 5^{\circ}$ , whose tangent is $1$ .
Sine, cosine and tangent in a right triangle Exact trigonometric ratios of the standard angles
3
Statement I: $h + x = h + d > d$ since $h > 0$
Why: A building has positive height, so adding it to $d$ must exceed $d$ . Statement I is TRUE.
4
Depression of $B$ from $A^{'}$ : $tan ϕ = \frac{h + x}{d} = \frac{h + d}{d} = 1 + \frac{h}{d} > 1$
Why: $B$ is at ground level and $A^{'}$ is at height $h + x$ , with $d$ between them horizontally.
5
$tan ϕ > 1 = tan 4 5^{\circ} \Rightarrow ϕ > 4 5^{\circ}$ , so Statement II is FALSE
Why: The tangent increases strictly across the first quadrant, so a tangent above $1$ means an angle above $4 5^{\circ}$ .
How the ratios move across the first quadrant

Shortcut

$4 5^{\circ}$ from $A$ forces $x = d$ ; then $A^{'}$ looks down on $B$ over the same horizontal run $d$ but from an extra height $h$ , so that depression must exceed $4 5^{\circ}$ .

Common mistakes

Option A

Took the depression of $B$ to be the $4 5^{\circ}$ alternate angle of the elevation of $A^{'}$ from $A$ , confusing $B$ with $A$ , and then read that as explaining Statement I.

Option B

Same confusion of $B$ with $A$ , judging Statement II true but unrelated to Statement I.

Option D

Concluded $h + x = d$ from the $4 5^{\circ}$ elevation, forgetting that the elevation is measured from $A$ at height $h$ , not from $B$ .

Skills used

Angles of elevation and depression Sine, cosine and tangent in a right triangle Exact trigonometric ratios of the standard angles How the ratios move across the first quadrant

54Trigonometry • Heights and Distances

HardIdeal time: about 2.5 min

A man, standing at a point X on the bank XY of a river that cannot be crossed, observes a tower to be N $α^{\circ}$ E on the opposite parallel bank. He then walks $200$ m along the bank to the point Y towards East, and finds the tower to be N $β^{\circ}$ W. From these observations, the breadth of the river will be

(Given that $tan α^{\circ} = 2$ and $tan β^{\circ} = 0 \cdot 5$ )

60

70

80

90

Show answer & solution

Option C

80

Approach

Put the bank on the east-west axis, let the breadth be $w$ , and write the tower's east offset from each of the two sightings.

Reading a compass bearing N

α^{\circ}

E as a right triangle

Step-by-step solution

1
Take $X$ at the origin with the bank running east, so $Y = (200, 0)$ and the tower is at $(t, w)$ with $w$ the breadth
Why: The banks are parallel, so the tower's distance north of the line $X Y$ is exactly the breadth of the river.
2
N $α^{\circ}$ E from $X$ : $tan α^{\circ} = \frac{t}{w} \Rightarrow t = 2 w$
Why: A bearing N $α^{\circ}$ E is measured from north towards east, so its tangent is the east offset over the north offset.
Sine, cosine and tangent in a right triangle Reading a compass bearing N $α^{\circ}$ E as a right triangle
3
N $β^{\circ}$ W from $Y$ : $tan β^{\circ} = \frac{200 - t}{w} \Rightarrow 200 - t = 0 \cdot 5 w$
Why: From $Y$ the tower lies to the west by $200 - t$ , over the same north offset $w$ .
4
$2 w + 0 \cdot 5 w = 200 \Rightarrow 2 \cdot 5 w = 200 \Rightarrow w = 80$
Why: Adding the two equations eliminates $t$ , because the two east offsets together span the $200$ m walked.
Turn a worded condition into a linear equation in one unknown

Shortcut

$200 = w (tan α^{\circ} + tan β^{\circ}) = w (2 + 0 \cdot 5)$ , so $w = \frac{200}{2 \cdot 5} = 80$ m.

Common mistakes

Option A

No natural route reaches $60$ m.

Option B

No natural route reaches $70$ m.

Option D

No natural route reaches $90$ m.

Skills used

Reading a compass bearing N

α^{\circ}

E as a right triangle Sine, cosine and tangent in a right triangle Turn a worded condition into a linear equation in one unknown

55Trigonometry • Trigonometric Identity • Degree and Radian

ModerateIdeal time: about 1.5 min

The value of $\frac{s i n 1 ^{\circ}}{s i n 1 ^{c}}$ where $1^{c}$ represents 1 radian is

AEqual to

1

BLess than

1

CGreater than

1

but less than

2

DGreater than

2

Show answer & solution

Option B

Less than

1

Approach

Estimate both sines by size rather than computing the ratio exactly: $1^{\circ}$ is a tiny angle and $1$ radian is a large one, so their sines are wildly different orders of magnitude.

Converting between degrees and radians

Step-by-step solution

1
$1^{\circ} = \frac{π}{180} \approx 0.01745$ radians, a small angle, so $sin 1^{\circ} \approx 0.01745$ (sine barely below its own radian measure for small angles)
Why: Convert to the same unit to compare sizes honestly.
Converting between degrees and radians
2
$1$ radian $\approx 57. 3^{\circ}$ , well into the first quadrant, so $sin 1^{c} \approx 0.841$
Why: A bare ' $1$ ' inside a trig ratio is $1$ radian, not $1^{\circ}$ — a nearly $6 0^{\circ}$ angle.
3
$\frac{sin 1 ^{\circ}}{sin 1 ^{c}} \approx \frac{0.01745}{0.841} \approx 0.0207$
4
$0.0207 < 1$ , and it is nowhere near $1$ or $2$
Why: Test option A ( $= 1$ ): fails by two orders of magnitude. Option C ( $1$ to $2$ ) and D ( $> 2$ ): fail, the ratio is far below $1$ .

Shortcut

No computation is really needed: $1^{\circ}$ is a much smaller angle than $1$ radian, and sine is increasing through the first quadrant, so $sin 1^{\circ}$ must be much smaller than $sin 1^{c}$ — ratio under $1$ .

Common mistakes

Option A

Read $1^{c}$ as if it also meant $1^{\circ}$ , making numerator and denominator identical and the ratio exactly $1$ .

Option C

Some intermediate rounding of both sines to comparable sizes without noticing the near- $50 \times$ gap.

Option D

Computed the reciprocal ratio $sin 1^{c} / sin 1^{\circ} \approx 48$ instead of the ratio actually printed, and reported it as 'greater than $2$ '.

Skills used

Converting between degrees and radians

56Mensuration • Plane Figures • Circle

ModerateIdeal time: about 1.5 min

The diameters of two given circles are in the ratio $12 : 5$ and the sum of their areas is equal to the area of a circle of diameter $65$ cm. What are their radii ?

12

cm and

5

cmB

24

cm and

10

cmC

60

cm and

25

cmD

30

cm and

12.5

Show answer & solution

Option D

30

cm and

12.5

Approach

Write the two radii as $6 k$ and $2 \cdot 5 k$ from the diameter ratio, add their areas, and match against the area of the given circle.

Area and circumference of a circle

Step-by-step solution

1
Diameters $12 k$ and $5 k$ , so radii $6 k$ and $2 \cdot 5 k$
Why: A ratio fixes the diameters only up to a common factor $k$ , and each radius is half its diameter.
Introduce a common factor $k$ for a ratio
2
$π (6 k)^{2} + π (2 \cdot 5 k)^{2} = π (32 \cdot 5)^{2}$
Why: The third circle has diameter $65$ cm, hence radius $32 \cdot 5$ cm, and the areas are to balance.
Area and circumference of a circle
3
$36 k^{2} + 6 \cdot 25 k^{2} = 1056 \cdot 25 \Rightarrow 42 \cdot 25 k^{2} = 1056 \cdot 25$
Why: Cancel $π$ throughout and collect.
4
$k^{2} = 25 \Rightarrow k = 5$ , so the radii are $30$ cm and $12 \cdot 5$ cm
Why: Only the positive root is meaningful for a length.
denotes the non-negative root

Shortcut

$12$ , $5$ , $13$ is a Pythagorean triple, so two circles with diameters $12 k$ and $5 k$ have the combined area of one of diameter $13 k$ ; matching $13 k = 65$ gives $k = 5$ and diameters $60$ and $25$ , i.e. radii $30$ and $12 \cdot 5$ .

Common mistakes

Option A

Read the ratio $12 : 5$ as the actual radii in centimetres, never using the $65$ cm circle at all.

Option B

No natural route reaches $24$ cm and $10$ cm.

Option C

Found the diameters $60$ cm and $25$ cm correctly but reported them instead of the radii.

Skills used

Area and circumference of a circle Introduce a common factor

k

for a ratio

denotes the non-negative root

57Mensuration • Solids • Cube

ModerateIdeal time: about 1.5 min

A hollow cube is formed by joining six identical squares. A rectangular cello tape of length $4$ cm and breadth $0.5$ cm is used for joining each pair of edges. What is the total area of cello tape used ?

12

square cmB

24

square cmC

36

square cmD

48

square cm

Show answer & solution

Option B

24

square cm

Approach

Count the joins — one per edge of the cube — and multiply by the area of one strip of tape.

A cube has

6

faces,

12

edges and

8

vertices

Step-by-step solution

1
A cube has $12$ edges, and each edge is where exactly two of the six squares meet
Why: The six squares are joined only along edges, and the finished hollow cube has twelve of them, so there are twelve pairs of edges to tape.
A cube has $6$ faces, $12$ edges and $8$ vertices
2
Area of one strip $= 4 \times 0 \cdot 5 = 2$ square cm
Why: Each strip is a rectangle of the stated length and breadth.
Perimeter and area of a rectangle
3
Total $= 12 \times 2 = 24$ square cm
Why: One strip per join, twelve joins.

Shortcut

Twelve edges at $2$ square cm of tape each is $24$ square cm.

Common mistakes

Option A

Counted one strip per face, $6 \times 2 = 12$ , instead of one per edge.

Option C

No natural route reaches $36$ square cm.

Option D

Counted $24$ joins by treating each edge once for each of the two squares meeting there, double-counting every edge.

Skills used

A cube has

6

faces,

12

edges and

8

vertices Perimeter and area of a rectangle

58Geometry • Circle • Tangents

HardIdeal time: about 2.5 min

Two straight lines AB and AC include an angle. A circle is drawn in this angle which touches both these lines. One more circle is drawn which touches both these lines as well as the previous circle. If the area of the bigger circle is $9$ times the area of the smaller circle, then what must be the angle A ?

4 5^{\circ}

6 0^{\circ}

7 5^{\circ}

9 0^{\circ}

Show answer & solution

Option B

6 0^{\circ}

Approach

Both centres lie on the bisector of angle $A$ at distance $\frac{radius}{sin ( A /2 )}$ ; equate the gap between them to the sum of the radii.

The two tangents from an external point

Step-by-step solution

1
Area ratio $9 \Rightarrow$ radius ratio $3$ , so $R = 3 r$
Why: Areas of similar figures go as the square of corresponding lengths, so a factor $9$ in area is a factor $3$ in radius.
Similar triangles have proportional sides Area and circumference of a circle
2
Each centre lies on the bisector of $∠ A$ , and if $2 θ = ∠ A$ then $sin θ = \frac{radius}{distance from A}$
Why: A circle touching both arms has its centre equidistant from them, hence on the bisector; the radius to a point of contact is perpendicular to that arm, giving a right triangle with angle $θ$ at $A$ .
The two tangents from an external point Sine, cosine and tangent in a right triangle
3
Distances from $A$ : $d_{1} = \frac{r}{sin θ}$ and $d_{2} = \frac{3 r}{sin θ}$
Why: Apply that relation to each circle in turn.
4
Circles touch externally: $d_{2} - d_{1} = r + R = 4 r$
Why: The centres of externally touching circles are apart by the sum of the radii, and both centres are on the same ray from $A$ .
Distance between the centres of touching circles
5
$\frac{2 r}{sin θ} = 4 r \Rightarrow sin θ = \frac{1}{2} \Rightarrow θ = 3 0^{\circ} \Rightarrow ∠ A = 6 0^{\circ}$
Why: Cancel $r$ , solve for the half-angle, then double it to get the angle asked for.
Exact trigonometric ratios of the standard angles

Shortcut

For two circles inscribed in an angle and touching each other, $sin \frac{A}{2} = \frac{R - r}{R + r} = \frac{3 r - r}{3 r + r} = \frac{1}{2}$ , so $A = 6 0^{\circ}$ .

Common mistakes

Option A

No natural route reaches $4 5^{\circ}$ .

Option C

No natural route reaches $7 5^{\circ}$ .

Option D

No natural route reaches $9 0^{\circ}$ .

Skills used

The two tangents from an external point Distance between the centres of touching circles Sine, cosine and tangent in a right triangle Similar triangles have proportional sides Area and circumference of a circle Exact trigonometric ratios of the standard angles

59Mensuration • Plane Figures • Triangle

ModerateIdeal time: about 2 min

An isosceles triangle is drawn outside on one of the sides of a square as base in such a way that the perimeter of the complete figure is $\frac{7}{6}$ times the perimeter of the original square. What is the ratio of area of the triangle to the area of the original square ?

1 : 1

2 : 3

1 : 2

1 : 3

Show answer & solution

Option D

1 : 3

Approach

Use the perimeter condition to find the triangle's equal sides in terms of the square's side, get its height by Pythagoras, then compare areas.

Pythagoras' theorem

Step-by-step solution

1
Let the square have side $a$ ; its perimeter is $4 a$ and the complete figure's perimeter is $3 a + 2 s$ , where $s$ is each equal side of the triangle
Why: The side carrying the triangle is interior to the finished figure, so only three sides of the square remain on its boundary, together with the triangle's two equal sides.
Perimeter and area of a square
2
$3 a + 2 s = \frac{7}{6} \times 4 a = \frac{14 a}{3} \Rightarrow 2 s = \frac{14 a}{3} - 3 a = \frac{5 a}{3} \Rightarrow s = \frac{5 a}{6}$
Why: Impose the given perimeter relation and solve for $s$ .
3
The altitude to the base bisects it, so $h = (\frac{5 a}{6})^{2} - (\frac{a}{2})^{2} = \frac{25 a ^{2} - 9 a ^{2}}{36} = \frac{4 a}{6} = \frac{2 a}{3}$
Why: In an isosceles triangle the foot of the altitude from the apex is the midpoint of the base, so Pythagoras applies to a right triangle with legs $h$ and $\frac{a}{2}$ .
Pythagoras' theorem Congruence criteria for triangles
4
Area of triangle $= \frac{1}{2} \times a \times \frac{2 a}{3} = \frac{a ^{2}}{3}$
Why: Half base times height.
Area of a triangle from base and height
5
$\frac{a ^{2} /3}{a ^{2}} = \frac{1}{3}$ , i.e. $1 : 3$
Why: Compare with the square's area $a^{2}$ ; the common $a^{2}$ cancels, so the ratio is independent of the size of the square.

Shortcut

Take $a = 6$ : the perimeter condition gives $18 + 2 s = 28$ , so $s = 5$ , and $(3, 4, 5)$ hands over the height $4$ at once, making the areas $12$ and $36$ .

Common mistakes

Option A

No natural route reaches $1 : 1$ .

Option B

No natural route reaches $2 : 3$ .

Option C

Took the triangle's height to equal the square's side, giving an area of $\frac{a ^{2}}{2}$ .

Skills used

Perimeter and area of a square Pythagoras' theorem Area of a triangle from base and height Congruence criteria for triangles

60Mensuration • Plane Figures • Triangle

ModerateIdeal time: about 1.5 min

What is the area of the triangle whose sides are $51$ cm, $37$ cm and $20$ cm ?

300

square cmB

305

square cmC

306

square cmD

307

square cm

Show answer & solution

Option C

306

square cm

Approach

Apply Heron's formula to the three given sides.

Heron's formula

Step-by-step solution

1
$s = \frac{51 + 37 + 20}{2} = \frac{108}{2} = 54$
Why: Heron's formula needs the semi-perimeter first.
Heron's formula
2
$s - a = 54 - 51 = 3$ , $s - b = 54 - 37 = 17$ , $s - c = 54 - 20 = 34$
Why: Form the three differences; they are all positive, confirming the triangle exists.
Triangle inequality
3
Area $= 54 \times 3 \times 17 \times 34 = 162 \times 578 = 93636$
Why: Multiply the four factors under the single radical.
4
$54 \times 3 \times 17 \times 34 = (27 \times 2) \times 3 \times 17 \times (17 \times 2) = 2^{2} \times 3^{4} \times 1 7^{2} \Rightarrow Area = 2 \times 9 \times 17 = 306$
Why: Pairing the factors into squares avoids the long multiplication and the square root of a five-digit number.
Prime factorisation $ab = a b$ and $a / b = a / b$

Shortcut

Group as $(54 \times 34) \times (3 \times 17) = 1836 \times 51$ , or better factorise to $2^{2} \times 3^{4} \times 1 7^{2}$ and read the root off as $2 \times 9 \times 17 = 306$ .

Common mistakes

Option A

Slip in the multiplication or the square root, landing on the round number $300$ .

Option B

Slip in the multiplication or the square root, landing just below $306$ .

Option D

Slip in the multiplication or the square root, landing just above $306$ .

Skills used

Heron's formula Prime factorisation

ab = a b

and

a / b = a / b

Triangle inequality