Elementary Mathematics 2020 (I)

Step-by-step solution

1. 1
   Write $a=\tan x$, $b=\tan y$, so $\cot x=\frac{1}{a}$ and $\cot y=\frac{1}{b}$

   Why: One letter per tangent makes the shared structure visible; cotangents are just reciprocals.

   Reciprocal and quotient relations between the ratios
2. 2
   First product $=\left(a+b\right)\left(1-\frac{1}{ab}\right)=\frac{\left(a+b\right)\left(ab-1\right)}{ab}$

   Why: Clearing the fraction inside the bracket puts the whole term over $ab$.

   Clear the denominators of an equation
3. 3
   Second product $=\left(\frac{1}{a}+\frac{1}{b}\right)\left(1-ab\right)=\frac{\left(a+b\right)\left(1-ab\right)}{ab}$

   Why: $\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$, so the second term carries the same $\frac{a+b}{ab}$ as the first.

4. 4
   Sum $=\frac{a+b}{ab}\left[\left(ab-1\right)+\left(1-ab\right)\right]=\frac{a+b}{ab}\times 0=0$

   Why: The two brackets are exact negatives, so the common factor is multiplied by zero — no values of $x$ and $y$ are needed.

   Difference of two squares

Common mistakes

Skills used

Step-by-step solution

1. 1
   ${\sec }^{2}x-{\tan }^{2}x=1$, so $\left(\sec x-\tan x\right)\left(\sec x+\tan x\right)=1$

   Why: The Pythagorean identity $1+{\tan }^{2}x={\sec }^{2}x$ factored as a difference of squares — this is the only fact the question needs.

   Pythagorean trigonometric identitiesDifference of two squares
2. 2
   $\sec x-\tan x=\frac{1}{\sec x+\tan x}$

   Why: Each of the two is the reciprocal of the other, so the numerator can be rewritten in terms of the denominator.

   Negative exponent is a reciprocal
3. 3
   $\frac{\sec x-\tan x}{\sec x+\tan x}=\frac{1}{{\left(\sec x+\tan x\right)}^2}$

   Why: Substituting the numerator turns the fraction into one over a square — exactly the shape a square root wants.

   $\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$
4. 4
   $\sqrt{\frac{1}{{\left(\sec x+\tan x\right)}^2}}=\frac{1}{\sec x+\tan x}$

   Why:  ​ denotes the non-negative root, and the expression under it is only defined where $\sec x+\tan x>0$.

    ​ denotes the non-negative root

Common mistakes

Skills used

Step-by-step solution

1. 1
   Adjacent $=63$, opposite $=16$

   Why: $\cot \theta =\frac{\text{adjacent}}{\text{opposite}}$; $\theta$ is in the first quadrant, so both legs are positive.

   Sine, cosine and tangent in a right triangle
2. 2
   Hypotenuse $=\sqrt{63^2+16^2}=\sqrt{3969+256}=\sqrt{4225}=65$

   Why: Pythagoras supplies the third side; $4225=65^2$, so the triangle is the Pythagorean triple $(16,63,65)$.

   Pythagoras' theorem
3. 3
   $\sin \theta =\frac{16}{65},\cos \theta =\frac{63}{65}$

   Why: Opposite over hypotenuse and adjacent over hypotenuse; both positive in the first quadrant.

4. 4
   $\sin \theta +\cos \theta =\frac{16+63}{65}=\frac{79}{65}$

   Why: Same denominator, so the numerators simply add.

   Reciprocal and quotient relations between the ratios

Common mistakes

Skills used

Step-by-step solution

1. 1
   ${\sin }^4\theta +{\cos }^4\theta ={\left({\sin }^2\theta +{\cos }^2\theta \right)}^2-2{\sin }^2\theta {\cos }^2\theta$

   Why: $a^2+b^2=(a+b{)}^2-2ab$ with $a={\sin }^2\theta$, $b={\cos }^2\theta$ — the move that connects the denominator to the numerator.

   Square of a sum and of a difference
2. 2
   $=1-2{\sin }^2\theta {\cos }^2\theta$

   Why: ${\sin }^2\theta +{\cos }^2\theta =1$, so the squared bracket is just $1$.

   Pythagorean trigonometric identities
3. 3
   Fraction $=\frac{1-2{\sin }^2\theta {\cos }^2\theta }{1-2{\sin }^2\theta {\cos }^2\theta }=1$

   Why: Numerator and denominator are the same expression, so the fraction is $1$ for every $\theta$ where it is defined.

   Evaluate a nested fraction from the inside out
4. 4
   Value $=1+4=5$

   Why: The $+4$ sits outside the fraction and must still be added — the step the question is testing.

Common mistakes

Skills used

Step-by-step solution

1. 1
   Diagonal $=\sqrt{48^2+14^2}=\sqrt{2304+196}=\sqrt{2500}=50$ cm

   Why: A rectangle's diagonal closes a right triangle on the two sides; $(14,48,50)$ is $2\times (7,24,25)$.

   Perimeter and area of a rectanglePythagoras' theorem
2. 2
   $\theta$ is at the longer side, so $\cos \theta =\frac{48}{50}$ and $\sin \theta =\frac{14}{50}$

   Why: The side the angle is measured from is the adjacent one; the width $14$ is then opposite.

   Sine, cosine and tangent in a right triangle
3. 3
   $\sec \theta =\frac{50}{48}=\frac{25}{24}$ and $\mathrm{cosec}\theta =\frac{50}{14}=\frac{25}{7}$

   Why: Secant and cosecant are the reciprocals of cosine and sine; cancelling the common $2$ first keeps the numbers small.

   Reciprocal and quotient relations between the ratios
4. 4
   $\frac{25}{24}+\frac{25}{7}=25\times \frac{7+24}{168}=\frac{775}{168}$

   Why: $\text{LCM}(24,7)=168$; pulling the $25$ out first means only one small addition, $7+24=31$, has to be right.

   LCM and HCF of fractions

Common mistakes

Skills used

Step-by-step solution

1. 1
   Let $u=\frac{x}{y},v=\frac{y}{z},w=\frac{z}{x}$, so $uvw=1$

   Why: The product of the three fractions cancels completely — that is the hidden structure the question is built on.

   Evaluate a nested fraction from the inside out
2. 2
   The sides are $a=v+w,b=w+u,c=u+v$

   Why: Each given side is a sum of two of the three fractions, with each fraction appearing in exactly two sides.

3. 3
   $s=\frac{a+b+c}{2}=u+v+w$, hence $s-a=u,s-b=v,s-c=w$

   Why: Each fraction appears twice in the perimeter, so the semi-perimeter is their plain sum and each Heron factor is a single fraction.

   Heron's formula
4. 4
   Area $=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=\sqrt{\left(u+v+w\right)uvw}=\sqrt{u+v+w}$

   Why: Heron's formula, with $uvw=1$ removing three of the four factors.

5. 5
   Area $=\sqrt{\frac{x}{y}+\frac{y}{z}+\frac{z}{x}}$

   Why: Translating back; note this is not $\frac{xy+yz+zx}{xyz}$, which is what $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ would give.

Common mistakes

Skills used

Step-by-step solution

1. 1
   Let $\angle B=30^{\circ }$, $\angle C=45^{\circ }$ and $BC=\sqrt{3}+1$ cm

   Why: 'Included side' means the side joining the two given angles, so $BC$ is the side between them.

   Angle sum of a triangle
2. 2
   Drop $AD\perp BC$ with $AD=h$; then $BD=h\cot 30^{\circ }=h\sqrt{3}$ and $DC=h\cot 45^{\circ }=h$

   Why: The altitude splits the triangle into two right triangles sharing $h$, and in each the base is $h$ times the cotangent of the base angle.

   Area of a triangle from base and heightSine, cosine and tangent in a right triangle
3. 3
   $h\sqrt{3}+h=\sqrt{3}+1\Rightarrow h\left(\sqrt{3}+1\right)=\sqrt{3}+1\Rightarrow h=1$ cm

   Why: $BD+DC=BC$; the bracket matches the given length exactly, so $h=1$ with no surd arithmetic at all.

   Exact trigonometric ratios of the standard anglesTurn a worded condition into a linear equation in one unknown
4. 4
   Area $=\frac{1}{2}\times BC\times h=\frac{1}{2}\left(\sqrt{3}+1\right)\times 1=\frac{1}{2}\left(\sqrt{3}+1\right)$ cm${}^2$

   Why: Base times height, halved, with $BC$ as the base since $h$ was measured to it.

Common mistakes

Skills used

Step-by-step solution

1. 1
   The diagonals of a parallelogram bisect each other, so $O$ is the midpoint of both $AC$ and $BD$

   Why: This is what makes $O$ a usable point: it is the centre of the figure, not just some interior point.

   The diagonals of a rectangle are equal and bisect each other
2. 2
   In triangle $DBC$, $O$ is the midpoint of $DB$ and $OE\parallel BC$, so $E$ is the midpoint of $DC$

   Why: $EF\parallel DA$ and $DA\parallel BC$, so $OE\parallel BC$; a line through the midpoint of one side parallel to another side bisects the third.

   Midpoint theoremParallel lines cut transversals proportionally
3. 3
   Let $DC=b$ and let $h$ be the distance between the parallel sides $DC$ and $AB$; the plate's area is $bh$

   Why: Naming the base and height of the parallelogram lets both areas be written in the same two letters.

   Perimeter and area of a rectangle
4. 4
   $O$ is the midpoint of $BD$, so its distance from $DC$ is $\frac{h}{2}$

   Why: $B$ lies on $AB$ at distance $h$ from $DC$ and $D$ lies on $DC$ itself, so their midpoint is halfway between.

5. 5
   $[DOE]=\frac{1}{2}\times DE\times \frac{h}{2}=\frac{1}{2}\times \frac{b}{2}\times \frac{h}{2}=\frac{bh}{8}$

   Why: Base $DE=\frac{b}{2}$ lies along $DC$, so the height to it is $O$'s distance from $DC$.

   Area of a triangle from base and height
6. 6
   Remaining $:$ whole $=\frac{bh-\frac{bh}{8}}{bh}=\frac{7}{8}$

   Why: The letters cancel, so the ratio is the same for every parallelogram — the shape of the plate never mattered.

   Introduce a common factor $k$ for a ratio

Common mistakes

Skills used

Step-by-step solution

1. 1
   The line joining the centres is the perpendicular bisector of the common chord, so each half is $12$ cm

   Why: Both centres are equidistant from the chord's two endpoints, so both lie on its perpendicular bisector.

   The perpendicular from the centre bisects a chord
2. 2
   From the $20$ cm centre: $\sqrt{20^2-12^2}=\sqrt{400-144}=16$ cm

   Why: Radius, half-chord and distance from the centre form a right triangle; $(12,16,20)$ is $4\times (3,4,5)$.

   Pythagoras' theorem
3. 3
   From the $16$ cm centre: $\sqrt{16^2-12^2}=\sqrt{256-144}=\sqrt{112}=4\sqrt{7}$ cm

   Why: Same right triangle on the other circle; $112$ is not a perfect square, so the answer will be an estimate.

4. 4
   The centres are on opposite sides of the chord, so $d=16+4\sqrt{7}$

   Why: Each centre sits on its own side of the chord for two circles crossing in the usual way, so the two distances add.

5. 5
   $10^2=100<112<121=11^2$, so $10<4\sqrt{7}<11$ and $26<d<27$

   Why: Bounding $\sqrt{112}$ between consecutive squares is enough — the options only ask which unit interval $d$ falls in. More precisely $4\sqrt{7}=10.58$, so $d=26.58$.

   Least perfect square not below a given bound

Common mistakes

Skills used

Step-by-step solution

1. 1
   Let $M$ be the midpoint of $BC$; since $AB=AC$, the line $AM$ is the perpendicular bisector of $BC$ and passes through the centre $O$

   Why: $A$ is equidistant from $B$ and $C$, and so is $O$ — two points determine the perpendicular bisector, so $A$, $O$, $M$ are collinear and $AM\perp BC$.

   The perpendicular from the centre bisects a chordBase angles of an isosceles triangle
2. 2
   Put $AM=p$ and $BM=m$. In triangle $ABM$: $p^2+m^2=8^2=64$

   Why: $AB$ is the hypotenuse of the right triangle $ABM$.

   Pythagoras' theorem
3. 3
   In triangle $OBM$: ${\left(p-5\right)}^2+m^2=5^2=25$

   Why: $O$ lies on $AM$ at distance $5$ from $A$, so $OM=p-5$, and $OB$ is the radius.

4. 4
   Subtracting: $p^2-{\left(p-5\right)}^2=39\Rightarrow 10p-25=39\Rightarrow p=6.4$

   Why: Subtracting kills $m^2$ and leaves one linear equation — the reason for setting both triangles up before solving either.

   Eliminate one variable from a linear pairDifference of two squares
5. 5
   $m^2=64-6.4^2=64-40.96=23.04\Rightarrow m=4.8$, so $BC=2m=9.6$ cm

   Why: $M$ is the midpoint, so the chord is twice the half-chord.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $R-r=4$

   Why: For two circles touching internally the centres are $\left|R-r\right|$ apart, and the larger radius is $R$.

   Distance between the centres of touching circles
2. 2
   $\pi R^2+\pi r^2=136\pi \Rightarrow R^2+r^2=136$

   Why: The sum of the two areas is given; dividing through by $\pi$ removes it from the working altogether.

   Area and circumference of a circle
3. 3
   $(r+4{)}^2+r^2=136\Rightarrow 2r^2+8r-120=0\Rightarrow r^2+4r-60=0$

   Why: Substituting $R=r+4$ leaves one unknown, and dividing by $2$ keeps the numbers small.

   Turn a worded condition into a linear equation in one unknown
4. 4
   $(r+10)(r-6)=0\Rightarrow r=6,R=10$

   Why: Split the middle term: $-60=10\times (-6)$ and $10+(-6)=4$. A radius cannot be negative, so $r=-10$ is rejected.

   Factorise a quadratic by splitting the middle termA product is zero only when a factor is zero

Common mistakes

Skills used

Step-by-step solution

1. 1
   $\pi r^2=a^2$

   Why: Let the circle have radius $r$ and the square side $a$; equal areas is the only link between the two figures.

   Area and circumference of a circlePerimeter and area of a square
2. 2
   $a=r\sqrt{\pi }$

   Why: Take the positive square root: a side length cannot be negative.

    ​ denotes the non-negative root
3. 3
   $\frac{2\pi r}{4a}=\frac{2\pi r}{4r\sqrt{\pi }}=\frac{\pi }{2\sqrt{\pi }}=\frac{\sqrt{\pi }}{2}$

   Why: Circumference is $2\pi r$ and the square's perimeter is $4a$; $r$ cancels, and $\pi /\sqrt{\pi }=\sqrt{\pi }$.

   $\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$

Common mistakes

Skills used

Step-by-step solution

1. 1
   $OA=4\sqrt{2}$

   Why: A circle touching both lines has its centre equidistant from them, hence on the bisector of the right angle. With radius $4$, the centre is $4$ from each line, so its distance from the corner $O$ is $\sqrt{4^2+4^2}=4\sqrt{2}$.

   Points equidistant from two intersecting lines form a pair of bisectorsPythagoras' theorem
2. 2
   $OB=\rho \sqrt{2}$

   Why: The small circle of radius $\rho$ sits in the same corner on the same bisector, so the identical calculation applies to it.

3. 3
   $AB=OA-OB=\sqrt{2}(4-\rho )$ and $AB=4+\rho$

   Why: $B$ lies between $O$ and $A$ on the bisector, and the two circles touch externally, so the centre distance is the sum of the radii.

   Distance between the centres of touching circles
4. 4
   $4\sqrt{2}-\sqrt{2}\rho =4+\rho \Rightarrow \rho =\frac{4(\sqrt{2}-1)}{\sqrt{2}+1}$

   Why: Collect the $\rho$ terms on one side and the constants on the other.

   Turn a worded condition into a linear equation in one unknown
5. 5
   $\rho =\frac{4(\sqrt{2}-1{)}^2}{(\sqrt{2}+1)(\sqrt{2}-1)}=4(3-2\sqrt{2})$, so the diameter is $8(3-2\sqrt{2})$

   Why: Multiply above and below by the conjugate $\sqrt{2}-1$; the denominator becomes $2-1=1$ and $(\sqrt{2}-1{)}^2=3-2\sqrt{2}$. The question asks for the diameter, so double the radius.

   Rationalise a denominator using the conjugateSquare of a sum and of a difference

Common mistakes

Skills used

Step-by-step solution

1. 1
   $r=3,R=3+1=4,h=7$

   Why: Thickness is the gap between the inner and outer walls, so the outer radius is the inner radius plus $1$.

   Unitary method: find one unit, then scale
2. 2
   $\text{inner curved}=2\pi rh=2\pi (3)(7)=42\pi$

   Why: This is the surface that one litre covers, so it is the unit of measurement for the whole question.

   Curved and total surface of a cylinder
3. 3
   $\text{outer curved}=2\pi Rh=2\pi (4)(7)=56\pi$

   Why: The outer wall is a cylinder of radius $4$ and the same height.

4. 4
   $\text{two rims}=2\times \pi (R^2-r^2)=2\pi (16-9)=14\pi$

   Why: The top and bottom of a hollow cylinder are rings, not discs: each is a circle of radius $4$ with a circle of radius $3$ removed.

   Area and circumference of a circleCut a compound plane figure into standard pieces
5. 5
   $\frac{42\pi +56\pi +14\pi }{42\pi }=\frac{112}{42}=\frac{8}{3}$

   Why: All surfaces total $112\pi$; paint is proportional to area, so divide by the $42\pi$ that one litre covers.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $\text{square’s area}=s^2$

   Why: Let the square have side $s$.

   Perimeter and area of a square
2. 2
   $\text{rectangle’s area}=s^2\times t$

   Why: One side of the rectangle is numerically $s^2$; call the other side $t$.

   Perimeter and area of a rectangle
3. 3
   $s^{2}t=s^2\Rightarrow t=1$

   Why: Equal areas, and $s^2\ne 0$, so the common factor cancels and the other side is $1$ — of unit length, whatever $s$ is.

   Turn a worded condition into a linear equation in one unknown

Common mistakes

Skills used

Step-by-step solution

1. 1
   $l=4k,b=3k$

   Why: A ratio fixes the sides only up to a common factor $k$.

   Introduce a common factor $k$ for a ratio
2. 2
   $\text{the diagonals bisect each other at }O$

   Why: In any parallelogram, and so in a rectangle, the diagonals cut each other in half; $O$ is therefore the midpoint of both.

   The diagonals of a rectangle are equal and bisect each other
3. 3
   $[\text{long-side triangle}]=\frac{1}{2}\times 4k\times \frac{3k}{2}=3k^2$

   Why: Its base is a long side and its apex is $O$, which lies halfway across the rectangle, so the height is half the breadth.

   Area of a triangle from base and height
4. 4
   $[\text{short-side triangle}]=\frac{1}{2}\times 3k\times \frac{4k}{2}=3k^2$, ratio $1:1$

   Why: Its base is a short side and its height is half the length. The longer base is matched by the shorter height, so the two areas are equal.