Elementary Mathematics 2018 (I)

Step-by-step solution

1. 1
   $\cos x+{\cos }^{2}x=1\Rightarrow {\cos }^{2}x=1-\cos x$

   Why: Rearranging the given condition.

2. 2
   ${\sin }^{2}x=1-{\cos }^{2}x=1-(1-\cos x)=\cos x$

   Why: Pythagoras, with the rearranged condition substituted for ${\cos }^{2}x$.

   Pythagorean trigonometric identities
3. 3
   ${\sin }^{4}x={\left({\sin }^{2}x\right)}^2={\cos }^{2}x$

   Why: Squaring both sides of the previous step.

   Power of a power, and the move to a common base
4. 4
   ${\sin }^{2}x+{\sin }^{4}x=\cos x+{\cos }^{2}x=1$

   Why: The sum is now the very expression the question gave as equal to $1$.

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Step-by-step solution

1. 1
   $p^2=(\sin A+\cos A{)}^2=1+2\sin A\cos A\Rightarrow \sin A\cos A=\frac{p^2-1}{2}$

   Why: Square the first given and use ${\sin }^{2}A+{\cos }^{2}A=1$.

   Square of a sum and of a differencePythagorean trigonometric identities
2. 2
   $q={\sin }^{3}A+{\cos }^{3}A=(\sin A+\cos A)\left({\sin }^{2}A-\sin A\cos A+{\cos }^{2}A\right)=p\left(1-\sin A\cos A\right)$

   Why: Sum of two cubes, with the Pythagorean identity collapsing the bracket.

   Sum and difference of two cubes
3. 3
   $q=p\left(1-\frac{p^2-1}{2}\right)=\frac{p\left(3-p^2\right)}{2}$

   Why: Substitute the expression already found for $\sin A\cos A$.

4. 4
   $2q=3p-p^3\Rightarrow p^3-3p+2q=0$

   Why: Clear the denominator and move every term to one side.

   Clear the denominators of an equation

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Step-by-step solution

1. 1
   With $t=\tan \theta$ and ${\sec }^2\theta =1+t^2$: $x=\frac{1+t^2-t}{1+t^2+t}=\frac{t^2-t+1}{t^2+t+1}$

   Why: Every admissible $\theta$ gives a real $t$, and every real $t$ arises from some $\theta$, so the range in $t$ is the range in $\theta$.

   Pythagorean trigonometric identities
2. 2
   $x\left(t^2+t+1\right)=t^2-t+1\Rightarrow (x-1)t^2+(x+1)t+(x-1)=0$

   Why: The denominator $t^2+t+1$ is never zero, so the fraction can be cleared safely.

   Clear the denominators of an equation
3. 3
   $x=1$ is attained: the equation becomes $2t=0$, i.e. $t=0$

   Why: The quadratic degenerates when $x=1$, and it still has a solution.

4. 4
   For $x\ne 1$ a real $t$ needs $(x+1{)}^2-4(x-1{)}^2\ge 0$

   Why: A quadratic with real coefficients has a real root exactly when its discriminant is non-negative.

   Quadratic formula
5. 5
   $(x+1{)}^2-4(x-1{)}^2=-3x^2+10x-3\ge 0\Rightarrow 3x^2-10x+3\le 0\Rightarrow (3x-1)(x-3)\le 0\Rightarrow \frac{1}{3}\le x\le 3$

   Why: Multiplying by $-1$ reverses the inequality; the solution of a downward product is the closed interval between the roots.

   Factorise a quadratic by splitting the middle termLocate roots by the sign of $f$ at test points
6. 6
   Both endpoints occur, at $t=1$ and $t=-1$, so the interval is closed

   Why: A non-strict discriminant condition means the boundary values are reached, which is what separates the two printed intervals.

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Step-by-step solution

1. 1
   The right angle is at $B$; write $AB=c$ (the height), $BC=a$ (the base) and $AC=b$ (the hypotenuse)

   Why: The base and the height meet at right angles, so the right angle is the vertex they share.

   Sine, cosine and tangent in a right triangle
2. 2
   Area two ways: $\frac{1}{2}ac=\frac{1}{2}bp$, so $p=\frac{ac}{b}$

   Why: The legs serve as base and height for each other, and $p$ is the height on base $AC$.

   Area of a triangle from base and height
3. 3
   $b=4p=\frac{4ac}{b}\Rightarrow b^2=4ac$, and $b^2=a^2+c^2$, so $a^2+c^2=4ac$

   Why: Substitute the given condition, then eliminate the hypotenuse with Pythagoras.

   Pythagoras' theorem
4. 4
   $\tan C=\frac{AB}{BC}=\frac{c}{a}=k$; dividing $a^2+c^2=4ac$ by $a^2$ gives $1+k^2=4k$, i.e. $k^2-4k+1=0$

   Why: In the right triangle, $\tan C$ is the side opposite $C$ over the side adjacent to it.

5. 5
   $k=\frac{4\pm \sqrt{16-4}}{2}=2\pm \sqrt{3}$

   Why: Quadratic formula; the two roots are reciprocals, which is what the symmetry of the condition predicts.

   Quadratic formula

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Step-by-step solution

1. 1
   $A{B}^2=a^2+b^2$, so $AB=\sqrt{a^2+b^2}$

   Why: The right angle is at $C$, so $AB$ is the hypotenuse.

   Pythagoras' theorem
2. 2
   $\text{area}=\frac{1}{2}ab=\frac{1}{2}pAB$

   Why: The legs are base and height for each other, while $p$ is the height on base $AB$.

   Area of a triangle from base and height
3. 3
   $ab=p\sqrt{a^2+b^2}$

   Why: Cancel the common factor $\frac{1}{2}$.

4. 4
   $a^2{b}^2=p^2\left(a^2+b^2\right)$

   Why: Square both sides to clear the square root, which is the form the options use.

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Step-by-step solution

1. 1
   $h=\sqrt{l^2-r^2}=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$ cm

   Why: The radius, the height and the slant height of a right circular cone are the two legs and the hypotenuse of one right triangle, so the height is the only missing piece of the volume formula.

   Slant height and curved surface of a conePythagoras' theorem
2. 2
   $V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi \times 5^2\times 12=100\pi$ cm${}^3$

   Why: With base radius and perpendicular height known, the volume follows directly.

   Volume of a right circular cone

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Step-by-step solution

1. 1
   Let the sheet have radius $R$, so $A=\pi R^2$, and let each cut circle have radius $\rho$.

   Why: Working with the radii turns the packing question into a single inequality.

   Area and circumference of a circle
2. 2
   Each centre lies within $R-\rho$ of the sheet's centre, so the two centres are at most $2(R-\rho )$ apart; not overlapping needs them at least $2\rho$ apart, so $2\rho \le 2(R-\rho )$, i.e. $\rho \le \frac{R}{2}$.

   Why: This is what fixes the greatest possible size: two circles of radius $\frac{R}{2}$ sitting on a diameter touch each other and touch the sheet, so the bound is attained.

3. 3
   Each cut circle then has area $\pi {\left(\frac{R}{2}\right)}^2=\frac{\pi R^2}{4}=\frac{A}{4}$.

   Why: Scaling the radius by $\frac{1}{2}$ scales the area by $\frac{1}{4}$.

4. 4
   Remaining area $=A-2\times \frac{A}{4}=A-\frac{A}{2}=\frac{A}{2}$.

   Why: The two discs together remove exactly half the sheet.

   Cut a compound plane figure into standard pieces

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Step-by-step solution

1. 1
   $r:l=1:3$, so put $r=k$ and $l=3k$.

   Why: A ratio fixes the radius and slant height only up to a common factor, and the factor will cancel at the end.

   Introduce a common factor $k$ for a ratio
2. 2
   Curved surface $=\pi rl=\pi \times k\times 3k=3\pi k^2$.

   Why: The curved surface of a cone is $\pi rl$.

   Slant height and curved surface of a cone
3. 3
   Total surface $=\pi rl+\pi r^2=3\pi k^2+\pi k^2=4\pi k^2$.

   Why: The total surface adds the circular base $\pi r^2$ to the curved surface.

   Area and circumference of a circle
4. 4
   $\frac{\text{total}}{\text{curved}}=\frac{4\pi k^2}{3\pi k^2}=\frac{4}{3}$, so the ratio is $4:3$.

   Why: The common factor $k^2$ and the $\pi$ cancel, so the answer is a pure number.

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Step-by-step solution

1. 1
   The cut is perpendicular to the axis, so the small cone is similar to the whole cone; call the linear ratio $k$.

   Why: A plane parallel to the base cuts off a cone with the same apex angle, so every length scales by one factor $k$.

   Similar triangles have proportional sides
2. 2
   Small cone $+$ frustum $=$ whole cone, so the volumes are in the ratio small : frustum : whole $=64:61:125$.

   Why: The two pieces partition the original solid, so their volumes add to it.

   Volume and surface of a composite solid
3. 3
   Volumes scale as $k^3$: $k^3=\frac{64}{125}$, so $k=\frac{4}{5}$.

   Why: For similar solids the volume ratio is the cube of the linear ratio.

4. 4
   Curved surfaces scale as $k^2$: small : whole $=16:25$, so frustum : whole $=(25-16):25=9:25$, and small : frustum $=16:9$.

   Why: Areas scale as the square of the linear ratio, and the frustum's curved surface is what is left of the whole cone's curved surface after the small cone's is removed.

   Slant height and curved surface of a cone

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Step-by-step solution

1. 1
   Floor area $=10^2=100$ m${}^2$.

   Why: The floor is a square of side $10$ m.

   Perimeter and area of a square
2. 2
   Table area $=\frac{\sqrt{3}}{4}\times 2^2=\sqrt{3}\approx 1.73$ m${}^2$.

   Why: The table top is an equilateral triangle of side $2$ m, and only its footprint matters.

   Area and altitude of an equilateral triangle
3. 3
   Each shelf stands on a $4$ m $\times$ $1$ m rectangle, so the four together cover $4\times 4=16$ m${}^2$; the $9$ m dimension is height and takes no extra floor.

   Why: A book-shelf occupies floor equal to its base, not to its whole surface or volume.

   Perimeter and area of a rectangle
4. 4
   Uncovered floor $=100-\sqrt{3}-16\approx 82.27$ m${}^2$, and half of it is $\approx 41.13$ m${}^2$.

   Why: The question carpets half of whatever floor is left after the furniture.

   Cut a compound plane figure into standard pieces
5. 5
   Cost $=41.13\times 100\approx$ Rs $4113$.

   Why: At Rs $100$ per square metre the cost in rupees is a hundred times the area in square metres.

   Unitary method: find one unit, then scale

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Step-by-step solution

1. 1
   The circle of radius $r_m$ encloses the innermost $m$ regions, so $\pi r_m^2=m\times \frac{A}{n}$.

   Why: Each of the $n$ regions has area $\frac{A}{n}$, and the disc of radius $r_m$ is made of the first $m$ of them.

   Area and circumference of a circle
2. 2
   $r_m=\sqrt{\frac{A}{n\pi }}\sqrt{m}$, so $r_m$ is proportional to $\sqrt{m}$.

   Why: Everything except $\sqrt{m}$ is a constant of the figure, so the radii grow like the square roots of the natural numbers.

   Direct variation as a power: $y\propto x^n$$\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$
3. 3
   $p_m=\frac{r_{m+1}}{r_m}=\sqrt{\frac{m+1}{m}}=\sqrt{1+\frac{1}{m}}$.

   Why: The constant cancels in the ratio, leaving a function of $m$ alone.

4. 4
   $\frac{1}{m}$ decreases as $m$ increases, so $1+\frac{1}{m}$ decreases, and since the square root preserves order, $p_m$ decreases.

   Why: A strictly decreasing quantity under a strictly increasing function stays strictly decreasing.

   Comparing two fractions

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Step-by-step solution

1. 1
   The base lies in a face of the cube, a square of side $2a$; the largest circle that fits in it is the inscribed circle, of radius $a$.

   Why: A circle inside a square cannot have diameter greater than the side, and the inscribed circle attains that.

   The largest circle that fits inside a square
2. 2
   The height can be taken as the full edge, $2a$, and enlarging either the radius or the height is impossible, so this cone is the one of maximum volume.

   Why: Volume $\frac{1}{3}\pi r^{2}h$ increases in both $r$ and $h$, so maximising each separately maximises the product.

3. 3
   $V=\frac{1}{3}\pi a^2\times 2a=\frac{2\pi a^3}{3}$.

   Why: Substituting $r=a$ and $h=2a$ into the cone volume formula.

   Volume of a right circular cone

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Step-by-step solution

1. 1
   For radii $r_1,r_2$ and centre distance $d$, the transverse (internal) common tangent has length $\sqrt{d^2-(r_1+r_2{)}^2}$.

   Why: The transverse tangent crosses the line of centres, so the two radii add; the direct tangent, which does not cross it, would use $r_1-r_2$ instead.

   Length of a common tangent to two circles
2. 2
   $r_1+r_2=4.5+3.5=8$ cm and $d=10$ cm, so the length is $\sqrt{10^2-8^2}=\sqrt{100-64}=\sqrt{36}=6$ cm.

   Why: Substituting the given values.

   Pythagoras' theorem

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Step-by-step solution

1. 1
   The condition says the two measures are numerically equal: $4\pi r^2=\frac{4}{3}\pi r^3$.

   Why: 'As many square centimetres in the surface as cubic centimetres in the volume' sets the two numbers equal, with $r$ in centimetres.

   Volume and surface area of a sphere
2. 2
   $r>0$, so divide both sides by $4\pi r^2$: $1=\frac{r}{3}$, hence $r=3$ cm.

   Why: The radius cannot be zero, so cancelling $r^2$ loses no solution.

   Clear the denominators of an equation

Common mistakes

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Step-by-step solution

1. 1
   Let CB $=x$. Since C lies between A and B with AB $=2$, AC $=2-x$ and $0<x<2$.

   Why: Naming the asked length directly avoids a conversion at the end, and the position of C bounds it.

   Turn a worded condition into a linear equation in one unknown
2. 2
   The condition AC${}^2=$ AB $\times$ CB becomes $(2-x{)}^2=2x$, i.e. $4-4x+x^2=2x$.

   Why: Expanding the square of a difference.

   Square of a sum and of a difference
3. 3
   $x^2-6x+4=0$, so $x=\frac{6\pm \sqrt{36-16}}{2}=\frac{6\pm 2\sqrt{5}}{2}=3\pm \sqrt{5}$.

   Why: The discriminant $20$ is not a perfect square, so the answer is a surd.

   Quadratic formula
4. 4
   $3+\sqrt{5}\approx 5.24>2$ is impossible for a part of a $2$ cm segment, so CB $=3-\sqrt{5}\approx 0.76$ cm.

   Why: Only the root inside $(0,2)$ describes a point of the segment.

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Step-by-step solution

1. 1
   A radius runs from the centre $O$ to a point $P$ of the circle, and here $OP=16$ cm for every such $P$.

   Why: The circle's radius is given as $16$ cm, so all the segments in question have the same length.

2. 2
   Its midpoint $M$ satisfies $OM=\frac{1}{2}OP=8$ cm, whatever direction $OP$ points in.

   Why: The midpoint halves the segment, and the halving is the same for every radius.

3. 3
   Conversely each point at distance $8$ cm from $O$ is the midpoint of exactly one radius, so the locus is the whole set of points $8$ cm from $O$: a circle of radius $8$ cm with the same centre, i.e. concentric with the given one.

   Why: The set of points at a fixed distance from a fixed point is a circle of that radius about that point, and 'same centre' is what 'concentric' means.

   A circle is the locus of points at a fixed distance from a fixed point

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Step-by-step solution

1. 1
   $1.76$ m${}^2=1.76\times 10^4=17600$ cm${}^2$, and the base radius is $r=\frac{140}{2}=70$ cm.

   Why: One square metre is $10^4$ square centimetres, and the options are in centimetres, so everything is put into centimetre units first.

   Unitary method: find one unit, then scale
2. 2
   $\pi rl=17600$ gives $\frac{22}{7}\times 70\times l=220l=17600$, so $l=80$ cm.

   Why: The curved surface of a cone is $\pi rl$, and $\frac{22}{7}\times 70=220$ exactly.

   Slant height and curved surface of a cone
3. 3
   $h=\sqrt{l^2-r^2}=\sqrt{80^2-70^2}=\sqrt{6400-4900}=\sqrt{1500}=10\sqrt{15}$ cm.

   Why: Radius, height and slant height form a right triangle with the slant height as hypotenuse, and $1500=100\times 15$.

   Pythagoras' theorem$\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$

Common mistakes

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Step-by-step solution

1. 1
   Statement 1 is false: in an obtuse triangle two of the altitudes meet the opposite sides only when produced, and the orthocentre falls outside the triangle.

   Why: 'Always' fails as soon as one counterexample exists, and the obtuse triangle is that counterexample.

   The orthocentre and the orthocentric system
2. 2
   Statement 2 is true: each median joins a vertex to a point of the opposite side and so lies inside the triangle, and the centroid lies on every median, dividing each internally.

   Why: A point interior to a segment whose ends are on or inside a convex region is interior to that region.

   The centroid divides each median in the ratio $2:1$
3. 3
   Statement 3 is true: in a right-angled triangle the two legs are themselves altitudes, so all three altitudes pass through the right-angle vertex and the orthocentre is that vertex, which lies on the triangle.

   Why: Each leg is perpendicular to the other, so each leg is the altitude from its own endpoint.

4. 4
   Statement 4 is false: nothing about a right angle moves the centroid out of the interior, so it lies inside, not on, the triangle.

   Why: Statement 2 already puts the centroid strictly inside every triangle.

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Step-by-step solution

1. 1
   Two intersecting lines make four angles at their common point, in two pairs of vertically opposite angles.

   Why: Every point of the plane other than the intersection lies inside one of these four angles.

   Angles on a straight line and vertically opposite angles
2. 2
   Inside one angle, a point is equidistant from the two arms exactly when it lies on that angle's bisector.

   Why: The perpendicular distances to the two arms are equal precisely on the bisector, which is why the incentre of a triangle is equidistant from its sides.

   Points equidistant from two intersecting lines form a pair of bisectors
3. 3
   Two vertically opposite angles share one bisector, so the four angles supply two bisecting lines, and they are perpendicular to each other; together they carry every equidistant point.

   Why: Bisecting one angle automatically bisects the one vertically opposite it, so four angles give two lines, not four.

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Step-by-step solution

1. 1
   Statement 1 is AAA. Equal angles fix the shape but not the size, so the triangles are similar and need not be congruent.

   Why: Nothing in three equal angles pins down a single length, so the triangles can be scaled copies of each other.

   Similar triangles have proportional sides
2. 2
   Statement 2 is SSS, statement 3 is SAS and statement 4 is ASA — all three are valid congruence criteria.

   Why: SSS, SAS, ASA (with AAS and RHS) are the accepted criteria; each fixes every remaining part of the triangle.

   Congruence criteria for triangles
3. 3
   So exactly statements 2, 3 and 4 are correct.

   Why: Only the AAA claim fails.