Elementary Mathematics 2018 (I)

Step-by-step solution

1. 1
   If every one of the $n$ interior angles is $90^{\circ }$, their sum is $90n$ degrees; the interior angles of an $n$-gon sum to $(n-2)\times 180^{\circ }$.

   Why: The angle sum of a polygon depends only on the number of sides, so the two expressions must agree.

   Interior and exterior angle sums of a polygon
2. 2
   $(n-2)180=90n\Rightarrow 180n-360=90n\Rightarrow 90n=360\Rightarrow n=4$. Statement-1 is true.

   Why: A single linear equation fixes $n$, so four sides is the only possibility.

   Turn a worded condition into a linear equation in one unknown
3. 3
   In right angles, the true sum is $\frac{(n-2)180^{\circ }}{90^{\circ }}=2n-4$ right angles, not $3n-8$. Statement-2 is false.

   Why: Dividing the degree sum by $90^{\circ }$ converts it into right angles.

4. 4
   Test $n=3$: the true sum is $2(3)-4=2$ right angles ($180^{\circ }$), while $3(3)-8=1$ right angle ($90^{\circ }$). The quoted formula fails.

   Why: One counterexample settles a general formula, and the triangle is the cheapest one.

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1. 1
   Let the side be $a$ cm. The enlarged square has side $a+8$, so $(a+8{)}^2-a^2=120$.

   Why: The area of a square is the square of its side, and the stated increase is the difference of the two areas.

   Perimeter and area of a squareTurn a worded condition into a linear equation in one unknown
2. 2
   $(a+8{)}^2-a^2=(a+8+a)(a+8-a)=8(2a+8)=16a+64$.

   Why: Factoring as a difference of two squares kills the $a^2$ term, leaving a linear equation.

   Difference of two squares
3. 3
   $16a+64=120\Rightarrow 16a=56\Rightarrow a=3.5$ cm.

   Why: Dividing through by $16$.

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1. 1
   $10=2\times 5$, so the largest power of $10$ dividing $25!$ is the smaller of the exponents of $2$ and of $5$ in its prime factorisation.

   Why: Each factor of $10$ needs one $2$ and one $5$, so whichever prime runs out first sets the limit.

   Prime factorisation
2. 2
   Multiples of $5$ up to $25$ are $5,10,15,20,25$ — five of them — and $25=5^2$ contributes a second $5$, so the exponent of $5$ is $5+1=6$.

   Why: Counting one $5$ for each multiple of $5$ and an extra $5$ for each multiple of $25$.

   Exponent of a prime in $n!$ (Legendre's formula)
3. 3
   The exponent of $2$ is $12+6+3+1=22$, far larger, so the $5$s are the binding constraint and the answer is $10^6$.

   Why: There are $12$ even numbers up to $25$, $6$ multiples of $4$, $3$ of $8$ and $1$ of $16$.

4. 4
   The largest power is $6$, which is not $2$, $4$ or $5$, so the correct option is 'None of the above'.

   Why: Only after the value is derived is it compared with the printed options.

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1. 1
   The four walls of one room have area $2(l+b)h=2(6+4)\times 2.5=50$ m${}^2$.

   Why: Roofs and floors are excluded, so only the lateral surface of the cuboid counts.

   Volume and wall area of a cuboid
2. 2
   Five identical rooms give $5\times 50=250$ m${}^2$.

   Why: The rooms are identical, so the total is a straight multiple.

3. 3
   At $20$ m${}^2$ per litre, the paint needed is $\frac{250}{20}=12.5$ L.

   Why: Dividing the area by the coverage of one litre.

   Unitary method: find one unit, then scale
4. 4
   Paint comes only in $1$ L cans, so $12$ cans are not enough and $13$ cans are required.

   Why: A fractional can cannot be bought, so the count is rounded up, not to the nearest whole number.

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1. 1
   Pathway area $=4.5\times 10=45$ m${}^2$.

   Why: The pathway is a rectangle, so its area is length times width.

   Perimeter and area of a rectangle
2. 2
   A tile of side $50$ cm $=0.5$ m has area $0.5^2=0.25$ m${}^2$.

   Why: Both measurements must be in the same unit before dividing.

   Perimeter and area of a square
3. 3
   Number of tiles $=\frac{45}{0.25}=180$.

   Why: Dividing the total area by the area of one tile.

   Unitary method: find one unit, then scale
4. 4
   Packets $=\frac{180}{20}=9$, costing $9\times 100=$ Rs $900$.

   Why: Each packet holds $20$ tiles and costs Rs $100$, and $180$ divides exactly, so no part packet is needed.

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1. 1
   Every corner touches the sphere, so the cube's space diagonal is a diameter. With edge $a$, a face diagonal is $a\sqrt{2}$ and the space diagonal is $\sqrt{a^2+(a\sqrt{2}{)}^2}=a\sqrt{3}$.

   Why: Pythagoras applied twice — once in a face, once in the right triangle formed by that face diagonal, an edge and the space diagonal.

   Pythagoras' theoremVolume and wall area of a cuboid
2. 2
   $a\sqrt{3}=2R$, so $a=\frac{2R}{\sqrt{3}}$.

   Why: The space diagonal joins two opposite corners through the centre, so it equals the diameter.

3. 3
   Cube volume $=a^3=\frac{8R^3}{3\sqrt{3}}$; sphere volume $=\frac{4}{3}\pi R^3$.

   Why: Cubing $\frac{2R}{\sqrt{3}}$ gives $\frac{8R^3}{3\sqrt{3}}$ since $(\sqrt{3}{)}^3=3\sqrt{3}$.

   Volume and surface area of a sphere
4. 4
   Ratio $=\frac{8R^3}{3\sqrt{3}}÷\frac{4}{3}\pi R^3=\frac{8}{3\sqrt{3}}\times \frac{3}{4\pi }=\frac{2}{\sqrt{3}\pi }$, that is $2:\sqrt{3}\pi$.

   Why: The $R^3$ cancels, so the ratio is a pure number independent of the sphere's size.

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

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1. 1
   $\frac{2\pi r}{h}=\frac{3}{1}$, so $h=\frac{2\pi r}{3}$.

   Why: The base circumference is $2\pi r$, and the given ratio is circumference to height.

   Area and circumference of a circleIntroduce a common factor $k$ for a ratio
2. 2
   $l=\sqrt{r^2+h^2}=\sqrt{r^2+\frac{4{\pi }^2{r}^2}{9}}=r\sqrt{\frac{9+4{\pi }^2}{9}}=\frac{r\sqrt{4{\pi }^2+9}}{3}$.

   Why: Radius, height and slant height form a right triangle, and $r^2$ factors out of the surd.

   Pythagoras' theorem$\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$
3. 3
   Curved surface $=\pi rl=\pi r\times \frac{r\sqrt{4{\pi }^2+9}}{3}=\frac{\pi r^2\sqrt{4{\pi }^2+9}}{3}$.

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

   Slant height and curved surface of a cone

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1. 1
   Length of wire $=2\pi r=2\times \frac{22}{7}\times 98=616$ cm.

   Why: The wire is bent into a circle of radius $98$ cm, so its length is that circumference; $\frac{98}{7}=14$ makes the arithmetic exact.

   Area and circumference of a circle
2. 2
   Reshaping does not change the wire's length, so the square has perimeter $616$ cm.

   Why: The same piece of wire forms both figures.

3. 3
   Side $=\frac{616}{4}=154$ cm.

   Why: A square's perimeter is four times its side.

   Perimeter and area of a square

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1. 1
   Double the median from $A$ to a point $A^{\prime }$, so $B{A}^{\prime }CA$ is a parallelogram; the triangle inequality in $△AB{A}^{\prime }$ gives $2p<b+c$, and similarly $2q<c+a$ and $2r<a+b$.

   Why: Doubling a median turns it into a diagonal of a parallelogram whose sides are two sides of the triangle, which is what makes the triangle inequality usable.

   Triangle inequality
2. 2
   Adding the three: $2(p+q+r)<2(a+b+c)$, so $p+q+r<a+b+c$.

   Why: Each side length appears twice on the right, once in each of two inequalities.

3. 3
   Hence $2(p+q+r)<2(a+b+c)<3(a+b+c)$, so option C holds for every triangle.

   Why: The perimeter is positive, so weakening the factor from $2$ to $3$ keeps the inequality true.

   The sum of the medians lies between $\frac{3}{4}$ of the perimeter and the perimeter
4. 4
   Option D fails: for an equilateral triangle of side $1$, each median is $\frac{\sqrt{3}}{2}$, so $11(p+q+r)=11\times \frac{3\sqrt{3}}{2}\approx 28.58$ while $10(a+b+c)=30$.

   Why: A statement offered as always correct is destroyed by one triangle, and the equilateral triangle is the cheapest test case.

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1. 1
   The centroid $G$ cuts each median in the ratio $2:1$ from the vertex, so $GA=\frac{2}{3}p$, $GB=\frac{2}{3}q$, $GC=\frac{2}{3}r$.

   Why: Everything about the medians can be moved to $G$, where the three of them meet.

   The centroid divides each median in the ratio $2:1$
2. 2
   Triangle inequality in $△GAB$: $GA+GB>AB$, i.e. $\frac{2}{3}(p+q)>c$; similarly $\frac{2}{3}(q+r)>a$ and $\frac{2}{3}(r+p)>b$.

   Why: $G$ is an interior point, so $A$, $B$, $G$ form a genuine triangle for each pair of vertices.

   Triangle inequality
3. 3
   Adding the three: $\frac{4}{3}(p+q+r)>a+b+c$, that is $4(p+q+r)>3(a+b+c)$ — option B.

   Why: Each of $p,q,r$ appears in two of the three inequalities.

4. 4
   The other three fail: A and D both contradict $4(p+q+r)>3(a+b+c)$ together with $p+q+r<a+b+c$, and C would need $p+q+r<\frac{2}{3}(a+b+c)$, below the $\frac{3}{4}$ floor just proved.

   Why: The pair of bounds $\frac{3}{4}(a+b+c)<p+q+r<a+b+c$ decides all four options at once.

   The sum of the medians lies between $\frac{3}{4}$ of the perimeter and the perimeter

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1. 1
   The largest circle inside a square has diameter equal to the side, so $2r=\frac{2}{\sqrt{\pi }}$ and $r=\frac{1}{\sqrt{\pi }}$.

   Why: A circle inside a square cannot be wider than the square, and the inscribed circle touches all four sides.

   The largest circle that fits inside a square
2. 2
   Area $=\pi r^2=\pi \times \frac{1}{\pi }=1$ square unit.

   Why: Squaring $\frac{1}{\sqrt{\pi }}$ gives $\frac{1}{\pi }$, which cancels the $\pi$ exactly — the side length was chosen to make this happen.

   Area and circumference of a circle

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1. 1
   A square's diagonals are equal; call each $d$. Their product is $d^2=50$, so $d=\sqrt{50}=5\sqrt{2}$ units.

   Why: Equal diagonals turn a product of two unknowns into a single square.

   The diagonals of a rectangle are equal and bisect each other$\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$
2. 2
   For a square of side $a$, $d=a\sqrt{2}$, so $a=\frac{5\sqrt{2}}{\sqrt{2}}=5$ units.

   Why: The diagonal is the hypotenuse of a right triangle with both legs equal to the side.

   Pythagoras' theoremPerimeter and area of a square

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1. 1
   Closed cylinder: total surface $=2\pi r(r+h)=352$ with $h=10$.

   Why: 'Closed' means both circular ends count, so the total surface is $2\pi rh+2\pi r^2$.

   Curved and total surface of a cylinder
2. 2
   $2\times \frac{22}{7}\times r(r+10)=352\Rightarrow r(r+10)=\frac{352\times 7}{44}=56$.

   Why: Clearing $\frac{44}{7}$ turns the equation into whole numbers.

3. 3
   $r^2+10r-56=0\Rightarrow (r+14)(r-4)=0$, and $r>0$ gives $r=4$ cm.

   Why: Splitting the middle term with $14$ and $-4$, whose product is $-56$ and sum $10$.

   Factorise a quadratic by splitting the middle term
4. 4
   Diameter $=2r=8$ cm.

   Why: The question asks for the diameter, not the radius.

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1. 1
   The square's diagonal is $a\sqrt{2}=6\sqrt{2}$, so its side is $a=6$ cm.

   Why: The diagonal of a square is $\sqrt{2}$ times its side.

   Perimeter and area of a squarePythagoras' theorem
2. 2
   Its perimeter is $4\times 6=24$ cm, and the equilateral triangle has the same perimeter, so its side is $\frac{24}{3}=8$ cm.

   Why: Equal perimeters is the link between the two figures.

   Area and altitude of an equilateral triangle
3. 3
   Area $=\frac{\sqrt{3}}{4}{s}^2=\frac{\sqrt{3}}{4}\times 64=16\sqrt{3}$ cm${}^2$.

   Why: The area of an equilateral triangle of side $s$ is $\frac{\sqrt{3}}{4}{s}^2$.

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1. 1
   A circle through all four corners of the square is its circumcircle, and the diagonal is a diameter: $2R=a\sqrt{2}$, so $R=\frac{a}{\sqrt{2}}$.

   Why: Opposite corners subtend a right angle at the other two, so the diagonal must be a diameter.

   Side of a regular polygon inscribed in a circlePythagoras' theorem
2. 2
   Circle area $=\pi R^2=\pi \times \frac{a^2}{2}=\frac{\pi a^2}{2}$.

   Why: Squaring $\frac{a}{\sqrt{2}}$ gives $\frac{a^2}{2}$.

   Area and circumference of a circle
3. 3
   Required region $=\frac{\pi a^2}{2}-a^2=\frac{(\pi -2)a^2}{2}$ square units.

   Why: The region is bounded outside by the circle and inside by the square, so it is the difference of the two areas.

   Cut a compound plane figure into standard piecesPerimeter and area of a square

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1. 1
   Each tangent is perpendicular to the radius at its point of contact, so $\angle OAX=\angle OBX=90^{\circ }$.

   Why: This is what lets the quadrilateral $OAXB$ be closed up.

   The two tangents from an external point
2. 2
   In quadrilateral $OAXB$: $\angle AOB=360^{\circ }-90^{\circ }-90^{\circ }-50^{\circ }=130^{\circ }$.

   Why: The angles of a quadrilateral sum to $360^{\circ }$.

   Interior and exterior angle sums of a polygon
3. 3
   $C$ lies on the arc $AB$ remote from $X$, so $\angle ACB=\frac{1}{2}\angle AOB=\frac{1}{2}\times 130^{\circ }=65^{\circ }$.

   Why: The angle at the centre is twice the angle the same arc subtends at a point of the remaining arc, and the figure places $C$ on that major arc.

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

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1. 1
   Three parallel lines cut off proportional intercepts on any transversal, so $\frac{AB}{BC}=\frac{DE}{EF}$.

   Why: The parallels $p$, $q$, $r$ cut $l$ at $A,B,C$ and $m$ at $D,E,F$; the intercept theorem is the parallel-line case of similarity.

   Angles made by a transversal of two parallel linesSimilar triangles have proportional sides
2. 2
   Cross-multiplying gives $AB\times EF=BC\times DE$, which is statement 2. So statement 2 is correct.

   Why: Cross-multiplication is an equivalence, so it says exactly the same thing.

3. 3
   Write $AB=\lambda DE$ and $BC=\lambda EF$ with the common factor $\lambda$. Then $AC=AB+BC=\lambda (DE+EF)=\lambda DF$, so $\frac{AB}{AC}=\frac{\lambda DE}{\lambda DF}=\frac{DE}{DF}$, which is statement 1. So statement 1 is correct too.

   Why: Introducing the common factor makes the whole intercepts $AC$ and $DF$ scale by the same $\lambda$, which is what turns the part-to-part ratio into a part-to-whole one.

   Introduce a common factor $k$ for a ratio
4. 4
   Both statements hold.

   Why: They are two forms of the one proportion.

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1. 1
   $D$ and $E$ are the midpoints of $AB$ and $AC$, so $DE$ is parallel to $BC$ and $DE=\frac{l}{2}$.

   Why: This is the midpoint theorem, and it is what makes $DECB$ a trapezium.

   Midpoint theorem
2. 2
   $△ADE$ is similar to $△ABC$ with linear ratio $\frac{1}{2}$, so its area is $\frac{1}{4}$ of $[ABC]$.

   Why: Areas of similar figures scale as the square of the linear ratio.

   Similar triangles have proportional sides
3. 3
   $[ABC]=\frac{\sqrt{3}}{4}{l}^2$ for an equilateral triangle of side $l$.

   Why: Standard area formula for an equilateral triangle.

   Area and altitude of an equilateral triangle
4. 4
   Shaded trapezium $=[ABC]-[ADE]=\left(1-\frac{1}{4}\right)\times \frac{\sqrt{3}}{4}{l}^2=\frac{3}{4}\times \frac{\sqrt{3}}{4}{l}^2=\frac{3\sqrt{3}{l}^2}{16}$.

   Why: The shaded region is everything below $DE$, that is the whole triangle less the small one at the apex.

   Cut a compound plane figure into standard pieces

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1. 1
   $OP\perp SPT$, since a tangent is perpendicular to the radius at the point of contact, so $\angle OPQ=90^{\circ }-\angle QPT=90^{\circ }-\alpha$.

   Why: $\angle OPT$ is split by the chord $PQ$ into $\angle OPQ$ and the given $\alpha$.

   The two tangents from an external point
2. 2
   $OP=OQ$ are radii, so $△OPQ$ is isosceles and $\angle OQP=\angle OPQ=90^{\circ }-\alpha$.

   Why: Angles opposite equal sides are equal.

   Base angles of an isosceles triangle
3. 3
   $\angle POQ=180^{\circ }-2(90^{\circ }-\alpha )=180^{\circ }-180^{\circ }+2\alpha =2\alpha$.

   Why: The angles of triangle $OPQ$ sum to $180^{\circ }$.

   Angle sum of a triangle