Elementary Mathematics 2019 (II)

Step-by-step solution

1. 1
   $R=3\text{ cm},r=2\text{ cm},\text{cone radius}=4\text{ cm}$

   Why: Every diameter given has to be halved first; the $6$, $4$ and $8$ in the question are all diameters.

   Volume and surface area of a sphere
2. 2
   $V_{\text{shell}}=\frac{4}{3}\pi \left(R^3-r^3\right)=\frac{4}{3}\pi (27-8)=\frac{76\pi }{3}{\text{ cm}}^3$

   Why: A hollow sphere's metal is the big sphere less the cavity, so the volumes subtract.

   Volume and surface of a composite solid
3. 3
   $\frac{1}{3}\pi (4{)}^{2}h=\frac{76\pi }{3}$

   Why: Melting conserves volume, so the cone made from the shell has exactly the shell's volume.

   Recasting preserves volume
4. 4
   $16h=76\Rightarrow h=\frac{19}{4}=4.75\text{ cm}$

   Why: The $\frac{\pi }{3}$ cancels from both sides, leaving one division.

   Volume of a right circular cone

Common mistakes

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Step-by-step solution

1. 1
   $V_{\text{cyl}}=\pi (6{)}^2(10)=360\pi {\text{ cm}}^3$

   Why: The volume of the cylinder is what the two cones share, since melting loses nothing.

   Volume of a right circular cylinder
2. 2
   $V_1=\frac{1}{3}(360\pi )=120\pi ,V_2=\frac{2}{3}(360\pi )=240\pi$

   Why: A ratio of $1:2$ has three parts, so the cones take a third and two thirds of the metal.

   Recasting preserves volume
3. 3
   $\frac{1}{3}\pi r_1^2(10)=120\pi \Rightarrow r_1^2=36;\frac{1}{3}\pi r_2^2(10)=240\pi \Rightarrow r_2^2=72$

   Why: Both cones have height $10$ cm, so only the squared radii differ — and these are all the flat-area calculation needs, so there is no reason to take the square roots.

   Volume of a right circular cone
4. 4
   $\text{flat area before}=2\pi (6{)}^2=72\pi {\text{ cm}}^2$

   Why: A closed cylinder has two flat circular ends; the curved wall is not flat and is excluded.

   Curved and total surface of a cylinder
5. 5
   $\text{flat area after}=\pi r_1^2+\pi r_2^2=\pi (36+72)=108\pi {\text{ cm}}^2$

   Why: Each cone contributes exactly one flat face, its base.

   Volume and surface of a composite solid
6. 6
   $\frac{108\pi -72\pi }{72\pi }=\frac{36}{72}=50\%$

   Why: Percentage increase is the rise divided by the original amount; $\pi$ cancels throughout.

   Percentage change as a multiplier

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Step-by-step solution

1. 1
   $\text{case 1: }15\text{ is the hypotenuse}\Rightarrow 9,12,15,\text{perimeter }36$

   Why: If $15$ were the hypotenuse the other two sides are bounded by it, so the perimeter can be at most $45$ — this case cannot be the maximum and is settled in one line.

   Pythagoras' theorem
2. 2
   $\text{case 2: }15^2+b^2=c^2\Rightarrow c^2-b^2=225$

   Why: Taking $15$ as a leg leaves the hypotenuse free to grow, which is where a large perimeter must come from.

3. 3
   $(c-b)(c+b)=225$

   Why: Factorising the difference of squares turns one equation in two unknowns into a short list of integer factor pairs.

   Difference of two squares
4. 4
   $\text{perimeter}=15+b+c=15+(c+b)$

   Why: Since $b+c$ is one of the two factors, maximising the perimeter means making $c+b$ as large as possible — so take the smallest allowed value of $c-b$.

   A product with a fixed sum is largest when the factors are equal
5. 5
   $c-b=1,c+b=225\Rightarrow c=113,b=112$

   Why: $225$ is odd, so every factor pair is odd-by-odd and $c,b$ come out as integers; $(1,225)$ is the most lopsided pair and so gives the largest $c+b$.

   Prime factorisation
6. 6
   $\text{perimeter}=15+112+113=240\text{ cm}$

   Why: Adding the three sides of the extreme triple.

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Step-by-step solution

1. 1
   $\text{a cube has }12\text{ edges}$

   Why: The skeleton is the wire frame, so one rod goes along each edge — not one per face and not one per vertex.

   A cube has $6$ faces, $12$ edges and $8$ vertices
2. 2
   $\text{edge}=\frac{24}{12}=2\text{ feet}$

   Why: Using the whole $24$ feet in $12$ equal pieces gives the largest edge possible: cutting shorter pieces would leave rod unused and build a smaller cube.

3. 3
   $\text{face area}=2^2=4\text{ square feet}$

   Why: Each face of the cube is a square on the edge.

   Perimeter and area of a square

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Step-by-step solution

1. 1
   $AD=2r$

   Why: $AB$ and $CD$ are parallel and both touch the circle, so the distance between them is a diameter; $AD$ is perpendicular to $AB$, hence is exactly that distance.

   The two tangents from an external point
2. 2
   $AE=r\text{ and }DF=r$

   Why: At $A$ the two sides $AB$ and $AD$ are perpendicular tangents, so the centre sits at distance $r$ from each and $AEOP$ is a square; the same holds at $D$, where $DC$ and $DA$ are perpendicular.

3. 3
   $BC=BE+CF=25+16=41\text{ cm}$

   Why: The two tangents from $B$ are equal, so the tangent along $BC$ is also $25$; likewise the tangent from $C$ along $CB$ is $16$, and together they make up $BC$.

4. 4
   $AB=r+25,DC=r+16$

   Why: Each parallel side splits at its point of contact into the tangent length from one end and from the other.

   Pitot's theorem for a tangential quadrilateral
5. 5
   $\text{horizontal gap}=AB-DC=(r+25)-(r+16)=9\text{ cm}$

   Why: Dropping a perpendicular from $C$ to $AB$ makes a rectangle with $AD$, so the foot is $DC$ along from $A$ and the remaining piece of $AB$ is the horizontal leg of the right triangle on $BC$. The $r$ cancels, which is what makes the next step solvable.

   Pythagoras' theorem
6. 6
   $41^2=9^2+(2r{)}^2\Rightarrow 1681-81=1600=(2r{)}^2\Rightarrow 2r=40\text{ cm}$

   Why: The right triangle has legs $9$ and $AD=2r$ and hypotenuse $BC=41$; solving gives the diameter directly, which is what was asked.

Common mistakes

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Step-by-step solution

1. 1
   $\frac{4}{3}\pi R^3=\frac{4}{3}\pi \left(3^3+4^3+5^3\right)$

   Why: The three spheres become one sphere of the same total volume, and every term carries the same $\frac{4}{3}\pi$.

   Recasting preserves volume
2. 2
   $R^3=27+64+125=216$

   Why: Cancelling $\frac{4}{3}\pi$ leaves the cubes of the radii — radii themselves never add.

   Volume and surface area of a sphere
3. 3
   $R=\sqrt[3]{216}=6\text{ cm}$

   Why: $216=6^3$, so the cube root is exact.

   Prime factorisation

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Step-by-step solution

1. 1
   $\frac{2}{3}\pi r^3=155232$

   Why: A solid hemisphere is half a sphere, so its volume is half of $\frac{4}{3}\pi r^3$.

   Volume and surface area of a sphere
2. 2
   $r^3=155232\times \frac{3}{2}\times \frac{7}{22}=\frac{155232}{22}\times \frac{21}{2}$

   Why: With $\pi =\frac{22}{7}$ the whole constant is $\frac{44}{21}$; dividing by $22$ first keeps the numbers small, and $155232÷22=7056$.

3. 3
   $r^3=7056\times \frac{21}{2}=3528\times 21=74088$

   Why: Halving $7056$ to $3528$ before multiplying by $21$ avoids a five-digit multiplication.

   Prime factorisation
4. 4
   $74088=42^3\Rightarrow r=42\text{ cm}$

   Why: The cube ends in $8$, so the radius ends in $2$; and $40^3=64000$ with $50^3=125000$, which pins the radius between $40$ and $50$ — only $42$ fits both.

   Unit digit of a power from its cycle

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Step-by-step solution

1. 1
   $R=6\text{ cm},r=3\text{ cm},h=7\text{ cm}$

   Why: The bucket's two diameters are $12$ cm and $6$ cm; the formula takes radii, and buckets in these questions are always described by diameters.

   Volume of a frustum of a cone
2. 2
   $V=\frac{\pi h}{3}\left(R^2+Rr+r^2\right)$

   Why: The capacity of a truncated cone is this formula, with $h$ the frustum's own height and not that of the cone it came from.

3. 3
   $R^2+Rr+r^2=36+18+9=63$

   Why: Computing the bracket first keeps the arithmetic in two digits; the middle term $Rr$ is the one candidates omit.

4. 4
   $V=\frac{1}{3}\times \frac{22}{7}\times 7\times 63=\frac{22\times 63}{3}=22\times 21=462{\text{ cm}}^3$

   Why: The height $7$ cancels the $7$ in $\frac{22}{7}$ and the $3$ divides into $63$, so the whole thing collapses to $22\times 21$.

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Step-by-step solution

1. 1
   $l=\sqrt{r^2+h^2}=\sqrt{36+64}=10\text{ cm}$

   Why: The radius, the vertical height and the slant height form a right triangle, so the slant height — not the height — is what the curved surface formula needs.

   Slant height and curved surface of a cone
2. 2
   $\text{curved surface}=\pi rl=\pi (6)(10)=60\pi {\text{ cm}}^2$

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

3. 3
   $\text{base}=\pi r^2=36\pi {\text{ cm}}^2$

   Why: 'Total' surface area includes the circular base; only 'curved' would exclude it.

   Area and circumference of a circle
4. 4
   $\text{total}=60\pi +36\pi =96\pi {\text{ cm}}^2$

   Why: Adding the two parts, which is the same as $\pi r(l+r)=\pi (6)(16)$.

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Step-by-step solution

1. 1
   $l=6\times 12=72\text{ cm},b=12\text{ cm},h=12\text{ cm}$

   Why: Joining the cubes end to end stacks them along one direction only, so one dimension is six edges long and the other two stay at one edge.

   Volume and surface of a composite solid
2. 2
   $S=2(lb+bh+hl)=2(72\times 12+12\times 12+12\times 72)$

   Why: The total surface of a cuboid is twice the sum of the three distinct face areas.

   Volume and wall area of a cuboid
3. 3
   $=2(864+144+864)=2\times 1872=3744{\text{ cm}}^2$

   Why: The two long faces are equal, which halves the multiplication work.

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Step-by-step solution

1. 1
   $x+\frac{1}{x}=\frac{26}{5}$, with $x\ne 0$

   Why: Name the real number $x$; "its reciprocal" is $\frac{1}{x}$, which already requires $x\ne 0$.

2. 2
   $5x^2-26x+5=0$

   Why: Multiply every term by $5x$ to clear both denominators; $x\ne 0$, so nothing is lost.

   Clear the denominators of an equation
3. 3
   $D=(-26{)}^2-4(5)(5)=676-100=576$

   Why: The question asks only how many such numbers exist, so the discriminant settles it without solving.

   Quadratic formula
4. 4
   $D>0$, so there are two distinct real roots, and neither is $0$

   Why: A positive discriminant gives two distinct real roots. The only excluded value was $x=0$, and $5(0{)}^2-26(0)+5=5\ne 0$, so both roots are admissible: two such numbers.

Common mistakes

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Step-by-step solution

1. 1
   $\mathrm{HCF}(p,q)=1$ and $\mathrm{HCF}(p,r)=1$: no prime divides $p$ and $q$, and none divides $p$ and $r$

   Why: Coprimality is most usable in its prime-factor form, so rewrite the hypothesis that way first.

   Coprime means no shared prime factor
2. 2
   Every prime factor of $qr$ is a prime factor of $q$ or of $r$, so no prime divides both $p$ and $qr$: $\mathrm{HCF}(p,qr)=1$

   Why: A prime shared by $p$ and $qr$ would already be shared by $p$ and $q$, or by $p$ and $r$ — both ruled out. Statement 1 is correct.

   Prime factorisation
3. 3
   Take $p=2$, $q=2$, $r=3$: $p\mid qr$ since $2\mid 6$, and $p\mid q$, yet $p\nmid r$

   Why: One counterexample refutes a universal claim, so statement 2 is wrong.

   Euclid's lemma
4. 4
   The true version is Euclid\'s lemma: $p\mid qr$ together with $\mathrm{HCF}(p,q)=1$ gives $p\mid r$

   Why: Statement 2 puts "$p$ divides $q$" where "$p$ is coprime to $q$" belongs. That swap is exactly what makes it false, and recognising it is what stops the statement looking familiar enough to accept.

Common mistakes

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Step-by-step solution

1. 1
   Let Rani be $r$ now and let the age gap be $d$, so Radha is $r+d$

   Why: The difference between two people\'s ages is the same at every date, so naming it directly is naming what the question asks for.

   Turn a worded condition into a linear equation in one unknown
2. 2
   Five years back: $r+d-5=3(r-5)\Rightarrow d=2r-10$

   Why: Radha was then three times Rani; collecting the terms leaves the gap in terms of $r$.

3. 3
   One year back: $r+d-1=2(r-1)\Rightarrow d=r-1$

   Why: The same translation at the second date.

4. 4
   $2r-10=r-1\Rightarrow r=9$, so $d=9-1=8$

   Why: Both lines describe the one gap, so equating them gives a single linear equation in $r$ — no simultaneous pair to manage.

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Step-by-step solution

1. 1
   $a$ apples and $b$ oranges with $a\ge 1$, $b\ge 1$: $5a+7b=500$

   Why: "Buy both" makes each count at least $1$, and "using total amount" makes the spend exactly Rs. $500$.

   Turn a worded condition into a linear equation in one unknown
2. 2
   $5a=500-7b$, so $7b$ is a multiple of $5$

   Why: Both $5a$ and $500$ are multiples of $5$, so what is left must be one too.

3. 3
   $\mathrm{HCF}(5,7)=1$, so $5$ divides $b$: write $b=5k$

   Why: A divisor coprime to one factor of a product must divide the other factor — $5$ cannot come out of the $7$, so it comes out of $b$.

   Euclid's lemma
4. 4
   $a\ge 1\Rightarrow 7b\le 495\Rightarrow b\le 70$, so $b=5,10,\dots ,70$

   Why: Bound $b$ first, then list the multiples of $5$ inside that bound.

   Enumerate a residue class within a bound
5. 5
   $k=1,2,\dots ,14$: there are $14$ ways

   Why: Each such $b$ gives exactly one $a=\frac{500-7b}{5}$, and that $a$ is a positive integer, so counting the values of $b$ counts the ways.

Common mistakes

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Step-by-step solution

1. 1
   $y=\frac{c}{\sqrt{x}}$

   Why: "Inversely proportional to $\sqrt{x}$" means $y\sqrt{x}$ is a constant, so it is $\sqrt{x}$ and not $x$ that carries the proportionality.

   Direct variation as a power: $y\propto x^n$
2. 2
   $36=\frac{c}{\sqrt{36}}=\frac{c}{6}\Rightarrow c=216$

   Why: One known pair determines the constant.

3. 3
   $54=\frac{216}{\sqrt{x}}\Rightarrow \sqrt{x}=\frac{216}{54}=4$

   Why: Substitute the new $y$ and solve for $\sqrt{x}$ first, not for $x$.

   Clear the denominators of an equation
4. 4
   $x=4^2=16$

   Why: Square to undo the radical.

   Square both sides to clear a radical

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Step-by-step solution

1. 1
   $16+6\sqrt{7}=16+2\sqrt{63}$

   Why: The denesting rule needs the middle term as $2\sqrt{b}$: $6\sqrt{7}=2\times 3\sqrt{7}=2\sqrt{9\times 7}$.

   Denest $\sqrt{a\pm 2\sqrt{b}}$$\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$
2. 2
   Find $p,q>0$ with $p+q=16$ and $pq=63$: $p=9$, $q=7$

   Why: This is the same search as splitting the middle term of a quadratic, which is why the rule is worth remembering in this form.

3. 3
   $16+6\sqrt{7}={\left(\sqrt{9}+\sqrt{7}\right)}^2={\left(3+\sqrt{7}\right)}^2$

   Why: Check by expanding: ${\left(3+\sqrt{7}\right)}^2=9+6\sqrt{7}+7=16+6\sqrt{7}$.

   Square of a sum and of a difference
4. 4
   $\sqrt{16+6\sqrt{7}}=3+\sqrt{7}$

   Why: Both parts are positive, so this is the non-negative root that the radical sign denotes.

    ​ denotes the non-negative root

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Step-by-step solution

1. 1
   ${\log }_{10}{7}^{25}=25\times 0.845=21.125$

   Why: The power rule brings the index down in front, and the given logarithm turns the rest into one multiplication.

   $\log a^n=n\log a$
2. 2
   Digits of $7^{25}=\lfloor 21.125\rfloor +1=22$

   Why: The characteristic $21$ places the number between $10^{21}$ and $10^{22}$, and a number in that range has $22$ digits.

   Number of digits from a common logarithm
3. 3
   ${\log }_{10}{8}^{23}=23{\log }_{10}{2}^3=69\times 0.301=20.769$, so $7^{25}$ aside, $8^{23}$ has $\lfloor 20.769\rfloor +1=21$ digits

   Why: ${\log }_{10}8$ is not given, so write $8=2^3$ to reach a logarithm that is.

   Power of a power, and the move to a common base
4. 4
   ${\log }_{10}{9}^{20}=20{\log }_{10}{3}^2=40\times 0.477=19.08$, so $9^{20}$ has $\lfloor 19.08\rfloor +1=20$ digits

   Why: The same move with $9=3^2$. The three counts are $22$, $21$, $20$.

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Step-by-step solution

1. 1
   $x=14k+7=7(2k+1)$

   Why: Remainder $7$ on division by $14$ means $x$ is $7$ more than a multiple of $14$ — that is, an odd multiple of $7$, which keeps the list short.

   Enumerate a residue class within a bound
2. 2
   $x=7,21,35,49,\dots$

   Why: List the class in increasing order, so the first member that also works is automatically the smallest $x$.

3. 3
   $7=15(0)+7$; $21=15(1)+6$; $35=15(2)+5$

   Why: Test the remainder on division by $15$ in turn. $35$ is the first to leave $5$.

4. 4
   $x=35$, and $30<35<40$

   Why: The smallest positive integer meeting both conditions is $35$, which falls in the second range.

Common mistakes

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Step-by-step solution

1. 1
   Take the full tank as $60$ units, since $\mathrm{LCM}(20,15)=60$

   Why: Choosing the LCM of the two stated times keeps every rate a whole number.

   Take total work as the LCM of the times
2. 2
   $X$ alone: $\frac{60}{20}=3$ units per minute. $X$ and $Y$ together: $\frac{60}{15}=4$ units per minute

   Why: Rate is total divided by time.

3. 3
   $Y$ alone: $4-3=1$ unit per minute

   Why: Both taps empty the tank, so they work in the same direction and the combined rate is their sum; $Y$\'s rate is therefore the combined rate less $X$\'s.

4. 4
   Time for $Y$ alone $=\frac{60}{1}=60$ minutes

   Why: Time is total work divided by rate.

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Step-by-step solution

1. 1
   $\sqrt{75}=\sqrt{25\times 3}=5\sqrt{3}$, which is irrational, so statement 1 is wrong

   Why: $75$ is not a perfect square, and taking the square factor out leaves $\sqrt{3}$ behind — an irrational number multiplied by $5$ stays irrational.

   $\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$
2. 2
   Statement 2: try $x=2$. $-\frac{4\times 2}{5}=-\frac{8}{5}$, and $\frac{8}{5}>\frac{7}{8}$ since $8\times 8=64>35=5\times 7$, so $-\frac{8}{5}<-\frac{7}{8}$. Statement 2 is correct

   Why: The statement only claims that at least one positive integer works, so one witness settles it; cross-multiplication compares the two magnitudes, and the negative signs reverse the order.

   Comparing two fractions
3. 3
   Statement 3: $\frac{x-2}{x}=1-\frac{2}{x}$, which is below $1$ only when $\frac{2}{x}>0$, i.e. $x>0$. At $x=-2$ it is $\frac{-4}{-2}=2$, and $2<1$ is false. Statement 3 is wrong

   Why: "For all real values of $x$" is a strong claim: it fails for every negative $x$, and $x=0$ is not even allowed.

4. 4
   Statement 4: $4.232323\dots =4+\frac{23}{99}=\frac{419}{99}$. Statement 4 is correct

   Why: Any recurring decimal converts to a ratio of integers — two digits under the bar give a denominator of two nines. So statements 2 and 4 are the correct pair.

   Convert a recurring decimal to a fraction