Elementary Mathematics 2019 (I)

Step-by-step solution

1. 1
   $x=\sqrt{2+\sqrt{2+\sqrt{\dots }}}$, so the value asked for is $2+x$

   Why: The leading $2$ sits outside every radical sign, so set it aside and work only on the nest.

   Value of an infinite nested radical
2. 2
   $x=\sqrt{2+x}\Rightarrow x^2=2+x$

   Why: Everything under the first root is the whole nest over again, so the nest contains a copy of itself; squaring clears the outer root.

3. 3
   $x^2-x-2=0\Rightarrow (x-2)(x+1)=0$

   Why: Split the middle term: $-2$ and $+1$ multiply to $-2$ and add to $-1$.

   Factorise a quadratic by splitting the middle termA product is zero only when a factor is zero
4. 4
   $x=2$, rejecting $x=-1$; hence $2+x=4$

   Why: A radical sign denotes the non-negative root, so the nest cannot be negative. Now restore the $2$ that was set aside.

    ​ denotes the non-negative root

Common mistakes

Skills used

Step-by-step solution

1. 1
   $n(E)=52$, $n(M)=42$, $n(E\cap M)=17$ out of a total of $100$

   Why: Percentages of one and the same total behave exactly like counts out of $100$, so no unknown total has to be carried.

   Inclusion-exclusion for two sets
2. 2
   $n(E\cup M)=52+42-17=77$

   Why: Adding the two failure figures counts the $17$ who failed both subjects twice, so one copy of them has to be removed.

3. 3
   passed in both $=100-n(E\cup M)=100-77=23$

   Why: Passing in both subjects is the same as failing in neither, which is the complement of the union.

Common mistakes

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Step-by-step solution

1. 1
   Let the wife's share be $x$

   Why: Name the quantity the question actually asks for, so nothing has to be converted back at the end.

   Turn a worded condition into a linear equation in one unknown
2. 2
   each daughter $=2x$, and each son $=3\times 2x=6x$

   Why: The two conditions chain: a daughter is pegged to the mother, and a son is pegged to a daughter — not to the mother.

   Join two ratios on their shared term
3. 3
   $x+5(6x)+4(2x)=x+30x+8x=39x$

   Why: There are five sons and four daughters, so each individual share is multiplied by the number of people drawing it.

   Introduce a common factor $k$ for a ratio
4. 4
   $39x=390000\Rightarrow x=10000$

   Why: The whole estate is distributed, so the $39$ units must account for all of Rs. $3,90,000$.

Common mistakes

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Step-by-step solution

1. 1
   $A=P{\left(1+\frac{40}{100}\right)}^n=P(1.4{)}^n$

   Why: Compound interest multiplies the balance by the same factor every year, so $n$ years multiply it $n$ times.

   Compound interest amount $A=P{\left(1+\frac{r}{100}\right)}^n$
2. 2
   'more than trebled' means $A>3P$, i.e. $(1.4{)}^n>3$

   Why: Trebling refers to the amount, not to the interest, and $P$ cancels — so the answer does not depend on the sum invested.

3. 3
   $(1.4{)}^2=1.96$, $(1.4{)}^3=1.96\times 1.4=2.744$, $(1.4{)}^4=1.96\times 1.96=3.8416$

   Why: Build the fourth power by squaring $1.96$ rather than multiplying by $1.4$ four times; it halves the arithmetic.

   Power of a power, and the move to a common base
4. 4
   $2.744<3<3.8416$, so the least number of complete years is $4$

   Why: At the end of three years the sum is still short of three times the principal; the fourth year carries it past.

Common mistakes

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Step-by-step solution

1. 1
   $\frac{P_1\times 5\times 2}{100}=\frac{P_2\times 6\times 2}{100}=\frac{P_3\times 9\times 2}{100}$

   Why: Simple interest is $\frac{PRT}{100}$, and the three interests are stated to be equal.

   Simple interest $=\frac{PRT}{100}$
2. 2
   $5P_1=6P_2=9P_3$

   Why: The common factor $\frac{2}{100}$ cancels, which is why the two years never enter the answer.

   Clear the denominators of an equation
3. 3
   $P_1:P_2:P_3=\frac{1}{5}:\frac{1}{6}:\frac{1}{9}=18:15:10$

   Why: Equal products mean the parts go inversely with the rates; multiplying the three reciprocals by $90$, the LCM of $5$, $6$ and $9$, clears the fractions.

   Introduce a common factor $k$ for a ratio
4. 4
   $18k+15k+10k=43k=17200\Rightarrow k=400$

   Why: The three parts together make up the whole sum.

5. 5
   money at $9\%=10k=\text{Rs. }4000$

   Why: The highest rate carries the smallest part — that is the whole point of the inverse ratio.

Common mistakes

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Step-by-step solution

1. 1
   Put $a=x-y$, $b=y-z$, $c=z-x$

   Why: Naming the brackets exposes the only structure that matters; nothing about $x$, $y$ and $z$ separately is ever needed.

   $a^3+b^3+c^3-3abc$, and the case $a+b+c=0$
2. 2
   $a+b+c=(x-y)+(y-z)+(z-x)=0$

   Why: Each of $x$, $y$ and $z$ appears once with a plus and once with a minus, so the sum vanishes identically — for all $x$, $y$, $z$, with no condition to check.

3. 3
   $a^3+b^3+c^3-3abc=(a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right)=0$, so $a^3+b^3+c^3=3abc$

   Why: Killing the factor $a+b+c$ kills the whole right-hand side, whatever the second bracket happens to be.

4. 4
   $\frac{(x-y{)}^3+(y-z{)}^3+(z-x{)}^3}{3(x-y)(y-z)(z-x)}=\frac{3abc}{3abc}=1$

   Why: Numerator and denominator are now literally the same expression.

Common mistakes

Skills used

Step-by-step solution

1. 1
   Let $a^x=b^y=c^z=k$

   Why: Three quantities are stated to be equal; naming that common value is what allows the three different bases to be compared at all.

   Power of a power, and the move to a common base
2. 2
   $a=k^{1/x}$, $b=k^{1/y}$, $c=k^{1/z}$

   Why: Raising $a^x=k$ to the power $\frac{1}{x}$ returns $a$, because a power of a power multiplies the indices.

3. 3
   $b^2=ac\Rightarrow k^{2/y}=k^{1/x}\times k^{1/z}=k^{\frac{1}{x}+\frac{1}{z}}$

   Why: Every term now sits on the single base $k$, so the given relation becomes a statement about indices alone.

4. 4
   $\frac{1}{x}+\frac{1}{z}=\frac{2}{y}$

   Why: Equal powers of one and the same base force the exponents to be equal.

Common mistakes

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Step-by-step solution

1. 1
   $p+q=15$ and $pq=r$

   Why: For $x^2-15x+r=0$ the sum of the roots is $-\frac{-15}{1}=15$ and their product is $\frac{r}{1}=r$.

   Sum and product of roots (Vieta)
2. 2
   $(p-q{)}^2=(p+q{)}^2-4pq$

   Why: This identity ties the difference to the sum and the product, so the roots themselves never have to be found.

   Square of a sum and of a difference
3. 3
   $1^2=15^2-4r\Rightarrow 4r=225-1=224\Rightarrow r=56$

   Why: Substitute $p-q=1$ and $p+q=15$ and solve the one linear equation in $r$.

Common mistakes

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Step-by-step solution

1. 1
   $x^2-7x+12=(x-3)(x-4)$

   Why: Split the middle term: $-3$ and $-4$ multiply to $12$ and add to $-7$.

   Factorise a quadratic by splitting the middle term
2. 2
   The roots are $3$ and $4$, and the coefficient of $x^2$ is $+1$

   Why: An upward parabola is negative only strictly between its roots and positive everywhere outside them, so the sign pattern is fixed once the roots are known.

   Locate roots by the sign of $f$ at test points
3. 3
   $(x-3)(x-4)>0\Rightarrow x<3$ or $x>4$, i.e. $-\infty <x<3$ or $4<x<\infty$

   Why: Both factors are negative when $x<3$ and both positive when $x>4$; between $3$ and $4$ they have opposite signs and the product is negative.

Common mistakes

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Step-by-step solution

1. 1
   $5^{2n}-2^{3n}={\left(5^2\right)}^n-{\left(2^3\right)}^n=25^n-8^n$

   Why: A power of a power multiplies the indices, which puts both terms into the single shape $a^n-b^n$ with the same exponent.

   Power of a power, and the move to a common base
2. 2
   $25-8=17$

   Why: For $a^n-b^n$ the difference of the bases is where a factor independent of $n$ can come from, so compute it first.

   $a-b$ divides $a^n-b^n$ for every $n$
3. 3
   $25$ and $8$ leave the same remainder on division by $17$, so $25^n$ and $8^n$ leave the same remainder as each other, and $25^n-8^n$ is a multiple of $17$ for every $n$

   Why: Replacing a base by another leaving the same remainder does not change the remainder of its powers; this is the remainder argument behind '$a-b$ divides $a^n-b^n$'.

   Reduce a large power by the cycle of its base modulo $n$
4. 4
   At $n=1$ the expression is $25-8=17$, which neither $3$ nor $7$ divides

   Why: A factor has to work for every $n$, so one counterexample disposes of an option — and the smallest $n$ is the cheapest place to look.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $\tan x=1$ with $0<x<90^{\circ }$ gives $x=45^{\circ }$

   Why: $45^{\circ }$ is the only acute angle whose opposite and adjacent sides are equal, so the tangent equals $1$ there and nowhere else in the range.

   Exact trigonometric ratios of the standard angles
2. 2
   $\sin 45^{\circ }=\cos 45^{\circ }=\frac{1}{\sqrt{2}}$

   Why: Read both off the half-square triangle with sides $1$, $1$, $\sqrt{2}$.

3. 3
   $2\sin x\cos x=2\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=2\times \frac{1}{2}=1$

   Why: Substitute both values; the factor $2$ is part of what is asked and must survive to the end.

Common mistakes

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Step-by-step solution

1. 1
   $44^{\circ }=90^{\circ }-46^{\circ }$, so $\cos 44^{\circ }=\sin 46^{\circ }$ and $\sin 44^{\circ }=\cos 46^{\circ }$

   Why: The two angles are complementary; the expression is designed so that bringing everything to one angle makes it collapse.

   Ratios of complementary angles
2. 2
   $\sin 46^{\circ }\cos 44^{\circ }+\cos 46^{\circ }\sin 44^{\circ }={\sin }^{2}46^{\circ }+{\cos }^{2}46^{\circ }$

   Why: Substituting both replacements turns each product into the square of a single ratio.

3. 3
   ${\sin }^{2}46^{\circ }+{\cos }^{2}46^{\circ }=1$

   Why: The Pythagorean identity holds for every angle, so no table of values is needed.

   Pythagorean trigonometric identities

Common mistakes

Skills used

Step-by-step solution

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

   Why: The expression and $4\sin \theta$ differ by a perfect square — recognise $4s^2-4s+1$ as $(2s-1{)}^2$ with $s=\sin \theta$.

   Square of a sum and of a difference
2. 2
   ${\left(2\sin \theta -1\right)}^2\ge 0$ for every $\theta$, so $4{\sin }^2\theta +1\ge 4\sin \theta$

   Why: A square is never negative, and equality occurs only where $\sin \theta =\frac{1}{2}$, i.e. at $\theta =30^{\circ }$ — which is why the option says 'greater than or equal to'.

3. 3
   As $\theta \to 0^{\circ }$ the expression tends to $1$, which is below $2$ and below $4\cos \theta \to 4$; and as $\theta \to 90^{\circ }$, $4\tan \theta$ grows without bound while the expression stays at $5$

   Why: 'For every $\theta$' means a single counterexample kills a claim, and the two ends of the range are where to look.

   Sine and cosine lie between $-1$ and $1$

Common mistakes

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Step-by-step solution

1. 1
   Let the hexagon have side $a$; $A$ and $B$ are adjacent vertices, so $AB=a$

   Why: The foot of the first tower is one side away from the observer, so its ground distance needs no work.

2. 2
   Each interior angle of a regular hexagon is $\frac{(6-2)\times 180^{\circ }}{6}=120^{\circ }$, so in $△ABC$ we have $AB=BC=a$ and $\angle ABC=120^{\circ }$

   Why: $A$ and $C$ are two vertices apart, and the triangle joining them through $B$ is isosceles with the hexagon's own interior angle at $B$.

   Interior and exterior angle sums of a polygonBase angles of an isosceles triangle
3. 3
   The base angles are $\frac{180^{\circ }-120^{\circ }}{2}=30^{\circ }$ each, and the perpendicular from $B$ bisects $AC$, so $AC=2a\cos 30^{\circ }=a\sqrt{3}$

   Why: Dropping the perpendicular from the apex of an isosceles triangle splits it into two right triangles, which is how the $120^{\circ }$ is turned into something a ratio can be taken of.

   Sine, cosine and tangent in a right triangleExact trigonometric ratios of the standard angles
4. 4
   $h_B=AB\tan 30^{\circ }=\frac{a}{\sqrt{3}}$

   Why: The tower is vertical and the elevation is taken from $A$ at ground level, so height $=$ ground distance $\times \tan (\text{angle})$.

   Angles of elevation and depression
5. 5
   $h_C=AC\tan 45^{\circ }=a\sqrt{3}\times 1=a\sqrt{3}$

   Why: The same right-triangle relation, but with the longer ground distance $AC$ — using $AB$ again here is the commonest error.

6. 6
   $h_B:h_C=\frac{a}{\sqrt{3}}:a\sqrt{3}=1:3$

   Why: Divide; $a$ cancels, so the side of the hexagon never had to be known.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $\tan \left(90^{\circ }-\theta \right)=\cot \theta$, so $\tan \theta \tan \left(90^{\circ }-\theta \right)=\tan \theta \cot \theta =1$

   Why: Complementary angles turn a tangent into a cotangent, and a ratio multiplied by its reciprocal is $1$.

   Ratios of complementary anglesReciprocal and quotient relations between the ratios
2. 2
   The $89$ factors pair as $\left(1^{\circ },89^{\circ }\right),\left(2^{\circ },88^{\circ }\right),\dots ,\left(44^{\circ },46^{\circ }\right)$ — that is $44$ pairs using $88$ of them, leaving $\tan 45^{\circ }$

   Why: The angles run from $1^{\circ }$ to $89^{\circ }$, which is symmetric about $45^{\circ }$; counting the pairs is what makes sure nothing is left unaccounted for.

3. 3
   Product $=1^{44}\times \tan 45^{\circ }=1\times 1=1$

   Why: $\tan 45^{\circ }=1$, so the one unpaired factor changes nothing.

   Exact trigonometric ratios of the standard angles

Common mistakes

Skills used

Step-by-step solution

1. 1
   First leg $=20\times \frac{15}{60}=5\text{ km}$; second leg $=30\times \frac{20}{60}=10\text{ km}$

   Why: Convert each time to hours before multiplying, so both distances come out in kilometres.

   Speed, distance and time
2. 2
   He faces north; turning $150^{\circ }$ to the right leaves him $150^{\circ }$ clockwise from north, which is $180^{\circ }-150^{\circ }=30^{\circ }$ east of due south

   Why: A turn is measured from the direction already being faced, and re-expressing the heading as an angle from the north-south line is what lets it be resolved.

   Reading a compass bearing N ${\alpha }^{\circ }$ E as a right triangle
3. 3
   Turning $60^{\circ }$ to the left brings the heading to $150^{\circ }-60^{\circ }=90^{\circ }$ clockwise from north, i.e. due east

   Why: A left turn subtracts from the clockwise angle. This is the step the question turns on: the last leg runs exactly perpendicular to the two streets.

4. 4
   Eastward part of the first leg $=5\sin 30^{\circ }=5\times \frac{1}{2}=2.5\text{ km}$

   Why: The heading was $30^{\circ }$ away from the north-south line, so the east-west offset is the leg times the sine of that angle.

   Exact trigonometric ratios of the standard angles
5. 5
   Distance between the streets $=2.5+10=12.5\text{ km}$

   Why: The second leg is due east, so all $10\text{ km}$ of it counts; the north-south parts of the journey are irrelevant because the streets run in that very direction.

Common mistakes

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Step-by-step solution

1. 1
   $3\tan \theta =\cot \theta =\frac{1}{\tan \theta }\Rightarrow 3{\tan }^2\theta =1$

   Why: Cotangent is the reciprocal of tangent, so multiplying through by $\tan \theta$ leaves a single unknown, ${\tan }^2\theta$.

   Reciprocal and quotient relations between the ratios
2. 2
   ${\tan }^2\theta =\frac{1}{3}\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}$

   Why: On $0\le \theta \le \frac{\pi }{2}$ the tangent is non-negative, so the negative square root is discarded.

3. 3
   $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$, so $\theta =30^{\circ }=\frac{\pi }{6}$

   Why: Read off the standard value, then convert to radians because the options are given in radians.

   Exact trigonometric ratios of the standard anglesConverting between degrees and radians

Common mistakes

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Step-by-step solution

1. 1
   $\sin 65^{\circ }=\sin \left(90^{\circ }-25^{\circ }\right)=\cos 25^{\circ }$

   Why: Angles adding to $90^{\circ }$ let a sine be rewritten as the cosine of the other one, which brings both terms to a single angle.

   Ratios of complementary angles
2. 2
   ${\sin }^{2}25^{\circ }+{\sin }^{2}65^{\circ }={\sin }^{2}25^{\circ }+{\cos }^{2}25^{\circ }=1$

   Why: The Pythagorean identity finishes it, with no table of values needed.

   Pythagorean trigonometric identities

Common mistakes

Skills used

Step-by-step solution

1. 1
   ${\left({\sin }^2\theta +{\cos }^2\theta \right)}^3=1^3=1$

   Why: The sixth powers wanted are the cubes of ${\sin }^2\theta$ and ${\cos }^2\theta$, so cube the one identity that already relates those two quantities.

   Pythagorean trigonometric identities
2. 2
   With $a={\sin }^2\theta$ and $b={\cos }^2\theta$, $(a+b{)}^3=a^3+b^3+3ab(a+b)$ gives $1={\sin }^6\theta +{\cos }^6\theta +3{\sin }^2\theta {\cos }^2\theta \times 1$

   Why: Use the grouped form of the cube, because here $a+b$ is known to be $1$ and the cross terms collapse at once.

   Cube of a sum and of a difference
3. 3
   ${\sin }^6\theta +{\cos }^6\theta +3{\sin }^2\theta {\cos }^2\theta =1$, so the expression is $1-1=0$

   Why: The question subtracts precisely the $1$ that the identity delivers, so the whole thing vanishes for every $\theta$.