Elementary Mathematics 2019 (I)

Step-by-step solution

1. 1
   Put $a=2\sqrt{2}$ and $b=2\sqrt{3}$. Then $ab=4\sqrt{6}$ and $a+b=2\left(\sqrt{2}+\sqrt{3}\right)$, so $x=\frac{ab}{\frac{a+b}{2}}=\frac{2ab}{a+b}$.

   Why: The two fractions pair $x$ against $2\sqrt{2}$ and $2\sqrt{3}$, and the numerator $4\sqrt{6}$ is exactly their product — that is the signal to name them rather than to rationalise.

   $\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$
2. 2
   $x-a=\frac{2ab-a(a+b)}{a+b}=\frac{a(b-a)}{a+b}$ and $x+a=\frac{2ab+a(a+b)}{a+b}=\frac{a(a+3b)}{a+b}$

   Why: Putting $x\pm a$ over the common denominator $a+b$ lets both the factor $a$ and that denominator cancel when the ratio is taken.

   Clear the denominators of an equation
3. 3
   $\frac{x+2\sqrt{2}}{x-2\sqrt{2}}=\frac{a+3b}{b-a}$, and interchanging $a$ and $b$ gives $\frac{x+2\sqrt{3}}{x-2\sqrt{3}}=\frac{3a+b}{a-b}$

   Why: $x$ is symmetric in $a$ and $b$, so the second fraction is the first with the letters swapped — no second computation is needed.

4. 4
   $\frac{a+3b}{b-a}+\frac{3a+b}{a-b}=\frac{(a+3b)-(3a+b)}{b-a}=\frac{2(b-a)}{b-a}=2$

   Why: $a-b=-(b-a)$, so flipping one sign gives the two fractions a common denominator, and $b-a$ then cancels outright.

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1. 1
   $x=\frac{30}{100}z=0.3z$ and $y=\frac{40}{100}z=0.4z$

   Why: Both are percentages of the same quantity $z$, so expressing both through $z$ is what makes them comparable.

   $x\%$ of a quantity
2. 2
   '$x$ is $p\%$ of $y$' means $x=\frac{p}{100}y$, so $p=\frac{x}{y}\times 100$

   Why: The base of the percentage is $y$, so it is $y$ that is divided by — not $x$.

3. 3
   $p=\frac{0.3z}{0.4z}\times 100=\frac{3}{4}\times 100=75$

   Why: $z$ cancels, so no value for $z$ is ever needed.

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1. 1
   The plane flies level, so $\tan (\text{elevation})=\frac{\text{height}}{\text{ground distance from P}}$; equal elevations at $t$ and $t+30$ therefore mean equal ground distances from P.

   Why: The height never changes, so the angle depends on the horizontal distance alone — that converts an angle condition into a distance condition.

   Angles of elevation and depression
2. 2
   The plane is vertically above P at some instant, so P lies on the circular path.

   Why: A level circular flight passes over only the points of its own circle, so this is what fixes P's position.

3. 3
   Two points of the circle equidistant from P are symmetric about the diameter through P, so they are equal arcs from P on either side.

   Why: Equal chords from P are equidistant from the centre, and the perpendicular from the centre bisects each chord, so equal chords cut off equal arcs.

   The perpendicular from the centre bisects a chord
4. 4
   One round is $180$ s, so $30$ s is $\frac{30}{180}\times 360^{\circ }=60^{\circ }$ of arc. If the plane is $\alpha$ past P at time $t$, it is $\alpha +60^{\circ }$ past P at $t+30$, and the symmetry needs $\alpha +60^{\circ }$ and $-\alpha$ to name the same point: $2\alpha =300^{\circ }$, so $\alpha =150^{\circ }$.

   Why: Uniform speed makes arc proportional to time. The condition $2\alpha =300^{\circ }$ comes from going once round; the other solution $2\alpha =-60^{\circ }$ gives $\alpha =-30^{\circ }$ and leads to $x=15$.

   Speed, distance and time
5. 5
   From $150^{\circ }$ past P the plane must fly a further $360^{\circ }-150^{\circ }=210^{\circ }$ to return to P, taking $\frac{210}{360}\times 180=105$ s.

   Why: The plane travels one way only, so the arc to P is the forward one, not the shorter backward one.

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Step-by-step solution

1. 1
   Coprime means $\text{HCF}(p,q)=1$, i.e. the two prime factorisations share no prime; neither number need itself be prime.

   Why: The definition is a condition on the pair, not on either number separately — that is what leaves all three cases open.

   Coprime means no shared prime factor
2. 2
   $p=2$, $q=3$: both prime, $\text{HCF}=1$. Statement 1 holds.

   Why: Two different primes can share no prime factor, so any two distinct primes are coprime.

3. 3
   $p=4$, $q=9$: $4=2^2$ and $9=3^2$ share no prime, so $\text{HCF}=1$. Statement 2 holds.

   Why: This is the case candidates reject; composite numbers built from different primes are coprime.

4. 4
   $p=2$, $q=9$: one prime, one composite, $\text{HCF}=1$. Statement 3 holds.

   Why: A prime $p$ fails to be coprime to a composite only when the composite is a multiple of $p$, and $9$ is not.

5. 5
   All three statements are correct.

   Why: Each has a witness, and none of them claims that something must always happen.

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Step-by-step solution

1. 1
   Let there be $g$ girls, hence $100-g$ boys

   Why: One unknown suffices because the class size is fixed, and taking $g$ as the quantity asked for avoids a conversion at the end.

   Turn a worded condition into a linear equation in one unknown
2. 2
   Total class weight $=100\times 30=3000$ kg

   Why: Mean times the number of observations recovers the total, which is the quantity the two groups must share.

   Mean of a frequency distribution
3. 3
   $24g+32(100-g)=3000$

   Why: This is the weighted-mean equation written out: each group contributes its own mean times its own size.

4. 4
   $3200-8g=3000\Rightarrow 8g=200\Rightarrow g=25$

   Why: Each girl replaces a boy and pulls the total down by $32-24=8$ kg, so $200$ kg of shortfall needs $25$ girls.

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1. 1
   $\sqrt{(a-b{)}^2}=|a-b|$ and $\sqrt{(b-a{)}^2}=|b-a|=|a-b|$

   Why: The radical sign denotes the non-negative root, so it returns the magnitude whichever of $a$ and $b$ is larger.

    ​ denotes the non-negative root
2. 2
   The sum is $|a-b|+|a-b|=2|a-b|$

   Why: The two terms are equal, not opposite — which is precisely what makes the 'always zero' reading wrong.

3. 3
   $2|a-b|=0$ exactly when $a=b$, and $2|a-b|>0$ otherwise

   Why: An absolute value vanishes only at zero, so the expression is positive precisely when $a\ne b$, with no condition on which is larger.

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1. 1
   $a:b=1:6\Rightarrow a=k$, $b=6k$; $c:d=1:6\Rightarrow c=m$, $d=6m$

   Why: Each ratio fixes its own pair only up to a factor, and the two pairs need not share the same factor — using one letter for both would be an extra assumption.

   Introduce a common factor $k$ for a ratio
2. 2
   $\frac{a^2+c^2}{b^2+d^2}=\frac{k^2+m^2}{36k^2+36m^2}$

   Why: $b^2=36k^2$ and $d^2=36m^2$, so $36$ is common to both terms underneath.

3. 3
   $=\frac{k^2+m^2}{36\left(k^2+m^2\right)}=\frac{1}{36}$

   Why: $k^2+m^2$ cancels, so the value depends on neither factor; it is simply ${\left(\frac{1}{6}\right)}^2$.

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Step-by-step solution

1. 1
   $0.\overline{53}=\frac{53}{99}$

   Why: Purely recurring with two digits under the bar, so the denominator is two nines.

   Convert a recurring decimal to a fraction
2. 2
   $0.5\overline{3}=\frac{53-5}{90}=\frac{48}{90}=\frac{8}{15}$

   Why: One digit after the point sits outside the bar and one under it, so the denominator is one nine followed by one zero, and the numerator is the digits up to the end of one period minus the digits before the bar.

3. 3
   $\frac{53}{99}+\frac{8}{15}=\frac{265}{495}+\frac{264}{495}=\frac{529}{495}$

   Why: $\text{LCM}(99,15)=495$, since $99=9\times 11$ and $15=3\times 5$.

   Clear the denominators of an equation
4. 4
   $\frac{529}{495}=1+\frac{34}{495}=1+\frac{68}{990}=1.0\overline{68}$

   Why: $\frac{68}{990}$ has denominator two nines after one zero, which by the same rule read backwards is $0.0\overline{68}$.

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Step-by-step solution

1. 1
   $N=1$: $3>1$, true. $N=2$: $9>8$, true. $N=3$: $27>27$, false. $N=4$: $81>64$, true.

   Why: The two sides are closest around $N=3$, so that is where the strict inequality can break — and at $N=3$ they are exactly equal, so $>$ fails.

   Compare $a^n$ with $n^k$ over the natural numbersPower of a power, and the move to a common base
2. 2
   For $N\ge 4$, each step multiplies $3^N$ by $3$ but multiplies $N^3$ by ${\left(1+\frac{1}{N}\right)}^3\le {\left(\frac{5}{4}\right)}^3\approx 1.95<3$.

   Why: Once $3^N$ is ahead at $N=4$ and grows by the larger factor at every further step, it stays ahead for ever — so no case beyond $N=4$ needs testing.

3. 3
   The inequality holds for $N=1,2$ and for every $N\ge 4$; it fails only at $N=3$.

   Why: That is exactly 'every natural number except $3$'.

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Step-by-step solution

1. 1
   $59049=3^{10}$, so $\sqrt{59049}=3^5=243$ — an integer, hence rational.

   Why: A radical sign does not make a number irrational; here the radicand happens to be a perfect square.

   Prime factorisation$\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$
2. 2
   $\frac{231}{593}$ is a ratio of integers, hence rational; its expansion is non-terminating but repeating, since $593$ is not of the form $2^m{5}^n$.

   Why: Every quotient of integers is rational by definition, however unpleasant its decimal looks.

   When a rational number's decimal expansion terminates
3. 3
   $0.45454545\dots =0.\overline{45}=\frac{45}{99}=\frac{5}{11}$, hence rational.

   Why: A repeating decimal always converts to a fraction, so repetition is evidence of rationality rather than against it.

   Convert a recurring decimal to a fraction
4. 4
   In $0.12112211122211112222\dots$ the blocks of $1$s and $2$s lengthen without limit, so no fixed block ever repeats: non-terminating and non-recurring, hence irrational.

   Why: A repeating decimal needs one fixed block repeated for ever; a pattern whose blocks grow has no period at all.

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Step-by-step solution

1. 1
   Part I: $9\times \frac{50}{60}=7.5$ km

   Why: Distance is speed times time, with the time in hours so that it matches km/hr.

   Speed, distance and time
2. 2
   Part II: $8\times \frac{80}{60}=\frac{32}{3}$ km; Part III: $7.5\times \frac{100}{60}=12.5$ km

   Why: Each leg has its own speed, so each distance has to be computed separately — there is no shortcut through the speeds alone.

3. 3
   Total distance $=7.5+\frac{32}{3}+12.5=20+\frac{32}{3}=\frac{92}{3}$ km; total time $=\frac{50+80+100}{60}=\frac{230}{60}=\frac{23}{6}$ h

   Why: Adding the two whole-number distances first leaves only one fraction to carry.

4. 4
   Average speed $=\frac{92/3}{23/6}=\frac{92}{3}\times \frac{6}{23}=8$ km/hr

   Why: $\frac{92}{23}=4$ and $\frac{6}{3}=2$, so the quotient is exactly $8$ — the mean of the three speeds is not the answer.

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Step-by-step solution

1. 1
   Let $\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=k$, so $a=k(b+c)$, $b=k(c+a)$, $c=k(a+b)$

   Why: Naming the common value converts three fractions into three linear equations that can be combined.

   Introduce a common factor $k$ for a ratio
2. 2
   Add $ka$, $kb$, $kc$ respectively: $a+ka=k(a+b+c)$, and similarly for $b$ and for $c$

   Why: Each right-hand side is 'everything except this letter', so adding the missing letter makes all three right-hand sides the same quantity $ks$, where $s=a+b+c$.

   Complete each equation to the common sum
3. 3
   Adding the three equations: $s(1+k)=3ks$, i.e. $s(1-2k)=0$

   Why: The left sides sum to $s+ks=s(1+k)$ and the right sides to $3ks$; a product equal to zero forces one factor to vanish.

   A product is zero only when a factor is zero
4. 4
   If $s\ne 0$ then $k=\frac{1}{2}$. If $s=0$ then $b+c=-a$, so $k=\frac{a}{-a}=-1$.

   Why: Both branches genuinely occur, so neither may be discarded: dividing by $s$ without checking is what loses the value $-1$.

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Step-by-step solution

1. 1
   $3^2=9\equiv 1(\mathrm{mod}8)$

   Why: The cycle closes after two steps, which is as short as it can get — so no large power ever needs computing.

   Reduce a large power by the cycle of its base modulo $n$
2. 2
   $521=2\times 260+1$, so $3^{521}={\left(3^2\right)}^{260}\times 3\equiv 1^{260}\times 3(\mathrm{mod}8)$

   Why: Splitting off the largest even part of the exponent leaves a factor of $1$ and a single leftover $3$.

3. 3
   $3^{521}\equiv 3(\mathrm{mod}8)$, so the remainder is $3$

   Why: A remainder on division by $8$ must lie between $0$ and $7$, and $3$ does.

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Step-by-step solution

1. 1
   A number ending in $0,2,4,6$ or $8$ is even, so $2$ is the only prime with an even unit digit — but $2$ itself is prime, so the digit $2$ does count.

   Why: Every even number above $2$ has $2$ as a proper factor; the single-digit prime $2$ is the exception that keeps its digit on the list.

   Write an odd number as $2m+1$
2. 2
   A number ending in $5$ is divisible by $5$, so $5$ is the only prime ending in $5$ — and it is prime, so $5$ counts too. A number ending in $0$ is divisible by $10$ and is never prime.

   Why: The same exception applies at $5$ as at $2$; only the digit $0$ has no prime at all behind it.

   Prime factorisation
3. 3
   Surviving digits, each with a witness: $1$ ($11$), $2$ ($2$), $3$ ($3$), $5$ ($5$), $7$ ($7$), $9$ ($19$)

   Why: The question asks how many digits are possible, so each needs one prime to prove it possible — nothing more.

4. 4
   $6$ digits are possible; the excluded ones are $0,4,6$ and $8$.

   Why: Counting the exclusions is the faster check: ten digits minus four leaves six.

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Step-by-step solution

1. 1
   At a gain of $6\%$: $S_1=1.06C$. At a loss of $6\%$: $S_2=0.94C$.

   Why: Gain and loss percentages are both taken on the cost price, so both selling prices are simple multiples of $C$.

   Gain percent is measured on the cost price
2. 2
   $S_1-S_2=0.12C=6\Rightarrow C=\frac{6}{0.12}=50$

   Why: The extra $\text{Rs }6$ is the whole gap between the two prices, which is $6\%+6\%=12\%$ of the cost — not $6\%$.

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Step-by-step solution

1. 1
   $12$ men and $18$ women each finish the job in $14$ days, so $12$ men and $18$ women have the same daily output: $2$ men $=3$ women.

   Why: Same job in the same time means the same rate, and that is the only way the two kinds of worker can be compared.

   The work-equivalence chain rule $M_1{D}_1{H}_1=M_2{D}_2{H}_2$
2. 2
   The job measures $12\times 14=168$ man-days.

   Why: $M\times D$ is invariant for a fixed job, so compute it once from the group already expressed in men.

3. 3
   $16$ women $=16\times \frac{2}{3}=\frac{32}{3}$ men, so the new team is $8+\frac{32}{3}=\frac{56}{3}$ men.

   Why: From $2$ men $=3$ women, one woman is $\frac{2}{3}$ of a man.

4. 4
   Days $=\frac{168}{56/3}=168\times \frac{3}{56}=9$

   Why: $\frac{168}{56}=3$, so the answer is $3\times 3=9$ days.

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Step-by-step solution

1. 1
   Let $3^x=4^y=12^z=k$, with $k>0$ and $k\ne 1$

   Why: Naming the common value is what lets three separate statements be compared; $k=1$ would force $x=y=z=0$ and is excluded.

   Logarithm to exponential form
2. 2
   Taking ${\log }_{10}$: $x\log 3=y\log 4=z\log 12=\log k$, so $\log 3=\frac{\log k}{x}$, $\log 4=\frac{\log k}{y}$, $\log 12=\frac{\log k}{z}$

   Why: Logarithms bring each unknown down out of its exponent, and each base is then written through its own unknown.

   $\log a^n=n\log a$
3. 3
   $12=3\times 4$, so $\log 12=\log 3+\log 4$, i.e. $\frac{\log k}{z}=\frac{\log k}{x}+\frac{\log k}{y}$

   Why: The multiplicative relation between the bases becomes an additive relation between their logarithms — this is the step the whole question is built on.

   $\log (ab)=\log a+\log b$ and $\log \frac{a}{b}=\log a-\log b$
4. 4
   Divide by $\log k\ne 0$: $\frac{1}{z}=\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$, so $z=\frac{xy}{x+y}$

   Why: It is the reciprocals that add, so $z$ is neither the sum nor the product of $x$ and $y$ but their 'parallel combination'.

   Clear the denominators of an equation

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Step-by-step solution

1. 1
   $(4a+7b)(4c-7d)=16ac-28ad+28bc-49bd$

   Why: Expand in full — which terms will cancel cannot be seen before they are written down.

2. 2
   $(4a-7b)(4c+7d)=16ac+28ad-28bc-49bd$

   Why: Only the two middle terms change sign between the two products; $16ac$ and $-49bd$ are common to both.

3. 3
   Equating and cancelling: $-28ad+28bc=28ad-28bc\Rightarrow 56bc=56ad\Rightarrow ad=bc$

   Why: Since the two sides differ only in the sign of the middle terms, the condition is exactly that those middle terms are equal.

4. 4
   $ad=bc\Rightarrow \frac{a}{b}=\frac{c}{d}$

   Why: Dividing both sides by $bd$ gives the proportion; $ad=bc$ is that proportion with its denominators already cleared.

   Clear the denominators of an equation

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Step-by-step solution

1. 1
   $f(x)=x^2+ax+b$. The remainder on dividing by $(x-1)$ is $f(1)=1+a+b$.

   Why: The remainder theorem replaces a division by a single substitution, so no long division is needed.

   Factor and remainder theorem
2. 2
   The remainder on dividing by $(x+1)$ is $f(-1)=1-a+b$.

   Why: For the divisor $(x+1)$, set $x+1=0$ and substitute $x=-1$.

3. 3
   $1+a+b=1-a+b\Rightarrow 2a=0\Rightarrow a=0$, with $b$ unrestricted.

   Why: The constant $1$ and the term $b$ cancel from both sides, so the condition says nothing whatever about $b$ — it constrains $a$ alone.

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Step-by-step solution

1. 1
   Total work $=\text{LCM}(6,10)=30$ units

   Why: Choosing the LCM makes both hourly rates integers, which keeps the arithmetic out of fractions until the last step.

   Take total work as the LCM of the times
2. 2
   Tushar's rate $=\frac{30}{6}=5$ units/hour; Amar's rate $=\frac{30}{10}=3$ units/hour

   Why: A rate is the total work divided by the time that worker needs alone.

3. 3
   Together $5+3=8$ units/hour, so the time is $\frac{30}{8}=\frac{15}{4}=3.75$ hours

   Why: Rates add when two people work on the same job at the same time; times never do.

4. 4
   $0.75$ hour $=45$ minutes, so the time is $3$ hours $45$ minutes

   Why: The options are in hours and minutes, so the fractional hour has to be converted.

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