Elementary Mathematics 2019 (II)

Step-by-step solution

1. 1
   Day $1$ is Saturday, so the Sundays are days $2,9,16,23,30$ — five of them

   Why: Sundays repeat every $7$ days from the first one, so list $2,2+7,2+14,\dots$ while the day number stays at most $30$. A $30$-day month starting on Saturday catches a fifth Sunday on its last day.

   Enumerate a residue class within a bound
2. 2
   The other $30-5=25$ days average $240$ visitors

   Why: Every day that is not one of the five Sundays counts as an "other day".

3. 3
   Total visitors $=5\times 510+25\times 240=2550+6000=8550$

   Why: An average times its count gives a total, and totals can be added where averages cannot.

   Mean of a frequency distribution
4. 4
   Average per day $=\frac{8550}{30}=285$

   Why: Divide the month\'s total by its $30$ days.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $\frac{36}{11}=3+\frac{3}{11}$, so $\frac{1}{x+\frac{1}{y+\frac{1}{z}}}=\frac{3}{11}$

   Why: Subtract the $3$ that is already printed on the right-hand side; $11\times 3=33$ leaves $3$.

   Evaluate a nested fraction from the inside out
2. 2
   $x+\frac{1}{y+\frac{1}{z}}=\frac{11}{3}=3+\frac{2}{3}$, so $x=3$

   Why: Invert both sides. Since $y+\frac{1}{z}>1$, the fraction added to $x$ is less than $1$, so $x$ must be the whole-number part.

3. 3
   $\frac{1}{y+\frac{1}{z}}=\frac{2}{3}\Rightarrow y+\frac{1}{z}=\frac{3}{2}=1+\frac{1}{2}$, so $y=1$ and $\frac{1}{z}=\frac{1}{2}$

   Why: The same move one level down.

4. 4
   $z=2$, so $x+y+z=3+1+2=6$

   Why: All three came out natural numbers, as the question requires.

Common mistakes

Skills used

Step-by-step solution

1. 1
   Profit item: $990=1.1C_1\Rightarrow C_1=\frac{990}{1.1}=900$

   Why: Selling price is cost price times the gain multiplier, so dividing by $1.1$ recovers the cost.

   Percentage change as a multiplierGain percent is measured on the cost price
2. 2
   Loss item: $990=0.9C_2\Rightarrow C_2=\frac{990}{0.9}=1100$

   Why: A $10\%$ loss is the multiplier $0.9$. Note the two cost prices are different, which is the whole point.

3. 3
   Total cost $=900+1100=2000$; total selling price $=2\times 990=1980$

   Why: Percentages on different bases cannot be added, but rupees can — so move to totals before taking any percentage.

4. 4
   Loss $=2000-1980=20$, and $\frac{20}{2000}\times 100=1\%$

   Why: The percentage is always measured on the cost price, so the base is $2000$: a $1\%$ loss.

Common mistakes

Skills used

Step-by-step solution

1. 1
   Train speed $t$, car speed $c$ km/h: $\frac{120}{t}+\frac{480}{c}=11$

   Why: Time is distance over speed, and the two stage times add to the journey time; $600-120=480$ km goes by car.

   Speed, distance and time
2. 2
   $\frac{200}{t}+\frac{400}{c}=11+\frac{40}{60}=\frac{35}{3}$

   Why: The second journey takes $40$ minutes more, which is $\frac{2}{3}$ hour — convert before writing it down, since every other quantity is in hours.

3. 3
   Put $u=\frac{1}{t}$, $v=\frac{1}{c}$: $120u+480v=11$ and $200u+400v=\frac{35}{3}$

   Why: Every term carries $\frac{1}{t}$ or $\frac{1}{c}$, so in those two letters the pair is ordinary and linear.

   Linearise a system by substituting reciprocals
4. 4
   $5\times$ first: $600u+2400v=55$. $3\times$ second: $600u+1200v=35$. Subtracting: $1200v=20\Rightarrow v=\frac{1}{60}$, so $c=60$

   Why: Scale both equations to the same coefficient of $u$ so that $u$ disappears in a single subtraction.

5. 5
   $120u=11-\frac{480}{60}=11-8=3\Rightarrow u=\frac{1}{40}$, so $t=40$

   Why: Back-substitute the car\'s speed into the first equation.

6. 6
   $c:t=60:40=3:2$

   Why: The question asks for car to train, in that order.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $x-x^2=-\left(x^2-x\right)$

   Why: Pull the minus out so that the bracket is a monic quadratic, which is easier to complete.

2. 2
   $x^2-x={\left(x-\frac{1}{2}\right)}^2-\frac{1}{4}$

   Why: Half the coefficient of $x$ is $-\frac{1}{2}$; completing the square adds and subtracts ${\left(\frac{1}{2}\right)}^2$.

   Range of a quadratic by completing the squareSquare of a sum and of a difference
3. 3
   $x-x^2=\frac{1}{4}-{\left(x-\frac{1}{2}\right)}^2$

   Why: A square is never negative, so the expression is largest exactly when that square is $0$ — the point of completing the square rather than guessing.

4. 4
   ${\left(x-\frac{1}{2}\right)}^2=0\Rightarrow x=\frac{1}{2}=0.5$, where $x-x^2=\frac{1}{4}$

   Why: That is the only value making the subtracted square vanish, so the maximum occurs there and nowhere else.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $6^{23}={\left(2\times 3\right)}^{23}=2^{23}\times 3^{23}$

   Why: Factorise each base into primes and carry the index through to every prime.

   Prime factorisationPower of a power, and the move to a common base
2. 2
   $75^9={\left(3\times 5^2\right)}^9=3^9\times 5^{18}$

   Why: $75=3\times 25$, and the index $2$ on the $5$ multiplies with the $9$ — this is where most of the fives come from.

3. 3
   $105^2={\left(3\times 5\times 7\right)}^2=3^2\times 5^2\times 7^2$

   Why: $105=3\times 5\times 7$.

4. 4
   Product $=2^{23}\times 3^{34}\times 5^{20}\times 7^2$

   Why: Collect each prime across the three factors: the threes total $23+9+2=34$ and the fives $18+2=20$. Only $6^{23}$ supplies twos.

5. 5
   $10^n=2^n{5}^n$ divides it only if $n\le 23$ and $n\le 20$, so the largest $n$ is $20$

   Why: Each power of $10$ consumes one $2$ and one $5$; the prime that runs out first fixes the answer, and here it is the $5$.

Common mistakes

Skills used

Step-by-step solution

1. 1
   Unit digits of $3^1,3^2,3^3,3^4$ are $3,9,7,1$, and then they repeat: a cycle of length $4$

   Why: Write out powers until the first unit digit reappears; that fixes the cycle and nothing beyond it is needed.

   Unit digit of a power from its cycle
2. 2
   $98=4\times 24+2$, so $3^{98}$ ends in the $2$nd entry of the cycle: $9$

   Why: The remainder on division by the cycle length gives the position, counted from $3^1$.

3. 3
   $89=4\times 22+1$, so $3^{89}$ ends in the $1$st entry: $3$

   Why: The same rule for the second power.

4. 4
   $3^{98}>3^{89}$ and $9-3=6$, so the difference ends in $6$

   Why: The difference is positive, and the first number\'s last digit is the larger, so the units column needs no borrowing: the last digit is simply $9-3$.

Common mistakes

Skills used

Step-by-step solution

1. 1
   Numbers $n,n+1,n+2,n+3$: $n^2+(n+1{)}^2+(n+2{)}^2+(n+3{)}^2=294$

   Why: Consecutive naturals differ by $1$, so one letter describes all four.

   Turn a worded condition into a linear equation in one unknownSquare of a sum and of a difference
2. 2
   $4n^2+12n+14=294\Rightarrow 4n^2+12n-280=0\Rightarrow n^2+3n-70=0$

   Why: Expand and collect, then divide through by $4$ so the factorisation stays in small numbers.

3. 3
   $n^2+3n-70=(n+10)(n-7)=0$

   Why: Two numbers with product $-70$ and sum $3$: $10$ and $-7$.

   Factorise a quadratic by splitting the middle term
4. 4
   $n=7$, so the numbers are $7,8,9,10$ and their sum is $34$

   Why: Set each factor to zero; $n=-10$ is not a natural number and is discarded.

   A product is zero only when a factor is zero

Common mistakes

Skills used

Step-by-step solution

1. 1
   For $x^2+px+q=0$ the roots sum to $-p$ and multiply to $q$

   Why: Vieta\'s relations with $a=1$, $b=p$, $c=q$.

   Sum and product of roots (Vieta)
2. 2
   The roots are $p$ and $q$, so $p+q=-p$ and $pq=q$

   Why: Substitute the stated roots into the two relations; now the coefficients and the roots are the same two letters.

3. 3
   $pq=q\Rightarrow q(p-1)=0$, and $q\ne 0$, so $p=1$

   Why: Collect rather than cancel the $q$, then use the zero-product property. This is exactly why the question bothers to say $q\ne 0$.

   A product is zero only when a factor is zero
4. 4
   $p+q=-p\Rightarrow q=-2p=-2$

   Why: Back-substitute $p=1$.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $a^2-b^2=(a-b)(a+b)=35$, with $a>b\ge 1$

   Why: The difference of squares turns one equation in two unknowns into a product of two whole numbers, which there are only finitely many of.

   Difference of two squares
2. 2
   $35=1\times 35=5\times 7$, and $a+b>a-b>0$, so $(a-b,a+b)$ is $(1,35)$ or $(5,7)$

   Why: Only pairs with the smaller factor first can occur, since $a+b$ exceeds $a-b$ whenever $b$ is positive.

   Prime factorisation
3. 3
   $(1,35)$: $a=\frac{35+1}{2}=18$, $b=\frac{35-1}{2}=17$

   Why: Adding and subtracting the two equations recovers $a$ and $b$; both come out whole because $35$ is odd, so both of its factors are odd.

4. 4
   $(5,7)$: $a=6$, $b=1$. So there are $2$ such pairs

   Why: Each admissible factor pair gives exactly one pair of natural numbers, and there are two such factor pairs.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $(b-6{)}^2-6(b-6)+b=0$

   Why: Putting the stated root into the equation is what "is a root" means, and it converts a statement about $x$ into one about $b$.

   Factor and remainder theorem
2. 2
   $b^2-12b+36-6b+36+b=0\Rightarrow b^2-17b+72=0$

   Why: Expand and collect; the two constants give $36+36=72$.

3. 3
   $b^2-17b+72=(b-8)(b-9)=0\Rightarrow b=8$ or $b=9$

   Why: Two numbers with product $72$ and sum $17$: $8$ and $9$. Both are integers, which the question requires.

   Factorise a quadratic by splitting the middle termA product is zero only when a factor is zero
4. 4
   $b^2$ is $64$ or $81$, so the maximum is $81$

   Why: Both values of $b$ are admissible, so "maximum" is asking which of the two gives the larger $b^2$ — the question is a count-both-cases question, not a single-answer one.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $7+4\sqrt{3}=7+2\sqrt{12}$

   Why: Put the middle term in the form $2\sqrt{b}$ so that the denesting rule applies.

   Denest $\sqrt{a\pm 2\sqrt{b}}$$\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$
2. 2
   $p+q=7$ and $pq=12$ give $p=4$, $q=3$, so $7+4\sqrt{3}={\left(2+\sqrt{3}\right)}^2$

   Why: Two numbers with sum $7$ and product $12$; the inside is then a perfect square.

   Square of a sum and of a difference
3. 3
   $a=2+\sqrt{3}$

   Why: Both parts are positive, so this is the principal root.

    ​ denotes the non-negative root
4. 4
   $\frac{1}{a}=\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$

   Why: Rationalise with the conjugate; the denominator becomes $2^2-{\left(\sqrt{3}\right)}^2=1$, which is what makes the reciprocal so clean.

   Rationalise a denominator using the conjugateDifference of two squares
5. 5
   $a+\frac{1}{a}=\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)=4$

   Why: The surds cancel, leaving a whole number.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $x^2+5x+10={\left(x+\frac{5}{2}\right)}^2+10-\frac{25}{4}$

   Why: Half of $5$ is $\frac{5}{2}$, and completing the square adds and subtracts ${\left(\frac{5}{2}\right)}^2=\frac{25}{4}$.

   Range of a quadratic by completing the squareSquare of a sum and of a difference
2. 2
   $={\left(x+\frac{5}{2}\right)}^2+\frac{15}{4}$

   Why: $10-\frac{25}{4}=\frac{40-25}{4}=\frac{15}{4}$.

3. 3
   The denominator is least at $x=-\frac{5}{2}$, where it equals $\frac{15}{4}$, and it is never zero or negative

   Why: A square is at least $0$, so the denominator is at least $\frac{15}{4}>0$ — which also tells us the fraction is defined for every real $x$ and always positive.

4. 4
   Maximum of $\frac{1}{x^2+5x+10}=\frac{1}{15/4}=\frac{4}{15}$

   Why: For a positive denominator the reciprocal is largest exactly where the denominator is smallest, so the answer is the reciprocal of the minimum.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $\frac{(x+2)(x-3)}{(x+4)(x-2)}=\frac{3}{4}$

   Why: With workers of equal efficiency the job is measured by (number of workers) $\times$ (number of days), so each side of the ratio is one such product.

   The work-equivalence chain rule $M_1{D}_1{H}_1=M_2{D}_2{H}_2$
2. 2
   $4(x+2)(x-3)=3(x+4)(x-2)$

   Why: Cross-multiply to clear both denominators.

   Clear the denominators of an equation
3. 3
   $4\left(x^2-x-6\right)=3\left(x^2+2x-8\right)\Rightarrow 4x^2-4x-24=3x^2+6x-24$

   Why: Expand each side. Both constants come to $-24$, so they will cancel — worth noticing before grinding on.

4. 4
   $x^2-10x=0\Rightarrow x(x-10)=0$

   Why: Collect on the left; with the constant gone, $x$ is a common factor.

   A product is zero only when a factor is zero
5. 5
   $x=10$

   Why: $x=0$ is rejected: it would mean $2$ workers taking $-3$ days. Both day counts must be positive, so $x>3$.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $1$ has exactly one divisor, so it is neither prime (two divisors) nor composite (more than two): the first statement is correct

   Why: Both definitions are counts of divisors, and $1$ meets neither count.

2. 2
   $0$ is neither greater than nor less than $0$, so it is neither positive nor negative: the second statement is correct

   Why: Sign is defined by comparison with $0$, so $0$ itself has none.

3. 3
   Take $p=2$, $q=3$: $pq=6$ is even, yet $q$ is odd

   Why: Writing $p=2m$ gives $pq=2mq$, which is even whatever $q$ is — one even factor is enough. So the claim that both must be even is false, and one counterexample is all it takes to say so.

   Write an odd number as $2m+1$
4. 4
   $\sqrt{2}$ is irrational: the fourth statement is correct. The statement that is NOT correct is the third

   Why: $2$ is not a perfect square, and its root is the standard example of an irrational number. With three statements correct, the odd one out is the parity claim.

Common mistakes

Skills used

Step-by-step solution

1. 1
   Let $n^2+19n+92=m^2$ with $m$ a non-negative integer

   Why: Naming the perfect square makes the condition an equation rather than a search.

2. 2
   $4\left(n^2+19n+92\right)=(2n+19{)}^2-361+368=(2n+19{)}^2+7$

   Why: Multiplying by $4$ first keeps everything in integers — completing the square on $n^2+19n$ directly would bring in $\frac{19}{2}$.

   Square of a sum and of a difference
3. 3
   $(2m{)}^2-(2n+19{)}^2=7$, so $\left(2m-2n-19\right)\left(2m+2n+19\right)=7$

   Why: Both sides are now differences of squares, and $7$ is prime, so only four integer factorisations are possible — the infinite search has become a finite one.

   Difference of two squares
4. 4
   $7=1\times 7=7\times 1=(-1)\times (-7)=(-7)\times (-1)$. Adding the two factors gives $4m=\pm 8$; subtracting gives $4n+38=\pm 6$

   Why: Each factor pair is solved by adding and subtracting the two linear expressions, exactly as in the difference-of-squares count.

5. 5
   $4n+38=6\Rightarrow n=-8$; $4n+38=-6\Rightarrow n=-11$. Sum $=-8+(-11)=-19$

   Why: The two signs give the only two integers, so their sum is the answer.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $x^3+3x^2+3x+1=(x+1{)}^3$

   Why: The coefficients $1,3,3,1$ are the expansion of a cube of a sum — worth recognising on sight rather than dividing.

   Cube of a sum and of a difference
2. 2
   $x^3+5x^2+5x+4$ at $x=-4$: $-64+80-20+4=0$, so $(x+4)$ is a factor, and $x^3+5x^2+5x+4=(x+4)\left(x^2+x+1\right)$

   Why: The factor theorem tests a candidate linear factor in one substitution; the candidates are $x=\pm 1,\pm 2,\pm 4$ from the constant term $4$.

   Factor and remainder theorem
3. 3
   $x^2+x+1$ has discriminant $1-4=-3<0$, so it does not factorise further

   Why: Factorisation must stop at an irreducible quadratic; carrying on would be looking for roots that are not real.

   Quadratic formula
4. 4
   $x^2+5x+4=(x+1)(x+4)$

   Why: Two numbers with product $4$ and sum $5$.

   Factorise a quadratic by splitting the middle term
5. 5
   LCM $=(x+1{)}^3(x+4)\left(x^2+x+1\right)$

   Why: The LCM takes each distinct irreducible factor to the greatest power in which it appears anywhere: $(x+1)$ to the third from the first polynomial, $(x+4)$ and $\left(x^2+x+1\right)$ to the first.

   HCF and LCM of polynomials from their complete factorisations

Common mistakes

Skills used

Step-by-step solution

1. 1
   Put $a=x-y$, $b=y-z$, $c=z-x$. Then $a+b+c=0$

   Why: The three differences cancel whatever $x$, $y$ and $z$ are — no condition is needed, and this is the whole point of how the question is built.

2. 2
   $a+b+c=0\Rightarrow a^3+b^3+c^3=3abc$

   Why: From $a^3+b^3+c^3-3abc=(a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right)$: the first bracket on the right is $0$, so the left side is $0$.

   $a^3+b^3+c^3-3abc$, and the case $a+b+c=0$
3. 3
   Numerator $=3(x-y)(y-z)(z-x)$, denominator $=9(x-y)(y-z)(z-x)$

   Why: Substitute back; the same product appears in both, which is why the question put a $9$ downstairs.

4. 4
   $\frac{3}{9}=\frac{1}{3}$

   Why: Cancel the common product. This needs $x$, $y$, $z$ distinct, otherwise the product is zero and the expression is undefined rather than equal to anything.

Common mistakes

Skills used