Elementary Mathematics 2016 (II)

Step-by-step solution

1. 1
   $\frac{125}{2.5}=50$ per kg of rice

   Why: Dividing a bulk price by the weight gives the unit price, which is what every later equivalence needs.

   Turn a worded condition into a linear equation in one unknownUnitary method: find one unit, then scale
2. 2
   $9\times 50=450=4r_p\Rightarrow r_p=112.5$

   Why: The cost of $9$ kg rice is a number now, and it equals the cost of $4$ kg pulses, giving one equation in the pulse price $r_p$.

3. 3
   $14\times 112.5=1575=1.5r_t\Rightarrow r_t=1050$

   Why: Same move one link along the chain: $14$ kg of pulses costs the same as $1.5$ kg of tea.

4. 4
   $2\times 1050=2100=5r_n\Rightarrow r_n=420$

   Why: The last equivalence converts tea into nuts.

5. 5
   $11\times 420=4620$

   Why: Scale the unit price of nuts up to the $11$ kg the question asks about.

Common mistakes

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Step-by-step solution

1. 1
   $1729=7\times 13\times 19$, so its positive divisors are $1,7,13,19,91,133,247,1729$.

   Why: Because $a+b$ divides $a^3+b^3$, the sum of the two integers must be one of these eight numbers — the search is finite and short before any cube is computed.

   Prime factorisation
2. 2
   $a^3+b^3=(a+b)\left[(a+b{)}^2-3ab\right]$.

   Why: Rewriting $a^2-ab+b^2$ as $(a+b{)}^2-3ab$ converts each candidate sum $s=a+b$ into a value of $ab$, and a sum together with a product determines the pair.

   Square of a sum and of a differenceSum and difference of two cubes
3. 3
   $s=1$ is impossible for positive integers; $s=7$ gives $49-3ab=247$, so $ab=-66<0$, rejected.

   Why: A negative product cannot come from two positive integers, so these sums are ruled out without any root extraction.

4. 4
   $s=13$: $169-3ab=133$, so $ab=12$ and $t^2-13t+12=0$ gives $\{1,12\}$.

   Why: The two integers are the roots of $t^2-st+ab=0$, which here factorises immediately.

   Sum and product of roots (Vieta)
5. 5
   $s=19$: $361-3ab=91$, so $ab=90$ and $t^2-19t+90=0$ gives $\{9,10\}$.

   Why: A second representation, so the count is already at least two and options A and B are dead.

6. 6
   $s=91,133,247,1729$: each forces $ab>\frac{s^2}{4}$, so $t^2-st+ab$ has negative discriminant and no real pair, let alone an integer one.

   Why: This closes the search: no divisor above $19$ can work, so the count cannot rise to three.

   Quadratic formula
7. 7
   Exactly two representations: $1729=1^3+12^3=9^3+10^3$.

   Why: A count cannot be tested by substitution — the range has to be walked — and walking it gives two.

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Step-by-step solution

1. 1
   $\text{MP}=\left(1+\frac{20}{100}\right)\text{CP}=1.2\text{CP}$

   Why: Marking up the cost price by $20\%$ multiplies it by $1.2$.

   Percentage change as a multiplier
2. 2
   $\text{SP}=\left(1-\frac{10}{100}\right)\text{MP}=0.9\times 1.2\text{CP}=1.08\text{CP}$

   Why: The discount is stated to be on the marked price, so its multiplier acts on $1.2\text{CP}$ and not on $\text{CP}$; successive changes multiply.

3. 3
   $\text{gain}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100=(1.08-1)\times 100=8$

   Why: Gain percent is always measured against the cost price, which is the base the $1.08$ factor is expressed in.

   Gain percent is measured on the cost price

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Step-by-step solution

1. 1
   $p=2,q=3$: prime factors $\{2\}$ and $\{3\}$ are disjoint, so $\text{HCF}=1$

   Why: Two distinct primes share no prime factor, so statement 1 is possible.

   Prime factorisationCoprime means no shared prime factor
2. 2
   $p=3,q=4=2^2$: prime factors $\{3\}$ and $\{2\}$ are disjoint, so $\text{HCF}=1$

   Why: One number prime, the other composite, and still no shared prime factor: statement 2 is possible.

3. 3
   $p=4=2^2,q=9=3^2$: prime factors $\{2\}$ and $\{3\}$ are disjoint, so $\text{HCF}=1$

   Why: $\text{HCF}=1$ forbids a shared prime factor; it does not forbid either number from having repeated factors of its own, so statement 3 is possible.

4. 4
   Statements 1, 2 and 3 are all correct

   Why: The question asks which statements may hold, and one witness settles each.

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Step-by-step solution

1. 1
   With $N$ terms the last summand is $\sqrt{1+\frac{1}{N^2}+\frac{1}{(N+1{)}^2}}$, and at $N=2007$ the options read $\text{A}=(N+1)-\frac{1}{N+1}$, $\text{B}=N-\frac{1}{N}$, $\text{C}=N-\frac{1}{N+1}$, $\text{D}=(N+1)-\frac{1}{N+2}$.

   Why: The four options differ only in where $2007$ and $2008$ sit, so reintroducing $N$ turns them into four formulas that a tiny case can separate — the full $2007$-term sum never has to be attempted.

2. 2
   $N=1$: the sum is $\sqrt{1+1+\frac{1}{4}}=\sqrt{\frac{9}{4}}=\frac{3}{2}$.

   Why: $\frac{9}{4}$ is a perfect square, so its principal root is exactly $\frac{3}{2}$ and the test is exact rather than numerical.

    ​ denotes the non-negative root
3. 3
   At $N=1$: $\text{A}=\frac{3}{2}$, $\text{B}=0$, $\text{C}=\frac{1}{2}$, $\text{D}=\frac{5}{3}$. Only A matches.

   Why: Three options are eliminated by a single one-term case, which is the whole economy of this route.

4. 4
   $N=2$: the next term is $\sqrt{1+\frac{1}{4}+\frac{1}{9}}=\sqrt{\frac{49}{36}}=\frac{7}{6}$, so the sum is $\frac{3}{2}+\frac{7}{6}=\frac{8}{3}$, while $\text{A}=3-\frac{1}{3}=\frac{8}{3}$ and $\text{D}=\frac{11}{4}$.

   Why: One surviving option is not proof; a second case confirms A and finishes off D, the only other option in the right neighbourhood.

5. 5
   The radicands are perfect squares in general: $1+\frac{1}{n^2}+\frac{1}{(n+1{)}^2}={\left(1+\frac{1}{n}-\frac{1}{n+1}\right)}^2$.

   Why: Expanding the right-hand side by the square-of-a-sum identity reproduces the left exactly, which is why both small cases came out rational.

   Square of a sum and of a difference
6. 6
   Each term is therefore $1+\frac{1}{n}-\frac{1}{n+1}$, and summing $n=1$ to $2007$ gives $2007+\left(1-\frac{1}{2008}\right)=2008-\frac{1}{2008}$.

   Why: The $1$s contribute $2007$ and the fractional parts cancel in pairs, leaving only the first and last — which is exactly the form option A prints.

   Telescoping a sum by writing each term as a difference

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Step-by-step solution

1. 1
   $u=\frac{20}{2}=10$ km/h

   Why: Downstream speed is the downstream distance divided by the time that trip took.

   Speed, distance and time
2. 2
   $v=\frac{4}{2}=2$ km/h

   Why: Upstream speed, from the second trip in the same way.

3. 3
   $u=b+c,v=b-c$

   Why: Downstream the current adds to the speed in still water; upstream it subtracts.

   Still-water speed and current from the two legs
4. 4
   $c=\frac{u-v}{2}=\frac{10-2}{2}=4$ km/h

   Why: Subtracting the two relations eliminates $b$ and leaves $2c$.

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Step-by-step solution

1. 1
   $x=2m+1,y=2n+1$ with $m,n\ge 0$

   Why: Every positive odd integer is one more than an even one.

   Write an odd number as $2m+1$
2. 2
   $x^2+y^2=(4m^2+4m+1)+(4n^2+4n+1)=4(m^2+m+n^2+n)+2$

   Why: Expanding each square isolates a multiple of $4$ plus a $1$; the two $1$s make the $+2$.

   Square of a sum and of a difference
3. 3
   $x^2+y^2$ is even, since it is $2$ times $\left(2(m^2+m+n^2+n)+1\right)$

   Why: Statement 1 holds: an odd square plus an odd square is always even.

4. 4
   $x^2+y^2\equiv 2(\mathrm{mod}4)$, so $4\nmid x^2+y^2$

   Why: Statement 2 fails, and it fails for every odd pair, not just for some.

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Step-by-step solution

1. 1
   $27^{\frac{x}{x-1}}=3^{\frac{3x}{x-1}}$, so $2^{x+2}\cdot 3^{\frac{3x}{x-1}}=3^2$ for $x\ne 1$

   Why: Writing $27=3^3$ leaves only the bases $2$ and $3$; $x=1$ is excluded because the exponent is undefined there.

   Power of a power, and the move to a common base
2. 2
   $2^{x+2}=3^{2-\frac{3x}{x-1}}=3^{-\frac{x+2}{x-1}}$

   Why: Divide by the power of $3$; the exponent simplifies because $2-\frac{3x}{x-1}=\frac{2x-2-3x}{x-1}=-\frac{x+2}{x-1}$.

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

   Why: Taking logarithms of both sides is the exponential-to-logarithmic reading of the definition of a logarithm, and it brings the unknown down from the exponents.

   Logarithm to exponential form$\log a^n=n\log a$
4. 4
   $(x+2)\left(\log 2+\frac{\log 3}{x-1}\right)=0$

   Why: Collect the common factor $(x+2)$ rather than cancelling it, so that the root $x=-2$ is not thrown away.

   A product is zero only when a factor is zero
5. 5
   $x=-2$, or $x-1=-\frac{\log 3}{\log 2}\Rightarrow x=1-\frac{\log 3}{\log 2}$

   Why: The second bracket gives $\frac{\log 3}{x-1}=-\log 2$, and neither root equals the excluded value $1$.

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Step-by-step solution

1. 1
   Index the friends $i=1$ for $A$ up to $i=12$ for $L$; investment $\propto i$, duration $\propto 13-i$

   Why: The investment ratio counts up $1$ to $12$ while the duration ratio counts down $12$ to $1$, so for each friend the two indices sum to $13$.

2. 2
   ${\text{share}}_i\propto i(13-i)$

   Why: When partners keep their money in for different lengths of time, profit is divided in the ratio of the capital-time products.

   Profit is shared in the ratio of capital $\times$ time
3. 3
   $i(13-i)=12,22,30,36,40,42,42,40,36,30,22,12$ for $i=1,\dots ,12$

   Why: The products are symmetric about $i=6.5$, so they rise to a peak in the middle of the list and fall back.

4. 4
   Maximum $=42$, attained at $i=6$ and at $i=7$

   Why: A product of two numbers with a fixed sum is largest when the factors are as nearly equal as possible, and $6$ and $7$ are equally close to $6.5$.

   A product with a fixed sum is largest when the factors are equal
5. 5
   $i=6$ is $F$ and $i=7$ is $G$, so both receive the maximum profit

   Why: The question asks who gets the maximum, and two friends tie for it.

Common mistakes

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Step-by-step solution

1. 1
   $4^{62}=2^{124},8^{42}=2^{126},4^{64}=2^{128}$

   Why: Using $4=2^2$ and $8=2^3$ puts all five terms on the common base $2$.

   Power of a power, and the move to a common base
2. 2
   $S=2^{122}+2^{124}+2^{126}+2^{128}+2^{130}$

   Why: The exponents now form the run $122,124,126,128,130$, stepping by $2$.

3. 3
   $S=2^{122}\left(1+2^2+2^4+2^6+2^8\right)=2^{122}(1+4+16+64+256)=2^{122}\times 341$

   Why: Factoring out the lowest power leaves a bracket of whole numbers.

4. 4
   $341=11\times 31$

   Why: The factor $2^{122}$ contributes only the prime $2$, so every odd divisor of $S$ must divide $341$; factorising $341$ therefore settles the question.

   Prime factorisation
5. 5
   $11\mid S$

   Why: $11$ is one of the two odd primes available, and it is among the options.

Common mistakes

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Step-by-step solution

1. 1
   $4p^2+4pq-3q^2=4p^2+6pq-2pq-3q^2=2p(2p+3q)-q(2p+3q)=(2p+3q)(2p-q)$

   Why: Splitting the middle term $4pq$ as $6pq-2pq$ exposes the common bracket $2p+3q$, which is exactly the quantity the first equation gives.

   Factorise a quadratic by splitting the middle term
2. 2
   $12(2p-q)=126\Rightarrow 2p-q=\frac{21}{2}$

   Why: Substituting the known value of the first factor turns the quadratic condition into a second linear equation.

3. 3
   $(2p+3q)-(2p-q)=4q=12-\frac{21}{2}=\frac{3}{2}\Rightarrow q=\frac{3}{8}$

   Why: Subtracting the two linear equations eliminates $p$.

4. 4
   $2p=12-3q=12-\frac{9}{8}=\frac{87}{8}\Rightarrow p=\frac{87}{16}$

   Why: Back-substitute into the simpler of the two equations.

5. 5
   $p+2q=\frac{87}{16}+\frac{3}{4}=\frac{87}{16}+\frac{12}{16}=\frac{99}{16}$

   Why: Assemble the quantity the question actually asks for.

Common mistakes

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Step-by-step solution

1. 1
   ${\log }_{10}{3}^{30}=30{\log }_{10}3=30\times 0.4771=14.313$

   Why: The power rule turns the exponent into a multiplier, which is why only ${\log }_{10}3$ is supplied.

   $\log a^n=n\log a$
2. 2
   $14<14.313<15\Rightarrow 10^{14}<3^{30}<10^{15}$

   Why: Raising $10$ to each part of the inequality is the exponential reading of the logarithm, and it brackets $3^{30}$ between two powers of ten.

   Logarithm to exponential form
3. 3
   A number in $[10^{14},10^{15})$ has $15$ digits

   Why: $10^{14}$ is the smallest $15$-digit number, so the digit count is the characteristic $14$ plus one.

   Number of digits from a common logarithm
4. 4
   $n=15$

   Why: This is the quantity the question names $n$.

Common mistakes

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Step-by-step solution

1. 1
   $7^1=7,7^2=49,7^3=343,7^4=2401$: unit digits $7,9,3,1$

   Why: Each unit digit depends only on the previous one, so once $1$ reappears the pattern repeats with period $4$.

   Unit digit of a power from its cycle
2. 2
   $139=4\times 34+3$

   Why: Divide the exponent by the cycle length; only the remainder decides the unit digit.

3. 3
   Remainder $3$ selects the third entry of the cycle

   Why: A remainder $r$ with $1\le r\le 3$ points at the $r$-th unit digit; a remainder of $0$ would point at the fourth.

4. 4
   Unit digit of $7^{139}$ is $3$

   Why: The third entry of $7,9,3,1$ is $3$.

Common mistakes

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Step-by-step solution

1. 1
   $4x+3a=0$ fixes only the ratio $x:a=-3:4$, so $a=4$, $x=-3$ is legal.

   Why: Any multiple of $(-3,4)$ satisfies the constraint, and whole numbers keep the cubes small enough to do in the head.

2. 2
   $a=4$, $x=-3$: $x^2+ax+a^2=9-12+16=13$ and $x^3-a^3=-27-64=-91$, so the first fraction is $\frac{13}{-91}=-\frac{1}{7}$.

   Why: The fraction collapses to a unit fraction, a sign that the numerator is a factor of the denominator.

3. 3
   $x^2-ax+a^2=9+12+16=37$ and $x^3+a^3=-27+64=37$, so the second fraction is $1$; the expression is $-\frac{1}{7}-1=-\frac{8}{7}$.

   Why: Both fractions are now numbers, so the whole expression is a number and can be compared with the options directly.

4. 4
   At $a=4$ the options evaluate to $-\frac{1}{7}$, $\frac{7}{4}$, $-\frac{8}{7}$ and $\frac{6}{7}$: only C matches.

   Why: Three options are eliminated by a single legal substitution.

5. 5
   Repeat with $a=8$, $x=-6$: the fractions are $\frac{52}{-728}=-\frac{1}{14}$ and $\frac{148}{296}=\frac{1}{2}$, total $-\frac{4}{7}$, while option C gives $-\frac{32}{56}=-\frac{4}{7}$.

   Why: One value cannot separate options that differ only in a constant over $a$; a second value confirms the $\frac{1}{a}$ dependence rather than merely a numerical coincidence.

6. 6
   Algebraically: $x^3-a^3=(x-a)\left(x^2+ax+a^2\right)$ and $x^3+a^3=(x+a)\left(x^2-ax+a^2\right)$, so the expression is $\frac{1}{x-a}-\frac{1}{x+a}=\frac{2a}{x^2-a^2}$, and $x^2-a^2=\frac{9a^2}{16}-a^2=-\frac{7a^2}{16}$ gives $-\frac{32}{7a}$.

   Why: The common denominator is a difference of two squares, which is what collapses the whole expression to a single term and confirms the substitutions generalise.

   Difference of two squaresSum and difference of two cubes

Common mistakes

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Step-by-step solution

1. 1
   Incomes $=k$ and $2k$; expenses $=m$ and $3m$, with $k,m>0$

   Why: Each ratio fixes its quantities only up to a common factor, and the two ratios are independent, so they need two different factors.

   Introduce a common factor $k$ for a ratio
2. 2
   Mahesh saves $k-m$; Kamal saves $2k-3m$

   Why: Saving is income minus expenditure for each person separately.

3. 3
   $(2k-3m)-(k-m)=k-2m$

   Why: Who saves more turns entirely on the sign of $k-2m$, and nothing in the question fixes that sign.

4. 4
   $k=1.6m$: savings $0.6m$ and $0.2m$; $k=2m$: savings $m$ and $m$; $k=3m$: savings $2m$ and $3m$

   Why: All three are consistent with incomes in $1:2$, expenses in $1:3$ and both savings non-negative, yet they give three different verdicts.

5. 5
   The comparison cannot be settled

   Why: A question with three mutually exclusive outcomes all reachable has no determinate answer.

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Step-by-step solution

1. 1
   $X$ runs at Rs. $700$ for months $1$ to $3$

   Why: Nothing changes before the first withdrawal at the three-month mark.

2. 2
   Withdrawal $=\frac{2}{7}\times 700=200$, so $X$ runs at Rs. $500$ for months $4$ to $6$

   Why: The fraction is of his stock, i.e. of the Rs. $700$ he put in.

3. 3
   He returns $\frac{3}{5}\times 200=120$, so $X$ runs at Rs. $620$ for months $7$ to $12$

   Why: "After $3$ months" is measured from the withdrawal, putting the restoration at the six-month mark with half the year still to run.

4. 4
   $X$: $700\times 3+500\times 3+620\times 6=2100+1500+3720=7320$; $Y$: $600\times 12=7200$

   Why: With unequal durations the profit is divided in the ratio of capital $\times$ time, so each spell is weighted by its own length.

   Profit is shared in the ratio of capital $\times$ time
5. 5
   $7320:7200=61:60$, and $726=121\times 6$, so $X$ gets $61\times 6=366$

   Why: The ratio has $121$ parts in total and the profit divides into $121$ whole rupees of $6$ each.

   Introduce a common factor $k$ for a ratio

Common mistakes

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Step-by-step solution

1. 1
   Total $=\text{LCM}(2,6)=6$ units; $X$ fills $3$ units/hour and $Y$ fills $1$ unit/hour

   Why: Taking the tank as $6$ units instead of $1$ keeps both rates whole and avoids fractions later.

   Take total work as the LCM of the times
2. 2
   10 a.m. to 11 a.m.: $X$ alone adds $3$ units, leaving $3$

   Why: $Y$ is shut for that hour, so only $X$ contributes.

3. 3
   From 11 a.m. the combined rate is $3+1=4$ units/hour

   Why: Rates add when both pipes feed the same tank.

4. 4
   $\frac{3}{4}$ hour $=45$ minutes to finish

   Why: Time is the remaining work divided by the combined rate.

5. 5
   11 a.m. $+45$ minutes $=$ 11:45 a.m.

   Why: The question asks for a clock time, not a duration.