Elementary Mathematics 2018 (I)

Step-by-step solution

1. 1
   Let the price be $P$ and the consumption $C$, so the budget is $PC$. After the rise the price is $1.25P$.

   Why: A rise of $25\%$ multiplies the price by $1+\frac{25}{100}=1.25$.

   Percentage change as a multiplier
2. 2
   For the budget to be unchanged, the new consumption $C^{\prime }$ must satisfy $1.25P\cdot C^{\prime }=PC$, so $C^{\prime }=\frac{C}{1.25}=0.8C$.

   Why: The two multipliers must be reciprocals, because their product is the unchanged expenditure.

3. 3
   $C^{\prime }=0.8C$ is a fall of $0.2C$, i.e. a reduction of $20\%$.

   Why: The reduction is measured against the original consumption $C$, not against the new one.

   Consumption cut that absorbs a price rise

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Step-by-step solution

1. 1
   Registered $=2000$; those who did not appear $=\frac{1}{25}\times 2000=80$.

   Why: The first fraction is taken on the registered total, which is the only number given.

   $x\%$ of a quantity
2. 2
   Appeared $=2000-80=1920$

   Why: Equivalently $\frac{24}{25}\times 2000$; the survivors of the first stage are the base for the second.

3. 3
   Passed $=\frac{11}{20}\times 1920=11\times 96=1056$

   Why: The second fraction is explicitly 'of those who appeared', so it is taken on $1920$, not on $2000$.

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Step-by-step solution

1. 1
   Let $N=0.\overline{9}$. One digit recurs, so $10N=9.\overline{9}$.

   Why: Multiplying by $10^q$ with $q=1$ shifts the decimal past exactly one period, so the tails of $10N$ and $N$ are identical.

   Convert a recurring decimal to a fraction
2. 2
   $10N-N=9.\overline{9}-0.\overline{9}=9$, so $9N=9$ and $N=1$.

   Why: The identical infinite tails cancel exactly, leaving a whole number; $0.\overline{9}$ is therefore not close to $1$, it is $1$.

3. 3
   $0.\overline{9}-0.9=1-0.9=0.1$

   Why: The second number is the terminating decimal $0.9=\frac{9}{10}$, not a recurring one.

   When a rational number's decimal expansion terminates

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Step-by-step solution

1. 1
   $B:C=3:4$ and $C:D=2:3$. Scale the second by $2$ to make $C$ read $4$ in both: $C:D=4:6$.

   Why: Two ratios can be joined only when the term they share carries the same number in each.

   Introduce a common factor $k$ for a ratioJoin two ratios on their shared term
2. 2
   Hence $B:C:D=3:4:6$.

   Why: With $C$ agreed at $4$, the three quantities sit on one scale.

3. 3
   $D:E=3:4$. Scale by $2$ to make $D$ read $6$: $D:E=6:8$.

   Why: Same move again, now matching $D$.

4. 4
   $B:C:D:E=3:4:6:8$, so $B:E=3:8$.

   Why: The chain is complete and $B:E$ is read directly; $A:B=1:2$ is never needed, since $A$ lies outside the pair asked for.

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Step-by-step solution

1. 1
   $10$ women in $12$ days: the job is $10\times 12=120$ woman-days. $5$ men in $8$ days: the same job is $5\times 8=40$ man-days.

   Why: The chain rule $M_1{D}_1=M_2{D}_2$ measures one fixed job in two different currencies.

   The work-equivalence chain rule $M_1{D}_1{H}_1=M_2{D}_2{H}_2$
2. 2
   Take the whole job as $120$ units. One woman does $\frac{120}{120}=1$ unit per day; one man does $\frac{120}{40}=3$ units per day.

   Why: Choosing the job to be $120$ units makes both rates whole numbers, so a man works at three times a woman's rate.

   Take total work as the LCM of the times
3. 3
   $6$ women and $3$ men do $6\times 1+3\times 3=6+9=15$ units per day.

   Why: Rates add when people work together on the same job.

4. 4
   Time $=\frac{120}{15}=8$ days.

   Why: Total work divided by the combined rate.

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Step-by-step solution

1. 1
   In $50$ days, $200$ men completed $\frac{1}{4}$ of the work, i.e. $200\times 50=10000$ man-days bought a quarter of the job.

   Why: Man-days done is the only measure of progress available, and it fixes the size of the whole job.

   The work-equivalence chain rule $M_1{D}_1{H}_1=M_2{D}_2{H}_2$
2. 2
   The whole job is $4\times 10000=40000$ man-days, so the remaining $\frac{3}{4}$ is $30000$ man-days.

   Why: Scaling a quarter up to the whole, then taking what is left.

3. 3
   Days remaining $=150-50=100$. Men needed $=\frac{30000}{100}=300$.

   Why: The deadline is the original $150$ days, so only $100$ days are left — this is the step the question is really testing.

4. 4
   Additional men $=300-200=100$.

   Why: The question asks how many more to employ, not how many will be on site.

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Step-by-step solution

1. 1
   A pole has no length, so the distance covered in the $30$ seconds is precisely the length of the train.

   Why: This is what makes a pole the cheapest object to cross: nothing is added to the train's own length.

   What a train covers when it crosses something
2. 2
   $60$ km/h $=60\times \frac{5}{18}$ m/s $=\frac{50}{3}$ m/s

   Why: One km/h is $\frac{1000}{3600}=\frac{5}{18}$ m/s; the answer is wanted in metres, so the speed must be in metres per second.

   km/hour to m/second
3. 3
   Length $=$ speed $\times$ time $=\frac{50}{3}\times 30=500$ metres

   Why: Distance is speed times time, with both in the same system of units.

   Speed, distance and time

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Step-by-step solution

1. 1
   Let A's share be Rs $x$. Then B $=$ Rs $(x-20)$ and C $=$ Rs $(x+20)$.

   Why: A is Rs $20$ more than B, so B is Rs $20$ less than A; A is Rs $20$ less than C, so C is Rs $20$ more than A.

   Turn a worded condition into a linear equation in one unknown
2. 2
   $(x-20)+x+(x+20)=120$

   Why: The three shares exhaust the Rs $120$.

3. 3
   $3x=120\Rightarrow x=40$

   Why: The two $20$s cancel, leaving A's share directly.

4. 4
   B $=$ Rs $(40-20)=$ Rs $20$

   Why: The question asks for B's share, not A's.

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Step-by-step solution

1. 1
   Inverse variation gives $N=\frac{k}{M}$, i.e. $MN=k$ for a constant $k$.

   Why: This is the inverse case of the variation law, and the constant is what the table is built on.

   Direct variation as a power: $y\propto x^n$
2. 2
   From the one complete column, $M=15$ and $N=-4$: $k=15\times (-4)=-60$.

   Why: Only that column has both entries, so it is the only place $k$ can be read from.

3. 3
   $M=-6\Rightarrow A=N=\frac{-60}{-6}=10$

   Why: Two negatives give a positive, so $A$ is positive even though $M$ is negative.

4. 4
   $M=2\Rightarrow B=N=\frac{-60}{2}=-30$

   Why: Same constant, next column.

5. 5
   $N=60\Rightarrow C=M=\frac{-60}{60}=-1$

   Why: Here the unknown is $M$, not $N$, so divide $k$ by the given $N$ instead.

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Step-by-step solution

1. 1
   Interest paid over $2$ years at $5\%$ on Rs $5000$: $5000\times \frac{5}{100}\times 2=$ Rs $500$.

   Why: Simple interest is the stated percentage of the principal, taken once for each year.

   $x\%$ of a quantitySimple interest $=\frac{PRT}{100}$
2. 2
   Interest received over $2$ years at $5.5\%$ on Rs $5000$: $5000\times \frac{5.5}{100}\times 2=$ Rs $550$.

   Why: The principal is the same Rs $5000$, lent immediately, and the period is the same $2$ years.

3. 3
   Gain $=550-500=$ Rs $50$

   Why: The loan is settled with the borrowed sum itself, so the whole gain is the difference in interest.

4. 4
   Equivalently, gain $=5000\times \frac{5.5-5}{100}\times 2=5000\times \frac{0.5}{100}\times 2=$ Rs $50$

   Why: Only the $0.5$ percentage-point gap matters, which is the fast way to see the answer.

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Step-by-step solution

1. 1
   $F+S=2\times 25=50$

   Why: An average of $25$ years over two people means the two ages total $50$.

   Mean of a frequency distribution
2. 2
   $S+7=17\Rightarrow S=10$

   Why: The son is $17$ seven years from now, so his present age is seven less.

   Turn a worded condition into a linear equation in one unknown
3. 3
   $F=50-10=40$

   Why: The father's present age is whatever the total leaves after the son's.

4. 4
   $F+10=40+10=50$

   Why: Ten years from now every age is ten years larger; the father's is what is asked.

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Step-by-step solution

1. 1
   $5$ tractors for $5$ days is $25$ tractor-days, and that ploughs $5$ hectares

   Why: Work done is proportional to the number of machines times the number of days.

   The work-equivalence chain rule $M_1{D}_1{H}_1=M_2{D}_2{H}_2$
2. 2
   One hectare needs $\frac{25}{5}=5$ tractor-days

   Why: Divide by the area to reach one unit of work.

   Unitary method: find one unit, then scale
3. 3
   $100$ hectares need $100\times 5=500$ tractor-days

   Why: Multiply the unit rate back up to the area wanted.

4. 4
   $\frac{500}{50}=10$ tractors

   Why: The $500$ tractor-days are to be spread over $50$ days.

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Step-by-step solution

1. 1
   $\text{factor}=1+\frac{25}{100}=\frac{5}{4}$

   Why: A gain of $25\%$ on the running capital is a multiplication by $\frac{5}{4}$.

   Percentage change as a multiplier
2. 2
   $P\times {\left(\frac{5}{4}\right)}^3=10000$

   Why: Three years of the same gain compound, so the factor is applied three times.

3. 3
   $P=10000\times \frac{64}{125}=80\times 64=5120$

   Why: Reverse a multiplication by dividing by the factor, that is by multiplying by its reciprocal.

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Step-by-step solution

1. 1
   $999=3^3\times 37$, so $\frac{1}{37}=\frac{27}{999}$ has period $3$

   Why: The period of a fraction is the smallest number of nines its denominator divides, and $37$ first divides $999$.

   Prime factorisation
2. 2
   $0.459459459\dots =0.\overline{459}=\frac{459}{999}$

   Why: A purely recurring block of three digits sits over three nines.

   Convert a recurring decimal to a fraction
3. 3
   $459=3^3\times 17$, so $\frac{459}{999}=\frac{17}{37}$

   Why: Cancelling the common $27$ leaves a denominator of exactly $37$.

4. 4
   $0.459459459=\frac{459459459}{10^9}$, whose reduced denominator has only the primes $2$ and $5$

   Why: A terminating decimal can never carry a $37$ in its reduced denominator.

   When a rational number's decimal expansion terminates
5. 5
   $0.0\overline{459}=\frac{459}{9990}=\frac{17}{370}$ and $0.00\overline{459}=\frac{459}{99900}=\frac{17}{3700}$

   Why: Each zero between the point and the recurring block multiplies the denominator by $10$.

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Step-by-step solution

1. 1
   $\text{income at }4\%=\frac{4P}{100}$ and $\text{income at }3.75\%=\frac{3.75P}{100}$, where $P$ is the sum invested

   Why: A year's interest is that percentage of the sum earning it.

   $x\%$ of a quantity
2. 2
   $\frac{4P}{100}-\frac{3.75P}{100}=64$

   Why: The drop in income is the gap between two percentages of the same sum.

   Turn a worded condition into a linear equation in one unknown
3. 3
   $\frac{0.25P}{100}=64\Rightarrow P=64\times 400=25600$

   Why: A quarter of one percent is $\frac{1}{400}$ of the sum, so multiply by $400$.

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Step-by-step solution

1. 1
   ${\log }_{10}m<0$, since $m<1=10^0$

   Why: A logarithm is negative exactly when its argument is below the base raised to zero.

   Logarithm to exponential form
2. 2
   $0<m^2<m$

   Why: Multiplying a positive number by $m<1$ makes it smaller, so squaring shrinks a proper fraction.

3. 3
   $m<1<\frac{1}{m}=m^{-1}$

   Why: The reciprocal of a number between $0$ and $1$ is greater than $1$.

   Negative exponent is a reciprocal
4. 4
   ${\log }_{10}m<m^2<m<m^{-1}$

   Why: Chaining the three comparisons gives the complete order.

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Step-by-step solution

1. 1
   $\text{widow}=w$, $\text{each daughter}=2w$, $\text{each son}=3\times 2w=6w$

   Why: Each daughter gets twice her mother's share and each son three times a daughter's.

   Introduce a common factor $k$ for a ratio
2. 2
   $w+5(6w)+4(2w)=w+30w+8w=39w$

   Why: Five sons and four daughters take their shares alongside the widow.

   Turn a worded condition into a linear equation in one unknown
3. 3
   $39w=39000\Rightarrow w=1000$

   Why: The estate is fixed, so the unit share follows by division.

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Step-by-step solution

1. 1
   Let the numbers be $a$, $b$, $c$ with $ab=286$ and $bc=770$

   Why: 'The first two' and 'the last two' overlap in the middle number.

   Turn a worded condition into a linear equation in one unknown
2. 2
   $286=2\times 11\times 13$ and $770=2\times 5\times 7\times 11$

   Why: Factorising both products shows what the middle number can be.

   Prime factorisation
3. 3
   $b$ divides both products, so $b\mid \gcd (286,770)=2\times 11=22$, leaving $b\in \{1,2,11,22\}$

   Why: Only a common divisor of the two products can sit in both of them.

4. 4
   $b=2$ gives $a=143=11\times 13$ and $c=385=5\times 7\times 11$, which share $11$; $b=11$ gives $a=26$ and $c=70$, which share $2$; $b=1$ gives $a=286$ and $c=770$, which share $22$

   Why: Each of these fails the requirement that the three be coprime to each other.

   Coprime means no shared prime factor
5. 5
   $b=22$ gives $a=13$ and $c=35$, and $13$, $22$, $35$ have no shared prime; $13+22+35=70$

   Why: This is the only split with disjoint prime sets, so the sum is determined.

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Step-by-step solution

1. 1
   $\text{woman}=10a+b$ and $\text{husband}=10b+a$, with $b>a$ since the husband is elder

   Why: Reversing a two-digit numeral interchanges its digits.

   Place-value expansion of a numeral
2. 2
   $\text{sum}=11(a+b)$ and $\text{difference}=9(b-a)$

   Why: These are the only two combinations of the digits the condition needs.

3. 3
   $9(b-a)=\frac{1}{11}\times 11(a+b)=a+b$

   Why: The difference is one-eleventh of the sum, and the $11$ cancels.

   Turn a worded condition into a linear equation in one unknown
4. 4
   $9b-9a=a+b\Rightarrow 8b=10a\Rightarrow 4b=5a$

   Why: Collecting the digits leaves a single proportion between them.

5. 5
   $a=4$, $b=5$ is the only digit solution, so the difference is $9(5-4)=9$

   Why: $a=8$ would force $b=10$, which is not a digit.

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Step-by-step solution

1. 1
   $\text{distance covered in either crossing}=L_p+L_g=L$

   Why: One train passes another by covering both their lengths, in the same direction or opposite.

   What a train covers when it crosses something
2. 2
   $t_{\text{same}}=\frac{L}{p-g}$ and $t_{\text{opp}}=\frac{L}{p+g}$

   Why: Speeds subtract when the trains run alike and add when they run opposed.

   Relative speed: add when opposed, subtract when alikeSpeed, distance and time
3. 3
   $\frac{L}{p-g}=3\times \frac{L}{p+g}\Rightarrow p+g=3(p-g)$

   Why: The same-direction crossing takes three times as long, and the common length $L$ cancels.

   Clear the denominators of an equation
4. 4
   $p+g=3p-3g\Rightarrow 4g=2p\Rightarrow \frac{p}{g}=\frac{2}{1}$

   Why: Collecting terms gives the ratio of the speeds directly.

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