CAT 2022 Quant Questions — Slot 1

Think of this as: we're given the perimeter (which tells us the sum of the sides) and the circle's radius (which tells us the sum of the squares of the sides). We need to find the area, which is the product of the sides. The bridge between these is a basic algebra identity.

Step 1 — Relate the sides to the circle. Let the sides of the rectangle be $a$ and $b$. Since all four vertices lie on the circle, the diagonal of the rectangle is the diameter of the circle.

💡 Teacher tip: Whenever a rectangle is inscribed in a circle, its diagonal is exactly the diameter of the circle. This immediately gives you $a^2+b^2 = (2R)^2$, a powerful relation connecting the sides directly to the radius.

So, the diagonal $= \sqrt{a^2 + b^2} = 2R$, which means: $a^2 + b^2 = 4R^2$

Step 2 — Relate the sides to the perimeter. The perimeter of the rectangle is $P = 2(a+b)$. Dividing by 2 gives: $a + b = \frac{P}{2}$

Step 3 — Connect the two to find the area. The area of the rectangle is $ab$. To introduce $ab$ into our equations, we use the identity: $(a+b)^2 = a^2 + b^2 + 2ab$

Substitute the results from Step 1 and Step 2 into this identity: $\left(\frac{P}{2}\right)^2 = 4R^2 + 2ab$ $\frac{P^2}{4} = 4R^2 + 2ab$

Now, isolate $2ab$: $2ab = \frac{P^2}{4} - 4R^2$

Divide by 2 to find the area $ab$: $ab = \frac{P^2}{8} - 2R^2$

Answer: $\dfrac{P^{2}}{8}-2R^{2}$

💡 Why this works: The identity $(a+b)^2 = a^2 + b^2 + 2ab$ perfectly bridges the gap between what we're given (the sum $a+b$ from the perimeter, and the sum of squares $a^2+b^2$ from the radius) and what we want (the product $ab$, which is the area).

Framing the problem: Three nuts whose prices are linked by one shared cost, a mixture with a marked price, and two different profit scenarios. Break it into three layers — (cost of nuts) → (marked price per kg) → (revenue from selling) — and each layer untangles cleanly.

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Step 1 — Link the three nut prices through one common value.

The problem states that 7 kg of cashews costs the same as 30 kg of peanuts or 9 kg of almonds. Let this common cost be $k$:

$7c = 30p = 9a = k$

where $c$, $p$, $a$ are the per-kg costs of cashews, peanuts, and almonds respectively. Solving for each:

$c = \frac{k}{7}, \quad p = \frac{k}{30}, \quad a = \frac{k}{9}$

💡 Teacher tip: When a problem says "the cost of $X$ of A equals the cost of $Y$ of B equals the cost of $Z$ of C," introduce a single common value $k$ for that shared cost. It collapses three unknowns into one.

Step 2 — Express total cost in terms of $k$.

Ankita buys 4 kg cashews, 14 kg peanuts, and 6 kg almonds. Her total cost is:

$\text{Cost} = 4 \cdot \frac{k}{7} + 14 \cdot \frac{k}{30} + 6 \cdot \frac{k}{9}$

Simplify each fraction: $\frac{4k}{7} + \frac{7k}{15} + \frac{2k}{3}$. The common denominator is $105$:

$\text{Cost} = \frac{60k + 49k + 70k}{105} = \frac{179k}{105}$

Step 3 — Set up two equations from the two profit statements.

Let $M$ be the marked price per kg. The total mixture weighs $4 + 14 + 6 = 24$ kg.

• Planned scenario: If all 24 kg were sold at $M$/kg, the profit would be ₹1752: $24M = \text{Cost} + 1752 \quad \cdots (1)$

• Actual scenario: 4 kg sold at $M$/kg; the remaining 20 kg sold at a 20% discount, i.e., at $0.8M$/kg. $\text{Revenue} = 4M + 20 \times 0.8M = 4M + 16M = 20M$ Profit is ₹744: $20M = \text{Cost} + 744 \quad \cdots (2)$

Step 4 — Eliminate $M$ and solve for Cost.

From (1) and (2), both equal $M$:

$\frac{\text{Cost} + 1752}{24} = \frac{\text{Cost} + 744}{20}$

Let $C = \text{Cost}$ and cross-multiply:

$20(C + 1752) = 24(C + 744)$ $20C + 35040 = 24C + 17856$ $4C = 17184$ $C = 4296$

So Ankita's total cost is ₹4296.

Step 5 — Back-solve for $k$, then find the almond spend.

$4296 = \frac{179k}{105} \implies k = 4296 \times \frac{105}{179}$

Notice that $179 \times 24 = 4296$, so the 179 cancels perfectly:

$k = 24 \times 105 = 2520$

Now, the amount spent on almonds is $6 \times a$:

$6 \times \frac{k}{9} = 6 \times \frac{2520}{9} = 6 \times 280 = 1680$

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Verification table — confirming every number is consistent:

Check on marked price: $M = \frac{4296 + 1752}{24} = \frac{6048}{24} = 252$/kg. Actual revenue $= 20 \times 252 = 5040$. Profit $= 5040 - 4296 = 744$ ✓.

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Alternative approach — use the profit difference directly.

The only difference between the two scenarios is that 20 kg which would have sold at $M$ now sells at $0.8M$ — a drop of $0.2M$ per kg on those 20 kg. That drop equals the profit difference:

$20 \times 0.2M = 1752 - 744 = 1008 \implies 4M = 1008 \implies M = 252$

Then from the actual scenario: $20M = \text{Cost} + 744 \implies \text{Cost} = 5040 - 744 = 4296$.

This reaches the same Cost in fewer steps; Step 5 (finding $k$ and the almond spend) is identical.

💡 Teacher tip: The two profit statements are really just two linear equations in two unknowns (Cost and $M$). Solve that system first — the individual nut prices only matter at the very end.

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Answer: $\boxed{1680}$

The key structural insight: the problem layers a pricing constraint (one common $k$), a cost expression, and a selling scenario (two equations in Cost and $M$). Solving the selling scenario first gives you the total cost, which then feeds back into the pricing constraint to isolate the almond spend — no need to find individual nut prices until the final step.

Think of this as a two-stage dilution: we start with a 50–50 lemon–syrup blend, then pour in a large amount of pure syrup. The lemon amount never changes — only the syrup grows.

Step 1 — Understand the original mixture.

The original mixture has lemon juice and sugar syrup in equal proportion:

$\text{Lemon}: \text{Syrup} = 1: 1$

So in any quantity of this mixture, half is lemon juice and half is sugar syrup.

Step 2 — Pin down a convenient amount.

The new mixture is made by combining the old mixture and pure sugar syrup in the ratio $1: 3$. That means for every 1 unit of old mixture, we add 3 units of pure sugar syrup.

Take 1 unit of the old mixture and split it into its two components:

Step 3 — Add the pure sugar syrup.

We now add 3 units of pure sugar syrup. Since it is pure syrup, it contributes syrup only — no lemon juice.

$\text{Lemon juice} = \frac{1}{2} \quad \text{(unchanged — no lemon was added)}$

$\text{Sugar syrup} = \frac{1}{2} + 3 = \frac{1}{2} + \frac{6}{2} = \frac{7}{2}$

💡 Teacher tip: The key insight is that lemon juice appears only in the old mixture. Once you fix the old mixture at 1 unit, the lemon is locked at $\frac{1}{2}$ and never grows. Only the syrup side changes.

Step 4 — Write the final ratio.

$\text{Lemon}: \text{Syrup} = \frac{1}{2}: \frac{7}{2}$

Multiply both parts by 2 to clear fractions:

$\text{Lemon}: \text{Syrup} = 1: 7$

Answer: 1: 7 (Option D)

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Alternative approach — Weighted average (alligation)

We can track the syrup concentration instead of absolute amounts.

• Old mixture: syrup is $\frac{1}{2} = 50\%$ of the mixture.
• Pure sugar syrup: syrup is $100\%$.
• They are mixed in the ratio $1: 3$ (old: pure).

The syrup fraction in the new mixture is the weighted average:

$\text{Syrup fraction} = \frac{1 \times 50\% + 3 \times 100\%}{1 + 3} = \frac{50 + 300}{4} = \frac{350}{4} = 87.5\% = \frac{7}{8}$

So syrup $= \frac{7}{8}$ of the new mixture, which means lemon $= 1 - \frac{7}{8} = \frac{1}{8}$.

$\text{Lemon}: \text{Syrup} = \frac{1}{8}: \frac{7}{8} = 1: 7$

Both methods agree. The part-tracking method is faster under exam time; the alligation method is a good cross-check.

💡 Why this works so cleanly: the ratio $1:3$ and the original $1:1$ split conspire to give a total syrup of $\frac{7}{2}$ against a lemon of $\frac{1}{2}$ — a factor of 7. Whenever a pure substance is added to a mixture, fix the mixture's size first, then let the pure addition do the work.

Think of this as: two chunks of money earn the same interest, but under different rates and time periods. We need to find what fraction of the total went into the first chunk.

Step 1 — Set up the simple interest for each part.

Let the amount invested in the first part be $P_1$ and in the second part be $P_2$.

Recall the simple interest formula: $\text{SI} = \frac{P \times R \times T}{100}$

For the first part ($R = 15\%$, $T = 4$ years): $\text{SI}_1 = \frac{P_1 \times 15 \times 4}{100} = \frac{60\,P_1}{100} = 0.6\,P_1$

For the second part ($R = 12\%$, $T = 3$ years): $\text{SI}_2 = \frac{P_2 \times 12 \times 3}{100} = \frac{36\,P_2}{100} = 0.36\,P_2$

Step 2 — Equate the two interests.

We are told the interests are equal: $0.6\,P_1 = 0.36\,P_2$

Solving for the ratio: $\frac{P_1}{P_2} = \frac{0.36}{0.6} = \frac{36}{60} = \frac{3}{5}$

So $P_1: P_2 = 3: 5$.

💡 Teacher tip: When two simple interests are equal, the principals must be in the inverse ratio of their $(R \times T)$ values. Here $R \times T$ for the two parts are $15 \times 4 = 60$ and $12 \times 3 = 36$, i.e. $60: 36 = 5: 3$. Inverting gives the principal ratio $3: 5$ — exactly what we got.

Step 3 — Find the percentage in the first part.

The total savings is $P_1 + P_2$. Using the ratio $3: 5$:

$\frac{P_1}{P_1 + P_2} = \frac{3}{3 + 5} = \frac{3}{8}$

Converting to a percentage: $\frac{3}{8} \times 100 = 37.5\%$

Quick check: The first part earns interest at a higher rate and for longer, so it needs less principal to match the second part's interest. A percentage below 50% makes sense. ✓

Answer: 37.5% → Option A

💡 Why this works so cleanly: "Equal SI" collapses directly into a ratio of principals once you fold in $R \times T$ for each part. No need to know the actual amounts — the ratio does all the work.

Think of this as: two of the three absolute values have fixed signs thanks to the ordering $0 \le a \le x \le 100$, so they collapse into a constant — leaving just one absolute value to maximise.

Step 1 — Simplify using the given order.

Since $a \le x$ and $x \le 100$, we immediately know the signs inside two of the absolute values:

• $|x - a| = x - a$ (because $x \ge a$)
• $|x - 100| = 100 - x$ (because $x \le 100$)

The third term $|x - a - 50|$ cannot be resolved yet because $x - a - 50$ could be positive, negative, or zero depending on $x$.

Substituting:

$f(x) = (x - a) + (100 - x) + |x - a - 50| = 100 - a + |x - a - 50|.$

Notice how neatly the $x$-terms cancelled: $(x - a) + (100 - x) = 100 - a$. So $f(x)$ is a constant plus one absolute value.

Step 2 — Maximise the surviving absolute value.

Let $u = x - a$. Since $a \le x \le 100$, we have:

$0 \le u \le 100 - a.$

So we need to maximise $|u - 50|$ on the interval $[0,\; 100 - a]$.

💡 Teacher tip: The maximum of $|u - c|$ on a closed interval always occurs at one of the endpoints — the expression is farthest from $c$ at the boundary, not in the interior.

Evaluating at both endpoints:

Now, since $0 \le a \le 100$:

• If $a \le 50$: $|50 - a| = 50 - a \le 50$.
• If $a > 50$: $|50 - a| = a - 50 \le 50$ (since $a \le 100$).

In both cases, $|50 - a| \le 50$. Therefore:

$\max\,|u - 50| = 50.$

Step 3 — Compute the maximum of $f(x)$ and solve for $a$.

$\max f(x) = (100 - a) + 50 = 150 - a.$

We want this maximum to equal $100$:

$150 - a = 100 \implies a = 50.$

Step 4 — Verify.

When $a = 50$:

• $f(x) = 100 - 50 + |x - 100| = 50 + (100 - x) = 150 - x$ (since $x \le 100$).
• $x$ ranges from $50$ to $100$, so $f(x)$ ranges from $150 - 100 = 50$ up to $150 - 50 = 100$.
• Maximum $= 100$ ✓

Answer: C. 50 $\quad\boxed{50}$

💡 Why $a = 50$ works: The third absolute value $|x - a - 50|$ can swing at most $50$ away from its centre. The constant part $100 - a$ must therefore drop by exactly $50$ (from $100$ to $50$) to bring the overall ceiling down to $100$ — and that happens precisely when $a = 50$.

Think of this as: one big word problem that ultimately reduces to one linear equation in one unknown. Track total cost and total revenue separately, then use the overall profit percentage to link them.

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Step 1 — Set up the unit costs with a single variable.

Let the cost price of juice be $j$ rupees per kg.

Syrup is 20% less costly than juice, so syrup costs 80% of what juice costs:

$\text{Syrup cost per kg} = 0.8j.$

💡 Teacher tip: Resist the urge to introduce separate variables for syrup and juice. Since their costs are linked by a ratio, one variable captures both. Everything in this problem is linear in $j$.

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Step 2 — Compute the total cost price.

Amal buys 110 kg of syrup and 120 kg of juice:

$\text{Total Cost} = 110 \times 0.8j + 120 \times j = 88j + 120j = 208j.$

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Step 3 — Compute the total revenue (three separate sales).

Amal sells in three batches:

1. 10 kg of syrup at 10% profit: $\text{Revenue}_1 = 10 \times 0.8j \times 1.10 = 8.8j.$

2. 20 kg of juice at 20% profit: $\text{Revenue}_2 = 20 \times j \times 1.20 = 24j.$

3. Remaining mixture sold at ₹308.32 per kg:

• Syrup remaining: $110 - 10 = 100$ kg.
• Juice remaining: $120 - 20 = 100$ kg.
• Total mixture: $100 + 100 = 200$ kg.

$\text{Revenue}_3 = 200 \times 308.32 = 61{,}664.$

Adding all three:

$\text{Total Revenue} = 8.8j + 24j + 61{,}664 = 32.8j + 61{,}664.$

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Step 4 — Apply the overall profit condition.

Overall profit is 64%, meaning total revenue equals 1.64 times the total cost:

$\text{Total Revenue} = 1.64 \times \text{Total Cost}.$

Substituting:

$32.8j + 61{,}664 = 1.64 \times 208j.$

Compute the right side:

$1.64 \times 208 = 341.12 \quad\Rightarrow\quad 341.12j.$

So the equation becomes:

$32.8j + 61{,}664 = 341.12j.$

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Step 5 — Solve for $j$, then find the syrup cost.

$61{,}664 = 341.12j - 32.8j = 308.32j.$

$j = \frac{61{,}664}{308.32} = 200.$

So juice costs ₹200 per kg. The syrup cost is:

$0.8 \times 200 = 160.$

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Sanity check.

Everything checks out.

Answer: $\boxed{160}$

💡 Closing insight: Notice that the coefficient on the left ($32.8$) and the mixture price ($308.32$) are not random — the difference $341.12 - 32.8 = 308.32$ is exactly the per-kg mixture price, which is what makes the equation collapse cleanly. When a problem gives you a "messy" decimal like 308.32, it's usually engineered to cancel beautifully once you set up the right equation.

Two trains start simultaneously towards each other. We know train A's total journey time (10 min) and train B's time after the meeting (9 min). We need train B's total journey time from Y to X.

Step 1 — Set up variables.

Let $v_A$ and $v_B$ be the speeds of trains A and B. Let $t$ be the time (in minutes) from start until they meet. Let $D$ be the distance between stations X and Y.

Step 2 — Read each given time carefully.

This is the crucial step where most errors happen:

• "Train A reaches station Y in 10 minutes" → A's total time = 10 min, so $D = 10\,v_A$.
• "Train B takes 9 minutes to reach station X after meeting train A" → B's post-meeting time = 9 min.

💡 Teacher tip: Before writing any equation, label each given time as either "total" or "after meeting." This problem deliberately mixes the two — misreading is the most common error.

Step 3 — Write the meeting-time equation.

Both trains travel towards each other for $t$ minutes until they meet, so together they cover the full distance $D$: $D = (v_A + v_B)\,t$

Since $D = 10\,v_A$: $10\,v_A = (v_A + v_B)\,t \quad \cdots (1)$

Step 4 — Use B's post-meeting time.

At the meeting point, the distance B still needs to cover to reach X is exactly the distance A has already covered, which is $v_A\,t$. B covers this remaining distance at speed $v_B$ in 9 minutes: $\frac{v_A\,t}{v_B} = 9 \quad \Longrightarrow \quad \frac{v_A}{v_B} = \frac{9}{t} \quad \cdots (2)$

Step 5 — Solve for $t$.

From (2): $\dfrac{v_B}{v_A} = \dfrac{t}{9}$. Substitute into (1): $10 = t\left(1 + \frac{t}{9}\right) = t + \frac{t^2}{9}$

Multiply through by 9: $90 = 9t + t^2 \quad \Longrightarrow \quad t^2 + 9t - 90 = 0$

Factor: $(t + 15)(t - 6) = 0$, so $t = 6$ (taking the positive root, since time must be positive).

Step 6 — Find B's total time.

B travels for $t = 6$ minutes until the meeting, then 9 minutes more to reach X: $T_B = 6 + 9 = 15 \text{ minutes}$

Quick check: If $t = 6$, then A's post-meeting time is $10 - 6 = 4$ min. We can verify: $t^2 = 36$ and the product of post-meeting times $= 4 \times 9 = 36$. ✓ Consistent.

Answer: $\boxed{15}$

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Alternative approach — The meeting-time property.

There is an elegant result for two objects starting simultaneously towards each other: if they meet after time $t$, and after meeting take $t_1$ and $t_2$ to reach the other's starting point, then: $t^2 = t_1 \cdot t_2$

Why it works: At the meeting, A has covered $v_A t$ and B has covered $v_B t$. After meeting, A must cover B's distance ($v_B t$) at speed $v_A$, taking $t_1 = \frac{v_B t}{v_A}$. Similarly, $t_2 = \frac{v_A t}{v_B}$. Multiplying: $t_1 \cdot t_2 = \frac{v_B t}{v_A} \cdot \frac{v_A t}{v_B} = t^2$.

Here:

• $t_1$ = A's post-meeting time = $10 - t$
• $t_2$ = B's post-meeting time = $9$

So: $t^2 = (10 - t) \cdot 9 \Rightarrow t^2 + 9t - 90 = 0 \Rightarrow t = 6$.

Then $T_B = t + 9 = 6 + 9 = 15$ min. Same answer, fewer variables.

💡 Why this works so cleanly: the product $t_1 \cdot t_2$ always equals $t^2$ regardless of the speeds — the speed ratio cancels out. That is why knowing one "total" time and one "after meeting" time is enough to pin everything down.

Think of this as: the new students are heavier than the old average, so they pull the class average up. The amount of pull depends on how many new students there are relative to the old ones. Set up the total-weight equation and the ratio falls out.

Step 1 — Name the variables.

Let the original class have $n$ students with average weight $A$ kg.

Let $m$ new students join. Their average weight is $A + 3$ kg (given: 3 kg more than the original average).

After they join, the combined average becomes $A + 0.6$ kg (given: the average rises by 600 gm $= 0.6$ kg).

Step 2 — Write the total-weight equation.

Total weight is conserved — the sum of the old total and the new students' total equals the new combined total:

So the equation is:

$nA + m(A+3) = (n+m)(A+0.6)$

Step 3 — Expand and cancel.

Left side: $nA + mA + 3m$

Right side: $nA + 0.6n + mA + 0.6m$

Setting them equal:

$nA + mA + 3m = nA + 0.6n + mA + 0.6m$

The $nA$ and $mA$ terms appear on both sides, so they cancel:

$3m = 0.6n + 0.6m$

Step 4 — Solve for the ratio.

$3m - 0.6m = 0.6n$ $2.4m = 0.6n$ $\frac{n}{m} = \frac{2.4}{0.6} = 4$

So the ratio of original students to new students is $n: m = 4: 1$.

💡 Teacher tip: Notice that the actual value of $A$ never mattered — it cancelled out. This is the hallmark of a weighted average problem: only the gaps between the averages and the group sizes matter, not the averages themselves.

Step 5 — Verify with concrete numbers.

Take $n = 4$, $m = 1$, and let $A = 50$ kg.

• Original total weight $= 4 \times 50 = 200$ kg.
• New student's weight $= 53$ kg.
• Combined total $= 200 + 53 = 253$ kg.
• Combined average $= \frac{253}{5} = 50.6$ kg.
• Increase $= 0.6$ kg $= 600$ gm. ✅

The numbers check out.

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Alternative approach — Alligation (10-second method)

The combined average $A + 0.6$ sits between the original average $A$ and the new students' average $A + 3$. By the rule of alligation, the ratio of the two group sizes is the inverse of the distances from each group's average to the combined average:

$n: m = 2.4: 0.6 = 4: 1$

Same answer, far faster once you spot the alligation structure.

Answer: 4: 1 (Option C)

💡 Why this works so cleanly: The 0.6 kg rise is the new students' 3 kg excess, diluted by the full class. The dilution factor is $\frac{m}{n+m}$, so $0.6 = 3 \cdot \frac{m}{n+m}$, giving $\frac{m}{n+m} = \frac{1}{5}$ — i.e., new students are $\frac{1}{5}$ of the total, so $n:m = 4:1$. Whenever a weighted-average shift is given, think "excess × share = shift."

What we want: the value of $N$ for which $\displaystyle\sum_{n=1}^{N}\left\lfloor \tfrac15 + \tfrac{n}{25}\right\rfloor = 25$, where $\lfloor\ \rfloor$ is the greatest-integer (floor) function.

Framing: a floor of a straight-line expression in $n$ stays constant over stretches of $n$, then jumps by 1. So instead of adding 25 terms one by one, find where the value jumps and add block by block.

Step 1 — Tidy the floor argument. $\left\lfloor \frac15 + \frac{n}{25}\right\rfloor = \left\lfloor \frac{5 + n}{25}\right\rfloor = \left\lfloor \frac{n+5}{25}\right\rfloor.$

Step 2 — Find the constant blocks. $\left\lfloor \tfrac{n+5}{25}\right\rfloor = t$ exactly when $25t \le n + 5 < 25(t+1), \quad\text{i.e.}\quad 25t - 5 \le n < 25t + 20.$

Step 3 — Accumulate to 25. The first $19$ terms each contribute $0$, so the running sum is still $0$ at $n = 19$. The next block contributes $1$ per term. Exactly $25$ ones bring the sum to $25$, and the $25$th of them sits at the end of the floor-$1$ block, $n = 44$. $N = 44.$

Check. $\displaystyle\sum_{n=1}^{44} = \underbrace{0 + \cdots + 0}_{19} + \underbrace{1 + \cdots + 1}_{25} = 25$ ✓. At $N = 45$ the next term is $2$, so the sum would overshoot to $27$.

Answer: $\boxed{44}$

💡 Teacher tip: For any $\sum \lfloor \text{linear} \rfloor$, never add term by term. Locate the blocks where the floor is constant — the answer almost always lands on a clean block boundary.

Think of this as: we're given a ratio of people ahead to behind Pinky, a cap on the total queue size, and we need to squeeze out the maximum possible count ahead of her. The key is to let a single multiplier absorb the ratio, then push it as high as the cap allows.

Step 1 — Translate the ratio into a variable model.

The ratio of people ahead of Pinky to people behind her is $3: 5$. Let a common multiplier be $k$, so:

• People ahead of Pinky $= 3k$
• People behind Pinky $= 5k$

Here's the subtle but crucial part: Pinky herself is in the queue too. She is neither ahead of nor behind herself — she is the reference point. So the total count must include her separately.

$\text{Total} = 3k + 1 + 5k = 8k + 1$

💡 Teacher tip: "Ahead: behind" ratios in a queue always come with a $+1$ for the person themself. Forgetting that $+1$ is the classic off-by-one trap on these questions.

Step 2 — Apply the total-size cap.

We're told the total number of persons is less than 300:

$8k + 1 < 300$ $8k < 299$ $k < \frac{299}{8} = 37.375$

Since $k$ must be a whole number (you can't have a fraction of a person), the largest integer $k$ can take is:

$k = 37$

Let's verify the boundary values to be sure:

So $k = 37$ is indeed the maximum valid multiplier.

Step 3 — Maximise the number of people ahead of Pinky.

People ahead $= 3k$. This is maximised when $k$ is maximised, i.e., at $k = 37$:

$\text{Ahead} = 3 \times 37 = 111$

Final check: With $k = 37$, the total is $8(37) + 1 = 297 < 300$ ✓, and the ahead-behind ratio is $3(37): 5(37) = 111: 185 = 3: 5$ ✓.

Answer: $\boxed{111}$

💡 Why this works so cleanly: The ratio $3:5$ sums to $8$, and the cap $300$ is just above a multiple of $8$ (since $8 \times 37 = 296$). That leaves room for Pinky's own $+1$ without spilling over $300$. Whenever a ratio and a total cap appear together, anchor on the total — express it in terms of the multiplier, bound the multiplier, then read off the quantity you need.

Think of this as: 20 identical items going into 4 distinct boxes, where each box must receive a positive even number. Two restrictions sit on top of each other — "positive" and "even" — but one clean substitution dissolves both at once.

Step 1 — Translate the conditions into variables.

Let $x_i$ be the number of balloons the $i$-th child gets ($i = 1,2,3,4$). Then:

$x_1 + x_2 + x_3 + x_4 = 20.$

The conditions are:

• "each child gets some balloons" → $x_i \ge 1$ (positive),
• "no child gets an odd number" → $x_i$ is even.

Combined: each $x_i$ is a positive even integer.

Step 2 — Kill both restrictions with one substitution.

Every positive even number is twice a positive integer, so write

$x_i = 2y_i, \qquad y_i \ge 1.$

Substituting into the total:

$2y_1 + 2y_2 + 2y_3 + 2y_4 = 20 \;\Rightarrow\; y_1 + y_2 + y_3 + y_4 = 10.$

Now the $y_i$ are just positive integers summing to 10 — an ordinary stars-and-bars problem.

💡 Teacher tip: "Even and positive" is best handled by substituting $x_i = 2y_i$ once. Don't try to memorise a separate "even-only" formula — the substitution reduces it to standard stars-and-bars and is far less error-prone.

Step 3 — Apply stars and bars (positive case).

The number of positive integer solutions of $y_1 + \cdots + y_n = S$ is $\binom{S-1}{n-1}$. Here $S = 10$, $n = 4$:

$\binom{10-1}{4-1} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84.$

Step 4 — Quick sanity check on the range.

The smallest each child can get is $2$ balloons (positive even), giving a minimum total of $4 \times 2 = 8 \le 20$. ✓ The largest any one child can get is $20 - 3\times 2 = 14$, also even. ✓ So valid distributions genuinely exist, and our count stands.

Answer: $\boxed{84}$

Alternative approach — remove positivity first, then evenness.

Give each child the minimum $2$ balloons upfront: set $x_i = 2 + z_i$ with $z_i \ge 0$. This uses up $8$ balloons, leaving

$z_1 + z_2 + z_3 + z_4 = 12,$

where each $z_i$ is a non-negative even integer. Write $z_i = 2w_i$ with $w_i \ge 0$:

$2(w_1 + w_2 + w_3 + w_4) = 12 \;\Rightarrow\; w_1 + w_2 + w_3 + w_4 = 6.$

Non-negative stars-and-bars gives $\binom{6+4-1}{4-1} = \binom{9}{3} = 84$ — same answer. The first method is slightly faster since it folds both restrictions into a single substitution.

We have two equations that look quadratic, but they share a hidden common factor. Factor it out and the system collapses into something simple.

Step 1 — Spot the common factor in each equation.

Look closely at each left-hand side — each one can be factored by pulling out a common variable:

$a^2 + ab + a = a(a + b + 1) = 14$

$b^2 + ab + b = b(a + b + 1) = 28$

The expression $(a + b + 1)$ appears in both equations. That is the key structural observation.

💡 Teacher tip: Whenever two equations share a common sub-expression, substituting a single variable for it often turns a messy system into a clean one. Here, that shared piece is $(a+b+1)$.

Step 2 — Introduce a substitution.

Let $s = a + b + 1$. Then the two equations become:

$as = 14 \quad \text{and} \quad bs = 28$

Step 3 — Find the ratio $\frac{b}{a}$.

Divide the second equation by the first (both sides are positive since $a, b$ are natural numbers, so $s > 0$):

$\frac{bs}{as} = \frac{28}{14} \implies \frac{b}{a} = 2 \implies b = 2a$

Step 4 — Solve for $a$.

Now express $s$ in terms of $a$ alone. Since $b = 2a$:

$s = a + b + 1 = a + 2a + 1 = 3a + 1$

Substitute into $as = 14$:

$a(3a + 1) = 14$

$3a^2 + a - 14 = 0$

Apply the quadratic formula:

$a = \frac{-1 \pm \sqrt{1 + 168}}{6} = \frac{-1 \pm \sqrt{169}}{6} = \frac{-1 \pm 13}{6}$

This gives $a = 2$ or $a = -\frac{7}{3}$. Since $a$ must be a natural number, we take $a = 2$.

Step 5 — Find $b$ and compute $2a + b$.

$b = 2a = 2(2) = 4$

$2a + b = 2(2) + 4 = 8$

Step 6 — Verify.

Both equations are satisfied.

Answer: $\boxed{8}$

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Alternative approach — Add the two equations.

Adding the original equations directly:

$a^2 + ab + a + b^2 + ab + b = 14 + 28 = 42$

$a^2 + 2ab + b^2 + a + b = 42$

$(a+b)^2 + (a+b) = 42$

Let $t = a + b$:

$t^2 + t - 42 = 0 \implies (t+7)(t-6) = 0$

Since $a, b$ are natural numbers, $t = a + b = 6$.

Now use the first equation: $a(a+b) + a = 14 \implies a \cdot 6 + a = 14 \implies 7a = 14 \implies a = 2$, so $b = 4$ and $2a + b = 8$. ✓

💡 Why this works: Adding the equations causes the cross-term $ab$ to double, which completes the perfect square $(a+b)^2$. The factoring method and the addition method exploit the same structure from two different angles — both collapse the system because the shared factor $(a+b+1)$ is doing the heavy lifting.

Think of this as: every group in the village (literate males, literate females, illiterate males, illiterate females) is described by a ratio, and we're given exactly one real count. We need to anchor a multiplier on the ratio that contains that count, then let everything else fall out.

Step 1 — Organise the four sub-groups.

The village splits into two categories — literate and illiterate — and each category has its own male: female ratio. Let's introduce a separate multiplier for each ratio:

So the total number of males is $2m + 4p$ and the total number of females is $3m + 3p$.

Step 2 — Use the overall male: female ratio.

We're told the ratio of total males to total females is $5: 4$:

$\frac{2m + 4p}{3m + 3p} = \frac{5}{4}$

Cross-multiplying:

$4(2m + 4p) = 5(3m + 3p)$

$8m + 16p = 15m + 15p$

$p = 7m$

💡 Teacher tip: This relation $p = 7m$ is the bridge between the two ratio groups. We found it without using the absolute number yet — it comes purely from the overall $5:4$ constraint.

Step 3 — Bring in the one absolute fact.

We're told 3600 males are literate. From our table, literate males $= 2m$:

$2m = 3600 \Rightarrow m = 1800$

Then from Step 2:

$p = 7m = 7 \times 1800 = 12600$

Step 4 — Compute total females.

Total females $= 3m + 3p$:

$3(1800) + 3(12600) = 5400 + 37800 = 43200$

Quick verification.

All three ratios check out.

Answer: $\boxed{43200}$

💡 Why this works: When a problem gives you several ratios plus one actual count, anchor a multiplier on the ratio containing that count. The overall ratio then gives you a linear relation between the multipliers, and one substitution pins everything down — no simultaneous equations needed.

Think of this as: we know three vertices of a parallelogram, and we need the fourth. The cleanest route is to use the single most powerful fact about parallelograms in coordinate geometry — the diagonals bisect each other.

Step 1 — Recall the key property.

In any parallelogram, the two diagonals cut each other exactly in half. So the midpoint of diagonal $\overline{AC}$ is the same point as the midpoint of diagonal $\overline{BD}$.

This means: if we know $A$, $B$, $C$, we can find $D$ by forcing the midpoint of $BD$ to equal the midpoint of $AC$.

Step 2 — Find the midpoint of $AC$.

With $A=(1,1)$ and $C=(-2,8)$, the midpoint is:

$M_{AC}=\left(\frac{1+(-2)}{2},\frac{1+8}{2}\right)=\left(-\frac{1}{2},\frac{9}{2}\right).$

This point must also be the midpoint of $\overline{BD}$.

Step 3 — Set the midpoint of $BD$ equal to this point.

Let $D=(x,y)$, with $B=(3,4)$. The midpoint of $\overline{BD}$ is:

$M_{BD}=\left(\frac{3+x}{2},\frac{4+y}{2}\right).$

Setting $M_{BD}=M_{AC}$:

$\left(\frac{3+x}{2},\frac{4+y}{2}\right)=\left(-\frac{1}{2},\frac{9}{2}\right).$

Step 4 — Solve for $x$ and $y$.

Equating the $x$-coordinates:

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

Equating the $y$-coordinates:

$\frac{4+y}{2}=\frac{9}{2}\implies 4+y=9\implies y=5.$

So $D=(-4,5)$.

Step 5 — Verify against the options.

💡 Teacher tip: "Diagonals bisect each other" is the single most useful parallelogram fact for coordinate problems — it avoids any side-vector bookkeeping and gives $D$ in one midpoint equation.

Answer: $(-4,\,5)$ — option D.

Alternative approach — using equal opposite sides.

In a parallelogram, $\overrightarrow{AB}=\overrightarrow{DC}$. Now $\overrightarrow{AB}=(3-1,4-1)=(2,3)$. If $D=(x,y)$, then $\overrightarrow{DC}=(-2-x,8-y)$. Setting these equal: $-2-x=2\implies x=-4$, and $8-y=3\implies y=5$. Same answer, $D=(-4,5)$. The diagonal-midpoint method is usually faster because it involves only one midpoint computation.