CAT 2024 Quant Questions — Slot 3

What we want: the number of integer pairs $(x,y)$ with $|x+y| + |x-y| = 2$.

Step 1 — Use a clean identity. For any real $x,y$, $|x+y| + |x-y| = 2\max(|x|,|y|).$ (Quick reason: if $|x| \ge |y|$, both $x+y$ and $x-y$ carry the sign of $x$, so their absolute values add to $2|x|$; symmetrically it's $2|y|$ when $|y|$ is larger.) So the equation becomes $2\max(|x|,|y|) = 2 \Rightarrow \max(|x|,|y|) = 1.$

Step 2 — Read off the integer condition. We need both $|x| \le 1$ and $|y| \le 1$, with at least one equal to $1$. So $x, y \in \{-1, 0, 1\}$, but not both zero.

Step 3 — Count. The grid $\{-1,0,1\}\times\{-1,0,1\}$ has $9$ points; drop $(0,0)$, leaving $8$.

Answer: 8

💡 Teacher tip: splitting $|x+y|+|x-y|$ into four sign-cases is slow and error-prone. Collapsing it to $2\max(|x|,|y|)$ turns the whole thing into counting points on a tiny grid.

What we want: the value of $abcdef$, where $3^a = 4,\ 4^b = 5,\ 5^c = 6,\ 6^d = 7,\ 7^e = 8,\ 8^f = 9$.

Step 1 — Rewrite each equation as a logarithm ratio. From $3^a = 4$ we get $a = \log_3 4 = \dfrac{\ln 4}{\ln 3}$. Doing the same for all: $a = \frac{\ln 4}{\ln 3},\ \ b = \frac{\ln 5}{\ln 4},\ \ c = \frac{\ln 6}{\ln 5},\ \ d = \frac{\ln 7}{\ln 6},\ \ e = \frac{\ln 8}{\ln 7},\ \ f = \frac{\ln 9}{\ln 8}.$

Step 2 — Multiply them and watch the cancellation. $abcdef = \frac{\ln 4}{\ln 3}\cdot\frac{\ln 5}{\ln 4}\cdot\frac{\ln 6}{\ln 5}\cdot\frac{\ln 7}{\ln 6}\cdot\frac{\ln 8}{\ln 7}\cdot\frac{\ln 9}{\ln 8} = \frac{\ln 9}{\ln 3}.$ Every middle term cancels with its neighbour — a clean telescoping chain.

Step 3 — Simplify. $\frac{\ln 9}{\ln 3} = \log_3 9 = \log_3(3^2) = 2.$

Answer: 2

💡 Teacher tip: whenever you see a chain $a^x = b,\ b^y = c, \ldots$, convert each to a log ratio and multiply — the inner terms always cancel in pairs.

What we want: the distance, given how the travel time responds to two speed changes.

Let distance $= D$, usual speed $= v$, usual time $= t$, so $D = vt$.

Step 1 — Write each scenario as a distance equation.

• $6$ km/h faster takes $4$ hours less: $\;D = (v+6)(t-4)$.
• $6$ km/h slower takes $6$ hours more: $\;D = (v-6)(t+6)$.

Step 2 — Expand and cancel the common $vt = D$.

First equation: $vt = vt - 4v + 6t - 24 \Rightarrow 4v - 6t = -24 \Rightarrow 2v - 3t = -12.$ … (1)

Second equation: $vt = vt + 6v - 6t - 36 \Rightarrow 6v - 6t = -36 \Rightarrow v - t = -6 \Rightarrow t - v = 6.$ … (2)

Step 3 — Solve the two linear equations. From (2), $t = v + 6$. Substitute into (1): $2v - 3(v+6) = -12 \Rightarrow -v - 18 = -12 \Rightarrow v = -6?$ Let me recompute cleanly using $v = t - 6$ from (2): $2(t-6) - 3t = -12 \Rightarrow -t - 12 = -12 \Rightarrow t = 24$, so $v = 30$.

Step 4 — Distance. $D = vt = 30 \times 24 = 720$.

Step 5 — Check. At $30$: $720/30 = 24$ h. At $36$: $720/36 = 20 = 24 - 4$ ✓. At $24$: $720/24 = 30 = 24 + 6$ ✓.

Answer: 720 (Choice A)

💡 Teacher tip: for a fixed distance with varied speed and time, always expand and cancel the $vt$ term — both equations turn linear at once.

What we want: the area of $\triangle XYZ$, where $M, N, P$ are the midpoints of the sides of $\triangle ABC$, and the medians from each vertex meet the opposite mid-segments at $X, Y, Z$.

Step 1 — Recall the "medial triangle". Joining the three midpoints $M, N, P$ makes a triangle whose sides are parallel to $ABC$'s sides and half their length. Its area is exactly $\frac14$ of $\triangle ABC$.

Step 2 — Find where the medians cross the mid-segments. The median from $A$ runs to $N$ (midpoint of $BC$) and crosses $MP$, which is parallel to $BC$. A line from a vertex to the far midpoint is cut in half by the mid-segment parallel to that side — so $X$ is the midpoint of $MP$. The same holds all around: $X = \text{midpoint of } MP, \quad Y = \text{midpoint of } MN, \quad Z = \text{midpoint of } NP.$

💡 Teacher tip: $X, Y, Z$ are simply the midpoints of the medial triangle's sides — so $\triangle XYZ$ is the medial triangle of the medial triangle.

Step 3 — Apply the $\frac14$ rule twice. $[XYZ] = \frac14[MNP] = \frac14\cdot\frac14[ABC] = \frac1{16}[ABC].$

Step 4 — Plug in. $[XYZ] = \dfrac{1440}{16} = 90$.

Answer: 90

💡 Why $\frac1{16}$: each "midpoint triangle" step shrinks the area to a quarter, and we do it twice: $\frac14 \times \frac14 = \frac1{16}$.

What we want: the area of the square $ACEG$ formed by joining every second corner of a regular octagon with side $6$.

Step 1 — Put the octagon on a circle. All eight corners lie on a circle of radius $R$. Each side spans $\frac{360^\circ}{8} = 45^\circ$ at the centre, so the side length is $s = 2R\sin(22.5^\circ) = 6.$

Step 2 — Find a diagonal like $AC$. $A$ and $C$ are two corners apart, spanning $90^\circ$ at the centre: $AC = 2R\sin 45^\circ = 2R\cdot\frac{\sqrt2}{2} = R\sqrt2.$

Step 3 — Take the ratio $AC/s$ so $R$ cancels. $\frac{AC}{s} = \frac{R\sqrt2}{2R\sin 22.5^\circ} = \frac{\sqrt2}{2\sin 22.5^\circ}.$ Using $\sin 22.5^\circ = \dfrac{\sqrt{2-\sqrt2}}{2}$: $\frac{AC}{s} = \frac{\sqrt2}{\sqrt{2-\sqrt2}} = \sqrt{\frac{2}{2-\sqrt2}} = \sqrt{\frac{2(2+\sqrt2)}{(2-\sqrt2)(2+\sqrt2)}} = \sqrt{\frac{2(2+\sqrt2)}{2}} = \sqrt{2+\sqrt2}.$ So $AC = 6\sqrt{2+\sqrt2}$.

Step 4 — Area of the square. By the octagon's symmetry $ACEG$ is a square, so its area is $AC^2$: $AC^2 = 36(2+\sqrt2).$

Answer: $36(2+\sqrt2)$ (Choice D)

💡 Teacher tip: fixing the circumradius $R$ once from the side length lets you read off every diagonal with a sine — no messy coordinate geometry.

What we want: how many positive integers up to $500$ have all-different digits.

Step 1 — One-digit numbers ($1$ to $9$). Every one qualifies: $9$ numbers.

Step 2 — Two-digit numbers. Tens digit: $9$ choices ($1$–$9$). Units digit: $9$ choices (any of $0$–$9$ except whatever's in the tens place). That's $9 \times 9 = 81$.

Step 3 — Three-digit numbers up to $500$. Split by the hundreds digit.

• Hundreds $= 1, 2, 3,$ or $4$: the number is automatically below $500$. Hundreds: $4$ choices. Tens: $9$ (any digit except the hundreds). Units: $8$ (any except the two used). That's $4 \times 9 \times 8 = 288$.
• Hundreds $= 5$: the only value $\le 500$ is $500$ itself, and it repeats the digit $0$. So $0$ valid numbers here.

Step 4 — Add up.

Answer: 378

💡 Teacher tip: for "up to $N$ with distinct digits", build by number-length and, for the top length, split at the leading digit of $N$. The leading-$5$ case gave nothing precisely because $500$ repeats a digit.

What we want: the first increment percentage, given two successive raises take the salary to $187.5\%$ and the second raise (in percent) is double the first.

Let the first raise be $x\%$; then the second is $2x\%$. Two successive raises multiply, so the salary becomes $1.875$ times the original.

Step 1 — Write the equation. $\left(1 + \frac{x}{100}\right)\!\left(1 + \frac{2x}{100}\right) = 1.875.$ Let $u = \frac{x}{100}$: $(1+u)(1+2u) = 1.875 \Rightarrow 2u^2 + 3u + 1 = 1.875 \Rightarrow 2u^2 + 3u - 0.875 = 0.$

Step 2 — Solve. Multiply by $8$: $16u^2 + 24u - 7 = 0$. Discriminant $= 576 + 448 = 1024 = 32^2$: $u = \frac{-24 \pm 32}{32} \Rightarrow u = \frac14 \ \text{ or } \ u = -\frac{56}{32} \ (\text{reject, negative}).$

Step 3 — Convert back. $x = 100u = 100 \times \frac14 = 25$.

Step 4 — Check. First raise $25\% \to 1.25$; second raise $50\% \to 1.5$; product $1.25 \times 1.5 = 1.875 = 187.5\%$ ✓.

Answer: 25 (Choice C)

💡 Teacher tip: successive percentage changes multiply, they don't add. Substituting $u = x/100$ keeps the algebra clean.

What we want: the sum of all real $x$ with $10^x + \dfrac{4}{10^x} = \dfrac{81}{2}$.

Step 1 — Substitute to hide the exponential. Let $y = 10^x$ (so $y > 0$). The equation becomes $y + \frac4y = \frac{81}{2} \Rightarrow 2y^2 - 81y + 8 = 0.$

Step 2 — Use the relationships between roots and coefficients. If the two roots are $y_1, y_2$, then their product is $y_1 y_2 = \frac{8}{2} = 4,$ and their sum $\frac{81}{2}$ is positive, so both roots are positive (a positive product and positive sum). The discriminant $81^2 - 4\cdot 2\cdot 8 = 6561 - 64 = 6497 > 0$, so there really are two distinct positive values.

Step 3 — Go back to $x$. Each $y_i$ gives $x_i = \log_{10} y_i$. The sum of the $x$-values is $x_1 + x_2 = \log_{10} y_1 + \log_{10} y_2 = \log_{10}(y_1 y_2) = \log_{10} 4 = 2\log_{10} 2.$

💡 Teacher tip: you never needed the actual roots. For any equation $y + \frac{k}{y} = c$, the product of the two roots is $k$ — and adding logs turns a product into what you want.

Answer: $2\log_{10} 2$ (Choice A)

Picture it: a circle cut by a chord. The chord makes a $120^\circ$ angle at the centre, and we want the area of the smaller piece it cuts off.

Step 1 — Find the radius from the chord. A chord that subtends angle $\theta$ at the centre has length $2r\sin(\theta/2)$. Here $\theta = 120^\circ$: $\text{chord} = 2r\sin 60^\circ = 2r\cdot\frac{\sqrt3}{2} = r\sqrt3.$ We're told the chord is $10\sqrt3$, so $r\sqrt3 = 10\sqrt3 \Rightarrow r = 10$.

Step 2 — Which piece is smaller? The angle $120^\circ$ is less than $180^\circ$, so the piece cut off (the minor segment) is the smaller region. Its area $=$ (sector) $-$ (triangle).

Step 3 — Sector area (a $120^\circ$ slice of the circle): $\frac{120}{360}\,\pi r^2 = \frac13\,\pi(100) = \frac{100\pi}{3}.$

Step 4 — Triangle area (two radii of $10$ with a $120^\circ$ angle between them): $\frac12 r^2 \sin 120^\circ = \frac12(100)\cdot\frac{\sqrt3}{2} = 25\sqrt3.$

Step 5 — Subtract. $\frac{100\pi}{3} - 25\sqrt3 = 25\!\left(\frac{4\pi}{3} - \sqrt3\right).$

Answer: $25\!\left(\dfrac{4\pi}{3} - \sqrt3\right)$ (Choice D)

💡 Why "sector minus triangle": for any angle under $180^\circ$, the small piece is the pie-slice with its straight-edged triangle removed. Always check which side of the chord is smaller before you subtract.

What we want: the minimum and maximum possible number of students who chose both swimming and running, as the number of girls varies within its allowed range.

Step 1 — Set the population limits. Total is 250. Girls $G$ satisfy $44\% \le \tfrac{G}{250} \le 60\%$, i.e. $110 \le G \le 150$. Boys $B = 250 - G$, so $100 \le B \le 140$.

Step 2 — Express "both" for each group. Everyone chose at least one activity, so for any group, "both" $=$ swim $+$ run $-$ total.

• Boys: $0.5B + 0.7B - B = 0.2B$ choose both.
• Girls: $0.8G + 0.6G - G = 0.4G$ choose both.

Step 3 — Total choosing both. $\text{Both} = 0.2B + 0.4G = 0.2(250 - G) + 0.4G = 50 + 0.2G.$

Step 4 — Push $G$ to each extreme. Since "Both" rises with $G$:

💡 Teacher tip: Because the both-count is a straight-line function of $G$, its smallest and largest values sit at the ends of the allowed range — no in-between value can beat them.

Answer: 72 and 80, respectively — Choice C

What we want: how many real values of $x$ satisfy $\max\{x,2\} - \min\{x,2\} = |x+2| - |x-2|$.

Step 1 — Simplify the left side. For any two numbers, the bigger minus the smaller is just the distance between them: $\max\{a,b\} - \min\{a,b\} = |a-b|$. With $a=x,\ b=2$: $\max\{x,2\} - \min\{x,2\} = |x-2|.$ So the equation becomes $|x-2| = |x+2| - |x-2| \Rightarrow 2|x-2| = |x+2|.$

Step 2 — Solve it. Both sides are non-negative, so we can square safely: $4(x-2)^2 = (x+2)^2 \Rightarrow 4(x^2 - 4x + 4) = x^2 + 4x + 4 \Rightarrow 3x^2 - 20x + 12 = 0.$

Step 3 — Roots. Discriminant $= 400 - 144 = 256 = 16^2$, so $x = \frac{20 \pm 16}{6} \Rightarrow x = 6 \ \text{ or } \ x = \frac23.$

Step 4 — Check both really work.

Two distinct valid solutions.

Answer: 2

💡 Teacher tip: squaring is safe here only because both sides were already non-negative. Whenever you square, back-substitute to rule out fake roots.

What we want: a condition that must hold whenever the system $px - 4y = 2$ and $3x + ky = a$ has no solution.

Step 1 — When do two lines have no common point? They must be parallel but not identical. In ratio form: $\frac{p}{3} = \frac{-4}{k} \neq \frac{2}{a}.$

Step 2 — Translate each part into plain equations.

• Parallel ($\frac{p}{3} = \frac{-4}{k}$): cross-multiplying gives $pk = -12$, i.e. $pk + 12 = 0$.
• Not identical means the constant term breaks the pattern, giving two "not-equal" facts:
• $\frac{2}{a} \neq \frac{p}{3} \Rightarrow ap \neq 6$.
• $\frac{2}{a} \neq \frac{-4}{k} \Rightarrow 2k \neq -4a \Rightarrow 2a + k \neq 0$.

So no solution means: $pk + 12 = 0$ and $ap \neq 6$ and $2a + k \neq 0$.

Step 3 — Check the options against those facts.

• (a) $ap + 6 = 0$ means $ap = -6$ — not forced (we only know $ap \neq 6$). ✗
• (b) $2a + k \neq 0$ — this is exactly one of our required conditions. ✅
• (c) $ap - 6 = 0$ means $ap = 6$ — this contradicts a requirement. ✗
• (d) $kp + 12 \neq 0$ — contradicts the parallel condition $kp + 12 = 0$. ✗

Answer: $2a + k \neq 0$ (Choice B)

💡 Teacher tip: read the word "necessary" carefully — it means "must be true whenever there's no solution", which is weaker than "guarantees no solution". Options (c) and (d) point the wrong way and are traps.

What we want: the fraction of the whole job that Sam personally does under the repeating pattern of pairs.

Step 1 — Turn the speed statements into daily rates. Sam finishes alone in $20$ days, so Sam does $\frac1{20}$ of the job per day. Mohit is twice as fast: $\frac2{20} = \frac1{10}$ per day. Mohit is thrice as fast as Ayna, so Ayna does $\frac13 \times \frac1{10} = \frac1{30}$ per day.

Step 2 — Work done on each day of the 3-day cycle. (I'll write everything over $60$ to compare easily.)

• Day 1, Sam + Mohit: $\frac1{20} + \frac1{10} = \frac{3+6}{60} = \frac9{60}$.
• Day 2, Sam + Ayna: $\frac1{20} + \frac1{30} = \frac{3+2}{60} = \frac5{60}$.
• Day 3, Mohit + Ayna: $\frac1{10} + \frac1{30} = \frac{6+2}{60} = \frac8{60}$.

Step 3 — Track the running total until the job is done (the whole job is $\frac{60}{60}$):

After Day 8 we're at $58/60$; Day 9 (Mohit + Ayna) finishes it. Sam does not work on Day 9.

Step 4 — Add up Sam's days. Sam worked on days $1,2,4,5,7,8$ — six full days — at $\frac1{20}$ each: $6 \times \frac1{20} = \frac6{20} = \frac3{10}.$

💡 Teacher tip: the job ends on a day Sam sits out, so all his contributing days are complete — no messy partial-day fraction to chase.

Answer: $\dfrac{3}{10}$ (Choice B)

The setup: a $300$ L container starts with some water $W$ and the rest milk. Then $2W$ litres of the mix are removed and replaced by water. The final mix is $72\%$ milk. Find $W$.

Initially: water $= W$, milk $= 300 - W$, so the milk fraction is $\dfrac{300-W}{300}$.

Step 1 — How much milk survives the removal. When $2W$ litres of the uniform mix are drawn off, the milk left is the original milk scaled by the fraction kept: $\text{milk left} = (300 - W)\left(1 - \frac{2W}{300}\right) = (300-W)\cdot\frac{300 - 2W}{300}.$ Refilling with water adds no milk, so this is the final milk.

Step 2 — Set it to $72\%$. Final milk $= 72\%$ of $300 = 216$ L: $\frac{(300-W)(300-2W)}{300} = 216 \Rightarrow (300-W)(300-2W) = 64{,}800.$

Step 3 — Solve the quadratic. Expanding: $90{,}000 - 900W + 2W^2 = 64{,}800$, so $2W^2 - 900W + 25{,}200 = 0 \Rightarrow W^2 - 450W + 12{,}600 = 0.$ $W = \frac{450 \pm \sqrt{202{,}500 - 50{,}400}}{2} = \frac{450 \pm \sqrt{152{,}100}}{2} = \frac{450 \pm 390}{2}.$ So $W = 420$ or $W = 30$.

Step 4 — Reject the impossible one. You can't remove $2W$ litres unless $2W \le 300$, i.e. $W \le 150$. So $W = 420$ is out, leaving $W = 30$.

Step 5 — Check. $W = 30$: milk $= 270$ ($90\%$). Remove $60$ L → lose $54$ L milk → $216$ L left $= 72\%$ ✓.

Answer: 30

What we want: the profit, in rupees, Gopi makes selling to Ravi at the discounted price.

Let Gopi's cost price be $C$.

Step 1 — Marked price for a $20\%$ profit. Gopi marks up by $20\%$, so the marked price is $M = 1.2C$.

Step 2 — Ravi's discount and his saving. Ravi gets $10\%$ off the marked price, so his saving is $10\%$ of $M$: $0.1M = 0.1(1.2C) = 0.12C.$ This saving is Rs $15$: $\;0.12C = 15 \Rightarrow C = 125$.

Step 3 — What Ravi actually pays, and the profit. Ravi pays $90\%$ of $M$: $\;0.9 \times 1.2 \times 125 = 1.08 \times 125 = 135$. Profit is $135 - 125 = 10.$

Step 4 — Check. Cost $125$, marked $150$, discount $15$ ✓, Ravi pays $135$, profit $10$ ✓.

Answer: 10 (Choice A)

💡 Teacher tip: chain the multipliers — cost $\xrightarrow{\times 1.2}$ marked $\xrightarrow{\times 0.9}$ selling gives a net $\times 1.08$, i.e. an $8\%$ profit on cost straight away.

What we want: the remainder when $10^{68}$ is divided by $13$.

Step 1 — Watch how the powers of $10$ behave, modulo $13$. $10^1 \equiv 10, \qquad 10^2 = 100 = 7\cdot 13 + 9 \equiv 9, \qquad 10^3 \equiv 9 \times 10 = 90 = 78 + 12 \equiv 12 \equiv -1.$ So $10^3 \equiv -1$, which means $10^6 = (10^3)^2 \equiv (-1)^2 = 1$.

💡 Teacher tip: landing on $-1$ is the golden moment. Once $10^3 \equiv -1$, you instantly know $10^6 \equiv 1$, so the powers repeat in a cycle of length $6$.

Step 2 — Reduce the exponent using that cycle. Divide $68$ by $6$: $68 = 6\times 11 + 2$. So $10^{68} = (10^{6})^{11}\times 10^{2} \equiv 1^{11}\times 9 = 9 \pmod{13}.$

Step 3 — Answer. The remainder is $9$.

Answer: 9 (Choice C)

Alternative approach. Since $13$ is prime, Fermat's little theorem gives $10^{12} \equiv 1$. Writing $68 = 12\times 5 + 8$, we get $10^{68} \equiv 10^{8}$. Then $10^8 = 10^6 \times 10^2 \equiv 1 \times 9 = 9$. Same answer — but spotting the $-1$ at $10^3$ is quicker under exam pressure.

What we want: the wheat-to-mustard ratio on Rajesh's $20$-hectare land.

Step 1 — Split Vimal's land. Vimal has $30$ ha with wheat: mustard $= 5:3$ (that's $8$ parts): $\text{wheat} = \frac58 \times 30 = 18.75, \qquad \text{mustard} = \frac38 \times 30 = 11.25.$

Step 2 — Split the combined total. Together they farm $20 + 30 = 50$ ha with overall wheat: mustard $= 11:9$ ($20$ parts): $\text{total wheat} = \frac{11}{20}\times 50 = 27.5, \qquad \text{total mustard} = \frac{9}{20}\times 50 = 22.5.$

Step 3 — Subtract Vimal's share to get Rajesh's. $\text{Rajesh wheat} = 27.5 - 18.75 = 8.75, \qquad \text{Rajesh mustard} = 22.5 - 11.25 = 11.25.$

Step 4 — Form and simplify the ratio. $8.75: 11.25 = 875: 1125 = 35: 45 = 7: 9.$

Answer: $7: 9$ (Choice B)

💡 Teacher tip: when both a part (Vimal) and the whole (total) are given as ratios, turn each into actual quantities and subtract — fighting it out purely in ratios is far messier.

Setting up: call the three numbers, in increasing order, $s < m < \ell$. Their average is $28$, so $s + m + \ell = 84.$

Step 1 — Use the "new average" clue. The smallest goes up by $7$ and the largest goes down by $10$; the middle is untouched. So the total changes by $+7 - 10 = -3$, making the new sum $84 - 3 = 81$. The new average is $\frac{81}{3} = 27.$ We're told this new average is $2$ more than the middle number: $27 = m + 2$, so $m = 25$.

💡 Teacher tip: don't recompute each number to get the new average — just track how the total shifts. A change of $-3$ in the sum is a change of $-1$ in the average of three numbers.

Step 2 — Use the "new difference" clue. After the change, largest $-$ smallest $= 64$: $(\ell - 10) - (s + 7) = 64 \Rightarrow \ell - s = 81.$

Step 3 — Solve for the original largest. We know $s + \ell = 84 - m = 84 - 25 = 59$, and $\ell - s = 81$. Adding: $2\ell = 59 + 81 = 140 \Rightarrow \ell = 70, \qquad s = 59 - 70 = -11.$

Step 4 — Check everything. Original order: $-11 < 25 < 70$ ✓. After the change: $-4 < 25 < 60$, order kept ✓; new average $27 = 25+2$ ✓; new difference $60-(-4) = 64$ ✓.

Answer: 70

What we want: the value of $\dfrac1{t_2} + \dfrac1{t_4} + \dfrac1{t_6} + \cdots + \dfrac1{t_{2024}}$, where $t_1 = 1$, $t_2 = -1$, and $t_n = \dfrac{n-3}{n-1}\,t_{n-2}$.

Step 1 — Compute a few even terms and look for a pattern. $t_4 = \frac{1}{3}\,t_2 = -\frac13, \qquad t_6 = \frac{3}{5}\,t_4 = -\frac15, \qquad t_8 = \frac{5}{7}\,t_6 = -\frac17.$ The pattern is clear: for the term with index $2k$, $t_{2k} = -\frac{1}{2k-1}.$

💡 Teacher tip: always compute three or four terms before trusting a formula. Here each new factor cancels neatly against the denominator two steps back, which is why every term is a tidy unit fraction.

Step 2 — Flip each term. $\dfrac1{t_{2k}} = -(2k-1)$, an odd number with a minus sign. As $2k$ runs from $2$ to $2024$, $k$ runs from $1$ to $1012$.

Step 3 — Add them up. $\sum_{k=1}^{1012}\frac1{t_{2k}} = -\big(1 + 3 + 5 + \cdots + (2\cdot 1012 - 1)\big) = -(\text{sum of first } 1012 \text{ odd numbers}).$ The sum of the first $n$ odd numbers is $n^2$, so this is $-1012^2$.

Step 4 — Evaluate. $1012^2 = (1000+12)^2 = 1{,}000{,}000 + 24{,}000 + 144 = 1{,}024{,}144$.

Answer: $-1{,}024{,}144$ (Choice A)

💡 Worth memorising: "sum of the first $n$ odd numbers $= n^2$" turns a 1012-term sum into one squaring. It shows up again and again.

What we want: the sum of all $x$ with $f(x) = 3$, given $f(x) + 2f\!\left(\frac1x\right) = 3x$ for every non-zero $x$.

Step 1 — Make a second equation. Replace $x$ with $\frac1x$ throughout: $f\!\left(\frac1x\right) + 2f(x) = \frac3x.$ Now we have two equations in the two unknowns $f(x)$ and $f\!\left(\frac1x\right)$: $f(x) + 2f\!\left(\tfrac1x\right) = 3x \quad (1), \qquad 2f(x) + f\!\left(\tfrac1x\right) = \tfrac3x \quad (2).$

Step 2 — Solve for $f(x)$. Double (1): $2f(x) + 4f\!\left(\frac1x\right) = 6x$. Subtract (2): $3f\!\left(\tfrac1x\right) = 6x - \tfrac3x \Rightarrow f\!\left(\tfrac1x\right) = 2x - \tfrac1x.$ Put this back into (1): $f(x) = 3x - 2\!\left(2x - \tfrac1x\right) = -x + \frac2x.$

Step 3 — Set $f(x) = 3$. $-x + \frac2x = 3 \Rightarrow -x^2 + 2 = 3x \Rightarrow x^2 + 3x - 2 = 0.$

Step 4 — Sum of the roots. For $x^2 + 3x - 2 = 0$ the sum of roots is $-3$ (discriminant $9 + 8 = 17 > 0$, so two real roots).

Answer: $-3$ (Choice C)

💡 Teacher tip: any equation of the form $f(x) + k\,f(1/x) = g(x)$ yields to the "swap $x$ and $1/x$" trick — it always produces a $2\times2$ system you can solve for $f(x)$.

What we're really being asked: we have $(a+b\sqrt3)^2 = 52 + 30\sqrt3$ with $a,b$ natural numbers, and we want $a+b$. The trick is that a number written as "rational part $+$ (something)$\sqrt3$" can only match another such number part-by-part.

Step 1 — Expand the left side. $(a+b\sqrt3)^2 = a^2 + 2ab\sqrt3 + 3b^2 = (a^2+3b^2) + (2ab)\sqrt3.$

Step 2 — Match the two sides separately. The plain (rational) parts must be equal, and the $\sqrt3$ parts must be equal: $a^2 + 3b^2 = 52, \qquad 2ab = 30 \Rightarrow ab = 15.$

💡 Teacher tip: $\sqrt3$ is irrational, so it can never "help balance" the plain number part. That's exactly why one equation splits cleanly into two — this is the whole idea behind these surd problems.

Step 3 — List the natural-number pairs with $ab=15$. They are $(1,15),(3,5),(5,3),(15,1)$.

Step 4 — Test each in $a^2+3b^2=52$.

So $a=5,\ b=3$.

Step 5 — Answer. $a+b = 5+3 = 8$.

Answer: 8 (Choice B)

What we want: the fewest whole years for Rs $4000$ to grow past Rs $20{,}000$, with annual compounding.

Step 1 — Find the yearly growth factor. Let each year multiply the money by $g$. After $3$ years we have $4000g^3$; after $5$ years, $4000g^5$. The ratio of the 3-year value to the 5-year value is $25:36$: $\frac{4000g^3}{4000g^5} = \frac{25}{36} \Rightarrow \frac{1}{g^2} = \frac{25}{36} \Rightarrow g^2 = \frac{36}{25} \Rightarrow g = \frac65.$ So the money grows by a factor $\frac65$ each year — a rate of $20\%$.

Step 2 — Set up the target. After $n$ years the value is $4000\left(\frac65\right)^n$, and we need this above $20{,}000$: $4000\left(\tfrac65\right)^n > 20{,}000 \Rightarrow \left(\tfrac65\right)^n > 5.$

Step 3 — Try years one at a time (each step multiplies by $1.2$):

Step 4 — Answer. The value first crosses Rs $20{,}000$ in year $9$.

Answer: 9

💡 Teacher tip: for "which year does it cross a threshold" questions, pull out the growth factor first, then just march year by year — far faster than reaching for logarithms.