CAT 2025 Quant Questions — Slot 1

What we want: the number of integer pairs $(x,y)$ with $x > y \ge 3$ and $x + y < 14$.

Step 1 — Fix $y$, then count the valid $x$. For a given integer $y \ge 3$:

• $x > y$ means $x \ge y + 1$.
• $x + y < 14$ means $x \le 13 - y$.

So $x$ runs from $y+1$ up to $13 - y$, giving $(13 - y) - (y + 1) + 1 = 13 - 2y$ values. This is positive only while $y \le 6$.

Step 2 — Add up over $y = 3, 4, 5, 6$.

Answer: 16

💡 Teacher tip: for "count the lattice points" problems, anchor on the tightest variable (here $y \ge 3$) and write the other variable's range in terms of it. The shrinking counts $7, 5, 3, 1$ are a nice sanity check.

Think of it as: we never actually need the two diagonals themselves — only their difference. Two standard rhombus facts hand us $d_1 d_2$ and $d_1^2 + d_2^2$, and an algebraic identity turns those into the difference directly.

Side $s = 36$, area $= 396$. Let the diagonals be $d_1$ and $d_2$.

Step 1 — Two facts about a rhombus.

• Area $= \dfrac{1}{2} d_1 d_2$.
• The diagonals meet at right angles and bisect each other, so each side is the hypotenuse of a right triangle with legs $\dfrac{d_1}{2}, \dfrac{d_2}{2}$. Hence $s^2 = \dfrac{d_1^2 + d_2^2}{4}$.

Step 2 — Plug in the numbers. $\frac{1}{2} d_1 d_2 = 396 \;\Rightarrow\; d_1 d_2 = 792.$ $36^2 = \frac{d_1^2 + d_2^2}{4} \;\Rightarrow\; d_1^2 + d_2^2 = 4 \times 1296 = 5184.$

Step 3 — Use the difference identity. $(d_1 - d_2)^2 = d_1^2 + d_2^2 - 2 d_1 d_2 = 5184 - 2(792) = 5184 - 1584 = 3600.$ $|d_1 - d_2| = \sqrt{3600} = 60.$

Answer: 60

💡 Teacher tip: whenever a problem gives you a product and a sum-of-squares, reach for $(x \pm y)^2 = x^2 + y^2 \pm 2xy$. It sidesteps solving for the individual quantities entirely.

What we want: the number of shares of stock B in a portfolio worth Rs $3300$.

Let $b$ = shares of B and $c$ = shares of C.

Step 1 — Translate the holdings. She holds $10$ shares of A, and "$20$ shares of B and C together" means $b + c = 20$.

Step 2 — Translate the total value. Prices are $120, 90, 150$: $10(120) + 90b + 150c = 3300 \Rightarrow 1200 + 90b + 150c = 3300 \Rightarrow 90b + 150c = 2100.$ Divide by $30$: $\;3b + 5c = 70$.

Step 3 — Solve. Using $b = 20 - c$: $3(20 - c) + 5c = 70 \Rightarrow 60 + 2c = 70 \Rightarrow c = 5, \ \ b = 15.$

Step 4 — Check. $10(120) + 15(90) + 5(150) = 1200 + 1350 + 750 = 3300$ ✓, and $15 + 5 = 20$ ✓.

Answer: 15

Think of it as: two equations, three unknowns — so you can't find $a,b,c$ separately. The trick is that the answer only needs one combination, $b - c$, which the equations do pin down.

We are given $a - 6b + 6c = 4, \qquad 6a + 3b - 3c = 50.$

Step 1 — Rewrite the target. Notice $2a + 3b - 3c = 2a + 3(b - c)$. So if we know $a$ and $b - c$, we're done.

Step 2 — Squeeze $b - c$ out of the second equation. Factor: $6a + 3(b - c) = 50$, so $2a + (b - c) = \frac{50}{3}. \qquad (\star)$

Step 3 — Bring in the first equation. From $a - 6(b - c) = 4$ we get $a = 4 + 6(b - c)$. Substitute into $(\star)$: $2\big(4 + 6(b - c)\big) + (b - c) = \frac{50}{3} \;\Rightarrow\; 8 + 13(b - c) = \frac{50}{3}.$ $13(b - c) = \frac{50}{3} - 8 = \frac{26}{3} \;\Rightarrow\; b - c = \frac{2}{3}.$

Step 4 — Get $a$, then the target. $a = 4 + 6 \cdot \tfrac{2}{3} = 8$. Therefore $2a + 3(b - c) = 2(8) + 3\left(\frac{2}{3}\right) = 16 + 2 = 18.$

Answer: 18 (Choice C)

💡 Teacher tip: when unknowns outnumber equations, don't panic — group the target around the combination the equations actually control. Here everything hinges on $b - c$.

Think of it as: the customer pays for 12 chairs but carries home 13. So the shop's revenue is "12 discounted prices," while its cost is "13 chairs." The profit condition ties those together.

Cost price $= ₹100$ per chair; discount $= 22\%$ on the marked price $M$.

Step 1 — Revenue side of the deal. The customer pays the discounted price for $12$ chairs, so $\text{Revenue} = 12 \times (\text{discounted price per chair}).$

Step 2 — Cost and the profit condition. The shop actually hands over $13$ chairs, costing $13 \times 100 = ₹1300$. A $26\%$ profit on cost means $\text{Revenue} = 1.26 \times 1300 = ₹1638.$

Step 3 — Discounted price per chair. From Step 1, $12 \times (\text{discounted price}) = 1638 \;\Rightarrow\; \text{discounted price} = \frac{1638}{12} = 136.5.$

Step 4 — Undo the discount to reach $M$. The discounted price is $(1 - 0.22)M = 0.78M$: $0.78M = 136.5 \;\Rightarrow\; M = \frac{136.5}{0.78} = 175.$

Step 5 — Verify. Marked $= 175$; after $22\%$ off, $136.5$. Customer pays $12 \times 136.5 = 1638$ for $13$ chairs costing $1300$; profit $= 338$, which is exactly $26\%$ of $1300$ ✓.

Answer: ₹175

💡 Teacher tip: in "give 13 for the price of 12" deals, the free chair is a cost to the shop, not a discount to the price. Count revenue over the paid units and cost over all units handed out.

What we want: the smallest possible starting class size.

Let boys $= B > 10$ and girls $= G$.

Step 1 — Use the "remaining" condition. After $40\%$ of girls and $60\%$ of boys leave, remaining girls $= 0.6G$ and remaining boys $= 0.4B$, with $0.6G = 0.4B + 8$: $3G = 2B + 40 \Rightarrow G = \frac{2B + 40}{3}.$

Step 2 — The numbers leaving must be whole people. "$40\%$ of the girls" $= \frac{2G}{5}$ must be an integer, so $G$ is a multiple of $5$. Likewise "$60\%$ of the boys" forces $B$ to be a multiple of $5$.

Step 3 — Minimise $B + G$. For $G$ to be a whole number, $2B + 40$ must be divisible by $3$, which needs $B \equiv 1 \pmod 3$. Combined with $B$ being a multiple of $5$ and $B > 10$:

• $B \equiv 0 \pmod 5$ and $B \equiv 1 \pmod 3$ together give $B \equiv 10 \pmod{15}$.
• The smallest such $B$ above $10$ is $B = 25$.

Then $G = \frac{2(25) + 40}{3} = \frac{90}{3} = 30$ (a multiple of $5$ ✓).

Step 4 — Check. $B = 25 > 10$ ✓. Girls leaving $= 12$, remaining $= 18$. Boys leaving $= 15$, remaining $= 10$. And $18 - 10 = 8$ ✓. Total $= 55$.

Answer: 55

💡 Teacher tip: a smaller $B$ might satisfy the algebra alone, but if $40\%$ of the girls isn't a whole number it's not a real class. Always check that a "percentage of a count" comes out to an integer.

What we want: the number $k$ in $\{1, 3, 5, \ldots, 57\}$ where everything below it sums to the same as everything above it.

Step 1 — Count and total the set. The odd numbers $1$ to $57$: since $57 = 2(29) - 1$, there are $29$ terms, and the sum of the first $29$ odd numbers is $29^2 = 841$.

Step 2 — Write the balance condition. Let the sum below $k$ equal the sum above $k$, each $= S$. Then $S + k + S = 841 \Rightarrow 2S + k = 841.$

Step 3 — Express $S$ through $k$. The numbers below $k$ are $1, 3, \ldots, k-2$, i.e. the first $\frac{k-1}{2}$ odd numbers, which sum to $\left(\frac{k-1}{2}\right)^2$. So $S = \left(\frac{k-1}{2}\right)^2$, and $2\left(\frac{k-1}{2}\right)^2 + k = 841 \Rightarrow \frac{(k-1)^2}{2} + k = 841.$ Multiply by $2$: $(k-1)^2 + 2k = 1682 \Rightarrow k^2 + 1 = 1682 \Rightarrow k^2 = 1681 \Rightarrow k = 41.$

Step 4 — Check. $k = 41$ is the $21$st odd number. Sum below $= 20^2 = 400$; sum above $= 841 - 400 - 41 = 400$ ✓.

Answer: 41 (Choice A)

💡 Teacher tip: the identity "sum of the first $n$ odd numbers $= n^2$" does all the work — it turns both sides of the balance into clean squares.

What we want: the set $S$ of $x \in [3,6]$ with $[x^2] = [x]^2$ (where $[\ ]$ is the floor, the greatest integer not exceeding the number), then which option is a subset of $S$.

For $x \in [3,6]$, $[x]$ can be $3, 4, 5,$ or $6$. Handle each band.

Step 1 — $[x] = 3$, i.e. $x \in [3,4)$. Need $[x^2] = 9$, so $x^2 \in [9, 10)$, i.e. $x \in [3, \sqrt{10})$.

Step 2 — $[x] = 4$, i.e. $x \in [4,5)$. Need $[x^2] = 16$, so $x^2 \in [16, 17)$, i.e. $x \in [4, \sqrt{17})$.

Step 3 — $[x] = 5$, i.e. $x \in [5,6)$. Need $[x^2] = 25$, so $x^2 \in [25, 26)$, i.e. $x \in [5, \sqrt{26})$.

Step 4 — $x = 6$. $[36] = 36 = 6^2$ ✓.

So $S = [3, \sqrt{10}) \cup [4, \sqrt{17}) \cup [5, \sqrt{26}) \cup \{6\}$.

💡 Teacher tip: mind the right endpoints. At $x = \sqrt{10}$, $x^2 = 10$ exactly, so $[x^2] = 10 \neq 9$ — the band is open there. Mishandling these edges is the classic trap.

Step 5 — "A possible subset" — check the options. An option is correct only if every point in it lies in $S$.

• $\sqrt{10} \notin S$ and $\sqrt{17} \notin S$ (checked above), so any option including them is out — this kills (b) and (c).
• Option (a) $(3, \sqrt{10}) \cup [5, \sqrt{26}) \cup \{6\}$: each piece sits inside $S$. It leaves out the $[4,\sqrt{17})$ band, but a subset is allowed to omit parts. ✅

Answer: $(3, \sqrt{10}) \cup [5, \sqrt{26}) \cup \{6\}$ (Choice A)

💡 Teacher tip: "a possible subset" is far weaker than "the set itself". After finding $S$, you must check which option is contained in $S$, not which one equals it.

What we want: the final acid percentage after three replacements on a $200$ L, $30\%$-acid solution.

Start: acid $= 30\%$ of $200 = 60$ L.

Step 1 — Replace $20\%$ with water. Removing $20\%$ of the mix keeps $80\%$ of the acid: $\text{acid} = 60 \times 0.8 = 48 \text{ L}.$

Step 2 — Replace $10\%$ with pure acid. Taking out $10\%$ removes $10\%$ of the acid ($4.8$ L); then $10\%$ of $200 = 20$ L of pure acid is poured in: $\text{acid} = 48 - 4.8 + 20 = 63.2 \text{ L}.$

💡 Teacher tip: two different replacement types appear here. "Refill with water" just scales the acid down by the fraction kept. "Refill with pure acid" scales and adds — handle the addition separately.

Step 3 — Replace $15\%$ with water. Keeps $85\%$ of the acid: $\text{acid} = 63.2 \times 0.85 = 53.72 \text{ L}.$

Step 4 — Final percentage. $\frac{53.72}{200}\times 100 = 26.86\% \approx 27\%.$

Answer: 27 (Choice D)

Think of it as: the roots must be two integers that add to $5$; each such pair produces a $k$. The catch is that different pairs can land on the same $k$, and we're counting distinct $k$ values.

The equation is $x^2 - 5x + k = 0$.

Step 1 — Vieta's relations. If the roots are integers $r_1, r_2$, then $r_1 + r_2 = 5, \qquad r_1 r_2 = k.$

Step 2 — List every integer pair summing to $5$. Write $r_1 = n$, $r_2 = 5 - n$; then $k = n(5 - n) = 5n - n^2$.

Step 3 — Impose $k \ge 0$. We need $n(5 - n) \ge 0$, which holds for $0 \le n \le 5$. So $n \in \{0, 1, 2, 3, 4, 5\}$.

Step 4 — Compute and dedupe.

The distinct values are $k \in \{0, 4, 6\}$ — three of them.

Answer: 3

💡 Why 3, not 6: the table is symmetric ($n$ and $5 - n$ give the same product), so the six pairs collapse to three distinct $k$ values. The question counts values of $k$, so remove duplicates.

What we want: how many integers $n$ satisfy $\log_{1/4}(n^2 - 7n + 11) > 0$.

Step 1 — Handle the fractional base carefully. The base is $\frac14$, which lies between $0$ and $1$. For such a base, $\log > 0$ exactly when the argument is between $0$ and $1$: $0 < n^2 - 7n + 11 < 1.$

💡 Teacher tip: with a base under $1$, the log reverses: "greater than $0$" wants the inside value squeezed into $(0,1)$, not made large. Always re-derive this rather than guessing.

Step 2 — Lower part: $n^2 - 7n + 11 > 0$. Roots are $n = \frac{7 \pm \sqrt5}{2} \approx 2.38, 4.62$, so this holds for integers $n \le 2$ or $n \ge 5$.

Step 3 — Upper part: $n^2 - 7n + 11 < 1$. That is $n^2 - 7n + 10 < 0$, i.e. $(n-2)(n-5) < 0$, which holds for integers $n = 3$ or $n = 4$.

Step 4 — Combine. We need $n$ that is both ($\le 2$ or $\ge 5$) and ($3$ or $4$). No integer can be in both — the two ranges don't overlap.

Answer: 0 (Choice D)

Think of it as: two parallel chords on opposite sides of the centre form a trapezium. We need both chord lengths (the parallel sides) and the gap between them (the height).

Circle: centre $C$, radius $r = 6\sqrt{2}$. Chords $PQ$ and $SR$ are parallel, on opposite sides of $C$. Given $\angle PQC = 45^\circ$ and the perpendicular distances of the two chords from $C$ are in ratio $3: 2$.

Step 1 — Chord $PQ$: its distance from $C$ and its length. Drop $CM \perp PQ$, where $M$ is the midpoint of $PQ$. Triangle $CMQ$ is right-angled at $M$, with $\angle CQM = 45^\circ$ and hypotenuse $CQ = r = 6\sqrt{2}$. A $45$–$45$–$90$ triangle gives $CM = QM = \frac{6\sqrt{2}}{\sqrt{2}} = 6.$ So the distance of $PQ$ from the centre is $d_1 = 6$, and $PQ = 2 \times QM = 12$.

Step 2 — Chord $SR$: its distance and length. The ratio $d_1: d_2 = 3: 2$ with $d_1 = 6$ gives $d_2 = 4$. Its half-length is $\sqrt{r^2 - d_2^2} = \sqrt{72 - 16} = \sqrt{56} = 2\sqrt{14}, \quad\text{so } SR = 4\sqrt{14}.$

Step 3 — Height of the trapezium. Because the chords sit on opposite sides of $C$, the distance between them is $d_1 + d_2 = 6 + 4 = 10$.

Step 4 — Area. With parallel sides $PQ = 12$ and $SR = 4\sqrt{14}$ and height $10$: $\text{Area} = \frac{1}{2}(12 + 4\sqrt{14})(10) = 5(12 + 4\sqrt{14}) = 60 + 20\sqrt{14} = 20(3 + \sqrt{14}).$

Answer: $20(3 + \sqrt{14})$ (Choice C)

💡 Teacher tip: a chord at distance $d$ from the centre has half-length $\sqrt{r^2 - d^2}$. If two parallel chords straddle the centre, their gap is $d_1 + d_2$; if they're on the same side, it's $d_1 - d_2$. Reading "separated by a diameter" as opposite sides is the whole game here.

What we want: the smallest natural number $m$ for which the right-hand product first beats the left-hand one, where $a_k = 3^k$.

Step 1 — Simplify the left side. $(a_1)^1 (a_2)^2 \cdots (a_{20})^{20} = \prod_{k=1}^{20}(3^k)^k = 3^{\,\sum_{k=1}^{20}k^2}.$ The exponent is $\sum_{k=1}^{20}k^2 = \dfrac{20\cdot 21\cdot 41}{6} = 2870.$

Step 2 — Simplify the right side. $a_{21}a_{22}\cdots a_{20+m} = 3^{\,\sum_{k=21}^{20+m}k}.$ The exponent is $\sum_{k=21}^{20+m}k = \dfrac{(20+m)(21+m)}{2} - 210.$

Step 3 — Compare exponents. Both are powers of $3$ (and $3 > 1$), so the left is smaller exactly when its exponent is smaller: $2870 < \frac{(20+m)(21+m)}{2} - 210 \Rightarrow (20+m)(21+m) > 6160.$

Step 4 — Find the smallest $m$. Let $n = 20 + m$; we need $n(n+1) > 6160$.

So $m = 58$.

Answer: 58 (Choice A)

💡 Teacher tip: when two products share the same base, ignore the base and compare exponents only. The formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$ do the rest.

Think of it as: three piles of money — stocks, bonds, gold — that add to ₹1,00,000, with bonds pinned to a quarter of gold. Two facts (total invested, total gained) are enough to unlock everything.

Step 1 — Name the piles. Let gold $= G$. Then bonds $= 0.25G$ and let stocks $= S$. Since the whole investment is ₹1,00,000: $S + 0.25G + G = 100000 \;\Rightarrow\; S + 1.25G = 100000. \qquad (1)$

Step 2 — Write the total-gain equation. The returns are $10\%$ on stocks, $6\%$ on bonds, $8\%$ on gold, and the total gain is ₹8,200: $0.10S + 0.06(0.25G) + 0.08G = 8200 \;\Rightarrow\; 0.10S + 0.095G = 8200. \qquad (2)$

Step 3 — Eliminate $S$. From $(1)$, $S = 100000 - 1.25G$. Put this into $(2)$: $0.10(100000 - 1.25G) + 0.095G = 8200 \;\Rightarrow\; 10000 - 0.125G + 0.095G = 8200,$ $10000 - 0.030G = 8200 \;\Rightarrow\; 0.030G = 1800 \;\Rightarrow\; G = 60000.$

Step 4 — Read off the bond gain. Bonds $= 0.25 \times 60000 = 15000$, so the gain from bonds is $0.06 \times 15000 = 900$.

Step 5 — Verify. Stocks $= 100000 - 1.25(60000) = 25000$. Total gain $= 0.10(25000) + 0.06(15000) + 0.08(60000) = 2500 + 900 + 4800 = 8200$ ✓.

Answer: ₹900

💡 Teacher tip: the question asks only for the bond gain, but you still need both equations — the total-invested line is what lets you eliminate stocks and expose gold.

Think of it as: we know three corners of a parallelogram; the fourth corner is forced by the fact that a parallelogram's diagonals cut each other exactly in half. Once we have that corner, the diagonal is just a line, and we ask where it crosses the $x$-axis.

Given $P(-3, -2)$, $Q(1, -5)$, $R(9, 1)$, in parallelogram $PQRS$.

Step 1 — Find $S$ from the "diagonals bisect" property. In $PQRS$ the diagonals are $PR$ and $QS$, and they share a midpoint. Midpoint of $PR$: $\left(\frac{-3 + 9}{2}, \frac{-2 + 1}{2}\right) = \left(3, -\tfrac{1}{2}\right).$ This must also be the midpoint of $QS$. With $Q(1, -5)$ and $S(x, y)$: $\frac{1 + x}{2} = 3 \Rightarrow x = 5, \qquad \frac{-5 + y}{2} = -\tfrac{1}{2} \Rightarrow y = 4.$ So $S = (5, 4)$.

Step 2 — Walk along diagonal $SQ$ and find $y = 0$. Travel from $S(5, 4)$ toward $Q(1, -5)$: $(x, y) = (5, 4) + t\big[(1, -5) - (5, 4)\big] = (5 - 4t,\; 4 - 9t).$ Set the height to zero: $4 - 9t = 0 \Rightarrow t = \dfrac{4}{9}$. Then $x = 5 - 4 \cdot \frac{4}{9} = 5 - \frac{16}{9} = \frac{29}{9}.$

Step 3 — Sanity check. $t = \tfrac{4}{9}$ lies between $0$ and $1$, so the crossing genuinely sits on segment $SQ$ ✓.

Answer: $\boxed{\dfrac{29}{9}}$ (Choice D)

💡 Teacher tip: to recover a missing parallelogram vertex, never bother with slopes or side lengths — just equate the midpoints of the two diagonals. One equation per coordinate.

What we want: the number of factors of the smallest three-digit number that fits all the digit rules.

Step 1 — Decide which digits are allowed. Digits are non-zero, so from $1$–$9$. None may be a perfect square, so throw out $1, 4, 9$. That leaves $\{2, 3, 5, 6, 7, 8\}.$ Among these, the primes are $\{2, 3, 5, 7\}$ and the non-primes (composites) are $\{6, 8\}$.

Step 2 — Apply "exactly one prime digit". We need three distinct digits, exactly one prime — so the other two must be non-primes. But the only non-primes available are $6$ and $8$, so both must be used, plus one prime.

Step 3 — Make the number as small as possible. Use the smallest prime, $2$. With digits $\{2, 6, 8\}$, the smallest arrangement puts the smallest digit first: $268$.

Step 4 — Count its factors. $268 = 2^2 \times 67 \quad (67 \text{ is prime}).$ Number of factors $= (2+1)(1+1) = 6$.

Answer: 6

💡 Teacher tip: the "two non-prime slots" plus "only $6$ and $8$ qualify" left no choice for two of the digits — constraints that force a unique digit set make these problems short.

What we want: the total number of distinct sandwich orders.

Step 1 — List the independent choices (independent choices multiply):

• Sandwich type: $5$.
• Bread: $4$.
• Size, small or large: $2$.
• Sauces: up to $2$ of the $6$ available.

Step 2 — Count the sauce options. "Up to $2$" means $0$, $1$, or $2$ sauces: $\binom60 + \binom61 + \binom62 = 1 + 6 + 15 = 22.$

Step 3 — Multiply everything. $5 \times 4 \times 2 \times 22 = 40 \times 22 = 880.$

Answer: 880 (Choice A)

💡 Teacher tip: "up to $k$" means add the combinations $\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{k}$ only up to $k$. Here it stops at $2$, so it's $22$, not the full $2^6 = 64$.

Think of it as: three snapshots of two shifts. At each stage, turn the ratio into real headcounts before moving on — chaining ratios in your head is where students slip.

Step 1 — Start from the first ratio. Let morning $= 13x$ and afternoon $= 9x$. When $21$ students move from morning to afternoon, the ratio becomes $19: 14$: $\frac{13x - 21}{9x + 21} = \frac{19}{14} \;\Rightarrow\; 14(13x - 21) = 19(9x + 21).$ $182x - 294 = 171x + 399 \;\Rightarrow\; 11x = 693 \;\Rightarrow\; x = 63.$

Step 2 — Actual counts after the move. Morning $= 13(63) - 21 = 798$, afternoon $= 9(63) + 21 = 588$. Check: $798: 588 = 19: 14$ ✓.

Step 3 — New students join in ratio $3: 8$. Let $3y$ join morning and $8y$ join afternoon. The ratio becomes $5: 4$: $\frac{798 + 3y}{588 + 8y} = \frac{5}{4} \;\Rightarrow\; 4(798 + 3y) = 5(588 + 8y).$ $3192 + 12y = 2940 + 40y \;\Rightarrow\; 252 = 28y \;\Rightarrow\; y = 9.$

Step 4 — Total new students. $3y + 8y = 11y = 11(9) = 99$.

Step 5 — Verify. Morning $= 798 + 27 = 825$, afternoon $= 588 + 72 = 660$, and $825: 660 = 5: 4$ ✓.

Answer: 99 (Choice D)

💡 Teacher tip: the moving $21$ students is a transfer (one side loses what the other gains); the joining students are fresh additions to both sides. Keeping those two mechanisms distinct is what keeps the arithmetic honest.

What we want: the distance of the fourth part, given the four speeds form one arithmetic progression (AP) and the four times form another.

First convert the first part to standard units: $960$ m/min $= 57.6$ km/h, for $30$ min $= 0.5$ h. So $s_1 = 57.6$, $t_1 = 0.5$.

Step 1 — Nail down all four times. The four times form an AP summing to $3$ h. In a 4-term AP, the outer pair and the inner pair have equal sums, and the whole sum is $2(t_1 + t_4)$. So $2(t_1 + t_4) = 3 \Rightarrow t_4 = 1.5 - 0.5 = 1.0.$ The common difference is $\frac{t_4 - t_1}{3} = \frac{0.5}{3} = \frac16$, giving times $\frac12, \frac23, \frac56, 1$.

Step 2 — Set up the speeds as an AP. $s_i = 57.6 + (i-1)\Delta$, with $\Delta$ the common speed step.

Step 3 — Use total distance $= 224$ km. Since each part's distance is speed $\times$ time, $\sum s_i t_i = 57.6\sum t_i + \Delta\sum (i-1)t_i = 224.$ Here $\sum t_i = 3$ and $\sum (i-1)t_i = 0\cdot\tfrac12 + 1\cdot\tfrac23 + 2\cdot\tfrac56 + 3\cdot 1 = \tfrac23 + \tfrac53 + 3 = \tfrac{16}{3}.$ So $57.6(3) + \Delta\cdot\frac{16}{3} = 224 \Rightarrow 172.8 + \frac{16}{3}\Delta = 224 \Rightarrow \Delta = 9.6$.

Step 4 — Fourth part. $s_4 = 57.6 + 3(9.6) = 86.4$ km/h, $t_4 = 1$ h, so distance $= 86.4$ km $= 86{,}400$ m.

Answer: 86400 (Choice D)

💡 Teacher tip: exploit "these values form an AP" first — it collapses several unknown speeds down to a single step $\Delta$ once the times are fixed.

What we want: the interest earned when the same sum is invested for $2$ years, compounded every $6$ months.

Step 1 — Extract the principal and rate from the simple-interest facts. Simple interest grows the amount by the same amount each year.

• Amount after $3$ years $= 13920$; after $6.5$ years $= 18960$.
• Interest over the $3.5$-year gap $= 18960 - 13920 = 5040$, so yearly interest $= 5040 / 3.5 = 1440$.
• Interest over the first $3$ years $= 3 \times 1440 = 4320$, so principal $= 13920 - 4320 = 9600$.
• Rate $= \frac{1440}{9600} = 15\%$ per year.

Step 2 — Reinvest at $15\%$, compounded half-yearly, for $2$ years. That's $4$ half-year periods, each at half the annual rate, $7.5\%$: $A = 9600(1.075)^4.$

Step 3 — Compute. $(1.075)^2 = 1.155625$, so $(1.075)^4 \approx 1.33547$. Then $A \approx 9600 \times 1.33547 \approx 12820.5, \qquad \text{interest} \approx 12820.5 - 9600 = 3220.5.$

Step 4 — Nearest option. $3220.5$ is closest to $3221$.

Answer: 3221 (Choice A)

💡 Teacher tip: for two simple-interest amounts at different times, their difference hands you the yearly interest directly — no need to solve for principal and rate separately first.

What we want: the cheapest way to finish the job in at most $10$ days, where any day a worker is used — even a part-day — is charged in full.

Step 1 — Find each worker's cost per unit of work. (Rate per day $\times$ full price for the days to do the whole job.)

• Arun: $\frac1{24}$/day at Rs $2160$ → $2160 \times 24 = 51{,}840$ per whole job.
• Varun: $\frac1{21}$/day at Rs $2400$ → $2400 \times 21 = 50{,}400$ per whole job.
• Tarun: $\frac1{15}$/day at Rs $2160$ → $2160 \times 15 = 32{,}400$ per whole job.

Tarun is both the cheapest per unit of work and the fastest — so use him as much as possible.

Step 2 — How much can Tarun do inside $10$ days? $\frac{10}{15} = \frac23$ of the job. So a second worker must cover the remaining $\frac13$.

Step 3 — Pick the cheaper partner for the last third. Between Arun and Varun, Varun is cheaper per unit, so use him. His time for $\frac13$ of the job is $\frac{1/3}{1/21} = 7$ days ($\le 10$ ✓).

Step 4 — Total cost. Tarun $10$ days, Varun $7$ days — both whole numbers, so no part-day waste: $10 \times 2160 + 7 \times 2400 = 21{,}600 + 16{,}800 = 38{,}400.$

💡 Teacher tip: swapping any Tarun day for Varun would need $1.4$ days → $2$ charged days (Rs $4800$) for the same work — strictly worse. So loading Tarun to the max really is optimal.

Answer: 38400 (Choice A)

What we want: a value of $c$ for which the lowest point of $f(x) = x^2 - 4cx + 8c$ sits above the highest point of $g(x) = -x^2 + 3cx - 2c$.

Step 1 — Minimum of $f$. $f$ opens upward, with its lowest point at $x = 2c$: $f_{\min} = f(2c) = 4c^2 - 8c^2 + 8c = -4c^2 + 8c.$

Step 2 — Maximum of $g$. $g$ opens downward, with its highest point at $x = \frac{3c}{2}$: $g_{\max} = -\frac{9c^2}{4} + \frac{9c^2}{2} - 2c = \frac{9c^2}{4} - 2c.$

Step 3 — Require $f_{\min} > g_{\max}$. $-4c^2 + 8c > \frac{9c^2}{4} - 2c.$ Multiply by $4$: $-16c^2 + 32c > 9c^2 - 8c \Rightarrow 25c^2 - 40c < 0 \Rightarrow 5c(5c - 8) < 0.$ This holds when $0 < c < \frac85 = 1.6$.

Step 4 — Pick the option in that range. Of $2,\ \frac12,\ -\frac12,\ -2$, only $\frac12$ lies in $(0, 1.6)$.

Answer: $\dfrac12$ (Choice B)

💡 Teacher tip: for $ax^2 + bx + c$, the peak/trough value is $\dfrac{-D}{4a}$ with $D = b^2 - 4ac$ — you don't have to find the vertex's $x$ first.