CAT 2022 Quant Questions — Slot 3

What we want: how many integer values the fourth side of the quadrilateral can take, given three sides $1, 2, 4$.

The key rule: four lengths form a genuine quadrilateral exactly when the longest side is strictly less than the sum of the other three. (A quadrilateral is floppy — unlike a triangle, it has only this one condition.)

Let the fourth side be $s$.

Case 1 — $s$ is the longest ($s \ge 4$). $s < 1 + 2 + 4 = 7 \;\Rightarrow\; s \in \{4, 5, 6\}.$

Case 2 — $4$ is the longest ($s \le 4$). $4 < 1 + 2 + s \;\Rightarrow\; s > 1 \;\Rightarrow\; s \in \{2, 3, 4\}.$

Combine. $s \in \{2, 3, 4, 5, 6\}$ — five values. A quick verification of the boundary cases:

Answer: 5 (Choice D)

💡 Teacher tip: A quadrilateral needs only "longest < sum of the other three", not the three pairwise checks of a triangle. That single, looser condition is why the count comes out larger than you might first expect.

What we want: the area of $\triangle EOD$, where medians $BD$ and $CE$ of $\triangle ABC$ meet at $O$, and area$(ABC) = 108$.

Since $BD$ and $CE$ are medians, $D$ is the midpoint of $AC$, $E$ is the midpoint of $AB$, and $O$ is the centroid.

Method 1 — Pure area ratios (clean and quick). As $E$ and $D$ are midpoints, segment $ED$ is a midsegment, so $ED \parallel BC$ and $\triangle AED \sim \triangle ABC$ with ratio $1: 2$. Similar triangles have areas in the square of the ratio: $\text{Area}(AED) = \frac{108}{4} = 27.$ The centroid $O$ divides the figure so that within $\triangle AED$, the sliver $\triangle EOD$ nearest $ED$ is exactly one-third of $\triangle AED$: $\text{Area}(EOD) = \frac13 \times 27 = 9.$

Method 2 — Coordinates (a reliable cross-check). Put $A$ at the origin, $B = (b_1, b_2)$, $C = (c_1, c_2)$. Then $E = \tfrac{B}{2}$, $D = \tfrac{C}{2}$, $O = \tfrac{B + C}{3}$. Working out the area of $\triangle EOD$ with the determinant formula gives $\text{Area}(EOD) = \frac{1}{24}\,|b_1 c_2 - b_2 c_1|.$ But area$(ABC) = \tfrac12 |b_1 c_2 - b_2 c_1| = 108$, so $|b_1 c_2 - b_2 c_1| = 216$, and $\text{Area}(EOD) = \frac{216}{24} = 9.$

Answer: $\boxed{9}$

💡 Teacher tip: Medians are full of fixed ratios — the centroid always sits two-thirds along each median. That's why these areas come out as tidy fractions of the whole ($\triangle EOD$ is exactly $\tfrac{1}{12}$ of $\triangle ABC$: $108/12 = 9$).

What we want: the interest Nitu would earn if the whole ₹20,000 were placed in bank C.

Step 1 — Find bank C's rate $x$. The amount in C is $20000 - 8000 - 5000 = 7000$. Her total interest equals $5\%$ of $20000 = ₹1000$: $\underbrace{8000 \times 0.055}_{440} + \underbrace{5000 \times 0.056}_{280} + 7000 \times \frac{x}{100} = 1000.$ $720 + 70x = 1000 \;\Rightarrow\; 70x = 280 \;\Rightarrow\; x = 4.$

Step 2 — Put everything in C at $4\%$. $20000 \times 0.04 = 800.$

Answer: ₹800 (Choice B)

💡 Teacher tip: The combined-interest condition is a weighted average in disguise — the overall $5\%$ is the weighted mean of $5.5\%,\ 5.6\%,\ 4\%$. Once $x = 4\%$ falls out, the final step is a single multiplication.

What we want: the value that $c = \dfrac{16x}{y} + \dfrac{49y}{x}$ can never equal, for non-zero reals $x, y$.

Step 1 — Collapse to one variable. Let $t = \dfrac{x}{y}$, which can be any non-zero real number. Then $c = 16t + \frac{49}{t}.$

Step 2 — Positive branch ($t > 0$). Both terms are positive, so by AM–GM: $16t + \frac{49}{t} \ge 2\sqrt{16t \cdot \frac{49}{t}} = 2\sqrt{16 \cdot 49} = 2 \times 28 = 56.$ Equality at $16t = \tfrac{49}{t} \Rightarrow t = \tfrac74$. So when $t > 0$, $c \ge 56$.

Step 3 — Negative branch ($t < 0$). Write $t = -u$ with $u > 0$: $c = -\left(16u + \frac{49}{u}\right) \le -56.$ So when $t < 0$, $c \le -56$.

Step 4 — The forbidden gap. Combining, $c \in (-\infty, -56] \cup [56, \infty)$. Everything strictly between $-56$ and $56$ is impossible. Now test the options:

Only $-50$ lands inside the forbidden gap.

Answer: −50 (Choice B)

💡 Teacher tip: Any expression of the form $at + \tfrac{b}{t}$ with $a, b > 0$ has a forbidden band $\left(-2\sqrt{ab},\ 2\sqrt{ab}\right)$ — the positive branch is trapped above $+2\sqrt{ab}$ and the negative branch below $-2\sqrt{ab}$. Here $2\sqrt{16 \cdot 49} = 56$ settles it instantly.

What we want: how far the slower ship is from the port when the faster ship arrives.

Let the port be $P$, and let the ships start at $S_1$ (slower) and $S_2$ (faster). Initially $PS_1 = PS_2 = S_1S_2 = 24$ — an equilateral triangle — so the two straight routes meet at $P$ at a $60^\circ$ angle.

Step 1 — Where does the right angle appear? After the slower ship sails $8$ km, it is $16$ km from $P$; let the faster ship be $d$ km from $P$. The angle at $P$ stays fixed at $60^\circ$, so the right angle must be at one of the ships.

• Right angle at $S_1$: then the side opposite (which is $PS_2 = d$) is the hypotenuse, forcing $d = 32$ — impossible, as a ship can't be farther than its $24$ km start.
• Right angle at $S_2$: then $PS_1 = 16$ is the hypotenuse. Using the cosine rule for the third side $S_1S_2$ and Pythagoras: $16^2 = d^2 + (16^2 + d^2 - 16d) \;\Rightarrow\; 2d^2 - 16d = 0 \;\Rightarrow\; d = 8.$

So when the slower ship is $16$ km out, the faster ship is $8$ km from $P$.

Step 2 — Speed ratio. In the same time, the slower ship covered $8$ km while the faster covered $24 - 8 = 16$ km — the faster ship is twice as fast.

Step 3 — Finish. The faster ship needs $8$ more km to dock. In that time the slower ship (half the speed) covers $4$ km, moving from $16$ km to $16 - 4 = 12$ km from the port.

Answer: 12 km (Choice B)

💡 Teacher tip: The angle at $P$ is locked at $60^\circ$ by the two routes, so the right angle must sit at a ship. Ruling out the impossible case leaves it on the faster ship's corner — after which it's a plain speed-ratio finish.

What we want: the total number of days Alex worked before the job finished.

Step 1 — Convert the speed comparisons into daily rates. Bob alone finishes in $40$ days, so Bob's rate is $\tfrac{1}{40}$ per day.

• Alex is twice as fast as Bob: $\tfrac{1}{20}$ per day.
• Alex is thrice as fast as Cole, so Cole is $\tfrac13$ of Alex: $\tfrac13 \times \tfrac{1}{20} = \tfrac{1}{60}$ per day.

Step 2 — Work done in one 3-day cycle.

Per full cycle: $\tfrac{3}{40} + \tfrac{1}{24} + \tfrac{1}{15} = \tfrac{9 + 5 + 8}{120} = \tfrac{22}{120} = \tfrac{11}{60}$.

Step 3 — Track the cycles. After $5$ full cycles: $5 \times \tfrac{11}{60} = \tfrac{55}{60} = \tfrac{11}{12}$ done, leaving $\tfrac{1}{12}$.

• Day 16 (Alex + Bob) adds $\tfrac{3}{40} = \tfrac{9}{120}$. Remaining: $\tfrac{1}{12} - \tfrac{3}{40} = \tfrac{10 - 9}{120} = \tfrac{1}{120}$.
• Day 17 (Bob + Cole) finishes the tiny remainder — Alex does not work this day.

Step 4 — Count Alex's days. Alex works on Day 1 and Day 3 of every cycle: $5 \text{ cycles} \times 2 + \underbrace{1}_{\text{Day 16}} = 11 \text{ days}.$

Answer: $\boxed{11}$

💡 Teacher tip: Compute the per-cycle work first, then handle the leftover day by day. Alex's tally is $2$ of every $3$ days, plus the one extra "Day 1" before the job closes on a "Day 2".

What we want: the maximum number of teams of $12$ that can be formed.

Step 1 — The "4 always left over" condition. If splitting into teams of $9, 10, 12, 25$ always leaves exactly $4$, then $N - 4$ is a common multiple of all four, hence a multiple of their LCM: $\text{lcm}(9, 10, 12, 25) = 2^2 \cdot 3^2 \cdot 5^2 = 900.$ So $N = 900k + 4$ for some whole number $k$.

Step 2 — The "teams of 11 leave nobody" condition. $N$ must be a multiple of $11$, with $N < 5000$, so $k \le 5$ (as $900 \times 5 + 4 = 4504$).

Only $N = 1804$ works.

Step 3 — Maximum teams of 12. Set aside the $4$ left over: $\frac{1804 - 4}{12} = \frac{1800}{12} = 150.$

Answer: $\boxed{150}$

💡 Teacher tip: "Same remainder $r$ for several divisors" means $N \equiv r \pmod{\text{lcm}}$. The extra condition (here divisibility by $11$) then singles out the one valid value of $N$.

What we want: the original number of workers, $N$.

The key principle: work done is proportional to (number of people) $\times$ (hours worked), since everyone works at the same rate.

Step 1 — Write the work in each phase.

Step 2 — Match the ratio of person-hours to the ratio of work. $\frac{65}{35} = \frac{140(N - 10)}{70N} = \frac{2(N - 10)}{N}.$ So $\frac{13}{7} = \frac{2(N - 10)}{N} \;\Rightarrow\; 13N = 14(N - 10) \;\Rightarrow\; N = 140.$

Check.: $140 \times 70 = 9800$ person-hours for $35\%$.: $130 \times 140 = 18200$ for $65\%$. Ratio $18200/9800 = 13/7$ ✓.

Answer: 140 (Choice D)

💡 Teacher tip: In "some workers leave mid-job" problems, set the person-hour ratio equal to the work-fraction ratio. The per-person rate cancels out, so you never need its actual value.

What we want: the distance the slower car covers before the two cars meet head-on.

Let the speeds be $60$ (slower) and $v$ (faster) km/h, and let the starting gap be $D$ km.

Step 1 — Two equations from the two scenarios.

• Towards each other, the gap closes at $60 + v$: $\dfrac{D}{60 + v} = 1.5$.
• Same direction, the faster car gains at $v - 60$: $\dfrac{D}{v - 60} = 10.5$.

Step 2 — Eliminate $D$. $1.5(60 + v) = 10.5(v - 60) \;\Rightarrow\; 90 + 1.5v = 10.5v - 630 \;\Rightarrow\; 720 = 9v \;\Rightarrow\; v = 80.$ So the faster car does $80$ km/h.

Step 3 — Distance of the slower car in $1.5$ hours. $60 \times 1.5 = 90 \text{ km}.$

Check. $D = 1.5(60 + 80) = 210$ km; in the same-direction case the gap closes at $20$ km/h, taking $210/20 = 10.5$ h ✓.

Answer: 90 km (Choice B)

💡 Teacher tip: For "meet head-on vs. same direction", the starting gap $D$ is the same in both, so $D = t_1(v_1 + v_2) = t_2(v_2 - v_1)$. The two times pin down the speeds directly.

What we want: the largest possible number of ₹500 cheques.

Let $a, b, c$ be the counts of ₹100, ₹250, ₹500 cheques.

Step 1 — Two equations. $a + b + c = 100, \qquad 100a + 250b + 500c = 15250.$ Divide the second by $50$: $2a + 5b + 10c = 305$.

Step 2 — Eliminate $a$. From the first, $a = 100 - b - c$: $2(100 - b - c) + 5b + 10c = 305 \;\Rightarrow\; 3b + 8c = 105.$ So $3b = 105 - 8c$. For $b$ to be a whole number, $8c$ must be a multiple of $3$, hence $c$ is a multiple of $3$. Write $c = 3m$: $b = \frac{105 - 24m}{3} = 35 - 8m, \qquad a = 100 - b - c = 65 + 5m.$

Step 3 — Non-negativity.

• $c = 3m \ge 0 \Rightarrow m \ge 0$,
• $b = 35 - 8m \ge 0 \Rightarrow m \le 4$,
• $a = 65 + 5m \ge 0$ (always true).

So $m \in \{0, 1, 2, 3, 4\}$, giving $c \in \{0, 3, 6, 9, 12\}$.

Step 4 — Maximum. The largest is $c = 12$ (with $a = 85,\ b = 3$).

Check. $85 + 3 + 12 = 100$ ✓; $100(85) + 250(3) + 500(12) = 8500 + 750 + 6000 = 15250$ ✓.

Answer: $\boxed{12}$

💡 Teacher tip: To carry the most ₹500 cheques, keep the constraint $b = 35 - 8m \ge 0$ in view — it caps $m$ at $4$, and hence $c$ at $12$.

What we want: the average of every 3-digit term in the arithmetic progression $38, 55, 72, \dots$.

Step 1 — Identify the progression. First term $38$, common difference $17$, so the $n$-th term is $a_n = 38 + 17(n - 1).$

Step 2 — First 3-digit term. $38 + 17(n - 1) \ge 100 \;\Rightarrow\; 17(n - 1) \ge 62 \;\Rightarrow\; n - 1 \ge 3.65 \;\Rightarrow\; n \ge 5.$ So $a_5 = 38 + 17 \times 4 = 106$ — the first 3-digit term.

Step 3 — Last 3-digit term. $38 + 17(n - 1) \le 999 \;\Rightarrow\; 17(n - 1) \le 961 \;\Rightarrow\; n - 1 \le 56.5 \;\Rightarrow\; n \le 57.$ So $a_{57} = 38 + 17 \times 56 = 990$ — the last 3-digit term.

Step 4 — Average. The 3-digit terms $106, 123, \dots, 990$ form an AP of their own, and the average of any AP is the average of its first and last terms: $\frac{106 + 990}{2} = \frac{1096}{2} = 548.$

Answer: $\boxed{548}$

💡 Teacher tip: A block of consecutive AP terms is itself an AP, so its mean is just $\tfrac{\text{first} + \text{last}}{2}$ — no need to count or sum the terms.

What we want: all real $x$ for which $f(x) = f(f(x))$, where $f(x) = \begin{cases} 2x - r, & x \ge r, \\ r, & x < r. \end{cases}$

Step 1 — Case $x < r$. Here $f(x) = r$. Now feed that in: $f(f(x)) = f(r)$, and since $r \ge r$, the top rule gives $f(r) = 2r - r = r$. So $f(x) = r = f(f(x)),$ which holds for every $x < r$.

Step 2 — Case $x \ge r$. Here $f(x) = 2x - r \ge 2r - r = r$, so the inner output is also $\ge r$ and the top rule applies again: $f(f(x)) = f(2x - r) = 2(2x - r) - r = 4x - 3r.$ Set the two equal: $2x - r = 4x - 3r \;\Rightarrow\; 2x = 2r \;\Rightarrow\; x = r.$ So on this side it holds only at $x = r$.

Step 3 — Combine. All $x < r$, together with the single point $x = r$, is exactly $x \le r.$

Answer: $x \le r$ (Choice B)

💡 Teacher tip: On the lower branch $f$ is constant ($=r$), and applying any function to a constant leaves it unchanged — so the equation is automatically satisfied there. On the upper branch $f$ is truly linear, and "apply once = apply twice" holds only at the meeting point $x = r$.

What we want: the number of integers $k$ for which $2x^2 + kx + 5 = 0$ has no real roots and $x^2 + (k-5)x + 1 = 0$ has two distinct real roots.

Step 1 — First condition (no real roots ⇒ discriminant $< 0$). $k^2 - 4(2)(5) < 0 \;\Rightarrow\; k^2 < 40 \;\Rightarrow\; -\sqrt{40} < k < \sqrt{40}.$ Since $\sqrt{40} \approx 6.32$, the integers are $k \in \{-6, -5, \dots, 5, 6\}$ — $13$ values.

Step 2 — Second condition (two distinct roots ⇒ discriminant $> 0$). $(k-5)^2 - 4 > 0 \;\Rightarrow\; (k-5)^2 > 4 \;\Rightarrow\; k - 5 > 2 \text{ or } k - 5 < -2.$ So $k > 7$ or $k < 3$, i.e. $k \le 2$ or $k \ge 8$.

Step 3 — Intersect the two conditions. We need $k \in \{-6, \dots, 6\}$ and ($k \le 2$ or $k \ge 8$). The branch $k \ge 8$ contributes nothing inside $\{-6, \dots, 6\}$, leaving $k \le 2$: $k \in \{-6, -5, -4, -3, -2, -1, 0, 1, 2\}.$

Step 4 — Count. That is $9$ values.

Answer: 9 (Choice A)

💡 Teacher tip: "Two distinct real roots" is a strict $>0$ discriminant; "no real roots" is $<0$. Sketch both allowed ranges on a number line and the overlap is obvious.

What we want: the area of the lens-shaped region shared by the two circles.

Step 1 — Read the geometry. Since $A$ lies on the circle with diameter $BC$, the angle in a semicircle gives $\angle BAC = 90^\circ$. With $AB = AC = 8$, triangle $ABC$ is right-angled and isosceles at $A$, so $BC = \sqrt{8^2 + 8^2} = 8\sqrt2.$

The distance between centres is $AM = \tfrac12 BC = 4\sqrt2$ (the midpoint of a right triangle's hypotenuse is equidistant from all three vertices). Both circles pass through $B$ and $C$, which are the two crossing points.

Step 2 — Apply the two-circle overlap formula with $d = 4\sqrt2$, $r_1 = 4\sqrt2$, $r_2 = 8$: $\text{Overlap} = r_1^2\cos^{-1}\!\left(\tfrac{d^2 + r_1^2 - r_2^2}{2 d r_1}\right) + r_2^2\cos^{-1}\!\left(\tfrac{d^2 + r_2^2 - r_1^2}{2 d r_2}\right) - \tfrac12\sqrt{(-d + r_1 + r_2)(d + r_1 - r_2)(d - r_1 + r_2)(d + r_1 + r_2)}.$

• First term: $\tfrac{32 + 32 - 64}{2 \times 32} = 0$, so $\cos^{-1}(0) = \tfrac{\pi}{2}$, giving $32 \times \tfrac{\pi}{2} = 16\pi$.
• Second term: $\tfrac{32 + 64 - 32}{2 \times 4\sqrt2 \times 8} = \tfrac{1}{\sqrt2}$, so $\cos^{-1}\!\left(\tfrac{1}{\sqrt2}\right) = \tfrac{\pi}{4}$, giving $64 \times \tfrac{\pi}{4} = 16\pi$.
• Root term: $\tfrac12 \sqrt{8(8\sqrt2 - 8)\,8\,(8\sqrt2 + 8)} = \tfrac12 \times 8\sqrt{(8\sqrt2)^2 - 8^2} = 4\sqrt{128 - 64} = 4 \times 8 = 32$.

$\text{Overlap} = 16\pi + 16\pi - 32 = 32\pi - 32 = 32(\pi - 1).$

Answer: $\boxed{32(\pi - 1)}$ (Choice C)

💡 Teacher tip: The right-isosceles symmetry makes every angle in the formula a neat multiple of $\tfrac{\pi}{4}$ or $\tfrac{\pi}{2}$, which is exactly why the messy overlap formula collapses to the clean $32(\pi - 1)$.

What we want: the value of $x + y$ from two index (exponent) equations.

Step 1 — Simplify the first equation. Break $\tfrac{875}{2401}$ into primes: $\frac{875}{2401} = \frac{5^3 \cdot 7}{7^4} = \frac{5^3}{7^3} = \left(\frac{5}{7}\right)^3 = \left(\frac{7}{5}\right)^{-3}.$ Also $\sqrt{\tfrac75} = \left(\tfrac75\right)^{1/2}$, so the first equation becomes \left(\frac75\right)^{(3x - y)/2} = \left(\frac75\right)^{-3} \;\Rightarrow\; \frac{3x - y}{2} = -3 \;\Rightarrow\; 3x - y = -6. \tag{1}

Step 2 — Simplify the second equation. It must hold for all non-zero $a, b$. Let $r = \tfrac{2a}{b}$ (any non-zero real); then $\tfrac{4a}{b} = 2r$, and the equation reads $(2r)^{6x - y} = r^{y - 6x} = r^{-(6x - y)}.$ Put $A = 6x - y$. Then $(2r)^A = r^{-A}$, i.e. $2^A r^A = r^{-A}$, so $2^A = r^{-2A}$. For this to be true for every $r$, the power of $r$ must vanish: $-2A = 0$, giving 6x - y = 0 \;\Rightarrow\; y = 6x. \tag{2}

Step 3 — Solve (1) and (2). Substitute $y = 6x$ into (1): $3x - 6x = -6 \Rightarrow x = 2$, so $y = 12$. $x + y = 2 + 12 = 14.$

Check. $3x - y = 6 - 12 = -6$ ✓, and $6x - y = 12 - 12 = 0$ ✓.

Answer: $\boxed{14}$

💡 Teacher tip: When a power identity must hold for all values of a base, the base is powerless — the exponents have to arrange themselves so the base drops out. Here that forces $6x - y = 0$.

What we want: the gap between the largest and smallest possible number of students in section A, given section A averages $32$, section B averages $60$, section A has $10$ fewer students than B, and the combined average is a whole number.

Let section A have $n$ students; then section B has $n + 10$.

Step 1 — Combined average. $\text{avg} = \frac{32n + 60(n + 10)}{n + (n + 10)} = \frac{92n + 600}{2n + 10} = \frac{46n + 300}{n + 5}.$ Split off the whole part: $\frac{46n + 300}{n + 5} = 46 + \frac{300 - 46 \times 5}{n + 5} = 46 + \frac{70}{n + 5}.$

Step 2 — Force it to be an integer. Since $46$ is already whole, we need $\dfrac{70}{n + 5}$ to be an integer, i.e. $(n + 5)$ must divide $70$. The divisors of $70$ are $1, 2, 5, 7, 10, 14, 35, 70$, so $n + 5 \in \{1, 2, 5, 7, 10, 14, 35, 70\} \;\Rightarrow\; n \in \{-4, -3, 0, 2, 5, 9, 30, 65\}.$ A section must have students, so drop the non-positive values: $n \in \{2, 5, 9, 30, 65\}$.

Step 3 — Range. Largest $n = 65$, smallest $n = 2$, difference $= 63$.

Answer: $\boxed{63}$

💡 Teacher tip: Rewriting $\tfrac{46n + 300}{n + 5}$ as $46 + \tfrac{70}{n + 5}$ turns "when is this an integer?" into "what are the divisors of 70?" The constant leftover, $70$, is exactly what bounds the answer set.

Framing: six distinct natural numbers, with the two smallest averaging $14$ and the two largest averaging $28$. We want the largest possible average of all six.

Order them $a < b < c < d < e < f$.

• Two smallest: $\tfrac{a + b}{2} = 14 \Rightarrow a + b = 28$.
• Two largest: $\tfrac{e + f}{2} = 28 \Rightarrow e + f = 56$.

The total is $a + b + c + d + e + f = 28 + (c + d) + 56 = 84 + (c + d)$. To maximise the average, maximise the middle pair $c + d$.

Step 1 — Push $e$ as high as possible. Since $e < f$ and $e + f = 56$, the largest $e$ can be is $27$ (then $f = 29$; $e = 28$ would force $f = 28$, not distinct).

Step 2 — Push $c, d$ up. With $e = 27$, the two largest distinct values below it are $d = 26$ and $c = 25$.

Step 3 — Check the small end fits. We need $a < b < c = 25$ with $a + b = 28$. Take $b = 24,\ a = 4$: indeed $4 < 24 < 25$ ✓.

The set $\{4, 24, 25, 26, 27, 29\}$ works. Its total is $84 + (25 + 26) = 135$, so the average is $\frac{135}{6} = 22.5.$

Answer: 22.5 (Choice D)

💡 Teacher tip: When both ends are pinned by fixed sums and you want the middle large, first raise the upper boundary ($e$) as high as it will go — that lifts the ceiling on $c$ and $d$. The small-pair sum of $28$ is what drags the average down to $22.5$ rather than $24$.

What we want: the minimum of $\dfrac{x^2 - 6x + 10}{3 - x}$ for $x < 3$.

Step 1 — Substitute $t = 3 - x$. Since $x < 3$, we have $t > 0$, and $x = 3 - t$. The numerator becomes $x^2 - 6x + 10 = (3 - t)^2 - 6(3 - t) + 10 = 9 - 6t + t^2 - 18 + 6t + 10 = t^2 + 1.$ So the whole expression is $\frac{t^2 + 1}{t} = t + \frac1t, \qquad t > 0.$

Step 2 — Minimise $t + \tfrac1t$. By AM–GM, for $t > 0$: $t + \frac1t \ge 2\sqrt{t \cdot \frac1t} = 2,$ with equality at $t = 1$, i.e. $x = 2$ (which satisfies $x < 3$ ✓).

Answer: 2 (Choice B)

💡 Teacher tip: Completing the square, $x^2 - 6x + 10 = (x - 3)^2 + 1$, is a signpost pointing straight to $t = 3 - x$. After the swap, the ungainly fraction becomes the textbook $t + \tfrac1t$, whose minimum is $2$.

What we want: after a two-way exchange, the ratio of (water now in the glass) to (milk now in the cup).

Start: glass has $500$ cc milk; cup has $500$ cc water.

Step 1 — First transfer (150 cc milk → cup).

• Glass: $500 - 150 = 350$ cc milk, no water.
• Cup: $500$ cc water $+ 150$ cc milk $= 650$ cc. Milk fraction $= \tfrac{150}{650} = \tfrac{3}{13}$; water fraction $= \tfrac{10}{13}$.

Step 2 — Second transfer (150 cc of the mixture → glass). In that $150$ cc: $\text{milk} = 150 \times \frac{3}{13} = \frac{450}{13}, \qquad \text{water} = 150 \times \frac{10}{13} = \frac{1500}{13}.$

Step 3 — Read off the two quantities asked for.

• Water now in the glass $= \dfrac{1500}{13}$.
• Milk now in the cup $= 150 - \dfrac{450}{13} = \dfrac{1950 - 450}{13} = \dfrac{1500}{13}$.

They are equal, so the ratio is $1: 1$.

Answer: 1: 1 (Choice A)

💡 Teacher tip: Whenever you move an equal volume out and then back ($150$ cc each way), the water that ends up in the milk container exactly equals the milk that ends up in the water container — a conservation fact, true regardless of the numbers.

What we want: $\dfrac{b}{m} + \dfrac{c - 2b}{n}$, where two quadratics with the same leading coefficient $a$ have the given irrational roots and all coefficients are integers.

Step 1 — Use the conjugate-root fact. A quadratic with rational (here integer) coefficients cannot have an irrational root $p + q\sqrt{d}$ without also having its conjugate $p - q\sqrt{d}$. So:

• Roots of $ax^2 + bx + c$: $3 + 2\sqrt2$ and $3 - 2\sqrt2$. Sum $= 6 = -\tfrac{b}{a} \Rightarrow b = -6a$. Product $= 9 - 8 = 1 = \tfrac{c}{a} \Rightarrow c = a$.
• Roots of $ay^2 + my + n$: $4 + 2\sqrt3$ and $4 - 2\sqrt3$. Sum $= 8 = -\tfrac{m}{a} \Rightarrow m = -8a$. Product $= 16 - 12 = 4 = \tfrac{n}{a} \Rightarrow n = 4a$.

Step 2 — Substitute (the common $a$ cancels). $\frac{b}{m} = \frac{-6a}{-8a} = \frac34, \qquad \frac{c - 2b}{n} = \frac{a - 2(-6a)}{4a} = \frac{13a}{4a} = \frac{13}{4}.$

Step 3 — Add. $\frac34 + \frac{13}{4} = \frac{16}{4} = 4.$

Answer: 4 (Choice D)

💡 Teacher tip: Every coefficient here is a multiple of $a$, so the final expression depends only on ratios of the roots' sums and products — the actual value of $a$ never matters.