CAT 2024 Quant Questions — Slot 1

What we want: the total number of unsold fruits.

Step 1 — Name the quantities. Apples and mangoes are in the ratio $5: 2$, so let apples $= 5k$ and mangoes $= 2k$. Since everything totals $187$, oranges $= 187 - 7k$.

After the day's sales:

• unsold apples $= 5k - 75$,
• unsold mangoes $= 2k - 26$,
• unsold oranges $= \dfrac{187 - 7k}{2}$ (half the oranges are sold).

Step 2 — Apply the unsold apples: oranges $= 3: 2$ condition. $\frac{5k - 75}{(187 - 7k)/2} = \frac{3}{2} \;\Rightarrow\; 4(5k - 75) = 3(187 - 7k).$ $20k - 300 = 561 - 21k \;\Rightarrow\; 41k = 861 \;\Rightarrow\; k = 21.$

Step 3 — Fill in the actual counts. Apples $= 105$, mangoes $= 42$, oranges $= 187 - 147 = 40$. So unsold apples $= 105 - 75 = 30$, unsold mangoes $= 42 - 26 = 16$, unsold oranges $= 40/2 = 20$. $\text{Total unsold} = 30 + 16 + 20 = 66.$

Check: unsold apples: oranges $= 30: 20 = 3: 2$ ✓.

💡 Teacher tip: When a ratio and a grand total are both given, write the ratio parts with a single unknown $k$ and let the leftover category ($187 - 7k$ here) soak up the rest. One condition then pins down $k$.

Answer: $\boxed{66}$

Setting the scene: one condition describes a ring-shaped region; the other is a straight line. We want the area of the part of the ring on one side of the line.

Step 1 — Tidy up the quadratic condition by completing the square. $x^2 + 4x + y^2 - 4y = (x+2)^2 - 4 + (y-2)^2 - 4 = (x+2)^2 + (y-2)^2 - 8.$ So $-4 \le (x+2)^2 + (y-2)^2 - 8 \le 0$ turns into $4 \le (x+2)^2 + (y-2)^2 \le 8.$ This is a ring (annulus) centred at $(-2, 2)$: outside a circle of radius $2$ and inside a circle of radius $2\sqrt2$.

Step 2 — See where the line sits. The line $y = x + 4$ passes through $(-2, 2)$ — exactly the centre of the ring. A straight line through the centre slices any ring into two equal halves.

Step 3 — Take half the ring's area. Ring area $= \pi(2\sqrt2)^2 - \pi(2)^2 = 8\pi - 4\pi = 4\pi$. The region we want is the half satisfying $y \ge x+4$, so its area is $2\pi$.

💡 Teacher tip: Completing the square on any $x^2 + y^2 + Dx + Ey$ expression instantly reveals a circle's centre and radius. And always check whether a cutting line runs through the centre — if it does, a symmetric shape is halved exactly.

Answer: $\boxed{2\pi}$ — Choice A

The setup: three people's September incomes are $8: 6: 5$, and they pay fixed percentages of income as rent. In October incomes rise but the rent (in rupees) stays the same. Find October rent as a percentage of October total income.

Let the September incomes be $8x, 6x, 5x.$

Step 1 — Rent (fixed in rupees). In September, rent $= 15\%$ of Kamal $+ 12\%$ of Amal $+ 18\%$ of Vimal: $0.15(8x) + 0.12(6x) + 0.18(5x) = 1.2x + 0.72x + 0.9x = 2.82x.$ This is the same rupee amount in October.

Step 2 — October total income (up $10\%$, $12\%$, $15\%$ respectively): $8x(1.10) + 6x(1.12) + 5x(1.15) = 8.80x + 6.72x + 5.75x = 21.27x.$

Step 3 — Rent as a fraction of October income. $\frac{2.82x}{21.27x} \times 100\% \approx 13.26\%.$

💡 Teacher tip: Carry a placeholder $x$ for the incomes — it cancels in the final ratio, so you never need actual rupee figures. Only the percentage multipliers matter.

Answer: $\boxed{13.26\%}$ — Choice B

What we want: how many years the money stays in bank A.

Step 1 — Interest earned in bank A. On ₹10,000 at 5% simple interest for $T$ years, $I_A = 10000 \times 0.05 \times T = 500T,$ so the maturity amount from A is $10000 + 500T$.

Step 2 — Interest earned in bank B. That whole maturity amount is invested at 6% simple interest for 5 years: $I_B = (10000 + 500T)\times 0.06 \times 5 = (10000 + 500T)\times 0.3 = 3000 + 150T.$

Step 3 — Apply the ratio $I_A: I_B = 10: 13$. $\frac{500T}{3000 + 150T} = \frac{10}{13} \;\Rightarrow\; 13 \times 500T = 10(3000 + 150T).$ $6500T = 30000 + 1500T \;\Rightarrow\; 5000T = 30000 \;\Rightarrow\; T = 6.$

💡 Teacher tip: In simple interest there is no compounding, so bank A's maturity amount feeds straight into bank B's principal. Keep everything in terms of $T$ and the ratio finishes the job.

Answer: $\boxed{6}$ — Choice D

The setup: the remainder when $10^{100}$ is divided by $7$. Big powers with a modulus are handled by finding the repeating cycle.

Step 1 — Reduce the base. $10 \equiv 3 \pmod 7$, so $10^{100} \equiv 3^{100} \pmod 7.$

Step 2 — Find the cycle of $3^k \bmod 7$.

The powers repeat every $6$ (since $3^6 \equiv 1$).

Step 3 — Reduce the exponent mod $6$. $100 = 6 \times 16 + 4$, so $3^{100} \equiv 3^4 \equiv 4 \pmod 7.$

💡 Teacher tip: When the base and modulus share no common factor, Fermat's little theorem gives $a^{p-1} \equiv 1 \pmod p$. Here $3^6 \equiv 1 \pmod 7$, so only the exponent's remainder on division by $6$ matters.

Answer: $\boxed{4}$ — Choice B

What we want: Anil's speed, given the two towns are 45 km apart.

Let Anil's speed be $v_a$ and Sunil's be $v_s$ (km/h).

Step 1 — They meet after 1 h 30 min. Between them they cover the whole 45 km in $1.5$ hours: $(v_a + v_s)\times 1.5 = 45 \;\Rightarrow\; v_a + v_s = 30.$

Step 2 — The arrival gap. Anil reaches B a full $1$ h $15$ min $= 1.25$ hours after Sunil reaches A: $\frac{45}{v_a} - \frac{45}{v_s} = 1.25.$

Step 3 — Solve. With $v_s = 30 - v_a$, clearing denominators leads to $v_a^2 - 21v_a + 108 = 0 = (v_a - 9)(v_a - 12).$ The two roots, $9$ and $12$, simply swap the roles of Anil and Sunil. Testing $v_a = 12$ (so $v_s = 18$): Anil takes $45/12 = 3.75$ h, Sunil takes $45/18 = 2.5$ h, and the gap is $3.75 - 2.5 = 1.25$ h ✓. So Anil is the slower traveller at $12$ km/h.

💡 Teacher tip: In "meet, then arrive" problems the meeting time fixes the sum of speeds and the arrival gap fixes the difference of the two travel times. In an MCQ, plugging in an option is usually faster than grinding the quadratic.

Answer: $\boxed{12}$ — Choice D

What we want: the largest value of $a + b + n$, where $a + b\sqrt n = \sqrt{29 - 12\sqrt5}$ with $a, b$ integers and $n$ a natural number.

Step 1 — Un-nest the square root. Try to write $\sqrt{29 - 12\sqrt5}$ as $\sqrt x - \sqrt y$. Squaring: $x + y = 29$ and $2\sqrt{xy} = 12\sqrt5$, so $xy = 180$. Then $x, y$ are roots of $t^2 - 29t + 180 = 0 = (t - 20)(t - 9),$ giving $x = 20$, $y = 9$. So $\sqrt{29 - 12\sqrt5} = \sqrt{20} - \sqrt{9} = 2\sqrt5 - 3.$ This equals $a + b\sqrt n = -3 + 2\sqrt5$, which is positive ($\approx 1.47$) ✓.

Step 2 — All valid integer forms. We need $b\sqrt n = 2\sqrt5$, keeping $n$ a natural number and $a = -3$ fixed. That means $n = 20/b^2$:

For $|b| \ge 3$, $n$ is no longer a whole number, so only these two work.

Step 3 — Take the maximum. $\max(a + b + n) = 18$.

💡 Teacher tip: The quantity $2\sqrt5$ can be written as $2\sqrt5$ (giving $n = 5$) or as $1\cdot\sqrt{20}$ (giving $n = 20$). "Maximum" simply favours pushing the square factor inside the root to make $n$ as big as possible.

Answer: $\boxed{18}$ — Choice A

What we want: the gap between the largest and smallest of four numbers.

Call the numbers $a, b, c, d$ and translate each sentence.

Step 1 — "Average of the first two is 1 more than the first number." $\frac{a + b}{2} = a + 1 \;\Rightarrow\; a + b = 2a + 2 \;\Rightarrow\; b = a + 2.$

Step 2 — "Average of the first three is 2 more than the average of the first two." The average of the first two is $a + 1$, so $\frac{a + b + c}{3} = (a + 1) + 2 = a + 3 \;\Rightarrow\; a + b + c = 3a + 9.$ Using $b = a + 2$, this gives $c = a + 7$.

Step 3 — "Average of the first four is 3 more than the average of the first three." The average of the first three is $a + 3$, so $\frac{a + b + c + d}{4} = a + 6 \;\Rightarrow\; a + b + c + d = 4a + 24.$ Since $a + b + c = 3a + 9$, we get $d = a + 15$.

Step 4 — Read off the spread. The four numbers are $a,\; a+2,\; a+7,\; a+15$. The largest minus the smallest is $(a + 15) - a = 15.$

💡 Teacher tip: Each condition pins the next number as a fixed shift from $a$. The gaps come out $2, 7, 15$ — and since both extremes shift with $a$ together, the answer doesn't depend on $a$ at all.

Answer: $\boxed{15}$

What we want: the starting quantity of grain.

Each customer buys "half of the current stock, plus 3 kg more", and after the third customer nothing is left. Working backwards is cleanest.

Step 1 — Just before the third customer. If the stock was $r_3$, the customer took $\tfrac12 r_3 + 3$, leaving $\tfrac12 r_3 - 3 = 0$, so $r_3 = 6$.

Step 2 — Just before the second customer. After that customer, 6 kg remained: $\tfrac12 r_2 - 3 = 6 \Rightarrow r_2 = 18$.

Step 3 — Just before the first customer. After that customer, 18 kg remained: $\tfrac12 r_1 - 3 = 18 \Rightarrow r_1 = 42$.

Forward check. Start 42 → customer 1 takes $21 + 3 = 24$, leaves 18 → customer 2 takes $9 + 3 = 12$, leaves 6 → customer 3 takes $3 + 3 = 6$, leaves 0 ✓.

💡 Teacher tip: For "halve, then subtract" processes that end at zero, reverse each step: first undo the "$+3$" by adding 3, then undo the halving by doubling. Running that backwards: $0 \to 3 \to 6 \to 9 \to 18 \to 21 \to 42$.

Answer: $\boxed{42}$ — Choice C

The setup: $a_n = \lfloor \sqrt n \rfloor$, the largest integer not exceeding $\sqrt n$. We want $a_1 + a_2 + \cdots + a_{50}$.

Step 1 — Group by the value of the floor. $\lfloor \sqrt n \rfloor = j$ exactly when $j^2 \le n \le (j+1)^2 - 1$, a block of $(j+1)^2 - j^2 = 2j + 1$ consecutive integers, each contributing $j$ to the sum.

Step 2 — List the blocks up to $50$.

The last block is cut short: normally $j = 7$ would run $49$–$63$, but we stop at $50$, giving only $2$ terms.

Step 3 — Add up. $3 + 10 + 21 + 36 + 55 + 78 + 14 = 217.$

💡 Teacher tip: For floor-of-square-root sums, split by the value of the floor — each value $j$ covers a block of $2j + 1$ numbers. The only care needed is the top block, which is usually truncated by the upper limit.

Answer: $\boxed{217}$

What we want: the volume of the sphere in which the box is inscribed.

Let the box have dimensions $l, w, h$.

Step 1 — Use the total edge length. A box has four edges of each dimension, so $4(l + w + h) = 144 \;\Rightarrow\; l + w + h = 36.$

Step 2 — Use the surface area. $2(lw + wh + hl) = 846 \;\Rightarrow\; lw + wh + hl = 423.$

Step 3 — Find the space diagonal without finding $l, w, h$. Use the identity $(l + w + h)^2 = l^2 + w^2 + h^2 + 2(lw + wh + hl).$ $36^2 = (l^2 + w^2 + h^2) + 2(423) \;\Rightarrow\; 1296 = (l^2 + w^2 + h^2) + 846,$ so $l^2 + w^2 + h^2 = 450$ and the space diagonal is $d = \sqrt{450} = 15\sqrt2$.

Step 4 — The sphere. For a box inscribed in a sphere, the space diagonal is the diameter, so the radius is $R = \tfrac{15\sqrt2}{2}$. Then $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi\left(\frac{15\sqrt2}{2}\right)^3 = \frac{4}{3}\pi \cdot \frac{3375 \cdot 2\sqrt2}{8} = 1125\sqrt2\,\pi.$

💡 Teacher tip: The two symmetric sums $l+w+h$ and $lw+wh+hl$, plugged into the square-of-a-sum identity, hand you $l^2+w^2+h^2$ — the space diagonal — without ever solving for the individual sides.

Answer: $\boxed{1125\pi\sqrt2}$ — Choice C

What we want: the original selling price of the product.

Let the original cost price be $C$.

Step 1 — The first situation. A 40% profit means the selling price is $1.4C$.

Step 2 — The second situation. The cost is 40% lower, so the new cost is $0.6C$. A 50% profit on that is $1.5 \times 0.6C = 0.9C.$ But this new selling price is also described as ₹5 less than the original, i.e. $1.4C - 5$. Setting the two descriptions equal: $1.4C - 5 = 0.9C \;\Rightarrow\; 0.5C = 5 \;\Rightarrow\; C = 10.$

Step 3 — Original selling price. $1.4C = 1.4 \times 10 = 14.$

💡 Teacher tip: Anchor everything on the cost price. "Cost 40% less" scales $C$ by $0.6$; "sold for ₹5 less" subtracts $5$ from the original price. Equate the two ways of writing the new price and $C$ pops out.

Answer: $\boxed{14}$ — Choice B

Setting the scene: $E$ is the midpoint of the top side of the rectangle, and we want the radius of the circle that fits snugly inside triangle $ADE$.

Step 1 — Read off the three sides of $\triangle ADE$. Place the rectangle with $A$ at the origin: $A=(0,0)$, $B=(56,0)$, $C=(56,45)$, $D=(0,45)$. Since $E$ is the midpoint of $CD$, it sits at $E=(28,45)$. Now:

• $AD = 45$ (the full left side),
• $DE = 28$ (half of the top side $CD = 56$),
• $AE = \sqrt{28^2 + 45^2} = \sqrt{784 + 2025} = \sqrt{2809} = 53.$

Step 2 — Notice the right angle. Because $28^2 + 45^2 = 2809 = 53^2$, the sides $AD$ and $DE$ meet at a right angle at $D$. (They are the vertical and horizontal sides of the rectangle, so this is no accident.)

Step 3 — Use the right-triangle inradius shortcut. For a right triangle with legs $p$ and $q$ and hypotenuse $h$, $r = \frac{p + q - h}{2} = \frac{28 + 45 - 53}{2} = \frac{20}{2} = 10.$

💡 Teacher tip: That formula isn't magic. Area $= \tfrac12 pq$ also equals $r \times s$ where $s = \tfrac{p+q+h}{2}$ is the semi-perimeter. Solving for $r$ gives $\tfrac{pq}{p+q+h}$, which simplifies to $\tfrac{p+q-h}{2}$.

Answer: $\boxed{10}$ cm

The setup: an equation with fractional powers. Everything in sight is a power of $2$, so rewrite it that way.

Note $8 = 2^3$ and $32768 = 2^{15}$, so $\tfrac18 = 2^{-3}$ and $\tfrac{1}{32768} = 2^{-15}$.

Step 1 — Rewrite both sides. $\text{LHS} = (2^{-3})^k (2^{-15})^{1/3} = 2^{-3k} \cdot 2^{-5} = 2^{-3k - 5}.$ $\text{RHS} = 2^{-3} (2^{-15})^{1/k} = 2^{-3} \cdot 2^{-15/k} = 2^{-3 - 15/k}.$

Step 2 — Equate exponents. $-3k - 5 = -3 - \frac{15}{k}.$ Multiply through by $k$: $-3k^2 - 5k = -3k - 15$, i.e. $3k^2 + 2k - 15 = 0.$

Step 3 — Sum of the roots (Vieta). For $3k^2 + 2k - 15 = 0$, the sum of roots is $-\tfrac{b}{a} = -\tfrac{2}{3}$. (Both roots, $\tfrac{-1 \pm \sqrt{46}}{3}$, are real and non-zero, so both are valid.)

💡 Teacher tip: If a question asks for the sum of all roots of a quadratic, Vieta's formula ($\alpha + \beta = -b/a$) gives it instantly — no need to solve for the roots themselves.

Answer: $\boxed{-\tfrac{2}{3}}$ — Choice C

The setup: Renu and Seema each have a known pace, then agree to a pair of "doubling" arrangements. The clean way is to measure the task in worker-hours.

Step 1 — Rates.

• Renu: $15 \text{ days} \times 4 \text{ h/day} = 60$ hours for the whole task → rate $\tfrac{1}{60}$ task per hour.
• Seema: $8 \text{ days} \times 5 \text{ h/day} = 40$ hours → rate $\tfrac{1}{40}$ task per hour.

Step 2 — Translate the agreements.

• "Seema works double the hours per day as Renu": Renu works $2$ h/day, so Seema works $4$ h/day.
• "Renu works double the number of days as Seema": if Seema works $D$ days, Renu works $2D$ days.

Step 3 — Work done by each. $\text{Renu} = 2D \times 2 \times \tfrac{1}{60} = \frac{4D}{60} = \frac{D}{15}, \qquad \text{Seema} = D \times 4 \times \tfrac{1}{40} = \frac{4D}{40} = \frac{D}{10}.$

Step 4 — Together they complete exactly one task. $\frac{D}{15} + \frac{D}{10} = 1 \;\Rightarrow\; \frac{2D + 3D}{30} = 1 \;\Rightarrow\; 5D = 30 \;\Rightarrow\; D = 6.$

💡 Teacher tip: Convert "X days at Y hours/day" into a single number of hours first — then the rate is just $\tfrac{1}{\text{hours}}$. The doubling clauses only change the day/hour multipliers, not the underlying rates.

Answer: $\boxed{6}$

The setup: $A$ has $6$ elements, $B$ has $3$, and every element of $B$ must be hit — so we're counting onto functions from $A$ to $B$.

Step 1 — Count all functions. Each of the $6$ elements of $A$ can map to any of the $3$ elements of $B$: $3^6 = 729$ functions in total.

Step 2 — Remove the ones that miss part of $B$ (inclusion–exclusion). A function fails to be onto if its outputs skip at least one element of $B$.

• Skipping one chosen element of $B$ (map into the other $2$): $2^6 = 64$ ways, and $\binom{3}{1} = 3$ choices of which to skip → subtract $3 \times 64 = 192.$
• Functions that skip two elements (all outputs equal) were subtracted twice, so add them back: $\binom{3}{2} \times 1^6 = 3.$
• Skipping all three is impossible.

$\text{onto} = 3^6 - \binom{3}{1}2^6 + \binom{3}{2}1^6 = 729 - 192 + 3 = 540.$

Step 3 — Confirm. Non-onto count $= 729 - 540 = 189$, which matches $192 - 3 = 189.$ ✓

💡 Teacher tip: The onto-function count is $\displaystyle \sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-k)^m$. Here it self-truncates after three terms because avoiding all three targets is impossible.

Answer: $\boxed{540}$ — Choice D

The setup: three quadratics share one common negative root. Call it $-r$ with $r > 0$, and substitute $x = -r$ into each.

x^2 + mx + 9 = 0:\quad r^2 - mr + 9 = 0 \;\Rightarrow\; m = r + \tfrac{9}{r}. \tag{1} x^2 + nx + 17 = 0:\quad r^2 - nr + 17 = 0 \;\Rightarrow\; n = r + \tfrac{17}{r}. \tag{2} x^2 + (m+n)x + 35 = 0:\quad r^2 - (m+n)r + 35 = 0. \tag{3}

Step 1 — Find $r$. Add (1) and (2): $m + n = 2r + \tfrac{26}{r}$. Put this into (3): $r^2 - \left(2r + \tfrac{26}{r}\right)r + 35 = 0 \;\Rightarrow\; r^2 - 2r^2 - 26 + 35 = 0 \;\Rightarrow\; -r^2 + 9 = 0 \;\Rightarrow\; r = 3.$

Step 2 — Get $m$ and $n$. $m = 3 + \tfrac{9}{3} = 6, \qquad n = 3 + \tfrac{17}{3} = \tfrac{26}{3}.$

Step 3 — Compute $2m + 3n$. $2m + 3n = 2(6) + 3\left(\tfrac{26}{3}\right) = 12 + 26 = 38.$

Check the shared root $x = -3$ in the third quadratic ($m + n = \tfrac{44}{3}$): $9 - \tfrac{44}{3}(3) + 35 = 9 - 44 + 35 = 0.$ ✓

💡 Teacher tip: The combination $2m + 3n$ is chosen so the "$3$" cancels the denominator of $n = \tfrac{26}{3}$ — the fractional $n$ still lands on a clean integer answer.

Answer: $\boxed{38}$

The setup: two equations, $4(x^2 + y^2 + z^2) = a$ and $4(x - y - z) = 3 + a$. Two equations in three unknowns usually leave a free variable — but a sum-of-squares trick pins everything.

Step 1 — Eliminate $a$. From the second equation, $a = 4(x - y - z) - 3$. Substitute into the first: $4(x^2 + y^2 + z^2) = 4(x - y - z) - 3.$ Divide by $4$ and rearrange: $x^2 - x + y^2 + y + z^2 + z = -\tfrac34.$

Step 2 — Complete the square in each variable. $\left(x - \tfrac12\right)^2 + \left(y + \tfrac12\right)^2 + \left(z + \tfrac12\right)^2 - \tfrac34 = -\tfrac34,$ so the three constants cancel and $\left(x - \tfrac12\right)^2 + \left(y + \tfrac12\right)^2 + \left(z + \tfrac12\right)^2 = 0.$ A sum of squares is zero only when each square is zero: $x = \tfrac12,\; y = -\tfrac12,\; z = -\tfrac12.$

Step 3 — Compute $a$. $a = 4(x^2 + y^2 + z^2) = 4\left(\tfrac14 + \tfrac14 + \tfrac14\right) = 4 \cdot \tfrac34 = 3.$ Check with the other equation: $4(x - y - z) = 4\left(\tfrac12 + \tfrac12 + \tfrac12\right) = 6 = 3 + a.$ ✓

💡 Teacher tip: Whenever a sum of squares equals something suspiciously small, try completing the square — if the leftover constant makes the whole thing collapse to zero, every variable is forced to a single value.

Answer: $\boxed{3}$ — Choice A

The setup: an AP with $x_5 = -4$ and a second condition. Write $x_k = a + (k-1)d$ and translate everything into $a$ and $d$.

Step 1 — Turn both conditions into equations.

• $x_5 = a + 4d = -4.$
• $2x_6 + 2x_9 = x_{11} + x_{13}$: $2(a + 5d) + 2(a + 8d) = (a + 10d) + (a + 12d) \;\Rightarrow\; 4a + 26d = 2a + 22d \;\Rightarrow\; 2a + 4d = 0 \;\Rightarrow\; a = -2d.$

Step 2 — Solve. Substitute $a = -2d$ into $a + 4d = -4$: $-2d + 4d = -4 \;\Rightarrow\; 2d = -4 \;\Rightarrow\; d = -2, \quad a = 4.$

Step 3 — The 100th term. $x_{100} = a + 99d = 4 + 99(-2) = 4 - 198 = -194.$

💡 Teacher tip: In any AP relation, replace each $x_k$ by $a + (k-1)d$ and the whole thing collapses to an equation in just $a$ and $d$. Two conditions, two unknowns, done.

Answer: $\boxed{-194}$ — Choice A

What we want: the greatest integer not exceeding $x$, where $4\log_{10}x + 4\log_{100}x + 8\log_{1000}x = 13.$

Step 1 — Put every logarithm on the same base. Since $100 = 10^2$ and $1000 = 10^3$, the change-of-base rule gives $\log_{100}x = \frac{\log_{10}x}{2}, \qquad \log_{1000}x = \frac{\log_{10}x}{3}.$ Let $t = \log_{10}x$. The equation becomes $4t + 4\cdot\frac{t}{2} + 8\cdot\frac{t}{3} = 13 \;\Rightarrow\; 4t + 2t + \frac{8t}{3} = 13.$

Step 2 — Solve the linear equation. $6t + \frac{8t}{3} = \frac{18t + 8t}{3} = \frac{26t}{3} = 13 \;\Rightarrow\; t = \frac{3}{2}.$

Step 3 — Recover $x$ and take the floor. $x = 10^{3/2} = 10\sqrt{10} \approx 10 \times 3.162 = 31.62,$ so the greatest integer not exceeding $x$ is $31$.

💡 Teacher tip: When several logs share a related base, rewrite them all in one base. Here the messy three-log equation collapses to a single linear equation in $t = \log_{10}x$.

Answer: $\boxed{31}$

What we want: the final milk-to-water ratio after the replacement is done four times in total.

Step 1 — What one step does to the milk. You pour out two-thirds of the contents, so one-third of whatever was there stays. The liquid you add back is water, which adds no milk. So each step keeps exactly $\tfrac13$ of the milk that was present.

Step 2 — After four steps. The fraction of the glass that is still milk is $\left(\frac13\right)^4 = \frac{1}{81}.$ So milk makes up $\tfrac{1}{81}$ of the glass and water makes up $1 - \tfrac{1}{81} = \tfrac{80}{81}$.

Step 3 — The ratio. $\text{milk: water} = \frac{1}{81}: \frac{80}{81} = 1: 80.$

💡 Teacher tip: The general rule: if you remove a fraction $f$ of the pure liquid and top up, $k$ times, the pure liquid left is $(1-f)^k$ of the whole. Here $f = \tfrac23$ and $k = 4$. The starting amount never matters — only the surviving fraction does.

Answer: $\boxed{1: 80}$ — Choice B

The setup: all four-digit numbers using four distinct non-zero digits $a, b, c, d$ (each digit once per number) are added; the total is $153310 + n$ for a single-digit $n$. Find $a + b + c + d + n$.

Step 1 — Sum of all such numbers. With $4$ distinct digits there are $4! = 24$ numbers. In any one place value, each digit shows up in exactly $3! = 6$ of them. So each place contributes $6(a + b + c + d)$ times its place value, and $\text{total} = 6(a + b + c + d)(1000 + 100 + 10 + 1) = 6(a+b+c+d)(1111) = 6666(a+b+c+d).$ Let $S = a + b + c + d.$

Step 2 — Match the given form. $6666S = 153310 + n$ with $n$ a single digit. Try $S = 23$: $6666 \times 23 = 153318 \;\Rightarrow\; n = 153318 - 153310 = 8,$ a valid single digit. So $S = 23$ and $n = 8.$

Step 3 — The requested sum. $a + b + c + d + n = S + n = 23 + 8 = 31.$

💡 Teacher tip: Memorise the template: the sum of all numbers formed from $r$ distinct digits is $(r-1)! \times (\text{sum of digits}) \times \underbrace{111\ldots1}_{r \text{ ones}}$. For four digits that's $6 \times S \times 1111$.

Answer: $\boxed{31}$