CAT 2023 Quant Questions — Slot 1

What we want: the value of $\sqrt{10x + 9}$, given $\sqrt{5x + 9} + \sqrt{5x - 9} = 3(2 + \sqrt2)$.

Let $a = \sqrt{5x + 9}$ and $b = \sqrt{5x - 9}$, so $a + b = 3(2 + \sqrt2)$.

Step 1 — Exploit $a^2 - b^2$. $a^2 - b^2 = (5x + 9) - (5x - 9) = 18.$ Since $a^2 - b^2 = (a - b)(a + b)$: $a - b = \frac{18}{a + b} = \frac{18}{3(2 + \sqrt2)} = \frac{6}{2 + \sqrt2}.$ Rationalise: $a - b = \frac{6(2 - \sqrt2)}{(2 + \sqrt2)(2 - \sqrt2)} = \frac{6(2 - \sqrt2)}{2} = 3(2 - \sqrt2) = 6 - 3\sqrt2.$

Step 2 — Solve for $a$. Add $a + b = 6 + 3\sqrt2$ and $a - b = 6 - 3\sqrt2$: $2a = 12 \;\Rightarrow\; a = 6 \;\Rightarrow\; \sqrt{5x + 9} = 6 \;\Rightarrow\; 5x = 27 \;\Rightarrow\; x = \frac{27}{5}.$

Check: $b = 3\sqrt2$, and $\sqrt{5x - 9} = \sqrt{18} = 3\sqrt2$ ✓.

Step 3 — Compute the target. $\sqrt{10x + 9} = \sqrt{10 \cdot \tfrac{27}{5} + 9} = \sqrt{54 + 9} = \sqrt{63} = 3\sqrt7.$

Answer: $\boxed{3\sqrt7}$ (Choice C)

💡 Teacher tip: With two separate surds, $a^2 - b^2$ wipes out both $x$-terms in one stroke and hands you $a - b$. Combined with the given $a + b$, both surds resolve by simple addition and subtraction.

What we want: the first moment after 8:48 when the angle between the two hands has grown by half again (a $50\%$ increase).

Step 1 — The angle at 8:48. The hour hand crawls at $0.5^\circ$ per minute, the minute hand races at $6^\circ$ per minute.

• Hour hand: at 8:00 it sits at $8 \times 30 = 240^\circ$; by 8:48 it has moved to $240 + 48(0.5) = 264^\circ$.
• Minute hand: $48 \times 6 = 288^\circ$.

So the gap is $|288 - 264| = 24^\circ$.

Step 2 — The target angle. A $50\%$ increase means $24 \times 1.5 = 36^\circ$.

Step 3 — Time to open the gap to $36^\circ$. The minute hand pulls away from the hour hand at $6 - 0.5 = 5.5^\circ$ per minute. Starting $24^\circ$ ahead: $24 + 5.5\,t = 36 \;\Rightarrow\; 5.5\,t = 12 \;\Rightarrow\; t = \frac{12}{5.5} = \frac{24}{11}\text{ minutes}.$

Since $36^\circ$ is still well under $180^\circ$, it genuinely is the minor angle — no wrap-around to worry about.

Answer: D

💡 Teacher tip: The two hands close or open at a steady $5.5^\circ$ per minute. Once you have the starting gap and the target gap, the time is simply $\dfrac{\text{change in angle}}{5.5}$.

What we want: the value of $x - 2y$, given $x^2 + (x - 2y - 1)^2 = -4y(x + y)$.

Step 1 — Bring everything to one side. $x^2 + (x - 2y - 1)^2 + 4y(x + y) = 0.$

Step 2 — Expand. Using $(x - 2y - 1)^2 = x^2 - 4xy + 4y^2 - 2x + 4y + 1$: $x^2 + (x^2 - 4xy + 4y^2 - 2x + 4y + 1) + 4xy + 4y^2 = 0,$ which simplifies to $2x^2 - 2x + 8y^2 + 4y + 1 = 0.$

Step 3 — Complete the square in each variable. $2\left(x - \tfrac12\right)^2 - \tfrac12 + 8\left(y + \tfrac14\right)^2 - \tfrac12 + 1 = 0.$ The constants cancel ($-\tfrac12 - \tfrac12 + 1 = 0$), leaving $2\left(x - \tfrac12\right)^2 + 8\left(y + \tfrac14\right)^2 = 0.$ A sum of two squares is zero only if each is zero, so $x = \tfrac12$ and $y = -\tfrac14$.

Step 4 — Compute. $x - 2y = \tfrac12 - 2\left(-\tfrac14\right) = \tfrac12 + \tfrac12 = 1.$

Answer: 1 (Choice B)

💡 Teacher tip: An equation of the form "(square) + (square) = 0" forces both squares to vanish. The examiners tune the constants so the leftover terms cancel exactly — spotting that shape is the whole solution.

The setup: start with $2$ organisms; each day the count is "$3$ more than twice the previous day," i.e. $a_1 = 2$ and $a_k = 2a_{k-1} + 3$. Find the smallest $n$ with $a_n$ over one million.

Step 1 — Tame the recurrence. The "$+3$" is the awkward part. Add $3$ to both sides: $a_k + 3 = 2a_{k-1} + 6 = 2(a_{k-1} + 3).$ So $b_k = a_k + 3$ doubles each day — a clean geometric progression with $b_1 = 5$: $b_n = 5 \cdot 2^{n-1} \;\Rightarrow\; a_n = 5 \cdot 2^{n-1} - 3.$

Step 2 — Solve $a_n > 1{,}000{,}000$. $5 \cdot 2^{n-1} - 3 > 1{,}000{,}000 \;\Rightarrow\; 2^{n-1} > 200{,}000.$ Since $2^{17} = 131072$ and $2^{18} = 262144$, we need $n - 1 \ge 18$, i.e. $n \ge 19$.

Step 3 — Confirm.

• $a_{18} = 5 \cdot 2^{17} - 3 = 655357 < 1{,}000{,}000$ ✗
• $a_{19} = 5 \cdot 2^{18} - 3 = 1310717 > 1{,}000{,}000$ ✓

Answer: $\boxed{19}$

💡 Teacher tip: For any recurrence $a_k = 2a_{k-1} + c$, absorb the constant: $a_k + c = 2(a_{k-1} + c)$. That converts a messy rule into a pure doubling you can compare to powers of $2$.

What we want: how many natural numbers from $1$ to $1000$ have all different digits. Split the count by how many digits the number has.

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

Two-digit numbers: the tens digit has $9$ choices ($1$–$9$); the units digit can be anything $0$–$9$ except the tens digit, so $9$ choices: $9 \times 9 = 81.$

Three-digit numbers: hundreds digit $9$ choices ($1$–$9$); tens digit $9$ choices (any digit except the hundreds one, $0$ now allowed); units digit $8$ choices (excluding the two already used): $9 \times 9 \times 8 = 648.$

The number $1000$: it has repeated zeros, so it is excluded.

Answer: C

💡 Teacher tip: Build the number one place at a time, subtracting the already-used digits at each step. The leading digit is special — it can never be $0$.

The setup: from a point outside a circle, two tangents are drawn making a $60^\circ$ angle. As that external point moves, it traces a locus $L$. We want where $L$ meets the vertical line $x = 6$.

Step 1 — Understand the circle. Complete the squares in $x^2 + y^2 + 4x - 6y - 3 = 0$: $(x+2)^2 + (y-3)^2 = 16,$ so the centre is $(-2, 3)$ and the radius is $4$.

Step 2 — Find the locus. From an external point at distance $d$ from the centre, the two tangents make an angle $\theta$ with $\sin\dfrac{\theta}{2} = \dfrac{r}{d}$. Here $\theta = 60^\circ$: $\sin 30^\circ = \frac{1}{2} = \frac{4}{d} \;\Rightarrow\; d = 8.$ So $L$ is a circle sharing the same centre, radius $8$: $(x+2)^2 + (y-3)^2 = 64$.

Step 3 — Meet the line $x = 6$. $(6+2)^2 + (y-3)^2 = 64 \;\Rightarrow\; 64 + (y-3)^2 = 64 \;\Rightarrow\; (y-3)^2 = 0 \;\Rightarrow\; y = 3.$

The line just grazes $L$ at the single point $(6, 3)$.

Answer: B

💡 Teacher tip: The line $x = 6$ lies exactly $8$ units — the radius of $L$ — from the centre, so it touches $L$ rather than cutting it, and the quadratic in $y$ collapses to a perfect square.

The setup: for positive distinct reals $x, y, z$, we are told $\dfrac{1}{yz}$ is the arithmetic mean of $\dfrac{1}{xz}$ and $\dfrac{1}{xy}$. Which ordering must form an AP?

Step 1 — Write the mean condition. $\frac{1}{yz} = \frac{1}{2}\left(\frac{1}{xz} + \frac{1}{xy}\right).$

Step 2 — Simplify the right side over the common denominator $xyz$: $\frac{1}{xz} + \frac{1}{xy} = \frac{y + z}{xyz}, \qquad\text{so}\qquad \frac{1}{yz} = \frac{y + z}{2xyz}.$

Step 3 — Solve. Multiply both sides by $xyz$ (positive, so safe): $x = \frac{y + z}{2} \;\Rightarrow\; 2x = y + z.$

Step 4 — Read the meaning. $2x = y + z$ says $x$ is the arithmetic mean of $y$ and $z$ — that is, $y, x, z$ are in arithmetic progression.

Answer: B

💡 Teacher tip: Put the fractions over the common denominator $xyz$ and the equation collapses to $x = \dfrac{y+z}{2}$. Whichever variable ends up alone in the numerator is the middle term of the AP.

What we want: $x + y$, given $\log_x(x^2 + 12) = 4$ and $3\log_y x = 1$, with $x, y > 0$.

Step 1 — Solve for $x$. By the definition of a logarithm, $\log_x(x^2 + 12) = 4$ means $x^4 = x^2 + 12$: $x^4 - x^2 - 12 = 0 \;\Rightarrow\; (x^2 - 4)(x^2 + 3) = 0.$ Since $x^2 + 3 > 0$, we need $x^2 = 4$, and $x > 0$ gives $x = 2$.

Step 2 — Solve for $y$. $3\log_y x = 1 \;\Rightarrow\; \log_y x = \frac13 \;\Rightarrow\; y^{1/3} = x = 2 \;\Rightarrow\; y = 2^3 = 8.$

Step 3 — Add. $x + y = 2 + 8 = 10.$

Answer: 10 (Choice C)

💡 Teacher tip: Rewrite each log statement in its exponential form first — $\log_b a = c \iff b^c = a$. After that, both unknowns fall out as ordinary powers.

What we want: $m + n$, where $n$ is the least positive integer with $168 \mid 1134^n$, and $m$ is the least with $1134^n \mid 168^m$.

Step 1 — Prime factorisations. $168 = 2^3 \cdot 3 \cdot 7, \qquad 1134 = 2 \cdot 3^4 \cdot 7.$

Step 2 — Find $n$ so that $168 \mid 1134^n$. For each prime, the power in $1134^n$ must reach the power in $168$:

The binding prime is $2$, so $n = 3$.

Step 3 — Find $m$ so that $1134^3 \mid 168^m$. Now each prime's power in $168^m$ must reach its power in $1134^3$:

The binding prime is $3$, so $m = 12$.

Step 4 — Answer. $m + n = 12 + 3 = 15.$

Answer: 15 (Choice B)

💡 Teacher tip: "Is a factor of" between two numbers is just an exponent comparison, prime by prime. The value you need is set by the prime where the required exponents are most demanding.

The setup in plain English: one quadratic gives us two numbers, $\alpha$ and $\beta$. Their sum and product then become the two roots of a second quadratic. We chase both unknown constants, $k$ and $p$, and finish with $8(k-p)$.

Step 1 — Read off the sum and product of the first equation. For $2x^2 - 6x + k = 0$ with roots $\alpha,\beta$, the sum of roots is $-\dfrac{-6}{2}$ and the product is $\dfrac{k}{2}$: $\alpha + \beta = 3, \qquad \alpha\beta = \frac{k}{2}.$

Step 2 — Feed these into the second equation. We are told $(\alpha+\beta)$ and $\alpha\beta$ are the two roots of $x^2 + px + p = 0$. So their sum is $-p$ and their product is $p$: $3 + \frac{k}{2} = -p, \qquad 3 \cdot \frac{k}{2} = p.$

Step 3 — Two equations, two unknowns. The product line gives $p = \dfrac{3k}{2}$. Put that into the sum line: $3 + \frac{k}{2} = -\frac{3k}{2} \;\Rightarrow\; 3 = -2k \;\Rightarrow\; k = -\frac{3}{2}.$ Then $p = \dfrac{3}{2}\left(-\dfrac{3}{2}\right) = -\dfrac{9}{4}$.

Step 4 — Compute the target. $8(k - p) = 8\left(-\frac{3}{2} + \frac{9}{4}\right) = 8 \cdot \frac{3}{4} = 6.$

Answer: $\boxed{6}$

💡 Teacher tip: Whenever roots of one equation become the sum/product feeding another, write both "sum of roots" and "product of roots" relations — that always hands you exactly enough equations to pin the unknown constants.

The setup: Amal, Sunil and Kamal's daily work-rates are in harmonic progression. Kamal takes twice as long as Amal. Amal works 4 days, Sunil 9 days, and then Kamal needs 16 days to finish. How many days would Sunil alone take?

Step 1 — Translate "rates in HP". If the daily rates are in harmonic progression, then their reciprocals — the number of days each takes alone — are in arithmetic progression. Call those solo days $D_A, D_S, D_K$.

Step 2 — Use Kamal's condition. Kamal takes twice as long as Amal: $D_K = 2D_A$. With $D_A, D_S, D_K$ in AP, the middle term is their average: $D_S = \frac{D_A + 2D_A}{2} = 1.5\,D_A.$

Step 3 — Write the work equation (whole job $= 1$). Rates are $\dfrac{1}{D_A}, \dfrac{1}{1.5D_A}, \dfrac{1}{2D_A}$: $\frac{4}{D_A} + \frac{9}{1.5\,D_A} + \frac{16}{2D_A} = 1 \;\Rightarrow\; \frac{4 + 6 + 8}{D_A} = 1 \;\Rightarrow\; D_A = 18.$

Step 4 — Sunil's solo time. $D_S = 1.5 \times 18 = 27 \text{ days}.$

Answer: $\boxed{27}$

💡 Teacher tip: "Rates in HP" is exactly "solo times in AP." That one translation turns a fuzzy harmonic condition into a clean linear relationship you can solve directly.

The classic road-meeting problem: two people start at the same instant from opposite towns, meet on the way, then finish. After meeting, Arvind needs $6$ hours to reach B and Surbhi needs $24$ hours to reach A. Arvind's speed is $54$ km/h. We want the distance between the towns.

Method 1 — The meeting-time shortcut. When two travellers start together and, after meeting, take $t_1$ and $t_2$ hours to reach each other's start, the time until they met is $T = \sqrt{t_1 \cdot t_2} = \sqrt{6 \times 24} = \sqrt{144} = 12 \text{ hours}.$ Arvind's total travel time is $T + t_1 = 12 + 6 = 18$ hours, so $\text{distance} = 54 \times 18 = 972 \text{ km}.$

Method 2 — From first principles. Let Surbhi's speed be $v$ and the time to meeting be $T$.

• Before meeting Arvind covers $54T$; that equals what Surbhi still has to cover after meeting, $24v$. So $24v = 54T$.
• Before meeting Surbhi covers $vT$; that equals what Arvind covers after meeting, $6 \times 54 = 324$. So $vT = 324$.

Divide the two: $\dfrac{24}{T} = \dfrac{T}{6} \Rightarrow T^2 = 144 \Rightarrow T = 12$, giving distance $= 54(12 + 6) = 972$.

Answer: $\boxed{972}$

💡 Teacher tip: The formula $T = \sqrt{t_1 t_2}$ falls straight out of $\dfrac{v_{\text{Surbhi}}}{v_{\text{Arvind}}} = \dfrac{t_1}{T} = \dfrac{T}{t_2}$. Memorise it and the answer is one line under time pressure.

The setup: 4 girls (each scoring $g$) and 6 boys (each scoring $b$) average $24$. Each girl's mark is at least a boy's and at most double it. We count the distinct integer values of (2 girls + 6 boys).

Step 1 — The total. $4g + 6b = 10 \times 24 = 240$, which simplifies to $2g + 3b = 120$.

The quantity we want is $2g + 6b$. Since $6b = 240 - 4g$: $2g + 6b = 2g + (240 - 4g) = 240 - 2g.$

Step 2 — Turn the word constraints into inequalities. From $2g + 3b = 120$, $b = \dfrac{120 - 2g}{3}$.

• $g \ge b$: $\;g \ge \dfrac{120 - 2g}{3} \Rightarrow 5g \ge 120 \Rightarrow g \ge 24$.
• $g \le 2b$: $\;g \le \dfrac{2(120 - 2g)}{3} \Rightarrow 7g \le 240 \Rightarrow g \le 34.29$.

So $g$ runs over $[24,\ 34.29]$.

Step 3 — Count the integer values of $240 - 2g$. As $g$ sweeps that interval, $240 - 2g$ covers $[171.4,\ 192]$. The integers inside are $172, 173, \dots, 192$: $192 - 172 + 1 = 21.$

Endpoint check: $240 - 2g = 192 \Rightarrow g = 24, b = 24$ ✓; $= 172 \Rightarrow g = 34$, and $g \le 2b$ still holds ✓.

Answer: A

💡 Teacher tip: Once $g$ is chosen, $b$ is fixed — so drive everything through the single variable $g$. The two inequalities carve out an interval, and you just count the integers of $240 - 2g$ inside it.

What we want: the smallest non-negative integer $r$ such that $-2$ is a root of $x^3 + (2r + 1)x^2 + (4r - 1)x + 2 = 0$ and the other two roots are real.

Step 1 — Confirm $-2$ is always a root. Substitute $x = -2$: $-8 + (2r + 1)(4) + (4r - 1)(-2) + 2 = -8 + 8r + 4 - 8r + 2 + 2 = 0.$ This is $0$ for every $r$, so $(x + 2)$ divides the cubic no matter what $r$ is.

Step 2 — Factor out $(x + 2)$. Dividing gives $x^3 + (2r + 1)x^2 + (4r - 1)x + 2 = (x + 2)\bigl(x^2 + (2r - 1)x + 1\bigr).$

Step 3 — Make the other two roots real. The quadratic $x^2 + (2r - 1)x + 1$ needs a non-negative discriminant: $(2r - 1)^2 - 4 \ge 0 \;\Rightarrow\; (2r - 1)^2 \ge 4 \;\Rightarrow\; 2r - 1 \ge 2 \text{ or } 2r - 1 \le -2.$ So $r \ge \tfrac32$ or $r \le -\tfrac12$.

Step 4 — Smallest non-negative integer. The smallest non-negative integer with $r \ge \tfrac32$ is $r = 2$.

Check: at $r = 2$ the quadratic is $x^2 + 3x + 1$, discriminant $9 - 4 = 5 > 0$ ✓. (At $r = 0$ and $r = 1$ the discriminant is $-3 < 0$.)

Answer: $\boxed{2}$

💡 Teacher tip: Because $-2$ is a root for all $r$, the only real constraint is that the remaining quadratic have real roots. Factoring hands you that quadratic directly, and it's a plain discriminant inequality from there.

The setup: a quadrilateral $ABCD$ is inscribed in a circle with $AB: CD = 2: 1$ and $BC: AD = 5: 4$. The diagonals cross at $E$. Find $AE: CE$.

Let $AB = 2k,\ CD = k$ and $BC = 5m,\ AD = 4m$.

Step 1 — The cyclic-quadrilateral diagonal ratio. For a cyclic quadrilateral, the diagonals divide each other so that $\frac{AE}{CE} = \frac{AB \cdot AD}{CB \cdot CD}.$ (Why: triangles $ABE$ and $DCE$ are similar, giving $\dfrac{AE}{DE} = \dfrac{AB}{CD}$; triangles $ADE$ and $CBE$ are similar, giving $\dfrac{DE}{CE} = \dfrac{AD}{BC}$. Multiply the two.)

Step 2 — Substitute. $\frac{AE}{CE} = \frac{(2k)(4m)}{(5m)(k)} = \frac{8}{5}.$

Sanity check: for a square (all sides equal) the formula gives $1:1$ — the diagonals bisect each other, exactly right.

Answer: C

💡 Teacher tip: In a cyclic quadrilateral, $\dfrac{AE}{CE} = \dfrac{AB \cdot AD}{CB \cdot CD}$ — the whole problem collapses to one multiplication of the side ratios.

The setup: right triangle $ABC$ with the right angle at $B$, $AB = 5$, $BC = 12$. Points $P$ and $Q$ sit on $BC$; the areas of $\triangle ABP$, $\triangle ABQ$, $\triangle ABC$ form an arithmetic progression, and $\triangle ABC$'s area is $1.5$ times $\triangle ABP$'s. Find $PQ$.

Setting up areas. The full triangle has area $\tfrac{1}{2}(5)(12) = 30$. For any point on $BC$ at distance $d$ from $B$, the triangle with apex $A$ has area $\tfrac{1}{2}(5)(d) = 2.5d$ — proportional to the base length.

Step 1 — Find $BP$. Given area$(ABC) = 1.5\,$area$(ABP)$: $30 = 1.5 \times 2.5 \times BP \;\Rightarrow\; 30 = 3.75\,BP \;\Rightarrow\; BP = 8.$

Step 2 — Use the AP condition. Since each area is $2.5 \times$ its base, "areas in AP" means the bases $BP, BQ, BC$ are also in AP: $8,\ BQ,\ 12 \text{ in AP} \;\Rightarrow\; BQ = \frac{8 + 12}{2} = 10.$

Step 3 — Find $PQ$. Both points lie on $BC$: $PQ = |BQ - BP| = |10 - 8| = 2.$

Answer: $\boxed{2}$

💡 Teacher tip: Multiplying every term of an AP by the same constant keeps it an AP. So "areas in AP" becomes "bases in AP," and the middle base is just the average of the outer two.

The setup: Gita sells A and B at one common price — profit $20\%$ on A, loss $10\%$ on B. Then she raises the price (still common to both) so that B now shows a $10\%$ profit. We want the new profit on A.

Let the first common selling price be $S$.

Step 1 — Recover the cost prices.

• A at $20\%$ profit: $S = 1.2\,C_A \Rightarrow C_A = \dfrac{S}{1.2}$.
• B at $10\%$ loss: $S = 0.9\,C_B \Rightarrow C_B = \dfrac{S}{0.9}$.

Step 2 — The new common price. Now B earns $10\%$ profit: $S_{\text{new}} = 1.1\,C_B = 1.1 \cdot \frac{S}{0.9} = \frac{11S}{9}.$

Step 3 — New profit on A. $\frac{S_{\text{new}} - C_A}{C_A} = \frac{11S/9}{S/1.2} - 1 = \frac{11}{9}\times 1.2 - 1 = \frac{13.2}{9} - 1 \approx 0.467.$

That is about $46.7\%$, nearest to $47\%$.

Answer: C

💡 Teacher tip: Both selling prices are equal, so the actual value of $S$ cancels out everywhere. Set $S = 1$ (or keep it as a symbol) and read the profit straight off the multipliers.

The picture: we start with a jar of pure coffee. We scoop out some, top up with cocoa, mix — that is mixture P. Do it a second time — that is mixture Q. We know the coffee-to-cocoa ratio in Q and want to compare the cocoa in P versus Q.

Let the jar hold $1$ unit, and let $f$ be the fraction removed and replaced each time.

Step 1 — Track coffee after each round. Removing a fraction $f$ of a well-mixed jar leaves the coffee multiplied by $(1-f)$ each time.

• After round 1 (this is P): coffee $= 1-f$, cocoa $= f$.
• After round 2 (this is Q): coffee $= (1-f)^2$, cocoa $= 1-(1-f)^2$.

Step 2 — Use the ratio in Q. Coffee: cocoa $= 16:9$, so coffee is $\dfrac{16}{25}$ of the jar: $(1-f)^2 = \frac{16}{25} \;\Rightarrow\; 1-f = \frac{4}{5} \;\Rightarrow\; f = \frac{1}{5}.$

Step 3 — Cocoa in each mixture. $\text{cocoa in P} = f = \frac{1}{5}, \qquad \text{cocoa in Q} = 1 - \frac{16}{25} = \frac{9}{25}.$

Step 4 — Take the ratio. $\frac{1}{5}: \frac{9}{25} = \frac{5}{25}: \frac{9}{25} = 5: 9.$

Answer: C

💡 Teacher tip: The one fact that cracks these problems: after $n$ replace-rounds each of fraction $f$, the original liquid left is $(1-f)^n$. That turns a wordy two-stage story into a single equation.

The picture: the odometer reads a palindrome $x$ before the trip and $26862$ (also a palindrome) after an 8-hour drive, with speed never above $110$ km/h. We want the greatest possible average speed.

Step 1 — Set the bounds. Trip distance $= 26862 - x$, average speed $= \dfrac{26862 - x}{8}$.

• The cap: speed $\le 110 \Rightarrow$ distance $\le 8 \times 110 = 880 \Rightarrow x \ge 26862 - 880 = 25982$.
• To make the speed as large as possible, make the distance as large as possible — that means making $x$ as small as possible.

So we need the smallest palindrome $x$ in the range $[25982,\ 26861]$.

Step 2 — Hunt the smallest palindrome. Near $26000$, a 5-digit palindrome looks like $\overline{2\,b\,c\,b\,2}$, worth $20002 + 1010b + 100c$.

• $b = 5$ tops out at $25952 < 25982$ — too small.
• $b = 6$ starts at $26062$, which sits inside the range. That is the smallest valid one.

Step 3 — The speed. $\text{distance} = 26862 - 26062 = 800, \qquad \text{speed} = \frac{800}{8} = 100 \text{ km/h}.$

Answer: C

💡 Teacher tip: Smallest allowed starting reading $\Rightarrow$ largest distance $\Rightarrow$ highest speed. Always let the speed cap set the floor on $x$ first, then find the smallest palindrome above it.

The setup: a scheme pays $4\%$ per year compounded half-yearly — that is $2\%$ every six months. Anil invests $22000$ for 6 years. Sunil invests some amount $P$ for 5 years in the same scheme, then rolls the whole balance into one more year at $10\%$ simple interest. Both end up with the same amount after 6 years. Find $P$.

Step 1 — Anil after 6 years (that's $12$ half-years): $A_{\text{Anil}} = 22000 \times (1.02)^{12}.$

Step 2 — Sunil after 6 years. Five years in the scheme is $10$ half-years, then one flat year at $10\%$ (a $\times 1.10$): $A_{\text{Sunil}} = P \times (1.02)^{10} \times 1.10.$

Step 3 — Set them equal and solve. $P \times (1.02)^{10} \times 1.10 = 22000 \times (1.02)^{12}.$ Most of the powers cancel: $P = \frac{22000 \times (1.02)^2}{1.10} = \frac{22000 \times 1.0404}{1.10} = \frac{22888.8}{1.10} = 20808.$

Answer: $\boxed{20808}$

💡 Teacher tip: Because both investors share the same scheme over overlapping periods, nearly all the compounding factors cancel. Only $(1.02)^2 / 1.10$ survives — no need to ever compute $(1.02)^{12}$.

What we want: the number of integer $x$ satisfying $2|x|(x^2 + 1) = 5x^2$.

Step 1 — Test $x = 0$. Both sides equal $0$, so $x = 0$ is a solution.

Step 2 — For $x \ne 0$, divide by $|x|$. Since $x^2 = |x|^2$: $2(x^2 + 1) = \frac{5x^2}{|x|} = 5|x|.$ Let $t = |x| > 0$. Then $2(t^2 + 1) = 5t$, i.e. $2t^2 - 5t + 2 = 0 \;\Rightarrow\; (2t - 1)(t - 2) = 0 \;\Rightarrow\; t = \tfrac12 \text{ or } t = 2.$

Step 3 — Keep only integer $x$.

• $t = \tfrac12$: $x = \pm\tfrac12$ — not integers, reject.
• $t = 2$: $x = \pm 2$ — both integers ✓.

So the integer solutions are $x \in \{-2, 0, 2\}$ — three values.

Answer: $\boxed{3}$

💡 Teacher tip: The equation depends only on $|x|$, so solve for $t = |x|$ first, and each positive $t$ then gives $\pm x$. The trap is $t = \tfrac12$, which produces non-integer $x$ and must be discarded.

The setup: salaries start at $5: 6: 7$ (Sita, Gita, Mita). Year 1 hikes: $20\%, 25\%, 20\%$. Year 2: Sita $+40\%$, Mita $+25\%$, and Gita's new salary equals the average of all three. We want Gita's year-2 hike.

Take the starting salaries as $5k,\ 6k,\ 7k$.

Step 1 — End of year 1.

• Sita: $5k \times 1.20 = 6k$.
• Gita: $6k \times 1.25 = 7.5k$.
• Mita: $7k \times 1.20 = 8.4k$.

Step 2 — Year 2 for Sita and Mita.

• Sita: $6k \times 1.40 = 8.4k$.
• Mita: $8.4k \times 1.25 = 10.5k$.

Step 3 — Pin Gita's year-2 salary. Call it $G$; it equals the mean of the three: $G = \frac{8.4k + G + 10.5k}{3} \;\Rightarrow\; 3G = 18.9k + G \;\Rightarrow\; G = 9.45k.$

Step 4 — Gita's hike. From $7.5k$ to $9.45k$: $\frac{9.45k - 7.5k}{7.5k} = \frac{1.95}{7.5} = 0.26 = 26\%.$

Answer: C

💡 Teacher tip: "Equals the mean" is the equation that unlocks the unknown — write the average out explicitly and solve. Carrying the common $k$ throughout keeps the arithmetic tidy.