CAT 2023 Quant Questions — Slot 2

What we want: which statement is forced to be true whenever $\dfrac{x}{y} < \dfrac{x+3}{y-3}$ (with $y \neq 0, 3$).

Step 1 — Bring to one side. $\frac{x}{y} - \frac{x+3}{y-3} < 0 \;\Rightarrow\; \frac{x(y-3) - y(x+3)}{y(y-3)} < 0.$

Step 2 — Simplify the numerator. $x(y-3) - y(x+3) = xy - 3x - xy - 3y = -3(x + y).$ So the inequality becomes $\frac{-3(x+y)}{y(y-3)} < 0 \;\Rightarrow\; \frac{x+y}{y(y-3)} > 0.$

Any valid $(x, y)$ must satisfy $(x + y)\, y\, (y - 3) > 0$.

Step 3 — Test option (B): "if $y < 0$ then $-x < y$." When $y < 0$, both $y < 0$ and $y - 3 < 0$, so $y(y-3) > 0$. For the fraction to stay positive we then need $x + y > 0$, i.e. $-x < y$ — exactly what (B) claims. ✓

Step 4 — Why the others fail. For (C), when $y > 10$ we again have $y(y-3) > 0$, forcing $x + y > 0$, i.e. $-x < y$ — the opposite of (C). Options (A) and (D) don't follow from the sign condition in general.

Answer: B

💡 Teacher tip: Never cross-multiply an inequality by something whose sign you don't know. Move everything to one side, factor, and read off the sign of the whole expression instead.

The setup: Ravi drives right at $40$ km/h. Vijay is $54$ m behind him, same direction. Ashok comes from the opposite side at $50$ km/h, currently $225$ m ahead of Ravi. At what speed must Vijay drive so all three meet at one instant?

Put Ravi at position $0$. In metres per hour: Ravi $40000$, Ashok $50000$. Vijay starts at $-54$ m; Ashok at $+225$ m moving left.

Step 1 — When and where do Ravi and Ashok meet? At time $t$ hours: $40000t = 225 - 50000t \;\Rightarrow\; 90000t = 225 \;\Rightarrow\; t = 0.0025 \text{ h}.$ Meeting point: $40000 \times 0.0025 = 100$ m from Ravi's start.

Step 2 — Force Vijay to that same point at that time. Vijay must go from $-54$ m to $+100$ m in $0.0025$ h. If his speed is $V$ km/h ($= 1000V$ m/h): $-54 + 1000V \times 0.0025 = 100 \;\Rightarrow\; 2.5V = 154.$

Step 3 — Solve. $V = \frac{154}{2.5} = 61.6 \text{ km/h}.$

Answer: C

💡 Teacher tip: "All three cross at once" fixes the meeting point. Find it from the two whose data is complete (Ravi and Ashok), then back out the unknown speed from the distance the third must cover in that exact time.

The setup: a tank holds $40$ L of milk. Repeatedly, $4$ L is removed and replaced by $4$ L of water. How many rounds until the milk drops below the water?

Step 1 — Model the decay. Each round removes $\dfrac{4}{40} = 10\%$ of the milk, so after $n$ rounds the milk left is $40 \times (0.9)^n \text{ litres}.$

Step 2 — Set the condition. Milk below water means milk below half the tank, i.e. below $20$ L: $40(0.9)^n < 20 \;\Rightarrow\; (0.9)^n < 0.5.$ Taking logs: $n > \dfrac{\log 0.5}{\log 0.9} \approx 6.58$.

Step 3 — Check the integers around it.

So the smallest number of rounds is $7$.

Answer: $\boxed{7}$

💡 Teacher tip: Replacing a fixed fraction each time is pure geometric decay: liquid left $= \text{start} \times (1 - \text{fraction})^n$. Don't simulate step by step — use the formula, then test the integer just above the log cut-off.

The setup: two arithmetic progressions whose step sizes are both prime numbers, tied together by a few equalities. We need $a_{11}$.

Write the $n$-th terms as $a_n = a_1 + (n-1)d_a$ and $b_n = b_1 + (n-1)d_b$, where $d_a$ and $d_b$ are prime.

Step 1 — Use $b_2 = 0$ to fix $b_1$. $b_2 = b_1 + d_b = 0 \quad\Rightarrow\quad b_1 = -d_b.$

Step 2 — Turn each equality into an equation for $a_1$.

From $a_5 = b_9$: $\;a_1 + 4d_a = b_1 + 8d_b = -d_b + 8d_b = 7d_b$, so $a_1 = 7d_b - 4d_a.$

From $a_{19} = b_{19}$: $\;a_1 + 18d_a = -d_b + 18d_b = 17d_b$, so $a_1 = 17d_b - 18d_a.$

Step 3 — Match the two expressions for $a_1$. $7d_b - 4d_a = 17d_b - 18d_a \quad\Rightarrow\quad 14d_a = 10d_b \quad\Rightarrow\quad 7d_a = 5d_b.$

💡 Teacher tip: This is where the word "prime" does all the work. Since $7d_a = 5d_b$ and $5, 7$ are themselves prime, the only way to balance the equation with prime $d_a, d_b$ is to let $d_a$ carry the factor $5$ and $d_b$ carry the factor $7$: $d_a = 5$, $d_b = 7$.

Step 4 — Compute the answer. $a_1 = 7(7) - 4(5) = 49 - 20 = 29,$ $a_{11} = a_1 + 10d_a = 29 + 10(5) = 29 + 50 = 79.$

Answer: B — $79$

The setup: rectangle $ABCD$ with $AB = 9$, $BC = 6$. Points $P, Q$ lie on $BC$; the areas of figures $ABP$, $APQ$, $AQCD$ are in geometric progression, and $AQCD = 4 \times ABP$. Find $BP: PQ: QC$.

Let $BP = p$, $PQ = q$, $QC = s$, with $p + q + s = 6$. Every figure spans the full width $9$ from $BC$ across to $AD$.

Step 1 — The three areas.

• $\triangle ABP$: $\tfrac{1}{2}(9)(p) = 4.5p$.
• $\triangle APQ$: $\tfrac{1}{2}(9)(q) = 4.5q$.
• $AQCD$: the rectangle minus the two triangles $= 54 - 4.5p - 4.5q$.

Step 2 — Apply the two conditions.

GP: $(4.5q)^2 = (4.5p)(AQCD)$.

$AQCD = 4 \times ABP$: $\;54 - 4.5(p + q) = 18p$. So the GP's right side is $4.5p \times 18p = 81p^2$, giving $(4.5q)^2 = 81p^2 \;\Rightarrow\; 20.25q^2 = 81p^2 \;\Rightarrow\; q = 2p.$

Step 3 — Solve for the pieces. Put $q = 2p$ into $54 - 4.5(p + q) = 18p$: $54 - 4.5(3p) = 18p \;\Rightarrow\; 54 = 31.5p \;\Rightarrow\; p = \frac{12}{7}.$ Then $q = \tfrac{24}{7}$ and $s = 6 - \tfrac{12}{7} - \tfrac{24}{7} = \tfrac{6}{7}$.

Step 4 — Ratio. $BP: PQ: QC = \frac{12}{7}: \frac{24}{7}: \frac{6}{7} = 2: 4: 1.$

Answer: C

💡 Teacher tip: With three areas in GP and one a clean multiple of another, the middle term gets pinned fast — here $(4.5q)^2 = 81p^2$ forces $q = 2p$ immediately, and the rest is arithmetic.

The setup: $20\%$ of employees are in manufacturing, and their combined salary is one-sixth of the company's total salary. Find the ratio of the average manufacturing salary to the average non-manufacturing salary.

A quick concrete model. Imagine $100$ employees and a total salary of Rs $600$.

• Manufacturing: $20$ people share $\tfrac{1}{6} \times 600 = 100$, so average $= \dfrac{100}{20} = 5$.
• Non-manufacturing: $80$ people share the other $500$, so average $= \dfrac{500}{80} = 6.25$.

Ratio. $5: 6.25 = 4: 5.$

(In symbols: with $E$ people and salary $S$, manufacturing average $= \dfrac{S/6}{0.2E}$ and non-manufacturing average $= \dfrac{5S/6}{0.8E}$; their ratio is $\dfrac{4.8}{6} = \dfrac{4}{5}$.)

Answer: B

💡 Teacher tip: When one side gives headcount percentages and the other gives salary fractions, plug in a friendly $100$-person, round-total company and read the averages straight off — far faster than juggling symbols.

The setup: a three-way equality $p^2 + q^2 - 29 = 2pq - 20 = 52 - 2pq$. Any chain equality like this is really two separate equations — equate the pieces two at a time.

Step 1 — The last two pieces give $pq$. $2pq - 20 = 52 - 2pq \quad\Rightarrow\quad 4pq = 72 \quad\Rightarrow\quad pq = 18.$

Step 2 — The first two pieces give $(p-q)^2$. $p^2 + q^2 - 29 = 2pq - 20 \quad\Rightarrow\quad p^2 + q^2 - 2pq = 9 \quad\Rightarrow\quad (p-q)^2 = 9,$ so $p - q = \pm 3$.

Step 3 — Find every $(p, q)$ pair.

If $p - q = 3$ with $pq = 18$: substituting $p = q + 3$ gives $q^2 + 3q - 18 = 0 = (q+6)(q-3)$.

• $(p, q) = (6, 3)$: $\;p^3 - q^3 = 216 - 27 = 189.$
• $(p, q) = (-3, -6)$: $\;p^3 - q^3 = -27 + 216 = 189.$

If $p - q = -3$ with $pq = 18$: substituting $p = q - 3$ gives $(q-6)(q+3) = 0$.

• $(p, q) = (3, 6)$: $\;p^3 - q^3 = 27 - 216 = -189.$
• $(p, q) = (-6, -3)$: $\;p^3 - q^3 = -216 + 27 = -189.$

Step 4 — Spread between largest and smallest. The possible values of $p^3 - q^3$ are just $+189$ and $-189$, so $\text{max} - \text{min} = 189 - (-189) = 378.$

💡 Teacher tip: The symmetry between $p$ and $q$ is why the cube values arrive in a neat $\pm$ pair — swapping $p$ and $q$ flips the sign of $p^3 - q^3$ but keeps both $pq$ and $(p-q)^2$ intact.

Answer: C — $378$

The setup: price $\propto$ (weight)$^2$, so price $= k w^2$. A stone of weight $18$ is broken into four pieces of distinct positive integer weights. The gap between the highest and lowest possible total price of the pieces is $288000$. Find the original stone's price.

The original stone is worth $k \cdot 18^2 = 324k$. For four pieces summing to $18$, the total price is $k \times (\text{sum of squares})$, so the price spread comes entirely from the spread in the sum of squares.

Step 1 — Maximum sum of squares. Make one piece as big as possible: smallest three are $1, 2, 3$, leaving $12$. $1 + 4 + 9 + 144 = 158.$

Step 2 — Minimum sum of squares. Since squaring is convex, the sum is smallest when the pieces are as equal as possible: $3, 4, 5, 6$. $9 + 16 + 25 + 36 = 86.$

Step 3 — Find $k$. The price gap is $k(158 - 86) = 72k = 288000$, so $k = 4000$.

Step 4 — Original price. $324k = 324 \times 4000 = 1{,}296{,}000.$

Answer: D

💡 Teacher tip: Because $x^2$ is convex, the most unequal split maximises the sum of squares and the most equal split minimises it. That single idea gives both extremes without listing every partition.

The setup: a sum of money split equally among $n$ people gives Rs $352$ each. Instead, if two people take Rs $506$ each and the rest is shared equally among the others, each of those others gets at most Rs $330$. Find the largest possible $n$.

Step 1 — Total money $= 352n$.

Step 2 — Set up the inequality. Two people take Rs $506$ each (Rs $1012$ total), leaving $352n - 1012$ for the remaining $n - 2$ people, each getting at most Rs $330$: $\frac{352n - 1012}{n - 2} \le 330.$

Step 3 — Solve (the denominator is positive since $n > 2$): $352n - 1012 \le 330(n - 2) = 330n - 660 \;\Rightarrow\; 22n \le 352 \;\Rightarrow\; n \le 16.$

Step 4 — Check $n = 16$. Total $= 5632$; two take $1012$, leaving $4620$ for $14$ people $= 330$ each. Since $330 \le 330$, it works.

Answer: $\boxed{16}$

💡 Teacher tip: "At most / at least" phrasing signals an inequality, and "maximum possible value" means solve the inequality and then verify the boundary case actually holds — here it lands exactly on the limit.

The setup: Anil borrows Rs $2{,}00{,}000$ at $8\%$ per year compounded half-yearly (so $4\%$ per half-year). He repays Rs $10{,}320$ at the end of year 1, then clears the whole balance at the end of year 3. Find the total interest paid, to the nearest option.

Step 1 — Balance at the end of year 1 (two half-years): $200000 \times (1.04)^2 = 200000 \times 1.0816 = 216320.$

Step 2 — After the Rs $10{,}320$ repayment: $216320 - 10320 = 206000.$

Step 3 — This grows for two more years (four half-years): $(1.04)^4 = 1.16985856, \qquad 206000 \times 1.16985856 \approx 240991.$ So he pays about Rs $240{,}991$ at the end of year 3.

Step 4 — Total paid and interest. $\text{Total paid} = 10320 + 240991 = 251311, \qquad \text{Interest} = 251311 - 200000 = 51311.$

Answer: B

💡 Teacher tip: For half-yearly compounding, halve the annual rate and double the periods: $8\%$ per year over $3$ years becomes $4\%$ over $6$ half-years. A mid-way repayment simply splits the timeline into two growth blocks.

The setup: Jayant buys white shirts at Rs $1000$ and blue at Rs $1125$ each. He marks every shirt at $25\%$ above the average cost, sells all at a $10\%$ discount, and profits Rs $51000$. He buys at least one of each colour. Find the maximum total number of shirts.

Let $w$ white and $b$ blue shirts. The average cost per shirt is $\dfrac{1000w + 1125b}{w + b}$. The selling price per shirt is $0.9 \times 1.25 = 1.125$ times that average — so he makes a flat $12.5\%$ profit on cost.

Step 1 — Turn profit into an equation. $0.125(1000w + 1125b) = 51000 \;\Rightarrow\; 1000w + 1125b = 408000.$ Divide by $125$: $\;8w + 9b = 3264.$

Step 2 — Maximise $w + b$. Since $w + b = \dfrac{w + 3264}{9}$, maximising the total means maximising $w$, i.e. minimising $b$.

Step 3 — Integer condition. For $w$ to be a whole number we need $b$ to be a multiple of $8$ (because $3264$ is a multiple of $8$ and $9 \equiv 1 \pmod 8$). With $b \ge 1$, the smallest valid $b$ is $8$. Then $w = \frac{3264 - 72}{8} = \frac{3192}{8} = 399.$

Total $= 399 + 8 = 407.$

Answer: $\boxed{407}$

💡 Teacher tip: Because both the markup and the discount apply to the average cost, the colour mix never changes the profit rate — it's a flat $12.5\%$. That collapses the whole thing to one linear equation you solve in integers.

Two parts: first find the integer $k$, then minimise an expression in $y$.

Step 1 — Find the largest $k$ giving no real roots. Expand $(x-1)^2 + 2kx + 11 = x^2 + (2k - 2)x + 12$. "No real roots" means the discriminant is negative: $(2k - 2)^2 - 48 < 0 \;\Rightarrow\; (k - 1)^2 < 12.$ So $1 - \sqrt{12} < k < 1 + \sqrt{12}$, roughly $-2.46 < k < 4.46$. The largest integer is $k = 4$.

Step 2 — Minimise with $k = 4$. The expression becomes $\frac{4}{4y} + 9y = \frac{1}{y} + 9y.$ By AM–GM (for $y > 0$), $\frac{1}{y} + 9y \ge 2\sqrt{\frac{1}{y} \cdot 9y} = 2\sqrt{9} = 6,$ with equality when $\dfrac{1}{y} = 9y$, i.e. $y = \tfrac{1}{3}$.

Check: at $y = \tfrac{1}{3}$, $\;3 + 3 = 6$. ✓

Answer: $\boxed{6}$

💡 Teacher tip: Any expression $\dfrac{a}{y} + by$ with $y > 0$ bottoms out at $2\sqrt{ab}$ by AM–GM — no calculus needed, because the two terms multiply to the constant $ab$.

The setup: a triangle has its vertices on a circle of radius $r$, one side is a diameter, and the other two sides are in the ratio $a: b$. Find the triangle's area.

Step 1 — Spot the right angle. A triangle with one side as the diameter of its circumcircle is right-angled (Thales' theorem), the right angle facing the diameter. So the diameter is the hypotenuse, of length $2r$, and the other two sides are the legs.

Let the legs be $ak$ and $bk$.

Step 2 — Find $k$ from Pythagoras. $(ak)^2 + (bk)^2 = (2r)^2 \;\Rightarrow\; k^2(a^2 + b^2) = 4r^2 \;\Rightarrow\; k^2 = \frac{4r^2}{a^2 + b^2}.$

Step 3 — Compute the area (half the product of the legs): $\text{Area} = \frac{1}{2}(ak)(bk) = \frac{1}{2}ab\,k^2 = \frac{1}{2}ab \cdot \frac{4r^2}{a^2 + b^2} = \frac{2ab\,r^2}{a^2 + b^2}.$

Answer: B

💡 Teacher tip: "One side is a diameter" is the standard signal for a right angle — reach for it at once. After that, the area is just $\tfrac{1}{2}$(leg)(leg), with the legs coming from the ratio plus Pythagoras.

The setup: two sequences, $a_n = 13 + 6(n-1)$ and $b_n = 15 + 7(n-1)$. We want the largest three-digit number that appears in both.

First tidy them up: $a_n = 6n + 7$ and $b_n = 7n + 8$.

Step 1 — Find one number common to both. A shared value needs $6n + 7 = 7m + 8$, i.e. $6n - 7m = 1$. Trying small numbers, $n = 6, m = 5$ works: $36 - 35 = 1$. So the value $6(6) + 7 = 43$ is common to both lists.

Step 2 — The common terms form their own AP. When two arithmetic progressions overlap, their shared terms are themselves an AP, spaced apart by the LCM of the two step sizes. Here $\operatorname{lcm}(6, 7) = 42$, so the common terms are $43,\; 85,\; 127,\; \ldots,\; 43 + 42k.$

Step 3 — Push up to the largest three-digit member. Require $43 + 42k \le 999$: $42k \le 956 \quad\Rightarrow\quad k \le 22.76\ldots \quad\Rightarrow\quad k = 22.$ $43 + 42(22) = 43 + 924 = 967.$ The next one, $967 + 42 = 1009$, has four digits — so $967$ is the last three-digit value.

💡 Teacher tip: Find one common term by trial, then ride the LCM-sized step to the range you want. You never have to list the whole sequence.

Answer: $\boxed{967}$

What we want: the area boxed in by the $Y$-axis ($x = 0$), the line $x = 5$, and the curve $|x - y| - |x - 5| = 2$.

Step 1 — Decode the curve. Absolute-value curves are piecewise-linear; split at where each modulus turns over.

For $x \le 5$: $|x - 5| = 5 - x$, so $|x - y| = 7 - x$, giving two lines: $y = 2x - 7 \qquad\text{and}\qquad y = 7.$

Step 2 — Read the region inside the strip $0 \le x \le 5$. Here the top boundary is $y = 7$ and the bottom is $y = 2x - 7$.

• At $x = 0$: top $7$, bottom $-7$.
• At $x = 5$: top $7$, bottom $3$.

Step 3 — Integrate the vertical gap. $\text{Area} = \int_0^5 \big[7 - (2x - 7)\big]\,dx = \int_0^5 (14 - 2x)\,dx = \big[14x - x^2\big]_0^5 = 70 - 25 = 45.$

Answer: $\boxed{45}$

💡 Teacher tip: For absolute-value equations, split at the points where each modulus's inside is zero (here $x = y$ and $x = 5$). Inside the strip only two lines survive, turning a scary equation into a trapezium you can integrate directly.

What we want: given that $k$ divides both $m + 2n$ and $3m + 4n$, which pair must $k$ divide? The key idea: any whole-number combination of two things $k$ divides is also divisible by $k$.

Step 1 — Knock out $m$. Take $2(m + 2n) - (3m + 4n)$: $2m + 4n - 3m - 4n = -m.$ So $k \mid m$.

Step 2 — Knock out $m$ the other way for $n$. Take $3(m + 2n) - (3m + 4n)$: $3m + 6n - 3m - 4n = 2n.$ So $k \mid 2n$.

Step 3 — Conclude. $k$ divides both $m$ and $2n$, so it is a common divisor of $m$ and $2n$.

Answer: C

💡 Teacher tip: When you're told a number divides two linear combinations, build fresh combinations of those to isolate each variable. The right multipliers (here $2$ and $3$) line the terms up to cancel.

What we want: how many positive integers below $50$ have exactly two factors other than $1$ and themselves — that is, exactly $4$ divisors in total.

Numbers with exactly $4$ divisors come in just two shapes:

• $p^3$ (divisors $1, p, p^2, p^3$), or
• $p \cdot q$ with $p, q$ distinct primes (divisors $1, p, q, pq$).

Shape 1 — $p^3 < 50$. Only $2^3 = 8$ and $3^3 = 27$ qualify ($5^3 = 125$ is too big). That's 2.

Shape 2 — $pq < 50$, $p < q$.

That's $8 + 4 + 1 = 13$.

Total: $2 + 13 = 15$.

Answer: $\boxed{15}$

💡 Teacher tip: "Exactly $k$ proper divisors" always reduces to counting total divisors. Memorise the two templates that give exactly four — $p^3$ and $p \cdot q$ — and you can sweep any range fast.

The setup: A and C fill a tank, B drains it. B empties a full tank in one hour less than A takes to fill it ($b = a - 1$). Together A, B, C fill an empty tank in $2$ hours. Separately, B and C run for one hour, then B is shut and C alone finishes in $1$ h $15$ min. Given A takes under $5$ hours, find C's fill time.

Let A, B, C take $a, b, c$ hours. Rates: A $= \tfrac{1}{a}$, B $= -\tfrac{1}{b}$, C $= \tfrac{1}{c}$, with $b = a - 1$.

Condition 1 (all three, $2$ hours): \frac{1}{a} - \frac{1}{b} + \frac{1}{c} = \frac{1}{2}. \tag{1}

Condition 2 (B+C for $1$ h, then C for $1.25$ h): \left(\frac{1}{c} - \frac{1}{b}\right) + 1.25 \cdot \frac{1}{c} = 1 \;\Rightarrow\; \frac{2.25}{c} - \frac{1}{b} = 1. \tag{2}

Step 1 — From (2), with $b = a - 1$: \frac{2.25}{c} = 1 + \frac{1}{a - 1} = \frac{a}{a - 1} \;\Rightarrow\; \frac{1}{c} = \frac{a}{2.25(a - 1)}. \tag{3}

Step 2 — Substitute into (1) and clear denominators. After simplifying, this reduces to $0.125a^2 - 1.125a + 2.25 = 0 \;\Rightarrow\; a^2 - 9a + 18 = 0 \;\Rightarrow\; (a - 3)(a - 6) = 0.$

Step 3 — Pick $a < 5$: so $a = 3$. From (3), $\dfrac{1}{c} = \dfrac{3}{2.25 \times 2} = \dfrac{2}{3}$, so $c = 1.5$ hours $= 90$ minutes.

Check: $a = 3, b = 2, c = 1.5$. Equation (1): $\tfrac{1}{3} - \tfrac{1}{2} + \tfrac{2}{3} = \tfrac{1}{2}$ ✓. Condition 2: one hour of B+C fills $\tfrac{1}{6}$, the remaining $\tfrac{5}{6}$ at rate $\tfrac{2}{3}$ takes $1.25$ h ✓.

Answer: A

💡 Teacher tip: In pipe problems, always work with rates (tank per hour) and let a drain be a negative rate. The "extra $5$ hours" clue is there to pick between the two roots — never ignore such a constraint.

What we want: the value of $\log_3(3x^2)$, given $\log_{\sqrt3}(x) + \frac{\log_x(25)}{\log_x(0.008)} = \frac{16}{3}.$

Step 1 — Simplify the fraction. By change of base, $\dfrac{\log_x(25)}{\log_x(0.008)} = \log_{0.008}(25)$. Now $0.008 = \dfrac{1}{125} = 5^{-3}$ and $25 = 5^2$, so $\log_{5^{-3}}(5^2) = \frac{2}{-3} = -\frac{2}{3}.$

Step 2 — Simplify the other logarithm. Since $\sqrt3 = 3^{1/2}$, $\log_{\sqrt3}(x) = \frac{\log_3(x)}{1/2} = 2\log_3(x).$

Step 3 — Solve for $\log_3(x)$. $2\log_3(x) - \frac{2}{3} = \frac{16}{3} \;\Rightarrow\; 2\log_3(x) = 6 \;\Rightarrow\; \log_3(x) = 3.$

Step 4 — Compute the target. $\log_3(3x^2) = \log_3(3) + 2\log_3(x) = 1 + 2(3) = 7.$

Answer: $\boxed{7}$

💡 Teacher tip: A ratio of two logs with the same base is itself a single log: $\dfrac{\log_x A}{\log_x B} = \log_B A$. Spotting that instantly clears the messiest-looking term.

What we want: the sum of all $x$ solving $2^{4x^2} - 2^{2x^2 + x + 16} + 2^{2x + 30} = 0$.

Step 1 — Rearrange and divide. Rewrite as $2^{4x^2} + 2^{2x + 30} = 2^{2x^2 + x + 16}$, then divide every term by $2^{2x^2 + x + 16}$: $2^{2x^2 - x - 16} + 2^{14 + x - 2x^2} = 1.$ Let $t = 2x^2 - x$. The second exponent is $14 - t$, so $2^{t - 16} + 2^{14 - t} = 1.$

Step 2 — Make it a quadratic. Let $u = 2^{t - 16}$. Then $2^{14 - t} = \dfrac{1}{4u}$, and $u + \frac{1}{4u} = 1 \;\Rightarrow\; 4u^2 - 4u + 1 = 0 \;\Rightarrow\; (2u - 1)^2 = 0.$ So $u = \tfrac{1}{2}$, giving $2^{t - 16} = 2^{-1}$, hence $t = 15$.

Step 3 — Solve for $x$. $2x^2 - x = 15 \Rightarrow 2x^2 - x - 15 = 0 \Rightarrow (2x + 5)(x - 3) = 0$: $x = 3 \quad\text{or}\quad x = -\frac{5}{2}.$

Step 4 — Sum. $3 + \left(-\tfrac{5}{2}\right) = \tfrac{1}{2}$.

Answer: D

💡 Teacher tip: When three exponential terms aren't a neat "$a^2 - ab + b^2$" pattern, divide through by the middle term so the two survivors multiply to a constant — that turns everything into one quadratic.

The setup: Minu buys sunglasses for Rs $1000$, sells to Kanu at $20\%$ profit, buys them back at $20\%$ loss, then sells to Tanu. Her total profit across everything is Rs $500$. Find her profit percentage on the final sale to Tanu.

Track every rupee. Let the final price to Tanu be $P$.

Step 1 — Total profit is Rs $500$. Add up the cash flows: $-1000 + 1200 - 960 + P = 500 \;\Rightarrow\; P = 1260.$

Step 2 — Profit percentage on the Tanu sale. She last bought them for Rs $960$ and sold for Rs $1260$: $\frac{1260 - 960}{960} \times 100 = \frac{300}{960} \times 100 = 31.25\%.$

Answer: C

💡 Teacher tip: "Profit from all her transactions" means total cash in minus total cash out over the whole chain — not the profit on a single sale. Lay out the cash column first so you don't mix up cost bases.

What we want: the largest possible value of $n - m$ where $a^m b^n = 144^{145}$, with $a, b, m, n$ natural numbers and $a, b > 1$. To make $n - m$ big, push $n$ up and $m$ down.

Step 1 — Prime-factorise. $144 = 2^4 \cdot 3^2$, so $144^{145} = 2^{580} \cdot 3^{290}.$

Step 2 — Smallest $m$. As a natural number, $m$ can be as small as $1$. With $m = 1$ we need $a \cdot b^n = 2^{580} \cdot 3^{290}$.

Step 3 — Largest $n$. Choose $b$ to be the smallest available prime, $b = 2$, so $b^n = 2^n$ stretches as far as the power of $2$ allows: $n \le 580$. Take $n = 580$; then $a = \frac{2^{580} \cdot 3^{290}}{2^{580}} = 3^{290} > 1. \checkmark$

Step 4 — Answer. With $m = 1$, $n = 580$: $n - m = 580 - 1 = 579.$

Answer: D

💡 Teacher tip: Pick the smallest base for $b$, because $2$ carries the highest exponent — that lets $n$ reach the furthest. A composite base like $6$ would cap $n$ at $\min(580, 290) = 290$, far worse.