CAT 2023 Quant Questions — Slot 3

The setup: A and B collect coins per week in the ratio $3: 4$. Write A's weekly count as $3k$ and B's as $4k$ for some positive integer $k$.

Step 1 — Turn each condition into a divisibility requirement.

• A over $5$ weeks $= 15k$ is a multiple of $7$. Since $15$ and $7$ share no factor, this forces $7 \mid k$.
• B over $3$ weeks $= 12k$ is a multiple of $24$. Since $24 = 2 \times 12$, this forces $2 \mid k$.

Step 2 — Smallest $k$. $k$ must be a multiple of both $7$ and $2$, so the least value is $k = \operatorname{lcm}(7, 2) = 14.$

Step 3 — A's weekly collection. $3k = 3 \times 14 = 42.$

💡 Teacher tip: A "per week" ratio lets you write the weekly amounts as $3k$ and $4k$ with one shared $k$. Each "is a multiple of …" clause becomes a divisibility rule on $k$, and you take the LCM of all of them.

Answer: $\boxed{42}$

The setup: two sequences, $a_n = 46 + 8n$ and $b_n = 98 + 4n$ for $n \le 100$. Find the sum of all terms common to both.

Step 1 — Find the common terms. A shared value needs $46 + 8n = 98 + 4m \;\Rightarrow\; 8n - 4m = 52 \;\Rightarrow\; 2n - m = 13 \;\Rightarrow\; m = 2n - 13.$

Step 2 — Which $n$ are valid? We need $1 \le m \le 100$:

• $m = 2n - 13 \ge 1 \Rightarrow n \ge 7.$
• $m = 2n - 13 \le 100 \Rightarrow n \le 56$ (since $n$ is a whole number).

So $n = 7, 8, \ldots, 56$ — that's $56 - 7 + 1 = 50$ common terms.

Step 3 — The common terms form an AP. As $n$ rises by $1$, the value $46 + 8n$ rises by $8$, so: $\text{first} = 46 + 8(7) = 102, \qquad \text{last} = 46 + 8(56) = 494.$

Step 4 — Sum the AP. $\text{Sum} = \frac{N}{2}(\text{first} + \text{last}) = \frac{50}{2}(102 + 494) = 25 \times 596 = 14900.$

💡 Teacher tip: The terms shared by two APs form another AP. Solve $a_n = b_m$ for one index in terms of the other, read off the valid range, then use $\tfrac{N}{2}(\text{first} + \text{last})$.

Answer: A — $14900$

The setup: an infinite sum whose $k$-th term (starting at $k = 0$) is a small geometric sum on top of a power of $4$: $T_k = \frac{1 + \tfrac13 + \cdots + \left(\tfrac13\right)^k}{4^k}.$

Step 1 — Close the numerator. The numerator is a finite geometric series: $1 + \tfrac13 + \cdots + \left(\tfrac13\right)^k = \frac{1 - (1/3)^{k+1}}{1 - 1/3} = \frac32\left(1 - \frac{1}{3^{k+1}}\right).$ So $T_k = \frac32\left(\frac{1}{4^k} - \frac{1}{3^{k+1} \cdot 4^k}\right).$

Step 2 — Sum each piece over $k \ge 0$. Both are infinite geometric series: $\sum_{k=0}^{\infty} \frac{1}{4^k} = \frac{1}{1 - 1/4} = \frac43,$ $\sum_{k=0}^{\infty} \frac{1}{3^{k+1} \cdot 4^k} = \frac13 \sum_{k=0}^{\infty} \frac{1}{12^k} = \frac13 \cdot \frac{1}{1 - 1/12} = \frac13 \cdot \frac{12}{11} = \frac{4}{11}.$

Step 3 — Combine. $\text{Sum} = \frac32\left(\frac43 - \frac{4}{11}\right) = \frac32 \cdot 4\left(\frac13 - \frac{1}{11}\right) = 6 \cdot \frac{8}{33} = \frac{48}{33} = \frac{16}{11}.$

💡 Teacher tip: A term that is a geometric sum divided by a geometric denominator splits into two infinite geometric series — sum each separately with $\tfrac{a}{1 - r}$. The nested fractions look intimidating but unravel into two clean pieces.

Answer: D — $\dfrac{16}{11}$

The setup: the largest-area rectangle that fits inside a semicircle of radius $2$, with its base on the diameter. Find the ratio of its longer to shorter side.

Step 1 — Set up coordinates. Place the semicircle with its flat side on the $x$-axis, centre at the origin. Let the rectangle have half-width $x$ (so full width $2x$) and height $y$. Its top corner $(x, y)$ lies on the circle: x^2 + y^2 = 2^2 = 4. \tag{1}

Step 2 — Maximise the area. Area $A = 2xy$. Working with $A^2$ to avoid the square root and using (1): $A^2 = 4x^2 y^2 = 4x^2(4 - x^2).$ Let $u = x^2$; then $A^2 = 4(4u - u^2)$, a downward parabola in $u$ peaking at $u = 2$. So $x^2 = 2$, giving $x = \sqrt2$ and $y^2 = 4 - 2 = 2$, i.e. $y = \sqrt2.$

Step 3 — Read off the sides. Width $= 2x = 2\sqrt2$, height $= y = \sqrt2.$

Step 4 — Ratio. $\text{longer}: \text{shorter} = 2\sqrt2: \sqrt2 = 2: 1.$

💡 Teacher tip: For a rectangle inscribed in a semicircle, the area-maximising corner always sits where $x^2 = y^2$ — that's why the optimal rectangle is exactly $2: 1$ (width: height) no matter the radius.

Answer: A — $2: 1$

The setup: find the first two natural numbers with exactly $15$ divisors, then add them.

Step 1 — What shapes give exactly $15$ divisors? If a number factors as $p^a q^b \cdots$, its divisor count is $(a+1)(b+1)\cdots$. We need this to equal $15$. Since $15 = 15 \times 1 = 5 \times 3$, the two possible shapes are

• $p^{14}$ (from $15 = 15$), or
• $p^4 q^2$ (from $15 = 5 \times 3$), with $p, q$ distinct primes.

The $p^{14}$ form is enormous ($2^{14} = 16384$), so the smallest such numbers come from $p^4 q^2$.

Step 2 — Sweep the small $p^4 q^2$ candidates. To keep the value small, put the larger exponent on the smaller prime:

The two smallest are $144$ and $324$.

Step 3 — Add them. $144 + 324 = 468.$

💡 Teacher tip: To make $p^4 q^2$ as small as possible, give the bigger exponent to the smaller prime. That single rule is why $2^4 \cdot 3^2 = 144$ beats $2^2 \cdot 3^4 = 324$.

Answer: $\boxed{468}$

The setup: for $x^2 + bx + c = 0$ with roots $\alpha, \beta$, we know two facts about the reciprocals of the roots. Recall $\alpha + \beta = -b$ and $\alpha\beta = c$. Let $S = \alpha + \beta$ and $P = \alpha\beta$.

Step 1 — "Difference of reciprocals $= \tfrac13$." \left|\frac1\alpha - \frac1\beta\right| = \frac13 \;\Rightarrow\; \frac{|\beta - \alpha|}{|\alpha\beta|} = \frac13 \;\Rightarrow\; \frac{(\beta - \alpha)^2}{(\alpha\beta)^2} = \frac19 \;\Rightarrow\; \frac{S^2 - 4P}{P^2} = \frac19. \tag{1}

Step 2 — "Sum of reciprocal squares $= \tfrac59$." \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{S^2 - 2P}{P^2} = \frac59. \tag{2}

Step 3 — Subtract (1) from (2). The $S^2$ terms cancel: $\frac{(S^2 - 2P) - (S^2 - 4P)}{P^2} = \frac59 - \frac19 \;\Rightarrow\; \frac{2P}{P^2} = \frac49 \;\Rightarrow\; \frac2P = \frac49 \;\Rightarrow\; P = \frac92.$ So $c = P = \tfrac92$.

Step 4 — Find $S$. Put $P = \tfrac92$ into (1): $\frac{S^2 - 18}{(9/2)^2} = \frac19 \;\Rightarrow\; S^2 - 18 = \frac{81/4}{9} = \frac94 \;\Rightarrow\; S^2 = \frac{81}{4} \;\Rightarrow\; S = \pm\frac92.$

Step 5 — Maximise $b + c = -S + c$. With $c = \tfrac92$:

• $S = +\tfrac92$: $\;b + c = -\tfrac92 + \tfrac92 = 0.$
• $S = -\tfrac92$: $\;b + c = \tfrac92 + \tfrac92 = 9.$

The largest value is $9$.

💡 Teacher tip: Converting conditions on the roots into the symmetric quantities $S = \alpha+\beta$ and $P = \alpha\beta$ is the standard first move. Here subtracting the two conditions kills $S^2$ and hands you $P$ in one line.

Answer: $\boxed{9}$

The setup: a population falls by $y\%$, then rises by $x\%$, with $x - y = 10$, and the final population beats the starting one. We want the lowest possible population after the fall.

Let the 2020 population be $P = 100000$. Then $\text{2021} = P\left(1 - \tfrac{y}{100}\right), \qquad \text{2022} = P\left(1 - \tfrac{y}{100}\right)\left(1 + \tfrac{x}{100}\right).$ The condition "2022 exceeds 2020" is $\left(1 - \tfrac{y}{100}\right)\left(1 + \tfrac{x}{100}\right) > 1.$

Step 1 — Substitute $x = y + 10$ and clear denominators. Multiplying by $100^2$: $(100 - y)(110 + y) > 10000.$ Expanding: $11000 - 10y - y^2 > 10000$, i.e. $y^2 + 10y - 1000 < 0.$

Step 2 — Find the ceiling on $y$. The positive root of $y^2 + 10y - 1000 = 0$ is about $27.0$. Checking integers: $y = 27$ gives $729 + 270 - 1000 = -1 < 0$ ✓, while $y = 28$ gives $784 + 280 - 1000 = 64 > 0$ ✗. So the largest allowed natural number is $y = 27$.

Step 3 — Lowest 2021 population means the biggest drop, so take $y = 27$. $\text{2021} = 100000\left(1 - \tfrac{27}{100}\right) = 100000 \times \tfrac{73}{100} = 73000.$ Sanity check on 2022 (with $x = 37$): $\tfrac{73}{100} \times \tfrac{137}{100} = \tfrac{10001}{10000} > 1$ ✓.

💡 Teacher tip: "Lowest population after a fall" means the largest percentage drop — so push $y$ as high as the "2022 beats 2020" rule allows. Writing that rule as one quadratic in $y$ pins the ceiling cleanly.

Answer: C — $73000$

The setup: an isosceles triangle with $AB = AC$. $AD$ is the altitude from $A$, $BE$ the altitude from $B$, and they meet at $O$ (the orthocentre) with $\angle AOB = 105^\circ$. We want $AD / BE$.

Step 1 — Find the angles. A standard orthocentre fact: the angle between the altitudes from $A$ and $B$ equals $180^\circ - C$. So $\angle AOB = 180^\circ - C = 105^\circ \;\Rightarrow\; C = 75^\circ.$ Since $AB = AC$, the base angles are equal: $B = C = 75^\circ$, hence $A = 180^\circ - 150^\circ = 30^\circ.$

Step 2 — Relate the two altitudes through area. The triangle's area can be written with either altitude: $\tfrac12 \cdot BC \cdot AD = \tfrac12 \cdot AC \cdot BE \;\Rightarrow\; \frac{AD}{BE} = \frac{AC}{BC}.$

Step 3 — Get $AC / BC$ from the sine rule. $\frac{AC}{BC} = \frac{\sin B}{\sin A} = \frac{\sin 75^\circ}{\sin 30^\circ} = \frac{\sin 75^\circ}{1/2} = 2\sin 75^\circ.$

Step 4 — Simplify. Since $75^\circ = 90^\circ - 15^\circ$, we have $\sin 75^\circ = \cos 15^\circ$, so $\frac{AD}{BE} = 2\cos 15^\circ.$

💡 Teacher tip: The ratio of two altitudes always equals the reciprocal ratio of the sides they drop onto ($\tfrac{h_a}{h_b} = \tfrac{b}{a}$) — straight from equating the area two ways. Combined with $\angle AOB = 180^\circ - C$, the problem falls out in a couple of lines.

Answer: C — $2\cos 15^\circ$

The setup: two boats and a river. Let the river speed be $s$, the distance A→B be $d$, and the still-water speeds be $u$ (boat 1) and $v$ (boat 2).

Step 1 — Get $d$ and $u$ from boat 1. Downstream takes $2$ h, upstream $3$ h, so $d = 2(u + s)$ and $d = 3(u - s)$. Equating: $2(u + s) = 3(u - s) \;\Rightarrow\; 2u + 2s = 3u - 3s \;\Rightarrow\; u = 5s,$ and then $d = 2(5s + s) = 12s.$

Step 2 — Get $v$ from boat 2's $6$-hour round trip. $\frac{d}{v - s} + \frac{d}{v + s} = 6 \;\Rightarrow\; 12s\left(\frac{1}{v-s} + \frac{1}{v+s}\right) = 6 \;\Rightarrow\; \frac{12s \cdot 2v}{v^2 - s^2} = 6.$ So $4sv = v^2 - s^2$, i.e. $v^2 - 4sv - s^2 = 0$, giving (positive root) $v = s(2 + \sqrt5).$

Step 3 — Which boat is slower? $v = s(2 + \sqrt5) \approx 4.24s$, which is less than $u = 5s$. So boat 2 is slower.

Step 4 — Time for boat 2 downstream, A→B. $T = \frac{d}{v + s} = \frac{12s}{s(2 + \sqrt5) + s} = \frac{12}{3 + \sqrt5}.$ Rationalise by multiplying top and bottom by $3 - \sqrt5$: $T = \frac{12(3 - \sqrt5)}{(3)^2 - (\sqrt5)^2} = \frac{12(3 - \sqrt5)}{9 - 5} = \frac{12(3 - \sqrt5)}{4} = 3(3 - \sqrt5).$

💡 Teacher tip: For river round-trips, first nail down $s$ and $d$ from whichever boat gives complete data, then use the round-trip-time equation for the other boat. The $\sqrt5$ is the fingerprint of $v^2 - 4sv - s^2 = 0$.

Answer: D — $3(3 - \sqrt5)$

The setup: a chain of mixtures. The reliable way through is to carry the actual amount of sugar forward, letting only the total change.

• Mixture A (cocoa: sugar $= 3: 2$, total $5$ parts): sugar fraction $= \tfrac25.$
• Mixture B (coffee: sugar $= 7: 3$, total $10$ parts): sugar fraction $= \tfrac{3}{10}.$

Step 1 — Sugar in mixture C ($A: B = 2: 3$). Take $2$ units of A and $3$ units of B: $\text{sugar} = 2 \times \tfrac25 + 3 \times \tfrac{3}{10} = \tfrac45 + \tfrac{9}{10} = \tfrac{17}{10}.$ Mixture C has $2 + 3 = 5$ units total, so its sugar fraction is $\frac{17/10}{5} = \frac{17}{50}.$

Step 2 — Dilute with an equal amount of milk. Milk has no sugar, so mixing C with an equal volume of milk doubles the total and halves the sugar fraction: $\frac{17}{50} \times \frac12 = \frac{17}{100} = 17\%.$

💡 Teacher tip: In chain-mixture problems, track "amount of the substance" rather than re-averaging percentages at each stage. Here $\tfrac{17}{10}$ units of sugar survive every step; only the denominator (total volume) grows.

Answer: A — $17\%$

The setup: Gautam and Suhani finish together in $20$ days. There's also a "compensation" clause linking their rates. Let their daily work be $g$ and $s$ (fractions of the job per day).

Step 1 — Together in $20$ days. g + s = \frac{1}{20}. \tag{1}

Step 2 — Decode the compensation clause. If Gautam does $60\%$ of his usual work, Suhani must do $150\%$ of hers so the day still completes the same total $g + s$: $0.6g + 1.5s = g + s \;\Rightarrow\; 0.5s = 0.4g \;\Rightarrow\; 5s = 4g \;\Rightarrow\; g = \tfrac54 s.$ So Gautam is the faster worker.

Step 3 — Solve for the rates. Put $g = \tfrac54 s$ into (1): $\tfrac54 s + s = \tfrac94 s = \frac{1}{20} \;\Rightarrow\; s = \frac{4}{180} = \frac{1}{45}, \qquad g = \tfrac54 \times \tfrac{1}{45} = \frac{1}{36}.$

Step 4 — Faster worker alone. Gautam's solo time is $1/g = 36$ days.

💡 Teacher tip: "Doing $150\%$ of usual work" just scales that person's rate by $1.5$ for the day. Setting the modified day's output equal to a normal day gives a clean ratio between the two rates — no simultaneous equations needed for that step.

Answer: $\boxed{36}$

The setup: two parts. First figure out which $n$ satisfy $5^{n-1} < 3^{n+1}$, then find the smallest $m$ that makes $3^{n+1} < 2^{n+m}$ hold for every such $n$.

Step 1 — Which $n$ satisfy $5^{n-1} < 3^{n+1}$? Just test values:

The left side grows by a factor of $5$ each step versus $3$ on the right, so once it overtakes it never comes back. Valid values are $n = 1, 2, 3, 4, 5$.

Step 2 — Smallest $m$ that beats every valid $n$. We need $3^{n+1} < 2^{n+m}$, i.e. $m > (n+1)\log_2 3 - n$. With $\log_2 3 \approx 1.585$:

Step 3 — Take the worst case. "For each such $n$" means $m$ must clear the largest lower bound, which is $4.51$ at $n = 5$. The least integer above it is $m = 5$.

Check: at $n = 5$, $3^6 = 729 < 2^{10} = 1024$ ✓, and smaller $n$ leave even more room.

💡 Teacher tip: "For each such $n$" is a signal to compute the bound for every valid $n$ and keep the maximum. The boundary case ($n = 5$, where $5^{n-1}$ nearly catches $3^{n+1}$) is the one that decides the answer.

Answer: $\boxed{5}$

The setup: exactly $41$ powers of $2$ lie strictly between $8^m$ and $8^n$. Find the smallest possible $n + m$.

The key move: everything is secretly a power of $2$. Since $8 = 2^3$, we have $8^m = 2^{3m}$ and $8^n = 2^{3n}$. The powers of $2$ strictly between them are $2^{3m+1},\; 2^{3m+2},\; \ldots,\; 2^{3n-1}.$

Step 1 — Count them. These are consecutive exponents from $3m+1$ up to $3n-1$, a count of $(3n - 1) - (3m + 1) + 1 = 3(n - m) - 1.$

Step 2 — Set the count equal to $41$. $3(n - m) - 1 = 41 \quad\Rightarrow\quad 3(n - m) = 42 \quad\Rightarrow\quad n - m = 14.$

Step 3 — Minimise $n + m$. With the gap $n - m = 14$ fixed and $m, n$ positive integers, the sum $n + m$ shrinks as $m$ shrinks. The smallest allowed is $m = 1$, giving $n = 15$, so $n + m = 1 + 15 = 16.$

💡 Teacher tip: "Powers of $2$ between two numbers" becomes plain counting of consecutive integers once you rewrite the bounds with base $2$. After that it's just $n - m = 14$ and pushing $m$ as low as it can go.

Answer: D — $16$

The setup: we're told $x^8 + x^{-8} = 47$ and asked for $x^9 + x^{-9}$. The trick is one identity used in both directions — go down to $x + \tfrac1x$, then climb back up.

The identity is $\left(x^k + x^{-k}\right)^2 = x^{2k} + 2 + x^{-2k}$, which rearranges to $x^{2k} + x^{-2k} = \left(x^k + x^{-k}\right)^2 - 2.$ Read backwards, it also lets us halve the power by taking a square root and subtracting.

Step 1 — Halve the power three times.

(We take the positive roots since $x > 0$.)

Step 2 — Build $x^3 + x^{-3}$. Using $\left(x + \tfrac1x\right)^3 = x^3 + x^{-3} + 3\left(x + \tfrac1x\right)$: $x^3 + x^{-3} = (\sqrt5)^3 - 3\sqrt5 = 5\sqrt5 - 3\sqrt5 = 2\sqrt5.$

Step 3 — Reach the ninth powers. First, $x^6 + x^{-6} = \left(x^3 + x^{-3}\right)^2 - 2 = (2\sqrt5)^2 - 2 = 20 - 2 = 18.$

Now use $\left(x^3 + x^{-3}\right)\left(x^6 + x^{-6}\right) = x^9 + x^{-9} + \left(x^3 + x^{-3}\right)$, i.e. $x^9 + x^{-9} = \left(x^3 + x^{-3}\right)\left(x^6 + x^{-6} - 1\right) = 2\sqrt5 \,(18 - 1) = 2\sqrt5 \times 17 = 34\sqrt5.$

💡 Teacher tip: The single identity $\left(x^k + x^{-k}\right)^2 = x^{2k} + 2 + x^{-2k}$ is a two-way street — square-and-subtract to climb up, root-and-subtract to come down. Reducing a scary 8th-power equation all the way to $x + \tfrac1x$ makes everything else routine.

Answer: D — $34\sqrt5$

The setup: the same trio A, B, C is joined by either D or E. We compare the two average shifts. Let the sum of the three weights be $S$, so their average is $\tfrac{S}{3}$.

Step 1 — Write the new averages.

• D joins (four people now): $\dfrac{S + D}{4} = \dfrac{S}{3} - x.$
• E joins instead: $\dfrac{S + E}{4} = \dfrac{S}{3} + 2x.$

Step 2 — Solve each for the newcomer's weight. $D = 4\left(\tfrac{S}{3} - x\right) - S = \tfrac{S}{3} - 4x, \qquad E = 4\left(\tfrac{S}{3} + 2x\right) - S = \tfrac{S}{3} + 8x.$

Step 3 — Use $E - D = 12$. Subtracting wipes out $S$ entirely: $E - D = \left(\tfrac{S}{3} + 8x\right) - \left(\tfrac{S}{3} - 4x\right) = 12x = 12 \;\Rightarrow\; x = 1.$

💡 Teacher tip: When one fixed group is joined by two different people, subtracting the two "new average" equations cancels the unknown group total. You're left with a direct link between the weight gap and $x$.

Answer: C — $x = 1$

The setup: three quantities are in arithmetic progression. Once we simplify the logs, the AP condition gives one clean equation.

Step 1 — Clean each term with change of base. The rule $\dfrac{\log_c A}{\log_c B} = \log_B A$ turns the messy terms into base-$4$ logs: $\frac{\log_3(2^x - 9)}{\log_3 4} = \log_4(2^x - 9), \qquad \frac{\log_5\!\left(2^x + \tfrac{17}{2}\right)}{\log_5 4} = \log_4\!\left(2^x + \tfrac{17}{2}\right).$ So the three AP terms are $\;\tfrac12,\;\; \log_4(2^x - 9),\;\; \log_4\!\left(2^x + \tfrac{17}{2}\right).$

Step 2 — Apply "middle = average of the outer two". Let $u = 2^x$. The AP condition says twice the middle equals the sum of the ends: $2\log_4(u - 9) = \tfrac12 + \log_4\!\left(u + \tfrac{17}{2}\right).$ Using $2\log_4(u-9) = \log_4(u-9)^2$ and $\tfrac12 = \log_4 2$: $\log_4 (u - 9)^2 = \log_4\!\left[2\left(u + \tfrac{17}{2}\right)\right] = \log_4(2u + 17).$ So $(u - 9)^2 = 2u + 17$, giving $u^2 - 20u + 64 = 0 = (u - 16)(u - 4)$, hence $u = 16$ or $u = 4$.

Step 3 — Reject the invalid root. The term $\log_4(u - 9)$ needs $u - 9 > 0$, i.e. $u > 9$. So $u = 4$ is out and $u = 16$ stands.

Step 4 — Common difference. $d = \log_4(u - 9) - \tfrac12 = \log_4(16 - 9) - \log_4 2 = \log_4 7 - \log_4 2 = \log_4\!\left(\tfrac72\right).$

💡 Teacher tip: When several logs with different bases appear, converting them all to one base collapses the algebra. The AP idea "middle is the average" then becomes a single equation you can solve.

Answer: D — $\log_4(7/2)$

The setup: a work problem with inequalities, ending in "cannot be." Questions like this are really about the range of the answer.

Let the daily rates be $p$ (Rahul), $g$ (Gurmeet), $k$ (Rakshita), with the whole job counted as $1$.

Step 1 — Translate each statement.

• All three take more than $7$ days: $\dfrac{1}{p+g+k} > 7 \Rightarrow p + g + k < \tfrac17.$
• Rahul and Gurmeet take less than $15$ days: $\dfrac{1}{p+g} < 15 \Rightarrow p + g > \tfrac{1}{15}.$
• Actual work done: all three for $6$ days, then Rakshita $3$ more days: $6(p+g+k) + 3k = 1.$

Step 2 — Let $s = p + g$ and solve for $k$. From the work equation, $6s + 9k = 1$, so $k = \dfrac{1 - 6s}{9}.$

Step 3 — Bound $s$. Put this into $p + g + k < \tfrac17$: $s + \frac{1 - 6s}{9} < \frac17 \;\Rightarrow\; \frac{3s + 1}{9} < \frac17 \;\Rightarrow\; 21s < 2 \;\Rightarrow\; s < \frac{2}{21}.$ Together with $s > \tfrac{1}{15}$: $\;\dfrac{1}{15} < s < \dfrac{2}{21}.$

Step 4 — Convert to Rakshita's solo time $T = \tfrac1k$. As $s$ increases, $k$ decreases, so $T$ increases. At the two (excluded) endpoints:

Because $s$ is strictly between these, $T$ is strictly between $15$ and $21$. The whole-number possibilities are $16, 17, 18, 19, 20$. So $T$ cannot be $21$.

💡 Teacher tip: "Cannot be" on a rate problem means "find the open interval." The strict inequalities from "more than $7$" and "less than $15$" are exactly what exclude the endpoints $15$ and $21$.

Answer: D — $21$

The setup: a merchant plays two tricks (free length when buying, short measure when selling) plus a discount. The clean way through is to track the cost per cm and the revenue per cm; the profit is just their ratio.

Step 1 — Effective cost per cm. He pays for $100$ cm but receives $105$ cm (the extra $5$ free). So Rs $100$ buys $105$ cm: $\text{cost per cm} = \frac{100}{105} = \frac{20}{21}.$

Step 2 — Effective revenue per cm. The list price is Rs $110$ per metre with a $5\%$ discount, but for every $100$ cm the customer pays for, only $95$ cm is handed over. For each $95$ cm actually given, the customer is charged the discounted price of $100$ cm: $\text{charged} = 110 \times 0.95 = 104.5 \text{ per } 95 \text{ cm} \;\Rightarrow\; \text{revenue per cm} = \frac{104.5}{95} = \frac{209}{190}.$

Step 3 — Profit percentage. $\frac{\text{revenue per cm}}{\text{cost per cm}} - 1 = \frac{209/190}{20/21} - 1 = \frac{209 \times 21}{190 \times 20} - 1 = \frac{4389}{3800} - 1 = \frac{589}{3800} \approx 0.155.$ That is $15.5\%$.

💡 Teacher tip: Both a "free length" and a "short measure" are just different ways of changing the true rate per cm. Reduce everything to "revenue per cm ÷ cost per cm" and every twist folds into one division.

Answer: C — $15.5\%$

The setup: mangoes are $40\%$ of the stock; the seller sells half the mangoes, $96$ bananas, and $40\%$ of the apples, which turns out to be exactly half the total stock. Find the smallest possible starting total.

Let the total be $T$. Then mangoes $= 0.4T$; write apples $= a$, bananas $= b$, with $a + b = 0.6T.$

Step 1 — Use "sold $=$ half the stock". $\underbrace{0.2T}_{\text{half the mangoes}} + 96 + 0.4a = 0.5T \;\Rightarrow\; 0.4a = 0.3T - 96 \;\Rightarrow\; a = 0.75T - 240.$ Then $b = 0.6T - a = 240 - 0.15T.$

Step 2 — Force whole numbers. Every count must be a non-negative integer.

• Mangoes $= 0.4T = \tfrac{2T}{5}$ integer needs $5 \mid T$. Write $T = 5t$.
• Then $a = \tfrac{15t}{4} - 240$ and $b = 240 - \tfrac{3t}{4}$ need $4 \mid t$. Write $t = 4s$, so $T = 20s.$

Now mangoes $= 8s$, $\;a = 15s - 240$, $\;b = 240 - 3s.$

Step 3 — Positivity and "enough to sell" bounds.

• $a \ge 1$: $\;15s \ge 241 \Rightarrow s \ge 17.$
• $b \ge 96$ (he sells $96$ bananas): $\;240 - 3s \ge 96 \Rightarrow s \le 48.$

So $17 \le s \le 48$. To minimise $T = 20s$, take $s = 17$.

Step 4 — Check $s = 17$. $T = 340$: mangoes $= 136$ (which is $40\%$ ✓), $a = 15$, $b = 189$. Sold $= 68 + 96 + 6 = 170 = 50\%$ of $340$ ✓, and every count is a positive integer.

💡 Teacher tip: "Smallest total" with percentage conditions is a divisibility hunt — chase each "percent-of-an-integer is an integer" into a factor of $T$, parametrise ($T = 20s$), then take the smallest $s$ that keeps every count at least what's sold.

Answer: $\boxed{340}$