CAT 2025 Quant Questions — Slot 2

Think of it as: two rowers on the same river. Sneha's timing gap tells us how fast the river flows; once we know that, Rita's round trip is straightforward.

Step 1 — River speed $v$ from Sneha. Sneha rows $6$ km/h in still water, so upstream she makes $6 - v$ and downstream $6 + v$. Over $14$ km each way, upstream takes $48$ minutes ($\tfrac{4}{5}$ h) longer: $\frac{14}{6 - v} - \frac{14}{6 + v} = \frac{4}{5}.$ Combining the fractions, $14 \cdot \frac{2v}{36 - v^2} = \frac{4}{5} \;\Rightarrow\; \frac{28v}{36 - v^2} = \frac{4}{5} \;\Rightarrow\; 140v = 144 - 4v^2.$ $v^2 + 35v - 36 = 0 \;\Rightarrow\; (v + 36)(v - 1) = 0 \;\Rightarrow\; v = 1 \text{ km/h}.$

Step 2 — Rita's round trip. Rita rows $5$ km/h in still water, so with the river at $1$ km/h her speeds are $4$ upstream and $6$ downstream. She goes a distance $D$ up and the same $D$ back in $100$ minutes $= \tfrac{5}{3}$ h: $\frac{D}{4} + \frac{D}{6} = \frac{5}{3} \;\Rightarrow\; \frac{5D}{12} = \frac{5}{3} \;\Rightarrow\; D = 4.$ Total distance $= 2D = 8$ km.

Step 3 — Verify. Sneha: $\tfrac{14}{5} - \tfrac{14}{7} = 2.8 - 2 = 0.8$ h $= 48$ min ✓. Rita: $\tfrac{4}{4} + \tfrac{4}{6} = 1 + \tfrac{2}{3} = \tfrac{5}{3}$ h ✓.

Answer: $\boxed{8}$

💡 Teacher tip: the river speed is a shared, hidden constant. Extract it from whichever rower you're given full data for, then reuse it — that's the whole structure of two-boat problems.

Think of it as: every "average" is just a hidden total. Convert each average into a sum, and the daily figures fall out by subtraction.

Step 1 — Turn averages into totals. Total $=$ average $\times$ number of days.

Step 2 — Copies on day 8. The 8-day total exceeds the 7-day total by exactly day 8: $\text{Day 8} = 504 - 420 = 84.$

Step 3 — Copies on day 9. Day 9 is $11$ fewer than day 8: $\text{Day 9} = 84 - 11 = 73.$

Step 4 — Use the days 2–9 average. Average over days $2$ to $9$ (that's $8$ days) is $66$, so $\text{Sum of days 2–9} = 66 \times 8 = 528.$ Also, sum of days $1$–$9$ $=$ sum of days $1$–$8$ plus day 9 $= 504 + 73 = 577$. Since days $2$–$9$ is the whole thing minus day 1: $\text{Day 1} = 577 - 528 = 49.$

Step 5 — Verify. Days $2$–$7$ sum to $420 - 49 = 371$; adding day 8 ($84$) and day 9 ($73$) gives $371 + 84 + 73 = 528 = 66 \times 8$ ✓.

Answer: $\boxed{49}$

💡 Teacher tip: never average averages. Push every average back to a running total, and overlapping windows reveal single-day values just by subtracting.

Think of it as: we want divisors that leave remainder $1$ on division by $3$. Any factor of $3$ would ruin that, so the $3$s are out — then it's a parity count on the leftover exponents.

A divisor of $2^6 \times 3^5 \times 5^3 \times 7^2$ looks like $2^a\,3^b\,5^c\,7^d$.

Step 1 — No factor of $3$. If $b \ge 1$ the divisor is a multiple of $3$, so it can't be $\equiv 1 \pmod 3$. Hence $b = 0$.

Step 2 — Reduce the rest mod $3$. Note $2 \equiv -1$, $5 \equiv -1$, $7 \equiv 1 \pmod 3$. So $2^a\,5^c\,7^d \equiv (-1)^a (-1)^c (1)^d = (-1)^{a+c} \pmod 3.$ For this to equal $+1$ we need $a + c$ even.

Step 3 — Count the valid exponents. Ranges: $a \in \{0,\dots,6\}$, $c \in \{0,1,2,3\}$, $d \in \{0,1,2\}$.

Valid $(a, c)$ pairs $= 8 + 6 = 14$. The exponent $d$ is free ($3$ choices), so $14 \times 3 = 42.$

Answer: $\boxed{42}$ (Choice D)

💡 Teacher tip: reducing each prime to $\pm 1 \pmod 3$ turns a divisor-counting problem into a simple even/odd tally — far quicker than listing divisors.

Think of it as: the second condition secretly shares a factor with the first. Divide one by the other and the geometric progression's ratio pops out almost for free.

Step 1 — Write the two conditions in $a$ and $r$. With $a_1 = a$, $a_2 = ar$, $a_3 = ar^2$: $a(1 + r + r^2) = 52, \qquad (1)$ $a_1 a_2 + a_2 a_3 + a_3 a_1 = a^2 r(1 + r + r^2) = 624. \qquad (2)$

Step 2 — Divide $(2)$ by $(1)$. The common factor $(1 + r + r^2)$ cancels: $\frac{a^2 r(1 + r + r^2)}{a(1 + r + r^2)} = ar = \frac{624}{52} = 12.$ So $ar = 12$, i.e. $a = \dfrac{12}{r}$.

Step 3 — Solve for $r$. Substitute into $(1)$: $\frac{12}{r}(1 + r + r^2) = 52 \;\Rightarrow\; 1 + r + \frac{1}{r} = \frac{13}{3} \;\Rightarrow\; r + \frac{1}{r} = \frac{10}{3}.$ $3r^2 - 10r + 3 = 0 \;\Rightarrow\; (3r - 1)(r - 3) = 0 \;\Rightarrow\; r = \tfrac{1}{3} \text{ or } r = 3.$ The progression is decreasing, so $r = \tfrac13$ and $a = 12 \times 3 = 36$.

Step 4 — Infinite sum. $S_\infty = \frac{a}{1 - r} = \frac{36}{1 - \tfrac13} = \frac{36}{2/3} = 54.$

Answer: $\boxed{54}$ (Choice B)

💡 Teacher tip: in a geometric progression, the pairwise-product sum $a_1a_2 + a_2a_3 + a_3a_1$ always factors as $a^2 r(1 + r + r^2)$. Dividing by the plain sum isolates $ar$ in one step.

Think of it as: everything is quoted against expenditures, so anchor on those. One variable $k$ carries the whole chain; income, savings, and the final ratio all fall out of it.

Step 1 — Set expenditures. Let Lakshmi's expenditure $E_L = 2k$ and Meenakshi's $E_M = 3k$ (the $2: 3$ ratio).

Step 2 — Lakshmi's income. Given income of Lakshmi to expenditure of Meenakshi is $6: 7$: $\frac{I_L}{3k} = \frac{6}{7} \;\Rightarrow\; I_L = \frac{18k}{7}.$

Step 3 — Savings, then Meenakshi's income. Lakshmi saves $S_L = I_L - E_L = \frac{18k}{7} - 2k = \frac{4k}{7}.$ Since savings are in ratio $4: 9$, $S_M = \frac{9}{4} \cdot \frac{4k}{7} = \frac{9k}{7}, \qquad I_M = S_M + E_M = \frac{9k}{7} + 3k = \frac{30k}{7}.$

Step 4 — Income ratio. $I_L: I_M = \frac{18k}{7}: \frac{30k}{7} = 18: 30 = 3: 5.$

Answer: $\boxed{3: 5}$ (Choice A)

💡 Teacher tip: in income–expenditure–savings chains, pick the quantity that every ratio refers to (here, expenditure) as your anchor. One scaling variable then threads cleanly through all the conditions.

Think of it as: one equation ties three numbers together, and we want the smallest weighted sum. Solve for $b$, then sweep the smallest values of $a$ — a divisibility rule filters what survives.

Given distinct natural numbers with $3ac = 8(a + b)$.

Step 1 — Isolate $b$. $b = \frac{a(3c - 8)}{8}.$ For $b$ to be a natural number we need $3c - 8 > 0$ (so $c \ge 3$) and $a(3c - 8)$ divisible by $8$.

Step 2 — Try small $a$ to push the sum $3a + 2b + c$ down.

Step 3 — Confirm the winner. For $(a, b, c) = (2, 1, 4)$: $3ac = 3 \cdot 2 \cdot 4 = 24$ and $8(a + b) = 8 \cdot 3 = 24$ ✓; the three numbers $2, 1, 4$ are distinct ✓. Larger $a$ only inflates $3a$, and the divisibility rule blocks $a = 1$ from beating $15$, so $12$ is the floor.

Answer: $\boxed{12}$

💡 Teacher tip: when an equation forces a divisibility condition, don't chase inequalities blindly — list the smallest admissible cases in a short table and let the constraint eliminate the rest.

Think of it as: two quadratics share one root. Vieta's rules give the sum of each quadratic's roots instantly; the only real task is finding the shared root in terms of $p$ and $q$.

Step 1 — Name the roots. Let $\alpha$ be the common root. Let $\beta$ be the other root of the first equation and $\gamma$ the other root of the second. By Vieta's sum rule, $3x^2 - 5x + p = 0: \quad \alpha + \beta = \tfrac{5}{3},$ $2x^2 - 2x + q = 0: \quad \alpha + \gamma = 1.$

Step 2 — Add to get the two "other" roots. $\beta + \gamma = \left(\tfrac{5}{3} - \alpha\right) + \left(1 - \alpha\right) = \tfrac{8}{3} - 2\alpha. \qquad (\star)$

Step 3 — Find $\alpha$. Since $\alpha$ satisfies both equations, $3\alpha^2 - 5\alpha + p = 0, \qquad 2\alpha^2 - 2\alpha + q = 0.$ Kill the $\alpha^2$ term: multiply the first by $2$, the second by $3$, and subtract: $(6\alpha^2 - 10\alpha + 2p) - (6\alpha^2 - 6\alpha + 3q) = 0 \;\Rightarrow\; -4\alpha + 2p - 3q = 0 \;\Rightarrow\; \alpha = \frac{2p - 3q}{4}.$

Step 4 — Substitute into $(\star)$. $\beta + \gamma = \frac{8}{3} - 2 \cdot \frac{2p - 3q}{4} = \frac{8}{3} - \frac{2p - 3q}{2} = \frac{8}{3} - p + \frac{3}{2}q.$

Answer: $\boxed{\dfrac{8}{3} - p + \dfrac{3}{2}q}$ (Choice A)

💡 Teacher tip: with a shared root, get its value by eliminating the squared term between the two equations — that leaves a single linear equation in the root, no messy quadratic formula needed.

Think of it as: a monstrous product that collapses the moment you break the bases into $2$s and $5$s — because every pair of one $2$ and one $5$ becomes a harmless $10$.

Step 1 — Prime-factorise the bases. $625 = 5^4$ and $128 = 2^7$, so $(625)^{65} \times (128)^{36} = 5^{4 \times 65} \times 2^{7 \times 36} = 5^{260} \times 2^{252}.$

Step 2 — Pair up $2$s and $5$s into $10$s. There are $252$ twos, so pair each with a five: $5^{260} \times 2^{252} = 5^{260 - 252} \times (5 \times 2)^{252} = 5^{8} \times 10^{252}.$

Step 3 — Read off the digits. $5^8 = 390625$, and multiplying by $10^{252}$ just appends $252$ zeros: $390625\underbrace{00\ldots0}_{252 \text{ zeros}}.$ Zeros contribute nothing to a digit sum, so $3 + 9 + 0 + 6 + 2 + 5 = 25.$

Answer: $\boxed{25}$

💡 Teacher tip: whenever a huge number is built from powers, hunt for matched $2$s and $5$s. Each pair becomes a $10$ — a trailing zero — leaving only a small leading number whose digits you actually have to add.

Think of it as: each installment, discounted back to today, must add up to the original loan. Set up that present-value balance and test the given rates.

Step 1 — The present-value equation. A loan of ₹1000 is cleared by ₹530 at the end of year 1 and ₹594 at the end of year 2. If the annual compound rate is $r$, $1000 = \frac{530}{1 + r} + \frac{594}{(1 + r)^2}.$

Step 2 — Test the option $r = 8\%$. With $1 + r = 1.08$: $\frac{530}{1.08} + \frac{594}{1.08^2} = 490.74 + \frac{594}{1.1664} = 490.74 + 509.26 = 1000.00.$ The two present values sum to exactly ₹1000, so $r = 8\%$.

Step 3 — Why the others fail. A higher rate discounts the future installments more heavily, dropping the right side below ₹1000; a lower rate pushes it above. Only $8\%$ balances.

Answer: $\boxed{8}$ (Choice D)

💡 Teacher tip: for two-installment repayment questions, plugging each integer-rate option into the present-value identity is faster than solving the quadratic — the correct rate makes the arithmetic land on a round figure.

Think of it as: three logarithms whose bases ($64, 8, 512$) are all powers of $2$. Convert to base $2$ and they melt into one condition on the product $xyz$ — perfect setup for AM–GM.

Step 1 — Rewrite every term in base $2$, using $\log_{2^k}(N) = \dfrac{\log_2 N}{k}$: $\log_{64} x^2 = \frac{2\log_2 x}{6} = \frac{\log_2 x}{3}, \quad \log_8 y = \frac{\log_2 y}{3}, \quad 3\log_{512} z = \frac{3\log_2 z}{9} = \frac{\log_2 z}{3}.$ Adding them, $\frac{\log_2 x + \log_2 y + \log_2 z}{3} = \frac{\log_2(xyz)}{3} = 4 \;\Rightarrow\; \log_2(xyz) = 12 \;\Rightarrow\; xyz = 2^{12} = 4096.$

Step 2 — Minimise the sum for a fixed product. By AM–GM on the three positive numbers, $\frac{x + y + z}{3} \ge \sqrt[3]{xyz} = \sqrt[3]{4096} = 16,$ with equality when $x = y = z$. Since $16^3 = 4096$, equality is achievable at $x = y = z = 16$. Hence $x + y + z \ge 3 \times 16 = 48.$

Answer: $\boxed{48}$ (Choice A)

💡 Teacher tip: logs with bases that are powers of one number always collapse to a single sum-of-logs — that is, a product condition. A fixed product plus "minimise the sum" is the classic AM–GM trigger.

Think of it as: a composed function breaks wherever either the inner function blows up or the inner output hits the outer function's forbidden value. List every such $x$.

$f(x) = \frac{x}{2x - 1} \ \text{(undefined at } x = \tfrac12), \qquad g(x) = \frac{x}{x - 1} \ \text{(undefined at } x = 1).$

We need the domain of $h(x) = f(g(x)) + g(f(x))$, so we collect every value of $x$ that breaks any piece.

Step 1 — Inner functions must exist. $g(x)$ requires $x \ne 1$; $f(x)$ requires $x \ne \tfrac12$.

Step 2 — $f(g(x))$ needs $g(x) \ne \tfrac12$ (the pole of $f$): $\frac{x}{x - 1} = \frac{1}{2} \;\Rightarrow\; 2x = x - 1 \;\Rightarrow\; x = -1.$

Step 3 — $g(f(x))$ needs $f(x) \ne 1$ (the pole of $g$): $\frac{x}{2x - 1} = 1 \;\Rightarrow\; x = 2x - 1 \;\Rightarrow\; x = 1,$ which is already excluded in Step 1.

Step 4 — Gather all exclusions.

Answer: all reals except $\boxed{-1,\ \tfrac12,\ \text{and } 1}$ (Choice A)

💡 Teacher tip: for a composition, check two layers of danger — the inner function's own pole, and the input value that sends the inner output onto the outer function's pole. Missing the second layer is the usual slip.

Think of it as: the expression is smallest when $b, c, d$ huddle as close to $a$ as possible. With everything an integer and the total fixed at $46$, they can't all coincide — so the question is how close they can get.

We minimise $(a - b)^2 + (a - c)^2 + (a - d)^2$ subject to $a + b + c + d = 46$, all integers.

Step 1 — Aim for near-equal values. The four numbers average $\tfrac{46}{4} = 11.5$, so the tightest cluster uses values around $11$ and $12$.

Step 2 — Try $a = 12$. Then $b + c + d = 34$, split as $11, 11, 12$: $(12 - 11)^2 + (12 - 11)^2 + (12 - 12)^2 = 1 + 1 + 0 = 2.$

Step 3 — Rule out $0$ and $1$.

• A value of $0$ needs $b = c = d = a$, forcing $4a = 46$ — impossible for an integer.
• A value of $1$ needs two of $b, c, d$ equal to $a$ and the third off by $1$, giving a total $4a \pm 1$, which is odd and can never equal the even number $46$.

So $2$ cannot be beaten, and Step 2 attains it.

Answer: $\boxed{2}$

💡 Teacher tip: sums of squared differences shrink as the numbers bunch together. When integers and a fixed total block perfect equality, just test the two nearest integers to the mean.

Think of it as: the product of two whole numbers equals $27$, so there are only a handful of factor pairs to try. The real work is keeping $m$ and $n$ integers.

Step 1 — Rename the two brackets. Let $a = m + 2n$ and $b = 2m + n$, so $ab = 27$. Solving back, $n = \frac{2a - b}{3}, \qquad m = \frac{-a + 2b}{3}.$ So $m, n$ are integers only when $2a - b$ is a multiple of $3$, i.e. $b \equiv 2a \pmod{3}$.

Step 2 — Write the target in $a, b$. $2m - 3n = \frac{2(-a + 2b) - 3(2a - b)}{3} = \frac{-8a + 7b}{3}.$ We want this as large as possible over the factor pairs of $27$ (both factors share a sign since $27 > 0$).

Step 3 — Test the pairs.

The winner is $(a, b) = (-9, -3)$.

Step 4 — Recover $m, n$ and confirm. $m = \dfrac{9 - 6}{3} = 1$, $n = \dfrac{-18 + 3}{3} = -5$. Check: $(1 + 2(-5))(2(1) + (-5)) = (-9)(-3) = 27$ ✓, and $2m - 3n = 2 + 15 = 17$.

Answer: $\boxed{17}$

💡 Teacher tip: after switching to the product variables $a, b$, carry the integer condition ($b \equiv 2a \pmod 3$) with you. It quietly rules out the tempting pair $(1, 27)$ that would otherwise look best.

Think of it as: the modulus $|x + 9|$ behaves differently on either side of $x = -9$. Split there, solve each half as an ordinary quadratic inequality, then glue the pieces.

We want all $x$ with $x^2 - |x + 9| + x > 0$.

Step 1 — Case $x \ge -9$, where $|x + 9| = x + 9$: $x^2 - (x + 9) + x > 0 \;\Rightarrow\; x^2 - 9 > 0 \;\Rightarrow\; x < -3 \text{ or } x > 3.$ Intersecting with $x \ge -9$ gives $x \in [-9, -3) \cup (3, \infty)$.

Step 2 — Case $x < -9$, where $|x + 9| = -(x + 9)$: $x^2 + (x + 9) + x > 0 \;\Rightarrow\; x^2 + 2x + 9 > 0.$ Its discriminant is $4 - 36 = -32 < 0$ with a positive leading coefficient, so this is true for every real $x$. Thus all $x < -9$ qualify: $x \in (-\infty, -9)$.

Step 3 — Union the two cases.

$(-\infty, -9) \cup [-9, -3) \cup (3, \infty) = (-\infty, -3) \cup (3, \infty).$

Answer: $\boxed{(-\infty, -3) \cup (3, \infty)}$ (Choice A)

💡 Teacher tip: always test whether the "far" branch of a modulus problem gives a quadratic with no real roots — if it does, that entire half-line is included, and the two pieces often merge into one clean interval.

Think of it as: turn "twice as efficient" into a daily work rate for each person, count how much work gets done in each stretch, and the total tells you how long the slowest worker needs alone.

Step 1 — Set the rates. Let Chandan do $1$ unit/day. Bipin is twice as efficient, so $2$ units/day; Ankita is twice Bipin, so $4$ units/day.

Step 2 — Add up the work in two phases.

Total work $= 140 + 200 = 340$ units.

Step 3 — Chandan alone. At $1$ unit/day, Chandan needs $\dfrac{340}{1} = 340$ days.

Answer: $\boxed{340}$

💡 Teacher tip: fixing the slowest worker's rate at $1$ makes every "twice as efficient" a clean integer, and the whole job's size emerges as a plain unit count — no fractions until the very end.

Think of it as: two clues about a marked price and a discount, then a final question about a special discount. Nail down the marked price first; the last part is then a one-line equation.

Cost price $= ₹1650$.

Step 1 — Selling price at $20\%$ profit. $\text{SP}_1 = 1650 \times 1.20 = 1980.$

Step 2 — Selling price when the discount is doubled. "The profit would have been ₹110" means the selling price is $1650 + 110 = 1760$. Let marked price $= M$ and original discount rate $= d$ (a decimal): $M(1 - d) = 1980, \qquad M(1 - 2d) = 1760.$ Subtracting, $M \cdot d = 220$, so $M = 1980 + 220 = 2200$ and $d = \dfrac{220}{2200} = 10\%$. (Check: doubling to $20\%$ gives $2200 \times 0.80 = 1760$ ✓.)

Step 3 — Find the discount where profit-% equals discount-%. Call that common value $p$. The selling price is $2200\left(1 - \frac{p}{100}\right)$, and profit as a percentage of cost equals $p$, i.e. profit $= 16.5p$: $2200\left(1 - \tfrac{p}{100}\right) - 1650 = 16.5p.$ $2200 - 22p - 1650 = 16.5p \;\Rightarrow\; 550 = 38.5p \;\Rightarrow\; p = \frac{550}{38.5} \approx 14.29.$

Answer: $\boxed{14}$ (nearest, Choice C)

💡 Teacher tip: "profit ₹110" is an amount, not a shortfall — add it to cost. Misreading that single phrase is the most common way this question goes wrong.

Think of it as: a scary-looking equation in powers of $9$ and $3$ that is secretly a plain quadratic. Two substitutions strip away the disguise.

Step 1 — Absorb the repeated exponent. The block $x^2 + 2x$ appears everywhere, so let $u = x^2 + 2x$. Then $x^2 + 2x - 3 = u - 3$ and $x^2 + 2x - 2 = u - 2$. Since $9 = 3^2$, the equation becomes $3^{2(u-3)} - 4 \cdot 3^{\,u-2} + 27 = 0.$

Step 2 — Substitute again to get a quadratic. Let $t = 3^{\,u-3}$. Then $3^{\,u-2} = 3t$ and $3^{2(u-3)} = t^2$, so $t^2 - 12t + 27 = 0 \;\Rightarrow\; (t - 3)(t - 9) = 0 \;\Rightarrow\; t = 3 \text{ or } t = 9.$

Step 3 — Translate each $t$ back into an equation for $x$.

For $x^2 + 2x - c = 0$, the product of the roots is $-c$ by Vieta's rule. Both discriminants ($20$ and $24$) are positive, so all four roots are real.

Step 4 — Multiply the products. $(-4) \times (-5) = 20$.

Answer: $\boxed{20}$ (Choice B)

💡 Teacher tip: when the same expression sits in every exponent, name it once. The double substitution ($u$ then $t$) turns an exponential mess into a quadratic you can factor by eye.

Think of it as: two cevians of a triangle crossing at $T$, with each split ratio given. Mass points convert those splits straight into the side ratio $BD: CD$.

Step 1 — Read the splits at $T$. From $AD: AT = 4: 3$ we get $AT: TD = 3: 1$. From $BE: BT = 5: 4$ we get $BT: TE = 4: 1$.

Step 2 — Assign masses so $T$ balances. A balance point satisfies mass $\times$ arm-length equal on each side: $\frac{m_A}{m_D} = \frac{TD}{AT} = \frac{1}{3}, \qquad \frac{m_B}{m_E} = \frac{TE}{BT} = \frac{1}{4}.$ Set $m_A = 1$, so $m_D = 3$. Since $D$ lies on $BC$, $m_D = m_B + m_C = 3$. Since $E$ lies on $AC$, $m_E = m_A + m_C = 1 + m_C$, and $m_E = 4m_B$, giving $m_C = 4m_B - 1$.

Step 3 — Solve the little system. Substituting into $m_B + m_C = 3$: $m_B + (4m_B - 1) = 3 \;\Rightarrow\; 5m_B = 4 \;\Rightarrow\; m_B = \tfrac{4}{5}, \quad m_C = \tfrac{11}{5}.$

Step 4 — Read off the side ratio. For $D$ on $BC$, the split is the inverse ratio of the endpoint masses: $BD: DC = m_C: m_B = \frac{11}{5}: \frac{4}{5} = 11: 4.$ (The point $F$ with $DF \parallel BE$ is consistent with this but isn't needed for the ratio.)

Answer: $\boxed{11: 4}$ (Choice B)

💡 Teacher tip: mass points turn cevian split-ratios into simple see-saw arithmetic. A mass at a vertex is inversely proportional to the segment on the far side — so the side ratio reads off directly as $m_C: m_B$.

Think of it as: the money is handed out in stages, and each percentage applies to whatever is left at that moment — not to the original total. Track the running remainder carefully.

Let the total $= 100$.

Step 1 — Pinu and Meena.

• Pinu gets $20\%$ of the total $= 20$. Remaining: $80$.
• Meena gets $40\%$ of the remaining $80 = 32$. Remaining: $80 - 32 = 48$.

Step 2 — Seema and Rinu split the last $48$. Seema receives $20\%$ less than Pinu: $\text{Seema} = 0.8 \times 20 = 16, \qquad \text{Rinu} = 48 - 16 = 32.$

Step 3 — Ratio. $\text{Pinu}: \text{Rinu} = 20: 32 = 5: 8.$

Answer: $\boxed{5: 8}$ (Choice C)

💡 Teacher tip: the phrase "of the remaining amount" resets the base. Meena's $40\%$ is of $80$, not $100$ — pin down what each percentage is measured against before you multiply.

Think of it as: three lines — $PQ$, $PR$, and $AB$ — all touch the same circle. So the circle is tucked into triangle $PAB$, and there's a clean angle rule connecting the centre-angle to the corner at $P$.

Step 1 — See how the circle sits. $P$ is outside the circle; the tangency points $Q$ and $R$ lie beyond $A$ and $B$ on the two rays. So the circle touches side $AB$ directly but touches the other two sides only on their extensions — it is the excircle opposite $P$ of triangle $PAB$, with $O$ as its centre.

Step 2 — Apply the excentre angle rule. For the excircle opposite $P$, the angle its centre subtends across the opposite side is $\angle AOB = 90^\circ - \frac{\angle APB}{2}.$

Step 3 — Solve. $50^\circ = 90^\circ - \frac{\angle APB}{2} \;\Rightarrow\; \frac{\angle APB}{2} = 40^\circ \;\Rightarrow\; \angle APB = 80^\circ.$

Answer: $\boxed{80}$

💡 Teacher tip: for the incircle the rule is $90^\circ + \tfrac{P}{2}$; for an excircle it flips to $90^\circ - \tfrac{P}{2}$. Deciding which circle you have — by checking whether the tangent points fall on the sides or their extensions — is the crux.