CAT 2021 Quant Questions — Slot 3

Think of this as: two isosceles triangles are built on the two pieces of side $AB$, and we want the angle trapped between them at $D$. The trick is to let the unknown angle at $A$ be a variable, compute each isosceles angle in terms of it, and then watch the variable cancel.

Step 1 — Label the angles of $\triangle ABC$.

Let $\angle BAC = \alpha$. We are given $\angle BCA = 50^\circ$, so by the angle-sum property of a triangle: $\angle ABC = 180^\circ - 50^\circ - \alpha = 130^\circ - \alpha.$

Step 2 — Use the isosceles triangle $ADE$.

Since $D$ lies on $AB$ and $E$ lies on $AC$, the angle $\angle DAE$ is exactly $\angle BAC = \alpha$.

We are told $AD = DE$, so $\triangle ADE$ is isosceles with the equal sides meeting at $D$. Therefore the base angles are equal: $\angle DAE = \angle DEA = \alpha.$

The angle at $D$ inside this triangle is $\angle ADE = 180^\circ - \alpha - \alpha = 180^\circ - 2\alpha.$

Step 3 — Use the isosceles triangle $BDF$.

Since $D$ lies on $AB$ and $F$ lies on $BC$, the angle $\angle DBF$ is exactly $\angle ABC = 130^\circ - \alpha$.

We are told $BD = DF$, so $\triangle BDF$ is isosceles with the equal sides meeting at $D$. Therefore the base angles are equal: $\angle DBF = \angle DFB = 130^\circ - \alpha.$

The angle at $D$ inside this triangle is $\angle BDF = 180^\circ - 2(130^\circ - \alpha) = 180^\circ - 260^\circ + 2\alpha = 2\alpha - 80^\circ.$

Step 4 — Combine the three angles at point $D$.

The point $D$ sits on the segment $AB$. On the side of $AB$ that contains $C$, the three angles $\angle ADE$, $\angle EDF$, and $\angle FDB$ together form a straight angle: $\angle ADE + \angle EDF + \angle FDB = 180^\circ.$

Substituting the expressions from Steps 2 and 3: $(180^\circ - 2\alpha) + \angle FDE + (2\alpha - 80^\circ) = 180^\circ.$

The $-2\alpha$ and $+2\alpha$ cancel, leaving $100^\circ + \angle FDE = 180^\circ \implies \angle FDE = 80^\circ.$

💡 Teacher tip: The $\alpha$-terms cancel against each other — a strong hint that the answer does not depend on the exact shape of $\triangle ABC$. Whenever a parameter cancels cleanly, the result is a fixed angle determined only by the given data.

Answer: $\boxed{80^\circ}$ (Choice B)

💡 Closing insight: Two isosceles triangles built back-to-back on the two pieces of side $AB$ produce an angle $\angle FDE$ that depends only on $\angle C$. In fact the relation generalises to $\angle FDE = 180^\circ - 2\angle C$, which gives $180^\circ - 2(50^\circ) = 80^\circ$ here.

Think of this as: we're given a fraction whose numerator is a sum of two logs and whose denominator is their product. That structure — $\frac{x+y}{xy}$ — is a strong signal to split it into reciprocals.

Step 1 — Split the fraction.

Let $x = \log_{15} a$ and $y = \log_{32} a$. The left-hand side is: $\frac{x + y}{xy} = \frac{1}{y} + \frac{1}{x}.$

Substituting back: $\frac{1}{\log_{32} a} + \frac{1}{\log_{15} a}.$

Step 2 — Flip each reciprocal using the base-switch identity.

The identity $\dfrac{1}{\log_b a} = \log_a b$ (which is just the change-of-base formula rearranged) turns each term around: $\frac{1}{\log_{32} a} = \log_a 32, \qquad \frac{1}{\log_{15} a} = \log_a 15.$

So the equation becomes: $\log_a 32 + \log_a 15 = 4.$

💡 Teacher tip: Whenever you see $\frac{1}{\log_b a}$, immediately read it as $\log_a b$. This single swap — exchanging the base and the argument — is what converts a messy two-log expression into something you can combine.

Step 3 — Combine the logs.

Since both terms now share the same base $a$, use the product rule $\log_a p + \log_a q = \log_a(pq)$: $\log_a(32 \times 15) = 4 \implies \log_a 480 = 4.$

Step 4 — Solve for $a$.

Rewrite the logarithmic equation in exponential form: $\log_a 480 = 4$ means $a^4 = 480$, so $a = 480^{1/4}.$

Step 5 — Pin $480^{1/4}$ between two consecutive integers.

We compare $480$ with nearby fourth powers:

Since $4^4 = 256 < 480 < 625 = 5^4$ and the fourth-root function is strictly increasing, taking fourth roots throughout preserves the inequalities: $4 < 480^{1/4} < 5.$

Therefore $4 < a < 5$.

Answer: $\boxed{4 < a < 5}$ (Choice C)

💡 Closing insight: The entire problem collapses once you spot that $\frac{x+y}{xy} = \frac{1}{x}+\frac{1}{y}$. After that, the base-switch identity does the heavy lifting, turning two logarithms with different bases into a single $\log_a 480$ — and the rest is just comparing fourth powers.

Think of this as: we have a fixed area to enclose, but one side is twice as expensive to fence as the other three. Where should we place the expensive side — long or short — to keep the total bill as low as possible?

Setting up the cost equation

Let the side fenced at ₹200/ft have length $x$, and the other side have length $y$.

The four sides and their costs are:

So the total cost is: $C = 200x + 100x + 100y + 100y = 300x + 200y$

We're also told the area is fixed: $xy = 60{,}000$

Step 1 — Reduce to one variable.

From $xy = 60{,}000$, we get $y = \dfrac{60{,}000}{x}$. Substituting: $C(x) = 300x + 200 \cdot \frac{60{,}000}{x} = 300x + \frac{1{,}20{,}00{,}000}{x}$

Step 2 — Minimise using AM–GM.

Both terms $300x$ and $\dfrac{1{,}20{,}00{,}000}{x}$ are positive, so we can apply the AM–GM inequality: $300x + \frac{1{,}20{,}00{,}000}{x} \geq 2\sqrt{300x \cdot \frac{1{,}20{,}00{,}000}{x}}$

The $x$ cancels inside the square root: $= 2\sqrt{300 \times 1{,}20{,}00{,}000} = 2\sqrt{3{,}60{,}00{,}00{,}000} = 2 \times 60{,}000 = 1{,}20{,}000$

So the minimum cost is ₹1,20,000, achieved when the two terms are equal.

💡 Teacher tip: AM–GM equality ($A = B$) is the fastest route to the minimum of a "$\text{term} + \dfrac{\text{constant}}{\text{term}}$" expression — no calculus needed. The minimum always occurs when the two terms are equal.

Step 3 — Verify the dimensions at the optimum.

Set the two terms equal: $300x = \frac{1{,}20{,}00{,}000}{x} \implies x^2 = \frac{1{,}20{,}00{,}000}{300} = 40{,}000 \implies x = 200$

Then: $y = \frac{60{,}000}{200} = 300$

Step 4 — Confirm the cost.

$C = 300(200) + 200(300) = 60{,}000 + 60{,}000 = 1{,}20{,}000$

Answer: ₹1,20,000 (Choice A)

💡 Closing insight: At the optimum, the two contributions to the cost — $300x$ (the total spent on the two sides of length $x$) and $200y$ (the total spent on the two sides of length $y$) — are equal. This balance is the signature of an AM–GM minimum: you spend the same on the "expensive-side pair" as on the "cheap-side pair." Notice also that the expensive side ($x = 200$) is kept shorter than the cheap side ($y = 300$) — a natural intuition, since you want less length at the higher rate.

Alternative approach (Calculus)

Differentiate $C(x) = 300x + \dfrac{1{,}20{,}00{,}000}{x}$: $C'(x) = 300 - \frac{1{,}20{,}00{,}000}{x^2}$

Set $C'(x) = 0$: $300 = \frac{1{,}20{,}00{,}000}{x^2} \implies x^2 = 40{,}000 \implies x = 200$

The second derivative $C''(x) = \dfrac{2{,}40{,}00{,}000}{x^3} > 0$, confirming a minimum. The rest follows identically, giving $C = 1{,}20{,}000$. The AM–GM method is faster under exam conditions since it directly gives the minimum value without computing $x$ first.

We need the smallest positive integer $n$ for which the product $\left(\sqrt[7]{10}\right)\left(\sqrt[7]{10}\right)^{2}\cdots\left(\sqrt[7]{10}\right)^{n}$ exceeds $999$.

Step 1 — Rewrite the product using index laws.

Since $\sqrt[7]{10} = 10^{1/7}$, each factor is a power of $10^{1/7}$:

$\left(10^{1/7}\right)^{1}\cdot\left(10^{1/7}\right)^{2}\cdots\left(10^{1/7}\right)^{n}$

When multiplying powers with the same base, we add the exponents:

$= \left(10^{1/7}\right)^{1+2+\cdots+n}$

The sum $1+2+\cdots+n = \dfrac{n(n+1)}{2}$, so:

$= \left(10^{1/7}\right)^{n(n+1)/2} = 10^{\,n(n+1)/(14)}$

💡 Teacher tip: The entire chain of $n$ factors collapses to a single power of 10. The problem is no longer about multiplication — it's about pinning down the exponent.

Step 2 — Translate "$> 999$" into a bound on the exponent.

We need $10^{\,n(n+1)/14} > 999$.

Since $999 < 1000 = 10^{3}$, the product clears $999$ as soon as the exponent reaches $3$ (at exactly $3$, the product is $10^3 = 1000 > 999$). So a clean sufficient condition is:

$\frac{n(n+1)}{14} \geq 3 \quad\Longrightarrow\quad n(n+1) \geq 42$

We should also confirm that no smaller $n$ sneaks past $999$ with an exponent slightly below $3$ — we'll check that in the table below.

Step 3 — Test integer values of $n$.

At $n=5$ the exponent is only $\approx 2.14$, far below the $\approx 2.9996$ threshold needed to exceed $999$. At $n=6$ the exponent hits exactly $3$, giving $1000 > 999$.

Step 4 — Verify the boundary.

• At $n = 6$: product $= 10^{42/14} = 10^{3} = 1000 > 999$ ✓
• At $n = 5$: product $= 10^{30/14} \approx 10^{2.14} \approx 139 < 999$ ✓

The smallest $n$ is therefore $6$.

Answer: $\boxed{6}$

💡 Closing insight: The identity $a^{1}\cdot a^{2}\cdots a^{n} = a^{n(n+1)/2}$ turns a long chain of multiplications into a single exponent, converting the problem into "when does the exponent cross $3$?" — a simple inequality check.

Think of this as: first build the composite function $f(g(x))$, then subtract $3x$, and finally minimise the resulting expression.

Step 1 — Compose $f(g(x))$.

Since $g(x) = x + 3$, we substitute $x + 3$ into $f$: $f(g(x)) = (x+3)^2 - 7(x+3).$

Expand each piece:

• $(x+3)^2 = x^2 + 6x + 9$
• $-7(x+3) = -7x - 21$

So: $f(g(x)) = x^2 + 6x + 9 - 7x - 21 = x^2 - x - 12.$

Step 2 — Subtract $3x$.

$f(g(x)) - 3x = (x^2 - x - 12) - 3x = x^2 - 4x - 12.$

💡 Teacher tip: Always simplify $f(g(x))$ fully before optimising. If you try to minimise $f(g(x))$ and $-3x$ separately, you'll miss the interaction between them — the $-x$ from the composition combines with $-3x$ to give $-4x$, which shifts the vertex.

Step 3 — Minimise the quadratic by completing the square.

We have $x^2 - 4x - 12$. Take half the coefficient of $x$ (which is $-4 \div 2 = -2$), square it to get $4$, then add and subtract: $x^2 - 4x - 12 = (x^2 - 4x + 4) - 4 - 12 = (x-2)^2 - 16.$

Since $(x-2)^2 \ge 0$ for every real $x$, the smallest the expression can be is when $(x-2)^2 = 0$, i.e. at $x = 2$.

Step 4 — Read off the minimum.

At $x = 2$: $(2-2)^2 - 16 = 0 - 16 = -16.$

Verification check:

The value at $x = 2$ is lower than at neighbouring points, confirming the minimum.

Answer: $\boxed{-16}$ (Choice D)

💡 Closing insight: After composition and simplification, the expression collapses to a plain quadratic whose vertex form $(x-2)^2 - 16$ reveals the minimum instantly. The entire problem is really just "simplify, then complete the square."

Alternative approach — Vertex formula. For a quadratic $ax^2 + bx + c$ with $a > 0$, the minimum occurs at $x = -\frac{b}{2a}$ and equals $c - \frac{b^2}{4a}$. Here $a = 1,\; b = -4,\; c = -12$: $\text{Min} = -12 - \frac{(-4)^2}{4(1)} = -12 - 4 = -16.$ This is faster if you're comfortable with the formula, though completing the square is more reliable under pressure since it's harder to mix up signs.

Think of this as: we know how the team has done so far, and we're told what would happen if they won 60% of the rest. That "what if" lets us pin down how many remaining matches there are — and once we know that, the 90% case is just arithmetic.

Step 1 — Count the wins so far.

The team has played 40 matches and won 30% of them: $0.30 \times 40 = 12 \text{ wins.}$ So they currently have 12 wins out of 40 matches.

Step 2 — Use the 50% condition to find how many matches remain.

Let $R$ be the number of remaining matches. If they win 60% of those, their overall win percentage becomes 50%.

• Wins from remaining matches: $0.6R$
• Total wins: $12 + 0.6R$
• Total matches: $40 + R$

Setting the overall win rate to 50%: $\frac{12 + 0.6R}{40 + R} = 0.5$

Cross-multiply: $12 + 0.6R = 0.5(40 + R)$ $12 + 0.6R = 20 + 0.5R$ $0.1R = 8$ $R = 80$

So there are 80 remaining matches.

💡 Teacher tip: The 50% here is a weighted average of 30% (over 40 matches) and 60% (over $R$ matches). By alligation, the ratio of matches is $(60-50):(50-30) = 10:20 = 1:2$. Since the 40 matches form 1 part, $R = 80$ — the same result in one line.

Step 3 — Apply the 90% condition.

Now suppose the team wins 90% of the 80 remaining matches: $0.9 \times 80 = 72 \text{ wins from the remaining matches.}$

Step 4 — Add the wins from both phases. $\text{Total wins} = 12 + 72 = 84.$

Quick check:

Answer: $\boxed{84}$ (Choice C)

💡 Closing insight: The key move is realising the 50% condition is just a tool to discover $R$ — the number of remaining matches. Once $R$ is known, swapping 60% for 90% is a straightforward recalculation. Whenever a problem gives you a hypothetical outcome to pin down a hidden quantity, solve for that quantity first, then answer the actual question.

Think of this as: we're given two different schedules for the same job, and each schedule tells us how long each person worked. Two equations, two unknowns — the rates of Rahul and Gautam.

What we want: the time Rahul alone would take to finish the work.

Let Rahul's work rate be $r$ (fraction of the job per hour) and Gautam's rate be $g$. The whole job $= 1$.

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Step 1 — Translate each scenario into an equation.

Scenario 1 (what actually happened): Rahul started at 9 AM and worked until 5 PM — that's $8$ hours. Gautam joined two hours later, i.e. at 11 AM, and also worked until 5 PM — that's $6$ hours.

$8r + 6g = 1$

Scenario 2 (the hypothetical): Both start together at 9 AM. The work finishes 30 minutes earlier than 5 PM, i.e. at 4:30 PM — so they work together for $7.5$ hours.

$7.5(r + g) = 1$

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Step 2 — Extract the combined rate from Scenario 2.

$r + g = \frac{1}{7.5} = \frac{1}{\frac{15}{2}} = \frac{2}{15}$

So Gautam's rate in terms of Rahul's is:

$g = \frac{2}{15} - r$

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Step 3 — Substitute into Scenario 1 and solve for $r$.

$8r + 6\left(\frac{2}{15} - r\right) = 1$

$8r + \frac{12}{15} - 6r = 1$

$2r + \frac{4}{5} = 1$

$2r = 1 - \frac{4}{5} = \frac{1}{5}$

$r = \frac{1}{10}$

So Rahul completes $\frac{1}{10}$ of the job per hour.

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Step 4 — Find Rahul's solo time.

Time $= \dfrac{\text{Total work}}{\text{Rate}} = \dfrac{1}{r} = \dfrac{1}{\frac{1}{10}} = 10$ hours.

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Quick verification. If $r = \frac{1}{10}$, then $g = \frac{2}{15} - \frac{1}{10} = \frac{4}{30} - \frac{3}{30} = \frac{1}{30}$.

• Scenario 1: $8 \cdot \frac{1}{10} + 6 \cdot \frac{1}{30} = \frac{4}{5} + \frac{1}{5} = 1$ ✓
• Scenario 2: $7.5 \cdot \left(\frac{1}{10} + \frac{1}{30}\right) = 7.5 \cdot \frac{4}{30} = 7.5 \cdot \frac{2}{15} = 1$ ✓

Both scenarios check out.

Answer: $\boxed{10}$ (Choice B)

💡 Closing insight: The 30-minute gap between the two scenarios is exactly the work Gautam didn't do during his first two absent hours. In fact, $2g = 2 \cdot \frac{1}{30} = \frac{1}{15}$, which equals half an hour of combined work: $\frac{1}{2} \cdot \frac{2}{15} = \frac{1}{15}$. The two scenarios self-check once you have both rates.

What we want: the value of $x + 2y$, given $3x + 2|y| + y = 7$ and $x + |x| + 3y = 1$.

The two equations carry $|x|$ and $|y|$, so the honest way through is to remove the modulus signs by testing each combination of signs. Each variable is either $\ge 0$ or $< 0$, giving four cases.

Case 1 — $x \ge 0,\ y \ge 0$. Then $|x| = x$ and $|y| = y$: $3x + 3y = 7, \qquad 2x + 3y = 1.$ Subtracting, $x = 6$, so $3y = 7 - 18 = -11$ — but that makes $y < 0$, contradicting $y \ge 0$. Rejected.

Case 2 — $x \ge 0,\ y < 0$. Then $|x| = x$ and $|y| = -y$: $3x - y = 7, \qquad 2x + 3y = 1.$ From the first, $y = 3x - 7$. Substituting into the second: $2x + 3(3x - 7) = 1 \;\Rightarrow\; 11x = 22 \;\Rightarrow\; x = 2, \quad y = -1.$ Both assumptions hold: $x = 2 \ge 0$ and $y = -1 < 0$. This case survives.

Step — Verify in the original equations.

• Equation 1: $3(2) + 2|-1| + (-1) = 6 + 2 - 1 = 7$ ✓
• Equation 2: $2 + |2| + 3(-1) = 2 + 2 - 3 = 1$ ✓

The other two cases fail.

• $x < 0,\ y \ge 0$: equation 2 becomes $3y = 1$, so $y = \tfrac13$, then equation 1 gives $3x + 1 = 7 \Rightarrow x = 2$, contradicting $x < 0$.
• $x < 0,\ y < 0$: equation 2 becomes $3y = 1 \Rightarrow y = \tfrac13 > 0$, contradicting $y < 0$.

So the unique solution is $x = 2,\ y = -1$.

Step — Compute the target. $x + 2y = 2 + 2(-1) = 0.$

Answer: $\boxed{0}$ (Choice C)

💡 Teacher tip: With two modulus variables you always have four sign cases, but the equations themselves usually rule out all but one. Fix the signs first; what's left is an ordinary pair of linear equations that solves in two lines.

Think of this as: we know the total number of shirts, the price gap between the two sizes, and the total spent on each size — we need to back out the individual prices and add them.

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Step 1 — Set up variables.

Let:

• $p_L$ = price of one large shirt, $p_S$ = price of one small shirt
• $L$ = number of large shirts bought, $S$ = number of small shirts bought

Translating every piece of given information:

We want: $p_L + p_S$.

Step 2 — Express prices in terms of quantities.

From the spending equations: $p_L = \frac{5000}{L}, \qquad p_S = \frac{1800}{S}$

Now use $S = 64 - L$ and substitute into the price-gap equation $p_S = p_L - 50$:

$\frac{1800}{64 - L} = \frac{5000}{L} - 50$

Step 3 — Clear denominators and solve for $L$.

Multiply both sides by $L(64 - L)$:

$1800L = (64 - L)(5000 - 50L)$

Expand the right side:

$1800L = 320000 - 3200L - 5000L + 50L^2$ $1800L = 320000 - 8200L + 50L^2$

Bring everything to one side:

$50L^2 - 10000L + 320000 = 0$

Divide through by $50$:

$L^2 - 200L + 6400 = 0$

Step 4 — Apply the quadratic formula.

$L = \frac{200 \pm \sqrt{200^2 - 4 \cdot 6400}}{2} = \frac{200 \pm \sqrt{40000 - 25600}}{2} = \frac{200 \pm \sqrt{14400}}{2} = \frac{200 \pm 120}{2}$

So $L = 160$ or $L = 40$.

Since $L + S = 64$, we need $L \le 64$. The root $L = 160$ is physically impossible (you can't buy 160 large shirts out of 64 total), so we reject it.

$L = 40$, and therefore $S = 64 - 40 = 24$.

💡 Teacher tip: A quadratic in a word problem often yields two roots. Always check each root against the real-world constraints — here, the total count of 64 shirts eliminates $L = 160$ immediately.

Step 5 — Find the prices.

$p_L = \frac{5000}{40} = 125, \qquad p_S = \frac{1800}{24} = 75$

Quick check: $p_L - p_S = 125 - 75 = 50$ ✓ (matches the given price gap).

Step 6 — Compute the answer.

$p_L + p_S = 125 + 75 = 200$

Answer: $\boxed{200}$ (Choice C)

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Alternative approach — Target the answer directly.

Instead of solving for $L$ first, define $T = p_L + p_S$ (exactly what we want) and use the given gap $p_L - p_S = 50$ to write both prices in terms of $T$:

$p_L = \frac{T + 50}{2}, \qquad p_S = \frac{T - 50}{2}$

Now express $L$ and $S$ in terms of $T$:

$L = \frac{5000}{p_L} = \frac{10000}{T + 50}, \qquad S = \frac{1800}{p_S} = \frac{3600}{T - 50}$

Since $L + S = 64$:

$\frac{10000}{T + 50} + \frac{3600}{T - 50} = 64$

Multiply through by $(T + 50)(T - 50)$:

$10000(T - 50) + 3600(T + 50) = 64(T^2 - 2500)$ $10000T - 500000 + 3600T + 180000 = 64T^2 - 160000$ $13600T - 320000 = 64T^2 - 160000$ $64T^2 - 13600T + 160000 = 0$

Divide by 32:

$2T^2 - 425T + 5000 = 0$

Quadratic formula:

$T = \frac{425 \pm \sqrt{425^2 - 4 \cdot 2 \cdot 5000}}{4} = \frac{425 \pm \sqrt{180625 - 40000}}{4} = \frac{425 \pm \sqrt{140625}}{4} = \frac{425 \pm 375}{4}$

So $T = \frac{800}{4} = 200$ or $T = \frac{50}{4} = 12.5$.

Since $T = 12.5$ would give $p_S = \frac{12.5 - 50}{2} = -18.75$ (a negative price), we reject it.

$T = 200$, confirming our answer.

💡 Why this is elegant: by defining $T = p_L + p_S$ upfront, we collapse a two-stage problem (find quantities → find prices → add them) into a single equation in the exact quantity we want. The price gap $p_L - p_S = 50$ is the bridge that lets both prices be written in terms of $T$ alone.

Think of this as: we know the area and the perimeter of a rhombus, and we need the sum of its two diagonals — because the wire runs along both. The trick is to avoid solving for each diagonal individually.

What's given:

• Area of the rhombus park = $96 \text{ sq m}$
• Fencing needed = $40 \text{ m}$, which is the perimeter.
• Wire runs along both diagonals; rate = ₹$125$ per metre.

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Step 1 — Find the side length.

A rhombus has four equal sides, so $4s = 40 \quad\Rightarrow\quad s = 10 \text{ m}.$

Step 2 — Translate the two given facts into equations involving the diagonals.

Let the diagonals be $d_1$ and $d_2$. Two standard properties of a rhombus connect them to what we know:

1. Area = half the product of the diagonals: $\tfrac{1}{2}\,d_1 d_2 = 96 \quad\Rightarrow\quad d_1 d_2 = 192.$

2. Diagonals bisect each other at right angles, so each side is the hypotenuse of a right triangle with legs $\tfrac{d_1}{2}$ and $\tfrac{d_2}{2}$: $\left(\frac{d_1}{2}\right)^{2}+\left(\frac{d_2}{2}\right)^{2}=s^{2}=100.$ Multiplying through by $4$: $d_1^{2}+d_2^{2}=400.$

💡 Teacher tip: Notice we now have $d_1 d_2$ and $d_1^2 + d_2^2$. That's enough to get $d_1 + d_2$ without ever finding $d_1$ or $d_2$ on their own — the algebra hands us exactly the combination we need.

Step 3 — Find $d_1 + d_2$, the total length of wire.

Use the identity $(d_1+d_2)^{2}=d_1^{2}+d_2^{2}+2\,d_1 d_2.$ Substitute the two values: $(d_1+d_2)^{2}=400+2(192)=400+384=784.$ $d_1+d_2=\sqrt{784}=28 \text{ m}.$

So the wire must cover $28$ m in total.

Step 4 — Compute the cost.

$\text{Cost}=28 \times 125 = 3500 \text{ INR}.$

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Answer: $\boxed{3500}$

💡 Closing insight: We never needed the individual diagonals — only their sum, which drops straight out of $(d_1+d_2)^2 = d_1^2 + d_2^2 + 2d_1d_2$ once the area and perimeter pin down the two symmetric combinations $d_1 d_2$ and $d_1^2 + d_2^2$. Whenever a question asks for a sum of unknowns, look for identities that produce that sum directly.

Think of this as: we know one side, the diagonal opposite a known angle, and that angle — so we can pin down the missing side with the cosine rule, then read off the area as base × height.

What we want: the area of parallelogram $ABCD$, given $AD = 10$, diagonal $AC = 20$, and $\angle ADC = 30°$.

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Step 1 — Identify the triangle that holds all the data

The diagonal $AC$ splits the parallelogram into two triangles. In $\triangle ADC$ we know:

• $AD = 10$
• $AC = 20$
• $\angle ADC = 30°$

This is enough to find the third side $DC$ (which is a side of the parallelogram, since $AB = DC$ and $AD = BC$).

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Step 2 — Find $DC$ using the cosine rule

Applying the cosine rule in $\triangle ADC$ at vertex $D$:

$AC^{2} = AD^{2} + DC^{2} - 2\cdot AD \cdot DC \cdot \cos 30^{\circ}$

Substituting the known values ($\cos 30^{\circ} = \frac{\sqrt{3}}{2}$):

$400 = 100 + DC^{2} - 2\cdot 10 \cdot DC \cdot \frac{\sqrt{3}}{2}$

$400 = 100 + DC^{2} - 10\sqrt{3}\, DC$

$DC^{2} - 10\sqrt{3}\, DC - 300 = 0$

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Step 3 — Solve the quadratic

Using the quadratic formula with $a = 1$, $b = -10\sqrt{3}$, $c = -300$:

$DC = \frac{10\sqrt{3} \pm \sqrt{(10\sqrt{3})^{2} + 4\cdot 300}}{2} = \frac{10\sqrt{3} \pm \sqrt{300 + 1200}}{2} = \frac{10\sqrt{3} \pm \sqrt{1500}}{2}$

Since $\sqrt{1500} = \sqrt{100 \times 15} = 10\sqrt{15}$:

$DC = \frac{10\sqrt{3} \pm 10\sqrt{15}}{2} = 5\left(\sqrt{3} \pm \sqrt{15}\right)$

Now we must pick the physically valid root — a length is positive.

So $DC = 5\left(\sqrt{3} + \sqrt{15}\right)$.

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Step 4 — Compute the area

The area of a parallelogram is $\text{base} \times \text{height}$. Taking $DC$ as the base, the height from $A$ down to $DC$ is:

$h = AD \cdot \sin(\angle ADC) = 10 \cdot \sin 30^{\circ} = 10 \cdot \tfrac{1}{2} = 5$

Therefore:

$\text{Area} = DC \times h = 5\left(\sqrt{3} + \sqrt{15}\right) \times 5 = 25\left(\sqrt{3} + \sqrt{15}\right)$

Answer: $\boxed{25(\sqrt{3}+\sqrt{15})}$ (Choice B)

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Alternative approach — Sine rule in $\triangle ADC$

By the sine rule: $\frac{AC}{\sin D} = \frac{AD}{\sin C} = \frac{DC}{\sin A}$.

$\frac{20}{\sin 30^{\circ}} = \frac{10}{\sin C} \implies \frac{20}{1/2} = 40 = \frac{10}{\sin C} \implies \sin C = \frac{1}{4}$

So $\cos C = \sqrt{1 - \frac{1}{16}} = \frac{\sqrt{15}}{4}$, and $\angle A = 180° - 30° - C = 150° - C$.

$\sin A = \sin(150° - C) = \sin 150°\cos C - \cos 150°\sin C = \frac{1}{2}\cdot\frac{\sqrt{15}}{4} - \left(-\frac{\sqrt{3}}{2}\right)\cdot\frac{1}{4} = \frac{\sqrt{15} + \sqrt{3}}{8}$

$DC = 40 \times \sin A = 40 \times \frac{\sqrt{3}+\sqrt{15}}{8} = 5(\sqrt{3}+\sqrt{15})$

The area is then $DC \times AD \sin 30° = 5(\sqrt{3}+\sqrt{15}) \times 5 = 25(\sqrt{3}+\sqrt{15})$, confirming the result. The cosine-rule method is faster under exam conditions.

💡 Closing insight: Two pieces of data ($AD$ and $\angle D$) give the height directly via $\sin 30°$; the diagonal $AC$ then fixes the base $DC$ through the cosine rule. The surds never collapse to a single radical — the answer is genuinely a sum of $\sqrt{3}$ and $\sqrt{15}$.

Think of this as: we have one unknown alloy (we don't know its weight or its silver percentage), and we're given two mixing experiments. Each experiment gives us one equation, so two experiments = two equations = enough to solve for both unknowns. We only need the weight.

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Let $W$ = weight (in kg) of the initial alloy, and let $s$ = fraction of silver in it.

Then the total silver in the initial alloy is $A = Ws$.

We'll treat $A$ and $W$ as our two unknowns — the silver fraction $s$ will take care of itself.

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Method 1 — Two linear equations

Step 1 — Translate the first mixing.

We add $3$ kg of pure silver ($100\%$ silver, so $3$ kg of silver) to the initial alloy. The new total weight is $W + 3$, and the new silver amount is $A + 3$. The result is $90\%$ silver:

$\frac{A + 3}{W + 3} = 0.90$

$A + 3 = 0.90(W + 3) = 0.90W + 2.70$

$\boxed{A - 0.90W = -0.30} \quad \cdots (1)$

Step 2 — Translate the second mixing.

We add $2$ kg of an alloy that is $90\%$ silver. The silver contributed by this added alloy is $0.90 \times 2 = 1.8$ kg. The new total weight is $W + 2$, and the new silver amount is $A + 1.8$. The result is $84\%$ silver:

$\frac{A + 1.8}{W + 2} = 0.84$

$A + 1.8 = 0.84(W + 2) = 0.84W + 1.68$

$\boxed{A - 0.84W = -0.12} \quad \cdots (2)$

💡 Teacher tip: Both equations are linear in $A$ and $W$. By keeping $A = Ws$ as a single symbol instead of substituting $Ws$ everywhere, we avoid a messy nonlinear system. The fraction $s$ is hidden inside $A$ and never causes trouble.

Step 3 — Eliminate $A$.

Subtract equation $(2)$ from equation $(1)$:

$(A - 0.90W) - (A - 0.84W) = (-0.30) - (-0.12)$

$-0.06W = -0.18$

$W = \frac{-0.18}{-0.06} = 3$

Step 4 — Verify both conditions.

From equation $(1)$: $A = 0.90(3) - 0.30 = 2.4$, so the initial alloy is $\frac{2.4}{3} = 80\%$ silver.

Both conditions check out.

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Alternative approach — Alligation

Alligation links the concentrations directly to the ratio of quantities mixed.

Mix 1: Pure silver ($100\%$) + initial alloy ($s\%$) → $90\%$.

$\frac{\text{pure silver}}{\text{initial alloy}} = \frac{90 - s}{100 - 90} = \frac{90 - s}{10}$

Since $3$ kg of pure silver is mixed with $W$ kg of the alloy:

$\frac{3}{W} = \frac{90 - s}{10} \;\Longrightarrow\; W = \frac{30}{90 - s} \quad \cdots (1')$

Mix 2: $90\%$ alloy + initial alloy ($s\%$) → $84\%$.

\frac{\text{90% alloy}}{\text{initial alloy}} = \frac{84 - s}{90 - 84} = \frac{84 - s}{6}

Since $2$ kg of the $90\%$ alloy is mixed with $W$ kg of the initial alloy:

$\frac{2}{W} = \frac{84 - s}{6} \;\Longrightarrow\; W = \frac{12}{84 - s} \quad \cdots (2')$

Equate the two expressions for $W$:

$\frac{30}{90 - s} = \frac{12}{84 - s}$

$30(84 - s) = 12(90 - s)$

$2520 - 30s = 1080 - 12s$

$1440 = 18s \;\Longrightarrow\; s = 80$

So the initial alloy is $80\%$ silver. Plug back:

$W = \frac{30}{90 - 80} = \frac{30}{10} = 3$

Same answer, reached through concentrations instead of silver amounts.

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

💡 Closing insight: The key move is noticing that both mixings share the same unknown alloy, so they form a $2 \times 2$ linear system. Whether you solve it with silver-amount equations (Method 1) or alligation ratios (Method 2), the silver fraction $s$ cancels out of the weight calculation — you never need to know it to find $W$. Method 1 is slightly faster under exam time since it goes straight to elimination.