CAT 2021 Quant Questions — Slot 2

Think of this as: we have four unknown digits and three equations linking them — so one digit stays free, and we scan its possible values to find the biggest number.

Let the 4-digit number be $\overline{abcd}$, where $a$ is the thousands digit, $b$ the hundreds, $c$ the tens, and $d$ the units. The three conditions are:

1. $a + b + c = 14$ (thousands + hundreds + tens)
2. $b + c + d = 15$ (hundreds + tens + units)
3. $c = d + 4$ (tens digit is 4 more than the units digit)

Step 1 — Express every digit in terms of $d$.

From condition (3), we already have $c$ in terms of $d$: $c = d + 4.$

Substitute this into condition (2): $b + (d+4) + d = 15 \;\Longrightarrow\; b + 2d + 4 = 15 \;\Longrightarrow\; b = 11 - 2d.$

Now subtract condition (1) from condition (2): $(b + c + d) - (a + b + c) = 15 - 14 \;\Longrightarrow\; d - a = 1 \;\Longrightarrow\; a = d - 1.$

So all four digits are pinned to a single free parameter $d$: $a = d-1, \quad b = 11-2d, \quad c = d+4, \quad d = d.$

💡 Teacher tip: Three independent equations in four unknowns leave exactly one degree of freedom. Choosing $d$ as the free variable is natural here because condition (3) already hands us $c$ directly from $d$.

Step 2 — Apply the digit rules to bound $d$.

Every digit must be a single decimal digit: $a \in \{1,\dots,9\}$ (it's the thousands digit, so it cannot be $0$), and $b, c, d \in \{0,\dots,9\}$.

• Thousands digit: $a = d - 1 \geq 1 \;\Longrightarrow\; d \geq 2$.
• Tens digit: $c = d + 4 \leq 9 \;\Longrightarrow\; d \leq 5$.
• Hundreds digit: $b = 11 - 2d \geq 0 \;\Longrightarrow\; d \leq 5$ (and $b \leq 9$ gives $d \geq 1$, which is already weaker than $d \geq 2$).

Combining all bounds: $d \in \{2, 3, 4, 5\}$.

Step 3 — List every candidate and pick the largest.

Quick verification of the largest candidate, $4195$:

• $a + b + c = 4 + 1 + 9 = 14$ ✓
• $b + c + d = 1 + 9 + 5 = 15$ ✓
• $c = 9 = 5 + 4 = d + 4$ ✓

As $d$ increases, the thousands digit $a = d - 1$ also increases, so the largest number corresponds to the largest allowed $d$, which is $d = 5$.

Answer: $\boxed{4195}$

💡 Why this works: With three independent conditions on four unknowns, exactly one free parameter survives. Driving everything through $d$ turns the search into a single column scan — and since the leading digit grows with $d$, the maximum sits at the top of the range.

Think of this as: two scenarios give us two equations in two unknowns (the fill rate and the drain rate). We only need the fill rate to answer the question.

Step 1 — Define variables and translate each scenario into an equation.

Let $a$ = pipe A's filling rate (in tanks per hour), and $b$ = pipe B's draining rate (in tanks per hour). Both are positive numbers. "Filling 1 tank" means the net work done equals 1.

• Case 1: A opens at 2 pm, B opens at 3 pm, tank is full at 10 pm.

• A runs from 2 pm to 10 pm = 8 hours.

• B runs from 3 pm to 10 pm = 7 hours.

• Net work: $8a - 7b = 1$.

• Case 2: A opens at 2 pm, B opens at 4 pm, tank is full at 6 pm.

• A runs from 2 pm to 6 pm = 4 hours.

• B runs from 4 pm to 6 pm = 2 hours.

• Net work: $4a - 2b = 1$.

💡 Teacher tip: The key is to compute how long each pipe runs in each scenario — not just the total elapsed time. Since A and B start at different times, their running durations differ even within the same case.

Step 2 — Solve the system.

From Case 2: $4a - 2b = 1 \Rightarrow 2a - b = \frac{1}{2} \Rightarrow b = 2a - \frac{1}{2}$.

Substitute into Case 1: $8a - 7\left(2a - \frac{1}{2}\right) = 1$ $8a - 14a + \frac{7}{2} = 1$ $-6a = 1 - \frac{7}{2} = -\frac{5}{2}$ $a = \frac{5}{12} \text{ tanks per hour.}$

Step 3 — Convert to the time for A alone.

Time to fill 1 tank with A alone $= \frac{1}{a} = \frac{12}{5}$ hours.

Convert to minutes: $\frac{12}{5} \times 60 = 144$ minutes.

Verification — check both scenarios with $a = \frac{5}{12}$ and $b = \frac{1}{3}$.

First confirm $b$: $b = 2 \cdot \frac{5}{12} - \frac{1}{2} = \frac{10}{12} - \frac{6}{12} = \frac{4}{12} = \frac{1}{3}$.

Both scenarios check out.

Answer: $\boxed{144}$ (Choice A)

💡 Closing insight: We never needed the value of $b$ to answer the question — solving for $a$ alone was enough. The two scenarios pin down both rates, but the question only asks about pipe A.

Alternative approach — Elimination.

Subtract Case 2 from Case 1 (both right-hand sides equal 1, so they cancel): $(8a - 7b) - (4a - 2b) = 0$ $4a - 5b = 0 \Rightarrow b = \frac{4a}{5}.$

Substitute into Case 2: $4a - 2 \cdot \frac{4a}{5} = 1 \Rightarrow \frac{20a - 8a}{5} = 1 \Rightarrow \frac{12a}{5} = 1 \Rightarrow a = \frac{5}{12}.$

Same result, slightly fewer algebraic steps. Under exam time, elimination is marginally faster here because both equations equal the same constant.

Think of this as a two-stage filter: first find every integer $n$ that satisfies the given bound, then keep only those $n$ for which $3 + 3^{n+1}$ comes out as a whole number.

Step 1 — Solve the inequality for $n$.

Start with $2.25 \le 2 + 2^{n+2} \le 202.$

Subtract $2$ from all three parts: $0.25 \le 2^{n+2} \le 200.$

Rewrite $0.25$ as a power of $2$: $0.25 = \frac{1}{4} = 2^{-2},$ so the inequality becomes $2^{-2} \le 2^{n+2} \le 200.$

Lower bound. Since $2^x$ is increasing, $2^{n+2} \ge 2^{-2}$ gives $n + 2 \ge -2 \quad\Rightarrow\quad n \ge -4.$

Upper bound. We need the largest power of $2$ not exceeding $200$. Compute: $2^7 = 128 \le 200 < 256 = 2^8.$ So $2^{n+2} \le 200$ forces $n + 2 \le 7$, i.e. $n \le 5.$

Combining both ends: $-4 \le n \le 5.$

So the integer values of $n$ are $n \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4, 5\},$ ten values in total.

Step 2 — Keep only the $n$ that make $3 + 3^{n+1}$ an integer.

The expression $3 + 3^{n+1}$ is an integer exactly when $3^{n+1}$ is an integer. A power of $3$ is an integer precisely when its exponent is non-negative: $n + 1 \ge 0 \quad\Rightarrow\quad n \ge -1.$

💡 Teacher tip: A negative exponent produces a fraction — for example $3^{-1} = \tfrac{1}{3}$, $3^{-2} = \tfrac{1}{9}$, and so on. Adding such a fraction to $3$ can never yield an integer, so only non-negative exponents survive the filter.

Intersecting $n \ge -1$ with $-4 \le n \le 5$ leaves $n \in \{-1, 0, 1, 2, 3, 4, 5\},$ seven values.

Step 3 — Verify each surviving $n$ gives a distinct integer.

The seven integer values $4, 6, 12, 30, 84, 246, 732$ are all distinct, so the count is $7$.

Answer: $\boxed{7}$

💡 Why this is neat: ten values of $n$ satisfy the original bound, but only the seven with $n \ge -1$ keep the power-of-$3$ term whole — the integrality filter, not the inequality, does the real work of reducing the count.

Think of this as: a trapezium is split into a parallelogram and a triangle, and we're told how their areas differ. Find the link between the two parallel sides, express both areas, and then scale up to the whole trapezium.

Let the parallel sides be $AB = a$ (the smaller side) and $CD = c$ (the larger side), and let $h$ be the perpendicular distance between them — the common height of both pieces.

Step 1 — Use the midpoint and the parallelogram to relate $c$ and $a$.

Since $P$ is the midpoint of $CD$, $CP = PD = \frac{c}{2}.$

Since $ABPD$ is a parallelogram, its opposite sides are equal. In particular, $AB = PD.$ So $a = \frac{c}{2}$, which gives $c = 2a.$

💡 Teacher tip: This is the key deduction of the problem. The midpoint condition alone only tells us $PD = \frac{c}{2}$; it is the parallelogram condition that forces this to equal $AB = a$. Together they lock $c = 2a$.

Step 2 — Write the areas of the two pieces.

Both the parallelogram $ABPD$ and the triangle $BPC$ sit inside the trapezium and share the same height $h$.

• Parallelogram $ABPD$: base $AB = a$, so area $= a \cdot h$.
• Triangle $BPC$: base $PC = \frac{c}{2} = a$, so area $= \frac{1}{2} \cdot a \cdot h = \frac{1}{2}ah$.

💡 Teacher tip: Notice both pieces have a base of length $a$ — one uses $AB$, the other uses $PC = \frac{c}{2}$. That is why their areas fall into a clean $1: \frac{1}{2}$ ratio.

Step 3 — Apply the given area difference.

The parallelogram's area minus the triangle's area is $10$: $ah - \frac{1}{2}ah = 10.$ $\frac{1}{2}ah = 10 \implies ah = 20.$

Step 4 — Compute the area of the trapezium.

Using the trapezium area formula with $a$ and $c = 2a$: $\text{Area}(ABCD) = \frac{1}{2}(a + c)\,h = \frac{1}{2}(a + 2a)\,h = \frac{3}{2}ah.$ Substituting $ah = 20$: $\text{Area}(ABCD) = \frac{3}{2} \times 20 = 30.$

Quick check. With $ah = 20$: parallelogram area $= 20$, triangle area $= 10$, difference $= 10$ ✓, and total $= 20 + 10 = 30$ ✓.

Answer: $\boxed{30}$ (Choice A)

💡 Closing insight: The parallelogram and the triangle tile the trapezium exactly. Because $c = 2a$, their areas are $ah$ and $\frac{1}{2}ah$, so the trapezium is precisely $\frac{3}{2}$ of the parallelogram. A difference of $10$ therefore scales up to a total of $30$.

Think of this as: two triangles $ADE$ and $ABC$ sit inside each other sharing the corner at $A$. We know the small one's area; we need the big one's. The only bridge we need is how the two sides from $A$ scale.

Step 1 — Spot the shared angle.

Both triangles $ADE$ and $ABC$ have the vertex $A$ in common, and the sides $AD$, $AE$ lie along $AB$, $AC$ respectively. So the included angle at $A$ is the same for both triangles.

Recall the area formula using two sides and the included angle $\theta$: $\text{Area} = \tfrac{1}{2}\, (\text{side 1})(\text{side 2})\sin\theta.$

Because $\sin\theta$ is identical for both triangles, it cancels when we take the ratio: $\frac{\text{Area}(ADE)}{\text{Area}(ABC)} = \frac{\tfrac12 \cdot AD \cdot AE \cdot \sin A}{\tfrac12 \cdot AB \cdot AC \cdot \sin A} = \frac{AD}{AB}\cdot\frac{AE}{AC}.$

💡 Teacher tip: Whenever two triangles share an included angle, their area ratio is just the product of the ratios of the two adjacent sides. The angle itself never needs to be known.

Step 2 — Convert the given part-ratios into fractions of the full sides.

• $AD: BD = 2: 1$, so $AB = AD + BD = 2 + 1 = 3$ parts. Hence $\frac{AD}{AB} = \frac{2}{3}.$

• $AE: CE = 2: 3$, so $AC = AE + CE = 2 + 3 = 5$ parts. Hence $\frac{AE}{AC} = \frac{2}{5}.$

💡 Teacher tip: A common slip is to use $\frac{AD}{BD}$ or $\frac{AE}{CE}$ directly. Always add the parts to get the whole side first — the ratio we need is part-to-whole, not part-to-part.

Step 3 — Form the area ratio and solve.

$\frac{\text{Area}(ADE)}{\text{Area}(ABC)} = \frac{AD}{AB}\cdot\frac{AE}{AC} = \frac{2}{3}\times\frac{2}{5} = \frac{4}{15}.$

We are given $\text{Area}(ADE) = 8$, so $\frac{8}{\text{Area}(ABC)} = \frac{4}{15}\;\Longrightarrow\;\text{Area}(ABC) = 8 \times \frac{15}{4} = 2 \times 15 = 30.$

Quick check. $\frac{4}{15}$ of $30$ is $\frac{4 \times 30}{15} = 8$ ✓ — matches the given area of $ADE$.

Answer: $\boxed{30}$

💡 Closing insight: For any problem where two points lie on two sides of a triangle and the inner triangle shares the vertex, only the two part-to-whole side fractions matter — not the angle, not the shape of the triangle. Multiply the fractions, set equal to the area ratio, and solve.

Think of this as: two equations, each messy on its own, but designed so that adding them collapses everything into a single quadratic in $x-y$.

What we want: the value of $x - y$, given $x > y$.

The two equations are: $x^2 - xy - x = 22 \qquad \text{and} \qquad y^2 - xy + y = 34.$

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Step 1 — Add the two equations.

Adding the left-hand sides and the right-hand sides: $(x^2 - xy - x) + (y^2 - xy + y) = 22 + 34.$

Group the terms carefully: $x^2 - 2xy + y^2 \;\;\text{(the quadratic terms)} \quad - \quad x - y \;\;\text{(the linear terms)} \quad = 56.$

Wait — check the linear terms: we have $-x$ from the first equation and $+y$ from the second, so the linear part is $-x + y = -(x - y)$. Thus: $x^2 - 2xy + y^2 - (x - y) = 56.$

Step 2 — Recognise the perfect square.

The first three terms form a perfect square: $x^2 - 2xy + y^2 = (x - y)^2.$

So the entire equation becomes: $(x - y)^2 - (x - y) = 56.$

💡 Teacher tip: When you see $x^2 - xy$ in one equation and $y^2 - xy$ in another, adding them is almost always the intended move — the $-xy$ terms combine into $-2xy$, which completes the square $(x-y)^2$.

Step 3 — Substitute $d = x - y$ and solve.

Let $d = x - y$. The equation becomes: $d^2 - d = 56 \quad\Longrightarrow\quad d^2 - d - 56 = 0.$

Factor the quadratic: $d^2 - d - 56 = (d - 8)(d + 7) = 0.$

So $d = 8$ or $d = -7$.

Step 4 — Use the condition $x > y$ to pick the right root.

Since $x > y$, we know $x - y > 0$, i.e. $d > 0$. Therefore: $d = 8.$

Step 5 — Verify a real solution exists (not just an algebraic artefact).

With $d = 8$, set $x = y + 8$ and substitute into the first equation: $(y+8)^2 - (y+8)y - (y+8) = 22.$

Expand: $y^2 + 16y + 64 - y^2 - 8y - y - 8 = 22,$ $7y + 56 = 22 \quad\Longrightarrow\quad y = -\frac{34}{7}, \qquad x = y + 8 = \frac{22}{7}.$

Check in the second equation: $y^2 - xy + y = \frac{1156}{49} + \frac{748}{49} - \frac{34}{7} = \frac{1156 + 748 - 238}{49} = \frac{1666}{49} = 34. \;\checkmark$

Both equations are satisfied, so $x - y = 8$ is genuinely achievable.

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

💡 Closing insight: The two equations are individually hard but collectively simple. Adding them turns the cross-term $xy$ into part of a perfect square, reducing a system of two equations to a single quadratic in $x - y$. The condition $x > y$ then acts as a sign filter — it tells you which root to keep without solving for $x$ and $y$ explicitly.

Think of this as: two possible futures are described, and they differ by exactly one goal. That single extra goal lifts the overall average by a known amount — and that is enough to pin down the total number of matches.

Let $n$ = matches already played, and $g$ = goals already scored. Ten more matches are still to be played.

The overall average is always

$\text{average} = \frac{\text{total goals}}{\text{total matches}}.$

Step 1 — Translate each scenario into an equation.

• Scenario A: one goal in the next 10 matches, average becomes $0.15$. $\frac{g+1}{n+10} = 0.15$

• Scenario B: two goals in the next 10 matches, average becomes $0.2$. $\frac{g+2}{n+10} = 0.2$

Step 2 — Subtract to eliminate the unknown $g$.

We don't actually know $g$, the goals already scored. But notice that both equations share the same denominator $n+10$ and the numerators differ by exactly $1$. Subtract Scenario A from Scenario B:

$\frac{(g+2)-(g+1)}{n+10} = 0.2 - 0.15$

$\frac{1}{n+10} = 0.05$

💡 Teacher tip: Whenever two averaging scenarios differ by a fixed amount of the numerator, subtracting the equations cancels everything unknown and leaves a clean relation for the denominator. This is the fastest route — no need to solve for $g$ first.

Step 3 — Solve for $n$.

$\frac{1}{n+10} = 0.05 = \frac{1}{20}$

$n+10 = 20 \implies n = 10$

Step 4 — Verify with both scenarios.

With $n=10$, total matches $= 20$. From Scenario A: $\frac{g+1}{20} = 0.15 \implies g+1 = 3 \implies g = 2.$

Now check Scenario B with $g=2$: $\frac{2+2}{20} = \frac{4}{20} = 0.2. \checkmark$

Both scenarios are consistent, so $n=10$ is confirmed.

Answer: $\boxed{10}$

💡 Why this is elegant: the goals already scored ($g$) never needed to be found first. The difference of one goal between the two futures maps directly to the difference of $0.05$ in average, immediately yielding the total number of matches.

Think of this as: we know one irrational root of a rational-coefficient quadratic — that alone pins down both roots, which pins down $b$ and $c$ in terms of $a$. The extra condition $b = c^3$ then becomes a single equation in $a$.

Step 1 — Use the conjugate-root principle.

The equation $ax^2 - bx + c = 0$ has rational coefficients ($a, b, c$ are rational). When a quadratic with rational coefficients has an irrational root of the form $p + \sqrt{q}$ (with $p, q$ rational and $\sqrt{q}$ irrational), its conjugate $p - \sqrt{q}$ must also be a root.

So if one root is $2 + \sqrt{3}$, the other root is $2 - \sqrt{3}$.

💡 Teacher tip: This is the conjugate-root principle for rational-coefficient polynomials. Irrational roots involving square roots always arrive in conjugate pairs — otherwise the coefficients couldn't all stay rational.

Step 2 — Read off the sum and product of the roots.

For $ax^2 - bx + c = 0$, Vieta's formulas give:

• Sum of roots $= \dfrac{b}{a}$
• Product of roots $= \dfrac{c}{a}$

Computing from the roots: $\text{Sum} = (2+\sqrt{3}) + (2-\sqrt{3}) = 4 \quad\Longrightarrow\quad \frac{b}{a} = 4 \quad\Longrightarrow\quad b = 4a.$

$\text{Product} = (2+\sqrt{3})(2-\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \quad\Longrightarrow\quad \frac{c}{a} = 1 \quad\Longrightarrow\quad c = a.$

So we have expressed both $b$ and $c$ as simple multiples of $a$: $b = 4a, \qquad c = a.$

Step 3 — Apply the condition $b = c^3$.

Substitute $b = 4a$ and $c = a$: $4a = (a)^3.$

Since $a \neq 0$, divide both sides by $a$: $a^2 = 4 \quad\Longrightarrow\quad a = 2 \text{ or } a = -2.$

Either way, $|a| = 2$.

Step 4 — Verify both values of $a$ satisfy all conditions.

Both values produce a valid equation with the correct roots and satisfy $b = c^3$.

Answer: $\boxed{2}$ (Choice B)

💡 Closing insight: The condition $b = c^3$ looks intimidating, but the conjugate-root principle reduces it to $b = 4a$ and $c = a$ — so $b = c^3$ collapses to $4a = a^3$, i.e. $a^2 = 4$. Recognising the conjugate pair first is the entire battle; the rest is substitution.

Alternative approach — factor and expand.

Since the roots are $2 \pm \sqrt{3}$, the quadratic (up to a constant factor $a$) is: $a\bigl(x - (2+\sqrt{3})\bigr)\bigl(x - (2-\sqrt{3})\bigr) = a\bigl(x^2 - 4x + 1\bigr) = ax^2 - 4ax + a.$

Comparing with $ax^2 - bx + c$, we read off $b = 4a$ and $c = a$ directly — the same result as Vieta's, obtained by expanding the factored form. The rest proceeds identically: $4a = a^3 \Rightarrow |a| = 2$. Under exam conditions, Vieta's is faster; the factor-expand route is a good cross-check.

Think of this as: three numbers in arithmetic progression have a product tied to their sum. The AP structure will let us collapse three unknowns into one clean equation.

Step 1 — Use the AP structure to simplify.

Since $x, y, z$ are in arithmetic progression, let the common difference be $d > 0$. Then: $x = y - d, \quad z = y + d$

Two useful consequences fall out immediately:

• The difference we want is $z - x = (y+d) - (y-d) = 2d$.
• The sum is $x + y + z = (y-d) + y + (y+d) = 3y$.
• The product is $xyz = (y-d)\,y\,(y+d) = y(y^2 - d^2)$.

Step 2 — Substitute into the given condition.

We are told $xyz = 5(x + y + z)$. Plugging in the AP forms: $y(y^2 - d^2) = 5 \cdot 3y = 15y$

Since $y$ is a positive integer, $y \neq 0$, so we can safely divide both sides by $y$: $y^2 - d^2 = 15$

Step 3 — Factor and interpret.

Recognise the difference of squares: $y^2 - d^2 = (y - d)(y + d) = 15$

But $y - d = x$ and $y + d = z$, so this is simply: $x \cdot z = 15$

💡 Teacher tip: The AP structure made the sum $x+y+z$ collapse to $3y$, which let us cancel $y$ from both sides. That single move turned a messy product-of-three-variables equation into the tiny fact $xz = 15$.

Step 4 — List the factor pairs of 15.

Since $x < z$ are positive integers, the only factor pairs of 15 are $(1, 15)$ and $(3, 5)$. For each, the middle AP term is $y = \frac{x+z}{2}$, and we must check the constraint $y - x > 2$.

Only the pair $(x, z) = (1, 15)$ survives the constraint.

Step 5 — Compute $z - x$.

$z - x = 15 - 1 = 14$

(Equivalently, $d = 7$, so $z - x = 2d = 14$.)

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

💡 Closing insight: Whenever three numbers are in AP, anchoring on the middle term removes the asymmetry — the sum becomes $3y$ and the product becomes $y(y^2-d^2)$. That is the move to look for whenever an AP meets a product/sum condition.

Think of this as: since profit is split in proportion to investment, each partner's profit is simply their own investment × the overall profit rate. A known change in one partner's payout therefore directly reveals that partner's investment.

Step 1 — Turn the profit-sharing rule into a usable formula.

Let the total investment be $T$. Anil owns $70\%$ of it, so Anil's investment is $0.7T$. If the overall profit rate is $r$ (written as a decimal), then each person's profit is:

💡 Teacher tip: Because profit is shared in proportion to investment, the total investment $T$ cancels inside the proportion. What drives each person's payout is their individual stake multiplied by the profit rate.

Step 2 — Use Anil's profit drop to find the total investment $T$.

The overall profit rate falls from $18\%$ to $15\%$ — a drop of $3$ percentage points. Anil's profit falls by ₹$420$:

$0.03 \times (0.7T) = 420$ $0.021T = 420$ $T = \frac{420}{0.021} = 20{,}000$

So the total investment is ₹$20{,}000$, and Anil's investment is $0.7 \times 20{,}000 = \text{₹}14{,}000$.

Step 3 — Use Chintu's profit rise to find Chintu's investment.

The overall profit rate rises from $15\%$ to $17\%$ — a rise of $2$ percentage points. Chintu's profit rises by ₹$80$:

$0.02 \times C = 80$ $C = \frac{80}{0.02} = 4{,}000$

So Chintu's investment is ₹$4{,}000$.

Step 4 — Bobby's investment is what remains.

$\text{Bobby} = T - \text{Anil} - \text{Chintu} = 20{,}000 - 14{,}000 - 4{,}000 = 2{,}000$

Step 5 — Verify with a full profit table.

• Anil's drop from 18% to 15%: $2{,}520 - 2{,}100 = 420$ ✓
• Chintu's rise from 15% to 17%: $680 - 600 = 80$ ✓
• Each column's total matches $r \times 20{,}000$ ✓

Answer: $\boxed{2000}$ (Choice A)

💡 Closing insight: Each one-percentage-point change in the overall profit rate moves a partner's payout by exactly $1\%$ of their own investment. So a known payout change directly reveals that partner's investment — two equations pin down Anil and Chintu, and Bobby is simply the remainder.

Alternative approach — Solve for individual investments directly.

Instead of finding $T$ first, let Anil's investment be $A$. A $3$-point drop in profit rate costs Anil ₹$420$:

$0.03 \times A = 420 \implies A = 14{,}000$

Since $A = 0.7T$, we recover $T = 14{,}000 / 0.7 = 20{,}000$.

Chintu's investment comes from the $2$-point rise: $0.02 \times C = 80 \implies C = 4{,}000$.

Bobby $= 20{,}000 - 14{,}000 - 4{,}000 = \text{₹}2{,}000$.

This is the same logic but enters through Anil's investment rather than the total — marginally faster since Anil's data already involves Anil directly.

Think of this as: a rhombus is built from two perpendicular diagonals that bisect each other. If we know the side and the area, we actually know enough to pin down both diagonals exactly.

What we want: the longer diagonal of a rhombus with area $12$ sq cm and side $5$ cm.

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Step 1 — Translate the two given facts into equations in the diagonals.

Let the diagonals be $d_1$ and $d_2$. A key property of a rhombus: its diagonals bisect each other at right angles. So the rhombus splits into four right-angled triangles, each with legs $\frac{d_1}{2}$ and $\frac{d_2}{2}$ and hypotenuse equal to the side, $5$.

Using Pythagoras on one such triangle: $\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = 5^2 = 25.$ Multiplying through by $4$: $d_1^2 + d_2^2 = 100. \quad\cdots (1)$

The area of a rhombus is half the product of its diagonals: $\frac{1}{2}d_1 d_2 = 12 \;\Longrightarrow\; d_1 d_2 = 24. \quad\cdots (2)$

💡 Teacher tip: Equation (1) gives us the sum of squares of the diagonals; equation (2) gives us their product. With sum-of-squares and product known, we can recover the diagonals — this is the heart of the problem.

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Step 2 — Find $d_1 + d_2$ and $d_1 - d_2$.

Use the two algebra identities $(a+b)^2 = a^2+b^2+2ab$ and $(a-b)^2 = a^2+b^2-2ab$.

For the sum: $(d_1+d_2)^2 = d_1^2 + d_2^2 + 2d_1 d_2 = 100 + 2(24) = 100 + 48 = 148.$ Since diagonals are positive, $d_1 + d_2 = \sqrt{148} = \sqrt{4\times 37} = 2\sqrt{37}.$

For the difference: $(d_1-d_2)^2 = d_1^2 + d_2^2 - 2d_1 d_2 = 100 - 48 = 52.$ Taking $d_1 > d_2$ (so the difference is positive), $d_1 - d_2 = \sqrt{52} = \sqrt{4\times 13} = 2\sqrt{13}.$

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Step 3 — Solve for the longer diagonal $d_1$.

We now have a simple linear system: $d_1 + d_2 = 2\sqrt{37}, \qquad d_1 - d_2 = 2\sqrt{13}.$

Adding the two equations eliminates $d_2$: $2d_1 = 2\sqrt{37} + 2\sqrt{13},$ $d_1 = \sqrt{37} + \sqrt{13}.$

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Quick sanity check.

Both diagonals are positive, and $d_1 > d_2$, consistent with $d_1$ being the longer one. ✓

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Answer: $\boxed{\sqrt{37}+\sqrt{13}}$ (Choice A)

💡 Closing insight: A rhombus is completely determined by its side length and its area — together they fix the sum of squares and the product of the diagonals, which is enough to recover each diagonal individually. Notice also that the $\frac{1}{2}$ in the area formula is what stops the answers from collapsing to half-expressions, which is exactly why options C and D are there to catch that slip.

Think of this as: three people chip in different amounts of work to finish one house. Pay follows work, so once we know what fraction of the house Charu painted, his money is just that fraction of ₹21,000.

Step 1 — Write down each person's daily work rate.

If someone can finish a job alone in $n$ days, their daily work rate is $\frac{1}{n}$ of the job per day.

• Anil's rate: $\frac{1}{60}$ house/day
• Bimal's rate: $\frac{1}{84}$ house/day
• Charu's rate: let's call it $c$ house/day (unknown for now)

Step 2 — Track how many days each person works.

Step 3 — Use "one full house painted" to find Charu's contribution.

The total work done by all three equals 1 house: $\underbrace{\frac{24}{60}}_{\text{Anil}} + \underbrace{\frac{14}{84}}_{\text{Bimal}} + \underbrace{14c}_{\text{Charu}} = 1$

Simplify each known fraction: $\frac{24}{60} = \frac{2}{5}, \qquad \frac{14}{84} = \frac{1}{6}$

So: $\frac{2}{5} + \frac{1}{6} + 14c = 1$

Bring to a common denominator of 30: $\frac{12}{30} + \frac{5}{30} + 14c = 1 \implies \frac{17}{30} + 14c = 1$

Therefore: $14c = 1 - \frac{17}{30} = \frac{13}{30}$

💡 Teacher tip: Notice we never needed to solve for $c$ itself. The quantity $14c$ — Charu's total work — is exactly what the equation hands us. In work-and-pay problems, always chase the total work fraction, not the rate.

Step 4 — Charu's fraction of the total work.

Charu did $\frac{13}{30}$ of the house. Since payment is split in proportion to work done, his share of the money is the same fraction of ₹21,000: $\text{Charu's share} = \frac{13}{30} \times 21{,}000$

Step 5 — Compute. $\frac{21{,}000}{30} = 700 \implies \text{Charu's share} = 13 \times 700 = 9{,}100$

Quick check of the split:

The fractions add to 1 and the money adds to ₹21,000 — everything checks.

Answer: $\boxed{9100}$ (Choice C)

💡 Closing insight: The key move was writing the work equation with $14c$ left as a single block. Solving for $c$ would have given $\frac{13}{420}$, an ugly number — but we didn't need it. Keeping Charu's total contribution as one unit turned the arithmetic clean.