CAT 2018 Quant Questions — Slot 1

Think of this as: we know the total number of students and how many like each food, but we don't know the overlap. The question asks what the "burger-only" count can possibly be — so we need to find the full range of valid values, then check which option lands inside it.

Step 1 — Set up the four regions.

With 200 students split by their pizza/burger preferences, there are four disjoint groups:

• $a$ = pizza only
• $b$ = burger only
• $c$ = both pizza and burger
• $d$ = neither

From the data: $a + c = 105 \quad \text{(pizza lovers)}, \qquad b + c = 134 \quad \text{(burger lovers)}$ $a + b + c + d = 200 \quad \text{(everyone)}$

So we can express $a$ and $b$ in terms of $c$: $a = 105 - c, \qquad b = 134 - c.$

Step 2 — Find the allowable range for $c$ (the overlap).

Every region count must be non-negative.

• From $a \ge 0$: $\;105 - c \ge 0 \;\Rightarrow\; c \le 105$.
• From $b \ge 0$: $\;134 - c \ge 0 \;\Rightarrow\; c \le 134$ (already weaker than $c \le 105$).
• From $c \ge 0$: trivially $c \ge 0$.

Now use the "neither" group. Substitute $a$ and $b$: $d = 200 - a - b - c = 200 - (105 - c) - (134 - c) - c = 200 - 105 + c - 134 + c - c = c - 39.$

Since $d \ge 0$: $c - 39 \ge 0 \;\Rightarrow\; c \ge 39.$

Combining both bounds: $\boxed{39 \le c \le 105}.$

💡 Teacher tip: The "neither" region is the hidden constraint here. Even though the question never mentions students who like neither food, that group must still exist as a non-negative count — and that's what pins down the lower bound on the overlap $c$.

Step 3 — Convert the range of $c$ into a range for burger-only $b$.

Since $b = 134 - c$, burger-only is largest when the overlap $c$ is smallest, and vice versa:

So the number of burger-only students can be any integer in: $29 \le b \le 95.$

Step 4 — Check which option falls inside the range.

Only 93 lies inside the valid range.

Answer: D — $\boxed{93}$

💡 Why this works: "Can possibly be" is the key phrase — it signals a range question, not a unique-value question. Whenever you see it, derive the minimum and maximum of the requested quantity using non-negativity of every region, then test the options against that range.

Think of this as: two nested-log equations, each with a clean cube on the outside. Undo the outer log first, then the inner one — and the first equation hands you $a$, which feeds straight into the second.

Step 1 — Translate the first equation.

We are given: $\log_2(5 + \log_3 a) = 3$

The key move: if $\log_x(\text{stuff}) = k$, then $\text{stuff} = x^k$. Applying that here with base $2$: $5 + \log_3 a = 2^3 = 8$

Now subtract $5$: $\log_3 a = 3$

Apply the same move again with base $3$: $a = 3^3 = 27$

So $\boxed{a = 27}$.

Step 2 — Translate the second equation.

We are given: $\log_5(4a + 12 + \log_2 b) = 3$

Again, undo the outer log first — the argument equals $5^3$: $4a + 12 + \log_2 b = 5^3 = 125$

Now substitute $a = 27$ from Step 1: $4(27) + 12 + \log_2 b = 125$ $108 + 12 + \log_2 b = 125$ $120 + \log_2 b = 125$

Subtract $120$: $\log_2 b = 5$

Undo this inner log with base $2$: $b = 2^5 = 32$

So $\boxed{b = 32}$.

Step 3 — Add them. $a + b = 27 + 32 = 59$

💡 Teacher tip: For nested logarithms, always work outside-in: first rewrite $\log_x(\text{stuff}) = k$ as $\text{stuff} = x^k$, then deal with whatever log is left inside. Trying to handle both layers at once is where sign and base errors creep in.

Quick check.

Everything closes cleanly, so the values are consistent.

Answer: A — $\boxed{59}$

The structure of the question is a chain: the first equation isolates $a$, and $a$ is then planted inside the second equation to isolate $b$. Once you see that dependency, each equation is just a single logarithm to undo at a time.

Think of this as: the selling price and profit percentage together tell us the cost price of each blend; each blend's cost is a weighted average of the two ingredient costs — so two different blends give us two equations in two unknowns.

Step 1 — Recover each blend's cost price from the selling price.

The mixture sells at Rs. 40/kg in both cases. Profit percentage links selling price to cost price:

$\text{Cost Price} = \frac{\text{Selling Price}}{1 + \text{profit fraction}}$

• 10% profit (ratio 3: 2): Cost $= \dfrac{40}{1.10} = \dfrac{40}{11/10} = \dfrac{400}{11}$ Rs/kg
• 5% profit (ratio 2: 3): Cost $= \dfrac{40}{1.05} = \dfrac{40}{21/20} = \dfrac{800}{21}$ Rs/kg

Step 2 — Write the weighted-cost equations.

When A and B are mixed in ratio $3:2$, the mixture cost is the weighted average of the ingredient costs:

$\frac{3C_A + 2C_B}{5} = \frac{400}{11}$

Multiplying both sides by 5:

$3C_A + 2C_B = \frac{2000}{11} \quad \cdots (1)$

Similarly, for the ratio $2:3$ mixture:

$\frac{2C_A + 3C_B}{5} = \frac{800}{21}$

$2C_A + 3C_B = \frac{4000}{21} \quad \cdots (2)$

Step 3 — Eliminate $C_B$ and solve for $C_A$.

Multiply equation (1) by 3 and equation (2) by 2 so the $C_B$ coefficients match:

$9C_A + 6C_B = \frac{6000}{11} \quad \cdots (3)$ $4C_A + 6C_B = \frac{8000}{21} \quad \cdots (4)$

Subtract (4) from (3):

$5C_A = \frac{6000}{11} - \frac{8000}{21}$

Common denominator is $11 \times 21 = 231$:

$5C_A = \frac{6000 \times 21 - 8000 \times 11}{231} = \frac{126000 - 88000}{231} = \frac{38000}{231}$

$C_A = \frac{7600}{231}$

Step 4 — Back-substitute to find $C_B$.

From equation (1):

$2C_B = \frac{2000}{11} - 3C_A = \frac{2000}{11} - \frac{22800}{231}$

Convert $\frac{2000}{11}$ to denominator 231 (multiply by $\frac{21}{21}$):

$2C_B = \frac{42000 - 22800}{231} = \frac{19200}{231}$

$C_B = \frac{9600}{231}$

Step 5 — Take the ratio.

$C_A: C_B = \frac{7600}{231}: \frac{9600}{231} = 7600: 9600 = 19: 24$

💡 Teacher tip: The denominators (231) cancel the moment you take the ratio — you never need the actual cost prices, only their relative sizes. This is why keeping $C_A$ and $C_B$ symbolic until the final ratio step avoids unnecessary arithmetic noise.

Answer: C — $\boxed{19: 24}$

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Alternative approach — Divide the equations directly (faster).

Instead of solving for $C_A$ and $C_B$ individually, let $r = \dfrac{C_A}{C_B}$ and divide the two equations to eliminate the common scale factor $C_B$.

From (1): $C_B(3r + 2) = \dfrac{2000}{11}$

From (2): $C_B(2r + 3) = \dfrac{4000}{21}$

Dividing the first by the second:

$\frac{3r + 2}{2r + 3} = \frac{2000/11}{4000/21} = \frac{2000 \times 21}{11 \times 4000} = \frac{21}{22}$

Cross-multiply:

$22(3r + 2) = 21(2r + 3)$ $66r + 44 = 42r + 63$ $24r = 19$ $r = \frac{19}{24}$

So $C_A: C_B = 19: 24$ — confirmed.

💡 Why this is faster: by dividing the equations, $C_B$ cancels immediately and you solve one equation in one unknown ($r$) — no common denominators, no back-substitution. Under exam time, this is the method to reach for.

Think of this as: a sector is cut out of a circle, and inside that sector we place a small isosceles triangle whose area is exactly half the sector's. We need the side length of that triangle.

Step 1 — Find the area of the sector $R$.

The region $R$ is a sector of radius $1$ cm with central angle $60^\circ$. The area of a sector is the fraction of the full circle given by $\dfrac{\theta}{360^\circ}$:

$\text{Area}(R) = \frac{60^\circ}{360^\circ}\cdot \pi (1)^2 = \frac{1}{6}\pi = \frac{\pi}{6}\text{ cm}^2.$

Since the triangle $OCD$ must have half this area:

$\text{Area}(\triangle OCD) = \frac12 \cdot \frac{\pi}{6} = \frac{\pi}{12}\text{ cm}^2.$

Step 2 — Express the area of $\triangle OCD$ in terms of $OC$.

Let $OC = OD = x$ (we are told the two are equal). The angle between these two sides is the sector's central angle, $\angle COD = 60^\circ$.

Using the formula for the area of a triangle given two sides and the included angle, $\text{Area} = \tfrac12 ab\sin C$:

$\text{Area}(\triangle OCD) = \frac12 \cdot x \cdot x \cdot \sin 60^\circ = \frac12 x^2 \cdot \frac{\sqrt3}{2} = \frac{\sqrt3}{4}x^2.$

💡 Teacher tip: Whenever a triangle sits inside a sector with its vertex at the centre, the two radii become two sides of the triangle and the sector angle becomes the included angle — so $\tfrac12 ab\sin C$ is usually the fastest route.

Step 3 — Equate and solve for $x$.

$\frac{\sqrt3}{4}x^2 = \frac{\pi}{12}.$

Multiply both sides by $4$:

$\sqrt3\, x^2 = \frac{\pi}{3}.$

Now divide by $\sqrt3$:

$x^2 = \frac{\pi}{3\sqrt3}.$

Taking the positive square root (since $x$ is a length):

$x = \sqrt{\frac{\pi}{3\sqrt3}}.$

Step 4 — Match with the options.

This is exactly option C.

Answer: C — $\boxed{\sqrt{\dfrac{\pi}{3\sqrt3}}}$

💡 Why this works out neatly: the sector area brings in a factor of $\pi$ (from the curved arc), while the triangle area brings in $\sqrt3$ (from $\sin 60^\circ$). Equating them forces $\pi$ and $\sqrt3$ into the same expression, which is why the answer mixes both.

Think of this as: we know the mixture's total cost price and its per-litre cost. Paint A is Rs. 8/L dearer than paint B, and B can be at most half the mixture. We want to push paint B's price as high as possible — so we need to find the extreme blend that still satisfies every condition.

Step 1 — Find the cost price of the mixture.

The trader sells 10 litres for Rs. 264 at a 10% profit. So the selling price is 110% of the cost price:

$\text{Cost price of 10 L} = \frac{264}{1.10} = 240 \text{ Rs.}$

$\text{Cost price per litre} = \frac{240}{10} = 24 \text{ Rs/L}$

This Rs. 24/L is the weighted average of the costs of paint A and paint B.

Step 2 — Set up variables.

Let the cost of paint B be $b$ Rs/L. Then the cost of paint A is $b + 8$ Rs/L.

Let the amount of paint A be $a$ litres, so paint B is $10 - a$ litres.

The condition "amount of B does not exceed amount of A" gives:

$10 - a \le a \implies a \ge 5$

So paint A is at least 5 litres (i.e., A is at least half the mixture).

Step 3 — Write the cost equation.

The total cost of the mixture equals the cost of A plus the cost of B:

$a(b+8) + (10-a)b = 240$

Expand and simplify:

$ab + 8a + 10b - ab = 240$

$8a + 10b = 240$

Divide by 2:

$4a + 5b = 120$

Solve for $b$:

$b = \frac{120 - 4a}{5} = 24 - \frac{4a}{5}$

Step 4 — Maximise $b$ by minimising $a$.

Since $b = 24 - \frac{4a}{5}$, the cost of B decreases as $a$ increases. To make $b$ as large as possible, we need $a$ as small as possible.

The smallest allowed value is $a = 5$ (from Step 2):

$b = 24 - \frac{4 \times 5}{5} = 24 - 4 = 20$

💡 Teacher tip: "Highest possible cost of B" combined with "B ≤ A" pushes the mixture to its most B-heavy legal blend — exactly half A and half B, i.e., 5 litres each. Plug that extreme in directly and you get the answer in one line.

Verification check:

As $a$ rises from 5 to 10, $b$ falls from 20 to 16 — confirming the highest possible cost of B is Rs. 20/L.

Answer: C — $\boxed{20}$

💡 Why this works: The per-litre cost of the mixture (Rs. 24) is fixed. Since paint A is costlier, using more A would drag the average up — meaning B's price would have to drop to compensate. So the most B-friendly (and thus highest-$b$) scenario is the one with the least A allowed, which is $a = 5$.

Think of this as: two parallel chords sit on the same side of the centre, 1 cm apart. The chord–centre-distance formula links each chord's length to the radius — and since both chords share the same radius, we get two equations we can solve together.

Key property (chord–centre-distance relation). A chord of length $\ell$ whose perpendicular distance from the centre is $d$ satisfies $r^2 = d^2 + \left(\frac{\ell}{2}\right)^2.$ The half-chord, the perpendicular distance to the centre, and the radius form a right triangle.

Step 1 — Set up the two chords.

Both chords live in the same circle, so each gives an equation for the same $r^2$: $r^2 = d_1^2 + 3^2 = d_1^2 + 9, \qquad r^2 = d_2^2 + 2^2 = d_2^2 + 4.$

Step 2 — Decide which chord is closer to the centre. A longer chord is always closer to the centre (think of the diameter, the longest chord, passing through the centre). So the 6-cm chord is nearer the centre: $d_1 < d_2$.

Since the chords are on the same side of the diameter and 1 cm apart, $d_2 - d_1 = 1.$

💡 Teacher tip: "Same side" means the distances subtract; "opposite sides" would mean they add. Getting this sign right is the make-or-break step in chord-gap problems.

Step 3 — Subtract the two radius equations. From $d_1^2 + 9 = d_2^2 + 4$: $d_2^2 - d_1^2 = 5.$ Factor the left side as a difference of squares: $(d_2 - d_1)(d_2 + d_1) = 5.$ Using $d_2 - d_1 = 1$: $1 \cdot (d_2 + d_1) = 5 \;\Rightarrow\; d_2 + d_1 = 5.$

Step 4 — Solve for $d_1$ and $d_2$. $d_2 - d_1 = 1, \qquad d_2 + d_1 = 5.$ Adding: $2d_2 = 6 \Rightarrow d_2 = 3$. Subtracting: $2d_1 = 4 \Rightarrow d_1 = 2$.

Step 5 — Compute the radius. $r^2 = d_1^2 + 3^2 = 2^2 + 9 = 4 + 9 = 13 \;\Rightarrow\; r = \sqrt{13}.$

Quick cross-check with the other chord: $r^2 = d_2^2 + 2^2 = 9 + 4 = 13$. ✓

Answer: A — $\boxed{\sqrt{13}\text{ cm}}$

💡 Why this works so cleanly: subtracting the two $r^2$-equations turns the problem into a difference of squares, $(d_2-d_1)(d_2+d_1)$, which immediately yields the sum of the distances from the known gap. That single identity is the engine of every "two parallel chords with a given separation" problem.

Think of this as: a small cone is sliced off the top of a bigger cone; the leftover solid (a frustum) has volume = big cone volume − small cone volume.

Step 1 — Understand the geometry.

The full cone has height $12$ ft and base diameter $8$ ft, so its base radius is $R = 4$ ft.

The cutting plane is parallel to the base and $9$ ft above the base. Since the total height is $12$ ft, the plane sits $12 - 9 = 3$ ft below the apex. The piece removed — the tip — is itself a small cone of height $h = 3$ ft, and it is similar to the original cone.

Step 2 — Find the radius of the small cone.

When a cone is cut by a plane parallel to its base, the removed tip is similar to the original cone. The linear scale factor is the ratio of their heights:

$\text{scale factor} = \frac{3}{12} = \frac{1}{4}.$

So the small cone's base radius is

$r = 4 \times \frac{1}{4} = 1 \text{ ft}.$

💡 Teacher tip: "Slice a cone parallel to its base" always produces a similar small cone — find its scale factor from the heights, and every linear dimension (radius, slant height, etc.) scales by the same factor.

Step 3 — Volume of the full cone.

$V_{\text{big}} = \frac{1}{3}\pi R^{2}H = \frac{1}{3}\pi (4)^{2}(12) = \frac{1}{3}\pi \cdot 16 \cdot 12 = 64\pi.$

Step 4 — Volume of the small cone (the tip).

$V_{\text{small}} = \frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi (1)^{2}(3) = \frac{1}{3}\pi \cdot 1 \cdot 3 = \pi.$

Step 5 — Subtract to get the remaining frustum.

$V = V_{\text{big}} - V_{\text{small}} = 64\pi - \pi = 63\pi.$

Now substitute $\pi = \frac{22}{7}$:

$V = 63 \times \frac{22}{7} = 9 \times 22 = 198.$

Answer: $\boxed{198 \text{ cubic ft}}$

Alternative approach — Frustum formula.

The remaining part is a frustum of height $H_f = 9$ ft, lower radius $R = 4$ ft, and upper radius $r = 1$ ft. Using the frustum volume formula:

$V = \frac{1}{3}\pi H_f\left(R^{2} + Rr + r^{2}\right) = \frac{1}{3}\pi (9)\left(16 + 4 + 1\right) = \frac{1}{3}\pi \cdot 9 \cdot 21 = 3\pi \cdot 21 = 63\pi.$

With $\pi = \frac{22}{7}$, this again gives $63 \times \frac{22}{7} = 198$. The subtraction method is faster here because the small cone's volume works out to a clean $\pi$.

💡 Why this is elegant: the small cone is a $\frac{1}{4}$-scale copy of the big one, so its volume is $\left(\frac{1}{4}\right)^{3} = \frac{1}{64}$ of the big cone's volume — which is exactly why $V_{\text{small}} = \pi$ when $V_{\text{big}} = 64\pi$. Volume scales with the cube of the linear scale factor.

Think of this as: two trains head toward each other from opposite ends, one starts an hour earlier and is faster. They meet at a point that splits the total distance in a known ratio. We need the faster train's total journey time.

Step 1 — Set up distances and speeds.

Let the total distance $XY = D$.

Since $Z$ is the meeting point and $XZ = \tfrac{3}{5}$ of $XY$: $XZ = \frac{3D}{5}, \qquad ZY = D - \frac{3D}{5} = \frac{2D}{5}.$

Let train $T$'s speed be $v$. Then train $S$'s speed is $\tfrac{3}{4}v$.

Step 2 — Write each train's travel time to reach $Z$.

Train $T$ travels from $X$ to $Z$, a distance of $\tfrac{3D}{5}$, at speed $v$: $\text{Time of } T = \frac{3D/5}{v} = \frac{3D}{5v}.$ This time is measured from 3 pm, when $T$ started.

Train $S$ travels from $Y$ to $Z$, a distance of $\tfrac{2D}{5}$, at speed $\tfrac{3v}{4}$: $\text{Time of } S = \frac{2D/5}{3v/4} = \frac{2D}{5} \times \frac{4}{3v} = \frac{8D}{15v}.$ This time is measured from 4 pm, when $S$ started.

Step 3 — Use the fact that they reach $Z$ at the same instant.

💡 Teacher tip: "They pass at the same place at the same time" is your single equation. Write both travel times to that place and equate them — but keep careful track of who started when, because $T$ had a one-hour head start.

Since $T$ left one hour earlier than $S$, $T$'s travel time is one hour more than $S$'s: $\frac{3D}{5v} = \frac{8D}{15v} + 1.$

Multiply through by $15v$ to clear denominators: $9D = 8D + 15v.$

$\boxed{D = 15v}.$

Step 4 — Find $T$'s total journey time from $X$ to $Y$.

Train $T$ must cover the full distance $D$ at speed $v$: $\text{Total time} = \frac{D}{v} = \frac{15v}{v} = 15 \text{ hours}.$

Quick sanity check. If $T$ takes 15 hours total, its speed is $v = \tfrac{D}{15}$. Then $S$'s speed is $\tfrac{3v}{4} = \tfrac{D}{20}$. $T$ reaches $Z$ in $\tfrac{3D/5}{D/15} = 9$ hours (from 3 pm → midnight). $S$ reaches $Z$ in $\tfrac{2D/5}{D/20} = 8$ hours (from 4 pm → midnight). Same instant. ✓

Answer: $\boxed{15 \text{ hours}}$

💡 Why this works cleanly: the distance $D$ and speed $v$ always appear together as $\tfrac{D}{v}$, which is exactly the total time we want — so the equation solves itself without needing numerical values for either.

Think of this as: we have a GP whose three terms we can write using a common ratio $r$, and an AP condition that will hand us a quadratic equation in $r$.

Step 1 — Parametrise the geometric progression.

Since $x, y, z$ are in GP with $x < y < z$, let the common ratio be $r > 1$ and write

$x = a,\quad y = ar,\quad z = ar^{2}$

for some positive $a$. The condition $x < y < z$ is exactly what forces $r > 1$.

Step 2 — Translate the AP condition.

We are told $5x,\ 16y,\ 12z$ are in arithmetic progression. Three numbers are in AP when the middle one equals the average of the other two, i.e.

$2 \times (\text{middle}) = (\text{first}) + (\text{last}).$

So

$2(16y) = 5x + 12z.$

Substituting the GP parametrisation:

$32(ar) = 5a + 12(ar^{2}).$

Step 3 — Simplify to a quadratic in $r$.

Divide both sides by $a$ (which is positive, so safe):

$32r = 5 + 12r^{2}$

$\Rightarrow 12r^{2} - 32r + 5 = 0.$

Step 4 — Solve the quadratic.

Using the quadratic formula $r = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ with $a=12,\ b=-32,\ c=5$:

$r = \frac{32 \pm \sqrt{(-32)^{2} - 4(12)(5)}}{2(12)} = \frac{32 \pm \sqrt{1024 - 240}}{24} = \frac{32 \pm \sqrt{784}}{24} = \frac{32 \pm 28}{24}.$

This gives two candidates:

Step 5 — Pick the valid root.

Because $x < y < z$, the common ratio must satisfy $r > 1$. Only $r = \dfrac{5}{2}$ qualifies.

💡 Teacher tip: "Three terms in AP" $\Rightarrow$ "twice the middle $=$ sum of the ends." Write that single equation, substitute the GP parametrisation, and you almost always land on a clean quadratic in $r$. Then use the ordering ($r>1$ or $0<r<1$) to pick the right root.

Answer: C — $\boxed{\dfrac{5}{2}}$

The second root $\dfrac{1}{6}$ is not a distractor to discard blindly — it is a genuine solution of the AP equation, but it corresponds to a decreasing GP ($r<1$), which contradicts $x < y < z$. The ordering condition is what selects the answer.

Think of this as: we have one equation in two unknowns ($U$ and $V$). Normally that's impossible to solve uniquely — unless the equation is secretly a sum of squares equal to zero. Let's find out if it is.

Step 1 — Expand everything and bring all terms to one side.

Start with the given equation: $U^2 + (U - 2V - 1)^2 = -4V(U + V)$

First, expand $(U - 2V - 1)^2$ using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$: $(U - 2V - 1)^2 = U^2 + 4V^2 + 1 - 4UV - 2U + 4V$

Also expand the right side: $-4V(U + V) = -4UV - 4V^2$

Now substitute both back into the equation: $U^2 + U^2 + 4V^2 + 1 - 4UV - 2U + 4V = -4UV - 4V^2$

Bring every term to the left side: $U^2 + U^2 + 4V^2 + 1 - 4UV - 2U + 4V + 4UV + 4V^2 = 0$

The $-4UV$ and $+4UV$ cancel each other out, and we combine like terms: $2U^2 + 8V^2 - 2U + 4V + 1 = 0$

Step 2 — Complete the square for both $U$ and $V$.

Factor out the leading coefficients from each variable's terms: $2(U^2 - U) + 8\!\left(V^2 + \tfrac{V}{2}\right) + 1 = 0$

Now complete the square inside each parenthesis:

• For $U$: $U^2 - U = \left(U - \tfrac{1}{2}\right)^2 - \tfrac{1}{4}$
• For $V$: $V^2 + \tfrac{V}{2} = \left(V + \tfrac{1}{4}\right)^2 - \tfrac{1}{16}$

Substitute these in: $2\!\left[\left(U - \tfrac{1}{2}\right)^2 - \tfrac{1}{4}\right] + 8\!\left[\left(V + \tfrac{1}{4}\right)^2 - \tfrac{1}{16}\right] + 1 = 0$

Distribute the coefficients: $2\!\left(U - \tfrac{1}{2}\right)^2 - \tfrac{1}{2} + 8\!\left(V + \tfrac{1}{4}\right)^2 - \tfrac{1}{2} + 1 = 0$

The constants combine neatly: $-\tfrac{1}{2} - \tfrac{1}{2} + 1 = 0$, leaving: $2\!\left(U - \tfrac{1}{2}\right)^2 + 8\!\left(V + \tfrac{1}{4}\right)^2 = 0$

💡 Teacher tip: A sum of non-negative terms (squares) can equal zero only if each term is individually zero. This is the key principle that lets us solve for two unknowns from one equation.

Step 3 — Read off the values.

Since both squared terms must be zero: $U - \tfrac{1}{2} = 0 \implies U = \tfrac{1}{2}$ $V + \tfrac{1}{4} = 0 \implies V = -\tfrac{1}{4}$

Step 4 — Compute $U + 3V$. $U + 3V = \tfrac{1}{2} + 3\!\left(-\tfrac{1}{4}\right) = \tfrac{1}{2} - \tfrac{3}{4} = \tfrac{2}{4} - \tfrac{3}{4} = -\tfrac{1}{4}$

Answer: C — $\boxed{-\tfrac{1}{4}}$

💡 Why this works: One equation in two unknowns can pin down both values only when it collapses into a sum of squares $= 0$. The telltale sign is that all the constant terms cancel perfectly after completing the square — that's not a coincidence, it's the structure of the problem.

Think of this as: one equilateral triangle spawns a smaller one inside it, that one spawns an even smaller one, and so on forever — what is the total area of all of them put together?

Step 1 — Area of the first triangle $T_1$.

$T_1$ is equilateral with side $24$ cm. Using the standard area formula for an equilateral triangle,

$\text{Area}(T_1) = \frac{\sqrt{3}}{4}\times(\text{side})^2 = \frac{\sqrt{3}}{4}\times 24^2 = \frac{\sqrt{3}}{4}\times 576 = 144\sqrt{3}\text{ sq cm}.$

Step 2 — What happens when we join the midpoints?

When you join the midpoints of the sides of any triangle, the small triangle formed in the middle is similar to the original, and each of its sides is exactly half the corresponding side of the original.

Since area scales as the square of the side length, halving the side multiplies the area by $\left(\tfrac12\right)^2 = \tfrac14$.

So each new triangle has one-fourth the area of the one before it:

$\text{Area}(T_{k+1}) = \tfrac14\,\text{Area}(T_k).$

💡 Teacher tip: "Join the midpoints" is a fixed signal in geometry problems — the inner triangle is always a $\tfrac14$-area copy of the outer one. Spotting this instantly turns the problem into a geometric series.

Step 3 — Write the areas as a geometric series.

The areas form the sequence

$144\sqrt{3},\;\; 36\sqrt{3},\;\; 9\sqrt{3},\;\; \tfrac{9}{4}\sqrt{3},\;\; \dots$

This is a geometric progression with

• first term $a = 144\sqrt{3}$,
• common ratio $r = \tfrac14$.

Step 4 — Sum the infinite series.

For an infinite geometric series with $|r| < 1$, the sum is $\dfrac{a}{1-r}$. Here $r = \tfrac14 < 1$, so

$S = \frac{144\sqrt{3}}{1-\tfrac14} = \frac{144\sqrt{3}}{\tfrac34} = 144\sqrt{3}\times\frac{4}{3} = 192\sqrt{3}.$

Quick check — partial sums should climb toward $192\sqrt{3}$:

The partial sums approach $192\sqrt{3}$, confirming the infinite-sum result.

Answer: D — $\boxed{192\sqrt{3}\text{ sq cm}}$

💡 Why this is elegant: the side lengths halve each time ($24, 12, 6, \dots$), so the areas quarter each time. The total area is just $\tfrac{4}{3}$ of the very first triangle — a neat reminder that an infinite process can still land on a clean, finite number.

Think of this as a chain of transactions: wholesaler → shopkeeper → final customer. The shopkeeper's cost is what he paid the wholesaler, and his revenue is what he earned after the transit loss. The 25% profit links those two — and every rupee in the problem traces back to one variable: the peanut's cost price.

Step 1 — Set up the cost prices.

Let the wholesaler's cost price of peanuts be $p$ Rs/kg.

Since walnuts cost thrice as much per kg: $\text{Walnut cost} = 3p \text{ Rs/kg}.$

💡 Teacher tip: In a multi-stage problem, anchor on the cheapest item's price and express everything else as a multiple of it. Here, both nut prices and both selling prices will be multiples of $p$ — so one equation will close the whole problem.

Step 2 — Find the shopkeeper's cost (what he paid the wholesaler).

The wholesaler sells to the shopkeeper at a profit:

• Peanuts: $10\%$ profit → selling price $= 1.1p$ Rs/kg.
• Walnuts: $20\%$ profit → selling price $= 1.2 \times 3p = 3.6p$ Rs/kg.

The shopkeeper buys $8$ kg of peanuts and $16$ kg of walnuts, so his total cost is: $\text{Shopkeeper's cost} = 8(1.1p) + 16(3.6p) = 8.8p + 57.6p = 66.4p.$

Step 3 — Find the shopkeeper's revenue.

In transit, the shopkeeper loses $5$ kg of walnuts and $3$ kg of peanuts. What remains:

• Peanuts: $8 - 3 = 5$ kg
• Walnuts: $16 - 5 = 11$ kg
• Total mixture sold $= 5 + 11 = 16$ kg

He sells this entire mixture at Rs. $166$/kg: $\text{Revenue} = 16 \times 166 = 2656 \text{ Rs}.$

Step 4 — Use the 25% overall profit to solve for $p$.

A $25\%$ profit means revenue is $1.25$ times the cost: $2656 = 1.25 \times 66.4p.$

Divide both sides by $1.25$: $66.4p = \frac{2656}{1.25} = 2124.8.$

Now solve for $p$: $p = \frac{2124.8}{66.4} = 32.$

Step 5 — Find the walnut cost price.

$\text{Walnut cost} = 3p = 3 \times 32 = 96 \text{ Rs/kg}.$

Quick verification:

Answer: A — $\boxed{\text{Rs. } 96 \text{ per kg}}$

💡 Why this works cleanly: The transit loss doesn't change the shopkeeper's cost — it only reduces how much he can sell. That's why his cost ($66.4p$) uses the full $8+16$ kg he bought, while his revenue uses only the $16$ kg that survived transit. Keeping cost and revenue on their correct sides of the equation is the key to not slipping here.

We need to count how many integers $x$ satisfy both conditions: the size bound $0.25 \le 2^x \le 200$, and the divisibility condition that $2^x + 2$ is divisible by $3$ or $4$.

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Step 1 — Translate the size bound into a range for $x$.

Write $0.25$ as a power of $2$: $0.25 = \frac{1}{4} = 2^{-2}$.

For the upper end, compute consecutive powers of $2$: $2^7 = 128 \le 200 < 256 = 2^8.$

So the inequality $2^{-2} \le 2^x \le 2^7$ gives: $-2 \le x \le 7.$

That's the candidate set $x \in \{-2, -1, 0, 1, 2, 3, 4, 5, 6, 7\}$.

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Step 2 — Recognise the hidden integer requirement.

The phrase "perfectly divisible by $3$ or $4$" only makes sense when $2^x + 2$ is an integer. Since $2$ is already an integer, we need $2^x$ to be an integer too — which forces $x \ge 0$.

💡 Teacher tip: "Perfectly divisible" silently demands an integer dividend. That knocks out $x = -2$ and $x = -1$ before you even start testing divisibility.

So the effective candidate set is: $x \in \{0, 1, 2, 3, 4, 5, 6, 7\}.$

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Step 3 — Test each candidate.

We compute $2^x + 2$ for each value and check divisibility by $3$ and by $4$:

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Step 4 — Count the valid values.

The valid $x$-values are $\{0, 1, 2, 4, 6\}$ — that's $5$ integers.

Answer: $\boxed{5}$

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Alternative approach — use modular arithmetic to avoid the table.

Instead of testing all eight values, we can find the pattern directly.

Divisibility by $3$: We need $2^x + 2 \equiv 0 \pmod{3}$, i.e. $2^x \equiv 1 \pmod{3}$. Since $2 \equiv -1 \pmod{3}$, this becomes $(-1)^x \equiv 1 \pmod{3}$, which holds when $x$ is even.

Divisibility by $4$: We need $2^x + 2 \equiv 0 \pmod{4}$, i.e. $2^x \equiv 2 \pmod{4}$. This holds only for $x = 1$ (since $2^1 = 2$; for $x \ge 2$, $2^x$ is a multiple of $4$).

So the condition is satisfied when $x$ is even or $x = 1$.

From $\{0,1,2,3,4,5,6,7\}$:

• Even values: $\{0, 2, 4, 6\}$ — four values.
• Plus $x = 1$ — one more value.
• Total: $4 + 1 = 5$.

This confirms the answer without a full table. The modular approach is faster under exam conditions once you spot the $(-1)^x$ pattern for divisibility by $3$.

💡 Closing insight: The divisibility-by-$3$ condition reduces to "$x$ is even" because $2$ and $3$ are coprime — so powers of $2$ cycle through all non-zero residues mod $3$ with period $2$. This kind of cycling is the backbone of most modular-arithmetic shortcuts.