CAT 2020 Quant Questions — Slot 1

Think of this as: we have a horizontal strip $0 \le y \le 1$, and the inequality $|x| - y \le 1$ cuts away part of that strip. We need the area that survives all three conditions at once.

Step 1 — Rearrange the first inequality into a lower bound on $y$.

Starting from $|x| - y \le 1$, move $y$ to the right and $1$ to the left: $|x| - 1 \le y \quad\Longleftrightarrow\quad y \ge |x| - 1.$

So $y$ must be at least $|x| - 1$, and also inside the strip $0 \le y \le 1$. Putting it together: $\max(0,\; |x| - 1) \le y \le 1.$

💡 Teacher tip: Whenever a problem mixes an absolute-value inequality with simple bounds, rewrite the absolute-value condition as an explicit bound on one variable. Here $y$ is trapped between a V-shaped floor and a flat ceiling — that's the picture to hold in your head.

Step 2 — Understand the lower boundary $y = |x| - 1$.

This is a V-shape with vertex at $(0, -1)$, rising through $(\pm 1, 0)$ and $(\pm 2, 1)$. But $y$ is also required to stay in $[0, 1]$, so the effective lower boundary is whichever is higher: the V or the floor $y = 0$.

This splits the $x$-axis into three zones:

Step 3 — Compute the area in each zone.

Zone A ($|x| \le 1$): The region is a full strip from $y = 0$ to $y = 1$, running from $x = -1$ to $x = 1$. $\text{Area}_A = \text{width} \times \text{height} = 2 \times 1 = 2.$

Zone B ($1 \le |x| \le 2$): The height shrinks linearly from $1$ (at $|x| = 1$) to $0$ (at $|x| = 2$). On each side this is a triangle of base $1$ and height $1$: $\text{Area of one triangle} = \tfrac12 \times 1 \times 1 = \tfrac12.$ There are two identical triangles (one on the left, one on the right): $\text{Area}_B = 2 \times \tfrac12 = 1.$

Zone C ($|x| > 2$): No region — the lower bound $|x| - 1$ has risen above $y = 1$, so the strip is empty here.

Step 4 — Add the pieces. $\text{Total area} = \text{Area}_A + \text{Area}_B = 2 + 1 = 3.$

Answer: $\boxed{3}$

💡 Why the shape is what it is: The strip $0 \le y \le 1$ acts like a window. The V-shaped curve $y = |x| - 1$ only enters that window between $|x| = 1$ and $|x| = 2$, carving out two identical triangles from the otherwise rectangular strip of area $2 \times 1 = 2$. Those two triangles contribute exactly $1$ more, giving $3$.

Alternative approach — Integration. By symmetry about the $y$-axis, the area is $2\int_0^2 \bigl[1 - \max(0,\, x - 1)\bigr]\,dx = 2\left[\int_0^1 1\,dx + \int_1^2 (2 - x)\,dx\right].$ The first integral is $1$. The second is $\left[2x - \frac{x^2}{2}\right]_1^2 = (4 - 2) - (2 - \tfrac12) = 2 - \tfrac32 = \tfrac12$. So the total is $2\left(1 + \tfrac12\right) = 3$. Same answer — the geometry method is faster under exam time, but integration is a reliable cross-check.

We need the smallest value of $a+b+c$ where $a,b,c$ are positive integers satisfying $ab=432$, $bc=96$, and $c<9$.

Step 1 — Link $a$ and $c$ directly.

Divide the two product equations: $\frac{ab}{bc}=\frac{432}{96}\;\Rightarrow\;\frac{a}{c}=\frac{9}{2}\;\Rightarrow\;a=\frac{9c}{2}.$

Since $a$ must be an integer, $\frac{9c}{2}$ must be an integer, which means $c$ must be even.

Step 2 — List the allowed values of $c$.

We need $c$ to be a positive even integer with $c<9$: $c\in\{2,4,6,8\}.$

That's only four candidates — small enough to check each one.

Step 3 — Compute $a$, $b$, and the sum for each candidate.

For each $c$:

• $a=\dfrac{9c}{2}$
• $b=\dfrac{96}{c}$ (from $bc=96$)

💡 Teacher tip: Notice that $ab=432$ is automatically satisfied in every row — that's not a coincidence. Once we enforce $a/c=9/2$ (from dividing the two equations) and $bc=96$, the value $ab=\frac{9c}{2}\times\frac{96}{c}=\frac{9\times96}{2}=432$ holds for any valid $c$. So we only need to check divisibility, not re-verify $ab$ each time.

Step 4 — Pick the minimum.

The smallest sum is $\mathbf{46}$, occurring at $c=4$, $a=18$, $b=24$.

Answer: $\boxed{46}$ (Choice D)

💡 Why a table beats algebra here: with only four candidates satisfying the divisibility and bound constraints, direct enumeration is faster and less error-prone than trying to optimise $a+b+c$ symbolically. When the candidate set is tiny, just check them all.

We need to find the negative value of $y$ satisfying $2^{y^2 \log_3 5} = 5^{\log_2 3}$.

The key idea: both sides are exponentials with different bases, so taking a logarithm on both sides will bring the exponents down where we can work with them algebraically.

Step 1 — Take $\log_2$ of both sides.

Since both sides are positive, taking $\log_2$ is safe: $\log_2\!\left(2^{y^2 \log_3 5}\right) = \log_2\!\left(5^{\log_2 3}\right)$

Using $\log_b(b^x) = x$ on the left and $\log_b(a^x) = x\log_b a$ on the right: $y^2 \log_3 5 = (\log_2 3)(\log_2 5)$

Step 2 — Change of base on $\log_3 5$.

The change-of-base formula $\log_a b = \dfrac{\log_c b}{\log_c a}$ lets us rewrite $\log_3 5$ using base 2: $\log_3 5 = \frac{\log_2 5}{\log_2 3}$

Substituting: $y^2 \cdot \frac{\log_2 5}{\log_2 3} = (\log_2 3)(\log_2 5)$

💡 Teacher tip: Notice both sides now contain the factor $\log_2 5$. That's not a coincidence — taking $\log_2$ of the right side produced it, and the change of base produced it on the left. This shared factor is what will cancel cleanly.

Step 3 — Solve for $y^2$.

Divide both sides by $\log_2 5$ (which is positive, so non-zero): $y^2 \cdot \frac{1}{\log_2 3} = \log_2 3$

Multiply both sides by $\log_2 3$: $y^2 = (\log_2 3)^2$

Taking square roots: $y = \pm \log_2 3$

Step 4 — Apply the sign constraint.

The problem states $y$ is negative. Since $\log_2 3 > 0$ (because $3 > 2^1$, so $\log_2 3 > 1 > 0$), we choose the negative root: $y = -\log_2 3$

Step 5 — Match the options.

Using the power rule of logarithms in reverse, $-\log_2 3 = \log_2 3^{-1} = \log_2 \dfrac{1}{3}$.

Answer: $\log_2 \dfrac{1}{3}$ (Choice A)

💡 Why it works: The $\log_2 5$ factor appears on both sides after taking logs and changing base — its clean cancellation is the heart of the problem. The sign constraint then selects the negative root, which converts neatly to $\log_2 \frac{1}{3}$ via the power rule.

Think of this as a classic overlapping-groups problem: every person is either young or old, and either literate or illiterate. A $2 \times 2$ table is the cleanest way to keep track.

💡 Teacher tip: When a question mixes two classifications (age × literacy) and gives percentages of different bases, assume 100 people and build a $2 \times 2$ table. Fill what's given, then complete the rest by row and column sums.

Step 1 — Set up the base and the two classifications.

Assume the group has 100 members.

• Young = $28\%$ of $100 = 28$, so Old = $100 - 28 = 72$.
• Literate = $65\%$ of $100 = 65$, so Illiterate = $100 - 65 = 35$.

Step 2 — Decode the trickiest statement.

"25% of the literates are young" means the base is literates, not the whole group.

$\text{Young literates} = 25\% \times 65 = 0.25 \times 65 = 16.25.$

Step 3 — Fill the $2 \times 2$ table by subtraction.

So the number of old illiterates is $23.25$.

💡 Teacher tip: You can cross-check the $23.25$ two ways: down the Old row ($72 - 48.75$) or across the Illiterate column ($35 - 11.75$). Both must agree — if they don't, a cell was filled wrong.

Step 4 — Answer the exact question asked.

We need the percentage of old people among the illiterates, i.e., old illiterates as a fraction of all illiterates:

$\frac{\text{Old illiterate}}{\text{Illiterate}} \times 100 = \frac{23.25}{35} \times 100 \approx 66.4\%.$

The nearest option is 66.

Answer: $\boxed{66\%}$ (Choice D)

💡 Why this works cleanly: anchoring on 100 people lets every percentage become a headcount instantly, and the $2 \times 2$ table turns the overlapping condition into simple row/column subtraction — no simultaneous equations needed.

We're given $x = 4096^{(7+4\sqrt{3})}$ and need to find which expression in $x$ equals $64$. The strategy: express $x$ as a power of $64$, then invert that relationship to write $64$ as a power of $x$.

Step 1 — Rewrite 4096 in terms of 64.

Since $64 = 2^6$ and $4096 = 2^{12}$, we have: $4096 = 2^{12} = (2^6)^2 = 64^2.$

Now substitute into the given expression: $x = (64^2)^{7+4\sqrt{3}} = 64^{\,2(7+4\sqrt{3})} = 64^{\,14+8\sqrt{3}}.$

Step 2 — Invert to express 64 in terms of $x$.

If $x = 64^{\,14+8\sqrt{3}}$, then raising both sides to the power $\frac{1}{14+8\sqrt{3}}$ gives: $$64 = x^{,1/(14+8\sqrt{3})}.$

Step 3 — Rationalise the exponent.

The exponent $\frac{1}{14+8\sqrt{3}}$ has a surd in the denominator. Multiply numerator and denominator by the conjugate $14-8\sqrt{3}$: $\frac{1}{14+8\sqrt{3}} = \frac{14-8\sqrt{3}}{(14)^2 - (8\sqrt{3})^2} = \frac{14-8\sqrt{3}}{196 - 192} = \frac{14-8\sqrt{3}}{4} = \frac{7}{2} - 2\sqrt{3}.$

💡 Teacher tip: The denominator simplifies beautifully here — $196 - 192 = 4$ — because $14^2 = 196$ and $(8\sqrt{3})^2 = 192$ are deliberately close. This is the signal that the question was designed for rationalisation.

Step 4 — Write 64 as a single power of $x$, then match the options.

$64 = x^{\,\frac{7}{2} - 2\sqrt{3}}.$

Using the law of exponents $x^{a-b} = \frac{x^a}{x^b}$: $64 = \frac{x^{7/2}}{x^{2\sqrt{3}}}.$

Step 5 — Verify against each option.

Only Option C produces the correct exponent.

Answer: C. $\dfrac{x^{7/2}}{x^{2\sqrt{3}}}$

💡 Closing insight: The answer options are all written with separated rational and surd exponents (a fraction minus a surd). Rationalising $\frac{1}{14+8\sqrt{3}}$ is what produces exactly this separated form — that's the bridge between the compact exponent $14+8\sqrt{3}$ and the option format.

Think of this as: two cars start from opposite ends, meet somewhere in the middle, and then each finishes the remaining part of the road. We're told how long each takes after meeting, and one speed — find the other.

What we know:

• Car 1 (A→B) and Car 2 (B→A) start at the same instant.
• After they meet, Car 1 takes $t_1 = 45$ min to reach B; Car 2 takes $t_2 = 20$ min to reach A.
• Car 1's speed $v_1 = 60$ km/hr.
• Find Car 2's speed $v_2$.

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Step 1 — Label the meeting point and the two pieces of road.

Let the cars meet after time $T$ from the start.

• The distance from A to the meeting point is what Car 1 covered before meeting: $v_1 T$.
• The distance from B to the meeting point is what Car 2 covered before meeting: $v_2 T$.

After meeting, each car must cover the other car's pre-meeting distance:

• Car 1 still has to travel the B-side distance $v_2 T$ at speed $v_1$, taking time $t_1 = 45$ min.
• Car 2 still has to travel the A-side distance $v_1 T$ at speed $v_2$, taking time $t_2 = 20$ min.

Step 2 — Write the two distance equations.

$v_2 T = v_1 \cdot t_1 \quad \text{...(i)}$ $v_1 T = v_2 \cdot t_2 \quad \text{...(ii)}$

💡 Teacher tip: The key insight is that the distance Car 1 covers after meeting is exactly the distance Car 2 covered before meeting, and vice-versa. That swap is what links the two cars' speeds and post-meeting times.

Step 3 — Eliminate $T$ by dividing (i) by (ii).

$\frac{v_2 T}{v_1 T} = \frac{v_1 t_1}{v_2 t_2}$

$\frac{v_2}{v_1} = \frac{v_1 t_1}{v_2 t_2}$

$\left(\frac{v_1}{v_2}\right)^2 = \frac{t_2}{t_1}$

This is the classic meeting-point relation: the square of the speed ratio equals the inverse ratio of the post-meeting times.

Step 4 — Plug in the numbers.

$\left(\frac{60}{v_2}\right)^2 = \frac{20}{45} = \frac{4}{9}$

$\frac{60}{v_2} = \frac{2}{3}$

$v_2 = 60 \times \frac{3}{2} = 90 \text{ km/hr}$

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Sanity check — verify the meeting time is consistent.

With $v_1 = 60$ and $v_2 = 90$, suppose they meet after $T$ minutes.

• Car 1's post-meeting distance $= v_2 T = 90T$. Time for Car 1 to cover it at 60 km/hr $= \frac{90T}{60} = 1.5T$. This equals 45 min, so $T = 30$ min. ✓
• Car 2's post-meeting distance $= v_1 T = 60T$. Time for Car 2 to cover it at 90 km/hr $= \frac{60T}{90} = \frac{2T}{3}$. With $T = 30$: $\frac{2 \times 30}{3} = 20$ min. ✓

Both post-meeting times check out.

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Answer: $\boxed{90 \text{ km/hr}}$ — Choice B

💡 Memorise the shortcut: For two bodies starting simultaneously from opposite ends and meeting en route, $\left(\dfrac{v_1}{v_2}\right)^2 = \dfrac{t_2}{t_1}$, where $t_i$ is the time taken after meeting. The faster car finishes sooner after the meeting, so the times are inversely related to the speeds — and the square appears because each car's post-meeting distance is proportional to the other car's speed.

Think of this as: two equations link three positive integers, and we want just $A+B$. The key is that both equations contain $C$, so we can eliminate $C$ first and then use the "positive integer" condition to pin down the values.

What we're given: $A + \frac{B+C}{2} = 5, \qquad B + \frac{A+C}{2} = 7.$

Step 1 — Clear the denominators. Multiply each equation by 2: $2A + B + C = 10 \quad \text{……(1)}, \qquad A + 2B + C = 14 \quad \text{……(2)}.$

Step 2 — Eliminate $C$ by subtracting. Both equations contain a lone $C$ term, so subtracting removes it cleanly. Subtract (1) from (2): $(A + 2B + C) - (2A + B + C) = 14 - 10$ $-A + B = 4 \;\Rightarrow\; B = A + 4.$

💡 Teacher tip: Whenever two equations share an identical term (here, the $C$), subtracting is usually the fastest move — it collapses three variables into two in one line.

Step 3 — Substitute back to find $A$ and $C$. Plug $B = A + 4$ into equation (1): $2A + (A+4) + C = 10 \;\Rightarrow\; 3A + C = 6.$

Now use the fact that $A$ and $C$ are positive integers (each $\ge 1$). Test the possible values of $A$:

So the only valid solution is $A = 1$, giving $C = 3$ and $B = A + 4 = 5$.

Step 4 — Compute $A + B$. $A + B = 1 + 5 = 6.$

Sanity check.

• $A + \dfrac{B+C}{2} = 1 + \dfrac{5+3}{2} = 1 + 4 = 5$ ✅
• $B + \dfrac{A+C}{2} = 5 + \dfrac{1+3}{2} = 5 + 2 = 7$ ✅

Answer: $\boxed{6}$ (Choice C)

💡 Closing insight: The "positive integer" constraint is what makes the answer unique — without it, $3A + C = 6$ would have infinitely many real solutions. In CAT-style problems, always check whether an integer/positivity condition narrows a family of solutions down to a single point.

Think of this as: we know the rectangle's area, the circle's size is locked by the rectangle's width, and a ratio ties painted to unpainted area. That's enough to solve for both sides and then the perimeter.

Step 1 — Set up the rectangle and the circle.

Let the rectangle have length $L$ and width $W$, so $L \cdot W = 135.$

The circle touches two opposite sides. The largest circle that can do this has its diameter equal to the shorter side $W$, so its radius is $r = \frac{W}{2}.$

💡 Teacher tip: "Touches two opposite sides" means the circle is snug between those sides — the gap between them (the width) equals the diameter. This is the geometric fact that links the circle's size to the rectangle's dimensions.

Step 2 — Write the painted and unpainted areas.

Using the area formula $\pi r^2$: $\text{Painted area} = \pi\left(\frac{W}{2}\right)^2 = \frac{\pi W^2}{4}.$

The unpainted area is whatever is left of the sheet: $\text{Unpainted area} = 135 - \frac{\pi W^2}{4}.$

Step 3 — Apply the ratio condition.

We're told unpainted $= \frac{2}{3} \times$ painted: $135 - \frac{\pi W^2}{4} = \frac{2}{3}\cdot\frac{\pi W^2}{4}.$

Move the $\pi$-term to the right side: $135 = \frac{\pi W^2}{4} + \frac{2}{3}\cdot\frac{\pi W^2}{4}.$

Combine the two terms on the right (they share the factor $\frac{\pi W^2}{4}$): $135 = \left(1 + \frac{2}{3}\right)\frac{\pi W^2}{4} = \frac{5}{3}\cdot\frac{\pi W^2}{4}.$

Now solve for $W$: $\frac{\pi W^2}{4} = 135 \cdot \frac{3}{5} = 81,$ $\pi W^2 = 324,$ $W = \frac{18}{\sqrt{\pi}}.$

Step 4 — Find the length $L$.

Since $L \cdot W = 135$: $L = \frac{135}{W} = \frac{135}{18/\sqrt{\pi}} = \frac{135\sqrt{\pi}}{18} = \frac{15\sqrt{\pi}}{2}.$

Step 5 — Compute the perimeter.

$P = 2(L + W) = 2\left(\frac{15\sqrt{\pi}}{2} + \frac{18}{\sqrt{\pi}}\right) = 15\sqrt{\pi} + \frac{36}{\sqrt{\pi}}.$

Now factor out $3\sqrt{\pi}$ so the expression matches the form of the answer choices:

• $15\sqrt{\pi} = 3\sqrt{\pi} \cdot 5$
• $\dfrac{36}{\sqrt{\pi}} = 3\sqrt{\pi} \cdot \dfrac{36}{3\pi} = 3\sqrt{\pi} \cdot \dfrac{12}{\pi}$

So: $P = 3\sqrt{\pi}\left(5 + \frac{12}{\pi}\right).$

Step 6 — Verify the ratio holds.

Answer: $\boxed{3\sqrt{\pi}\!\left(5 + \dfrac{12}{\pi}\right)}$ — Choice A

💡 Why factor at the end: all four options are written in the form $k\sqrt{\pi}(\cdots)$, so recognising that $15\sqrt{\pi} + \frac{36}{\sqrt{\pi}} = 3\sqrt{\pi}\!\left(5 + \frac{12}{\pi}\right)$ is the key last step. Always glance at the option format before simplifying — it tells you what shape your answer should take.

Think of this as: the train overtakes two walkers at different relative speeds, and we use the two crossing times to back out the train's true speed and length — then apply that to a stationary pole.

Step 1 — Understand what "crossing" means here.

When a train passes a moving person, the distance covered is the train's own length $L$, but the speed at which that distance is closed is the relative speed — the train's speed minus the person's speed (since everyone moves in the same direction).

Let the train's speed be $v$ km/hr and its length be $L$ metres.

💡 Teacher tip: Convert km/hr to m/s by multiplying by $\dfrac{5}{18}$. The relative speed is smaller for the faster walker ($v-4 < v-2$), which is exactly why the train takes longer (100 s vs 90 s) to pass them — a good sanity check on the setup.

Step 2 — Write the two length equations.

Distance = Relative speed × Time, and in both cases the distance is the train's length $L$:

$L = \frac{5}{18}(v-2) \times 90 = 25(v-2)$

$L = \frac{5}{18}(v-4) \times 100 = \frac{250}{9}(v-4)$

Step 3 — Equate and solve for the train's speed.

Since both expressions equal $L$:

$25(v-2) = \frac{250}{9}(v-4)$

Divide both sides by $25$:

$v - 2 = \frac{10}{9}(v-4)$

Multiply through by $9$:

$9v - 18 = 10v - 40$

$40 - 18 = 10v - 9v$

$v = 22 \text{ km/hr}$

Step 4 — Find the train's length.

Substitute back into either length equation:

$L = 25(22 - 2) = 25 \times 20 = 500 \text{ metres}$

Step 5 — Time to cross an electric pole.

A pole is effectively a point — it has zero length. So the train only needs to cover its own length, at its full speed (no relative-speed adjustment needed since the pole doesn't move).

Convert the train's speed to m/s:

$22 \text{ km/hr} = 22 \times \frac{5}{18} = \frac{110}{18} \text{ m/s}$

Now compute the time:

$\text{Time} = \frac{L}{\text{speed}} = \frac{500}{\frac{110}{18}} = \frac{500 \times 18}{110} = \frac{9000}{110} \approx 81.8 \text{ seconds}$

Step 6 — Pick the nearest option.

$81.8$ rounds to $82$.

Answer: $\boxed{82 \text{ seconds}}$ (Choice B)

💡 Why a pole is "zero-length": Crossing a pole means the train covers exactly its own length at full speed. Crossing a moving walker also covers the train's length, but the closing speed changes because the walker is moving — that's why the two walkers gave different times even though the train's length never changed.

We need to find how many real values of $x$ satisfy $2^x + 2^{-x} = 2 - (x-2)^2.$

Think of this as: the left side is always "heavy" (at least 2), and the right side is always "light" (at most 2). Equality can only happen if both sides are exactly 2 at the same $x$. Let's pin down where each side hits 2.

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Step 1 — Find the minimum of the left-hand side.

Let $a = 2^x$ and $b = 2^{-x}$. Both are positive, so by the AM–GM inequality, $2^x + 2^{-x} \ge 2\sqrt{2^x \cdot 2^{-x}} = 2\sqrt{2^0} = 2.$ Equality holds only when $2^x = 2^{-x}$, i.e. $x = 0$.

So the LHS is $\ge 2$ for every real $x$, and it touches 2 only at $x = 0$.

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Step 2 — Find the maximum of the right-hand side.

The right side is $2 - (x-2)^2$. Since a square is never negative, $(x-2)^2 \ge 0 \implies 2 - (x-2)^2 \le 2.$ Equality holds only when $(x-2)^2 = 0$, i.e. $x = 2$.

So the RHS is $\le 2$ for every real $x$, and it touches 2 only at $x = 2$.

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Step 3 — Force the two sides to meet.

For the equation to hold, we need LHS $=$ RHS. But:

• LHS $\ge 2$ always,
• RHS $\le 2$ always.

The only way they can be equal is if both are exactly 2 at the same $x$.

The LHS reaches 2 at $x = 0$; the RHS reaches 2 at $x = 2$. These are different points, so no single $x$ makes both sides equal to 2 simultaneously.

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Step 4 — Sanity-check directly at the two candidate points.

• At $x = 0$: LHS $= 2^0 + 2^0 = 1 + 1 = 2$, but RHS $= 2 - (0-2)^2 = 2 - 4 = -2$. Not equal. ✗
• At $x = 2$: RHS $= 2 - 0 = 2$, but LHS $= 2^2 + 2^{-2} = 4 + 0.25 = 4.25$. Not equal. ✗

At every other $x$, the LHS is strictly greater than 2 while the RHS is strictly less than 2, so they can never agree.

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Answer: $\boxed{0}$ real solutions — Choice D

💡 Why this works so cleanly: No graphing or algebraic solving was needed. The left side has a built-in minimum of 2 and the right side has a built-in maximum of 2, but they hit that value at different $x$-values. Two curves trapped on opposite sides of the same line, never touching — so there is nothing to solve.

Think of this as: we're given two (speed, arrival-time) pairs for the same distance and same leaving time. Those two facts alone let us pin down the distance — and then everything else follows.

What we want: the speed needed to cover the home-to-office distance in exactly 50 minutes (9:10 am → 10:00 am).

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Step 1 — Translate each scenario into an equation.

Let the home-to-office distance be $D$ km, and let the usual leaving time be $T$ hours (measured from midnight, say).

• At $8$ km/hr, he reaches at $10{:}15$ am $= 10.25$ hours $= \frac{41}{4}$ hours. $\frac{D}{8} = \frac{41}{4} - T$
• At $15$ km/hr, he reaches at $9{:}40$ am $= 9 + \frac{40}{60} = \frac{29}{3}$ hours. $\frac{D}{15} = \frac{29}{3} - T$

Each equation simply says: travel time = arrival time − leaving time.

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Step 2 — Subtract the equations to eliminate $T$ and find $D$.

The unknown leaving time $T$ appears identically in both equations, so subtracting cancels it: $D\left(\frac{1}{8} - \frac{1}{15}\right) = \frac{41}{4} - \frac{29}{3}$

Compute each side: $\frac{1}{8} - \frac{1}{15} = \frac{15 - 8}{120} = \frac{7}{120}$ $\frac{41}{4} - \frac{29}{3} = \frac{123 - 116}{12} = \frac{7}{12}$

So: $D \cdot \frac{7}{120} = \frac{7}{12} \quad\Longrightarrow\quad D = \frac{7}{12} \times \frac{120}{7} = 10 \text{ km}$

💡 Teacher tip: Whenever two scenarios share the same distance and same start time but differ in speed and arrival time, subtracting the two time equations kills the unknown start time and hands you the distance directly. This is the standard move for "two speeds, two arrival times" problems.

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Step 3 — Find the usual leaving time (as a sanity check).

At $8$ km/hr, the travel time is $\frac{10}{8} = 1.25$ hr $= 75$ min. Reaching at $10{:}15$ am means he left at: $10{:}15 - 1{:}15 = 9{:}00 \text{ am}$

Verify with the other speed: $\frac{10}{15} = \frac{2}{3}$ hr $= 40$ min. Leaving at $9{:}00$ am gives arrival $9{:}40$ am ✓.

Both scenarios agree: the usual leaving time is $9{:}00$ am, and the distance is $10$ km.

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Step 4 — Compute the required speed for the new schedule.

Now he leaves at $9{:}10$ am and must arrive at $10{:}00$ am, so the available travel time is: $50 \text{ min} = \frac{50}{60} \text{ hr} = \frac{5}{6} \text{ hr}$

The distance is still $10$ km, so the required speed is: $\text{Speed} = \frac{D}{\text{time}} = \frac{10}{\,5/6\,} = 10 \times \frac{6}{5} = 12 \text{ km/hr}$

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Quick check of the answer.

Only $12$ km/hr lands him at the office at exactly $10{:}00$ am.

Answer: $\boxed{12 \text{ km/hr}}$ — Choice B

💡 Closing insight: The two given scenarios quietly encode three unknowns (distance, leaving time, and the relationship between them), but because the leaving time is the same in both, subtracting collapses the system to a single equation in $D$. Once the distance is known, any new schedule is just a one-line speed calculation.

Think of this as: 100 students must be placed into 12 monthly "buckets." We want to arrange them so that the fullest bucket is as small as possible.

Step 1 — The core idea: to minimise a maximum, spread things out evenly.

If we want the largest month-count to be as small as possible, we should avoid clustering students in any one month. The most balanced distribution is the one where every month has nearly the same number of students.

💡 Teacher tip: Whenever a problem asks you to minimise the maximum across several groups, your first instinct should be: distribute the total as evenly as possible across the groups.

Step 2 — Divide 100 by 12.

$100 \div 12 = 8 \text{ with remainder } 4.$

So the most even split is: 8 students in every month, plus 1 extra student in 4 of the months.

That gives us:

• 8 months with 8 students,
• 4 months with 9 students.

The maximum here is $9$.

Step 3 — Prove the maximum cannot be 8 (or less).

Suppose, for contradiction, that every month had at most 8 students. Then the total number of students would be at most:

$12 \times 8 = 96.$

But we have 100 students, and $96 < 100$. So it is impossible for every month to have 8 or fewer students. At least one month must contain 9 or more students.

Therefore: $x_0 \geq 9.$

Step 4 — Confirm 9 is actually achievable.

Distribute the students as 4 months of 9 and 8 months of 8:

$4 \times 9 + 8 \times 8 = 36 + 64 = 100. \checkmark$

The maximum month-count in this arrangement is $9$, and the total is exactly 100. So the lower bound of 9 is reachable.

Step 5 — Verification table.

Answer: $\boxed{9}$ (Choice B)

💡 Why this works: This is the pigeonhole principle in action. With 100 students (pigeons) and 12 months (holes), at least one month must contain $\lceil 100/12 \rceil = \lceil 8.33 \rceil = 9$ students. The even distribution shows this bound is tight — 9 is both necessary and sufficient.

We need the mean (average) of every 4-digit even number shaped like aabb, where a > 0.

Think of it as: the number is built from two independent digits — a (which can't be zero) and b (which must be even). We'll exploit that independence to avoid listing all 45 numbers.

Step 1 — Write the number in expanded form.

A number of the form aabb means:

• thousands digit = a
• hundreds digit = a
• tens digit = b
• units digit = b

So in place value: $\overline{aabb} = 1000a + 100a + 10b + b = 1100a + 11b.$

Step 2 — Identify the allowed values of each digit.

• Since it's a 4-digit number and a > 0, we have $a \in \{1, 2, \ldots, 9\}$ — 9 choices.
• The number must be even, so its last digit b must be even: $b \in \{0, 2, 4, 6, 8\}$ — 5 choices.

Total numbers: $9 \times 5 = 45$.

Step 3 — Use the key idea: average the digits independently.

💡 Teacher tip: Because a and b are chosen independently, the mean of $1100a + 11b$ splits cleanly: it equals $1100 \times (\text{mean of } a) + 11 \times (\text{mean of } b)$. Linearity of averages does the heavy lifting — no need to enumerate all 45 numbers.

• Mean of $a$ over $\{1, 2, \ldots, 9\}$: $\bar{a} = \frac{1 + 2 + \cdots + 9}{9} = \frac{45}{9} = 5.$

• Mean of $b$ over $\{0, 2, 4, 6, 8\}$: $\bar{b} = \frac{0 + 2 + 4 + 6 + 8}{5} = \frac{20}{5} = 4.$

Step 4 — Plug into the expanded form.

$\text{Mean} = 1100\,\bar{a} + 11\,\bar{b} = 1100 \times 5 + 11 \times 4 = 5500 + 44 = 5544.$

Step 5 — Sanity check with a couple of boundary values.

The midpoint of the smallest and largest values is exactly $5544$ — a reassuring confirmation that the mean sits in the sensible middle of the range.

Answer: $\boxed{5544}$ (Choice D)

💡 Why this works so cleanly: the digits a and b vary independently, so the average of the whole expression is just the expression evaluated at the average digits. Whenever a number is a linear combination of independently varying parts, average each part first — then combine.

Think of this as: we first decode the train's usual journey time from the delay clue, then figure out what speed is needed on the return trip to still finish within that usual time.

Step 1 — Find the usual journey time from the "one-third speed" clue.

Speed and time are inversely related for a fixed distance. If the train runs at $\tfrac{1}{3}$ of its usual speed, the time taken becomes $3$ times the usual time.

Let the usual journey time be $T$ minutes.

• Time at reduced speed $= 3T$
• Delay $= 3T - T = 2T$
• Given delay $= 30$ minutes

So $2T = 30 \Rightarrow T = 15$ minutes.

The scheduled (usual) journey time is 15 minutes.

💡 Teacher tip: Whenever a problem says "at $\tfrac{1}{n}$ of usual speed, the delay is $D$", the usual time is $\dfrac{D}{n-1}$. Here, $\tfrac{30}{3-1} = 15$ min.

Step 2 — Track the return journey against the 15-minute schedule.

The train has a total budget of 15 minutes to reach on time. On the return trip:

• It travels at usual speed for 5 minutes. Since the full trip at usual speed takes 15 minutes, in 5 minutes it covers $\dfrac{5}{15} = \dfrac{1}{3}$ of the distance.
• Remaining distance $= 1 - \dfrac{1}{3} = \dfrac{2}{3}$.
• It then stops for 4 minutes (emergency halt).
• Total time consumed so far $= 5 + 4 = 9$ minutes.
• Time remaining in the schedule $= 15 - 9 = 6$ minutes.

Step 3 — Compute the required new speed.

The remaining $\tfrac{2}{3}$ of the distance would normally take $\tfrac{2}{3} \times 15 = 10$ minutes at the usual speed. But only 6 minutes are left.

For the same distance, speed must scale inversely with time: $\frac{\text{New speed}}{\text{Usual speed}} = \frac{\text{time at usual speed}}{\text{time available}} = \frac{10}{6} = \frac{5}{3}.$

So the new speed is $\tfrac{5}{3}$ of the usual speed.

Step 4 — Convert to a percentage increase.

$\text{Percentage increase} = \left(\frac{5}{3} - 1\right) \times 100 = \frac{2}{3} \times 100 \approx 66.7\%.$

Step 5 — Pick the nearest option.

$66.7\%$ is closest to 67%.

Answer: $\boxed{67\%}$ — Choice C

💡 Why the absolute distance cancels: every step uses fractions of the trip (one-third of the distance, two-thirds remaining), so the route length never matters — the answer is a clean ratio regardless of how long the track actually is.

Think of this as: a cone is sliced parallel to its base, giving a small cone on top and a frustum below. We're told how much bigger the frustum is than the small cone — and we need to work backwards to the original cone's volume.

Step 1 — Picture the two pieces.

The original cone has height $27$ cm. The cut is made 18 cm above the base, so:

• The bottom piece (frustum) has height $18$ cm.
• The top piece (small cone) has height $27 - 18 = 9$ cm.

Step 2 — Relate the small cone to the original using similarity.

The cut is parallel to the base, so the small cone on top is similar to the original cone. The linear scale factor is: $\frac{\text{height of small cone}}{\text{height of original cone}} = \frac{9}{27} = \frac{1}{3}.$

Since volume scales as the cube of the linear scale factor: $V_{\text{top}} = \left(\frac{1}{3}\right)^3 \cdot V = \frac{V}{27}.$

💡 Teacher tip: You never need the actual radius. For cones cut parallel to the base, the volume ratio is fixed entirely by the height ratio — cubed. That's the key shortcut here.

Step 3 — Express the frustum volume.

The frustum is everything that's left after removing the small cone: $V_{\text{bottom}} = V - \frac{V}{27} = \frac{26V}{27}.$

Step 4 — Use the given volume difference.

The frustum is the larger piece, so: $V_{\text{bottom}} - V_{\text{top}} = \frac{26V}{27} - \frac{V}{27} = \frac{25V}{27}.$

We're told this difference equals $225$ cc: $\frac{25V}{27} = 225.$

Step 5 — Solve for $V$. $V = 225 \times \frac{27}{25} = 9 \times 27 = 243.$

Sanity check.

Everything matches.

Answer: $\boxed{243 \text{ cc}}$ — Choice A

💡 Why this is fast: The volume difference works out to a clean fraction $\frac{25}{27}$ of the whole cone. No radii, no $\pi$ — just the height ratio and a single equation.