CAT 2019 Quant Questions — Slot 1

What we want: the value of $xy$, given two logarithmic equations — one in base 5 involving $x+y$ and $x-y$, and one in base 2 relating $y$ to $x$. The plan: convert each log equation into a plain algebraic relation, then solve.

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Step 1 — Simplify the base-5 equation.

We are given: $\log_5(x+y) + \log_5(x-y) = 3$

Since both terms share the same base, apply the product rule $\log_b A + \log_b B = \log_b(AB)$: $\log_5\big((x+y)(x-y)\big) = 3$

Now convert from logarithmic to exponential form using $\log_b M = c \Rightarrow M = b^c$: $(x+y)(x-y) = 5^3 = 125$

The left side is a difference of squares: x^2 - y^2 = 125 \tag{1}

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Step 2 — Simplify the base-2 equation.

We are given: $\log_2 y - \log_2 x = 1 - \log_2 3$

Apply the quotient rule $\log_b A - \log_b B = \log_b\frac{A}{B}$ on the left, and rewrite $1$ as $\log_2 2$ on the right: $\log_2\frac{y}{x} = \log_2 2 - \log_2 3 = \log_2\frac{2}{3}$

💡 Teacher tip: Whenever a logarithmic equation has the same base on both sides, you can "drop the logs" — equal logs (same base) means equal arguments. This is because $\log_b$ is a one-to-one function.

So: \frac{y}{x} = \frac{2}{3} \quad\Longrightarrow\quad y = \frac{2x}{3} \tag{2}

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Step 3 — Substitute and solve for $x$ and $y$.

Plug equation (2) into equation (1): $x^2 - \left(\frac{2x}{3}\right)^2 = 125$ $x^2 - \frac{4x^2}{9} = 125$ $\frac{9x^2 - 4x^2}{9} = 125$ $\frac{5x^2}{9} = 125$ $x^2 = 125 \times \frac{9}{5} = 225$

Since $x > 0$ (given), we take the positive root: $x = 15$

Then from equation (2): $y = \frac{2 \times 15}{3} = 10$

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Step 4 — Compute $xy$.

$xy = 15 \times 10 = 150$

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Verification — confirm both original equations hold.

Both equations check out.

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Answer: $\boxed{150}$ (Choice A)

💡 Why this works so cleanly: The two equations decouple beautifully — the base-2 equation fixes the ratio $y/x = 2/3$, while the base-5 equation fixes the scale through $x^2 - y^2 = 125$. Once the ratio is known, a single substitution resolves everything. This is a common CAT pattern: one equation gives a proportion, the other gives a magnitude.

Think of this as a classic telescoping series: each fraction looks messy, but rationalising it turns it into a difference of two terms — and differences cancel beautifully when summed.

Step 1 — Rationalise a single term.

Take a typical summand, say the $k$-th one: $\dfrac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}$.

Multiply the numerator and denominator by the conjugate $\sqrt{a_{k+1}}-\sqrt{a_k}$:

$$\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} \cdot \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{\sqrt{a_{k+1}}-\sqrt{a_k}} = \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{(\sqrt{a_{k+1}})^2-(\sqrt{a_k})^2} = \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{a_{k+1}-a_k}.$$

Step 2 — Use the A.P. structure.

Since $a_1, a_2, \dots$ are in an arithmetic progression, the difference $a_{k+1}-a_k$ is the common difference $d$, which is the same for every $k$. So each term becomes:

$$\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}.$$

Now write out the entire sum $S$:

$$S = \frac{\sqrt{a_2}-\sqrt{a_1}}{d} + \frac{\sqrt{a_3}-\sqrt{a_2}}{d} + \cdots + \frac{\sqrt{a_{n+1}}-\sqrt{a_n}}{d}.$$

Factor out $\dfrac{1}{d}$:

$$S = \frac{1}{d}\Big[(\sqrt{a_2}-\sqrt{a_1}) + (\sqrt{a_3}-\sqrt{a_2}) + \cdots + (\sqrt{a_{n+1}}-\sqrt{a_n})\Big].$$

This is a telescoping sum: $-\sqrt{a_2}$ cancels $+\sqrt{a_2}$, $-\sqrt{a_3}$ cancels $+\sqrt{a_3}$, and so on. Everything vanishes except the first and last terms:

$$S = \frac{\sqrt{a_{n+1}}-\sqrt{a_1}}{d}.$$

💡 Teacher tip: Whenever you see an A.P. sitting inside a difference of square roots, the move is almost always the same: rationalise, then telescope. The common difference $d$ becomes a clean constant you can factor out.

Step 3 — Remove the $d$ from the denominator.

In an A.P. with common difference $d$, the $(n+1)$-th term satisfies:

$$a_{n+1} = a_1 + nd \quad\Longrightarrow\quad a_{n+1} - a_1 = nd \quad\Longrightarrow\quad d = \frac{a_{n+1}-a_1}{n}.$$

Substitute this into $S$:

$$S = \frac{\sqrt{a_{n+1}}-\sqrt{a_1}}{\dfrac{a_{n+1}-a_1}{n}} = \frac{n\,\big(\sqrt{a_{n+1}}-\sqrt{a_1}\big)}{a_{n+1}-a_1}.$$

Now recognise the denominator as a difference of squares in disguise:

$$a_{n+1}-a_1 = \big(\sqrt{a_{n+1}}\big)^2 - \big(\sqrt{a_1}\big)^2 = \big(\sqrt{a_{n+1}}-\sqrt{a_1}\big)\big(\sqrt{a_{n+1}}+\sqrt{a_1}\big).$$

Cancel the common factor $\big(\sqrt{a_{n+1}}-\sqrt{a_1}\big)$:

$$S = \frac{n}{\sqrt{a_{n+1}}+\sqrt{a_1}} = \frac{n}{\sqrt{a_1}+\sqrt{a_{n+1}}}.$$

Answer: $\boxed{\dfrac{n}{\sqrt{a_1}+\sqrt{a_{n+1}}}}$ (Choice A)

Why this works so cleanly: the A.P. guarantees that every denominator $a_{k+1}-a_k$ is the same constant $d$. That uniformity is exactly what lets us factor it out and collapse the sum to a single fraction involving only the first and last terms.

Think of this as: we're told how two mixed crews compare in speed, and we need to convert "machine-power" into "man-power" to find the equivalent crew size.

Step 1 — Set up work rates.

Let one man's daily work rate be $a$ (fraction of job per day). Let one machine's daily work rate be $b$ (fraction of job per day).

Step 2 — Translate the time comparison into a rate equation.

Crew 1: 3 men + 8 machines → rate $= 3a + 8b$. Crew 2: 3 machines + 8 men → rate $= 3b + 8a$.

We're told Crew 1 finishes in half the time of Crew 2. Since work rate is inversely proportional to time, finishing in half the time means working at twice the rate:

$3a + 8b = 2\,(3b + 8a)$

💡 Teacher tip: "Half the time" always means "double the rate." Time and rate are reciprocals — if one crew takes $\tfrac{1}{2}$ the time, it produces work at $2$ times the speed.

Step 3 — Solve for the man-to-machine rate ratio.

$3a + 8b = 6b + 16a$ $8b - 6b = 16a - 3a$ $2b = 13a$ $b = \frac{13}{2}\,a$

So one machine does as much work per day as $\frac{13}{2}$ men (i.e., 6.5 men).

Step 4 — Use the concrete data: 2 machines finish in 13 days.

Work done by 2 machines in 13 days $= 2b \times 13 = 1$ (one complete job):

$26b = 1 \;\Rightarrow\; b = \frac{1}{26}$

Now back-substitute to find $a$:

$a = \frac{2b}{13} = \frac{2}{13 \times 26} = \frac{2}{338} = \frac{1}{169}$

Step 5 — Find how many men finish the job in 13 days.

Let $N$ men finish the job in 13 days. Their total work must equal 1:

$N \times a \times 13 = 1$ $N \times \frac{13}{169} = 1$ $N = \frac{169}{13} = 13$

Answer: $\boxed{13}$

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Alternative approach — the shortcut (much faster).

From Step 3, we found $b = \frac{13}{2}\,a$, meaning 1 machine $\equiv$ $\frac{13}{2}$ men in work rate.

Therefore 2 machines $\equiv$ $2 \times \frac{13}{2} = 13$ men.

Since 2 machines finish the job in 13 days, and 2 machines have the same total work rate as 13 men, it follows immediately that 13 men also finish the job in 13 days.

No need to compute individual rates at all — the ratio does everything.

💡 Why this is elegant: The number 13 appears in the given data (days) and again in the answer (men). That's not a coincidence — it's because $2 \times \frac{13}{2} = 13$ collapses the conversion neatly. Whenever a problem gives you a rate ratio and a fixed time, check whether the ratio alone resolves the question before computing absolute rates.

Think of this as: two wheels of different sizes roll over the same distance. The smaller wheel must spin more times. We're told exactly how many more — and that's enough to pin down the distance, and hence B's speed.

Step 1 — Distance covered in one revolution.

A wheel covers a distance equal to its circumference in one full revolution.

For bicycle A (radius $30$ cm): $\text{Circumference}_A = 2\pi r_A = 2\pi(30) = 60\pi \text{ cm}$

For bicycle B (radius $40$ cm): $\text{Circumference}_B = 2\pi r_B = 2\pi(40) = 80\pi \text{ cm}$

Step 2 — Set up the revolution equation.

Let the common distance travelled be $D$ cm.

• Number of revolutions of each wheel of A $= \dfrac{D}{60\pi}$
• Number of revolutions of each wheel of B $= \dfrac{D}{80\pi}$

We are told A's wheel makes $5000$ more revolutions than B's: $\frac{D}{60\pi} - \frac{D}{80\pi} = 5000$

Step 3 — Solve for $D$.

Factor out $\dfrac{D}{\pi}$: $\frac{D}{\pi}\left(\frac{1}{60} - \frac{1}{80}\right) = 5000$

Compute the bracket with a common denominator of $480$: $\frac{1}{60} - \frac{1}{80} = \frac{80 - 60}{4800} = \frac{20}{4800} = \frac{1}{240}$

So: $\frac{D}{240\pi} = 5000 \implies D = 5000 \times 240\pi = 1{,}200{,}000\pi \text{ cm}$

Step 4 — Convert the distance to kilometres.

$1{,}200{,}000\pi \text{ cm} = 12{,}000\pi \text{ m} = 12\pi \text{ km}$

💡 Teacher tip: The "5000 more revolutions" clause is doing the heavy lifting here. Without an actual revolution count for either wheel, the difference in revolutions — combined with the difference in circumferences — is what uniquely determines the distance.

Step 5 — Compute B's speed.

Bicycle B covers $12\pi$ km in $45$ minutes. Converting time to hours: $45 \text{ min} = \frac{45}{60} \text{ h} = \frac{3}{4} \text{ h}$

$\text{Speed of B} = \frac{\text{Distance}}{\text{Time}} = \frac{12\pi}{3/4} = 12\pi \times \frac{4}{3} = 16\pi \text{ km/h}$

Answer: $\boxed{16\pi \text{ km/h}}$ (Choice C)

Closing insight: Notice the distance came out to a clean $12\pi$ km — the $5000$ extra revolutions and the $240\pi$ circumference-gap combined to give a round number. Whenever a problem gives a difference of revolutions rather than absolute counts, expect the distance to be recoverable from that difference alone.

Think of this as: we have three overlapping circles (football, tennis, cricket) covering all 256 members. We're told the sizes of each circle and each pairwise overlap, but not the triple overlap. First find the triple overlap, then peel away everything that isn't purely tennis.

Step 1 — Set up notation.

Let:

• $F$ = set of football players, $|F| = 144$
• $T$ = set of tennis players, $|T| = 123$
• $C$ = set of cricket players, $|C| = 132$
• $|F \cap T| = 58$, $\;|T \cap C| = 25$, $\;|F \cap C| = 63$
• $x = |F \cap T \cap C|$ (the unknown triple overlap)

Since every member plays at least one game, $|F \cup T \cup C| = 256$.

Step 2 — Apply the three-set inclusion-exclusion principle.

$|F \cup T \cup C| = |F| + |T| + |C| - |F \cap T| - |T \cap C| - |F \cap C| + |F \cap T \cap C|$

Substituting the known values:

$256 = 144 + 123 + 132 - 58 - 25 - 63 + x$

$256 = 399 - 146 + x$

$256 = 253 + x$

$x = 3$

So exactly 3 members play all three sports.

💡 Teacher tip: The pairwise numbers ($58, 25, 63$) each include the triple-overlap members. Inclusion-exclusion subtracts the three pairwise intersections, which removes the triple members three times — but they were added three times in the singles, so we add $x$ back once to balance.

Step 3 — Find the number who play only tennis.

"Only tennis" means: in $T$, but not in $F$ and not in $C$.

Start with all tennis players ($123$), then remove those who also play football ($58$) and those who also play cricket ($25$):

$\text{Only } T = |T| - |F \cap T| - |T \cap C| + x$

💡 Teacher tip: Why add $x$ back? The 3 triple-sport players are counted inside both $|F \cap T|$ and $|T \cap C|$, so they get subtracted twice. We add $x$ once to undo the double subtraction.

$\text{Only } T = 123 - 58 - 25 + 3 = 43$

Step 4 — Verify with a quick Venn-diagram sanity check.

The regions sum to exactly $256$, confirming our answer.

Answer: $\boxed{43}$ (Choice D)

Closing insight: The triple overlap ($x=3$) is small here, but it still matters — without adding it back, "only tennis" would come out to $40$ instead of $43$. Whenever a question asks for an "only" region, always account for the triple intersection explicitly.

Think of this as: the second car has one hour less to cover the same distance, so it must be faster. We want the largest possible speed gap — which happens when the first car takes the least allowed time.

What we want: the largest possible percentage by which the second car's speed exceeds the first's, given they cover the same distance and arrive together — but the second car starts an hour later.

Step 1 — Translate the timing into an equation. Let the common distance be $D$. Let the first car travel for $t_1$ hours and the second for $t_2$ hours.

The first car starts at 10:00 am, the second at 11:00 am, yet both arrive together. So the second car is on the road for exactly one hour less:

$t_2 = t_1 - 1.$

Step 2 — Write the two speeds. Since distance $=$ speed $\times$ time,

$v_1 = \frac{D}{t_1}, \qquad v_2 = \frac{D}{t_2} = \frac{D}{t_1-1}.$

Step 3 — Build the percentage-excess expression. The percentage by which $v_2$ exceeds $v_1$ is

$\frac{v_2 - v_1}{v_1} \times 100 = \left(\frac{v_2}{v_1} - 1\right) \times 100.$

Now $\dfrac{v_2}{v_1} = \dfrac{D/(t_1-1)}{D/t_1} = \dfrac{t_1}{t_1-1}$, so

$\text{excess } \% = \left(\frac{t_1}{t_1-1} - 1\right) \times 100 = \frac{1}{t_1-1} \times 100.$

💡 Teacher tip: The distance $D$ cancels out completely — the answer depends only on the times. This is why the problem never tells you the actual distance.

Step 4 — Maximise the expression. The percentage $\dfrac{100}{t_1-1}$ gets larger as $t_1 - 1$ gets smaller, i.e. as $t_1$ gets smaller. We are told the first car travelled at least 6 hours, so $t_1 \ge 6$. The smallest allowed value is therefore

$t_1 = 6.$

💡 Teacher tip: The phrase "at least 6 hours" is the key signal — it sets a floor on $t_1$, and to maximise a "one-over-something" expression we push that something down to its floor.

Step 5 — Substitute and compute.

$\text{excess } \% = \frac{100}{6 - 1} = \frac{100}{5} = 20.$

Quick check with concrete numbers. Take $D = 30$ km (chosen so the speeds come out clean). Then $t_1 = 6$ h gives $v_1 = 5$ km/h, and $t_2 = 5$ h gives $v_2 = 6$ km/h. The excess is $\dfrac{6-5}{5} \times 100 = 20\%$. ✓

Answer: $\boxed{20}$ (Option A)

💡 Why this is clean: the percentage excess collapses to the single fraction $\frac{100}{t_1-1}$, so the entire question reduces to finding the smallest legal value of $t_1$.

Think of this as: a brick has three edges and three face diagonals. If we know the ratio of the three face diagonals, can we back out the ratio of the edges themselves?

Step 1 — Set up the face diagonals.

Let the brick have edges $a, b, c$. The three rectangular faces have diagonals:

$\sqrt{a^2+b^2},\quad \sqrt{b^2+c^2},\quad \sqrt{a^2+c^2}$

These are given in the ratio $3: 2\sqrt{3}: \sqrt{15}$.

Step 2 — Square the ratio to remove the square roots.

$3^2: (2\sqrt{3})^2: (\sqrt{15})^2 \;=\; 9: 12: 15$

So the three quantities $a^2+b^2$, $b^2+c^2$, $a^2+c^2$ take the values $9, 12, 15$ in some order.

💡 Teacher tip: Squaring a ratio of square roots converts it into a ratio of plain integers — always do this first when face diagonals are involved, since each face diagonal squared is just a sum of two squared edges.

Step 3 — Use the sum to find each squared edge.

Add all three face-diagonal-squared values:

$(a^2+b^2)+(b^2+c^2)+(a^2+c^2) = 9+12+15 = 36$

$\Rightarrow\quad 2(a^2+b^2+c^2) = 36 \quad\Rightarrow\quad a^2+b^2+c^2 = 18$

Now each squared edge is the total $18$ minus the squared diagonal of the opposite face:

So the three edges are $\sqrt{9}=3$, $\sqrt{6}$, and $\sqrt{3}$.

💡 Teacher tip: The assignment of which diagonal goes with which pair of edges doesn't matter — we always recover the same set $\{3, 6, 9\}$ of squared edges, just with different labels.

Step 4 — Identify shortest and longest.

• Shortest edge $= \sqrt{3}$
• Longest edge $= 3$

$\text{Ratio} = \sqrt{3}: 3 = 1: \sqrt{3}$

Answer: $\boxed{1: \sqrt{3}}$ (Choice B)

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Alternative approach — order the edges first.

Assume $a \le b \le c$. Then the smallest face diagonal uses the two smallest edges and the largest uses the two largest:

$a^2+b^2 = 9,\quad a^2+c^2 = 12,\quad b^2+c^2 = 15$

Subtract the first from the second: $c^2 - b^2 = 3$.

Subtract the second from the third: $b^2 - a^2 = 3$.

So $b^2 = a^2+3$ and $c^2 = a^2+6$. Substituting into $a^2+b^2=9$:

$a^2 + (a^2+3) = 9 \;\Rightarrow\; 2a^2 = 6 \;\Rightarrow\; a^2 = 3$

This gives $a=\sqrt{3}$, $b=\sqrt{6}$, $c=3$, and the ratio $\sqrt{3}:3 = 1:\sqrt{3}$. Same result, reached by subtracting equations instead of summing them.

Closing insight: The three squared face diagonals $9, 12, 15$ and the total $18$ conspire to give squared edges $3, 6, 9$ — an arithmetic progression. That symmetry is what makes the shortest-to-longest ratio come out to the clean value $1:\sqrt{3}$.

Think of this as: the function $f$ uses one rule on even inputs and a different rule on odd inputs. Since $m$ and $m+1$ are always of opposite parity, the equation $8f(m+1)-f(m)=2$ will mix both rules — so we just need to test which parity of $m$ makes the algebra work out.

The two rules, restated:

$f(n)=\begin{cases} n(n+1), & n\text{ even}\\[4pt] n+3, & n\text{ odd}\end{cases}$

Since consecutive integers have opposite parity, exactly one of $f(m)$ and $f(m+1)$ uses the even-rule and the other uses the odd-rule.

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Case 1 — $m$ is even (so $m+1$ is odd).

Here $f(m)$ uses the even-rule and $f(m+1)$ uses the odd-rule:

$f(m)=m(m+1), \qquad f(m+1)=(m+1)+3=m+4.$

Substitute into $8f(m+1)-f(m)=2$:

$8(m+4)-m(m+1)=2.$

Expand and simplify:

$8m+32-m^2-m=2$

$-m^2+7m+30=0$

$m^2-7m-30=0.$

Factor the quadratic:

$(m-10)(m+3)=0 \;\Rightarrow\; m=10 \text{ or } m=-3.$

Since $m$ must be a positive integer, $m=10$. Note $10$ is indeed even ✓, so our case assumption holds.

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Case 2 — $m$ is odd (so $m+1$ is even).

Here $f(m)$ uses the odd-rule and $f(m+1)$ uses the even-rule:

$f(m)=m+3, \qquad f(m+1)=(m+1)(m+2).$

Substitute:

$8(m+1)(m+2)-(m+3)=2.$

Expand:

$8(m^2+3m+2)-m-3=2$

$8m^2+24m+16-m-3=2$

$8m^2+23m+11=0.$

The discriminant is $23^2-4(8)(11)=529-352=177$, which is not a perfect square, so this case gives no integer solution.

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Verification of $m=10$:

The equation is satisfied.

Answer: $\boxed{10}$

💡 Teacher tip: Whenever a piecewise function depends on parity and the argument shifts by 1, the two pieces always interact. Try the parity that makes the algebra cleanest first; if it yields a valid integer, you're done — but a quick check of the other case (or its discriminant) confirms uniqueness.

Think of this as: we need to count monotone (right/up only) lattice paths that must pass through a specific checkpoint. The key idea is to split the journey at that checkpoint into two independent legs.

Step 1 — Split at the via-point $(4,6)$.

Every valid path goes from $(1,1)$ to $(4,6)$, and then continues from $(4,6)$ to $(8,10)$. These two legs are independent: any path for the first leg can be combined with any path for the second leg.

So: $\text{Total paths} = \text{Paths}\,(1,1)\to(4,6) \times \text{Paths}\,(4,6)\to(8,10).$

💡 Teacher tip: Whenever a path must pass through a specific point, split there. The multiplication principle applies because the choice made in the first leg doesn't restrict the choice in the second leg.

Step 2 — Count paths on the first leg: $(1,1)\to(4,6)$.

To go from $(1,1)$ to $(4,6)$ using only right-steps $(x+1,y)$ and up-steps $(x,y+1)$:

• Right-steps needed: $4 - 1 = 3$
• Up-steps needed: $6 - 1 = 5$
• Total steps: $3 + 5 = 8$

A path is completely determined by choosing which 3 of the 8 steps are right-steps (the rest are automatically up-steps):

$\binom{8}{3} = \frac{8!}{3!\,5!} = 56.$

Step 3 — Count paths on the second leg: $(4,6)\to(8,10)$.

• Right-steps needed: $8 - 4 = 4$
• Up-steps needed: $10 - 6 = 4$
• Total steps: $4 + 4 = 8$

Choose which 4 of the 8 steps are right-steps:

$\binom{8}{4} = \frac{8!}{4!\,4!} = 70.$

Step 4 — Multiply the two legs.

$56 \times 70 = 3920.$

Quick verification of the arithmetic:

Answer: $\boxed{3920}$

💡 Closing insight: The general formula for monotone lattice paths from $(x_1,y_1)$ to $(x_2,y_2)$ is $\binom{(x_2-x_1)+(y_2-y_1)}{x_2-x_1}$. When a via-point is imposed, just evaluate this formula for each leg and multiply — the via-point acts as a natural factorisation of the counting problem.

Think of this as: we're slicing off three identical small equilateral triangles from the corners of a larger equilateral triangle, and the leftover shape is a regular hexagon. We need to figure out how much area survives.

Step 1 — Visualise the cut.

When you cut off each corner of an equilateral triangle with a straight line, the piece removed is itself a small equilateral triangle. After all three corners are cut, the remaining shape has six sides:

• 3 sides come from the cuts themselves (one per corner removed).
• 3 sides are the middle segments left on the original triangle's edges.

For this leftover shape to be a regular hexagon, all six sides must be equal.

Step 2 — Set up side lengths.

Let the original equilateral triangle $T$ have side length $1$.

Suppose each small corner triangle has side length $a$. Then on each original edge, two segments of length $a$ are removed (one at each end), leaving a middle segment of length:

$1 - 2a$

For a regular hexagon, the cut sides ($a$) must equal the middle segments ($1 - 2a$):

$a = 1 - 2a$

$3a = 1$

$a = \frac{1}{3}$

💡 Teacher tip: The key insight is that a regular hexagon needs all six sides equal. Three of those sides are the cut edges, and three are the leftover middle portions of the original triangle's sides. Setting them equal pins down exactly how big each cut must be.

Step 3 — Compute the areas.

The area of an equilateral triangle with side $s$ is $\dfrac{\sqrt{3}}{4}\,s^2$.

Original triangle (side $1$):

$\text{Area}(T) = \frac{\sqrt{3}}{4} \times 1^2 = \frac{\sqrt{3}}{4}$

Each removed corner triangle (side $\frac{1}{3}$):

$\text{Area of one corner} = \frac{\sqrt{3}}{4}\left(\frac{1}{3}\right)^2 = \frac{\sqrt{3}}{4} \times \frac{1}{9} = \frac{\sqrt{3}}{36}$

Three corners removed:

$3 \times \frac{\sqrt{3}}{36} = \frac{\sqrt{3}}{12}$

Hexagon area (original minus the three corners):

$\text{Area}(H) = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{12} = \frac{3\sqrt{3}}{12} - \frac{\sqrt{3}}{12} = \frac{2\sqrt{3}}{12} = \frac{\sqrt{3}}{6}$

Step 4 — Find the ratio.

$\frac{\text{Area}(H)}{\text{Area}(T)} = \frac{\sqrt{3}/6}{\sqrt{3}/4} = \frac{1/6}{1/4} = \frac{4}{6} = \frac{2}{3}$

Quick verification: Each small triangle has $\frac{1}{9}$ the area of the original (since area scales with the square of the side ratio $\frac{1}{3}$). Three of them remove $\frac{3}{9} = \frac{1}{3}$ of the total area, leaving $\frac{2}{3}$. ✓

Answer: $\boxed{2: 3}$ (Choice A)

💡 Closing insight: The ratio $\frac{2}{3}$ holds regardless of the original triangle's size — it depends only on the shape of the construction, not on any particular side length. That's why we could safely set the side to $1$ without loss of generality.

Think of this as: a diameter gives us two free right angles, so both triangles become straightforward Pythagoras applications — one to find $AP$, the other to find $QB$.

Step 1 — Spot the right angles (Thales' theorem).

Since $AB$ is a diameter, any point on the circle subtends a right angle with $AB$. Therefore: $\angle APB = 90^\circ \quad\text{and}\quad \angle AQB = 90^\circ.$

So $\triangle APB$ is right-angled at $P$, and $\triangle AQB$ is right-angled at $Q$.

💡 Teacher tip: Whenever you see a diameter in a circle problem, immediately look for points on the circle — Thales' theorem converts them into right-angled triangles for free. This is the single most useful circle fact in CAT geometry.

Step 2 — Find $AP$ using Pythagoras in $\triangle APB$.

The radius is $5$ cm, so the diameter is: $AB = 2 \times 5 = 10 \text{ cm}.$

Given $PB = 6$ cm, apply Pythagoras: $AP^2 + PB^2 = AB^2$ $AP^2 + 6^2 = 10^2$ $AP^2 + 36 = 100$ $AP^2 = 64 \;\Rightarrow\; AP = 8 \text{ cm}.$

Step 3 — Find $AQ$ from the given ratio.

We are told $AP = 2 \cdot AQ$: $8 = 2 \cdot AQ \;\Rightarrow\; AQ = 4 \text{ cm}.$

Step 4 — Find $QB$ using Pythagoras in $\triangle AQB$.

$AQ^2 + QB^2 = AB^2$ $4^2 + QB^2 = 10^2$ $16 + QB^2 = 100$ $QB^2 = 84$ $QB = \sqrt{84} = 2\sqrt{21} \approx 9.165 \text{ cm}.$

Step 5 — Match to the nearest option.

$9.165$ is closest to $9.1$.

Answer: $\boxed{9.1 \text{ cm}}$ (Choice C)

💡 Why this works so directly: the diameter hands us two right angles for free, so each triangle reduces to a one-line Pythagoras. The only piece of reasoning needed beyond that is using the ratio $AP = 2\cdot AQ$ to transfer information from one triangle to the other.

Think of this as: one chunk earns a fixed 6%, the rest earns a blended rate (since it's split between 4% and 3%), and the overall interest is known. Find the chunk sizes.

Step 1 — Set up the deposits.

Let the fixed deposit (FD) hold $x$ lakh. The remaining $(15 - x)$ lakh is split in the ratio $2:1$ between the 4% and 3% deposits:

Step 2 — Find the blended rate on the non-FD portion.

Since the non-FD money is split $2:1$, the effective rate on that entire $(15-x)$ lakh is the weighted average:

$\frac{2}{3}\times 4\% + \frac{1}{3}\times 3\% = \frac{8}{3}\% + \frac{3}{3}\% = \frac{11}{3}\%.$

💡 Teacher tip: Instead of computing interest from each of the three deposits separately, collapse the two non-FD deposits into one "bucket" earning $\tfrac{11}{3}\%$. This turns a three-part problem into a clean two-part equation.

Step 3 — Write the interest equation.

Total interest $=$ Rs $76000 = 0.76$ lakh. So:

$0.06\,x + \frac{0.11}{3}(15 - x) = 0.76.$

Multiply through by $3$ to clear the denominator:

$0.18\,x + 0.11(15 - x) = 2.28.$

Expand:

$0.18\,x + 1.65 - 0.11\,x = 2.28.$

Combine like terms:

$0.07\,x = 0.63 \quad\Longrightarrow\quad x = \frac{0.63}{0.07} = 9.$

Step 4 — Verify.

$0.76$ lakh $=$ Rs $76000$ ✓ — matches the given total interest.

Answer: $\boxed{9}$

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Alternative approach — Alligation (faster under exam time).

The FD earns $6\%$; the non-FD bucket earns $\tfrac{11}{3}\%$. The overall rate on the full Rs 15 lakh is:

$\frac{0.76}{15}\times 100 = \frac{76}{15}\%.$

Now apply alligation between the FD rate and the non-FD rate:

$\text{FD} \;\left(\frac{90}{15}\%\right) \qquad \text{Non-FD}\;\left(\frac{55}{15}\%\right)$

$\text{Mixture} = \frac{76}{15}\%$

Ratio of FD to Non-FD $= \left(\frac{76}{15}-\frac{55}{15}\right): \left(\frac{90}{15}-\frac{76}{15}\right) = \frac{21}{15}: \frac{14}{15} = 21: 14 = 3: 2.$

So FD $= \tfrac{3}{5}\times 15 = 9$ lakh — same answer, no algebraic equation needed.

💡 Why this works so cleanly: whenever a total is split into parts with different rates and the overall average is known, alligation gives the split ratio directly — no variable, no equation to solve.

Think of this as: two people work at their normal speeds and finish in 12 days; if one slows down and the other speeds up, they finish in 9 days. We need to back out each person's individual speed.

Step 1 — Set up rate equations.

Let A's usual daily work rate be $a$ (fraction of task per day) and B's usual daily work rate be $b$.

At usual efficiencies, together they finish in 12 days: $a + b = \frac{1}{12} \quad \cdots (1)$

In the modified scenario, A works at half her usual rate ($\frac{a}{2}$) and B works at triple his usual rate ($3b$), and together they finish in 9 days: $\frac{a}{2} + 3b = \frac{1}{9} \quad \cdots (2)$

💡 Teacher tip: In Time & Work problems, always express each person's contribution as a rate (work per day), not days. Rates add cleanly; days don't.

Step 2 — Solve the system.

From equation (1), express $a$ in terms of $b$: $a = \frac{1}{12} - b$

Substitute into equation (2): $\frac{1}{2}\left(\frac{1}{12} - b\right) + 3b = \frac{1}{9}$

Expand the first term: $\frac{1}{24} - \frac{b}{2} + 3b = \frac{1}{9}$

Combine the $b$ terms ($-\frac{b}{2} + 3b = \frac{5b}{2}$): $\frac{1}{24} + \frac{5b}{2} = \frac{1}{9}$

Isolate $\frac{5b}{2}$: $\frac{5b}{2} = \frac{1}{9} - \frac{1}{24}$

Find a common denominator (72): $\frac{5b}{2} = \frac{8}{72} - \frac{3}{72} = \frac{5}{72}$

Solve for $b$: $b = \frac{2}{5} \times \frac{5}{72} = \frac{1}{36}$

Step 3 — Find A's rate.

Plug $b$ back into equation (1): $a = \frac{1}{12} - \frac{1}{36} = \frac{3}{36} - \frac{1}{36} = \frac{2}{36} = \frac{1}{18}$

So A completes $\frac{1}{18}$ of the task per day at her usual efficiency.

Step 4 — Convert rate to days.

If A does $\frac{1}{18}$ of the task each day, she needs $\frac{1}{a} = 18$ days to finish the whole task alone.

Answer: $\boxed{18}$ (Choice C)

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Alternative approach — LCM (units) method.

Take the LCM of 12 and 9, which is 36. Let the total task = 36 units.

• At usual rates: $a + b = \frac{36}{12} = 3$ units/day.
• Modified rates: $\frac{a}{2} + 3b = \frac{36}{9} = 4$ units/day.

From the first equation, $a = 3 - b$. Substitute: $\frac{3 - b}{2} + 3b = 4 \implies \frac{3}{2} - \frac{b}{2} + 3b = 4 \implies \frac{5b}{2} = \frac{5}{2} \implies b = 1$

Then $a = 3 - 1 = 2$ units/day. A alone needs $\frac{36}{2} = 18$ days.

💡 The LCM method avoids fractions entirely and is faster under exam conditions — both methods give the same answer.

Think of this as: two chords cross inside a circle, and we know one chord's full length plus pieces of the other. The intersecting-chords theorem links the four segments — from there it's algebra.

Step 1 — Find the other piece of the diameter.

The radius is 11 cm, so the diameter is: $CD = 2 \times 11 = 22\text{ cm}.$

We're told $CE = 7$ cm. Since $E$ lies on $CD$ between $C$ and $D$: $ED = CD - CE = 22 - 7 = 15\text{ cm}.$

Step 2 — Apply the intersecting-chords theorem.

When two chords $AB$ and $CD$ meet at an interior point $E$, the theorem says: $AE \cdot EB = CE \cdot ED.$

Substituting: $AE \cdot EB = 7 \times 15 = 105.$

We also know the full chord length $AB = 20.5$ cm, and since $E$ lies on $AB$: $AE + EB = 20.5.$

So we have two equations:

• $AE + EB = 20.5$
• $AE \cdot EB = 105$

Step 3 — Find $|BE - AE|$ without solving for each piece.

💡 Teacher tip: Whenever you know the sum and product of two numbers, you can get the difference directly via the identity $(a-b)^2 = (a+b)^2 - 4ab$ — no need to solve a quadratic.

Using $(BE - AE)^2 = (AE + EB)^2 - 4\,AE\cdot EB$:

$(BE - AE)^2 = (20.5)^2 - 4(105) = 420.25 - 420 = 0.25.$

Taking the positive square root (since we want a length difference): $|BE - AE| = \sqrt{0.25} = 0.5\text{ cm}.$

Quick check. Solving $x^2 - 20.5x + 105 = 0$ gives $x = \frac{20.5 \pm 0.5}{2}$, i.e. $AE = 10$ and $EB = 10.5$ (or vice versa). Their product is $10 \times 10.5 = 105$ ✓ and their difference is $0.5$ ✓.

Answer: $\boxed{0.5}$ (Choice D)

💡 Closing insight: The numbers are engineered so that $20.5^2 = 420.25$ sits just $0.25$ above $4 \times 105 = 420$ — a tiny gap that produces the clean answer $0.5$. Whenever a chord-length-difference question gives you the sum and product of segments, reach for $(a-b)^2 = (a+b)^2 - 4ab$ first.