CAT 2020 Quant Questions — Slot 2

Think of this as: two separate ratios share a common person (Sunil), so the first job is to stitch them into one unified ratio — then the largest-minus-smallest gap tells us the scale.

Step 1 — Write down both ratios and identify the link.

We are given:

• Amal: Sunil $= 3: 2$
• Sunil: Mita $= 4: 5$

Sunil appears in both ratios, so he is the bridge. To combine the two ratios into one, Sunil's number must be the same in each.

Step 2 — Make Sunil's number equal across both ratios.

In the first ratio Sunil is $2$; in the second he is $4$. The least common multiple of $2$ and $4$ is $4$, so scale the first ratio by $2$:

$\text{Amal: Sunil} = 3: 2 = 6: 4$

The second ratio already has Sunil as $4$:

$\text{Sunil: Mita} = 4: 5$

Now Sunil is $4$ in both, so we can read off the combined ratio directly:

$\boxed{\text{Amal: Sunil: Mita} = 6: 4: 5}$

💡 Teacher tip: Whenever two ratios share a quantity, multiply each ratio so the shared quantity carries the same number in both. Then concatenate. The LCM of the two shared-numbers is the cleanest common value to aim for.

Step 3 — Parametrise the shares.

Let the common multiplier be $k$. Then:

• Amal's share $= 6k$
• Sunil's share $= 4k$
• Mita's share $= 5k$

Step 4 — Identify the largest and smallest shares.

Comparing $6k$, $4k$, and $5k$ (with $k > 0$):

• Largest $= 6k$ (Amal)
• Smallest $= 4k$ (Sunil)

Step 5 — Use the given gap to find $k$.

The difference between the largest and smallest share is Rs. 400:

$6k - 4k = 400$

$2k = 400 \implies k = 200$

Step 6 — Compute Sunil's share.

$\text{Sunil's share} = 4k = 4 \times 200 = 800$

Sanity check.

Largest $-$ smallest $= 1200 - 800 = 400$ ✓ — matches the given condition.

Answer: $\boxed{800}$

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Alternative approach — work in fractions of Sunil's share.

Let Sunil's share be $S$. From the two ratios:

• $\dfrac{A}{S} = \dfrac{3}{2} \implies A = \dfrac{3}{2}S$
• $\dfrac{M}{S} = \dfrac{5}{4} \implies M = \dfrac{5}{4}S$

So $A = 1.5S$, $M = 1.25S$, and $S = S$. The largest is Amal and the smallest is Sunil, so:

$A - S = \frac{3}{2}S - S = \frac{1}{2}S = 400 \implies S = 800.$

This skips the combined ratio entirely and is arguably faster under exam time — but the LCM method is more systematic and generalises better when the ratios get messier.

💡 Closing insight: The combined ratio $6:4:5$ is almost arithmetic — the gaps are $2$ and $1$. The largest-minus-smallest gap of $2k$ is exactly twice Sunil's distance from Amal, which is why Sunil's own share ($4k$) lands neatly at $800$.

Think of this as: a chord of the bigger circle just grazes the smaller circle. That grazing point fixes the chord's distance from the common centre, and the chord's length then pins down the radii.

Step 1 — Translate the diameter condition into radii.

Let $R$ be the radius of $C_1$ and $r$ be the radius of $C_2$.

The diameter of $C_1$ is $2R$, and the diameter of $C_2$ is $2r$. We are told the diameter of $C_1$ is $2$ cm longer than that of $C_2$: $2R - 2r = 2 \;\Rightarrow\; R - r = 1.$

So the radii differ by $1$ cm.

Step 2 — Use the fact that the chord is tangent to $C_2$.

Because the chord of $C_1$ is tangent to $C_2$, the perpendicular distance from the common centre to that chord equals the radius of $C_2$, namely $r$.

For a circle of radius $R$, a chord at perpendicular distance $d$ from the centre has length $2\sqrt{R^2 - d^2}$. Here $d = r$ and the chord length is $6$: $6 = 2\sqrt{R^2 - r^2} \;\Rightarrow\; \sqrt{R^2 - r^2} = 3 \;\Rightarrow\; R^2 - r^2 = 9.$

💡 Teacher tip: The chord-length formula $2\sqrt{R^2 - d^2}$ is just the Pythagoras theorem in disguise: half the chord, the perpendicular from the centre, and the radius form a right triangle. The tangent condition tells you that perpendicular equals $r$.

Step 3 — Factor the difference of squares.

$R^2 - r^2 = (R - r)(R + r) = 9.$

We already know $R - r = 1$, so: $1 \cdot (R + r) = 9 \;\Rightarrow\; R + r = 9.$

Now solve the two linear equations:

• $R - r = 1$
• $R + r = 9$

Adding them: $2R = 10 \;\Rightarrow\; R = 5$. Subtracting: $2r = 8 \;\Rightarrow\; r = 4$.

Step 4 — Find the diameter of $C_1$.

$\text{Diameter of } C_1 = 2R = 2 \times 5 = 10 \text{ cm}.$

Sanity check.

Answer: $\boxed{10 \text{ cm}}$

💡 Why this works out so cleanly: the diameter difference gives the small, clean value $R - r = 1$, which turns the difference of squares $R^2 - r^2$ into a single unknown $R + r$. Two linear equations then solve the system instantly — no quadratic needed.

Framing. Two runners start together from opposite ends of a path, meet somewhere in between, and then each one still has to cover the distance the other runner already covered before the meeting. The post-meeting times (1 min and 4 min) secretly encode the speed ratio — let's see exactly how.

Step 1 — Set up variables and locate the meeting point.

Let Ram's speed be $v_R$ and Rahim's speed be $v_H$. Suppose they meet at point $M$, $T$ minutes after they both start.

Before the meeting:

• Ram has covered $AM = v_R T$,
• Rahim has covered $BM = v_H T$.

💡 Teacher tip: The crucial insight is what happens after the meeting. Once they cross at $M$, Ram must finish the remaining part $MB$ — which is exactly the distance Rahim already ran before meeting. Likewise, Rahim must finish $MA$ — exactly the distance Ram already ran. Each runner "inherits" the other's pre-meeting distance.

Step 2 — Write the post-meeting time equations.

After crossing at $M$:

• Ram runs $MB = v_H T$ at speed $v_R$, taking $\dfrac{v_H T}{v_R} = 1$ minute.
• Rahim runs $MA = v_R T$ at speed $v_H$, taking $\dfrac{v_R T}{v_H} = 4$ minutes.

Step 3 — Divide the two equations to eliminate the unknown meeting time $T$.

$$\frac{\;\dfrac{v_R T}{v_H}\;}{\;\dfrac{v_H T}{v_R}\;} = \frac{4}{1} \;\Longrightarrow\; \frac{v_R T}{v_H} \cdot \frac{v_R}{v_H T} = 4 \;\Longrightarrow\; \left(\frac{v_R}{v_H}\right)^2 = 4.$$

Step 4 — Solve for the speed ratio.

$$\frac{v_R}{v_H} = \sqrt{4} = 2.$$

So Ram is twice as fast as Rahim.

Sanity check — verify the meeting time is consistent.

From $\dfrac{v_R T}{v_H} = 4$ and $\dfrac{v_R}{v_H} = 2$, we get $2T = 4$, so $T = 2$ minutes. Plug into the other equation:

Both post-meeting times match the given data, so the ratio is consistent.

💡 General takeaway: For two bodies starting together from opposite ends, meeting, and then taking $t_1$ and $t_2$ to finish, the speed ratio is $\dfrac{v_1}{v_2} = \sqrt{\dfrac{t_2}{t_1}}$. The square root appears because each runner's remaining distance is proportional to the other's speed — so the speed ratio enters twice.

Answer: $\boxed{2}$ — Choice C

Think of this as: we need 4-digit numbers with all distinct digits that contain both 7 and 3, with 7 sitting somewhere to the left of 3. The key tasks are (a) placing 7 and 3 in two of the four positions, and (b) filling the remaining two slots — while remembering the first digit can't be 0.

Step 1 — Choose positions for 7 and 3 (with 7 to the left).

There are 4 positions in the number. We pick 2 of them for the digits 7 and 3; since 7 must come before 3, once the two positions are chosen their assignment is fixed (7 goes in the left one, 3 in the right one).

$\text{Ways to choose 2 positions from 4} = \binom{4}{2} = 6.$

The six possible placements of $(7, 3)$ are: $(1,2),\;(1,3),\;(1,4),\;(2,3),\;(2,4),\;(3,4)$, where each pair lists the position of 7 first and the position of 3 second.

Step 2 — Fill the remaining two positions.

The other two digits are drawn from $\{0, 1, 2, 4, 5, 6, 8, 9\}$ — 8 digits, all distinct from 7 and 3. Order matters (different digits in different slots give different numbers), and the two digits must be distinct.

Here is the crucial subtlety: the first digit cannot be 0. This restriction only matters when position 1 is not already occupied by 7.

Why the counts differ:

• When 7 is in position 1, both remaining slots are free to take any of the 8 available digits (0 included), so we get $8 \times 7 = 56$.
• When 7 is not in position 1, position 1 must be filled from $\{1,2,4,5,6,8,9\}$ — only 7 choices (0 is excluded). The last remaining slot then takes any of the 7 digits left over, giving $7 \times 7 = 49$.

💡 Teacher tip: The leading-zero rule is the single most common trap in digit-counting problems. Always check: is position 1 already guaranteed to be non-zero by the digits you've placed? If yes, the restriction is already satisfied; if no, you must enforce it manually.

Step 3 — Add up.

$\text{Total} = 3 \times 56 + 3 \times 49 = 168 + 147 = 315.$

Answer: $\boxed{315}$

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Alternative approach — symmetry (faster under exam time).

Instead of splitting on whether 7 is in position 1, count all 4-digit numbers with distinct digits that contain both 7 and 3, then divide by 2.

Why divide by 2? For every valid number with 7 before 3, swapping the positions of 7 and 3 gives a valid number with 3 before 7 — and vice versa. Since neither 7 nor 3 is 0, the swap never creates a leading-zero problem, so the pairing is perfect: exactly half the numbers have 7 before 3.

Counting all such numbers (both orders of 7 and 3):

• 7 in position 1: Choose 1 of 3 remaining spots for 3, then fill 2 slots from 8 digits: $3 \times 8 \times 7 = 168$.
• 3 in position 1: Same structure: $3 \times 8 \times 7 = 168$.
• Neither 7 nor 3 in position 1: Position 1 has 7 choices ($\{1,2,4,5,6,8,9\}$). Choose 2 of the 3 remaining positions for 7 and 3 and arrange them: $\binom{3}{2}\times 2! = 6$. Fill the last slot from 7 remaining digits: $7 \times 6 \times 7 = 294$.

$\text{Total} = 168 + 168 + 294 = 630.$

$\text{With 7 before 3} = \frac{630}{2} = 315.$

Both methods agree, confirming the answer.

💡 Closing insight: The symmetry method works here because swapping 7 and 3 is a safe operation — it never touches the leading digit's zero-restriction. Whenever you spot such a clean pairing, dividing by 2 can save you a case split.

Think of this as: two people spend the same money on slightly different bundles — one has twice the pencils and 10 fewer sharpeners. That single equality will pin down a relation between the pencil's price and how many pencils Aron bought.

What we want: the minimum number of pencils Aron and Aditya bought together. If Aron bought $p$ pencils, Aditya bought $2p$, so together they bought $p + 2p = 3p$ pencils. Our job is to minimise $3p$, i.e. minimise $p$.

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Step 1 — Set up variables.

Let the cost of one pencil be $c$ (in ₹). Since a sharpener costs ₹ 2 more: $\text{Sharpener cost} = c + 2.$

Let Aron buy $p$ pencils and $s$ sharpeners. Then Aditya, spending the same amount, buys $2p$ pencils and $s - 10$ sharpeners.

Step 2 — Equate the two spends.

Aron's spend: $pc + s(c+2)$.

Aditya's spend: $2pc + (s-10)(c+2)$.

Since both spent the same amount: $pc + s(c+2) = 2pc + (s-10)(c+2).$

Step 3 — Expand and simplify.

Expand each side:

• Left: $pc + sc + 2s$.
• Right: $2pc + sc + 2s - 10c - 20$.

Equating: $pc + sc + 2s = 2pc + sc + 2s - 10c - 20.$

Cancel $sc + 2s$ from both sides: $pc = 2pc - 10c - 20.$

Rearrange: $0 = pc - 10c - 20 = c(p - 10) - 20,$ $\boxed{c(p - 10) = 20.}$

💡 Teacher tip: Notice that every term containing $s$ — the number of sharpeners — cancelled out completely. That's the beauty of the setup: the spend-equality leaves a clean relation purely between the pencil's price $c$ and Aron's pencil count $p$.

Step 4 — Find the valid values of $p$.

Since $c > 0$ (a cost must be positive) and the product $c(p-10) = 20 > 0$, we need $p - 10 > 0$, i.e. $p > 10$.

For $p$ to be a positive integer, $p - 10$ must be a positive integer that divides 20:

All six values of $p$ are valid (each gives a positive cost $c$ and allows $s \ge 10$ sharpeners).

Step 5 — Minimise the total pencils.

We want the minimum of $3p$. The smallest $p$ in the table is $p = 11$: $3p = 3 \times 11 = 33.$

Sanity check. With $p = 11$ and $c = 20$: Aron spends $11 \times 20 + s \times 22 = 220 + 22s$. Aditya spends $22 \times 20 + (s-10) \times 22 = 440 + 22s - 220 = 220 + 22s$. ✅ The spends match for any $s \ge 10$.

Answer: $\boxed{33}$ (Choice A)

💡 Closing insight: The sharpener count $s$ is a free variable — it cancels out of the spend equation, so it doesn't constrain $p$ at all. The entire problem reduces to the single Diophantine relation $c(p-10)=20$, and minimising $p$ means taking the smallest positive divisor of 20 for $p-10$, which is 1.

Think of this as: we're told the sum of three perpendicular distances from an interior point equals $s$; we need to connect that sum to the triangle's altitude, and then to its area.

What we want: the area of an equilateral triangle in which the sum of the three perpendicular distances from any interior point to the sides equals $s$.

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Step 1 — Connect the sum of perpendiculars to the altitude.

This is the key idea. In an equilateral triangle, the sum of the perpendicular distances from any interior point to the three sides is constant — it always equals the altitude $h$, no matter where the point sits.

💡 Teacher tip: This result is called Viviani's theorem. The cleanest way to see why it's true is to think in terms of area — we prove it from scratch in the Alternative approach below.

So the given sum $s$ is actually the altitude: $h = s.$

Step 2 — Find the side length from the altitude.

In an equilateral triangle with side $a$ and altitude $h$: $h = \frac{\sqrt{3}}{2}\,a \;\Rightarrow\; a = \frac{2h}{\sqrt{3}} = \frac{2s}{\sqrt{3}}.$

Step 3 — Compute the area.

Using the equilateral-triangle area formula: $\text{Area} = \frac{\sqrt{3}}{4}\,a^2 = \frac{\sqrt{3}}{4}\left(\frac{2s}{\sqrt{3}}\right)^2 = \frac{\sqrt{3}}{4}\cdot\frac{4s^2}{3} = \frac{s^2}{\sqrt{3}}.$

Cross-check using base × height. Since we already know the altitude $h = s$ and the side $a = \frac{2s}{\sqrt{3}}$: $\text{Area} = \frac{1}{2}\times\text{base}\times\text{height} = \frac{1}{2}\cdot\frac{2s}{\sqrt{3}}\cdot s = \frac{s^2}{\sqrt{3}}.\quad\checkmark$

Both formulas agree, which confirms the result.

Answer: $\boxed{\dfrac{s^2}{\sqrt{3}}}$ (Choice D)

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Alternative approach — Derive Viviani's theorem from area (first principles).

If you don't remember Viviani's theorem, you can derive the key fact on the spot.

Let the equilateral triangle have side $a$ and altitude $h$. Pick any interior point $P$ and drop perpendiculars of lengths $d_1, d_2, d_3$ to the three sides.

Joining $P$ to the three vertices divides the original triangle into three smaller triangles. Each smaller triangle has:

• base $= a$ (one full side of the original triangle),
• height $=$ the perpendicular from $P$ to that side ($d_1$, $d_2$, or $d_3$).

So the total area is: $\text{Area}_{\text{original}} = \frac{1}{2}a\,d_1 + \frac{1}{2}a\,d_2 + \frac{1}{2}a\,d_3 = \frac{1}{2}a(d_1+d_2+d_3).$

But the original area is also $\frac{1}{2}a\,h$. Equating the two expressions: $\frac{1}{2}a(d_1+d_2+d_3) = \frac{1}{2}a\,h \;\Rightarrow\; d_1+d_2+d_3 = h.$

So the sum of the perpendiculars is always the altitude, regardless of where $P$ sits. With $d_1+d_2+d_3 = s$, we again get $h = s$, and the rest follows exactly as in Steps 2–3.

💡 Why this is elegant: the position of the interior point never matters — the sum is "locked" to the altitude by a simple area argument. Once you spot that, the problem reduces to a one-line altitude-to-area conversion.

Think of this as: we know the sum of the lowest 9 and the sum of the highest 9. That pins down the gap between the top and bottom scores — and then we squeeze the middle 8 scores to either extreme to find the widest possible range for the overall mean.

Step 1 — Translate the two means into two equations.

Sort the ten scores as $s_1 \le s_2 \le \cdots \le s_{10}$, and let the total be $S = s_1 + s_2 + \cdots + s_{10}$.

• The lowest 9 are $s_1, s_2, \ldots, s_9$. Their mean is 42, so their sum is $9 \times 42 = 378$: S - s_{10} = 378. \tag{1}

• The highest 9 are $s_2, s_3, \ldots, s_{10}$. Their mean is 47, so their sum is $9 \times 47 = 423$: S - s_1 = 423. \tag{2}

Step 2 — Find the gap between the top and bottom scores.

Subtract equation (1) from equation (2): $(S - s_1) - (S - s_{10}) = 423 - 378$ $s_{10} - s_1 = 45.$

So the highest score always exceeds the lowest by exactly 45. Also, from (1): $S = 378 + s_{10}$, so maximising the mean means maximising $s_{10}$, and minimising the mean means minimising $s_1$.

💡 Teacher tip: The total $S$ is $378 + s_{10}$, so it grows with $s_{10}$ and shrinks with $s_1$. Our job reduces to: how large can $s_{10}$ get? and how small can $s_1$ get? — all while keeping the scores sorted.

Step 3 — Maximise the mean (push $s_{10}$ as high as possible).

Since $s_1 = s_{10} - 45$, the sum of the middle eight scores is: $s_2 + s_3 + \cdots + s_9 = 378 - s_1 = 378 - (s_{10} - 45) = 423 - s_{10}.$

These eight scores must satisfy $s_1 \le s_2 \le \cdots \le s_9 \le s_{10}$. To make $s_{10}$ as large as possible, we need the middle eight to be as small as possible — their minimum allowed value is $s_1$. Setting $s_2 = s_3 = \cdots = s_9 = s_1$: $8\,s_1 = 423 - s_{10}$ $8(s_{10} - 45) = 423 - s_{10}$ $8s_{10} - 360 = 423 - s_{10}$ $9s_{10} = 783 \implies s_{10} = 87.$

Then $s_1 = 87 - 45 = 42$, and all of $s_1$ through $s_9$ equal 42. Check: $s_9 = 42 \le s_{10} = 87$ ✓

$S_{\max} = 378 + 87 = 465, \qquad \text{mean}_{\max} = \frac{465}{10} = 46.5.$

Step 4 — Minimise the mean (push $s_1$ as low as possible).

By the same logic, $s_{10} = s_1 + 45$, and the middle eight sum to $378 - s_1$. To make $s_1$ as small as possible, we push the middle eight as high as possible — their maximum allowed value is $s_{10} = s_1 + 45$. Setting $s_2 = s_3 = \cdots = s_9 = s_{10}$: $8\,s_{10} = 378 - s_1$ $8(s_1 + 45) = 378 - s_1$ $8s_1 + 360 = 378 - s_1$ $9s_1 = 18 \implies s_1 = 2.$

Then $s_{10} = 2 + 45 = 47$, and all of $s_2$ through $s_9$ equal 47. Check: $s_1 = 2 \le s_2 = 47$ ✓

$S_{\min} = 423 + 2 = 425, \qquad \text{mean}_{\min} = \frac{425}{10} = 42.5.$

Step 5 — Take the difference.

$\text{Difference} = 46.5 - 42.5 = 4.$

Answer: $\boxed{4}$ (Choice B)

💡 Why extremes come from "clustering": to maximise the top score, cram the other nine at the bottom allowed value; to minimise the bottom score, cram the other nine at the top allowed value. The 45-point gap is fixed — the only freedom is where the middle eight sit, and that freedom spans exactly 4 points of mean.

We need to count the number of integer pairs $(x, a)$ satisfying: $x^2 - 2|x| + |a - 2| = 0$

Think of this as: the equation links $x$ and $a$ only through their absolute values, so we can isolate one absolute-value term and then enumerate.

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Step 1 — Isolate $|a - 2|$.

Rearrange the equation: $|a - 2| = 2|x| - x^2$

Now factor the right side. Since $x^2 = |x|^2$: $|a - 2| = |x|^2 - 2|x| \cdot (-1) = |x|\bigl(2 - |x|\bigr)$

Wait — let's be careful: $2|x| - x^2 = 2|x| - |x|^2 = |x|(2 - |x|)$. Yes.

Step 2 — Use the non-negativity of absolute value.

Since $|a - 2| \ge 0$, the right side must also be $\ge 0$: $|x|(2 - |x|) \ge 0$

Because $|x| \ge 0$ always, we need: $2 - |x| \ge 0 \quad\Longrightarrow\quad |x| \le 2$

💡 Teacher tip: An equation of the form $|\text{something}| = f(x)$ immediately tells you $f(x) \ge 0$. That constraint often shrinks the search space dramatically — here it limits $x$ to just five integer values.

Step 3 — Enumerate every integer $x$ with $|x| \le 2$.

$x \in \{-2,\ -1,\ 0,\ 1,\ 2\}$

Step 4 — For each $x$, solve for $a$.

Recall: if $|a - 2| = c$ with $c > 0$, then $a = 2 \pm c$ (two solutions). If $c = 0$, then $a = 2$ (one solution).

Step 5 — Add up all the pairs.

$1 + 2 + 2 + 1 + 1 = 7$

Each row gives distinct pairs because $x$ differs (or, when $x$ is the same, $a$ differs). So there is no double-counting.

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Answer: $\boxed{7}$ — Choice D

💡 Why $|a-2| = 0$ gives only one value of $a$: when the right side equals $0$, the equation $|a - 2| = 0$ forces $a - 2 = 0$, i.e., $a = 2$ — a single solution. For any positive right side $c$, the equation $|a - 2| = c$ yields two solutions: $a = 2 + c$ and $a = 2 - c$. This one-vs-two split is what makes the count $1 + 2 + 2 + 1 + 1 = 7$ rather than a uniform $2 \times 5 = 10$.

Think of this as: John buys the same quantities in May as April, but each item's price went up. So his spend on each item simply rises by the same percentage as its price. We need to find May's wheat spend.

Step 1 — Set up April and May spends.

Let April's spend on rice be $R$ and on wheat be $W$.

We are given $R = ₹450$.

Since John buys the same amount of each grain in May:

• May rice spend $= 1.20 \times R$ (a 20% price rise means a 20% spend rise)
• May wheat spend $= 1.12 \times W$ (a 12% price rise means a 12% spend rise)

💡 Teacher tip: When the quantity bought is unchanged, a $p\%$ price increase translates directly into a $p\%$ spend increase on that item. No need to track units, kilograms, or per-kg rates — the spend ratio equals the price ratio.

Step 2 — Use the "₹150 more in May" condition.

The extra amount spent in May is the difference between May's total and April's total: $(1.20R + 1.12W) - (R + W) = 150$ $0.20R + 0.12W = 150$

This makes sense: the extra spend is just 20% of the rice spend plus 12% of the wheat spend.

Step 3 — Substitute $R = 450$ and solve for $W$. $0.20(450) + 0.12W = 150$ $90 + 0.12W = 150$ $0.12W = 60$ $W = \frac{60}{0.12} = 500$

So April's wheat spend was ₹500.

Step 4 — Find May's wheat spend. $\text{May wheat} = 1.12 \times 500 = ₹560$

Step 5 — Sanity check.

The total increase is ₹150, matching the given condition.

Answer: $\boxed{\text{Rs. } 560}$ — Choice A

💡 Takeaway: Whenever a ratio and a total are both given, anchor on the total first. Here, "same quantity bought" is the key phrase — it lets you convert a price percentage change directly into a spend percentage change, turning a word problem into a single linear equation.

Think of this as: we have one linear equation in two variables, so there are infinitely many real solutions — but the integer constraint plus the inequality chain $x \ge y \ge -20$ will pin down a finite range of $y$ values, each giving exactly one $x$.

Step 1 — Express $x$ in terms of $y$ and find the parity constraint.

From $2x + 5y = 99$: $x = \frac{99 - 5y}{2}.$

For $x$ to be an integer, $99 - 5y$ must be even. Since $99$ is odd, $5y$ must be odd, which forces $y$ to be odd.

💡 Teacher tip: Whenever a linear Diophantine equation has a coefficient of $2$ on one variable, solving for that variable immediately reveals a parity condition on the other — that single observation cuts the candidate set in half.

Step 2 — Translate $x \ge y$ into a bound on $y$.

Substitute the expression for $x$: $\frac{99 - 5y}{2} \ge y \;\Rightarrow\; 99 - 5y \ge 2y \;\Rightarrow\; 99 \ge 7y \;\Rightarrow\; y \le \frac{99}{7} \approx 14.14.$

Since $y$ is an integer, $y \le 14$.

Step 3 — Combine with $y \ge -20$ and the oddness condition.

We need $y$ to be an odd integer with $-20 \le y \le 14$.

• The largest odd integer $\le 14$ is $\mathbf{13}$.
• The smallest odd integer $\ge -20$ is $\mathbf{-19}$.

So $y \in \{-19, -17, -15, \ldots, 11, 13\}$.

Step 4 — Count the valid values.

The number of odd integers from $-19$ to $13$ inclusive: $\frac{13 - (-19)}{2} + 1 = \frac{32}{2} + 1 = 16 + 1 = 17.$

Each value of $y$ produces exactly one integer $x$, so there are $17$ pairs.

Step 5 — Verify the endpoints.

Both boundary values work, confirming the range is correct.

Answer: $\boxed{17}$

Alternative approach — solve for $y$ instead.

From $5y = 99 - 2x$ we get $y = \frac{99 - 2x}{5}$. For $y$ to be an integer, $99 - 2x \equiv 0 \pmod{5}$, i.e. $2x \equiv 4 \pmod{5}$, so $x \equiv 2 \pmod{5}$.

Applying $y \ge -20$: $\frac{99-2x}{5} \ge -20 \Rightarrow x \le 99$. Applying $x \ge y$: $x \ge \frac{99-2x}{5} \Rightarrow x \ge \frac{99}{7} \approx 14.14$, so $x \ge 15$.

So $x \in \{17, 22, 27, \ldots, 97\}$ (multiples offset by $2$ mod $5$), giving $\frac{97-17}{5}+1 = 17$ values — the same answer. Both methods are equally fast; pick whichever variable gives the cleaner parity/modular condition.

💡 Closing insight: The answer $17$ arises because the inequality $x \ge y$ sets an upper bound on $y$ at $14$, while $y \ge -20$ sets the lower bound — and the odd-integer spacing of $2$ means the count is simply $\frac{\text{range}}{2}+1$.

Think of this as: we know the actual revenue (with discounts) and the actual profit %. From that we can recover both the labelled price and the cost price — then ask what profit % would have been with no discounts at all.

Step 1 — Set up variables. Let the labelled price of each toy be $L$ and the cost price of each toy be $C$.

Step 2 — Compute the actual revenue in terms of $L$.

• First 8 toys: sold at a 20% discount on the labelled price.

• Selling price each $= L - 0.20L = 0.8L$

• Revenue $= 8 \times 0.8L = 6.4L$

• Remaining 4 toys: sold at an additional 25% discount on the already-discounted price.

• Selling price each $= 0.8L \times (1 - 0.25) = 0.8L \times 0.75 = 0.6L$

• Revenue $= 4 \times 0.6L = 2.4L$

💡 Teacher tip: "Additional 25% discount on the discounted price" means successive discounts — they multiply, not add. So it is $0.8L \times 0.75 = 0.6L$, not $(0.8 - 0.25)L = 0.55L$. This is the most common slip in this question.

• Total revenue $= 6.4L + 2.4L = 8.8L$

We are told this equals Rs 2112: $8.8L = 2112 \;\Rightarrow\; L = \frac{2112}{8.8} = 240.$

So the labelled price is Rs 240 per toy.

Step 3 — Find the cost price using the 10% profit.

Total cost of 12 toys $= 12C$. The actual revenue (Rs 2112) gives a 10% profit, so: $\text{Revenue} = 1.10 \times \text{Total cost}$ $2112 = 1.10 \times 12C \;\Rightarrow\; 12C = \frac{2112}{1.10} = 1920$ $\Rightarrow\; C = \frac{1920}{12} = 160.$

So the cost price is Rs 160 per toy, and total cost $= 12 \times 160 = \text{Rs }1920$.

Step 4 — Compute the no-discount scenario.

If Anil had sold all 12 toys at the labelled price (no discounts):

• Revenue $= 12 \times L = 12 \times 240 = \text{Rs }2880$
• Cost $= \text{Rs }1920$
• Profit $= 2880 - 1920 = \text{Rs }960$

$\text{Profit \%} = \frac{960}{1920} \times 100 = 50\%.$

Quick check of the numbers:

Answer: $\boxed{50\%}$ (Choice A)

💡 Closing insight: The labelled price (240) is exactly 1.5 times the cost price (160) — so selling at the label with no discount naturally gives a 50% profit. The discounts in the actual scenario simply eroded that 50% down to 10%.

Think of this as: two runners start together at the only point where their tracks meet. They can only re-meet at that same point — so we need the first time both have completed a whole number of laps simultaneously.

Step 1 — Why the meeting must be at A.

The two circular tracks touch at exactly one point, $A$. Everywhere else on $T_1$, Ram is physically on a different circle from Rahim — there is no shared point. So the only place they can ever meet again is $A$.

💡 Teacher tip: This is the key insight. If the tracks overlapped or crossed elsewhere, we'd have to consider other meeting points. Here, the geometry forces the meeting to be at $A$.

Step 2 — Find each runner's lap time.

Speed conversions: $15$ km/hr $= \dfrac{15000}{60}=250$ m/min, and $5$ km/hr $= \dfrac{5000}{60}=\dfrac{250}{3}$ m/min.

Step 3 — Find the first time both are back at A together.

Ram returns to $A$ every $\dfrac{4\pi}{5}$ minutes (after $1, 2, 3, \ldots$ laps). Rahim returns to $A$ every $\dfrac{12\pi}{25}$ minutes (after $1, 2, 3, \ldots$ laps).

They meet at $A$ when these times coincide: $k_1 \cdot \frac{4\pi}{5} = k_2 \cdot \frac{12\pi}{25}$ where $k_1, k_2$ are positive integers (number of laps completed).

Cancel $\pi$ and cross-multiply: $\frac{4k_1}{5} = \frac{12k_2}{25} \;\Rightarrow\; 100\,k_1 = 60\,k_2 \;\Rightarrow\; 5\,k_1 = 3\,k_2.$

Step 4 — Smallest positive solution.

From $5k_1 = 3k_2$: since $\gcd(5,3)=1$, the smallest positive integers satisfying this are $k_1 = 3$ and $k_2 = 5$.

So Ram completes 3 full rounds and Rahim completes 5 full rounds before they next meet at $A$.

Verification:

Both times match — confirmed. ✅

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

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Alternative approach — ratio of rounds per unit time (faster).

Instead of computing lap times, compute each runner's rounds per minute:

• Ram: $\dfrac{250}{200\pi} = \dfrac{5}{4\pi}$ rounds/min
• Rahim: $\dfrac{250/3}{40\pi} = \dfrac{25}{12\pi}$ rounds/min

Their ratio is $\dfrac{5}{4\pi}: \dfrac{25}{12\pi} = \dfrac{5}{4}: \dfrac{25}{12} = 15: 25 = 3: 5$.

So in any common time interval, Ram completes $3$ rounds for every $5$ Rahim completes. Since $3$ and $5$ are coprime, the first time both have completed whole rounds is when Ram has done $3$ and Rahim has done $5$.

💡 Why this works: The ratio of rounds-per-unit-time directly gives the ratio of laps completed in the same elapsed time. Coprime parts mean the first simultaneous whole-lap moment is exactly those parts.

Closing insight: The $\pi$ cancels out entirely — the answer depends only on the ratio of speed-to-circumference for each runner, which simplifies to $3:5$. Whenever two objects loop with coprime round-ratios, their first meeting after start occurs at exactly those counts.

Think of this as: two specific terms of a GP are given, and the common ratio must be an integer. We need to find which integer ratio, together with the gap between the two term positions, gives the smallest value of $r + (n-m)$.

Step 1 — Set up the ratio of the two given terms.

The $k$-th term of a GP with first term $a$ and common ratio $r$ is $T_k = a\,r^{k-1}$. So:

$T_m = a\,r^{m-1} = \frac{3}{4}, \qquad T_n = a\,r^{n-1} = 12.$

Dividing $T_n$ by $T_m$ eliminates $a$:

$\frac{T_n}{T_m} = \frac{a\,r^{n-1}}{a\,r^{m-1}} = r^{(n-1)-(m-1)} = r^{\,n-m}.$

So we get the key equation:

$r^{\,n-m} = \frac{12}{3/4} = 16.$

💡 Teacher tip: Whenever two terms of a GP are given, always take their ratio — the first term cancels out and you're left with a pure relationship between the common ratio and the gap $n-m$.

Step 2 — Find all integer solutions of $r^{\,n-m} = 16$ with $n > m$.

Since $n > m$, the exponent $d = n - m$ is a positive integer ($d \ge 1$). We need every pair $(r, d)$ with $r \in \mathbb{Z}$, $d \ge 1$, and $r^d = 16$.

The factorisations of $16$ as an integer power are:

• $16 = 2^4$, so $r = 2,\ d = 4$.
• $16 = (-2)^4$ (even exponent → negative base works), so $r = -2,\ d = 4$.
• $16 = 4^2$, so $r = 4,\ d = 2$.
• $16 = (-4)^2$ (even exponent again), so $r = -4,\ d = 2$.
• $16 = 16^1$, so $r = 16,\ d = 1$.

💡 Teacher tip: The easy-to-miss cases are the negative bases. When the exponent is even, a negative $r$ still produces a positive result. Skipping $r = -4$ here would cost you the correct answer.

Step 3 — Compute $r + (n-m)$ for each candidate and pick the smallest.

The smallest value is $\mathbf{-2}$, achieved at $r = -4$ and $n - m = 2$.

Step 4 — Sanity check: a valid GP actually exists.

Take $m = 1$, so the first term is $a = \tfrac{3}{4}$. Then $n = m + 2 = 3$, and:

$T_3 = \frac{3}{4} \cdot (-4)^{2} = \frac{3}{4} \cdot 16 = 12. \quad\checkmark$

Both given terms are satisfied, so the case $r = -4,\ n-m = 2$ is genuinely realisable.

Answer: $\boxed{-2}$ (Choice D)

💡 Why negative $r$ matters: the winning case comes from a negative common ratio with an even exponent — $(-4)^2 = 16$. Always list negative integer roots alongside positive ones whenever the target value is positive and the exponent could be even.