CAT 2018 Quant Questions — Slot 2

We're given the running total of a sequence and asked to find which term equals 103. The key idea: if you know the partial sums, you can recover any individual term by subtraction.

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Step 1 — Set up notation

Let $S_n = t_1 + t_2 + \cdots + t_n = 2n^2 + 9n + 13$, valid for every $n \ge 2$.

Step 2 — Recover a single term from two partial sums

The $k$-th term is the difference between the sum of the first $k$ terms and the sum of the first $k-1$ terms:

$t_k = S_k - S_{k-1}$

This is valid whenever both $S_k$ and $S_{k-1}$ are defined by the given formula — i.e., when $k - 1 \ge 2$, or $k \ge 3$. (We'll check at the end that our answer satisfies this.)

Step 3 — Compute $S_{k-1}$

Substitute $n = k - 1$ into the formula:

$S_{k-1} = 2(k-1)^2 + 9(k-1) + 13$

Expand each piece:

• $(k-1)^2 = k^2 - 2k + 1$, so $2(k-1)^2 = 2k^2 - 4k + 2$
• $9(k-1) = 9k - 9$

Putting it together:

$S_{k-1} = (2k^2 - 4k + 2) + (9k - 9) + 13 = 2k^2 + 5k + 6$

Step 4 — Subtract to get $t_k$

$t_k = S_k - S_{k-1} = (2k^2 + 9k + 13) - (2k^2 + 5k + 6) = 4k + 7$

The $k^2$ terms cancel, leaving a clean linear expression — as expected when the partial sum is quadratic.

💡 Teacher tip: Whenever a sequence is defined through its partial sums, the $k$-th term is simply $S_k - S_{k-1}$. A quadratic $S_n$ always produces a linear $t_n$, because the $n^2$ terms cancel in the difference.

Step 5 — Solve for $k$

$4k + 7 = 103 \implies 4k = 96 \implies k = 24$

Since $24 \ge 3$, our formula $t_k = S_k - S_{k-1}$ applies. ✅

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Verification

where $S_{23} = 2(529) + 207 + 13 = 1278$.

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Alternative approach — Spot the arithmetic progression

Since $S_n$ is quadratic in $n$, the sequence $\{t_n\}$ (for $n \ge 3$) must be an arithmetic progression. We can confirm this by computing two consecutive terms directly:

• $S_2 = 2(4) + 18 + 13 = 39$
• $S_3 = 2(9) + 27 + 13 = 58$
• $S_4 = 2(16) + 36 + 13 = 81$

Then:

• $t_3 = S_3 - S_2 = 58 - 39 = 19$
• $t_4 = S_4 - S_3 = 81 - 58 = 23$

Common difference: $d = 23 - 19 = 4$.

So for $n \ge 3$:

$t_n = t_3 + (n - 3) \cdot 4 = 19 + 4n - 12 = 4n + 7$

Setting $4k + 7 = 103$ gives $k = 24$, confirming our answer.

This route is faster if you immediately recognise that a quadratic partial sum implies an arithmetic sequence — you can skip the algebra and just compute two terms to pin down the common difference.

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Answer: $\boxed{24}$

💡 Closing insight: The constant term $13$ in $S_n$ cancels completely when we take the difference $S_k - S_{k-1}$ — it shifts every partial sum by the same amount but has no effect on individual terms. Only the $n^2$ and $n$ coefficients shape the sequence itself.

Think of this as: three cars start at the same instant on a straight road — two from A heading toward B, one from B heading toward A. We're told where each pair meets, and we need to convert those meeting points into a speed ratio.

Step 1 — Lay out the geometry.

All distances are measured from A:

So the gaps on the road are:

• $AQ = 200$ km and $BQ = 300 - 200 = 100$ km.
• $AP = 100$ km and $BP = 300 - 100 = 200$ km.

All three cars start simultaneously. Cars 1 and 2 head from A toward B; car 3 heads from B toward A.

Step 2 — Use the meeting of car 3 and car 1 at Q.

When car 3 and car 1 meet at Q, they have been travelling for the same amount of time. In that time:

• Car 1 (starting at A) has covered $AQ = 200$ km.
• Car 3 (starting at B) has covered $BQ = 100$ km.

Since time = distance / speed, and the times are equal:

$\frac{200}{v_1} = \frac{100}{v_3}$

$\Rightarrow v_1 = 2\,v_3$

Step 3 — Use the meeting of car 3 and car 2 at P.

When car 3 and car 2 meet at P, again the travel times are equal. In that time:

• Car 2 (starting at A) has covered $AP = 100$ km.
• Car 3 (starting at B) has covered $BP = 200$ km.

$\frac{100}{v_2} = \frac{200}{v_3}$

$\Rightarrow v_3 = 2\,v_2 \quad\Rightarrow\quad v_2 = \frac{v_3}{2}$

💡 Teacher tip: The key phrase is "meets at the same instant." Whenever two objects start simultaneously and meet, their travel times are equal — so the ratio of distances covered is the ratio of their speeds. Set the two distance/speed expressions equal and the unknown time cancels automatically.

Step 4 — Combine to get the required ratio.

From Step 2: $v_1 = 2v_3$. From Step 3: $v_2 = \dfrac{v_3}{2}$.

$\frac{v_2}{v_1} = \frac{v_3 / 2}{2v_3} = \frac{1}{4}$

So the ratio of the speed of car 2 to that of car 1 is $1: 4$.

Quick sanity check. Car 1 is faster than car 3 ($v_1 = 2v_3$), so it pushes deep toward B and meets car 3 at Q — close to B. Car 2 is slower than car 3 ($v_2 = v_3/2$), so it barely gets away from A before car 3 catches it at P — close to A. The meeting points are consistent with the speeds, so the ratio makes physical sense.

Answer: D — $\boxed{1: 4}$

Alternative approach — ratio shortcut. When two objects start at the same time and move toward each other, the ratio of distances covered when they meet equals the ratio of their speeds.

• Car 1: Car 3 at Q → $200: 100 = 2: 1$, so $v_1: v_3 = 2: 1$.
• Car 2: Car 3 at P → $100: 200 = 1: 2$, so $v_2: v_3 = 1: 2$.

Writing all three in a common ratio with $v_3 = 2$: $v_2: v_3: v_1 = 1: 2: 4 \quad\Rightarrow\quad v_2: v_1 = 1: 4.$ This confirms the same result — and is the faster route under exam time.

Framing: Three days give us three clues — each day tells us which pump(s) ran and when the tank got full. The unknowns are the daily empty-time $T$ and each pump's solo fill-time. Wednesday's split-shift is what locks the relationship between A and B.

Step 1 — Translate Monday and Tuesday into fill-times.

Let $T$ be the clock time at which the tank is emptied each day. Let $a$ and $b$ be the number of hours pumps A and B respectively need to fill the tank from empty (working alone).

• Monday: A starts at $T$ and finishes at 8 pm, so $a = 8\text{ pm} - T$.
• Tuesday: B starts at $T$ and finishes at 6 pm, so $b = 6\text{ pm} - T$.

Subtracting: $a - b = 2$ hours. A takes 2 hours longer than B to fill the tank alone.

💡 Teacher tip: We don't yet know $T$, $a$, or $b$ individually — but the difference $a - b = 2$ is already pinned down. That's one equation for free.

Step 2 — Convert Wednesday's schedule into a fraction-of-tank equation.

On Wednesday, A worked from $T$ until 5 pm, then B worked from 5 pm to 7 pm (a 2-hour shift).

How long did A work? Since $a = 8\text{ pm} - T$, the time from $T$ to 5 pm is: $5\text{ pm} - T = (8\text{ pm} - T) - 3 = a - 3 \text{ hours}.$

Now convert each pump's running time into the fraction of the tank it filled:

• A filled $\dfrac{a-3}{a}$ of the tank,
• B filled $\dfrac{2}{b}$ of the tank.

Since the tank ended up full: $\frac{a-3}{a} + \frac{2}{b} = 1.$

Step 3 — Simplify and solve.

$\frac{a-3}{a} + \frac{2}{b} = 1 \;\Rightarrow\; 1 - \frac{3}{a} + \frac{2}{b} = 1 \;\Rightarrow\; \frac{2}{b} = \frac{3}{a}.$

Cross-multiplying: $2a = 3b$, i.e. $b = \dfrac{2a}{3}$.

Now use $a - b = 2$: $a - \frac{2a}{3} = 2 \;\Rightarrow\; \frac{a}{3} = 2 \;\Rightarrow\; a = 6.$

So $b = \dfrac{2 \times 6}{3} = 4$, and $T = 8\text{ pm} - a = 8\text{ pm} - 6\text{ h} = \textbf{2 pm}$.

Step 4 — Verify with Wednesday's data.

The tank is exactly full — our values are correct.

Step 5 — Thursday: both pumps together.

Combined rate: $\frac{1}{a} + \frac{1}{b} = \frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12} \text{ tank/hour}.$

Time to fill: $\frac{1}{5/12} = \frac{12}{5} \text{ hours} = 2 \text{ h } 24 \text{ min}.$

Starting from 2 pm: $2\text{ pm} + 2\text{ h }24\text{ min} = \textbf{4:24 pm}$.

Answer: C — $\boxed{4{:}24\text{ pm}}$

💡 Why this works: The key move is turning clock times into durations, then durations into fractions of the tank. Once Wednesday's split-shift gives us $b = \frac{2a}{3}$, the 2-hour gap between Monday and Tuesday fixes everything — and the rest is a straightforward combined-rate calculation.

Think of this as: a semicircle is scooped out of one side of a rectangle. The new perimeter loses the straight scooped edge but gains the curved arc.

Step 1 — Find the rectangle's dimensions. The area of the removed semicircle is $72\pi$. Using the formula for the area of a semicircle, $\frac{1}{2}\pi r^2$: $\frac{1}{2}\pi r^2 = 72\pi$ $r^2 = 144 \implies r = 12 \text{ cm}$ Since the semicircle has diameter $AB$, we have $AB = 2r = 24$ cm. We are given the area of rectangle $ABCD$ is $768$ sq cm. Thus: $AB \times BC = 768$ $24 \times BC = 768 \implies BC = 32 \text{ cm}$ Since $ABCD$ is a rectangle, $CD = AB = 24$ cm and $DA = BC = 32$ cm.

Step 2 — Build the new perimeter. When the semicircular part is removed, the straight edge $AB$ is taken away from the boundary, and the semicircular arc takes its place. The perimeter of the leftover portion is the sum of:

• $BC$ (32 cm)
• $CD$ (24 cm)
• $DA$ (32 cm)
• The semicircular arc

The length of the semicircular arc is $\pi r = 12\pi$ cm. So, the total perimeter is: $32 + 24 + 32 + 12\pi = 88 + 12\pi \text{ cm}$

💡 Teacher tip: When a piece is removed from a shape, trace the new boundary explicitly: the cut edge (the diameter) disappears from the perimeter, and the newly exposed curved boundary (the arc) takes its place. Don't accidentally include the diameter!

Answer: C — $\boxed{88 + 12\pi}$

💡 Why this works: The key trap here is including the diameter $AB$ in the final perimeter. By carefully tracking which edges are on the outside of the new shape, you avoid adding the extra $24$ cm.

We're given a chord and its central angle, and asked for a different chord in the same circle. So the plan is: first find the radius, then use it to compute the new chord.

Step 1 — Recall the chord-length formula.

A chord that subtends an angle $\theta$ at the centre of a circle of radius $r$ has length:

$\text{Chord} = 2r\sin\!\left(\frac{\theta}{2}\right)$

💡 Teacher tip: This formula comes straight from splitting the isosceles triangle (two radii + chord) down the middle into two right-angled triangles. Memorise it — it turns any "angle at centre → chord length" question into one line.

Step 2 — Find the radius using the first chord.

The first chord has length $5$ cm and subtends $60^\circ$:

$5 = 2r\sin\!\left(\frac{60^\circ}{2}\right) = 2r\sin 30^\circ = 2r \cdot \frac{1}{2} = r$

So the radius is $\boxed{r = 5\text{ cm}}$.

Step 3 — Compute the chord for $120^\circ$.

Now plug $\theta = 120^\circ$ and $r = 5$ into the same formula:

$\text{Chord} = 2(5)\sin\!\left(\frac{120^\circ}{2}\right) = 10\sin 60^\circ = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3}\text{ cm}$

Quick sanity check. The diameter ($10$ cm) subtends $180^\circ$ and is the longest possible chord. Our $120^\circ$ chord should be shorter than the diameter but longer than the $60^\circ$ chord ($5$ cm). Indeed $5 < 5\sqrt{3} \approx 8.66 < 10$. ✓

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Alternative approach — Equilateral triangle + cosine rule (no chord formula needed).

When a chord subtends $60^\circ$ at the centre, the triangle formed by the two radii and the chord is isosceles with the included angle $60^\circ$. A triangle with two equal sides and a $60^\circ$ angle between them is equilateral, so the chord equals the radius: $r = 5$ cm.

For the $120^\circ$ chord, apply the cosine rule to the same isosceles triangle (sides $r, r$, included angle $120^\circ$):

$\text{Chord}^2 = r^2 + r^2 - 2(r)(r)\cos 120^\circ = 25 + 25 - 2(25)\!\left(-\tfrac{1}{2}\right) = 50 + 25 = 75$

$\text{Chord} = \sqrt{75} = 5\sqrt{3}\text{ cm}$

Same answer, no formula required — just the cosine rule and the equilateral-triangle insight.

Answer: A — $\boxed{5\sqrt{3}\text{ cm}}$

💡 Closing insight: Doubling the central angle from $60^\circ$ to $120^\circ$ does not double the chord — it multiplies it by $\sin 60^\circ / \sin 30^\circ = \sqrt{3}$. Chord length grows with the sine of the half-angle, not linearly with the angle.

We need the smallest integer $n$ for which $n^3 - 11n^2 + 32n - 28 > 0$. The plan: factor the cubic, then read off where it is positive.

Step 1 — Find one root by testing small integers.

Try $n = 2$: $2^3 - 11(2^2) + 32(2) - 28 = 8 - 44 + 64 - 28 = 0.$ So $n = 2$ is a root, which means $(n - 2)$ is a factor.

Step 2 — Divide out the factor.

Dividing $n^3 - 11n^2 + 32n - 28$ by $(n - 2)$: $n^3 - 11n^2 + 32n - 28 = (n - 2)(n^2 - 9n + 14).$

Now factor the quadratic: $n^2 - 9n + 14 = (n - 7)(n - 2)$. Putting it together: $n^3 - 11n^2 + 32n - 28 = (n - 2)(n - 7)(n - 2) = (n - 2)^2(n - 7).$

💡 Teacher tip: A cubic with integer coefficients that vanishes at an integer is almost always factorable there. Find one rational root, divide out, and the rest unravels.

Step 3 — Sign analysis of $(n-2)^2(n-7)$.

The factor $(n-2)^2$ is a square, so it is always $\ge 0$ — and it equals $0$ only at $n = 2$. Everywhere else it is strictly positive, so the sign of the whole product is controlled by $(n - 7)$:

• $n < 7$, $n \neq 2$: $(n-2)^2 > 0$ but $(n-7) < 0$, so the product is negative.
• $n = 2$: the product is $0$ (not $> 0$).
• $n = 7$: the product is $0$ (not $> 0$).
• $n > 7$: both factors are positive, so the product is positive. ✅

Step 4 — Verify the boundary values.

The inequality first becomes true at $n = 8$.

Answer: $\boxed{8}$

Why this works cleanly: the double root at $n = 2$ never changes sign (squares don't), so it effectively drops out of the sign analysis. The entire behaviour is governed by the single simple factor $(n - 7)$, and the smallest integer strictly above $7$ is $8$.

Think of this as two successive mixing stages — first a $1:3$ blend, then a $1:1$ blend — and work backwards from the final $31.25\%$ to recover $S$.

What we want: the concentration of solution $S$.

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Step 1 — Set up the first blend.

Let the concentration of $S$ be $s\%$.

The first mix combines $1$ part of $20\%$ ethanol with $3$ parts of $s\%$ ethanol. The concentration of this first mixture is the weighted average:

$$\text{mix}_1 = \frac{1 \cdot 20 + 3 \cdot s}{1 + 3} = \frac{20 + 3s}{4}.$$

Step 2 — Set up the second blend.

This mixture ($\text{mix}_1$) is now combined with an equal volume of $20\%$ ethanol solution.

💡 Teacher tip: "Equal volumes" is the magic phrase — when two solutions are mixed in equal volume, the resulting concentration is simply the average of the two. No weighted-average formula needed.

So the final concentration is:

$$\frac{\text{mix}_1 + 20}{2} = 31.25.$$

Step 3 — Solve for $\text{mix}_1$.

$$\text{mix}_1 + 20 = 62.5 \quad\Rightarrow\quad \text{mix}_1 = 42.5.$$

So the first mixture had concentration $42.5\%$.

Step 4 — Solve for $s$.

Plug back into the expression from Step 1:

$$\frac{20 + 3s}{4} = 42.5$$ $$20 + 3s = 170$$ $$3s = 150 \quad\Rightarrow\quad s = 50.$$

Quick check:

The final concentration matches the given $31.25\%$, so our value of $s$ is correct.

Answer: C — $\boxed{50\%}$

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Alternative approach — Alligation

You can also solve the first blend using alligation instead of the weighted-average formula.

We know the first mixture is $42.5\%$, made from $20\%$ and $s\%$ in the ratio $1:3$.

By alligation, the ratio of quantities is the ratio of the opposite concentration gaps:

$$\frac{\text{quantity of } 20\%}{\text{quantity of } S} = \frac{s - 42.5}{42.5 - 20} = \frac{1}{3}.$$

So:

$$\frac{s - 42.5}{22.5} = \frac{1}{3} \quad\Rightarrow\quad s - 42.5 = 7.5 \quad\Rightarrow\quad s = 50.$$

Same answer, slightly faster once you're fluent with alligation. The weighted-average method is more reliable under exam pressure; alligation is the speed move when the numbers are clean.

💡 Closing insight: Working backwards from the final mixture is the key structural choice here. Once you spot that the second blend is a $1:1$ mix (a plain average), the entire problem collapses to one equation in $s$.

Think of this as: two workers share a job, one slows down mid-way, and we need to back out each person's individual speed from the delay that slowdown caused.

What we want: the time Ganesh alone would take to finish the remaining work after the first 7 days.

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Step 1 — Set up the rates.

Let Ramesh's daily work rate be $r$ and Ganesh's be $g$ (in "work per day").

Together they finish the job in 16 days, so: $r + g = \frac{1}{16} \quad \cdots (1)$

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Step 2 — Model what actually happened over the 17 days.

The job was completed in two phases:

The total work equals 1 (the whole job): $7(r+g) + 10(0.7r + g) = 1$

Expand and simplify: $7r + 7g + 7r + 10g = 1$ $14r + 17g = 1 \quad \cdots (2)$

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Step 3 — Solve the two equations.

From equation (1), multiply through by 16: $16r + 16g = 1$

Subtract equation (2) from this: $(16r + 16g) - (14r + 17g) = 0$ $2r - g = 0 \quad\Rightarrow\quad g = 2r$

So Ganesh is twice as fast as Ramesh. Substitute back into equation (1): $r + 2r = \frac{1}{16} \quad\Rightarrow\quad 3r = \frac{1}{16} \quad\Rightarrow\quad r = \frac{1}{48}$ $g = 2r = \frac{1}{24}$

💡 Teacher tip: Whenever efficiency changes mid-job, split the schedule into constant-rate phases and set total work $= 1$. The rates then fall out of two linear equations — no guessing required.

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Step 4 — Find the remaining work and Ganesh's solo time.

Work done in the first 7 days (both working together): $7 \times \frac{1}{16} = \frac{7}{16}$

Remaining work: $1 - \frac{7}{16} = \frac{9}{16}$

If Ganesh works alone at rate $g = \frac{1}{24}$: $\text{Time} = \frac{\text{Remaining work}}{\text{Ganesh's rate}} = \frac{9/16}{1/24} = \frac{9}{16} \times 24 = \frac{216}{16} = 13.5 \text{ days}$

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Alternative approach — the "lost work" insight.

Normally the job takes 16 days. It actually took 17 — one extra day. Why? Because in the last 9 of the original 16 days, Ramesh lost $30\%$ of his rate.

Work lost over those 9 days: $9 \times 0.3r = 2.7r$

This lost work was recovered by one extra day of both working (at the reduced rate): $2.7r = 0.7r + g \quad\Rightarrow\quad 2r = g$

Same relation, found by pure reasoning. From here, $r + g = \frac{1}{16}$ gives $r = \frac{1}{48}$, $g = \frac{1}{24}$, and the remaining-work calculation is identical.

💡 Teacher tip: When a delay is caused by a known efficiency drop, you can often shortcut straight to the rate ratio by equating lost work to extra work — no simultaneous equations needed.

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Answer: C — $\boxed{13.5 \text{ days}}$

The key insight is that the extra day encodes the rate relationship: Ganesh turns out to be exactly twice as fast as Ramesh, which is why the 30% slowdown only cost one day.

Think of this as: four variables are locked together by one shared value, and we need to express a product-and-log combination purely in terms of exponents.

Step 1 — Introduce a common value.

We are given $p^3 = q^4 = r^5 = s^6$. Let this common value be $K$:

$p^3 = q^4 = r^5 = s^6 = K.$

Now write each variable as a power of $K$ by inverting its exponent:

$p = K^{1/3},\quad q = K^{1/4},\quad r = K^{1/5},\quad s = K^{1/6}.$

Step 2 — Combine $p$, $q$, and $r$ into one power of $K$.

When multiplying powers of the same base, we add exponents:

$pqr = K^{1/3} \cdot K^{1/4} \cdot K^{1/5} = K^{\,1/3 + 1/4 + 1/5}.$

Compute the exponent with a common denominator of $60$:

$\frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{20 + 15 + 12}{60} = \frac{47}{60}.$

So $pqr = K^{47/60}$.

Step 3 — Take $\log_s$ of both sides.

Since $s = K^{1/6}$, we are computing

$\log_s(pqr) = \log_{K^{1/6}}\!\left(K^{47/60}\right).$

Using the change-of-base / power-base rule $\log_{a^m}(a^n) = \dfrac{n}{m}$:

$\log_{K^{1/6}}\!\left(K^{47/60}\right) = \frac{47/60}{1/6} = \frac{47}{60} \times \frac{6}{1} = \frac{47}{10}.$

💡 Teacher tip: Whenever you see a chain like $p^a = q^b = r^c = \dots$, set the whole chain equal to a single constant $K$ and write every variable as a power of $K$. All the logarithm work then collapses to plain arithmetic on exponents.

Answer: A — $\boxed{\dfrac{47}{10}}$

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Alternative approach — anchor directly on $s$.

Since $s^6 = K$, we have $K = s^6$. Substitute into each variable:

$p = K^{1/3} = s^{6/3} = s^2,\quad q = K^{1/4} = s^{6/4} = s^{3/2},\quad r = K^{1/5} = s^{6/5}.$

Then

$pqr = s^2 \cdot s^{3/2} \cdot s^{6/5} = s^{\,2 + 3/2 + 6/5} = s^{\,(20 + 15 + 12)/10} = s^{47/10}.$

Therefore $\log_s(pqr) = \log_s\!\left(s^{47/10}\right) = \dfrac{47}{10}$.

This route skips the explicit change-of-base step and is slightly faster under exam conditions; the $K$-method is more systematic when the chain has many variables.

Think of this as: we're given two weighted-average equations for the same three solutions A, B, C — and we need the strength of a third blend (B and C in ratio 2: 7) relative to A. The key is that we never need to pin down all three strengths individually; one free parameter will cancel out.

What we want: the ratio of the strength of solution D (a mix of B and C in ratio 2: 7) to the strength of solution A.

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Step 1 — Set up variables.

Let the strengths of A, B, C be $a$, $b$, $c$ (in %). Since strength is defined as grams of salt per 100 ml of solution, the strength of a mixture is simply the weighted average of the strengths of its components, weighted by their volumes.

Step 2 — Translate the two mixing conditions into equations.

Mixing A: B: C = 1: 2: 3 (total 6 parts) gives strength 20%: $\frac{a + 2b + 3c}{6} = 20 \implies a + 2b + 3c = 120 \quad \cdots (1)$

Mixing A: B: C = 3: 2: 1 (total 6 parts) gives strength 30%: $\frac{3a + 2b + c}{6} = 30 \implies 3a + 2b + c = 180 \quad \cdots (2)$

Step 3 — Relate $a$ and $c$.

Subtract equation (1) from equation (2): $(3a + 2b + c) - (a + 2b + 3c) = 180 - 120$ $2a - 2c = 60 \implies a - c = 30 \implies a = c + 30$

💡 Teacher tip: Notice that the $2b$ terms cancel when we subtract. This is no accident — the ratio of B is the same (2 parts) in both mixtures, so B contributes equally to both averages. The entire difference in resulting strength comes from swapping A and C.

Step 4 — Express $b$ in terms of $c$.

From equation (1): $2b = 120 - a - 3c = 120 - (c + 30) - 3c = 90 - 4c$ $b = 45 - 2c$

Step 5 — Compute the strength of D.

Solution D mixes B and C in the ratio 2: 7 (total 9 parts), so: $d = \frac{2b + 7c}{9} = \frac{2(45 - 2c) + 7c}{9} = \frac{90 - 4c + 7c}{9} = \frac{90 + 3c}{9} = 10 + \frac{c}{3}$

Step 6 — Find the ratio $d: a$.

$\frac{d}{a} = \frac{10 + \frac{c}{3}}{c + 30} = \frac{\frac{30 + c}{3}}{c + 30} = \frac{30 + c}{3(c + 30)} = \frac{1}{3}$

The parameter $c$ cancels completely — the ratio is fixed at $1: 3$ regardless of the individual strengths.

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Verification with a concrete value. Let $c = 10$. Then $a = 40$, $b = 45 - 20 = 25$.

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Alternative approach — Target the numerator directly.

Instead of solving for $b$ explicitly, directly express $2b + 7c$ (the numerator of $d$) in terms of $a$.

From equation (1): $2b = 120 - a - 3c$, so: $2b + 7c = (120 - a - 3c) + 7c = 120 - a + 4c$

Since $a = c + 30$, substitute $c = a - 30$: $2b + 7c = 120 - a + 4(a - 30) = 120 - a + 4a - 120 = 3a$

Therefore: $d = \frac{2b + 7c}{9} = \frac{3a}{9} = \frac{a}{3}$

So $d: a = 1: 3$ — found without ever computing $b$ or $c$ individually. This is faster under exam time.

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Answer: B — $\boxed{1: 3}$

💡 Why this is elegant: The problem gives two equations in three unknowns, so you might expect the answer to depend on a free parameter. But the specific ratio 2: 7 for D is exactly the combination that makes $2b + 7c = 3a$ — a hidden linear dependence that locks the ratio at $1: 3$ no matter what the individual strengths are.

Think of this as an interest-arbitrage problem: Gopal borrows at one rate, lends at a higher rate, and keeps the spread. We need to track his net retained interest under two different lending scenarios.

Since all transactions are for one year with annual compounding, the interest for one year is simply $\text{Principal} \times \text{Rate}$. (Compounding only changes things from the second year onward.)

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Step 1 — Write Gopal's net retained interest in the first scenario.

Gopal borrows $X$ from Ankit at $8\%$, so he owes Ankit: $\text{Interest paid to Ankit} = 0.08X.$

He adds his own $Y$ and lends $X+Y$ to Ishan at $10\%$, so he earns: $\text{Interest earned from Ishan} = 0.10(X+Y).$

His net retained interest is what he keeps after paying Ankit: $\text{Net retained} = 0.10(X+Y) - 0.08X = 0.02X + 0.10Y.$

We are told this equals the interest accrued to Ankit, i.e. $0.08X$: $0.02X + 0.10Y = 0.08X.$

Solving: 0.10Y = 0.06X \quad\Rightarrow\quad Y = 0.6X. \tag{1}

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Step 2 — Use the second scenario to get a second equation.

Now Gopal lends $X + 2Y$ to Ishan at $10\%$ (still paying Ankit $0.08X$): $\text{Net retained} = 0.10(X+2Y) - 0.08X = 0.02X + 0.20Y.$

This exceeds the first scenario's net retained by Rs. $150$: $(0.02X + 0.20Y) - (0.02X + 0.10Y) = 150.$

💡 Teacher tip: When two scenarios share the same fixed cost (here, the $0.08X$ paid to Ankit), subtracting the two net-retained expressions cancels the fixed term and isolates the variable part immediately.

The $0.02X$ terms cancel: 0.10Y = 150 \quad\Rightarrow\quad Y = 1500. \tag{2}

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Step 3 — Find $X$ and then $X + Y$.

From equation (1): $X = \frac{Y}{0.6} = \frac{1500}{0.6} = 2500.$

Therefore: $X + Y = 2500 + 1500 = 4000.$

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Quick verification.

• First scenario net retained $= 200$, which equals Ankit's interest $= 200$. ✅
• Difference between scenarios $= 350 - 200 = 150$. ✅

Both conditions are satisfied.

Answer: $\boxed{4000}$

💡 Closing insight: The key was recognising that for a single year, compounding doesn't introduce any extra factor — so the problem reduces to a clean linear system. The first condition fixes the ratio $Y/X$, and the second condition fixes the scale via the Rs. 150 difference.

We need the largest value of $2a - 6b$, where $a$ and $b$ are integers making two quadratics keep a fixed sign for every real $x$.

Step 1 — Translate "positive for all $x$" into a discriminant condition.

A quadratic $Ax^2 + Bx + C$ with positive leading coefficient ($A > 0$) has the following behaviour:

• It is strictly positive for all real $x$ if and only if its discriminant $B^2 - 4AC < 0$ (the parabola sits entirely above the $x$-axis, never touching it).
• It is non-negative for all real $x$ if and only if its discriminant $B^2 - 4AC \le 0$ (the parabola may just touch the $x$-axis at one point).

💡 Teacher tip: Whenever you see "positive for all $x$" or "$\ge 0$ for all $x$," convert it immediately to a discriminant inequality. The leading coefficient must also be positive — check that first.

Step 2 — Apply to the first inequality: $2x^2 - ax + 2 > 0$.

Here $A = 2$, $B = -a$, $C = 2$. The leading coefficient $2 > 0$ ✓. For strict positivity: $a^2 - 4(2)(2) < 0 \;\Rightarrow\; a^2 - 16 < 0 \;\Rightarrow\; a^2 < 16 \;\Rightarrow\; |a| < 4.$

Since $a$ is an integer: $a \in \{-3, -2, -1, 0, 1, 2, 3\}$.

Step 3 — Apply to the second inequality: $x^2 - bx + 8 \ge 0$.

Here $A = 1$, $B = -b$, $C = 8$. The leading coefficient $1 > 0$ ✓. For non-negativity: $b^2 - 4(1)(8) \le 0 \;\Rightarrow\; b^2 - 32 \le 0 \;\Rightarrow\; b^2 \le 32 \;\Rightarrow\; |b| \le \sqrt{32} \approx 5.66.$

Since $b$ is an integer: $b \in \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$.

Step 4 — Maximise $2a - 6b$.

The expression $2a - 6b$ grows when $a$ is large and when $b$ is small (since $-6b$ increases as $b$ decreases). So pick:

• The largest allowed $a$: $a = 3$
• The smallest allowed $b$: $b = -5$

$2a - 6b = 2(3) - 6(-5) = 6 + 30 = 36.$

Step 5 — Verify both conditions hold at $a = 3,\; b = -5$.

Both discriminants satisfy their required inequalities, so $a = 3,\; b = -5$ is valid.

Answer: $\boxed{36}$

💡 Closing insight: The two conditions are independent — $a$ and $b$ are constrained separately, so you optimise each one at its own extreme. The only subtlety is the strict vs. non-strict inequality: the first gives $|a| < 4$ (excluding $\pm 4$), while the second gives $|b| \le \sqrt{32}$ (including $\pm 5$). Getting that boundary distinction right is what unlocks the correct maximum.

We need to evaluate $7 \times 11 + 11 \times 15 + 15 \times 19 + \dots + 95 \times 99$.

Think of this as: each term is the product of two numbers that are 4 apart, and the first number in each product marches forward by 4 each time. Our plan is to expand each term into a polynomial in the index, then sum term-by-term.

Step 1 — Identify the structure of each term.

Look at the first factor in each product: $7, 11, 15, \dots, 95$. This is an arithmetic progression with first term $7$ and common difference $4$.

The second factor is always 4 more than the first: $11 = 7+4$, $15 = 11+4$, $19 = 15+4$, and so on. So every term has the form $a \times (a+4)$, where $a$ runs over $7, 11, 15, \dots, 95$.

Step 2 — Count the number of terms.

The first factors form the AP $7, 11, 15, \dots, 95$. Using the $n$-th term formula $a_n = a_1 + (n-1)d$: $95 = 7 + (n-1)(4) \Rightarrow 88 = 4(n-1) \Rightarrow n-1 = 22 \Rightarrow n = 23.$

So there are 23 terms in the sum.

Step 3 — Write the general term as a polynomial in the index.

Let $m = 0, 1, 2, \dots, 22$ (23 values). The first factor is $7 + 4m$, and the second is $11 + 4m$. The general term is: $(7+4m)(11+4m).$

Expand it: $(7+4m)(11+4m) = 77 + 28m + 44m + 16m^2 = 77 + 72m + 16m^2.$

Step 4 — Sum term-by-term using standard formulas.

We need: $S = \sum_{m=0}^{22} (77 + 72m + 16m^2) = 77\sum_{m=0}^{22}1 + 72\sum_{m=0}^{22}m + 16\sum_{m=0}^{22}m^2.$

Compute each piece:

• $\displaystyle\sum_{m=0}^{22} 1 = 23$
• $\displaystyle\sum_{m=0}^{22} m = \frac{22 \times 23}{2} = 253$
• $\displaystyle\sum_{m=0}^{22} m^2 = \frac{22 \times 23 \times 45}{6} = 3795$

💡 Teacher tip: The formula $\sum_{k=0}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ is used with $n=22$ here (not 23), because the index $m$ runs from $0$ to $22$.

Step 5 — Assemble the total. $S = 77 \times 23 + 72 \times 253 + 16 \times 3795.$

Compute each product:

• $77 \times 23 = 1771$
• $72 \times 253 = 18216$
• $16 \times 3795 = 60720$

Add them: $S = 1771 + 18216 + 60720 = 80707.$

Verification — check the first and last terms.

The general term correctly reproduces the first, second, and last terms of the given sum.

Answer: $\boxed{80707}$

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Alternative approach — completing the square.

Notice that $a(a+4) = (a+2)^2 - 4$. So each term can be rewritten as a perfect square minus 4: $7 \times 11 = 9^2 - 4, \quad 11 \times 15 = 13^2 - 4, \quad \dots, \quad 95 \times 99 = 97^2 - 4.$

The numbers $9, 13, 17, \dots, 97$ form an AP (23 terms, $d=4$). Writing $b = 9 + 4m$ for $m = 0, \dots, 22$: $\sum b^2 = \sum (9+4m)^2 = \sum (81 + 72m + 16m^2) = 23 \times 81 + 72 \times 253 + 16 \times 3795.$ $= 1863 + 18216 + 60720 = 80799.$

Subtracting $4 \times 23 = 92$: $S = 80799 - 92 = 80707.$

Both methods agree. The first method is slightly more direct; the second is a nice cross-check that exploits the algebraic identity $a(a+4) = (a+2)^2 - 4$.

💡 Closing insight: Whenever a series consists of products of two APs that differ by a constant, expanding into a polynomial in the index and summing with $\sum m$ and $\sum m^2$ is the most systematic route. The completing-the-square trick is faster only when the constant gap is small and even.