CAT 2022 Quant Questions — Slot 2

What we want: the length $AD$, where $D$ is on side $BC$ of an equilateral triangle of side $3$, and the area of $\triangle ADC$ is half the area of $\triangle ABD$.

Step 1 — Locate $D$ using areas. Triangles $ABD$ and $ADC$ share the same height — the perpendicular from $A$ to line $BC$. So their areas are in the ratio of their bases: $\frac{\text{Area}(ADC)}{\text{Area}(ABD)} = \frac{DC}{BD} = \frac12 \;\Rightarrow\; BD: DC = 2: 1.$ With $BC = 3$, this gives $BD = 2$ and $DC = 1$.

Step 2 — Set up coordinates. Place $B = (0, 0)$ and $C = (3, 0)$, so $D = (2, 0)$. For an equilateral triangle of side $3$, the apex is $A = \left(\tfrac32,\ \tfrac{3\sqrt3}{2}\right).$

Step 3 — Distance $AD$. $AD = \sqrt{\left(2 - \tfrac32\right)^2 + \left(0 - \tfrac{3\sqrt3}{2}\right)^2} = \sqrt{\tfrac14 + \tfrac{27}{4}} = \sqrt{\tfrac{28}{4}} = \sqrt{7}.$

Answer: $\boxed{\sqrt{7}}$ (Choice C)

💡 Teacher tip: Two triangles that share a height have areas in the ratio of their bases. That single fact converts an area condition into a position on the side, turning the rest into a one-line distance calculation.

Think of this as: we have three unknowns (correct, wrong, unattempted) tied together by two equations — the total-question count and the score. The condition "unattempted > attempted" then becomes an inequality that pins down the maximum possible value of "correct."

Step 1 — Set up variables and equations.

Let the number of correct answers be $c$, wrong answers be $w$, and unattempted questions be $u$.

Every question falls into exactly one of these three buckets, so: c + w + u = 75 \tag{1}

Scoring: $+3$ for each correct, $-1$ for each wrong, $+1$ for each unattempted: 3c - w + u = 97 \tag{2}

Step 2 — Eliminate $u$ to relate $c$ and $w$.

Subtract equation (1) from equation (2): $(3c - w + u) - (c + w + u) = 97 - 75$ $2c - 2w = 22$ c - w = 11 \quad\Rightarrow\quad c = w + 11 \tag{3}

💡 Teacher tip: When two equations share the same variable ($u$ here) with the same coefficient ($+1$), subtracting eliminates it in one line — always look for that shortcut.

Step 3 — Express $u$ in terms of $w$.

From equation (1): $u = 75 - c - w$. Substituting $c = w + 11$: u = 75 - (w + 11) - w = 64 - 2w \tag{4}

Step 4 — Translate "unattempted > attempted" into an inequality.

"Attempted" means correct + wrong, so the condition is $u > c + w$.

Substitute (3) and (4): $64 - 2w > (w + 11) + w$ $64 - 2w > 2w + 11$ $53 > 4w$ $w < 13.25$

Since $w$ must be a whole number, $w \le 13$.

Step 5 — Maximise $c$.

From $c = w + 11$, the largest $c$ comes from the largest $w$, which is $w = 13$: $c = 13 + 11 = 24$

Step 6 — Verify the solution.

All conditions are satisfied.

Answer: $\boxed{24}$

💡 Why this works: The two linear equations reduce three unknowns to a one-parameter family (everything in terms of $w$). The inequality then acts as a ceiling on $w$, and since $c$ increases with $w$, the ceiling on $w$ directly gives the maximum $c$.

Alternative approach — work directly in terms of $c$.

Since we want to maximise $c$, it's natural to make $c$ the free variable. From $c - w = 11$ we get $w = c - 11$, and then $u = 75 - c - w = 86 - 2c$.

The condition $u > c + w$ becomes: $86 - 2c > c + (c - 11) = 2c - 11$ $97 > 4c \quad\Rightarrow\quad c < 24.25$

So the largest integer $c$ is $\boxed{24}$. This route is slightly faster because it targets $c$ directly — no need to maximise $w$ and then convert back.

Framing: five students score distinct whole numbers from $0$ to $50$, averaging $38$; exactly three score above $32$; Amit scores the least. We want the gap between Amit's highest possible and lowest possible mark.

The total of all five marks is $5 \times 38 = 190$.

"Exactly three above 32" means exactly two are $\le 32$ — and since Amit is the lowest, those two are Amit and one other. So Amit $\le 31$ (the other $\le 32$ mark must be a larger distinct value).

Maximum Amit — make everyone else as small as possible. The three "above 32" scores are distinct, so the smallest they can be is $33, 34, 35$. Try Amit $= 31$: then the other $\le 32$ score is $32$, and the three high scores must total $190 - 31 - 32 = 127$. Can three distinct numbers above $32$, each $\le 50$, add to $127$? Yes: $33 + 44 + 50 = 127$. So $\max \text{Amit} = 31.$

Minimum Amit — make everyone else as large as possible. The one other $\le 32$ score is at most $32$; the three above-$32$ scores are at most $50, 49, 48$ (distinct), summing to $147$. So the other four total at most $32 + 147 = 179$, forcing $\text{Amit} \ge 190 - 179 = 11.$ Check Amit $= 11$: the set $\{11, 32, 48, 49, 50\}$ sums to $190$, all distinct, exactly three above $32$ ✓. So $\min \text{Amit} = 11.$

Difference. $31 - 11 = 20.$

Answer: 20 (Choice D)

💡 Teacher tip: To find an extreme value of the smallest element, push the other elements the opposite way — to maximise Amit, squeeze everyone else down; to minimise Amit, stretch everyone else up. The "exactly three above 32" rule fixes which side of 32 each other score sits on.

We need to count the integer values of $n$ for which the rational expression $\frac{4 - \log_2 n}{3 - \log_4 n}$ is negative. The key hurdle is that the two logarithms are in different bases — so our first job is to unify them.

Step 1 — Express both logs in the same base.

Using the change-of-base relationship between $\log_2 n$ and $\log_4 n$: since $4 = 2^2$, we have $\log_4 n = \frac{\log_2 n}{\log_2 4} = \frac{\log_2 n}{2} \quad\Longrightarrow\quad \log_2 n = 2\log_4 n.$

💡 Teacher tip: Whenever two logs of the same argument appear in different bases, unify them first. The relation $\log_{b^2} n = \frac{1}{2}\log_b n$ is worth remembering — it turns a two-base problem into a one-variable problem.

Step 2 — Substitute a single variable.

Let $y = \log_4 n$, so $\log_2 n = 2y$. The inequality becomes: $\frac{4 - 2y}{3 - y} < 0 \iff \frac{2(2 - y)}{3 - y} < 0 \iff \frac{2 - y}{3 - y} < 0.$

(The factor of $2$ is positive, so it doesn't affect the sign.)

Step 3 — Sign analysis of $\frac{2-y}{3-y}$.

The numerator is zero at $y = 2$; the denominator is zero at $y = 3$. These two critical points split the number line into three regions:

The quotient is negative only when the signs differ, which happens for $2 < y < 3$.

Step 4 — Convert back to $n$.

We have $2 < \log_4 n < 3$. Since $\log_4$ is an increasing function, we can exponentiate with base $4$: $4^2 < n < 4^3 \implies 16 < n < 64.$

Note: $n$ must be positive for the logarithms to be defined, and every value in $(16, 64)$ is positive, so the domain is automatically satisfied. Also, $n \neq 64$ (which would make the denominator zero) is guaranteed since $n < 64$.

Step 5 — Count the integers.

The integers strictly between $16$ and $64$ are $17, 18, 19, \ldots, 63$.

Verification of the endpoints:

The count is: $63 - 17 + 1 = 47.$

Answer: $\boxed{47}$

💡 Why this works cleanly: the two critical values $y = 2$ and $y = 3$ map to the perfect powers $n = 16$ and $n = 64$, so the open interval $(16, 64)$ contains exactly $64 - 16 - 1 = 47$ integers — no messy boundary cases to worry about.

We need the largest natural number $n$ such that $(n!)!$ divides $(15000)!$.

Think of this as: a factorial divides another factorial exactly when the first factorial's argument is not larger than the second's.

Step 1 — Use the factorial-divisibility rule.

For natural numbers $k$ and $m$,

$k! \mid m! \iff k \le m.$

This is because $m!$ contains every factor from $1$ to $k$ whenever $k \le m$, so all prime powers in $k!$ appear with at least as large an exponent in $m!$.

Here, $k = n!$ and $m = 15000$. So

$(n!)! \mid (15000)! \iff n! \le 15000.$

So the question reduces to: find the largest $n$ such that $n! \le 15000$.

Step 2 — Test values of $n!$.

Since $7! = 5040 \le 15000$ but $8! = 40320 > 15000$, the largest possible value of $n$ is $7$.

💡 Teacher tip: The condition "$a!$ divides $b!$" is essentially shorthand for "$a \le b$". For nested factorials like $(n!)! \mid (15000)!$, first replace the outer factorial divisibility by an inequality, then handle the remaining $n! \le 15000$.

Answer: 7

We need the average of the first 25 partial sums of the AP $3, 7, 11, \ldots$ — that is, $\frac{1}{25}\sum_{n=1}^{25} A_n$. The plan: get a clean formula for each partial sum $A_n$, then sum that formula using the two standard results $\sum n$ and $\sum n^2$.

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Step 1 — Write $A_n$ in closed form

The AP has first term $a = 3$ and common difference $d = 4$. The sum of the first $n$ terms is:

$A_n = \frac{n}{2}\bigl[2a + (n-1)d\bigr]$

Substituting $a = 3$ and $d = 4$:

$A_n = \frac{n}{2}\bigl[6 + 4(n-1)\bigr] = \frac{n}{2}\bigl[6 + 4n - 4\bigr] = \frac{n}{2}(4n + 2) = n(2n+1)$

Expanding:

$\boxed{A_n = 2n^2 + n}$

💡 Teacher tip: The partial sums of an AP always form a quadratic in $n$ (no constant term when $A_0 = 0$). Once you spot that, summing $A_n$ over $n = 1$ to $25$ reduces to using $\sum n$ and $\sum n^2$ — no term-by-term addition needed.

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Step 2 — Sum $A_n$ from $n = 1$ to $25$

$\sum_{n=1}^{25} A_n = \sum_{n=1}^{25}(2n^2 + n) = 2\sum_{n=1}^{25} n^2 + \sum_{n=1}^{25} n$

Recall the two standard formulas (with $N = 25$):

Now substitute:

$\sum_{n=1}^{25} A_n = 2(5525) + 325 = 11050 + 325 = 11375$

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Step 3 — Divide by 25

$\frac{1}{25}\sum_{n=1}^{25} A_n = \frac{11375}{25} = 455$

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Quick sanity check

For $n = 1$: $A_1 = 2(1) + 1 = 3$ ✓ (first term is 3). For $n = 2$: $A_2 = 2(4) + 2 = 10$ ✓ ($3 + 7 = 10$). For $n = 3$: $A_3 = 2(9) + 3 = 21$ ✓ ($3 + 7 + 11 = 21$).

The closed form is correct, so the final sum is trustworthy.

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Answer: $\boxed{455}$ — Option A

💡 Why this works so cleanly: The partial sums of any AP are quadratic in $n$. Summing a quadratic over $1$ to $N$ always boils down to $\sum n$ and $\sum n^2$ — two formulas worth memorising cold for the CAT.

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Alternative approach — Double-sum interchange

Instead of finding $A_n$ as a polynomial, think of $\sum_{n=1}^{25} A_n$ as a double sum: each $A_n$ adds up the first $n$ terms $t_k$ of the AP.

$\sum_{n=1}^{25} A_n = \sum_{n=1}^{25}\sum_{k=1}^{n} t_k$

Swap the order of summation. A term $t_k$ appears in every $A_n$ with $n \ge k$, i.e. for $n = k, k+1, \ldots, 25$ — that's $(25 - k + 1) = (26 - k)$ times:

$\sum_{n=1}^{25} A_n = \sum_{k=1}^{25} t_k \cdot (26 - k)$

The $k$-th term of the AP is $t_k = 3 + (k-1)(4) = 4k - 1$, so:

$\sum_{n=1}^{25} A_n = \sum_{k=1}^{25}(4k - 1)(26 - k) = \sum_{k=1}^{25}(-4k^2 + 105k - 26)$

$= -4(5525) + 105(325) - 26(25) = -22100 + 34125 - 650 = 11375$

Dividing by 25 gives $\frac{11375}{25} = 455$ — the same result. The polynomial method is faster under exam time; the double-sum interchange is a powerful cross-check when the closed form is messy.

Think of this as: two ships leave the same point along perpendicular paths (one south, one west). After 2 hours the line joining them is 60 km. We need the slower speed, given that the speeds differ by 6 km/h.

Step 1 — Recognise the hidden right angle.

South and West are perpendicular directions. So after 2 hours, the two ships sit at the ends of the two legs of a right triangle, and their separation — 60 km — is the hypotenuse.

💡 Teacher tip: Whenever a problem says one object goes north/south and another goes east/west, draw a right triangle. The travel distances are the legs; the straight-line gap between them is the hypotenuse.

Step 2 — Express the two legs.

Let the slower ship's speed be $v$ km/h. Then the faster ship's speed is $v + 6$ km/h.

In 2 hours:

• Slower ship travels $2v$ km.
• Faster ship travels $2(v+6)$ km.

Step 3 — Apply Pythagoras.

$(2v)^2 + \bigl[2(v+6)\bigr]^2 = 60^2$

$4v^2 + 4(v+6)^2 = 3600$

Divide both sides by $4$ immediately to keep the numbers small:

$v^2 + (v+6)^2 = 900$

Step 4 — Expand and solve.

$v^2 + v^2 + 12v + 36 = 900$

$2v^2 + 12v - 864 = 0$

$v^2 + 6v - 432 = 0$

Using the quadratic formula:

$v = \frac{-6 \pm \sqrt{36 + 1728}}{2} = \frac{-6 \pm \sqrt{1764}}{2} = \frac{-6 \pm 42}{2}$

Taking the positive root (speed must be positive):

$v = \frac{-6 + 42}{2} = \frac{36}{2} = 18$

So the slower ship is $18$ km/h and the faster is $24$ km/h.

Step 5 — Verify.

Check: $36^2 + 48^2 = 1296 + 2304 = 3600 = 60^2$. ✅

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Alternative approach — test the options.

For each option, the slower speed $v$ gives legs $2v$ and $2(v+6)$; we need $(2v)^2 + (2(v+6))^2 = 3600$.

Only Option C satisfies the Pythagorean condition.

Answer: C. 18

💡 Why this is elegant: The legs 36 and 48 with hypotenuse 60 are just the classic $3$-$4$-$5$ right triangle scaled up by $12$. Spotting that lets you jump to the answer almost instantly: the 2-hour distances are $36$ and $48$, so the speeds are $18$ and $24$.

Think of this as: two regular polygons have linked side counts and a fixed angle ratio, so one variable is enough to pin down both.

Step 1 — Assign sides using the given ratio.

The number of sides of A and B are in the ratio $1:2$. So let:

• sides of A $= n$
• sides of B $= 2n$

Also, a regular polygon with more sides has a larger interior angle, so B's angle being the larger $4$-part in the ratio $3:4$ is consistent with B having $2n$ sides.

Step 2 — Write each interior angle.

For a regular $n$-gon, each interior angle is

$\frac{(n-2)180^\circ}{n}.$

So:

$\text{Angle of A}=\frac{(n-2)180^\circ}{n}$

$\text{Angle of B}=\frac{(2n-2)180^\circ}{2n}=\frac{(n-1)180^\circ}{n}$

Step 3 — Use the angle ratio $3:4$.

$\frac{\frac{(n-2)180^\circ}{n}}{\frac{(n-1)180^\circ}{n}}=\frac{3}{4}$

The common factor $\frac{180^\circ}{n}$ cancels:

$\frac{n-2}{n-1}=\frac{3}{4}$

Cross-multiply:

$4(n-2)=3(n-1)$

$4n-8=3n-3$

$n=5$

Step 4 — Find the sides of B.

B has $2n$ sides:

$2n=2\times 5=10$

Check:

$108:144=3:4$ ✓

💡 Teacher tip: Memorise $\text{interior angle}=\frac{(n-2)180^\circ}{n}$. With sides in ratio $1:2$, the two angle expressions share a factor of $\frac{180^\circ}{n}$, which cancels in the ratio and leaves a simple linear equation.

Answer: $\boxed{10}$

Alternative approach — using exterior angles.

Each exterior angle of a regular $n$-gon is $\frac{360^\circ}{n}$, so each interior angle is $180^\circ-\frac{360^\circ}{n}$.

For A: $180^\circ-\frac{360^\circ}{n}$

For B: $180^\circ-\frac{360^\circ}{2n}=180^\circ-\frac{180^\circ}{n}$

So:

$\frac{180-\frac{360}{n}}{180-\frac{180}{n}}=\frac{3}{4}$

Cross-multiplying:

$4\left(180-\frac{360}{n}\right)=3\left(180-\frac{180}{n}\right)$

$720-\frac{1440}{n}=540-\frac{540}{n}$

$180=\frac{900}{n}$

$n=5$

Hence B again has $2n=10$ sides.

The key simplification came from the side ratio $1:2$: B's angle expression reduced cleanly and shared a factor with A's, turning the ratio equation into a short linear solve.

Think of this as: each day, the number of new particles is the current count divided by the day number (rounded down). We need to find which day the total first reaches 1000.

Step 1 — Translate the rule into a formula.

On day $n$ (for $n \ge 2$), one out of every $n$ particles produces a new particle. So if the count at the start of day $n$ is $C$, the number of new particles created that day is $\lfloor C/n \rfloor$, and the count at the end of day $n$ is: $C + \left\lfloor \frac{C}{n} \right\rfloor$

Step 2 — Compute the first few days to spot a pattern.

A pattern jumps out: the count rises by exactly 50 every day.

Step 3 — Prove the pattern holds for every day.

The end of day 1 is $100 = 50 \times 2 = 50(1+1)$.

Now assume the end of day $n-1$ is $50n$ — meaning the start of day $n$ is $50n$. Then on day $n$:

• New particles $= \left\lfloor \dfrac{50n}{n} \right\rfloor = \lfloor 50 \rfloor = 50$
• End of day $n = 50n + 50 = 50(n+1)$

So by induction, the count at the end of day $n$ is $50(n+1)$ for every $n \ge 1$.

💡 Teacher tip: The reason the floor function never causes trouble is that the count always lands on a multiple of the day number. Since the end of day $n-1$ is $50n$, dividing by $n$ gives exactly 50 with no remainder. That is what keeps the daily increment constant.

Step 4 — Solve for the day that reaches 1000.

We need the end-of-day count to equal 1000: $50(m+1) = 1000$ $m + 1 = 20$ $m = 19$

Step 5 — Verify the boundary.

The count reaches 1000 exactly at the end of day 19 — not before, not after.

Answer: $\boxed{19}$

💡 Why this works so cleanly: The starting value 100 is divisible by 2, and each day's increment of 50 keeps the running total on a multiple of the next day's number. The problem is designed so the division is always exact — the floor function never "bites" — which is what turns a messy recurrence into a simple linear formula.

What we want: the average of the largest and smallest possible values of $a + b$, where $a, b \ge 0$ and $a + 2b = 6$.

Step 1 — Reduce to one variable. From $a + 2b = 6$ we get $a = 6 - 2b$, so $a + b = (6 - 2b) + b = 6 - b.$

Step 2 — Find the range of $b$. Both variables are non-negative: $a \ge 0 \Rightarrow 6 - 2b \ge 0 \Rightarrow b \le 3, \qquad b \ge 0.$ So $b \in [0, 3]$.

Step 3 — Range of $a + b = 6 - b$. As $b$ moves from $0$ up to $3$, $6 - b$ falls from $6$ down to $3$: $\max(a + b) = 6 \ (\text{at } b = 0), \qquad \min(a + b) = 3 \ (\text{at } b = 3).$

Step 4 — Average the extremes. $\frac{6 + 3}{2} = 4.5.$

Answer: 4.5 (Choice D)

💡 Teacher tip: When one linear combination is fixed and you want the range of another, write the target in the single free variable and just read the endpoints off its allowed interval — no calculus needed.

Think of this as: we know the ratio of how fast each person works, and we know how long the job takes when all three work together. That's enough to pin down each person's exact rate — and then we just track how much work Anu and Tanu finish before Manu takes over.

Step 1 — Convert the time ratio into rates.

The solo times are in the ratio $5:8:10$. Let's write them as $5t$, $8t$, and $10t$ for some common multiplier $t$.

💡 Teacher tip: In work-rate problems, always convert times to rates (reciprocals) before combining people. You can add rates; you cannot add times.

So the hourly rates are: $\text{Anu} = \frac{1}{5t},\quad \text{Tanu} = \frac{1}{8t},\quad \text{Manu} = \frac{1}{10t}.$

Step 2 — Use the "all three together" fact to find $t$.

Together, they finish the job in $4$ days at $8$ hours/day, i.e. $32$ hours total. So their combined rate is $\frac{1}{32}$ of the job per hour: $\frac{1}{5t}+\frac{1}{8t}+\frac{1}{10t}=\frac{1}{32}.$

Bring the left side to a common denominator ($40$): $\frac{8+5+4}{40t}=\frac{1}{32}\;\Longrightarrow\;\frac{17}{40t}=\frac{1}{32}.$

Cross-multiply: $40t = 17 \times 32 = 544 \;\Longrightarrow\; t = \frac{544}{40} = \frac{68}{5}.$

Step 3 — Write each person's solo time and rate.

Step 4 — Work done by Anu and Tanu together.

Their combined rate: $\frac{1}{68}+\frac{5}{544} = \frac{8}{544}+\frac{5}{544} = \frac{13}{544}\text{ per hour.}$

They work for $6$ days at $6$ hours $40$ minutes per day. Converting: $6\text{ h }40\text{ min} = 6+\frac{40}{60} = 6+\frac{2}{3} = \frac{20}{3}\text{ hours.}$

Total hours worked: $6 \times \frac{20}{3} = 40\text{ hours.}$

Work completed: $40 \times \frac{13}{544} = \frac{520}{544} = \frac{65}{68}.$

Step 5 — Remaining work and Manu's time.

Remaining work: $1 - \frac{65}{68} = \frac{3}{68}.$

Manu's rate is $\frac{1}{136}$ per hour, so the time he needs: $\text{Time} = \frac{\text{Work}}{\text{Rate}} = \frac{\,3/68\,}{1/136} = \frac{3}{68}\times 136 = 3 \times 2 = 6\text{ hours.}$

Answer: $\boxed{6}$

💡 Why this works cleanly: The ratio $5:8:10$ and the single "32 hours together" fact are enough to fix every rate exactly — no extra information is wasted. The remaining fraction $\frac{3}{68}$ and Manu's rate $\frac{1}{136}$ are designed so the $68$ and $136$ cancel to give a tidy $6$.

Think of this as a two-phase budget: figure out how much Manu has already saved, then find the monthly income he needs for the remaining months to hit his annual target.

Step 1 — Convert the average target into a total target. Manu wants to save an average of ₹550 per month over 12 months. So his total savings target for the year is: $550 \times 12 = \text{₹}6600.$

Step 2 — Compute savings already banked in the first 9 months. For each of the first 9 months, he earns ₹4000 and spends ₹3500, so he saves: $4000 - 3500 = \text{₹}500 \text{ per month}.$ Over 9 months, total savings become: $500 \times 9 = \text{₹}4500.$

Step 3 — Find how much more he must save in the remaining 3 months. Total target minus savings already done: $6600 - 4500 = \text{₹}2100.$ This ₹2100 must be saved across the last 3 months (months 10, 11, and 12), so the required saving per month is: $2100 \div 3 = \text{₹}700 \text{ per month}.$

Step 4 — Determine the new monthly earnings needed. From the 10th month onward, his monthly expense rises to ₹3700, and he still needs to save ₹700 each month. Therefore, his required monthly earnings are: $3700 + 700 = \text{₹}4400.$

💡 Teacher tip: "Average savings target" is just total savings = average × number of months. Always convert an average target into a total before subtracting what's already done.

Answer: ₹4400 (Option A)

💡 Why this works cleanly: the problem splits neatly into "past" and "future" — bank what's already happened, then solve a simple monthly equation for the remaining months.

Think of this as: we know the gap between the top two candidates (2512 votes), and we know everyone's share as a percentage of cast votes. If we can express that gap as a percentage of registered voters, we can find the total.

Step 1 — Set up the shares of cast votes.

Let the number of cast votes be $V_c$ and the number of registered voters be $R$.

We are told $V_c = 80\% \text{ of } R = 0.80\,R$.

One candidate gets $30\%$ of $V_c$. The remaining $70\%$ of $V_c$ is split among the other three candidates in the ratio $1:2:3$.

The ratio $1:2:3$ has $6$ total parts, so the three candidates receive:

Step 2 — Identify the winner and runner-up.

💡 Teacher tip: Don't assume the candidate with the stated $30\%$ is the winner. After splitting the remainder in the ratio $1:2:3$, the largest share — the "$3$" part — is actually $35\%$, which is bigger than $30\%$. Always recompute the absolute shares before ranking.

• Winner: the "$3$" candidate with $35\%$ of cast votes.
• Runner-up (second highest): the $30\%$ candidate.

The gap between them is: $35\% - 30\% = 5\% \text{ of cast votes.}$

Step 3 — Translate the gap into registered voters.

Cast votes are $80\%$ of registered voters, so: $5\% \text{ of cast votes} = 0.05 \times 0.80 \times R = 0.04\,R = 4\% \text{ of registered voters.}$

We are told this gap equals $2512$ votes: $0.04\,R = 2512$

$R = \frac{2512}{0.04} = \frac{2512 \times 100}{4} = 62800.$

Step 4 — Verify.

The gap matches the given $2512$, so the answer is confirmed.

Answer: 62800 $\;\boxed{\text{D}}$

💡 Closing insight: The key move was converting a percentage of cast votes into a percentage of registered voters by multiplying by $0.80$. Whenever a problem chains percentages through different bases, always track which base each percentage refers to before combining them.

Think of this as: we're given a function that alternates between two values, and we need to figure out which value applies where.

What we know:

• $f(x) + f(x-1) = 1$ for all integers $x$.
• $g(x) = x^2$.
• $f(x^2 - x) = 5$.
• We need $f(g(5)) + g(f(5))$.

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Step 1 — Show that $f$ has period 2.

From the given equation: $f(x) = 1 - f(x-1)$

Now replace $x$ with $x-1$ in that same equation: $f(x-1) = 1 - f(x-2)$

Substitute this back into the first equation: $f(x) = 1 - \bigl(1 - f(x-2)\bigr) = f(x-2)$

So $f(x) = f(x-2)$ for every integer $x$. This means $f$ is periodic with period 2 — its value depends only on whether the input is even or odd.

Let $f(\text{even}) = E$ and $f(\text{odd}) = O$. From $f(x) + f(x-1) = 1$, consecutive integers have opposite parity, so: $E + O = 1$

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Step 2 — Use $f(x^2 - x) = 5$ to find $E$ and $O$.

The expression $x^2 - x$ factors as: $x^2 - x = x(x-1)$

This is the product of two consecutive integers, so one of them is always even. Therefore $x^2 - x$ is always even, regardless of $x$.

Since $f(x^2 - x) = 5$ and the input is always even: $E = 5$

Then from $E + O = 1$: $O = 1 - 5 = -4$

💡 Teacher tip: Whenever you see $f(x) + f(x-1) = \text{constant}$, the function must have period 2. Then look for an expression like $x^2 - x = x(x-1)$ — a product of consecutive integers is always even, which pins down the even value directly.

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Step 3 — Compute each piece of the target sum.

We need $f(g(5)) + g(f(5))$.

First term: $f(g(5))$ $g(5) = 5^2 = 25$ $f(g(5)) = f(25)$ Since $25$ is odd, $f(25) = O = -4$.

Second term: $g(f(5))$ $f(5) = O = -4 \quad (\text{since 5 is odd})$ $g(f(5)) = g(-4) = (-4)^2 = 16$

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Step 4 — Add them. $f(g(5)) + g(f(5)) = (-4) + 16 = 12$

Answer: $\boxed{12}$

The key insight is that a functional equation linking $f(x)$ and $f(x-1)$ forces a two-value alternation, and the cleverly chosen condition $f(x^2 - x) = 5$ exploits the fact that $x(x-1)$ is always even to hand you one of those two values directly.

We're told a cubic has two roots that are negatives of each other — $r$ and $-r$. That symmetry is powerful: it gives us two equations, and adding/subtracting them cleanly isolates the unknown $c$.

Step 1 — Translate the root condition into equations.

Let $f(x) = 5x^3 + cx^2 - 10x + 9$. Since $r$ is a root:

$f(r) = 5r^3 + cr^2 - 10r + 9 = 0$

Since $-r$ is also a root:

$f(-r) = -5r^3 + cr^2 + 10r + 9 = 0$

Notice what happens when we substitute $-r$: the odd-power terms ($x^3$ and $x$) flip sign, while the even-power terms ($x^2$ and the constant) stay the same.

Step 2 — Add the two equations to eliminate the odd-degree terms.

$f(r) + f(-r) = (5r^3 - 5r^3) + (cr^2 + cr^2) + (-10r + 10r) + (9 + 9) = 0$

$2cr^2 + 18 = 0$

$cr^2 = -9 \quad \Rightarrow \quad c = -\frac{9}{r^2}$

💡 Teacher tip: Adding $f(r) + f(-r)$ kills every odd power of $r$ because $r^n + (-r)^n = 0$ when $n$ is odd. What survives — the even powers and the constant — is exactly what we need to pin down $c$.

Step 3 — Subtract the two equations to find $r^2$.

$f(r) - f(-r) = (5r^3 + 5r^3) + (cr^2 - cr^2) + (-10r - 10r) + (9 - 9) = 0$

$10r^3 - 20r = 0$

$10r(r^2 - 2) = 0$

Now, $r \neq 0$ — if $r = 0$ were a root, then $f(0) = 9$ would have to equal $0$, which it doesn't. So we can divide by $10r$:

$r^2 = 2$

Step 4 — Compute $c$.

Substituting $r^2 = 2$ into $cr^2 = -9$:

$c \cdot 2 = -9 \quad \Rightarrow \quad c = -\frac{9}{2}$

Answer: $-\dfrac{9}{2}$

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Alternative approach — Factor the cubic directly.

Since $r$ and $-r$ are both roots, $(x - r)(x + r) = x^2 - r^2$ must be a factor of the cubic. So we can write:

$5x^3 + cx^2 - 10x + 9 = (x^2 - r^2)(5x + a)$

for some constant $a$. Expanding the right side:

$(x^2 - r^2)(5x + a) = 5x^3 + ax^2 - 5r^2 x - ar^2$

Now match coefficients on both sides:

From $r^2 = 2$ and $9 = -ar^2 = -c \cdot 2$:

$c = -\frac{9}{2}$

Both methods arrive at the same value. The factor method is slightly faster under exam conditions, but the add/subtract method is more mechanical and less error-prone.

💡 Closing insight: The key idea is that roots $r$ and $-r$ impose a symmetry on the polynomial. Whether you exploit it by adding $f(r) + f(-r)$ or by factoring out $x^2 - r^2$, that symmetry is what makes the problem solvable from a single cubic with one unknown.