CAT 2017 Quant Questions — Slot 2

Think of this as: first lock in divisibility by 3 (which digits we pick), then lock in divisibility by 2 (where the even digit sits), all while keeping 0 out of the lead.

Step 1 — What does divisibility by 6 require?

A number is divisible by 6 if and only if it is divisible by both 2 and 3:

• Divisible by 2 → the last digit must be even (0, 2, 4, or 6).
• Divisible by 3 → the sum of all four digits must be a multiple of 3.

We also have two structural rules: no digit repeats, and the first digit cannot be 0.

Step 2 — Find which 4-digit sets have a digit-sum that's a multiple of 3.

We choose 4 digits from $\{0, 2, 3, 4, 6\}$. The sum of all five available digits is: $0 + 2 + 3 + 4 + 6 = 15$

Since 15 is divisible by 3, dropping one digit leaves a sum divisible by 3 if and only if the dropped digit is itself divisible by 3.

💡 Teacher tip: This is modular arithmetic in disguise. If the total is $15 \equiv 0 \pmod{3}$, then $15 - d \equiv 0 \pmod{3}$ requires $d \equiv 0 \pmod{3}$. So the dropped digit must be a multiple of 3.

Among $\{0, 2, 3, 4, 6\}$, the multiples of 3 are 0, 3, and 6. Dropping 2 (leaving sum 13) or 4 (leaving sum 11) would break divisibility by 3, so those sets are discarded.

Only three 4-digit sets can possibly work. Now we count valid arrangements for each.

Step 3 — Count valid 4-digit numbers for each set.

For every set: last digit must be even, first digit cannot be 0, no repeats.

Set A: $\{2, 3, 4, 6\}$ — contains no 0, so the leading-zero constraint is automatically satisfied.

• Last digit (must be even): 2, 4, or 6 → 3 choices.
• First three positions: arrange the remaining 3 digits → $3! = 6$ ways.
• Count: $3 \times 6 = \mathbf{18}$.

Set B: $\{0, 2, 4, 6\}$ — all four digits are even, so every arrangement automatically ends in an even digit. The only constraint is no leading 0.

• First digit (cannot be 0): 2, 4, or 6 → 3 choices.
• Remaining three positions: arrange the other 3 digits → $3! = 6$ ways.
• Count: $3 \times 6 = \mathbf{18}$.

Set C: $\{0, 2, 3, 4\}$ — last digit must be even (0, 2, or 4), and first digit cannot be 0. We split by the last digit:

• Count: $6 + 4 + 4 = \mathbf{14}$.

💡 Teacher tip: When 0 is in the set, always handle the last-digit cases separately. If the last digit is 0, the leading-zero constraint is easy — any of the remaining non-zero digits can lead. If the last digit is a non-zero even digit, then 0 is still in the pool for the middle positions but must be excluded from the first position.

Step 4 — Add the counts.

$18 + 18 + 14 = 50$

Answer: 50

Why this works cleanly: The sum of all five available digits being 15 (a multiple of 3) is the key break — it means divisibility by 3 reduces to a simple test on the dropped digit, instantly narrowing five candidate sets down to three.

---

Alternative approach (cross-check by fixing the last digit).

Instead of grouping by digit-set, fix the last digit (must be even: 0, 2, 4, or 6), then choose 3 more digits from the remaining four so that the total sum is divisible by 3, and count arrangements with no leading 0.

For each last digit, we need the sum of the first three digits to satisfy a specific remainder mod 3 (so that total ≡ 0 mod 3). We list the valid 3-digit subsets and count arrangements (subtracting those with leading 0 when 0 is in the subset):

(Each subset with no 0 gives $3! = 6$ arrangements; each subset containing 0 gives $3! - 2! = 4$ valid arrangements after removing those with leading 0.)

Total: $12 + 14 + 14 + 10 = 50$ ✓ — same answer, confirming the count.

Think of this as: per capita production is just production divided by population. So if we know how the numerator (production) and the whole ratio (per capita) changed, we can back out how the denominator (population) changed.

Step 1 — Write the defining relationship.

Per capita production $= \dfrac{\text{total production}}{\text{population}}$.

Let the population increase by $p\%$, so the population becomes $(1+p)$ times its original value.

Step 2 — Translate each percentage into a growth factor.

• Total production rises by $40\%$ $\Rightarrow$ production is multiplied by $1.40$.
• Per capita production rises by $27\%$ $\Rightarrow$ per capita production is multiplied by $1.27$.
• Population is multiplied by $(1+p)$.

Since per capita $= \dfrac{\text{production}}{\text{population}}$, the per-capita growth factor is the ratio of the two growth factors: $\frac{1.40}{1+p} = 1.27.$

💡 Teacher tip: For any "$X$ per person" quantity, the growth factor is always $\dfrac{\text{numerator factor}}{\text{denominator factor}}$. Percentages combine multiplicatively through ratios, not additively.

Step 3 — Solve for the population factor.

$1+p = \frac{1.40}{1.27} = 1.1023\ldots$

So the population is multiplied by about $1.1023$, meaning the increase is $p \approx 0.1023 \approx 10.2\%.$

Step 4 — Match to the nearest option.

$10.2\%$ is closest to $10\%$.

Answer: C — 10

💡 Takeaway: When a ratio changes and you know how the numerator changed, solve for the denominator using $\text{new ratio} = \dfrac{\text{new numerator}}{\text{new denominator}}$. Convert each percentage to a factor first, then divide — never add or subtract the percentages directly.

Framing. We do not need the exact value of $x$; we only need a tight enough interval. The equation hands us $x$ as a power of $5$, and the exponent itself is a logarithm we can bound between two integers.

Step 1 — Rewrite the equation in exponential form.

Let $k = \log_3 5.$ Then the given equation $\log_5(2+x) = k$ means, by definition of logarithm, $2+x = 5^k \quad\Rightarrow\quad x = 5^k - 2.$ So everything depends on bounding $k$.

Step 2 — Bound the exponent $k$.

The number $k = \log_3 5$ answers: to what power must $3$ be raised to get $5$? Compare $5$ with nearby powers of $3$: $3^1 = 3 < 5 < 9 = 3^2.$ Since the base $3$ is greater than $1$, the function $\log_3$ is increasing, so $1 < k < 2.$

💡 Teacher tip: For range questions, compare the logarithm's argument with nearby powers of the base. That often gives exact bounds without any computation.

Step 3 — Convert the bound on $k$ into a bound on $x$.

Since the base $5$ is also greater than $1$, the function $5^k$ is increasing. Therefore from $1 < k < 2$, $5^1 < 5^k < 5^2 \quad\Rightarrow\quad 5 < 5^k < 25.$ Now subtract $2$: $3 < 5^k - 2 < 23 \quad\Rightarrow\quad 3 < x < 23.$

Step 4 — Match with the options.

Sanity check. Numerically, $k = \log_3 5 \approx 1.465$, so $5^k \approx 10.56$ and $x \approx 8.56$, comfortably inside $3 < x < 23$.

Answer: D — $3 < x < 23$

💡 Why this works so cleanly: the option boundaries $3$ and $23$ are exactly $5^1 - 2$ and $5^2 - 2$, so bounding the exponent between $1$ and $2$ is enough — no exact value needed.

Think of this as: we're given the product of three neighbouring integers and asked for the sum of their squares. The fastest way in is to use the fact that three consecutive numbers multiply to something very close to a perfect cube.

Method 1 — The Cube-Root Estimate

Step 1 — Set up the algebra. Let the middle number be $n$. The three consecutive positive integers are then $n-1$, $n$, and $n+1$.

Their product is: $(n-1) \times n \times (n+1) = n(n^2 - 1) = n^3 - n$

We are given that this product equals $15600$: $n^3 - n = 15600$

Step 2 — Estimate the middle number. For numbers in this range, $n^3$ is much larger than $n$, so $n^3 \approx 15600$. Taking the cube root, we need an integer close to $\sqrt[3]{15600}$.

Recall that $25^3 = 15625$, which is remarkably close to $15600$. Let's test $n = 25$: $25^3 - 25 = 15625 - 25 = 15600$

This matches perfectly. So the three integers are $24$, $25$, and $26$.

💡 Teacher tip: Whenever you have the product of $n$ consecutive integers, taking the $n$-th root lands you within one of the middle number. It turns a guessing game into a one-shot estimate.

Method 2 — Prime Factorization

We can also find the three numbers by breaking $15600$ into its prime factors: $15600 = 156 \times 100 = (12 \times 13) \times (4 \times 25)$ $= (3 \times 4 \times 13) \times (2^2 \times 5^2)$ $= 2^4 \times 3 \times 5^2 \times 13$

We need to group these into three consecutive integers. Since we have $5^2 = 25$ as a factor, $25$ is a strong candidate for the middle number. That means the other two numbers should be $24$ and $26$.

Let's verify:

• $24 = 2^3 \times 3$
• $25 = 5^2$
• $26 = 2 \times 13$

Multiplying them: $24 \times 25 \times 26 = (2^3 \times 3) \times (5^2) \times (2 \times 13) = 2^4 \times 3 \times 5^2 \times 13 = 15600$

This confirms our numbers are $24$, $25$, and $26$.

Step 3 — Calculate the sum of squares. $24^2 + 25^2 + 26^2 = 576 + 625 + 676$ $576 + 625 = 1201$ $1201 + 676 = 1877$

Answer: D — 1877

💡 Why this is elegant: The product $15600$ is just $25$ less than $25^3$. That tiny gap is exactly the $n$ in $n^3 - n$, making the cube-root estimate snap into place instantly.

Think of this as: we need the area of a square built on the segment $AC$, so the real task is simply finding $AC^2$ — the squared length of a diagonal of the regular hexagon.

Step 1 — Understand what's being asked.

A square with $AC$ as one of its sides has area $= AC^2$. So we don't even need $AC$ itself — just $AC^2$.

Step 2 — Locate $AC$ inside the hexagon.

Let $O$ be the centre of the regular hexagon $ABCDEF$. The vertices $A$ and $C$ are two steps apart (we go $A \to B \to C$). Since each side subtends $60°$ at the centre, the angle $\angle AOC$ is: $\angle AOC = 2 \times 60° = 120°.$

💡 Teacher tip: In a regular hexagon, the circumradius equals the side length. So $OA = OC = 1$ cm — the same as each side of the hexagon.

Step 3 — Apply the cosine rule in $\triangle OAC$.

With $OA = OC = 1$ and $\angle AOC = 120°$: $AC^2 = OA^2 + OC^2 - 2 \cdot OA \cdot OC \cdot \cos(\angle AOC)$ $AC^2 = 1^2 + 1^2 - 2(1)(1)\cos 120°$

Since $\cos 120° = -\tfrac{1}{2}$: $AC^2 = 1 + 1 - 2\left(-\tfrac{1}{2}\right) = 1 + 1 + 1 = 3.$

Step 4 — Compute the square's area.

The square built on side $AC$ has area $AC^2 = 3$ sq cm.

Answer: B — 3

💡 Why this is clean: the square's area is just $AC^2$, so the cosine rule delivers the answer in one step — no need to take a square root and then square it back.

Think of this as: we have two functions — one that squares its input and one that raises 2 to the power of its input — and we need to carefully feed values through a nested chain of compositions.

Given: $f(x) = x^2$ and $g(x) = 2^x$, for all real $x$.

To evaluate: $f\bigl[f(g(x)) + g(f(x))\bigr]$ at $x = 1$.

💡 Teacher tip: With nested function compositions, always work from the inside out — compute the innermost values first, then feed those results outward. The two most common mistakes here are (1) confusing which function squares and which exponentiates, and (2) misreading the brackets. Read the expression as: square of $\bigl[$ the sum of $f(g(x))$ and $g(f(x))$ $\bigr]$.

Step 1 — Evaluate the two innermost functions at $x = 1$.

$g(1) = 2^1 = 2, \qquad f(1) = 1^2 = 1.$

Step 2 — Compute each mixed composition.

First, $f(g(1))$: we already found $g(1) = 2$, so we apply $f$ to that result: $f(g(1)) = f(2) = 2^2 = 4.$

Next, $g(f(1))$: we already found $f(1) = 1$, so we apply $g$ to that result: $g(f(1)) = g(1) = 2^1 = 2.$

💡 Teacher tip: Notice that $f(g(1))$ and $g(f(1))$ are not the same thing. In the first, we exponentiate first then square; in the second, we square first then exponentiate. Function composition is generally not commutative — order matters.

Step 3 — Add the two results inside the outer $f$.

$f(g(1)) + g(f(1)) = 4 + 2 = 6.$

Step 4 — Apply the outermost $f$.

$f\bigl[f(g(1)) + g(f(1))\bigr] = f(6) = 6^2 = 36.$

Answer: C — $\boxed{36}$

Why the other options are tempting traps: If you had stopped at Step 2 and just added $4 + 2 = 6$ without applying the final squaring, you might look for $6$ — but that isn't even an option. If you had mistakenly computed $g(f(1)) = g(1) = 2$ but then treated the outer function as $g$ instead of $f$, you'd get $2^6 = 64$ (also not listed). The value $18$ could arise from computing $f(4) + f(2) = 16 + 2 = 18$ — i.e., distributing the outer $f$ over the sum, which is not valid since $f(a+b) \neq f(a) + f(b)$ in general.

Think of this as: we know one diagonal's endpoints, and we know the slope of the other diagonal. The single fact that links them is that a rectangle's diagonals bisect each other — they meet at the same midpoint.

Step 1 — Use the key property of rectangles.

A rectangle is a parallelogram, and in every parallelogram the diagonals bisect each other. So both diagonals pass through the same midpoint. If we find the midpoint of the diagonal we do know, the other diagonal must pass through that same point.

Step 2 — Find the midpoint of the given diagonal.

The endpoints are $(2, 5)$ and $(6, 3)$. Using the midpoint formula $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$:

$\text{Midpoint} = \left(\frac{2+6}{2},\; \frac{5+3}{2}\right) = (4,\; 4).$

Step 3 — Make the other diagonal pass through this midpoint.

The other diagonal has equation $y = 3x + c$. Since it must pass through $(4, 4)$, substitute $x = 4$ and $y = 4$:

$4 = 3(4) + c$ $4 = 12 + c$ $c = 4 - 12 = -8.$

💡 Teacher tip: Notice we never needed the side lengths or the slope of the first diagonal. The only fact we used is that the two diagonals share a midpoint. The equation $y = 3x + c$ already fixes the slope of the second diagonal; the midpoint pins down the intercept $c$.

Answer: D — $\boxed{-8}$

💡 Closing insight: In any parallelogram (rectangle, rhombus, square — all included), the diagonals bisect each other. So whenever a problem gives you one diagonal and asks about the other, your first move is almost always: find the midpoint.

Think of this as: two prime bases (3 and 2) appear on both sides of the equation. If we collect all powers of 3 on one side and all powers of 2 on the other, the equation should collapse into a single exponential equation.

Step 1 — Rewrite every term using prime bases.

Since $9 = 3^2$ and $4 = 2^2$:

• $9^{x - 1/2} = \left(3^2\right)^{x - 1/2} = 3^{2x - 1} = \dfrac{3^{2x}}{3}$
• $2^{2x - 2} = \dfrac{2^{2x}}{2^2} = \dfrac{4^x}{4}$
• $4^x = 2^{2x}$ (keep as $4^x$ for now)
• $3^{2x - 3} = \dfrac{3^{2x}}{3^3} = \dfrac{3^{2x}}{27}$

Step 2 — Substitute into the original equation.

$\frac{3^{2x}}{3} - \frac{4^x}{4} = 4^x - \frac{3^{2x}}{27}$

Step 3 — Group powers of 3 on the left, powers of 2 on the right.

Move $-\dfrac{3^{2x}}{27}$ to the left and $-\dfrac{4^x}{4}$ to the right:

$\frac{3^{2x}}{3} + \frac{3^{2x}}{27} = 4^x + \frac{4^x}{4}$

Factor each side:

$3^{2x}\left(\frac{1}{3} + \frac{1}{27}\right) = 4^x\left(1 + \frac{1}{4}\right)$

$3^{2x} \cdot \frac{10}{27} = 4^x \cdot \frac{5}{4}$

💡 Teacher tip: When two bases appear on both sides of an equation, collecting each base on one side is the cleanest path — it turns a messy four-term equation into a simple ratio.

Step 4 — Solve the ratio.

$\frac{3^{2x}}{4^x} = \frac{5}{4} \cdot \frac{27}{10} = \frac{135}{40} = \frac{27}{8}$

Now recognise both sides as powers of $\dfrac{3}{2}$:

$\left(\frac{3}{2}\right)^{2x} = \frac{27}{8} = \left(\frac{3}{2}\right)^3$

Since the bases match, equate exponents:

$2x = 3 \implies x = \frac{3}{2}$

Step 5 — Verify by substitution.

With $x = \dfrac{3}{2}$:

LHS = RHS ✓

Answer: A — $\dfrac{3}{2}$

💡 Why this works so cleanly: the equation was constructed so that after collecting like bases, both sides reduce to the same base $\frac{3}{2}$ raised to different powers — turning a surds/indices equation into a simple linear one.

Think of this as: since all three work together for the same 4 days, the amount each person contributes is in proportion to their daily work rate. Find each rate, turn them into a ratio, and split Rs 1000 accordingly.

Step 1 — Write down each person's daily work rate.

If someone finishes the whole job in $n$ days, their one-day work is $\frac{1}{n}$ of the job.

• Amal's rate: $\frac{1}{10}$ of the job per day
• Bimal's rate: $\frac{1}{8}$ of the job per day
• All three together finish in 4 days, so their combined rate: $\frac{1}{4}$ of the job per day

Step 2 — Find Kamal's rate by subtraction.

Since Amal + Bimal + Kamal together do $\frac{1}{4}$ per day, Kamal's share of the daily work is what remains after Amal and Bimal's contributions:

$\text{Kamal's rate} = \frac{1}{4} - \frac{1}{10} - \frac{1}{8}$

Bring to a common denominator of 40:

$\frac{1}{4} - \frac{1}{10} - \frac{1}{8} = \frac{10}{40} - \frac{4}{40} - \frac{5}{40} = \frac{1}{40}$

So Kamal does $\frac{1}{40}$ of the job per day.

💡 Teacher tip: Whenever people work together for the same duration, their work-done ratio equals their rate ratio. You don't need to multiply by the number of days — it cancels out for everyone equally.

Step 3 — Form the work ratio.

$\text{Amal}: \text{Bimal}: \text{Kamal} = \frac{1}{10}: \frac{1}{8}: \frac{1}{40}$

Multiply through by 40 to clear denominators:

$\text{Amal}: \text{Bimal}: \text{Kamal} = 4: 5: 1$

Step 4 — Split the Rs 1000 in this ratio.

Total parts $= 4 + 5 + 1 = 10$.

Kamal gets 1 part out of 10:

$\text{Kamal's share} = \frac{1}{10} \times 1000 = \text{Rs } 100$

Quick check: Amal gets $\frac{4}{10}\times 1000 = 400$, Bimal gets $\frac{5}{10}\times 1000 = 500$, Kamal gets $100$. Total $= 400+500+100 = 1000$. ✓

Answer: A — Rs 100

💡 Closing insight: Kamal is the slowest worker ($\frac{1}{40}$ per day vs. Amal's $\frac{1}{10}$ and Bimal's $\frac{1}{8}$), so his share is the smallest. The ratio $4:5:1$ makes that visible at a glance — Kamal does one-tenth of the total work, so he gets one-tenth of the money.

Think of this as: a rectangle has two side lengths, and any three of its sides must consist of two of one dimension and one of the other. We need to find which arrangement maximises the area, and then report the longer side.

Setting up the variables

Let the two side lengths of the rectangle be $x$ and $y$, with $x \ge y$ (so $x$ is the longer side). Three sides totalling 400 ft means one of two cases:

• Case A: $2x + y = 400$ (two longer sides + one shorter side)
• Case B: $x + 2y = 400$ (one longer side + two shorter sides)

We'll check both, since the question only asks for the longer side.

---

Case A: $2x + y = 400$

Solve for $y$: $y = 400 - 2x$

The area is: $A = x \cdot y = x(400 - 2x) = 400x - 2x^2$

This is a downward-opening quadratic in $x$. A quadratic $ax^2 + bx + c$ with $a < 0$ is maximised at $x = -\dfrac{b}{2a}$: $x = \frac{-400}{2(-2)} = \frac{400}{4} = 100$

Then: $y = 400 - 2(100) = 200$

So the sides are $x = 100$ and $y = 200$. Since we assumed $x \ge y$, the longer side here is $y = 200$ ft.

---

Case B: $x + 2y = 400$

Solve for $x$: $x = 400 - 2y$

The area is: $A = x \cdot y = y(400 - 2y) = 400y - 2y^2$

Same structure — maximised at: $y = \frac{-400}{2(-2)} = 100$

Then: $x = 400 - 2(100) = 200$

So the sides are $x = 200$ and $y = 100$. The longer side here is $x = 200$ ft.

---

Verification table

Both cases give the same maximum area ($20{,}000$ sq ft) and the same longer side of $200$ ft.

💡 Teacher tip: The symmetry here is no accident. Whichever dimension appears twice in the 400-ft constraint gets optimised to 100 ft, pushing the other dimension to 200 ft. The longer side is always 200 — the problem is symmetric under swapping which dimension is doubled.

Answer: $\boxed{200}$

Closing insight: The maximum area works out to $200 \times 100 = 20{,}000$ sq ft. Notice that the side appearing twice in the fence is exactly half the side appearing once — a clean 1:2 ratio that emerges naturally from optimising the quadratic.

Think of this as: two sequences are tied together by a single link — the middle term $a_3$. Find that link first, and both sequences fall into place.

Step 1 — Write the new even sequence in terms of $a_3$.

The new sequence is five consecutive even numbers ending with $2a_3$. Working backwards from $2a_3$, each previous term is smaller by 2:

$2a_3 - 8,\;\; 2a_3 - 6,\;\; 2a_3 - 4,\;\; 2a_3 - 2,\;\; 2a_3$

Step 2 — Use the sum to solve for $a_3$.

Adding all five terms: $\text{Sum} = (2a_3 - 8) + (2a_3 - 6) + (2a_3 - 4) + (2a_3 - 2) + 2a_3 = 10a_3 - 20$

We are told this sum equals 450: $10a_3 - 20 = 450 \implies 10a_3 = 470 \implies a_3 = 47$

💡 Teacher tip: For any five equally spaced numbers, the sum is always $5 \times (\text{middle term})$. Here the middle term of the even sequence is $2a_3 - 4$, so the sum is $5(2a_3 - 4) = 10a_3 - 20$ — the same equation, found faster.

Step 3 — Build the original odd sequence.

Since $a_1, a_2, a_3, a_4, a_5$ are five consecutive odd numbers with middle term $a_3 = 47$, we step by 2 on each side:

Quick check: The even sequence ends at $2 \times 47 = 94$, so it is $86, 88, 90, 92, 94$, which sums to $5 \times 90 = 450$. ✓

Answer: $\boxed{51}$

💡 Why this works so cleanly: both sequences are arithmetic progressions with common difference 2, so each is fully determined by its middle term. The single bridge $2a_3$ connects them, making $a_3$ the only variable you ever need to solve for.

Think of this as: each term is exactly 3 times everything that comes after it. Since the "everything after" part of a geometric progression is itself a geometric progression, this condition locks down the common ratio — and then the total sum pins the first term.

Step 1 — Express the tail sum as a GP.

Let the common ratio be $r$. The terms after $a_n$ are: $a_{n+1} + a_{n+2} + a_{n+3} + \cdots = a_n r + a_n r^2 + a_n r^3 + \cdots$

This is a geometric series with first term $a_n r$ and common ratio $r$. Since the progression is infinite and convergent (the total sum is 32, a finite number), we have $|r|<1$, and the tail sum is: $a_{n+1}+a_{n+2}+\cdots = \frac{a_n r}{1-r}.$

Step 2 — Use the given condition to find $r$.

We are told $a_n = 3\,(a_{n+1}+a_{n+2}+\cdots)$ for every $n$. Substituting the tail-sum formula: $a_n = 3\cdot\frac{a_n r}{1-r}.$

Since $a_n \neq 0$ (otherwise the whole sum would be 0, not 32), divide both sides by $a_n$: $1 = \frac{3r}{1-r}.$

Cross-multiply and solve: $1 - r = 3r \quad\Rightarrow\quad 1 = 4r \quad\Rightarrow\quad r = \frac{1}{4}.$

💡 Teacher tip: Any condition that links a term of a GP to its own tail immediately determines $r$, because the tail is just a scaled copy of the whole series. The first term cancels out — only the ratio survives.

Step 3 — Find $a_1$ from the total sum.

The sum of the entire infinite GP is $\dfrac{a_1}{1-r} = 32$: $\frac{a_1}{1-\frac{1}{4}} = 32 \quad\Rightarrow\quad \frac{a_1}{\frac{3}{4}} = 32 \quad\Rightarrow\quad a_1 = 32 \times \frac{3}{4} = 24.$

Step 4 — Compute $a_5$.

The $n$-th term of a GP is $a_n = a_1 r^{n-1}$, so: $a_5 = a_1 r^4 = 24 \times \left(\frac{1}{4}\right)^4 = 24 \times \frac{1}{256} = \frac{24}{256} = \frac{3}{32}.$

Quick check: With $a_1=24$ and $r=\frac14$, the total sum is $\frac{24}{1-1/4}=\frac{24}{3/4}=32$ ✓, and the tail after $a_1$ is $\frac{24\cdot\frac14}{1-\frac14}=\frac{6}{3/4}=8$, and indeed $24 = 3\times 8$ ✓.

Answer: C — $\dfrac{3}{32}$

💡 Closing insight: The ratio $r=\frac14$ means each term is $\frac14$ of the previous, so the tail after any term is $\frac{r}{1-r}=\frac{1}{3}$ of that term — which is exactly why the "3 times the tail" condition works. The structure is self-similar at every position.

Think of this as: two short toddlers leave, taking away very little total height, so the same big pool gets shared among fewer heads — and the per-head share rises by 2 inches. We need to find that new per-head share.

Step 1 — Set up the variable.

Let the average height of the original 22 toddlers be $A$ inches.

Then their total height is: $22A$

Step 2 — Height of the two toddlers who leave.

We're told their average is one-third of the original average: $\text{Average of the two leavers} = \frac{A}{3}$

So their combined height is: $2 \times \frac{A}{3} = \frac{2A}{3}$

Step 3 — Total height of the remaining 20.

Subtract the leavers' combined height from the original total: $22A - \frac{2A}{3} = \frac{66A - 2A}{3} = \frac{64A}{3}$

So the average of the remaining 20 is: $\frac{64A/3}{20} = \frac{64A}{60} = \frac{16A}{15}$

Step 4 — Use the "increase by 2" condition.

The new average is 2 inches more than the old average $A$: $\frac{16A}{15} = A + 2$

Multiply through by 15: $16A = 15A + 30$ $A = 30$

So the original average was 30 inches.

Step 5 — Find the average of the remaining 20.

$\text{New average} = A + 2 = 30 + 2 = 32 \text{ inches}$

Quick verification:

Answer: C — 32

---

Alternative approach — the redistribution view.

When the two toddlers leave, the average of the rest rises by 2. That rise comes from the deficit the two leavers carried below the group average.

• Each leaver is $A - \frac{A}{3} = \frac{2A}{3}$ below the group average.
• Total deficit removed $= 2 \times \frac{2A}{3} = \frac{4A}{3}$.
• This deficit gets redistributed across the remaining 20 toddlers, raising their average by 2: $\frac{4A/3}{20} = 2 \;\Rightarrow\; \frac{4A}{3} = 40 \;\Rightarrow\; A = 30$

New average $= 30 + 2 = 32$.

💡 Teacher tip: In any "people leave/join and the average changes" problem, anchor on the total — the average only moves because the total removed (or added) differs from what the group average would predict. The redistribution view makes this intuition explicit.

💡 Why this is elegant: the two leavers are so short (average only one-third of the group's) that removing them lifts everyone else's share by a clean 2 inches — the arithmetic collapses neatly because $\frac{16}{15}A - A = 2$ forces $A = 30$ directly.

Think of this as: two separate ratios share the middle quantity $b$, so we first need to "stitch" them into one combined ratio $a:b:c$ before we can say anything about their sum.

Step 1 — Write down what's given.

$a: b = 3: 4 \quad\text{and}\quad b: c = 2: 1$

The quantity $b$ appears in both ratios, but with different reference values: in the first ratio $b$ corresponds to $4$, while in the second it corresponds to $2$.

💡 Teacher tip: Whenever two ratios share a variable, combine them by making that shared variable the same number in both. The cleanest way is to use the LCM of the two reference values.

Step 2 — Equalise $b$ across both ratios.

The LCM of $4$ and $2$ is $4$. The first ratio already has $b = 4$, so we only need to scale the second ratio so that its $b$-value also becomes $4$.

Multiply $b: c = 2: 1$ by $2$:

$b: c = 2: 1 \;\longrightarrow\; 4: 2$

Now both ratios use $b = 4$.

Step 3 — Write the combined ratio.

Since $b$ is now the same in both, we can read the three quantities directly:

$a: b: c = 3: 4: 2$

Step 4 — Express the sum in terms of a common multiplier.

Let $a = 3k$, $b = 4k$, $c = 2k$ for some positive integer $k$. Then:

$a + b + c = 3k + 4k + 2k = 9k$

So the sum must always be a multiple of $9$.

💡 Teacher tip: Once a combined ratio is found, the sum is just (sum of ratio parts) $\times k$. Here that's $9k$, so divisibility by $9$ becomes the test.

Step 5 — Check which option is divisible by $9$.

A number is divisible by $9$ when the sum of its digits is divisible by $9$.

Only 207 passes the divisibility-by-9 test. As a quick confirmation, $207 = 9 \times 23$, so $k = 23$ gives $a = 69$, $b = 92$, $c = 46$, and indeed $69 + 92 + 46 = 207$.

Answer: C — 207

💡 Closing insight: The entire question reduces to one divisibility check. Once the combined ratio is $3:4:2$, the sum is locked to multiples of $9$ — no other option can ever work, regardless of the value of $k$.

Think of it this way: the rule $f(ab) = f(a)f(b)$ holds for every pair of positive integers — including the simplest possible pair, $a=b=1$. Testing that one case pins down $f(1)$ completely.

Step 1 — Feed in $a = 1$ and $b = 1$. The rule says $f(a\cdot b) = f(a)\,f(b)$, so $f(1\cdot 1) = f(1)\,f(1) \quad\Rightarrow\quad f(1) = \big(f(1)\big)^2.$

Step 2 — Solve this little equation. Let $t = f(1)$. Then $t = t^2$, i.e. $t^2 - t = 0 \quad\Rightarrow\quad t(t-1) = 0,$ so $t = 0$ or $t = 1$. That is, $f(1)$ can only be $0$ or $1$.

Step 3 — Pick the largest, and check it's allowed. The larger value is $1$. It genuinely works: for instance $f(n) = 1$ for every $n$ satisfies $f(ab)=f(a)f(b)$, since $1 = 1\times 1$.

Answer: $\boxed{1}$

💡 Takeaway: with a multiplicative rule like this, substituting $a=b=1$ almost always reveals $f(1)$ in one move.

Think of this as: we're given two "experiments" — one with only the inlet, one with both pipes together. The difference between those two rates tells us exactly how fast the outlet drains.

Step 1 — Find the inlet pipe's rate.

With the outlet closed, the inlet fills the entire tank in 8 hours. So the inlet's filling rate is: $\text{Inlet rate} = \frac{1}{8} \text{ tank per hour.}$

Step 2 — Find the net rate when both pipes are open.

With both pipes open, the tank still fills (just more slowly) — it takes 10 hours. So the net filling rate is: $\text{Net rate} = \frac{1}{10} \text{ tank per hour.}$

💡 Teacher tip: The net rate is slower than the inlet rate alone because the outlet is working against the inlet. The gap between these two rates is exactly the outlet's draining rate.

Step 3 — Subtract to get the outlet's rate.

The outlet removes water while the inlet adds it, so: $\text{Outlet rate} = \text{Inlet rate} - \text{Net rate} = \frac{1}{8} - \frac{1}{10}.$

Bringing to a common denominator of 40: $\frac{1}{8} - \frac{1}{10} = \frac{5}{40} - \frac{4}{40} = \frac{1}{40} \text{ tank per hour.}$

So the outlet drains $\frac{1}{40}$ of the tank every hour.

Step 4 — Time to go from full to half-full.

Going from a full tank to a half-full tank means draining exactly half the tank, i.e. $\frac{1}{2}$ of the tank.

$\text{Time} = \frac{\text{Work}}{\text{Rate}} = \frac{1/2}{1/40} = \frac{1}{2} \times \frac{40}{1} = 20 \text{ hours.}$

Quick check: At $\frac{1}{40}$ per hour, in 20 hours the outlet drains $20 \times \frac{1}{40} = \frac{1}{2}$ of the tank. A full tank minus half a tank = half-full. ✓

Answer: A — 20 hours

💡 Closing insight: Whenever a problem gives you one pipe's solo time and the combined time, the other pipe's rate is always the difference of the two rates. Anchor on rates (tanks per hour), not on times — times don't subtract cleanly, rates do.

Think of this as: each person in the chain adds a profit percentage to the price they paid, so the final customer price is the original cost multiplied by three markup factors. To find the original cost, undo those markups.

Step 1 — Write the price chain. Let the manufacturing cost be $C$.

• A 10% profit means the manufacturer sells for $C + 0.10C = 1.10C$.
• A 30% profit means the wholesaler sells for $1.30$ times the price he paid: $1.30(1.10C)$.
• A 50% profit means the retailer sells for $1.50$ times the price he paid: $1.50(1.30)(1.10C)$.

The customer pays Rs 4290, so: $1.10(1.30)(1.50)C = 4290.$

Step 2 — Multiply the markup factors. $1.10(1.30) = 1.43$, and then $1.43(1.50) = 2.145$. So the final price is $2.145C$.

Step 3 — Solve for the manufacturing cost. $2.145C = 4290.$ Therefore, $C = 4290 / 2.145 = 2000$.

💡 Teacher tip: Successive profits multiply, they do not add. Adding 10%, 30%, and 50% would suggest 90%, but the true combined factor is $1.10(1.30)(1.50) = 2.145$.

Check by moving forward from Rs 2000:

The final price matches Rs 4290, so the cost is correct.

Alternative approach — Work backwards. Each selling price is cost price plus profit, so to undo a profit we divide by the corresponding factor.

1. Retailer added 50% profit, so before that profit: $4290 / 1.50 = 2860$.
2. Wholesaler added 30% profit, so before that profit: $2860 / 1.30 = 2200$.
3. Manufacturer added 10% profit, so before that profit: $2200 / 1.10 = 2000$.

This gives the same manufacturing cost, Rs 2000, and is often quicker because the divisions come out clean.

Answer: B — Rs 2000

💡 Why this works: Each markup is applied to a larger base than the previous one, so simply adding the percentages would understate the final price. Multiplying the factors forward (or dividing them out backward) respects that compounding effect.

Think of this as: two mixtures have different milk strengths, and we want to blend them into one target strength. The question is only asking how much of each to pour.

Step 1 — Convert every ratio into a milk fraction.

The milk:water ratio tells us what part of each bottle is milk:

We track milk (we could equally track water — the answer is identical).

Step 2 — Apply alligation.

💡 Teacher tip: Alligation says the mixing ratio equals the gap between each ingredient's strength and the target strength. The ingredient farther from the target is needed in smaller amount, and the ratio of the two amounts is the ratio of the two gaps — crossed.

So, $\text{Bottle 1}: \text{Bottle 2} = \left|\frac{9}{13}-\frac{3}{4}\right|: \left|\frac{3}{4}-\frac{7}{9}\right|.$

Compute each gap: $\frac{9}{13}-\frac{3}{4} = \frac{36-39}{52} = -\frac{3}{52}, \qquad \frac{3}{4}-\frac{7}{9} = \frac{27-28}{36} = -\frac{1}{36}.$

Taking absolute values, the gaps are $\dfrac{3}{52}$ and $\dfrac{1}{36}$.

Step 3 — Form and simplify the ratio. $\text{B1}: \text{B2} = \frac{3}{52}: \frac{1}{36} = (3\times 36): (1\times 52) = 108: 52 = 27: 13.$

Step 4 — Verify the blend gives 3: 1.

Take 27 units of Bottle 1 and 13 units of Bottle 2.

• Milk $= 27\cdot\dfrac{7}{9} + 13\cdot\dfrac{9}{13} = 21 + 9 = 30$.
• Water $= 27\cdot\dfrac{2}{9} + 13\cdot\dfrac{4}{13} = 6 + 4 = 10$.
• Milk: Water $= 30: 10 = 3: 1$. ✅

Answer: B — $\boxed{27:13}$

---

Alternative approach (algebra, no alligation).

Let $x$ units of Bottle 1 and $y$ units of Bottle 2 be mixed. Then total milk is $\dfrac{7x}{9}+\dfrac{9y}{13}$ and total water is $\dfrac{2x}{9}+\dfrac{4y}{13}$. Setting their ratio to $3:1$: $\frac{7x}{9}+\frac{9y}{13} = 3\left(\frac{2x}{9}+\frac{4y}{13}\right) \;\Rightarrow\; \frac{x}{9} = \frac{3y}{13} \;\Rightarrow\; \frac{x}{y}=\frac{27}{13}.$ Same answer. Alligation is faster here; the algebra is a reliable cross-check.

💡 Closing insight: Alligation turns a mixture problem into a one-line ratio of gaps — once you convert everything to a single-ingredient fraction, the rest is arithmetic.