CAT 2017 Quant Questions — Slot 1

Think of this as: once we choose $y$ and $z$, the equation fixes $x$ automatically — so the entire problem reduces to counting admissible pairs $(y, z)$.

Step 1 — Express $x$ in terms of $y$ and $z$.

From $x - y - z = 25$: $x = 25 + y + z.$

Since $y$ and $z$ are positive integers ($y, z \ge 1$), the smallest $x$ can be is $25 + 1 + 1 = 27$, which already satisfies $x \le 40$. The real restriction comes from the upper cap: $25 + y + z \le 40 \implies y + z \le 15.$

So we need to count pairs $(y, z)$ with $1 \le y \le 12$, $1 \le z \le 12$, and $y + z \le 15$.

💡 Teacher tip: Whenever one variable is determined by the others, eliminate it first. Here $x$ is a "freebie" — it adds no combinatorial freedom, so we focus entirely on $(y, z)$.

Step 2 — Group pairs by their sum $s = y + z$.

For a fixed sum $s$, the number of positive-integer pairs $(y, z)$ with $y + z = s$ is $s - 1$ (since $y$ can take values $1, 2, \ldots, s-1$, and $z = s - y$ is then determined). However, we must also enforce $y \le 12$ and $z \le 12$.

Step 3 — Count for each range of $s$.

• $s = 2$ to $13$: The largest any single variable can be is $s - 1 \le 12$, so the cap of 12 never bites. The count is simply $s - 1$. $\text{Total} = 1 + 2 + 3 + \cdots + 12 = \frac{12 \times 13}{2} = 78.$

• $s = 14$: $y$ ranges from $1$ to $13$, but $z = 14 - y \le 12$ forces $y \ge 2$, and $y \le 12$. So $y = 2, 3, \ldots, 12$ → 11 pairs.

• $s = 15$: $y$ ranges from $1$ to $14$, but $z = 15 - y \le 12$ forces $y \ge 3$, and $y \le 12$. So $y = 3, 4, \ldots, 12$ → 10 pairs.

Step 4 — Add them up. $78 + 11 + 10 = 99.$

Answer: B — 99

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Alternative approach — Complementary counting.

Count all positive-integer pairs with $y + z \le 15$ (ignoring the cap of 12), then subtract those that violate $y \le 12$ or $z \le 12$.

• Total without caps: $\displaystyle\sum_{s=2}^{15}(s-1) = 1 + 2 + \cdots + 14 = \frac{14 \times 15}{2} = 105.$
• Subtract pairs with $y \ge 13$: $y = 13$ allows $z = 1, 2$ (2 pairs); $y = 14$ allows $z = 1$ (1 pair). Total = 3.
• Subtract pairs with $z \ge 13$: by symmetry, also 3.
• No pair violates both (that would need $y + z \ge 26 > 15$).

$105 - 3 - 3 = 99.$

Both methods agree. The complementary approach is faster under exam pressure — the direct method is more transparent.

💡 Closing insight: The cap of 12 only starts removing pairs at $s = 14$ and $s = 15$ — for smaller sums, the natural ceiling $s - 1$ already keeps both variables $\le 12$. Spotting where a constraint actually bites is often the key to a clean count.

Reframe: We have a product of two linear terms minus another linear term. The cleanest path is to collapse everything into a single quadratic, factor it, and read off where it is non-positive.

Step 1 — Expand and simplify.

$\begin{aligned}(n-5)(n-10) - 3(n-2) &= n^2 - 10n - 5n + 50 - 3n + 6 \\ &= n^2 - 18n + 56.\end{aligned}$

So the inequality becomes: $n^2 - 18n + 56 \le 0.$

Step 2 — Factor the quadratic.

We need two numbers that multiply to $56$ and add to $-18$: those are $-4$ and $-14$.

$n^2 - 18n + 56 = (n-4)(n-14).$

So the inequality is now: $(n-4)(n-14) \le 0.$

Step 3 — Determine where the product is non-positive.

💡 Teacher tip: A quadratic with a positive leading coefficient (here the coefficient of $n^2$ is $+1$) opens upward — it dips below zero between its two roots and equals zero at the roots. So "$\le 0$" means "between the roots, inclusive."

The roots are $n = 4$ and $n = 14$, so the inequality holds for: $4 \le n \le 14.$

Step 4 — Verify the boundaries.

The endpoints are included (equality holds), and everything between them satisfies the strict inequality.

Step 5 — Count the integers from 4 to 14 inclusive.

$14 - 4 + 1 = 11.$

Answer: $\boxed{11}$

Closing insight: The expression looked like it involved three separate linear factors, but expanding collapsed it into a single quadratic whose sign pattern is trivial to read. Whenever you see a product minus a linear term, simplify to one quadratic before analyzing the sign.

Alternative approach — Quadratic formula. If the factorization doesn't jump out, find the roots of $n^2 - 18n + 56 = 0$ directly: $n = \frac{18 \pm \sqrt{(-18)^2 - 4(1)(56)}}{2} = \frac{18 \pm \sqrt{324 - 224}}{2} = \frac{18 \pm \sqrt{100}}{2} = \frac{18 \pm 10}{2}.$ This gives $n = 4$ and $n = 14$. Since the parabola opens upward, the inequality $\le 0$ holds for $4 \le n \le 14$ — the same 11 integers. Use this as a reliable fallback whenever factoring isn't immediately obvious.

Think of this as: a round trip has a fast leg (downstream, with the current) and a slow leg (upstream, against it). Doubling the boat's speed changes both legs, and the fact that the total time drops to exactly one-quarter of the original is enough to pin down the boat-to-river ratio.

Step 1 — Set up variables.

Let the motor boat's speed in still water be $v$ and the river's speed be $u$. Let the one-way distance to the office be $D$.

• Downstream speed (with the current): $v + u$
• Upstream speed (against the current): $v - u$

Step 2 — Write the original round-trip time.

The total time is the sum of the downstream leg and the upstream leg: $T_{\text{old}} = \frac{D}{v+u} + \frac{D}{v-u}.$

Combine over a common denominator: $T_{\text{old}} = \frac{D(v-u) + D(v+u)}{(v+u)(v-u)} = \frac{2vD}{v^2 - u^2}.$

Step 3 — Write the new round-trip time after doubling the boat speed.

Now the boat speed becomes $2v$ (the river speed $u$ is unchanged): $T_{\text{new}} = \frac{D}{2v+u} + \frac{D}{2v-u}.$

Combine the same way: $T_{\text{new}} = \frac{D(2v-u) + D(2v+u)}{(2v+u)(2v-u)} = \frac{4vD}{4v^2 - u^2}.$

Step 4 — Use the 75% reduction condition.

A 75% reduction means the new time is $25\%$ of the old time, i.e. $T_{\text{new}} = \frac{1}{4}\,T_{\text{old}}$: $\frac{4vD}{4v^2 - u^2} = \frac{1}{4} \cdot \frac{2vD}{v^2 - u^2}.$

Cancel $D$ from both sides (the distance is the same either way): $\frac{4v}{4v^2 - u^2} = \frac{v}{2(v^2 - u^2)}.$

Since $v \neq 0$, cancel $v$: $\frac{4}{4v^2 - u^2} = \frac{1}{2(v^2 - u^2)}.$

Cross-multiply: $8(v^2 - u^2) = 4v^2 - u^2.$

Expand and collect terms: $8v^2 - 8u^2 = 4v^2 - u^2 \implies 4v^2 = 7u^2.$

Step 5 — Extract the ratio.

$\frac{v^2}{u^2} = \frac{7}{4} \implies \frac{v}{u} = \frac{\sqrt{7}}{2}.$

So the ratio of the motor boat's original speed to the river's speed is $v: u = \sqrt{7}: 2$.

💡 Teacher tip: The key structural move is that both legs of the round trip combine into a single fraction of the form $\frac{(\text{sum of speeds})\cdot D}{(\text{product term})}$. That is why doubling the boat speed gives such a clean equation — the downstream and upstream times don't need to be handled separately once combined.

Quick verification. With $v = \sqrt{7}\,k$ and $u = 2k$:

The ratio is indeed $\frac{1}{4}$, confirming a 75% reduction.

Answer: B — $\sqrt{7}: 2$

Alternative approach (ratio substitution). Let $r = v/u$ be the ratio we seek. Dividing every speed by $u$, the times become proportional to $\frac{2r}{r^2-1}$ (old) and $\frac{4r}{4r^2-1}$ (new). Setting $\frac{4r}{4r^2-1} = \frac{1}{4}\cdot\frac{2r}{r^2-1}$ and cancelling $r$ gives $8(r^2-1)=4r^2-1$, so $r^2=\frac{7}{4}$ and $r=\frac{\sqrt{7}}{2}$. This is the same algebra with one fewer variable — slightly faster if you immediately spot that only the ratio matters.

We need to evaluate $\log_{0.008}\sqrt{5} + \log_{\sqrt{3}}81 - 7$. The key idea: each base is a power of a small prime, so we rewrite every logarithm using the rule $\log_{a^m}(a^n) = \dfrac{n}{m}$.

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Step 1 — Simplify the first term: $\log_{0.008}\sqrt{5}$

Express both the base and the argument as powers of $5$.

• Base: $0.008 = \dfrac{8}{1000} = \dfrac{1}{125} = \dfrac{1}{5^3} = 5^{-3}$
• Argument: $\sqrt{5} = 5^{1/2}$

Now apply the power-of-a-base rule: $\log_{5^{-3}}\!\left(5^{1/2}\right) = \frac{1/2}{-3} = -\frac{1}{6}$

💡 Teacher tip: The rule $\log_{a^m}(a^n) = \dfrac{n}{m}$ comes straight from writing the log as $\dfrac{\ln a^n}{\ln a^m} = \dfrac{n\ln a}{m\ln a}$. It works whenever base and argument share the same underlying base $a$.

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Step 2 — Simplify the second term: $\log_{\sqrt{3}}81$

Express both as powers of $3$.

• Base: $\sqrt{3} = 3^{1/2}$
• Argument: $81 = 3^4$

Apply the same rule: $\log_{3^{1/2}}\!\left(3^4\right) = \frac{4}{1/2} = 4 \times 2 = 8$

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Step 3 — Combine all three pieces

$\log_{0.008}\sqrt{5} + \log_{\sqrt{3}}81 - 7 = \left(-\frac{1}{6}\right) + 8 - 7$

$= -\frac{1}{6} + 1 = \frac{-1 + 6}{6} = \frac{5}{6}$

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Quick verification of the arithmetic:

The value matches option C.

Answer: C — $\dfrac{5}{6}$

💡 Closing insight: Whenever a logarithm's base and argument are both powers of the same number, convert them to that common base immediately — the expression collapses to a simple fraction $\frac{n}{m}$. Recognising $0.008 = 5^{-3}$ is the main hurdle here.

Think of this as: two separate ratio statements share one company (C2). If we can express both ratios with the same number for C2, we can merge them into one combined ratio for all five companies.

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

We are given:

• $\text{C1}: \text{C2}: \text{C3} = 9: 10: 8$
• $\text{C2}: \text{C4}: \text{C5} = 18: 19: 20$

C2 appears in both. In the first ratio C2 is represented by $10$; in the second it is represented by $18$. These two numbers must describe the same actual profit, so we need to make them equal.

Step 2 — Equalise C2's share in both ratios.

To make C2 the same number in both ratios, find the LCM of $10$ and $18$: $\text{LCM}(10, 18) = 90.$

Now scale each ratio so that C2 becomes $90$:

• First ratio: multiply every term by $9$ (since $10 \times 9 = 90$): $\text{C1}: \text{C2}: \text{C3} = 9\times9: 10\times9: 8\times9 = 81: 90: 72.$
• Second ratio: multiply every term by $5$ (since $18 \times 5 = 90$): $\text{C2}: \text{C4}: \text{C5} = 18\times5: 19\times5: 20\times5 = 90: 95: 100.$

💡 Teacher tip: When two ratios share a term, scale both so that the shared term becomes the same number. Once that link is aligned, the two ratios snap together into one combined ratio.

Step 3 — Merge into one combined ratio.

Since C2 is now $90$ in both, we can chain them: $\text{C1}: \text{C2}: \text{C3}: \text{C4}: \text{C5} = 81: 90: 72: 95: 100.$

Step 4 — Use the given difference to find the value of one unit.

We are told C5 made Rs 19 crore more than C1. From the combined ratio: $\text{C5} - \text{C1} = 100 - 81 = 19 \text{ units}.$

So $19$ units $=$ Rs $19$ crore, which gives: $1 \text{ unit} = 1 \text{ crore}.$

Step 5 — Add up all five shares.

$81 + 90 + 72 + 95 + 100 = 438.$

Since each unit equals Rs $1$ crore, the total profit made by all five companies is Rs 438 crore.

Answer: A — 438 crore

💡 Why this works so cleanly: the difference in ratio units ($100 - 81 = 19$) exactly matches the given Rs 19 crore difference, making each unit worth exactly Rs 1 crore — no fractions, no messy arithmetic. That is usually a sign the combined ratio was set up correctly.

Think of this as: the fire doesn't reduce what Fatima paid — she still owns the cost of all 60 toys. To bank the same total profit as originally planned, the surviving toys must now earn enough to cover the cost of the lost ones too.

Step 1 — Set up variables. Let the cost price of each toy be $C$ and the printed (marked) price be $P$.

Step 2 — Decode the original plan. Selling at a 40% discount on $P$ means each toy sells for $0.6P$.

Total revenue from 60 toys $= 60 \times 0.6P = 36P$.

This gives a 20% profit on total cost. Total cost $= 60C$, so: $36P = 1.2 \times 60C = 72C \;\Rightarrow\; P = 2C.$

So the printed price is twice the cost price, and the original total profit is: $72C - 60C = 12C.$

💡 Teacher tip: The key trap here is treating the 10 burnt toys as "free" or ignoring their cost. They were already paid for, so their cost still sits on Fatima's books. The same total profit therefore requires the same total revenue as before.

Step 3 — What changes after the fire?

• 10 toys are destroyed, so only 50 survive.
• But she still paid for all 60, so total cost is still $60C$.
• She wants the same total profit of $12C$.

So the total revenue she must collect from the 50 surviving toys is: $60C + 12C = 72C.$

Step 4 — Selling price per surviving toy. $\text{Selling price per toy} = \frac{72C}{50} = 1.44C.$

Convert this to a fraction of the printed price $P = 2C$: $\frac{1.44C}{2C} = 0.72.$

So each toy is sold at 72% of the printed price, which means a discount of: $100\% - 72\% = 28\%.$

Quick check:

Answer: D — 28%

Alternative approach (faster — skip finding $C$). Since the total cost is unchanged and the desired profit is unchanged, the total revenue required is exactly the same as in the original plan: $36P$.

Now this same $36P$ must come from 50 toys instead of 60: $\text{Selling price per toy} = \frac{36P}{50} = 0.72P \;\Rightarrow\; \text{discount} = 28\%.$

This skips the cost-price step entirely — the answer falls straight out of the fact that same profit + same cost ⇒ same revenue, spread over fewer toys.

💡 Closing insight: Whenever part of a batch is destroyed after purchase, anchor on the total revenue target, not the per-toy economics. The lost units don't lower your cost — they raise the burden on the survivors.

Think of this as: we have 11 points on and inside a circle, and almost every group of three will form a triangle — our only job is to find the groups that don't.

Step 1 — Count the total points.

The five diameters $AB, CD, EF, GH, JK$ contribute $5 \times 2 = 10$ endpoints on the circle, plus the centre $O$. So there are $10 + 1 = 11$ points in total.

Step 2 — Start from all possible triples.

Any choice of 3 distinct points could form a triangle. The total number of ways to choose 3 points out of 11 is: $\binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165.$

Step 3 — Identify the triples that do NOT form a triangle.

Three points fail to form a triangle exactly when they are collinear — i.e., they lie on a single straight line.

💡 Teacher tip: In "how many triangles from a point set" problems, don't try to count triangles directly. Count all triples, then subtract the collinear ones. The collinear triples are the only failures.

Where can collinearity hide here? Each diameter is a straight line passing through $O$, and on that line sit exactly three of our eleven points: the two endpoints and the centre.

For example, on diameter $AB$ the three collinear points are $A, O, B$. These three cannot form a triangle. The same holds for each diameter:

Step 4 — Check there are no other collinear triples.

• No three of the ten circumference points are collinear, since a straight line meets a circle in at most two points.
• The only straight lines containing three of our points are the five diameters themselves.

So the collinear triples are exactly 5 in number.

Step 5 — Subtract the failures.

$\text{Triangles} = \binom{11}{3} - 5 = 165 - 5 = 160.$

Answer: $\boxed{160}$

💡 Why this works so cleanly: the circle guarantees that no three boundary points line up, so the only collinear triples come from the diameters through the centre. That makes the subtraction tiny — just 5 out of 165 — and the answer is almost the full $\binom{11}{3}$.

Think of this as: a ball too wide to fall into a cup sits wedged on the rim — how high does its top reach?

Step 1 — Find the cylinder's radius.

The cylinder has height $h = 3$ cm and volume $V = 9\pi$ cm³. Using the volume formula: $V = \pi r^2 h \implies 9\pi = \pi r^2 (3) \implies r^2 = 3 \implies r = \sqrt{3}\text{ cm}.$

So the circular rim at the top of the cylinder has radius $\sqrt{3} \approx 1.73$ cm and sits at height $3$ cm above the base.

Step 2 — Determine how the ball rests.

The ball has diameter $4$ cm, so its radius is $R = 2$ cm. Since $R = 2 > \sqrt{3} \approx 1.73$, the ball is wider than the cylinder's opening and cannot fall through. Instead, it sits wedged on the circular rim.

💡 Teacher tip: The key geometric insight is that the ball touches the rim — a circle of radius $\sqrt{3}$ at height $3$. By symmetry, the ball's centre lies directly above the cylinder's central axis.

Step 3 — Locate the ball's centre.

Set up coordinates with the base of the cylinder at height $0$. The rim is then a circle of radius $\sqrt{3}$ at height $3$. Let the ball's centre be at height $H$ on the cylinder's axis.

Since the ball just touches the rim, the distance from the centre to any point on the rim equals the ball's radius, $2$. Pick a rim point at horizontal distance $\sqrt{3}$ from the axis and height $3$: $\sqrt{(\sqrt{3})^2 + (H - 3)^2} = 2.$

Squaring both sides: $3 + (H-3)^2 = 4 \implies (H-3)^2 = 1 \implies H - 3 = \pm 1.$

This gives $H = 4$ or $H = 2$. Since the ball sits on top of the cylinder, its centre must be above the rim, so $H = 4$ cm.

💡 Teacher tip: The two solutions correspond to the ball sitting above the rim ($H=4$) or below it ($H=2$, dangling inside). The problem says "on top," so we pick $H = 4$.

Step 4 — Find the topmost point.

The topmost point of the ball is one full radius above the centre: $\text{Top height} = H + R = 4 + 2 = 6\text{ cm}.$

Verification check:

Answer: $\boxed{6}$

The elegance here is that the ball's centre floats exactly $1$ cm above the rim — a clean Pythagorean triple ($1^2 + (\sqrt{3})^2 = 2^2$) — making the final height a tidy integer.

Think of it this way: the condition $x+1=x^2$ is a recycling rule. Every time a power of $x$ shows up, we can trade it back down for something simpler, so we never need the messy value of $x$ until the very last step.

Step 1 — Write the condition cleanly. We are told $x + 1 = x^2$, i.e. $x^2 = x + 1.$ This is the only tool we need.

Step 2 — Climb to $x^4$ using that rule. Square both sides of $x^2 = x+1$: $x^4 = (x^2)^2 = (x+1)^2 = x^2 + 2x + 1.$ Now replace that $x^2$ using the rule again ($x^2 = x+1$): $x^4 = (x+1) + 2x + 1 = 3x + 2.$

💡 Teacher tip: Notice we never solved for $x$. The condition let us turn a fourth power into a simple straight-line expression $3x+2$.

Step 3 — Double it. $2x^4 = 2(3x + 2) = 6x + 4.$

Step 4 — Now bring in the actual value of $x$. From $x^2 - x - 1 = 0$ with $x>0$, $x = \frac{1 + \sqrt{5}}{2}.$ Substitute: $6x + 4 = 6\cdot\frac{1+\sqrt{5}}{2} + 4 = 3(1+\sqrt{5}) + 4 = 7 + 3\sqrt{5}.$

Answer: $\boxed{7 + 3\sqrt{5}}$ — option D.

💡 Why it's neat: a relation like $x^2 = x+1$ (the golden-ratio equation) lets you reach any power of $x$ by repeated substitution, keeping every step as a simple linear expression.

Think of this as: after giving every child the compulsory 1 eraser, we only have to distribute 3 spare erasers, with no child allowed to take more than 2 of those spares.

Step 1 — Translate to an equation. Let $x_i$ be the number of erasers kid $i$ gets. Since erasers are identical and kids are distinct, we count integer solutions of $x_1 + x_2 + x_3 + x_4 = 7,$ with each $x_i$ between 1 and 3 inclusive.

Step 2 — Remove the lower bound. Give each kid 1 eraser first. This uses 4 erasers, leaving 3. Write $y_i = x_i - 1$ for the extra erasers kid $i$ gets. Then each $y_i$ is a nonnegative integer, at most 2, and $y_1 + y_2 + y_3 + y_4 = 7 - 4 = 3.$ So we are distributing 3 identical extra erasers among 4 distinct kids, with each kid allowed at most 2 extras.

💡 Teacher tip: The lower bound is handled by giving everyone 1 eraser first. The upper bound then becomes a cap on the extras: at most 2 each.

Step 3 — Count without the upper cap. If we ignore the 'at most 2 extras' rule for a moment, the number of nonnegative integer solutions to $y_1 + y_2 + y_3 + y_4 = 3$ is given by stars-and-bars: $C(3+4-1, 4-1) = C(6,3) = 20.$

Step 4 — Subtract the over-limit cases. The cap is that each $y_i$ must be at most 2. Since only 3 extras are being distributed, the only way to break this cap is for one child to take all 3 extras ($y_i = 3$). That child would then have $1 + 3 = 4$ erasers in total, which is not allowed.

There are 4 choices for which child takes all 3 extras. No two children can break the cap simultaneously, because 3 extras cannot be split into two parts of size 3. So the number of invalid cases is 4.

Valid distributions = $20 - 4 = 16$.

Alternative approach — Direct enumeration of the extras. Since only 3 extras are being handed out, we can list the possible sorted patterns:

For (2,1,0,0): choose the child who gets 2 extras (4 ways), then a different child who gets 1 extra (3 ways). For (1,1,1,0): choose the three children who get one extra each, $C(4,3)=4$.

This matches the stars-and-bars result, so the count is confirmed.

Answer: A — 16

Why the answer is 16: The only case the upper cap removes is the one where a single child takes all 3 spare erasers. Subtracting those 4 cases from the 20 stars-and-bars arrangements leaves 16. Option B (20) is the trap if you forget the 'no more than 3' condition.

We need to evaluate $g(f(f(3)))$. The nesting looks heavy, but the function $f$ has a hidden time-saving property that makes the inner part collapse.

Step 1 — Compute the innermost value $f(3)$.

Substitute $x = 3$ into $f(x) = \dfrac{5x+2}{3x-5}$:

$f(3) = \frac{5(3)+2}{3(3)-5} = \frac{15+2}{9-5} = \frac{17}{4}.$

Step 2 — Compute $f(f(3)) = f\!\left(\tfrac{17}{4}\right)$.

Now feed $\dfrac{17}{4}$ back into $f$:

$f\!\left(\tfrac{17}{4}\right) = \frac{5\cdot\tfrac{17}{4}+2}{3\cdot\tfrac{17}{4}-5} = \frac{\tfrac{85}{4}+\tfrac{8}{4}}{\tfrac{51}{4}-\tfrac{20}{4}} = \frac{\tfrac{93}{4}}{\tfrac{31}{4}} = \frac{93}{31} = 3.$

So $f(f(3)) = 3$ — applying $f$ twice returned us to the original input.

💡 Teacher tip: This is not a coincidence. The function $f(x)=\dfrac{5x+2}{3x-5}$ is its own inverse: $f(f(x))=x$ for every admissible $x$. Once you spot this, any double application of $f$ collapses instantly and you skip the arithmetic entirely.

Step 3 — Evaluate $g$ at the result.

Since $f(f(3)) = 3$, we have:

$g(f(f(3))) = g(3).$

Now use $g(x) = x^2 - 2x - 1$:

$g(3) = 3^2 - 2(3) - 1 = 9 - 6 - 1 = 2.$

Answer: A — 2

💡 Closing insight: For a Möbius-type function $f(x)=\dfrac{ax+b}{cx-a}$ (note the matching diagonal coefficients $a$ and $-a$), the property $f(f(x))=x$ always holds. Recognising this structure turns a messy nested computation into a single substitution.

Think of this as: first draw the curve $y=|x-1|+|x+1|$, then find the nearest point on it to $(\tfrac12,1)$.

Step 1 — Remove the absolute values by splitting into cases.

The expression changes behaviour at $x=-1$ and $x=1$, so we consider three intervals.

• For $-1 \le x \le 1$:

• $|x-1| = 1-x$ (since $x-1 \le 0$)

• $|x+1| = x+1$ (since $x+1 \ge 0$)

• So $y = (1-x)+(x+1) = 2$.

• For $x > 1$:

• $|x-1| = x-1$ and $|x+1| = x+1$

• So $y = (x-1)+(x+1) = 2x$.

• For $x < -1$:

• $|x-1| = 1-x$ and $|x+1| = -x-1$

• So $y = (1-x)+(-x-1) = -2x$.

So the curve is a flat horizontal segment $y=2$ between $x=-1$ and $x=1$, with two upward-sloping arms outside that interval.

💡 Teacher tip: Whenever you see $|x-a|+|x-b|$ with $a<b$, the graph is flat between $a$ and $b$ and rises linearly outside. That flat middle piece is usually where the action is.

Step 2 — Locate the point relative to the curve.

The given point is $(\tfrac12,1)$.

Its $x$-coordinate, $\tfrac12$, lies inside $[-1,1]$, so it sits directly below the flat segment $y=2$.

The closest point on that flat segment is therefore straight above it: $\left(\tfrac12, 2\right).$

The distance is purely vertical: $2 - 1 = 1.$

Step 3 — Confirm the slanted arms are not closer.

On the right arm, $x>1$ and $y=2x>2$. Any point here has both a larger horizontal gap from $\tfrac12$ and a vertical gap bigger than $1$.

On the left arm, $x<-1$ and $y=-2x>2$. Here too the horizontal gap is at least $1.5$ and the vertical gap is bigger than $1$.

A quick check at the two corners of the flat segment:

So no point on the rising arms can beat the vertical distance of $1$.

Answer: A — $1$

💡 Why this works: The point lies directly underneath the flat part of the curve, so the shortest path is simply the perpendicular vertical drop up to $y=2$.

What we want: the area of the closed region enclosed by the curve $|x| + |y| = 2$. Because absolute values hide the sign of $x$ and $y$, the cleanest move is to figure out what this looks like in each quadrant.

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Step 1 — Find where the figure meets the axes.

The axes are the natural place to start because one absolute value vanishes on each.

• On the $x$-axis, $y = 0$, so $|x| + 0 = 2 \Rightarrow |x| = 2 \Rightarrow x = \pm 2$. Points: $(2,0)$ and $(-2,0)$.
• On the $y$-axis, $x = 0$, so $0 + |y| = 2 \Rightarrow |y| = 2 \Rightarrow y = \pm 2$. Points: $(0,2)$ and $(0,-2)$.

Step 2 — See the shape in each quadrant.

Inside each quadrant the signs of $x$ and $y$ are fixed, so we can drop the absolute values and get a straight-line piece:

Each piece is a line segment joining two adjacent axis-intercepts. Joining the four intercepts in order gives a square standing on a corner (a "diamond") with vertices

$(2,0),\;(0,2),\;(-2,0),\;(0,-2).$

💡 Teacher tip: Whenever you see $|x| + |y| = k$, picture a square rotated by $45°$, with its corners sitting on the axes. The intercepts you just found are the corners.

Step 3 — Compute the area.

The two diagonals of this square lie along the coordinate axes:

• Horizontal diagonal: from $(-2,0)$ to $(2,0)$, length $d_1 = 4$.
• Vertical diagonal: from $(0,-2)$ to $(0,2)$, length $d_2 = 4$.

A square's area is half the product of its diagonals:

$\text{Area} = \tfrac{1}{2}\, d_1\, d_2 = \tfrac{1}{2}(4)(4) = 8.$

Step 4 — Quick check by counting triangles.

In Quadrant I, the line $x + y = 2$ together with the axes forms a right triangle with base $2$ and height $2$, so its area is $\tfrac{1}{2}(2)(2) = 2$. By symmetry the same triangle appears in each of the four quadrants:

$\text{Area} = 4 \times 2 = 8. \checkmark$

This matches the diagonal method, so we can be confident.

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Answer: C — $\boxed{8}$

💡 General lesson: The equation $|x| + |y| = k$ always produces a square with vertices $(\pm k, 0)$ and $(0, \pm k)$, diagonals of length $2k$, and area $\tfrac{1}{2}(2k)(2k) = 2k^2$. Here $k = 2$, giving $2(2^2) = 8$.

Think of this as two separate jobs: first find the total area of triangle $ABC$, then figure out what fraction the centroid-based cut removes.

Step 1 — Find the area of $ABC$ using Heron's formula.

The sides are $40$, $25$, and $35$ ft, so the semi-perimeter is $s = \frac{40+25+35}{2} = 50.$

Applying Heron's formula: $\text{Area}(ABC) = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{50 \cdot 10 \cdot 25 \cdot 15}.$

Computing the product inside the root: $50 \times 10 = 500, \quad 25 \times 15 = 375, \quad 500 \times 375 = 187{,}500.$

Now simplify $\sqrt{187{,}500}$: $187{,}500 = 62{,}500 \times 3 = 250^2 \times 3 \implies \sqrt{187{,}500} = 250\sqrt{3}.$

So $\text{Area}(ABC) = 250\sqrt{3}$ sq ft.

Step 2 — Locate the centroid and its height above $BC$.

The centroid $G$ lies on the median from $A$ to the midpoint $M$ of $BC$. A centroid divides every median in the ratio $2:1$, with the longer part between the vertex and the centroid.

So $G$ sits $\tfrac{2}{3}$ of the way from $A$ down to $M$, which means it is only $\tfrac{1}{3}$ of the full height above $BC$.

💡 Teacher tip: Triangle $GBC$ shares the same base $BC$ as triangle $ABC$, but its height is only $\tfrac{1}{3}$ as large. Since area scales linearly with height (base fixed), $\text{Area}(GBC) = \tfrac{1}{3}\,\text{Area}(ABC)$.

Step 3 — Compute the area cut off and the area remaining.

$\text{Area}(GBC) = \frac{1}{3} \times 250\sqrt{3} = \frac{250\sqrt{3}}{3}.$

The remaining portion is the rest: $\text{Remaining} = 250\sqrt{3} - \frac{250\sqrt{3}}{3} = \frac{750\sqrt{3} - 250\sqrt{3}}{3} = \frac{500\sqrt{3}}{3}.$

Equivalently, since $GBC$ is one-third of the whole, the remaining portion is two-thirds: $\frac{2}{3} \times 250\sqrt{3} = \frac{500\sqrt{3}}{3}.$

Answer: B — $\dfrac{500\sqrt{3}}{3}$

💡 Why this works so cleanly: the three medians of any triangle divide it into six smaller triangles of equal area. Triangle $GBC$ is made up of exactly two of those six, so it is always $\tfrac{2}{6} = \tfrac{1}{3}$ of the total — regardless of the triangle's shape. Removing it always leaves $\tfrac{2}{3}$ behind.

Think of this as: two people age at the same rate (one year per year), so the gap between their ages never changes. We're told how that gap sits inside two different ratios — now and later — and we need the percentage increase in the older person's age.

Step 1 — Anchor Barun's present age at a friendly number.

Since the final answer is a percentage, picking Barun's present age as 100 lets the percentage read off directly at the end.

• Barun's present age = $100$
• Arun's present age = $40\%$ of $100 = 40$

Step 2 — Let $t$ years pass and write the new ages.

Both ages increase by the same amount $t$:

• Arun's new age = $40 + t$
• Barun's new age = $100 + t$

Step 3 — Apply the second condition: Arun becomes half of Barun.

$40 + t = \frac{100 + t}{2}$

Multiply both sides by $2$:

$80 + 2t = 100 + t$

Subtract $t$ from both sides, then subtract $80$:

$t = 20$

So 20 years pass.

Step 4 — Compute the percentage increase in Barun's age.

Barun goes from $100$ to $100 + 20 = 120$.

$\text{Percentage increase} = \frac{120 - 100}{100} \times 100 = 20\%$

💡 Teacher tip: Because we anchored Barun's present age at $100$, the number of years gained ($20$) is the percentage increase ($20\%$). No extra division needed — that's the payoff of a smart anchor.

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Alternative approach — using the constant age gap.

The difference between two people's ages never changes. Work in ratios.

• Present ratio: Arun: Barun = $40: 100 = 2: 5$, so the gap is $5 - 2 = 3$ parts.
• Future ratio: Arun: Barun = $1: 2$, so the gap is $2 - 1 = 1$ part.

For the gap to be the same, scale the future ratio by $3$:

• Future ratio = $1: 2 = 3: 6$
• Present ratio = $2: 5$

Now Barun goes from $5$ parts to $6$ parts — an increase of $1$ part.

$\text{Percentage increase} = \frac{6 - 5}{5} \times 100 = \frac{1}{5} \times 100 = 20\%$

Same answer, no assumption about Barun's actual age. The first method is faster under exam time; this one is a good cross-check.

Answer: 20

💡 Why this works so cleanly: both people age at the same rate, so the difference in their ages is an invariant. Once you fix that invariant, the two ratio conditions pin down exactly how much time must pass — and the percentage follows directly.