CAT 2021 Quant Questions — Slot 1

Think of this as: three people pair up in all possible ways, and each pair's combined speed is given. From those three pair-speeds we can recover each person's individual speed — then pick the middle one.

Let the monthly work-rates of Amar, Akbar, and Anthony be $a$, $k$, and $t$ respectively, measured in projects per month.

Step 1 — Convert every pair-time to a pair-rate.

Step 2 — Add all three equations to get the combined rate.

$2(a + k + t) = \frac{1}{12} + \frac{1}{16} + \frac{1}{24}$

The LCM of $12, 16, 24$ is $48$, so:

$\frac{1}{12} + \frac{1}{16} + \frac{1}{24} = \frac{4}{48} + \frac{3}{48} + \frac{2}{48} = \frac{9}{48} = \frac{3}{16}$

Therefore:

$2(a+k+t) = \frac{3}{16} \quad\Longrightarrow\quad a + k + t = \frac{3}{32}$

💡 Teacher tip: Adding the three pair equations always gives twice the total rate, because each person appears in exactly two of the three pairs. Halving it gives the rate of all three working together.

Step 3 — Subtract each pair-rate to isolate each solo rate.

Since $a + k + t = \frac{3}{32}$, subtracting any one pair-rate leaves the third person's rate:

• Amar: $a = (a+k+t) - (k+t) = \frac{3}{32} - \frac{1}{16} = \frac{3}{32} - \frac{2}{32} = \frac{1}{32}$

• Amar alone takes $\frac{1}{1/32} = 32$ months.

• Akbar: $k = (a+k+t) - (t+a) = \frac{3}{32} - \frac{1}{24} = \frac{9}{96} - \frac{4}{96} = \frac{5}{96}$

• Akbar alone takes $\frac{1}{5/96} = \frac{96}{5} = 19.2$ months.

• Anthony: $t = (a+k+t) - (a+k) = \frac{3}{32} - \frac{1}{12} = \frac{9}{96} - \frac{8}{96} = \frac{1}{96}$

• Anthony alone takes $\frac{1}{1/96} = 96$ months.

Step 4 — Rank the three solo times and pick the middle one.

The person who is neither the fastest nor the slowest is Amar, and he takes 32 months working alone.

Answer: $\boxed{32}$

💡 Takeaway: Whenever you're given all three pair-times, add the three pair-rates and halve — that instantly gives the combined rate. Subtracting each pair-rate back out then yields every individual rate. No simultaneous equations needed.

Think of this as a distance question: $|n-a|$ is how far $n$ is from $a$ on the number line. So the chain $|n-60|<|n-100|<|n-20|$ says $n$ is closer to 60 than to 100, and closer to 100 than to 20.

Step 1 — Closer to 60 than to 100. The points 60 and 100 have midpoint $\frac{60+100}{2}=80$. At $n=80$, the two distances are equal; to the left of 80, a point is closer to 60; to the right, closer to 100. Since the inequality is strict, we need $n<80$.

Step 2 — Closer to 100 than to 20. The points 20 and 100 have midpoint $\frac{20+100}{2}=60$. At $n=60$, the distances to 20 and 100 are equal; to the right of 60, a point is closer to 100. So we need $n>60$.

Step 3 — Combine the two conditions. Both must hold simultaneously: $60<n<80.$ Both ends are strict, so 60 and 80 themselves are excluded.

Step 4 — Count the integers. The integers in the open interval $(60,80)$ are $n\in\{61,62,\ldots,79\}.$ The count is $79-61+1=19.$

A quick boundary check:

💡 Teacher tip: A statement like closer to $a$ than to $b$ means lies on $a$'s side of the midpoint of $a$ and $b$. A chain of distance inequalities often collapses to one simple open interval.

Alternative approach — by algebra. Since absolute values are non-negative, square the inequalities to remove the absolute value signs.

First: $|n-60|<|n-100|$ gives $(n-60)^2<(n-100)^2$. $n^2-120n+3600<n^2-200n+10000$ $80n<6400\Rightarrow n<80.$

Second: $|n-100|<|n-20|$ gives $(n-100)^2<(n-20)^2$. $n^2-200n+10000<n^2-40n+400$ $-160n<-9600\Rightarrow n>60.$

Again $60<n<80$, giving 19 integers. The distance method is faster once you spot it.

Answer: $\boxed{19}$ (Option B)

The two midpoints, 60 and 80, are exactly where a pair of distances ties, which is why both endpoints are excluded and the count is the integers strictly between them.

We have a right triangle $ABC$ with the right angle at $B$, one leg $BC = 10$, and an inscribed circle (incircle) of diameter $8$ — so the inradius is $r = 4$. We need the area.

Think of this as: the incircle pins down the triangle's shape enough that knowing one leg fixes the other.

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Step 1 — Identify what we know and what we need.

• Right angle at $B$, so $AB$ and $BC$ are the legs and $AC$ is the hypotenuse.
• $BC = 10$.
• Incircle diameter $= 8 \Rightarrow$ inradius $r = 4$.
• Let $AB = a$. Then $AC = \sqrt{a^2 + 10^2} = \sqrt{a^2 + 100}$.
• Area $= \tfrac12 \cdot AB \cdot BC = \tfrac12 \cdot a \cdot 10 = 5a$.

So if we find $a$, we're done.

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Step 2 — Use the inradius formula for a right triangle.

For a right triangle with legs $p, q$ and hypotenuse $h$, the inradius is

$$r = \frac{p + q - h}{2}.$$

💡 Teacher tip: This formula comes from the general relation $r = \frac{\text{Area}}{s}$ (where $s$ is the semi-perimeter). For a right triangle, Area $= \frac{pq}{2}$ and $s = \frac{p+q+h}{2}$, so $r = \frac{pq}{p+q+h}$. Rationalising using $(p+q-h)(p+q+h) = (p+q)^2 - h^2 = 2pq$ gives the compact form $r = \frac{p+q-h}{2}$.

Here $p = a$, $q = 10$, $h = \sqrt{a^2+100}$, and $r = 4$:

$$\frac{a + 10 - \sqrt{a^2+100}}{2} = 4.$$

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Step 3 — Solve for $a$.

Multiply both sides by $2$:

$$a + 10 - \sqrt{a^2+100} = 8.$$

Rearrange to isolate the square root:

$$\sqrt{a^2+100} = a + 2.$$

Square both sides:

$$a^2 + 100 = (a+2)^2 = a^2 + 4a + 4.$$

The $a^2$ terms cancel:

$$100 = 4a + 4 \quad\Rightarrow\quad 4a = 96 \quad\Rightarrow\quad a = 24.$$

So $AB = 24$.

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Step 4 — Compute the area.

$$\text{Area} = \tfrac12 \cdot AB \cdot BC = \tfrac12 \cdot 24 \cdot 10 = 120.$$

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Step 5 — Verify the inradius checks out.

The inradius matches the given diameter of $8$ (radius $4$). ✓

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Answer: $\boxed{120}$

💡 Takeaway: For a right triangle, the inradius formula $r = \frac{p+q-h}{2}$ converts an "incircle + one leg" problem into a single equation in one unknown — much faster than working through $r = \Delta/s$ from scratch each time.

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Alternative approach — using $r = \frac{\Delta}{s}$ directly.

The general inradius formula is $r = \frac{\Delta}{s}$, where $\Delta$ is the area and $s$ is the semi-perimeter.

• $\Delta = 5a$ (from Step 1).
• $s = \frac{a + 10 + \sqrt{a^2+100}}{2}$.
• $r = 4$.

Set $\frac{\Delta}{s} = 4$:

$$\frac{5a}{\frac{a+10+\sqrt{a^2+100}}{2}} = 4 \quad\Rightarrow\quad \frac{10a}{a+10+\sqrt{a^2+100}} = 4.$$

So $10a = 4\bigl(a + 10 + \sqrt{a^2+100}\bigr)$, giving

$$10a = 4a + 40 + 4\sqrt{a^2+100} \quad\Rightarrow\quad 6a - 40 = 4\sqrt{a^2+100}.$$

Divide by $2$: $3a - 20 = 2\sqrt{a^2+100}$. Square both sides:

$$9a^2 - 120a + 400 = 4a^2 + 400 \quad\Rightarrow\quad 5a^2 - 120a = 0 \quad\Rightarrow\quad 5a(a - 24) = 0.$$

Since $a > 0$, we get $a = 24$, and the area is $5 \times 24 = 120$. ✓

Both methods reach the same answer; the right-triangle inradius formula is faster under exam conditions since it avoids squaring a binomial with a square root on both sides.

Think of this as: the family buys different amounts of onion each month because both the price and the monthly budget change. The "average expense per kg" is simply total money spent ÷ total kilograms bought — a quantity-weighted average, not a simple average of the five prices.

Step 1 — Set up the monthly budget.

Let the fixed monthly spend for the first three months be $A$ rupees. Then each of the last two months the family spends half that amount, i.e. $A/2$ rupees.

💡 Teacher tip: We don't know $A$, and we don't need to. It will cancel out in the final ratio. Using a symbol keeps the algebra clean and avoids committing to an arbitrary number.

Step 2 — Find the quantity bought each month (kg = spend ÷ price).

Step 3 — Compute total money spent over all 5 months.

$\text{Total spend} = A + A + A + \frac{A}{2} + \frac{A}{2} = 3A + A = 4A.$

Step 4 — Compute total kilograms bought.

$\text{Total kg} = A\!\left(\frac{1}{10} + \frac{1}{20} + \frac{1}{25} + \frac{1}{50} + \frac{1}{100}\right).$

Bring everything to a common denominator of $100$:

$\frac{1}{10} + \frac{1}{20} + \frac{1}{25} + \frac{1}{50} + \frac{1}{100} = \frac{10 + 5 + 4 + 2 + 1}{100} = \frac{22}{100} = 0.22.$

So total kg bought $= 0.22\,A$.

Step 5 — Average price per kg = total spend ÷ total quantity.

$\text{Average price} = \frac{4A}{0.22\,A} = \frac{4}{0.22} = \frac{400}{22} = 18.18\ldots$

The $A$ cancels completely — confirming that the actual budget value was irrelevant.

Step 6 — Pick the closest option.

$18.18\ldots$ is closest to $18$.

Answer: $\boxed{18}$ — Option B

💡 Takeaway: Whenever a problem says "average price/cost per unit" with varying quantities purchased, use total cost ÷ total quantity, not the arithmetic mean of the prices. This is the weighted (by quantity) average — and when the spend is fixed, it becomes a harmonic-mean flavour of problem. Setting the unknown spend to a symbol and letting it cancel is the cleanest route.

Think of this as: a lamp post gives us the faster train's speed for free, and the "crossing each other" event gives us an equation for the unknown length.

Step 1 — Find the faster train's speed from the lamp-post data.

A lamp post has effectively zero width, so crossing it means the train covers a distance equal to its own length.

$v_{\text{fast}} = \frac{\text{length}}{\text{time}} = \frac{160\text{ m}}{12\text{ s}} = \frac{40}{3}\text{ m/s}.$

Step 2 — Find the slower train's speed.

The other train is $6\text{ km/h}$ slower. First convert this gap to m/s:

$6\text{ km/h} = 6 \times \frac{5}{18} = \frac{5}{3}\text{ m/s}.$

So,

$v_{\text{slow}} = \frac{40}{3} - \frac{5}{3} = \frac{35}{3}\text{ m/s}.$

💡 Teacher tip: Convert km/h to m/s once, right at the start, and then do all arithmetic in m/s. Mixing units mid-problem is the most common source of silly errors here.

Step 3 — Relative speed when moving in opposite directions.

When two bodies move toward each other, their relative speed is the sum of their individual speeds:

$v_{\text{rel}} = \frac{40}{3} + \frac{35}{3} = \frac{75}{3} = 25\text{ m/s}.$

Step 4 — Use the "trains crossing each other" event.

When two trains pass each other completely, the relative distance covered is the sum of both lengths. Let the slower train's length be $L$ metres.

$\text{Distance} = \text{Relative speed} \times \text{Time}$

$160 + L = 25 \times 14 = 350.$

$L = 350 - 160 = 190\text{ m}.$

Quick check of the numbers:

Answer: $\boxed{190}$

💡 Takeaway: "Crosses a lamp post" hands you a train's speed directly (length ÷ time). "Crosses another train" hands you the sum of lengths = relative speed × time. Anchor on whichever piece the question gives you, then chase the unknown.

We need the value of $r$ for which $|x^2 - 4x - 13| = r$ has exactly three distinct real roots.

Think of this as: the expression inside the absolute value is a parabola; setting its absolute value equal to $r$ means we're slicing the parabola with two horizontal lines — one above the $x$-axis and one below. The total number of intersection points is what we're counting.

Step 1 — Rewrite the quadratic in vertex form.

$f(x) = x^2 - 4x - 13 = (x-2)^2 - 17$

This is an upward-opening parabola with vertex at $(2, -17)$ — its minimum value is $-17$.

Step 2 — Split the absolute value into two equations.

$|f(x)| = r$ means either $f(x) = r$ or $f(x) = -r$. We count the real roots of each.

Step 3 — Roots of $f(x) = r$ (the upper line).

$(x-2)^2 - 17 = r \implies (x-2)^2 = r + 17$

Since $r \ge 0$, we have $r + 17 > 0$, so this always gives two distinct real roots: $x = 2 \pm \sqrt{r+17}$.

Step 4 — Roots of $f(x) = -r$ (the lower line).

$(x-2)^2 - 17 = -r \implies (x-2)^2 = 17 - r$

The number of roots depends on the sign of $17 - r$:

• $r < 17$: $17 - r > 0$ → two distinct real roots.
• $r = 17$: $17 - r = 0$ → exactly one real root (the line is tangent to the parabola at the vertex).
• $r > 17$: $17 - r < 0$ → no real roots.

Step 5 — Match the "exactly three" requirement.

We always get 2 roots from Step 3. To have exactly 3 total, we need exactly 1 more from Step 4 — that's the tangent case $r = 17$.

Step 6 — Confirm the three roots are distinct.

• $f(x) = 17 \Rightarrow (x-2)^2 = 34 \Rightarrow x = 2 \pm \sqrt{34}$
• $f(x) = -17 \Rightarrow (x-2)^2 = 0 \Rightarrow x = 2$

All three values are different. ✓

💡 Teacher tip: For $|f(x)| = r$ where $f$ is a parabola, the equation $f(x) = -r$ (the lower line) is the one whose root-count changes with $r$. The "exactly three roots" case always occurs when this lower line is tangent to the parabola — i.e., when $r = |\text{minimum of } f|$.

Answer: $\boxed{17}$

The key insight is that the minimum of the parabola is $-17$, so $|{-17}| = 17$ is the critical value of $r$ that creates the tangent condition — giving one root from the lower equation and two from the upper, for a total of exactly three.

Think of this as: when you reverse a three-digit number, how much does it change — and which digit pairs produce exactly a change of 198?

Step 1 — Represent the number and its reverse.

Let the three-digit number be $\overline{abc} = 100a + 10b + c$, where:

• $a \in \{1, 2, \dots, 9\}$ (it can't be 0, or the number wouldn't be three-digit),
• $b, c \in \{0, 1, \dots, 9\}$.

When we reverse the digits we get $\overline{cba} = 100c + 10b + a$.

Step 2 — Compute the reversal gap.

$\overline{cba} - \overline{abc} = (100c + 10b + a) - (100a + 10b + c)$

$= 100c + a - 100a - c = 99c - 99a = 99(c - a).$

💡 Teacher tip: Notice the middle digit $b$ cancels out completely. The reversal gap of any three-digit number depends only on the first and last digits — never on the middle one. This is the key insight that makes the problem fast.

Step 3 — Set the gap equal to 198.

$99(c - a) = 198 \implies c - a = 2.$

So the units digit must be exactly 2 more than the hundreds digit.

Step 4 — Count the valid $(a, c)$ pairs.

We need $c = a + 2$, with $a \geq 1$ and $c \leq 9$.

That gives 7 valid pairs: $a$ can be any integer from 1 to 7.

Step 5 — Account for the free middle digit.

Since $b$ dropped out of the equation entirely, it can be anything from 0 to 9 — that's 10 choices.

Step 6 — Total count.

$7 \text{ pairs} \times 10 \text{ middle digits} = 70.$

Answer: $\boxed{70}$

💡 Why this works so cleanly: the reversal difference $99(c-a)$ is a fixed multiple of 9, so any reversal gap that's a multiple of 99 reduces to a simple digit-pair condition. The middle digit is always a "free" factor of 10 in every such problem.

Think of this as: the large box costs twice as much, but each gram inside it is cheaper per rupee. So how much more chocolate must the large box hold to make both facts true?

Step 1 — Set up variables.

Let the small box cost $P$ and weigh $w$ grams. Let the large box cost $2P$ (twice the price) and weigh $W$ grams.

Step 2 — Translate the "price per gram" condition.

Price per gram of the small box $= \dfrac{P}{w}$.

Price per gram of the large box $= \dfrac{2P}{W}$.

We are told the large box's price per gram is 12% less than the small box's. "12% less" means the large box's rate is $88\%$ of the small box's: $\frac{2P}{W} = 0.88 \times \frac{P}{w}.$

Step 3 — Solve for $W$ in terms of $w$.

Cancel $P$ from both sides: $\frac{2}{W} = \frac{0.88}{w}.$

Cross-multiply: $2w = 0.88\,W \quad\Rightarrow\quad W = \frac{2w}{0.88} = \frac{200w}{88} = \frac{25w}{11} \approx 2.2727\,w.$

So the large box holds about $2.27$ times the weight of the small box.

💡 Teacher tip: "12% cheaper per gram" does not mean the weight rises by 12%. It means the large box delivers more grams per rupee. The weight must compensate for both the doubled price and the 12% discount per gram.

Step 4 — Convert to a percentage excess.

The question asks: by what percentage does the large box's weight exceed the small box's? $\text{Percentage excess} = \frac{W - w}{w} \times 100 = \left(\frac{W}{w} - 1\right) \times 100.$

Substituting $\dfrac{W}{w} = \dfrac{25}{11}$: $\left(\frac{25}{11} - 1\right) \times 100 = \frac{14}{11} \times 100 \approx 127.27\%.$

Step 5 — Pick the nearest option.

$127.27\%$ is closest to $127\%$.

Answer: $\boxed{127}$

💡 Why this works cleanly: the two given facts — "double the price" and "12% cheaper per gram" — combine into a single multiplier $\frac{2}{0.88}$ on the weight. Subtract 1 and convert to a percentage, and you're done. No need to assume actual numbers for $P$ or $w$.

Think of this as: Vinu works every day, while Anu and Manu take turns on odd and even days. So the work happens in a repeating two-day pattern.

Step 1 — Convert each person's rate to convenient units.

Since $15 = 3 \times 5$, $12 = 4 \times 3$, and $20 = 4 \times 5$, the LCM is $4 \times 3 \times 5 = 60$. Treat the whole work as 60 units.

• Anu: $60/15 = 4$ units/day
• Vinu: $60/12 = 5$ units/day
• Manu: $60/20 = 3$ units/day

Step 2 — Write the two-day schedule.

Vinu works every day. Anu works on odd days (1, 3, 5,...), and Manu works on even days (2, 4, 6,...).

So every 2-day cycle completes $9 + 8 = 17$ units.

💡 Teacher tip: In alternating-day problems, first find the work done in one complete cycle. Then multiply the cycle by the number of full cycles and handle the leftover separately.

Step 3 — Accumulate full cycles.

After 3 cycles (6 days): $3 \times 17 = 51$ units.

Remaining work: $60 - 51 = 9$ units.

Step 4 — Finish the leftover on day 7.

Day 7 is odd, so Anu and Vinu work. Together they do $4 + 5 = 9$ units, which is exactly the remaining work. Hence the job is completed on day 7.

Answer: 7 days (Option D)

💡 Takeaway: Alternating-day problems are cleaner in a small table than in algebra — list the cycle, multiply it out, and read off the leftover for the final day.

Alternative approach (fractions instead of units).

Day 1 work = $1/15 + 1/12 = 4/60 + 5/60 = 9/60$ of the job. Day 2 work = $1/20 + 1/12 = 3/60 + 5/60 = 8/60$ of the job. So one 2-day cycle = $17/60$ of the job.

Three cycles = $3 \times 17/60 = 51/60$, leaving $9/60$. Day 7 (Anu + Vinu) = $9/60$, exactly the leftover.

Same answer: 7 days.

Reframe. Every logarithm here has a square root sitting inside it. The clean move is to convert each $\log\sqrt{\cdot}$ into $\tfrac12\log$, because then the $\log(1+x)$ pieces appear on both sides and cancel — leaving one log equation in one unknown.

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Step 1 — Lock down the domain first.

Every argument of a logarithm must be strictly positive:

• $1+x>0 \Rightarrow x>-1$
• $1-x>0 \Rightarrow x<1$
• $1-x^2>0 \Rightarrow -1<x<1$

So $x\in(-1,1)$. Keep this in mind — our final answer must land here.

Step 2 — Rewrite each log using $\log\sqrt{A}=\tfrac12\log A$.

The equation is $5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\log_{10}\frac{1}{\sqrt{1-x^2}}.$

Convert term by term:

• $\log_{10}\sqrt{1+x}=\tfrac12\log_{10}(1+x)$
• $4\log_{10}\sqrt{1-x}=4\cdot\tfrac12\log_{10}(1-x)=2\log_{10}(1-x)$
• RHS: $\log_{10}(1-x^2)^{-1/2}=-\tfrac12\log_{10}(1-x^2)$. Now factor $1-x^2=(1+x)(1-x)$ and split the log: $-\tfrac12\bigl[\log_{10}(1+x)+\log_{10}(1-x)\bigr].$

Step 3 — Substitute and watch the cancellation.

$5-\tfrac12\log_{10}(1+x)+2\log_{10}(1-x)=-\tfrac12\log_{10}(1+x)-\tfrac12\log_{10}(1-x).$

💡 Teacher tip: The $-\tfrac12\log_{10}(1+x)$ term sits identically on both sides — that is the whole point of converting to $\tfrac12\log$ form. It vanishes, and $\log(1+x)$ is gone from the problem entirely.

Adding $\tfrac12\log_{10}(1+x)$ to both sides: $5+2\log_{10}(1-x)=-\tfrac12\log_{10}(1-x).$

Step 4 — Collect the surviving log.

$5=-\tfrac12\log_{10}(1-x)-2\log_{10}(1-x)=-\tfrac52\log_{10}(1-x).$

$\log_{10}(1-x)=5\cdot\left(-\tfrac25\right)=-2.$

Step 5 — Solve for $x$.

By definition of $\log_{10}$: $1-x=10^{-2}=\frac{1}{100}.$ $x=1-\frac{1}{100}=\frac{99}{100}.$ $100x=99.$

This lies inside the domain $(-1,1)$, so it is valid.

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Quick check — plug $x=\tfrac{99}{100}$ back in.

With $1-x=0.01$ and $1+x=1.99$:

• LHS $=5-\tfrac12\log 1.99+2\log 0.01=5-\tfrac12\log 1.99+2(-2)=1-\tfrac12\log 1.99$.
• RHS $=-\tfrac12\log(1.99\times0.01)=-\tfrac12(\log 1.99-2)=1-\tfrac12\log 1.99$.

LHS $=$ RHS. ✅

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Alternative approach — combine into single logarithms, then equate arguments.

Write the constant $5$ as $\log_{10}10^5$ and merge each side into one logarithm:

LHS $=\log_{10}10^5+\log_{10}(1+x)^{-1/2}+\log_{10}(1-x)^{2}=\log_{10}\!\left(\dfrac{10^5(1-x)^2}{\sqrt{1+x}}\right)$.

RHS $=\log_{10}\!\left(\dfrac{1}{\sqrt{(1+x)(1-x)}}\right)$.

Since the logarithm is one-to-one, equate arguments: $\frac{10^5(1-x)^2}{\sqrt{1+x}}=\frac{1}{\sqrt{(1+x)(1-x)}}.$

Cross-multiply and cancel $\sqrt{1+x}$ (positive in the domain): $10^5(1-x)^2\sqrt{1-x}=1\quad\Longrightarrow\quad 10^5(1-x)^{5/2}=1.$

$(1-x)^{5/2}=10^{-5}\quad\Longrightarrow\quad 1-x=\left(10^{-5}\right)^{2/5}=10^{-2}=\frac{1}{100}.$

$x=\frac{99}{100}\quad\Longrightarrow\quad 100x=99.$

Both routes agree. The first is faster under exam time because the cancellation does most of the work for you.

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Answer: $\boxed{99}$

💡 Why this works so neatly: the equation is secretly built so that $\log(1+x)$ cancels completely — converting every $\log\sqrt{\cdot}$ to $\tfrac12\log$ is what exposes that cancellation. Whenever you see square roots inside logs, make that conversion your first move.

Think of this as: two scenarios share the same costs (buying pens + employee wage) but differ only in the selling price of the leftover pens. That difference is the key to unlocking everything.

What we want: the employee's fixed wage $W$.

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Step 1 — Set up variables.

Let the total number of pens Amal buys be $n$ (each costing ₹8), and the employee's fixed wage be $W$.

In both scenarios:

• Total cost = purchase cost + wage $= 8n + W$.
• The first 100 pens are always sold at ₹12, giving revenue $100 \times 12 = 1200$.
• The remaining $n - 100$ pens are sold at different prices.

Step 2 — Write the net-profit equation for each scenario.

Net profit = Total revenue $-$ Total cost.

Scenario 1 (profit of ₹300): remaining pens sold at ₹11 each. $1200 + 11(n - 100) - (8n + W) = 300$

Scenario 2 (loss of ₹300, i.e. profit $= -300$): remaining pens sold at ₹9 each. $1200 + 9(n - 100) - (8n + W) = -300$

Step 3 — Subtract the two equations to eliminate $W$.

Both equations contain the same $8n$ and $W$ terms, so subtracting cancels them cleanly:

$\bigl[11(n-100)\bigr] - \bigl[9(n-100)\bigr] = 300 - (-300)$

$2(n - 100) = 600$

$n - 100 = 300 \quad\Rightarrow\quad n = 400$

So Amal bought 400 pens.

💡 Teacher tip: When two scenarios differ in only one thing (here, the second selling price), subtracting the equations wipes out every shared unknown ($W$ and the $8n$ cost) and leaves a single equation in $n$. Always look for what stays the same — that's what disappears when you subtract.

Step 4 — Substitute back to find $W$.

Using Scenario 1 (profit = ₹300) with $n = 400$:

$1200 + 11(400 - 100) - 8(400) - W = 300$

$1200 + 11(300) - 3200 - W = 300$

$1200 + 3300 - 3200 - W = 300$

$1300 - W = 300$

$W = 1000$

Step 5 — Verify with both scenarios.

With $n = 400$ and $W = 1000$, total cost $= 8(400) + 1000 = 4200$.

Both match the given conditions perfectly.

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Answer: $\boxed{1000}$

💡 Takeaway: When two profit/loss scenarios share the same cost structure but differ in one selling price, subtract the equations first — the shared costs vanish and you get the quantity ($n$) immediately, then back-substitute for the wage ($W$).

Alternative approach — Solve each equation for $W$ and equate.

From Scenario 1: $W = 1200 + 11(n-100) - 8n - 300 = 3n - 200$

From Scenario 2: $W = 1200 + 9(n-100) - 8n + 300 = n + 600$

Setting them equal: $3n - 200 = n + 600 \Rightarrow 2n = 800 \Rightarrow n = 400$

Then $W = 400 + 600 = 1000$. Same answer, slightly more algebra — the subtraction method is faster under exam time.

We are told three different baskets all cost the same amount. We need to find how many mangoes alone would cost that same amount.

Let the price of one apple be $A$, one orange be $O$, and one mango be $M$. Let the common cost of each basket be $C$. Then: $2A + 4O + 6M \;=\; A + 4O + 8M \;=\; 8O + 7M \;=\; C.$

Step 1 — Compare the first two baskets to relate apple and mango prices.

Since Basket 1 and Basket 2 cost the same: $2A + 4O + 6M = A + 4O + 8M.$

Subtract the right side from the left: $(2A - A) + (4O - 4O) + (6M - 8M) = 0$ $A - 2M = 0$ $\boxed{A = 2M}.$

So one apple costs the same as two mangoes.

Step 2 — Compare Basket 2 and Basket 3 to relate orange and mango prices.

$A + 4O + 8M = 8O + 7M.$

Bring everything to one side: $A + 8M - 7M = 8O - 4O$ $A + M = 4O.$

Now substitute $A = 2M$ from Step 1: $2M + M = 4O$ $3M = 4O$ $\boxed{O = \tfrac{3}{4}M}.$

So one orange costs three-fourths the price of one mango.

💡 Teacher tip: With three "equal-cost" baskets, you get two independent equations — enough to express both $A$ and $O$ in terms of $M$. You never need actual prices; only the ratios matter.

Step 3 — Compute the common cost $C$ in terms of $M$.

Use Basket 3 (the simplest, since it has only oranges and mangoes): $C = 8O + 7M.$

Substitute $O = \tfrac{3}{4}M$: $C = 8\!\left(\tfrac{3}{4}M\right) + 7M = 6M + 7M = 13M.$

Step 4 — Interpret the result.

Since the common cost equals $13M$, and each mango costs $M$, a basket containing only mangoes at that same cost must hold 13 mangoes.

Verification check — confirming all three baskets equal $13M$:

All three baskets cost $13M$, confirming our answer.

Answer: $\boxed{13}$

💡 Takeaway: When multiple baskets share the same total cost, pick one item as your "unit of currency" and express every other item's price as a multiple of it. The common cost then reduces to a clean multiple of that unit — here, 13 mangoes.

What we want: the ratio of the largest daily earning to the smallest daily earning among Neeta, Geeta and Sita.

Let the daily earnings of Neeta, Geeta and Sita be $N$, $G$ and $S$ respectively.

Step 1 — Translate each statement into an equation.

• "Neeta and Geeta together earn in a day what Sita alone earns in 6 days": $N+G=6S$
• "Sita and Neeta together earn in a day what Geeta alone earns in 2 days": $S+N=2G$

💡 Teacher tip: Both statements compare one-day combined earnings with one person's earnings over several days. So multiply the single person's daily earning by the number of days to keep the same unit: earnings per day.

Step 2 — Express $N$ in terms of $G$ and $S$.

From the second equation: $N=2G-S.$

Substitute this into the first equation: $2G-S+G=6S$ $3G=7S$ $G=\frac{7}{3}S.$

Step 3 — Find $N$ as well.

Using $N=2G-S$ again:

ight)-S=\frac{14}{3}S-\frac{3}{3}S=\frac{11}{3}S.$$ **Step 4 — Order the three earnings.** Taking $S$ as the base: $$S:G:N = S:\frac{7}{3}S:\frac{11}{3}S.$$ Multiplying all three by $3$ to clear fractions: $$S:G:N = 3:7:11.$$ So Sita earns the least and Neeta earns the most. **Step 5 — Quick check.** - $N+G = \frac{11}{3}S+\frac{7}{3}S=\frac{18}{3}S=6S$ ✅ - $S+N = S+\frac{11}{3}S=\frac{14}{3}S=2\left(\frac{7}{3}S ight)=2G$ ✅ Both original conditions are satisfied. > **Answer:** $oxed{11:3}$, which corresponds to **Option C**. > 💡 **Why this works neatly:** once two linear relations connect three variables, fixing one variable as a base lets the other two fall out as multiples of it. The ratio then just collects those multiples.

Think of this as: two hospitals, two totals of recovery days, and a known gap between their averages. We need to find how many patients hospital A admitted.

What we know:

• Hospital A admitted 21 fewer patients than hospital B.
• Total recovery days: A = 200, B = 152.
• Average recovery days in A = Average recovery days in B + 3.

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Step 1 — Set up the variable.

Let hospital B admit $b$ patients. Then hospital A admits $b - 21$ patients.

Step 2 — Write both averages.

$\text{Average for A} = \frac{200}{b-21}, \qquad \text{Average for B} = \frac{152}{b}.$

The question says A's average is 3 more than B's:

$\frac{200}{b-21} = \frac{152}{b} + 3.$

💡 Teacher tip: Whenever a problem gives you a total and talks about an average, the average is just total ÷ count. Anchor on the count as your unknown — the totals are already fixed numbers.

Step 3 — Clear the denominators.

Multiply both sides by $b(b-21)$:

$200b = 152(b - 21) + 3b(b - 21).$

Expand each piece:

$200b = 152b - 3192 + 3b^2 - 63b.$

Bring everything to one side:

$200b - 152b + 3192 - 3b^2 + 63b = 0,$

$111b + 3192 - 3b^2 = 0.$

Rearranging:

$3b^2 - 111b - 3192 = 0.$

Divide throughout by 3:

$b^2 - 37b - 1064 = 0.$

Step 4 — Solve the quadratic.

Use the quadratic formula with $a=1$, $b=-37$, $c=-1064$:

$b = \frac{37 \pm \sqrt{(-37)^2 - 4(1)(-1064)}}{2} = \frac{37 \pm \sqrt{1369 + 4256}}{2} = \frac{37 \pm \sqrt{5625}}{2}.$

Since $\sqrt{5625} = 75$:

$b = \frac{37 + 75}{2} = 56 \quad \text{or} \quad b = \frac{37 - 75}{2} = -19.$

A patient count cannot be negative, so we reject $b = -19$.

$b = 56.$

Step 5 — Find hospital A's count.

Hospital A admitted $b - 21$ patients:

$56 - 21 = 35.$

Step 6 — Verify the averages.

Gap $\approx 5.71 - 2.71 = 3$ ✓ — exactly as stated.

Answer: $\boxed{35}$

💡 Closing insight: The key move was translating the word problem into a single equation linking the unknown count $b$ to the two known totals. Once the averages are expressed as fractions, the rest is just clearing denominators and solving a quadratic — and the verification step confirms the root is physically meaningful.

Think of this as: bottle 1 starts at 33% and we dilute it toward 17% by swapping out some of its contents with bottle 2's solution. We need to find how much was swapped — and then how much is left behind in bottle 2.

What we want: the volume left in the second bottle after the replacement.

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Step 1 — Set up the variable.

Let $v$ cc be the volume removed from bottle 1 and replaced by $v$ cc taken from bottle 2.

After the swap, bottle 1 still holds $800$ cc total (we removed $v$ and added $v$ back), but its composition has changed:

• $(800 - v)$ cc of the original 33% solution remains,
• $v$ cc of the 17% solution has been added.

Step 2 — Write the concentration equation for bottle 1.

The new strength is the weighted average of the two parts:

$\frac{(800-v)(0.33) + v(0.17)}{800} = 0.21$

The numerator is the total grams of indigo in bottle 1 after mixing; dividing by $800$ gives the new strength (grams per 100 cc… scaled to the whole bottle).

Step 3 — Solve for $v$.

Multiply both sides by $800$:

$(800-v)(0.33) + 0.17v = 0.21 \times 800 = 168$

Expand the first term:

$264 - 0.33v + 0.17v = 168$

Combine the $v$-terms:

$264 - 0.16v = 168$

$0.16v = 96$

$v = \frac{96}{0.16} = 600$

So $600$ cc was poured out of bottle 1 and replaced with $600$ cc from bottle 2.

💡 Teacher tip: A replacement keeps the total volume constant (still 800 cc) but mixes $(800-v)$ of the old with $v$ of the new. Set that weighted average equal to the target concentration and $v$ falls out in one equation.

Step 4 — Find what remains in bottle 2.

Bottle 2 started with $800$ cc. We poured $600$ cc out of it into bottle 1, so:

$800 - 600 = 200 \text{ cc}$

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Quick check.

Bottle 1 after replacement: $200$ cc at 33% + $600$ cc at 17%.

$\text{New strength} = \frac{200 \times 0.33 + 600 \times 0.17}{800} = \frac{66 + 102}{800} = \frac{168}{800} = 0.21 = 21\% \checkmark$

The target strength matches, confirming $v = 600$.

Answer: $\boxed{200}$

The key insight is that the replacement volume $v$ controls everything: it determines how much bottle 1 is diluted and how much bottle 2 loses. Once $v = 600$ falls out of the concentration equation, the remaining volume in bottle 2 is just $800 - 600 = 200$ cc.