CAT 2021 Quant Questions — Slot 3

The actual Quantitative Ability section from CAT 2021, Slot 3 — 22 questions in original paper order, each with a full worked solution. Tap an option (or type your answer) to check yourself before reading the solution.

22 questions14 MCQ · 8 TITA40 min section0/22 attempted

Q1TrianglesMCQ

If a triangle $ABC$ , $\angle BCA = 50^\circ$ . $D$ and $E$ are points on $AB$ and $AC$ , respectively, such that $AD = DE$ . If $F$ is a point on $BC$ such that $BD = DF$ , then $\angle FDE$ , in degrees, is equal to

Show solution

Correct answer: B

Think of this as: two isosceles triangles are built on the two pieces of side $AB$ , and we want the angle trapped between them at $D$ . The trick is to let the unknown angle at $A$ be a variable, compute each isosceles angle in terms of it, and then watch the variable cancel.

Step 1 — Label the angles of $\triangle ABC$ .

Let $\angle BAC = \alpha$ . We are given $\angle BCA = 50^\circ$ , so by the angle-sum property of a triangle: $\angle ABC = 180^\circ - 50^\circ - \alpha = 130^\circ - \alpha.$

Step 2 — Use the isosceles triangle $ADE$ .

Since $D$ lies on $AB$ and $E$ lies on $AC$ , the angle $\angle DAE$ is exactly $\angle BAC = \alpha$ .

We are told $AD = DE$ , so $\triangle ADE$ is isosceles with the equal sides meeting at $D$ . Therefore the base angles are equal: $\angle DAE = \angle DEA = \alpha.$

The angle at $D$ inside this triangle is $\angle ADE = 180^\circ - \alpha - \alpha = 180^\circ - 2\alpha.$

Step 3 — Use the isosceles triangle $BDF$ .

Since $D$ lies on $AB$ and $F$ lies on $BC$ , the angle $\angle DBF$ is exactly $\angle ABC = 130^\circ - \alpha$ .

We are told $BD = DF$ , so $\triangle BDF$ is isosceles with the equal sides meeting at $D$ . Therefore the base angles are equal: $\angle DBF = \angle DFB = 130^\circ - \alpha.$

The angle at $D$ inside this triangle is $\angle BDF = 180^\circ - 2(130^\circ - \alpha) = 180^\circ - 260^\circ + 2\alpha = 2\alpha - 80^\circ.$

Step 4 — Combine the three angles at point $D$ .

The point $D$ sits on the segment $AB$ . On the side of $AB$ that contains $C$ , the three angles $\angle ADE$ , $\angle EDF$ , and $\angle FDB$ together form a straight angle: $\angle ADE + \angle EDF + \angle FDB = 180^\circ.$

Substituting the expressions from Steps 2 and 3: $(180^\circ - 2\alpha) + \angle FDE + (2\alpha - 80^\circ) = 180^\circ.$

The $-2\alpha$ and $+2\alpha$ cancel, leaving $100^\circ + \angle FDE = 180^\circ \implies \angle FDE = 80^\circ.$

💡 Teacher tip: The $\alpha$ -terms cancel against each other — a strong hint that the answer does not depend on the exact shape of $\triangle ABC$ . Whenever a parameter cancels cleanly, the result is a fixed angle determined only by the given data.

Answer: $\boxed{80^\circ}$ (Choice B)

💡 Closing insight: Two isosceles triangles built back-to-back on the two pieces of side $AB$ produce an angle $\angle FDE$ that depends only on $\angle C$ . In fact the relation generalises to $\angle FDE = 180^\circ - 2\angle C$ , which gives $180^\circ - 2(50^\circ) = 80^\circ$ here.

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Q2LogarithmsMCQ

For a real number a, if $\dfrac{\log_{15} a + \log_{32} a}{(\log_{15} a)(\log_{32} a)} = 4$ then a must lie in the range

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Correct answer: C

Think of this as: we're given a fraction whose numerator is a sum of two logs and whose denominator is their product. That structure — $\frac{x+y}{xy}$ — is a strong signal to split it into reciprocals.

Step 1 — Split the fraction.

Let $x = \log_{15} a$ and $y = \log_{32} a$ . The left-hand side is: $\frac{x + y}{xy} = \frac{1}{y} + \frac{1}{x}.$

Substituting back: $\frac{1}{\log_{32} a} + \frac{1}{\log_{15} a}.$

Step 2 — Flip each reciprocal using the base-switch identity.

The identity $\dfrac{1}{\log_b a} = \log_a b$ (which is just the change-of-base formula rearranged) turns each term around: $\frac{1}{\log_{32} a} = \log_a 32, \qquad \frac{1}{\log_{15} a} = \log_a 15.$

So the equation becomes: $\log_a 32 + \log_a 15 = 4.$

💡 Teacher tip: Whenever you see $\frac{1}{\log_b a}$ , immediately read it as $\log_a b$ . This single swap — exchanging the base and the argument — is what converts a messy two-log expression into something you can combine.

Step 3 — Combine the logs.

Since both terms now share the same base $a$ , use the product rule $\log_a p + \log_a q = \log_a(pq)$ : $\log_a(32 \times 15) = 4 \implies \log_a 480 = 4.$

Step 4 — Solve for $a$ .

Rewrite the logarithmic equation in exponential form: $\log_a 480 = 4$ means $a^4 = 480$ , so $a = 480^{1/4}.$

Step 5 — Pin $480^{1/4}$ between two consecutive integers.

We compare $480$ with nearby fourth powers:

Integer $n$	$n^4$	Compare with 480
4	$256$	$256 < 480$ ✓
5	$625$	$480 < 625$ ✓

Since $4^4 = 256 < 480 < 625 = 5^4$ and the fourth-root function is strictly increasing, taking fourth roots throughout preserves the inequalities: $4 < 480^{1/4} < 5.$

Therefore $4 < a < 5$ .

Answer: $\boxed{4 < a < 5}$ (Choice C)

💡 Closing insight: The entire problem collapses once you spot that $\frac{x+y}{xy} = \frac{1}{x}+\frac{1}{y}$ . After that, the base-switch identity does the heavy lifting, turning two logarithms with different bases into a single $\log_a 480$ — and the rest is just comparing fourth powers.

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Q3Maxima & MinimaMCQ

The cost of fencing a rectangular plot is ₹ 200 per ft along one side, and ₹ 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in INR, is

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Correct answer: A

Think of this as: we have a fixed area to enclose, but one side is twice as expensive to fence as the other three. Where should we place the expensive side — long or short — to keep the total bill as low as possible?

Setting up the cost equation

Let the side fenced at ₹200/ft have length $x$ , and the other side have length $y$ .

The four sides and their costs are:

Side	Length	Rate (₹/ft)	Cost
Expensive side	$x$	200	$200x$
Opposite side	$x$	100	$100x$
Third side	$y$	100	$100y$
Fourth side	$y$	100	$100y$

So the total cost is: $C = 200x + 100x + 100y + 100y = 300x + 200y$

We're also told the area is fixed: $xy = 60{,}000$

Step 1 — Reduce to one variable.

From $xy = 60{,}000$ , we get $y = \dfrac{60{,}000}{x}$ . Substituting: $C(x) = 300x + 200 \cdot \frac{60{,}000}{x} = 300x + \frac{1{,}20{,}00{,}000}{x}$

Step 2 — Minimise using AM–GM.

Both terms $300x$ and $\dfrac{1{,}20{,}00{,}000}{x}$ are positive, so we can apply the AM–GM inequality: $300x + \frac{1{,}20{,}00{,}000}{x} \geq 2\sqrt{300x \cdot \frac{1{,}20{,}00{,}000}{x}}$

The $x$ cancels inside the square root: $= 2\sqrt{300 \times 1{,}20{,}00{,}000} = 2\sqrt{3{,}60{,}00{,}00{,}000} = 2 \times 60{,}000 = 1{,}20{,}000$

So the minimum cost is ₹1,20,000, achieved when the two terms are equal.

💡 Teacher tip: AM–GM equality ( $A = B$ ) is the fastest route to the minimum of a " $\text{term} + \dfrac{\text{constant}}{\text{term}}$ " expression — no calculus needed. The minimum always occurs when the two terms are equal.

Step 3 — Verify the dimensions at the optimum.

Set the two terms equal: $300x = \frac{1{,}20{,}00{,}000}{x} \implies x^2 = \frac{1{,}20{,}00{,}000}{300} = 40{,}000 \implies x = 200$

Then: $y = \frac{60{,}000}{200} = 300$

Step 4 — Confirm the cost.

$C = 300(200) + 200(300) = 60{,}000 + 60{,}000 = 1{,}20{,}000$

Check	Value
$x = 200$ , $y = 300$	✓
Area $= 200 \times 300 = 60{,}000$	✓
Cost $= 60{,}000 + 60{,}000 = 1{,}20{,}000$	✓

Answer: ₹1,20,000 (Choice A)

💡 Closing insight: At the optimum, the two contributions to the cost — $300x$ (the total spent on the two sides of length $x$ ) and $200y$ (the total spent on the two sides of length $y$ ) — are equal. This balance is the signature of an AM–GM minimum: you spend the same on the "expensive-side pair" as on the "cheap-side pair." Notice also that the expensive side ( $x = 200$ ) is kept shorter than the cheap side ( $y = 300$ ) — a natural intuition, since you want less length at the higher rate.

Alternative approach (Calculus)

Differentiate $C(x) = 300x + \dfrac{1{,}20{,}00{,}000}{x}$ : $C'(x) = 300 - \frac{1{,}20{,}00{,}000}{x^2}$

Set $C'(x) = 0$ : $300 = \frac{1{,}20{,}00{,}000}{x^2} \implies x^2 = 40{,}000 \implies x = 200$

The second derivative $C''(x) = \dfrac{2{,}40{,}00{,}000}{x^3} > 0$ , confirming a minimum. The rest follows identically, giving $C = 1{,}20{,}000$ . The AM–GM method is faster under exam conditions since it directly gives the minimum value without computing $x$ first.

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Q4Surds & IndicesTITA

If $n$ is a positive integer such that $(\sqrt[7]{10})(\sqrt[7]{10})^2 ...(\sqrt[7]{10})^n > 999$ , then the smallest value of $n$ is

Your answer:

Show solution

Correct answer: 6

We need the smallest positive integer $n$ for which the product $\left(\sqrt[7]{10}\right)\left(\sqrt[7]{10}\right)^{2}\cdots\left(\sqrt[7]{10}\right)^{n}$ exceeds $999$ .

Step 1 — Rewrite the product using index laws.

Since $\sqrt[7]{10} = 10^{1/7}$ , each factor is a power of $10^{1/7}$ :

$\left(10^{1/7}\right)^{1}\cdot\left(10^{1/7}\right)^{2}\cdots\left(10^{1/7}\right)^{n}$

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

$= \left(10^{1/7}\right)^{1+2+\cdots+n}$

The sum $1+2+\cdots+n = \dfrac{n(n+1)}{2}$ , so:

$= \left(10^{1/7}\right)^{n(n+1)/2} = 10^{\,n(n+1)/(14)}$

💡 Teacher tip: The entire chain of $n$ factors collapses to a single power of 10. The problem is no longer about multiplication — it's about pinning down the exponent.

Step 2 — Translate " $> 999$ " into a bound on the exponent.

We need $10^{\,n(n+1)/14} > 999$ .

Since $999 < 1000 = 10^{3}$ , the product clears $999$ as soon as the exponent reaches $3$ (at exactly $3$ , the product is $10^3 = 1000 > 999$ ). So a clean sufficient condition is:

$\frac{n(n+1)}{14} \geq 3 \quad\Longrightarrow\quad n(n+1) \geq 42$

We should also confirm that no smaller $n$ sneaks past $999$ with an exponent slightly below $3$ — we'll check that in the table below.

Step 3 — Test integer values of $n$ .

$n$	$n(n+1)$	Exponent $\frac{n(n+1)}{14}$	Product	$> 999$ ?
$4$	$20$	$\approx 1.43$	$\approx 27$	✗
$5$	$30$	$\approx 2.14$	$\approx 139$	✗
$6$	$42$	$= 3$	$= 1000$	✅

At $n=5$ the exponent is only $\approx 2.14$ , far below the $\approx 2.9996$ threshold needed to exceed $999$ . At $n=6$ the exponent hits exactly $3$ , giving $1000 > 999$ .

Step 4 — Verify the boundary.

At $n = 6$ : product $= 10^{42/14} = 10^{3} = 1000 > 999$ ✓
At $n = 5$ : product $= 10^{30/14} \approx 10^{2.14} \approx 139 < 999$ ✓

The smallest $n$ is therefore $6$ .

Answer: $\boxed{6}$

💡 Closing insight: The identity $a^{1}\cdot a^{2}\cdots a^{n} = a^{n(n+1)/2}$ turns a long chain of multiplications into a single exponent, converting the problem into "when does the exponent cross $3$ ?" — a simple inequality check.

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Q5Functions & GraphsMCQ

If $f(x) = x^2 - 7x$ and $g(x) = x + 3$ , then the minimum value of $f(g(x)) - 3x$ is:

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Correct answer: D

Think of this as: first build the composite function $f(g(x))$ , then subtract $3x$ , and finally minimise the resulting expression.

Step 1 — Compose $f(g(x))$ .

Since $g(x) = x + 3$ , we substitute $x + 3$ into $f$ : $f(g(x)) = (x+3)^2 - 7(x+3).$

Expand each piece:

$(x+3)^2 = x^2 + 6x + 9$
$-7(x+3) = -7x - 21$

So: $f(g(x)) = x^2 + 6x + 9 - 7x - 21 = x^2 - x - 12.$

Step 2 — Subtract $3x$ .

$f(g(x)) - 3x = (x^2 - x - 12) - 3x = x^2 - 4x - 12.$

💡 Teacher tip: Always simplify $f(g(x))$ fully before optimising. If you try to minimise $f(g(x))$ and $-3x$ separately, you'll miss the interaction between them — the $-x$ from the composition combines with $-3x$ to give $-4x$ , which shifts the vertex.

Step 3 — Minimise the quadratic by completing the square.

We have $x^2 - 4x - 12$ . Take half the coefficient of $x$ (which is $-4 \div 2 = -2$ ), square it to get $4$ , then add and subtract: $x^2 - 4x - 12 = (x^2 - 4x + 4) - 4 - 12 = (x-2)^2 - 16.$

Since $(x-2)^2 \ge 0$ for every real $x$ , the smallest the expression can be is when $(x-2)^2 = 0$ , i.e. at $x = 2$ .

Step 4 — Read off the minimum.

At $x = 2$ : $(2-2)^2 - 16 = 0 - 16 = -16.$

Verification check:

$x$	$(x-2)^2 - 16$	Value
$0$	$4 - 16$	$-12$
$2$	$0 - 16$	$-16$ ✓ (minimum)
$4$	$4 - 16$	$-12$

The value at $x = 2$ is lower than at neighbouring points, confirming the minimum.

Answer: $\boxed{-16}$ (Choice D)

💡 Closing insight: After composition and simplification, the expression collapses to a plain quadratic whose vertex form $(x-2)^2 - 16$ reveals the minimum instantly. The entire problem is really just "simplify, then complete the square."

Alternative approach — Vertex formula. For a quadratic $ax^2 + bx + c$ with $a > 0$ , the minimum occurs at $x = -\frac{b}{2a}$ and equals $c - \frac{b^2}{4a}$ . Here $a = 1,\; b = -4,\; c = -12$ : $\text{Min} = -12 - \frac{(-4)^2}{4(1)} = -12 - 4 = -16.$ This is faster if you're comfortable with the formula, though completing the square is more reliable under pressure since it's harder to mix up signs.

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Q6PercentagesMCQ

In a tournament, a team has played 40 matches so far and won 30% of them. If they win 60% of the remaining matches, their overall win percentage will be 50%. Suppose they win 90% of the remaining matches, then the total number of matches won by the team in the tournament will be

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Correct answer: C

Think of this as: we know how the team has done so far, and we're told what would happen if they won 60% of the rest. That "what if" lets us pin down how many remaining matches there are — and once we know that, the 90% case is just arithmetic.

Step 1 — Count the wins so far.

The team has played 40 matches and won 30% of them: $0.30 \times 40 = 12 \text{ wins.}$ So they currently have 12 wins out of 40 matches.

Step 2 — Use the 50% condition to find how many matches remain.

Let $R$ be the number of remaining matches. If they win 60% of those, their overall win percentage becomes 50%.

Wins from remaining matches: $0.6R$
Total wins: $12 + 0.6R$
Total matches: $40 + R$

Setting the overall win rate to 50%: $\frac{12 + 0.6R}{40 + R} = 0.5$

Cross-multiply: $12 + 0.6R = 0.5(40 + R)$ $12 + 0.6R = 20 + 0.5R$ $0.1R = 8$ $R = 80$

So there are 80 remaining matches.

💡 Teacher tip: The 50% here is a weighted average of 30% (over 40 matches) and 60% (over $R$ matches). By alligation, the ratio of matches is $(60-50):(50-30) = 10:20 = 1:2$ . Since the 40 matches form 1 part, $R = 80$ — the same result in one line.

Step 3 — Apply the 90% condition.

Now suppose the team wins 90% of the 80 remaining matches: $0.9 \times 80 = 72 \text{ wins from the remaining matches.}$

Step 4 — Add the wins from both phases. $\text{Total wins} = 12 + 72 = 84.$

Quick check:

Phase	Matches	Win %	Wins
Already played	40	30%	12
Remaining	80	90%	72
Total	120	—	84

Answer: $\boxed{84}$ (Choice C)

💡 Closing insight: The key move is realising the 50% condition is just a tool to discover $R$ — the number of remaining matches. Once $R$ is known, swapping 60% for 90% is a straightforward recalculation. Whenever a problem gives you a hypothetical outcome to pin down a hidden quantity, solve for that quantity first, then answer the actual question.

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Q7Time & WorkMCQ

One day, Rahul started a work at 9 AM and Gautam joined him two hours later. They then worked together and completed the work at 5 PM the same day. If both had started at 9 AM and worked together, the work would have been completed 30 minutes earlier. Working alone, the time Rahul would have taken, in hours, to complete the work is

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Correct answer: B

Think of this as: we're given two different schedules for the same job, and each schedule tells us how long each person worked. Two equations, two unknowns — the rates of Rahul and Gautam.

What we want: the time Rahul alone would take to finish the work.

Let Rahul's work rate be $r$ (fraction of the job per hour) and Gautam's rate be $g$ . The whole job $= 1$ .

Step 1 — Translate each scenario into an equation.

Scenario 1 (what actually happened): Rahul started at 9 AM and worked until 5 PM — that's $8$ hours. Gautam joined two hours later, i.e. at 11 AM, and also worked until 5 PM — that's $6$ hours.

$8r + 6g = 1$

Scenario 2 (the hypothetical): Both start together at 9 AM. The work finishes 30 minutes earlier than 5 PM, i.e. at 4:30 PM — so they work together for $7.5$ hours.

$7.5(r + g) = 1$

Step 2 — Extract the combined rate from Scenario 2.

$r + g = \frac{1}{7.5} = \frac{1}{\frac{15}{2}} = \frac{2}{15}$

So Gautam's rate in terms of Rahul's is:

$g = \frac{2}{15} - r$

Step 3 — Substitute into Scenario 1 and solve for $r$ .

$8r + 6\left(\frac{2}{15} - r\right) = 1$

$8r + \frac{12}{15} - 6r = 1$

$2r + \frac{4}{5} = 1$

$2r = 1 - \frac{4}{5} = \frac{1}{5}$

$r = \frac{1}{10}$

So Rahul completes $\frac{1}{10}$ of the job per hour.

Step 4 — Find Rahul's solo time.

Time $= \dfrac{\text{Total work}}{\text{Rate}} = \dfrac{1}{r} = \dfrac{1}{\frac{1}{10}} = 10$ hours.

Quick verification. If $r = \frac{1}{10}$ , then $g = \frac{2}{15} - \frac{1}{10} = \frac{4}{30} - \frac{3}{30} = \frac{1}{30}$ .

Scenario 1: $8 \cdot \frac{1}{10} + 6 \cdot \frac{1}{30} = \frac{4}{5} + \frac{1}{5} = 1$ ✓
Scenario 2: $7.5 \cdot \left(\frac{1}{10} + \frac{1}{30}\right) = 7.5 \cdot \frac{4}{30} = 7.5 \cdot \frac{2}{15} = 1$ ✓

Both scenarios check out.

Answer: $\boxed{10}$ (Choice B)

💡 Closing insight: The 30-minute gap between the two scenarios is exactly the work Gautam didn't do during his first two absent hours. In fact, $2g = 2 \cdot \frac{1}{30} = \frac{1}{15}$ , which equals half an hour of combined work: $\frac{1}{2} \cdot \frac{2}{15} = \frac{1}{15}$ . The two scenarios self-check once you have both rates.

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Q8InequalitiesMCQ

If $3x + 2|y| + y = 7$ and $x + |x| + 3y = 1$ then $x + 2y$ is:

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Correct answer: C

What we want: the value of $x + 2y$ , given $3x + 2|y| + y = 7$ and $x + |x| + 3y = 1$ .

The two equations carry $|x|$ and $|y|$ , so the honest way through is to remove the modulus signs by testing each combination of signs. Each variable is either $\ge 0$ or $< 0$ , giving four cases.

Case 1 — $x \ge 0,\ y \ge 0$ . Then $|x| = x$ and $|y| = y$ : $3x + 3y = 7, \qquad 2x + 3y = 1.$ Subtracting, $x = 6$ , so $3y = 7 - 18 = -11$ — but that makes $y < 0$ , contradicting $y \ge 0$ . Rejected.

Case 2 — $x \ge 0,\ y < 0$ . Then $|x| = x$ and $|y| = -y$ : $3x - y = 7, \qquad 2x + 3y = 1.$ From the first, $y = 3x - 7$ . Substituting into the second: $2x + 3(3x - 7) = 1 \;\Rightarrow\; 11x = 22 \;\Rightarrow\; x = 2, \quad y = -1.$ Both assumptions hold: $x = 2 \ge 0$ and $y = -1 < 0$ . This case survives.

Step — Verify in the original equations.

Equation 1: $3(2) + 2|-1| + (-1) = 6 + 2 - 1 = 7$ ✓
Equation 2: $2 + |2| + 3(-1) = 2 + 2 - 3 = 1$ ✓

The other two cases fail.

$x < 0,\ y \ge 0$ : equation 2 becomes $3y = 1$ , so $y = \tfrac13$ , then equation 1 gives $3x + 1 = 7 \Rightarrow x = 2$ , contradicting $x < 0$ .
$x < 0,\ y < 0$ : equation 2 becomes $3y = 1 \Rightarrow y = \tfrac13 > 0$ , contradicting $y < 0$ .

So the unique solution is $x = 2,\ y = -1$ .

Step — Compute the target. $x + 2y = 2 + 2(-1) = 0.$

Answer: $\boxed{0}$ (Choice C)

💡 Teacher tip: With two modulus variables you always have four sign cases, but the equations themselves usually rule out all but one. Fix the signs first; what's left is an ordinary pair of linear equations that solves in two lines.

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Q9Quadratic & Linear EquationsMCQ

A shop owner bought a total of 64 shirts from a wholesale market that came in two sizes, small and large. The price of a small shirt was INR 50 less than that of a large shirt. She paid a total of INR 5000 for the large shirts, and a total of INR 1800 for the small shirts. Then, the price of a large shirt and a small shirt together, in INR, is

Show solution

Correct answer: C

Think of this as: we know the total number of shirts, the price gap between the two sizes, and the total spent on each size — we need to back out the individual prices and add them.

Step 1 — Set up variables.

Let:

$p_L$ = price of one large shirt, $p_S$ = price of one small shirt
$L$ = number of large shirts bought, $S$ = number of small shirts bought

Translating every piece of given information:

Condition	Equation
Total shirts	$L + S = 64$
Price gap	$p_L - p_S = 50$
Total spent on large	$L \cdot p_L = 5000$
Total spent on small	$S \cdot p_S = 1800$

We want: $p_L + p_S$ .

Step 2 — Express prices in terms of quantities.

From the spending equations: $p_L = \frac{5000}{L}, \qquad p_S = \frac{1800}{S}$

Now use $S = 64 - L$ and substitute into the price-gap equation $p_S = p_L - 50$ :

$\frac{1800}{64 - L} = \frac{5000}{L} - 50$

Step 3 — Clear denominators and solve for $L$ .

Multiply both sides by $L(64 - L)$ :

$1800L = (64 - L)(5000 - 50L)$

Expand the right side:

$1800L = 320000 - 3200L - 5000L + 50L^2$ $1800L = 320000 - 8200L + 50L^2$

Bring everything to one side:

$50L^2 - 10000L + 320000 = 0$

Divide through by $50$ :

$L^2 - 200L + 6400 = 0$

Step 4 — Apply the quadratic formula.

$L = \frac{200 \pm \sqrt{200^2 - 4 \cdot 6400}}{2} = \frac{200 \pm \sqrt{40000 - 25600}}{2} = \frac{200 \pm \sqrt{14400}}{2} = \frac{200 \pm 120}{2}$

So $L = 160$ or $L = 40$ .

Since $L + S = 64$ , we need $L \le 64$ . The root $L = 160$ is physically impossible (you can't buy 160 large shirts out of 64 total), so we reject it.

$L = 40$ , and therefore $S = 64 - 40 = 24$ .

💡 Teacher tip: A quadratic in a word problem often yields two roots. Always check each root against the real-world constraints — here, the total count of 64 shirts eliminates $L = 160$ immediately.

Step 5 — Find the prices.

$p_L = \frac{5000}{40} = 125, \qquad p_S = \frac{1800}{24} = 75$

Quick check: $p_L - p_S = 125 - 75 = 50$ ✓ (matches the given price gap).

Step 6 — Compute the answer.

$p_L + p_S = 125 + 75 = 200$

Answer: $\boxed{200}$ (Choice C)

Alternative approach — Target the answer directly.

Instead of solving for $L$ first, define $T = p_L + p_S$ (exactly what we want) and use the given gap $p_L - p_S = 50$ to write both prices in terms of $T$ :

$p_L = \frac{T + 50}{2}, \qquad p_S = \frac{T - 50}{2}$

Now express $L$ and $S$ in terms of $T$ :

$L = \frac{5000}{p_L} = \frac{10000}{T + 50}, \qquad S = \frac{1800}{p_S} = \frac{3600}{T - 50}$

Since $L + S = 64$ :

$\frac{10000}{T + 50} + \frac{3600}{T - 50} = 64$

Multiply through by $(T + 50)(T - 50)$ :

$10000(T - 50) + 3600(T + 50) = 64(T^2 - 2500)$ $10000T - 500000 + 3600T + 180000 = 64T^2 - 160000$ $13600T - 320000 = 64T^2 - 160000$ $64T^2 - 13600T + 160000 = 0$

Divide by 32:

$2T^2 - 425T + 5000 = 0$

Quadratic formula:

$T = \frac{425 \pm \sqrt{425^2 - 4 \cdot 2 \cdot 5000}}{4} = \frac{425 \pm \sqrt{180625 - 40000}}{4} = \frac{425 \pm \sqrt{140625}}{4} = \frac{425 \pm 375}{4}$

So $T = \frac{800}{4} = 200$ or $T = \frac{50}{4} = 12.5$ .

Since $T = 12.5$ would give $p_S = \frac{12.5 - 50}{2} = -18.75$ (a negative price), we reject it.

$T = 200$ , confirming our answer.

💡 Why this is elegant: by defining $T = p_L + p_S$ upfront, we collapse a two-stage problem (find quantities → find prices → add them) into a single equation in the exact quantity we want. The price gap $p_L - p_S = 50$ is the bridge that lets both prices be written in terms of $T$ alone.

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Q10MensurationTITA

A park is shaped like a rhombus and has area 96 sq m. If 40 m of fencing is needed to enclose the park, the cost, in INR, of laying electric wires along its two diagonals, at the rate of ₹125 per m, is

Your answer:

Show solution

Correct answer: 3500

Think of this as: we know the area and the perimeter of a rhombus, and we need the sum of its two diagonals — because the wire runs along both. The trick is to avoid solving for each diagonal individually.

What's given:

Area of the rhombus park = $96 \text{ sq m}$
Fencing needed = $40 \text{ m}$ , which is the perimeter.
Wire runs along both diagonals; rate = ₹ $125$ per metre.

Step 1 — Find the side length.

A rhombus has four equal sides, so $4s = 40 \quad\Rightarrow\quad s = 10 \text{ m}.$

Step 2 — Translate the two given facts into equations involving the diagonals.

Let the diagonals be $d_1$ and $d_2$ . Two standard properties of a rhombus connect them to what we know:

Area = half the product of the diagonals: $\tfrac{1}{2}\,d_1 d_2 = 96 \quad\Rightarrow\quad d_1 d_2 = 192.$
Diagonals bisect each other at right angles, so each side is the hypotenuse of a right triangle with legs $\tfrac{d_1}{2}$ and $\tfrac{d_2}{2}$ : $\left(\frac{d_1}{2}\right)^{2}+\left(\frac{d_2}{2}\right)^{2}=s^{2}=100.$ Multiplying through by $4$ : $d_1^{2}+d_2^{2}=400.$

💡 Teacher tip: Notice we now have $d_1 d_2$ and $d_1^2 + d_2^2$ . That's enough to get $d_1 + d_2$ without ever finding $d_1$ or $d_2$ on their own — the algebra hands us exactly the combination we need.

Step 3 — Find $d_1 + d_2$ , the total length of wire.

Use the identity $(d_1+d_2)^{2}=d_1^{2}+d_2^{2}+2\,d_1 d_2.$ Substitute the two values: $(d_1+d_2)^{2}=400+2(192)=400+384=784.$ $d_1+d_2=\sqrt{784}=28 \text{ m}.$

So the wire must cover $28$ m in total.

Step 4 — Compute the cost.

$\text{Cost}=28 \times 125 = 3500 \text{ INR}.$

Answer: $\boxed{3500}$

💡 Closing insight: We never needed the individual diagonals — only their sum, which drops straight out of $(d_1+d_2)^2 = d_1^2 + d_2^2 + 2d_1d_2$ once the area and perimeter pin down the two symmetric combinations $d_1 d_2$ and $d_1^2 + d_2^2$ . Whenever a question asks for a sum of unknowns, look for identities that produce that sum directly.

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Q11Polygons & QuadrilateralsMCQ

Let $ABCD$ be a parallelogram. The lengths of the side $AD$ and the diagonal $AC$ are 10 cm and 20 cm, respectively. If the angle $\angle ADC$ is equal to $30^\circ$ then the area of the parallelogram, in sq.cm is

Show solution

Correct answer: B

Think of this as: we know one side, the diagonal opposite a known angle, and that angle — so we can pin down the missing side with the cosine rule, then read off the area as base × height.

What we want: the area of parallelogram $ABCD$ , given $AD = 10$ , diagonal $AC = 20$ , and $\angle ADC = 30°$ .

Step 1 — Identify the triangle that holds all the data

The diagonal $AC$ splits the parallelogram into two triangles. In $\triangle ADC$ we know:

$AD = 10$
$AC = 20$
$\angle ADC = 30°$

This is enough to find the third side $DC$ (which is a side of the parallelogram, since $AB = DC$ and $AD = BC$ ).

Step 2 — Find $DC$ using the cosine rule

Applying the cosine rule in $\triangle ADC$ at vertex $D$ :

$AC^{2} = AD^{2} + DC^{2} - 2\cdot AD \cdot DC \cdot \cos 30^{\circ}$

Substituting the known values ( $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ ):

$400 = 100 + DC^{2} - 2\cdot 10 \cdot DC \cdot \frac{\sqrt{3}}{2}$

$400 = 100 + DC^{2} - 10\sqrt{3}\, DC$

$DC^{2} - 10\sqrt{3}\, DC - 300 = 0$

Step 3 — Solve the quadratic

Using the quadratic formula with $a = 1$ , $b = -10\sqrt{3}$ , $c = -300$ :

$DC = \frac{10\sqrt{3} \pm \sqrt{(10\sqrt{3})^{2} + 4\cdot 300}}{2} = \frac{10\sqrt{3} \pm \sqrt{300 + 1200}}{2} = \frac{10\sqrt{3} \pm \sqrt{1500}}{2}$

Since $\sqrt{1500} = \sqrt{100 \times 15} = 10\sqrt{15}$ :

$DC = \frac{10\sqrt{3} \pm 10\sqrt{15}}{2} = 5\left(\sqrt{3} \pm \sqrt{15}\right)$

Now we must pick the physically valid root — a length is positive.

Root	Value	Sign	Valid?
$5(\sqrt{3} + \sqrt{15})$	$\approx 5(1.73 + 3.87) \approx 28$	Positive	✅
$5(\sqrt{3} - \sqrt{15})$	$\approx 5(1.73 - 3.87) \approx -10.7$	Negative	❌

So $DC = 5\left(\sqrt{3} + \sqrt{15}\right)$ .

Step 4 — Compute the area

The area of a parallelogram is $\text{base} \times \text{height}$ . Taking $DC$ as the base, the height from $A$ down to $DC$ is:

$h = AD \cdot \sin(\angle ADC) = 10 \cdot \sin 30^{\circ} = 10 \cdot \tfrac{1}{2} = 5$

Therefore:

$\text{Area} = DC \times h = 5\left(\sqrt{3} + \sqrt{15}\right) \times 5 = 25\left(\sqrt{3} + \sqrt{15}\right)$

Answer: $\boxed{25(\sqrt{3}+\sqrt{15})}$ (Choice B)

Alternative approach — Sine rule in $\triangle ADC$

By the sine rule: $\frac{AC}{\sin D} = \frac{AD}{\sin C} = \frac{DC}{\sin A}$ .

$\frac{20}{\sin 30^{\circ}} = \frac{10}{\sin C} \implies \frac{20}{1/2} = 40 = \frac{10}{\sin C} \implies \sin C = \frac{1}{4}$

So $\cos C = \sqrt{1 - \frac{1}{16}} = \frac{\sqrt{15}}{4}$ , and $\angle A = 180° - 30° - C = 150° - C$ .

$\sin A = \sin(150° - C) = \sin 150°\cos C - \cos 150°\sin C = \frac{1}{2}\cdot\frac{\sqrt{15}}{4} - \left(-\frac{\sqrt{3}}{2}\right)\cdot\frac{1}{4} = \frac{\sqrt{15} + \sqrt{3}}{8}$

$DC = 40 \times \sin A = 40 \times \frac{\sqrt{3}+\sqrt{15}}{8} = 5(\sqrt{3}+\sqrt{15})$

The area is then $DC \times AD \sin 30° = 5(\sqrt{3}+\sqrt{15}) \times 5 = 25(\sqrt{3}+\sqrt{15})$ , confirming the result. The cosine-rule method is faster under exam conditions.

💡 Closing insight: Two pieces of data ( $AD$ and $\angle D$ ) give the height directly via $\sin 30°$ ; the diagonal $AC$ then fixes the base $DC$ through the cosine rule. The surds never collapse to a single radical — the answer is genuinely a sum of $\sqrt{3}$ and $\sqrt{15}$ .

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Q12Mixtures & AlligationsMCQ

If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy will have 90% silver by weight. If the same weight of the initial alloy is mixed with 2 kg of another alloy which has 90% silver by weight, the resulting alloy will have 84% silver by weight. Then, the weight of the initial alloy, in kg, is

Show solution

Correct answer: C

Think of this as: we have one unknown alloy (we don't know its weight or its silver percentage), and we're given two mixing experiments. Each experiment gives us one equation, so two experiments = two equations = enough to solve for both unknowns. We only need the weight.

Let $W$ = weight (in kg) of the initial alloy, and let $s$ = fraction of silver in it.

Then the total silver in the initial alloy is $A = Ws$ .

We'll treat $A$ and $W$ as our two unknowns — the silver fraction $s$ will take care of itself.

Method 1 — Two linear equations

Step 1 — Translate the first mixing.

We add $3$ kg of pure silver ( $100\%$ silver, so $3$ kg of silver) to the initial alloy. The new total weight is $W + 3$ , and the new silver amount is $A + 3$ . The result is $90\%$ silver:

$\frac{A + 3}{W + 3} = 0.90$

$A + 3 = 0.90(W + 3) = 0.90W + 2.70$

$\boxed{A - 0.90W = -0.30} \quad \cdots (1)$

Step 2 — Translate the second mixing.

We add $2$ kg of an alloy that is $90\%$ silver. The silver contributed by this added alloy is $0.90 \times 2 = 1.8$ kg. The new total weight is $W + 2$ , and the new silver amount is $A + 1.8$ . The result is $84\%$ silver:

$\frac{A + 1.8}{W + 2} = 0.84$

$A + 1.8 = 0.84(W + 2) = 0.84W + 1.68$

$\boxed{A - 0.84W = -0.12} \quad \cdots (2)$

💡 Teacher tip: Both equations are linear in $A$ and $W$ . By keeping $A = Ws$ as a single symbol instead of substituting $Ws$ everywhere, we avoid a messy nonlinear system. The fraction $s$ is hidden inside $A$ and never causes trouble.

Step 3 — Eliminate $A$ .

Subtract equation $(2)$ from equation $(1)$ :

$(A - 0.90W) - (A - 0.84W) = (-0.30) - (-0.12)$

$-0.06W = -0.18$

$W = \frac{-0.18}{-0.06} = 3$

Step 4 — Verify both conditions.

From equation $(1)$ : $A = 0.90(3) - 0.30 = 2.4$ , so the initial alloy is $\frac{2.4}{3} = 80\%$ silver.

Mixing	Silver in mix	Total weight	Silver %	Matches?
Add 3 kg pure silver	$2.4 + 3 = 5.4$	$3 + 3 = 6$	$5.4/6 = 90\%$	✅
Add 2 kg of 90% alloy	$2.4 + 1.8 = 4.2$	$3 + 2 = 5$	$4.2/5 = 84\%$	✅

Both conditions check out.

Alternative approach — Alligation

Alligation links the concentrations directly to the ratio of quantities mixed.

Mix 1: Pure silver ( $100\%$ ) + initial alloy ( $s\%$ ) → $90\%$ .

$\frac{\text{pure silver}}{\text{initial alloy}} = \frac{90 - s}{100 - 90} = \frac{90 - s}{10}$

Since $3$ kg of pure silver is mixed with $W$ kg of the alloy:

$\frac{3}{W} = \frac{90 - s}{10} \;\Longrightarrow\; W = \frac{30}{90 - s} \quad \cdots (1')$

Mix 2: $90\%$ alloy + initial alloy ( $s\%$ ) → $84\%$ .

$\frac{\text{90% alloy}}{\text{initial alloy}} = \frac{84 - s}{90 - 84} = \frac{84 - s}{6}$

Since $2$ kg of the $90\%$ alloy is mixed with $W$ kg of the initial alloy:

$\frac{2}{W} = \frac{84 - s}{6} \;\Longrightarrow\; W = \frac{12}{84 - s} \quad \cdots (2')$

Equate the two expressions for $W$ :

$\frac{30}{90 - s} = \frac{12}{84 - s}$

$30(84 - s) = 12(90 - s)$

$2520 - 30s = 1080 - 12s$

$1440 = 18s \;\Longrightarrow\; s = 80$

So the initial alloy is $80\%$ silver. Plug back:

$W = \frac{30}{90 - 80} = \frac{30}{10} = 3$

Same answer, reached through concentrations instead of silver amounts.

Answer: $\boxed{3}$ (Choice C)

💡 Closing insight: The key move is noticing that both mixings share the same unknown alloy, so they form a $2 \times 2$ linear system. Whether you solve it with silver-amount equations (Method 1) or alligation ratios (Method 2), the silver fraction $s$ cancels out of the weight calculation — you never need to know it to find $W$ . Method 1 is slightly faster under exam time since it goes straight to elimination.

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Q13Simple & Compound InterestMCQ

Bank A offers 6% interest rate per annum compounded half-yearly. Bank B and Bank C offer simple interest but the annual interest rate offered by Bank C is twice that of Bank B. Raju invests a certain amount in Bank B for a certain period and Rupa invests ₹ 10,000 in Bank C for twice that period. The interest that would accrue to Raju during that period is equal to the interest that would have accrued had he invested the same amount in Bank A for one year. The interest accrued, in INR, to Rupa is

Show solution

Correct answer: B

Think of this as: we're given one equation linking Raju's interest to Bank A's one-year return, and we need to translate that into Rupa's interest — without ever finding the individual rate or time.

Step 1 — Set up the variables.

Let Bank B's simple-interest rate be $r\%$ per annum. Then Bank C's rate is $2r\%$ per annum (given as twice Bank B's).

Let Raju invest an amount $P$ in Bank B for a period of $t$ years.

Step 2 — Compute Bank A's one-year interest (the benchmark).

Bank A offers $6\%$ p.a. compounded half-yearly, so:

Half-yearly rate $= \frac{6\%}{2} = 3\%$
Number of compounding periods in one year $= 2$

If the same amount $P$ were invested in Bank A for one year: $\text{Amount} = P\,(1.03)^{2} = P \times 1.0609$ $\text{Interest}_{A} = P(1.0609 - 1) = 0.0609\,P$

💡 Teacher tip: Half-yearly compounding at $6\%$ p.a. does not give $6\%$ for the year — it gives slightly more, because the first half-year's interest itself earns interest. That's why $1.03^2 = 1.0609$ , not $1.06$ .

Step 3 — Equate Raju's Bank B interest to the Bank A benchmark.

Raju's simple interest in Bank B (principal $P$ , rate $r\%$ , time $t$ years): $\text{Interest}_{B} = \frac{P \cdot r \cdot t}{100}$

We're told this equals the Bank A interest: $\frac{P\,r\,t}{100} = 0.0609\,P$

Cancel $P$ from both sides: $\frac{r\,t}{100} = 0.0609 \quad\Longrightarrow\quad r\,t = 6.09$

💡 Teacher tip: Notice we only need the product $rt$ , not $r$ and $t$ separately. The problem is designed so that this product is enough.

Step 4 — Compute Rupa's interest.

Rupa invests ₹ $10{,}000$ in Bank C at rate $2r\%$ for a period of $2t$ years. Her simple interest is: $\text{Interest}_{\text{Rupa}} = \frac{10{,}000 \times (2r) \times (2t)}{100}$

$= \frac{10{,}000 \times 4\,(r\,t)}{100}$

Substitute $r\,t = 6.09$ : $= \frac{10{,}000 \times 4 \times 6.09}{100} = 10{,}000 \times 0.2436 = 2436$

Quick check of the arithmetic:

Quantity	Value
$rt$ from Step 3	$6.09$
Rupa's effective $\frac{\text{rate} \times \text{time}}{100}$	$\frac{4 \times 6.09}{100} = 0.2436$
Rupa's principal	₹ $10{,}000$
Interest $= 10{,}000 \times 0.2436$	₹ $2{,}436$

Answer: $\boxed{2436}$ (Choice B)

💡 Closing insight: Rupa's interest is $4rt\%$ of ₹ $10{,}000$ because both her rate ( $\times 2$ ) and her period ( $\times 2$ ) are doubled relative to Raju's setup — giving a factor of $4$ . The entire problem reduces to finding the single number $rt = 6.09$ , which Bank A's one-year compounded return hands us directly.

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Q14Polynomials & ProgressionsMCQ

Consider a sequence of real numbers, $x_1, x_2, x_3, ...$ such that $x_{n+1} = x_n + n - 1$ for all $n \geq 1$ . If $x_1 = -1$ then $x_{100}$ is equal to

Show solution

Correct answer: D

Think of this as: each step adds a number that grows by 1 each time — first we add 0, then 1, then 2, then 3, and so on. So $x_{100}$ is just $x_1$ plus the sum of all those additions.

Step 1 — Write the recurrence as a difference.

We are given: $x_{n+1} = x_n + n - 1 \quad\Longrightarrow\quad x_{n+1} - x_n = n - 1.$

This tells us the jump from one term to the next.

Step 2 — Unroll from $x_1$ to $x_{100}$ .

Write out the jumps for each step:

Step	Jump
$x_2 - x_1$	$1 - 1 = 0$
$x_3 - x_2$	$2 - 1 = 1$
$x_4 - x_3$	$3 - 1 = 2$
$\vdots$	$\vdots$
$x_{100} - x_{99}$	$99 - 1 = 98$

Adding all of these, the left side telescopes — every intermediate term cancels: $x_{100} - x_1 = 0 + 1 + 2 + \cdots + 98.$

Step 3 — Sum the arithmetic series.

The right side is the sum of integers from $0$ to $98$ : $0 + 1 + 2 + \cdots + 98 = \frac{98 \times 99}{2} = 49 \times 99 = 4851.$

Step 4 — Solve for $x_{100}$ .

Since $x_1 = -1$ : $x_{100} = x_1 + 4851 = -1 + 4851 = 4850.$

💡 Teacher tip: Whenever a recurrence has the form $x_{n+1} - x_n = f(n)$ , unroll it and telescope. The closed form is always $x_n = x_1 + \sum_{k=1}^{n-1} f(k)$ . Here $f(k) = k-1$ , so the sum is $0 + 1 + \cdots + (n-2) = \frac{(n-1)(n-2)}{2}$ , giving the general formula $x_n = -1 + \frac{(n-1)(n-2)}{2}$ .

Sanity check with small terms.

$n$	Recurrence value	Formula value
$1$	$-1$	$-1 + \frac{0 \cdot (-1)}{2} = -1$ ✓
$2$	$-1 + 0 = -1$	$-1 + \frac{1 \cdot 0}{2} = -1$ ✓
$3$	$-1 + 1 = 0$	$-1 + \frac{2 \cdot 1}{2} = 0$ ✓
$4$	$0 + 2 = 2$	$-1 + \frac{3 \cdot 2}{2} = 2$ ✓

Answer: $\boxed{4850}$ (Choice D)

💡 Why this is neat: the recurrence feeds in a steadily growing arithmetic sequence, so the closed form is quadratic in $n$ . Summing $1$ through $98$ and subtracting $1$ lands exactly on $4850$ .

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Q15Permutation & CombinationTITA

A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is

Your answer:

Show solution

Correct answer: 50

Think of this as: we have a universe of all four-digit numbers made from $\{1,2,3\}$ , and we want to keep only those that contain at least one $2$ and at least one $3$ . The cleanest way is to count everything, then throw away what we don't want.

Step 1 — Count the total universe.

Each of the $4$ positions can be filled with any of $\{1,2,3\}$ , independently. So the total number of four-digit numbers using only these digits is: $3^{4} = 81.$

Step 2 — Identify what must be removed.

We need both $2$ and $3$ to appear. So we must remove the numbers that are missing $2$ or missing $3$ .

Numbers missing $2$ use only $\{1,3\}$ : each position has $2$ choices, giving $2^{4} = 16$ .
Numbers missing $3$ use only $\{1,2\}$ : each position has $2$ choices, giving $2^{4} = 16$ .

A first pass gives $81 - 16 - 16 = 49$ , but this is not yet correct — we have subtracted something twice.

Step 3 — Fix the double subtraction.

💡 Teacher tip: When two sets overlap, anything in both gets subtracted twice. Inclusion–exclusion says: add that overlap back once.

Which numbers are missing $2$ and missing $3$ ? Those use only $\{1\}$ — there is exactly $1^{4} = 1$ such number: $1111$ . It was removed in both Step 2 subtractions, so we add it back once: $81 - 16 - 16 + 1 = 50.$

Step 4 — Verify with a direct count.

Let $(a,b,c)$ denote the counts of $1,2,3$ in the number, with $a+b+c=4$ and $b \ge 1, c \ge 1$ . Each valid composition gives $\frac{4!}{a!\,b!\,c!}$ arrangements.

$(a,b,c)$	Arrangements $\frac{4!}{a!\,b!\,c!}$
$(2,1,1)$	$\frac{24}{2}=12$
$(1,2,1)$	$\frac{24}{2}=12$
$(1,1,2)$	$\frac{24}{2}=12$
$(0,2,2)$	$\frac{24}{4}=6$
$(0,1,3)$	$\frac{24}{6}=4$
$(0,3,1)$	$\frac{24}{6}=4$

Total: $12+12+12+6+4+4 = 50$ . ✅

Both methods agree, confirming the count is complete.

Answer: $\boxed{50}$

💡 Why inclusion–exclusion wins here: the direct count requires listing six compositions and computing a multinomial for each. Inclusion–exclusion turns the same problem into three lines of arithmetic — count all, subtract the two exclusions, add back the overlap. Whenever a question says "each of these must appear at least once," reach for this pattern first.

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Q16Averages & Weighted AveragesTITA

The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is

Your answer:

Show solution

Correct answer: 92

Think of this as a budgeting problem: the total marks are fixed at $1250$ . The five toppers share an equal score $t$ , and the remaining $20$ students have distinct integer scores with the smallest being $30$ . To make $t$ as large as possible, we need to spend as little of the total as possible on those other $20$ students.

Step 1 — Lock down the total marks.

The mean of $25$ students is $50$ , so the total marks are fixed: $\text{Total} = 25 \times 50 = 1250.$

Step 2 — Set up the two groups.

Let each of the five toppers score $t$ . Then the five toppers contribute $5t$ marks, and the remaining $20$ students contribute the rest: $5t + (\text{sum of the other } 20 \text{ scores}) = 1250.$

To maximise $t$ , we must minimise the sum of the other $20$ scores.

Step 3 — Find the minimum possible sum of the other 20 scores.

The other $20$ students have distinct integer scores, and the lowest score is $30$ . To make their sum as small as possible, we pack them as tightly as possible just above $30$ : $30,\; 31,\; 32,\; \dots,\; 49.$

This is an arithmetic progression with $20$ terms, first term $30$ , last term $49$ : $\text{Sum} = \frac{20}{2}\,(30 + 49) = 10 \times 79 = 790.$

💡 Teacher tip: Whenever you need the minimum sum of several distinct integers with a fixed floor, use a block of consecutive integers starting at that floor. Any gap you insert only pushes later terms higher, increasing the sum.

Step 4 — Solve for the toppers' score.

The five toppers must account for the remaining marks: $5t = 1250 - 790 = 460 \quad\Longrightarrow\quad t = \frac{460}{5} = 92.$

Step 5 — Verify the toppers are genuinely on top.

We must confirm that $t$ actually exceeds every non-topper score — otherwise they wouldn't be toppers.

Check	Value	Valid?
Largest of the other 20 scores	$49$	—
Toppers' score $t$	$92$	—
Is $t > 49$ ?	$92 > 49$	✅
Are the five topper scores distinct from the others?	$92 \notin \{30,\dots,49\}$	✅

All conditions are satisfied.

Answer: $\boxed{92}$

💡 Closing insight: The key move was recognising that maximising one group's score means minimising everyone else's. And "20 distinct integers with a fixed minimum of 30" has a unique minimum-sum configuration — the consecutive block $30, 31, \dots, 49$ — which made the rest of the problem a simple subtraction and division.

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Q17InequalitiesTITA

The number of distinct pairs of integers (m,n), satisfying $|1 + mn| < |m + n| < 5$ is:

Your answer:

Show solution

Correct answer: 12

We need to count all integer pairs $(m,n)$ satisfying $|1+mn| < |m+n| < 5$ . Think of this as two filters stacked: the first inequality $|1+mn| < |m+n|$ restricts which integer pairs are even eligible, and the second inequality $|m+n| < 5$ then bounds how large the sum can be.

Step 1 — Convert the first inequality into an algebraic sign condition.

The inequality $|1+mn| < |m+n|$ involves absolute values, which are awkward to handle case-by-case. Instead, since both sides are non-negative, we can square both sides without changing the inequality:

$(1+mn)^2 < (m+n)^2$

Rearranging:

$(m+n)^2 - (1+mn)^2 > 0$

Now expand both squares:

$(m+n)^2 = m^2 + 2mn + n^2$ $(1+mn)^2 = 1 + 2mn + m^2n^2$

Subtracting:

$(m+n)^2 - (1+mn)^2 = m^2 + 2mn + n^2 - 1 - 2mn - m^2n^2 = m^2 + n^2 - 1 - m^2n^2$

Factor this expression:

$m^2 + n^2 - 1 - m^2n^2 = -(m^2n^2 - m^2 - n^2 + 1) = -(m^2-1)(n^2-1)$

So the first inequality becomes:

$-(m^2-1)(n^2-1) > 0 \quad\Longleftrightarrow\quad (m^2-1)(n^2-1) < 0$

💡 Teacher tip: Squaring both sides of an absolute-value inequality is valid when both sides are non-negative (which they always are here). The payoff is huge: a messy absolute-value condition collapses into a clean sign condition on a factored expression.

Step 2 — Interpret the sign condition for integers.

The product $(m^2-1)(n^2-1) < 0$ means exactly one factor is positive and the other is negative. Let's figure out when each factor is positive or negative, using the fact that $m$ and $n$ are integers:

$m^2 - 1 < 0 \Leftrightarrow m^2 < 1 \Leftrightarrow |m| < 1 \Leftrightarrow m = 0$ (the only integer with absolute value less than 1)
$m^2 - 1 > 0 \Leftrightarrow m^2 > 1 \Leftrightarrow |m| \geq 2$ (since $|m|=1$ gives $m^2-1=0$ , not positive)

The same logic applies to $n$ .

So the condition $(m^2-1)(n^2-1) < 0$ translates to:

Exactly one of $m, n$ equals $0$ , and the other has absolute value $\geq 2$ .

The pair $(0,0)$ is excluded because both factors would be $-1$ , making the product $+1 > 0$ .

Step 3 — Apply the second inequality $|m+n| < 5$ .

We now have two symmetric cases:

Case A: $n = 0$ and $|m| \geq 2$ .

Then $|m+n| = |m+0| = |m|$ . The second inequality gives $|m| < 5$ . Combined with $|m| \geq 2$ :

$|m| \in \{2, 3, 4\}$

This gives $m \in \{-4, -3, -2, 2, 3, 4\}$ — that's 6 pairs: $(-4,0), (-3,0), (-2,0), (2,0), (3,0), (4,0)$ .

Case B: $m = 0$ and $|n| \geq 2$ .

By symmetry, $|n| \in \{2, 3, 4\}$ , giving $n \in \{-4, -3, -2, 2, 3, 4\}$ — another 6 pairs: $(0,-4), (0,-3), (0,-2), (0,2), (0,3), (0,4)$ .

Step 4 — Verify boundary cases.

Pair $(m,n)$	$\\|1+mn\\|$	$\\|m+n\\|$	$\\|1+mn\\| < \\|m+n\\| < 5$ ?
$(2, 0)$	$1$	$2$	✅ $1 < 2 < 5$
$(4, 0)$	$1$	$4$	✅ $1 < 4 < 5$
$(-4, 0)$	$1$	$4$	✅ $1 < 4 < 5$
$(0, 3)$	$1$	$3$	✅ $1 < 3 < 5$
$(1, 0)$ — excluded	$1$	$1$	✗ $1 \not< 1$
$(5, 0)$ — excluded	$1$	$5$	✗ $5 \not< 5$

The boundary values $|m|=1$ (fails the first inequality) and $|m|=5$ (fails the second) are correctly excluded.

Step 5 — Count.

$\text{Total} = 6 + 6 = 12$

Answer: $\boxed{12}$

💡 Why this is elegant: The two absolute-value conditions look unrelated at first glance, but the difference-of-squares identity transforms the first inequality into a simple structural rule — exactly one variable must be zero. The second inequality then just bounds the non-zero partner's magnitude to $\{2, 3, 4\}$ , making the count a straightforward $6 + 6$ .

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Q18Boats Trains & Relative SpeedTITA

Mira and Amal walk along a circular track, starting from the same point at the same time. If they walk in the same direction, then in 45 minutes, Amal completes exactly 3 more rounds than Mira. If they walk in opposite directions, then they meet for the first time exactly after 3 minutes. The number of rounds Mira walks in one hour is

Your answer:

Show solution

Correct answer: 8

Think of this as: two walkers on a loop give us two equations — one from gaining laps (same direction) and one from closing a lap (opposite directions). Each equation is just "relative speed × time = relative distance."

Let Mira's speed be $m$ rounds/minute and Amal's speed be $a$ rounds/minute.

Step 1 — Same direction: Amal gains 3 rounds in 45 minutes.

When two people walk the same way on a circular track, the faster one pulls ahead at their relative speed $a - m$ . Every time the gap reaches one full round, the faster person has "lapped" the slower one.

Here, in 45 minutes Amal gains exactly 3 rounds on Mira, so the total relative distance covered is 3 rounds:

(a - m) \times 45 = 3

a - m = \frac{3}{45} = \frac{1}{15}\;\text{rounds/min}

Step 2 — Opposite directions: first meeting after 3 minutes.

When they walk towards each other, they close the gap at relative speed $a + m$ . They start together, so the first meeting happens when, between them, they have covered exactly one full round:

(a + m) \times 3 = 1

a + m = \frac{1}{3}\;\text{rounds/min}

💡 Teacher tip: Same-direction problems use $a - m$ (the gap opens); opposite-direction problems use $a + m$ (the gap closes). The "distance" that the relative speed acts on is always measured in rounds of the track.

Step 3 — Solve for Mira's speed.

We now have a clean pair:

a + m = \frac{1}{3}, \qquad a - m = \frac{1}{15}

Subtracting the second from the first eliminates $a$ :

(a + m) - (a - m) = \frac{1}{3} - \frac{1}{15}

2m = \frac{5}{15} - \frac{1}{15} = \frac{4}{15}

m = \frac{2}{15}\;\text{rounds/min}

Step 4 — Convert to rounds per hour.

One hour is 60 minutes, so Mira's rounds in an hour are:

60 \times \frac{2}{15} = \frac{120}{15} = 8

Quick check. If $m = \frac{2}{15}$ , then $a = \frac{1}{3} - \frac{2}{15} = \frac{3}{15} = \frac{1}{5}$ rounds/min. In 45 min: Amal $= 45 \times \frac{1}{5} = 9$ rounds, Mira $= 45 \times \frac{2}{15} = 6$ rounds — difference is exactly 3. ✓ In 3 min opposite: combined $= 3\left(\frac{1}{5} + \frac{2}{15}\right) = 3 \times \frac{1}{3} = 1$ round — first meeting. ✓

Answer: $\boxed{8}$

💡 Closing insight: the two scenarios hand us $a+m$ and $a-m$ directly — adding and subtracting recovers each walker's rate without any heavy algebra. The "3 more rounds" and "meet after 3 min" are simply total relative distance covered at the relative speed.

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Q19Quadratic & Linear EquationsTITA

A tea shop offers tea in cups of three different sizes. The product of the prices, in INR, of three different sizes is equal to 800. The prices of the smallest size and the medium size are in the ratio 2 : 5. If the shop owner decides to increase the prices of the smallest and the medium ones by INR 6 keeping the price of the largest size unchanged, the product then changes to 3200. The sum of the original prices of three different sizes, in INR, is

Your answer:

Show solution

Correct answer: 34

Think of this as: we have three unknown prices linked by two product equations and one ratio. The ratio lets us collapse two unknowns into one, and dividing the two product equations wipes out the third unknown entirely.

Let the original prices (in INR) of the small, medium, and large cups be $s$ , $m$ , and $L$ respectively.

Step 1 — Use the ratio to reduce unknowns.

We are told $s: m = 2: 5$ , so write $s = 2k, \qquad m = 5k$ for some positive number $k$ . Now the two product conditions become:

Original product: $s \cdot m \cdot L = 800$ , i.e. $2k \cdot 5k \cdot L = 800 \;\Longrightarrow\; 10k^{2}L = 800. \quad\cdots(1)$
After increasing the small and medium prices by INR 6 (large unchanged): $(s+6)(m+6)L = 3200,$ $\text{i.e.}\;(2k+6)(5k+6)\cdot L = 3200. \quad\cdots(2)$

Step 2 — Eliminate $L$ by dividing equation (2) by equation (1).

$\frac{(2k+6)(5k+6)\cdot L}{10k^{2}\cdot L} = \frac{3200}{800} = 4$

The $L$ cancels completely, leaving a single equation in $k$ : $\frac{(2k+6)(5k+6)}{10k^{2}} = 4 \;\Longrightarrow\; (2k+6)(5k+6) = 40k^{2}.$

💡 Teacher tip: Whenever two equations share a common factor you don't care about ( $L$ here), dividing them is the fastest way to make it disappear — no need to solve for it first.

Step 3 — Expand and solve the quadratic.

$10k^{2} + 12k + 30k + 36 = 40k^{2}$ $10k^{2} + 42k + 36 = 40k^{2}$ $30k^{2} - 42k - 36 = 0$

Divide throughout by 6: $5k^{2} - 7k - 6 = 0.$

Factorising: $5k^{2} - 7k - 6 = (5k+3)(k-2) = 0.$

So $k = -\tfrac{3}{5}$ or $k = 2$ . Since prices must be positive, we take $k = 2$ .

Step 4 — Recover the three original prices.

$s = 2k = 4, \qquad m = 5k = 10.$

From equation (1): $10k^{2}L = 800 \Rightarrow 10(4)L = 800 \Rightarrow L = 20$ .

Step 5 — Verify against both conditions.

Condition	Check
Original product	$4 \times 10 \times 20 = 800$ ✅
New product	$(4+6)\times(10+6)\times 20 = 10 \times 16 \times 20 = 3200$ ✅
Ratio $s:m$	$4:10 = 2:5$ ✅

Step 6 — Compute the required sum.

$s + m + L = 4 + 10 + 20 = 34.$

Answer: $\boxed{34}$

💡 Closing insight: The key move was dividing the two product equations — it turned a three-variable system into a single quadratic in $k$ . Once $k=2$ drops out of the factorisation, the rest is simple arithmetic.

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Q20PercentagesMCQ

The total of male and female populations in a city increased by 25% from 1970 to 1980. During the same period, the male population increased by 40% while the female population increased by 20%. From 1980 to 1990, the female population increased by 25%. In 1990, if the female population is twice the male population, then the percentage increase in the total of male and female populations in the city from 1970 to 1990 is

Show solution

Correct answer: B

Think of this as a two-decade journey: the first decade (1970→1980) fixes the gender split of the city, and the second decade (1980→1990) uses that split together with one extra condition to pin down the final total.

What we want: the percentage increase in the total (male + female) population from 1970 to 1990.

Let the 1970 male and female populations be $m$ and $f$ .

Step 1 — Find the 1970 gender ratio using alligation.

From 1970 to 1980:

the total population grew by $25\%$ ,
the male population grew by $40\%$ ,
the female population grew by $20\%$ .

The overall growth rate $25\%$ is the weighted average of the male growth $40\%$ and the female growth $20\%$ , weighted by the 1970 male and female shares. By alligation:

$\frac{m}{f} = \frac{25 - 20}{40 - 25} = \frac{5}{15} = \frac{1}{3}.$

So in 1970, males: females $= 1: 3$ . If the 1970 total is $T$ , then

$m = \frac{T}{4}, \qquad f = \frac{3T}{4}.$

💡 Teacher tip: Whenever a combined percentage change and the component percentage changes are all given, alligation instantly reveals the component ratio. No equations needed.

Step 2 — Pick a friendly number and track both decades.

Choose the 1970 total $T = 400$ (divisible by 4, so $m$ and $f$ are clean integers). Then $m = 100$ , $f = 300$ .

1970 → 1980: Male increases by $40\%$ : $100 \to 140$ . Female increases by $20\%$ : $300 \to 360$ . Total $= 140 + 360 = 500$ (which is indeed $25\%$ above 400 — a quick check).
1980 → 1990: Female increases by $25\%$ : $360 \to 360 \times 1.25 = 450$ . We are also told that in 1990 the female population is twice the male population, so male $= 450 / 2 = 225$ . Total $= 225 + 450 = 675$ .

Year	Male	Female	Total
$1970$	$100$	$300$	$400$
$1980$	$140$	$360$	$500$
$1990$	$225$	$450$	$675$

Step 3 — Compute the percentage increase from 1970 to 1990.

$\text{Percentage increase} = \frac{675 - 400}{400} \times 100\% = \frac{275}{400} \times 100\% = 68.75\%.$

Answer: $\boxed{68.75\%}$ (Choice B)

💡 Closing insight: Alligation fixes the 1970 gender split from the first decade's data; that split then carries forward through the second decade. Choosing $T = 400$ keeps every number an integer all the way to 1990, so the final percentage is just $\frac{275}{400}$ — no messy decimals until the very last step.

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Q21Time & WorkTITA

Anil can paint a house in 12 days while Barun can paint it in 16 days. Anil, Barun, and Chandu undertake to paint the house for ₹ 24000 and the three of them together complete the painting in 6 days. If Chandu is paid in proportion to the work done by him, then the amount in INR received by him is

Your answer:

Show solution

Correct answer: 3000

Think of this as: three people share ₹24,000 for a job they finished together in 6 days. Since pay is proportional to work done, we need to find what fraction of the total house Chandu actually painted.

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

If Anil finishes the house alone in 12 days, his rate is $\frac{1}{12}$ of the house per day.

If Barun finishes it alone in 16 days, his rate is $\frac{1}{16}$ of the house per day.

Let Chandu's rate be $c$ of the house per day.

Since all three together finish in 6 days, their combined rate is $\frac{1}{6}$ of the house per day.

Step 2 — Solve for Chandu's rate.

$\frac{1}{12} + \frac{1}{16} + c = \frac{1}{6}$

Bring the two known rates to a common denominator of 48:

$\frac{4}{48} + \frac{3}{48} + c = \frac{1}{6}$

$\frac{7}{48} + c = \frac{1}{6}$

Convert $\frac{1}{6}$ to 48ths: $\frac{1}{6} = \frac{8}{48}$ .

$\frac{7}{48} + c = \frac{8}{48} \implies c = \frac{1}{48}$

So Chandu paints $\frac{1}{48}$ of the house per day.

💡 Teacher tip: When three people work together and you know two of their rates plus the combined rate, the third person's rate is simply the combined rate minus the other two. No need to set up anything more complicated.

Step 3 — Find Chandu's total work contribution.

All three worked for 6 days. Chandu's contribution over those 6 days:

$6 \times \frac{1}{48} = \frac{6}{48} = \frac{1}{8}$

So Chandu painted $\frac{1}{8}$ of the house.

Step 4 — Calculate Chandu's payment.

Payment is proportional to work done. Chandu did $\frac{1}{8}$ of the total job, so he receives $\frac{1}{8}$ of the total payment:

$\frac{1}{8} \times 24{,}000 = 3{,}000$

Quick check: Anil's share would be $6 \times \frac{1}{12} = \frac{1}{2}$ → ₹12,000. Barun's share would be $6 \times \frac{1}{16} = \frac{3}{8}$ → ₹9,000. Adding all three: $12{,}000 + 9{,}000 + 3{,}000 = 24{,}000$ ✓

Answer: $\boxed{3000}$

💡 Closing insight: Once you have Chandu's rate, you don't need the others' work fractions — only his. The combined rate minus the two known rates isolates his contribution directly, and his 6-day work comes out to exactly one-eighth of the job.

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Q22Quadratic & Linear EquationsMCQ

One part of a hostel's monthly expenses is fixed, and the other part is proportional to the number of its boarders. The hostel collects ₹ 1600 per month from each boarder. When the number of boarders is 50, the profit of the hostel is ₹ 200 per boarder, and when the number of boarders is 75, the profit of the hostel is ₹ 250 per boarder. When the number of boarders is 80, the total profit of the hostel, in INR, will be

Show solution

Correct answer: B

Think of this as: the hostel's cost has a fixed piece and a per-boarder piece, and we're given two snapshots of profit-per-boarder. Convert those into total profits, set up a linear system, solve for the two unknowns, then evaluate at 80 boarders.

Step 1 — Set up the cost and profit model.

Let the monthly cost be $F + v\,n$ , where:

$F$ = fixed monthly expense,
$v$ = variable cost per boarder,
$n$ = number of boarders.

Revenue is $1600$ per boarder, so total revenue $= 1600\,n$ .

Total profit: $P(n) = 1600\,n - F - v\,n.$

Step 2 — Convert "profit per boarder" into total profit.

The question gives profit per boarder, so multiply by $n$ to get total profit.

At $n = 50$ : profit per boarder $= 200 \Rightarrow P(50) = 200 \times 50 = 10{,}000$ .
At $n = 75$ : profit per boarder $= 250 \Rightarrow P(75) = 250 \times 75 = 18{,}750$ .

💡 Teacher tip: "Profit per boarder" is total profit divided by $n$ . Always convert to total profit first — it removes the division and gives you clean linear equations in $F$ and $v$ .

Step 3 — Build two equations.

Using $P(n) = 1600\,n - F - v\,n$ :

At $n = 50$ : $10{,}000 = 1600 \times 50 - F - 50v = 80{,}000 - F - 50v$ $\Rightarrow F + 50v = 70{,}000 \quad \cdots (1)$
At $n = 75$ : $18{,}750 = 1600 \times 75 - F - 75v = 1{,}20{,}000 - F - 75v$ $\Rightarrow F + 75v = 1{,}01{,}250 \quad \cdots (2)$

Step 4 — Solve for $v$ and $F$ .

Subtract equation $(1)$ from equation $(2)$ : $25v = 31{,}250 \Rightarrow v = 1{,}250.$

Substitute back into $(1)$ : $F = 70{,}000 - 50 \times 1{,}250 = 70{,}000 - 62{,}500 = 7{,}500.$

So the fixed cost is ₹ $7{,}500$ and the variable cost per boarder is ₹ $1{,}250$ .

Step 5 — Compute total profit at $n = 80$ .

$P(80) = 1600 \times 80 - 7{,}500 - 1{,}250 \times 80$ $= 1{,}28{,}000 - 7{,}500 - 1{,}00{,}000$ $= 20{,}500.$

Quick check: Profit per boarder at $n=80$ is $1600 - \frac{7500}{80} - 1250 = 1600 - 93.75 - 1250 = 256.25$ , and $256.25 \times 80 = 20{,}500$ . ✓

Answer: $\boxed{20{,}500}$ (Choice B)

Alternative approach — work in profit-per-boarder directly.

Profit per boarder $= \dfrac{P(n)}{n} = 1600 - \dfrac{F}{n} - v$ .

At $n=50$ : $200 = 1600 - \dfrac{F}{50} - v \Rightarrow \dfrac{F}{50} + v = 1400 \quad \cdots (1')$
At $n=75$ : $250 = 1600 - \dfrac{F}{75} - v \Rightarrow \dfrac{F}{75} + v = 1350 \quad \cdots (2')$

Subtracting $(2')$ from $(1')$ : $F\left(\frac{1}{50} - \frac{1}{75}\right) = 50 \Rightarrow F \cdot \frac{1}{150} = 50 \Rightarrow F = 7{,}500.$

Then $v = 1400 - \frac{7500}{50} = 1400 - 150 = 1{,}250$ .

At $n=80$ : profit per boarder $= 1600 - \frac{7500}{80} - 1250 = 256.25$ , so total profit $= 256.25 \times 80 = 20{,}500$ .

Both methods arrive at the same answer; the first is cleaner because it avoids fractions until the very end.

💡 Closing insight: Two data points pin down a linear cost model ( $F$ and $v$ ) completely. Converting "profit per boarder" into "total profit" first removes the division and makes the linear system clean.

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Practice more questions like these ↗Ask a doubt in the forum ↗

Step 1 — Identify the triangle that holds all the data

Step 2 — Find DCDCDC using the cosine rule

Step 3 — Solve the quadratic

Step 4 — Compute the area

Alternative approach — Sine rule in △ADC\triangle ADC△ADC

Method 1 — Two linear equations

Alternative approach — Alligation

Step 2 — Find $DC$ using the cosine rule

Alternative approach — Sine rule in $\triangle ADC$