CAT 2017 Quant Questions — Slot 1

The actual Quantitative Ability section from CAT 2017, Slot 1 — 34 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.

34 questions23 MCQ · 11 TITA60 min section0/34 attempted

Q1Permutation & CombinationMCQ

The number of solutions (x, y, z) to the equation x − y − z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is

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Correct answer: B

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

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

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

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

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

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

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

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

Step 3 — Count for each range of $s$ .

$s = 2$ to $13$ : The largest any single variable can be is $s - 1 \le 12$ , so the cap of 12 never bites. The count is simply $s - 1$ . $\text{Total} = 1 + 2 + 3 + \cdots + 12 = \frac{12 \times 13}{2} = 78.$
$s = 14$ : $y$ ranges from $1$ to $13$ , but $z = 14 - y \le 12$ forces $y \ge 2$ , and $y \le 12$ . So $y = 2, 3, \ldots, 12$ → 11 pairs.
$s = 15$ : $y$ ranges from $1$ to $14$ , but $z = 15 - y \le 12$ forces $y \ge 3$ , and $y \le 12$ . So $y = 3, 4, \ldots, 12$ → 10 pairs.

Sum $s$	Valid values of $y$	Number of pairs
$2$ to $13$	$1$ to $s-1$ (all $\le 12$ )	$1+2+\cdots+12 = 78$
$14$	$2$ to $12$	$11$
$15$	$3$ to $12$	$10$

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

Answer: B — 99

Alternative approach — Complementary counting.

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

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

$105 - 3 - 3 = 99.$

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

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

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Q2InequalitiesTITA

For how many integers n, will the inequality (n − 5)(n − 10) − 3(n − 2) ≤ 0 be satisfied?

Your answer:

Show solution

Correct answer: 11

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

Step 1 — Expand and simplify.

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

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

Step 2 — Factor the quadratic.

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

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

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

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

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

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

Step 4 — Verify the boundaries.

$n$	$(n-4)(n-14)$	$\le 0$ ?
$3$ (just below)	$(-1)(-11) = 11$	✗
$4$ (root)	$0$	✓
$9$ (middle)	$(5)(-5) = -25$	✓
$14$ (root)	$0$	✓
$15$ (just above)	$(11)(1) = 11$	✗

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

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

$14 - 4 + 1 = 11.$

Answer: $\boxed{11}$

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

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

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Q3Boats Trains & Relative SpeedMCQ

A man travels by a motor boat down a river to his office and back. With the speed of the river unchanged, if he doubles the speed of his motor boat, then his total travel time gets reduced by 75%. The ratio of the original speed of the motor boat to the speed of the river is

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Correct answer: B

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

Step 1 — Set up variables.

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

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

Step 2 — Write the original round-trip time.

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

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

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

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

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

Step 4 — Use the 75% reduction condition.

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

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

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

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

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

Step 5 — Extract the ratio.

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

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

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

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

Quantity	Value
$v^2 - u^2$	$7k^2 - 4k^2 = 3k^2$
$4v^2 - u^2$	$28k^2 - 4k^2 = 24k^2$
$T_{\text{old}} \propto \frac{2v}{v^2-u^2}$	$\frac{2\sqrt{7}\,k}{3k^2} = \frac{2\sqrt{7}}{3k}$
$T_{\text{new}} \propto \frac{4v}{4v^2-u^2}$	$\frac{4\sqrt{7}\,k}{24k^2} = \frac{\sqrt{7}}{6k}$
Ratio $T_{\text{new}}/T_{\text{old}}$	$\frac{\sqrt{7}/(6k)}{2\sqrt{7}/(3k)} = \frac{1}{4}$ ✓

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

Answer: B — $\sqrt{7}: 2$

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

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Q4LogarithmsMCQ

The value of log₀.₀₀₈ √5 + log_√3 81 − 7 is equal to

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Correct answer: C

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

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

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

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

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

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

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

Express both as powers of $3$ .

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

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

Step 3 — Combine all three pieces

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

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

Quick verification of the arithmetic:

Piece	Value
$\log_{0.008}\sqrt{5}$	$-\tfrac{1}{6}$
$\log_{\sqrt{3}}81$	$8$
Constant term	$-7$
Sum	$\tfrac{5}{6}$

The value matches option C.

Answer: C — $\dfrac{5}{6}$

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

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Q5Ratio & ProportionMCQ

Suppose, C1, C2, C3, C4, and C5 are five companies. The profits made by C1, C2, and C3 are in the ratio 9 : 10 : 8 while the profits made by C2, C4, and C5 are in the ratio 18 : 19 : 20. If C5 has made a profit of Rs 19 crore more than C1, then the total profit (in Rs) made by all five companies is

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Correct answer: A

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

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

We are given:

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

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

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

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

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

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

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

Step 3 — Merge into one combined ratio.

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

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

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

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

Step 5 — Add up all five shares.

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

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

Answer: A — 438 crore

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

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Q6Profit Loss & DiscountMCQ

If Fatima sells 60 identical toys at a 40% discount on the printed price, then she makes 20% profit. Ten of these toys are destroyed in fire. While selling the rest, how much discount should be given on the printed price so that she can make the same amount of profit?

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Correct answer: D

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

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

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

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

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

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

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

Step 3 — What changes after the fire?

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

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

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

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

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

Quick check:

Quantity	Original plan	After fire
Toys sold	60	50
Total cost paid	$60C$	$60C$ (unchanged)
Total revenue needed	$72C$	$72C$ (same profit)
Price per toy	$0.6P = 1.2C$	$1.44C = 0.72P$
Discount	40%	28%

Answer: D — 28%

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

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

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

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

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Q7Permutation & CombinationTITA

Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?

Your answer:

Show solution

Correct answer: 160

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

Step 1 — Count the total points.

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

Step 2 — Start from all possible triples.

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

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

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

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

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

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

Diameter	Collinear triple
$AB$	$A, O, B$
$CD$	$C, O, D$
$EF$	$E, O, F$
$GH$	$G, O, H$
$JK$	$J, O, K$

Step 4 — Check there are no other collinear triples.

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

So the collinear triples are exactly 5 in number.

Step 5 — Subtract the failures.

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

Answer: $\boxed{160}$

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

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Q8MensurationTITA

A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. The height of the cylinder is 3 cm, while its volume is 9π cubic centimeters. Then the vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is

Your answer:

Show solution

Correct answer: 6

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

Step 1 — Find the cylinder's radius.

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

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

Step 2 — Determine how the ball rests.

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

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

Step 3 — Locate the ball's centre.

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

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

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

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

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

Step 4 — Find the topmost point.

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

Verification check:

Quantity	Value
Cylinder radius	$\sqrt{3} \approx 1.73$ cm
Ball radius	$2$ cm (wider ✓ — rests on rim)
Rim height	$3$ cm
Centre height above base	$4$ cm
Topmost point	$4 + 2 = 6$ cm

Answer: $\boxed{6}$

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

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Q9Quadratic & Linear EquationsMCQ

If x + 1 = x² and x > 0, then 2x⁴ is

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Correct answer: D

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

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

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

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

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

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

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

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

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Q10Permutation & CombinationMCQ

In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?

Show solution

Correct answer: A

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

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

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

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

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

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

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

Valid distributions = $20 - 4 = 16$ .

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

Extra pattern	Valid?	Number of assignments
(3,0,0,0)	No — someone gets 4 total	4 (excluded)
(2,1,0,0)	Yes	4 × 3 = 12
(1,1,1,0)	Yes	C(4,3) = 4
Total valid		16

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

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

Answer: A — 16

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

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Q11Functions & GraphsMCQ

If f(x) = (5x + 2)/(3x − 5) and g(x) = x² − 2x − 1, then the value of g(f(f(3))) is

Show solution

Correct answer: A

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

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

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

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

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

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

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

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

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

Step 3 — Evaluate $g$ at the result.

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

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

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

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

Answer: A — 2

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

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Q12Functions & GraphsMCQ

The shortest distance of the point (1/2, 1) from the curve y = |x − 1| + |x + 1| is

Show solution

Correct answer: A

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

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

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

For $-1 \le x \le 1$ :
$|x-1| = 1-x$ (since $x-1 \le 0$ )
$|x+1| = x+1$ (since $x+1 \ge 0$ )
So $y = (1-x)+(x+1) = 2$ .
For $x > 1$ :
$|x-1| = x-1$ and $|x+1| = x+1$
So $y = (x-1)+(x+1) = 2x$ .
For $x < -1$ :
$|x-1| = 1-x$ and $|x+1| = -x-1$
So $y = (1-x)+(-x-1) = -2x$ .

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

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

Step 2 — Locate the point relative to the curve.

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

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

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

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

Step 3 — Confirm the slanted arms are not closer.

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

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

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

Point on curve	Distance to $(\tfrac12,1)$	Closer than 1?
$(\tfrac12,2)$	$1$	✅ closest
$(1,2)$	$\sqrt{(1-\tfrac12)^2+(2-1)^2}=\sqrt{1.25}>1$	❌
$(-1,2)$	$\sqrt{(-1-\tfrac12)^2+(2-1)^2}=\sqrt{3.25}>1$	❌

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

Answer: A — $1$

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

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Q13Functions & GraphsMCQ

The area of the closed region bounded by the equation |x| + |y| = 2 in the two-dimensional plane is

Show solution

Correct answer: C

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

Step 1 — Find where the figure meets the axes.

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

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

Step 2 — See the shape in each quadrant.

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

Quadrant	Sign of $x, y$	Equation becomes
I	$x\ge0,\; y\ge0$	$x + y = 2$
II	$x\le0,\; y\ge0$	$-x + y = 2$
III	$x\le0,\; y\le0$	$-x - y = 2$
IV	$x\ge0,\; y\le0$	$x - y = 2$

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

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

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

Step 3 — Compute the area.

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

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

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

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

Step 4 — Quick check by counting triangles.

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

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

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

Answer: C — $\boxed{8}$

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

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Q14TrianglesMCQ

From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is

Show solution

Correct answer: B

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer: B — $\dfrac{500\sqrt{3}}{3}$

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

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Q15PercentagesTITA

Arun's present age in years is 40% of Barun's. In another few years, Arun's age will be half of Barun's. By what percentage will Barun's age increase during this period?

Your answer:

Show solution

Correct answer: 20

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

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

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

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

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

Both ages increase by the same amount $t$ :

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

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

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

Multiply both sides by $2$ :

$80 + 2t = 100 + t$

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

$t = 20$

So 20 years pass.

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

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

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

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

Alternative approach — using the constant age gap.

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

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

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

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

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

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

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

Answer: 20

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

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Q16PercentagesTITA

Ravi invests 50% of his monthly savings in fixed deposits. Thirty percent of the rest of his savings is invested in stocks and the rest goes into Ravi's savings bank account. If the total amount deposited by him in the bank (for savings account and fixed deposits) is Rs 59500, then Ravi's total monthly savings (in Rs) is

Your answer:

Show solution

Correct answer: 70000

Think of this as: a fixed portion of Ravi's savings goes to the bank (fixed deposits + savings account), and we're told that bank-bound portion equals Rs 59500. We need to find what fraction of his total savings lands in the bank, then scale up.

Step 1 — Let total monthly savings $= S$ .

Step 2 — Fixed deposits. Ravi puts $50\%$ of his total savings into fixed deposits: $\text{FD} = 0.5S.$ Amount remaining after fixed deposits $= S - 0.5S = 0.5S$ .

Step 3 — Stocks and savings account (from the rest). The phrase "thirty percent of the rest" is the key: the $30\%$ applies to the remaining $0.5S$ , not to the original $S$ .

Stocks $= 30\% \text{ of } 0.5S = 0.3 \times 0.5S = 0.15S.$
Savings account $=$ whatever is left of the remainder after stocks $= 0.5S - 0.15S = 0.35S.$

💡 Teacher tip: When a question says " $x\%$ of the rest", always multiply the percentages as fractions of the running remainder. Here, stocks take $0.3 \times 0.5 = 0.15$ , i.e. $15\%$ of the total savings, leaving the savings account with $0.5 - 0.15 = 0.35$ , i.e. $35\%$ of the total.

Step 4 — Total amount deposited in the bank. The bank receives both the fixed deposits and the savings-account money: $\text{Bank total} = \text{FD} + \text{Savings account} = 0.5S + 0.35S = 0.85S.$

Step 5 — Equate and solve. We're told this bank total is Rs 59500: $0.85S = 59500$ $S = \frac{59500}{0.85} = \frac{59500 \times 100}{85} = \frac{5950000}{85} = 70000.$

Quick check. If $S = 70000$ : FD $= 35000$ , remainder $= 35000$ , stocks $= 10500$ , savings account $= 24500$ . Bank total $= 35000 + 24500 = 59500$ . ✓

Answer: Rs 70000

💡 Closing insight: The only money not sent to the bank is the stocks portion ( $15\%$ of total savings). So the bank gets $100\% - 15\% = 85\%$ of Ravi's savings, and $59500 \div 0.85 = 70000$ . Spotting "what's left outside the bank" is often faster than adding up what's inside.

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Q17Surds & IndicesMCQ

If $9^{(2x − 1)}$ − 8 $1^{(x − 1)}$ = 1944, then x is

Show solution

Correct answer: B

Think of this as: two exponential terms that look different are actually powers of the same base. Unify the base, factor, and the equation collapses to a simple linear one.

Step 1 — Rewrite both terms with the same base. Notice that $81 = 9^2$ . We can use this to express both terms as powers of $9$ :

The second term: $81^{x-1} = (9^2)^{x-1} = 9^{2(x-1)} = 9^{2x-2}$ .
The first term: $9^{2x-1} = 9^{(2x-2)+1} = 9^1 \cdot 9^{2x-2} = 9 \cdot 9^{2x-2}$ .

Step 2 — Substitute to make the equation linear. Let $y = 9^{2x-2}$ . The original equation $9^{2x-1} - 81^{x-1} = 1944$ becomes: $9y - y = 1944$ $8y = 1944$ $y = \frac{1944}{8} = 243$

💡 Teacher tip: Whenever you see different bases in an exponential equation, check if they are powers of a common base. Converting to a single base and substituting turns a messy exponential equation into a simple linear one.

Step 3 — Solve for $x$ . We now have $9^{2x-2} = 243$ . Convert both sides to base $3$ (since $9 = 3^2$ and $243 = 3^5$ ): $(3^2)^{2x-2} = 3^5$ $3^{2(2x-2)} = 3^5$ $3^{4x-4} = 3^5$ Since the bases are equal, set the exponents equal: $4x - 4 = 5$ $4x = 9$ $x = \frac{9}{4}$

Step 4 — Verify the answer. Let's plug $x = \frac{9}{4}$ back into the original equation to be sure.

Term	Calculation	Value
$9^{2x-1}$	$9^{2(9/4)-1} = 9^{7/2} = (3^2)^{7/2} = 3^7$	$2187$
$81^{x-1}$	$81^{(9/4)-1} = 81^{5/4} = (3^4)^{5/4} = 3^5$	$243$
Difference	$2187 - 243$	$1944$ ✓

The difference matches the right-hand side of the equation.

Answer: B — $\frac{9}{4}$

💡 Closing insight: The entire equation hinged on spotting that $9$ and $81$ share a common base. Once unified, the terms cleanly factored into $9y - y$ , making the solution straightforward.

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Q18Polynomials & ProgressionsMCQ

Let a₁, a₂, ..., a₃ₙ be an arithmetic progression with a₁ = 3 and a₂ = 7. If a₁ + a₂ + ... + a₃ₙ = 1830, then what is the smallest positive integer m such that m(a₁ + a₂ + ... + aₙ) > 1830?

Show solution

Correct answer: B

Think of this as: the grand total of all $3n$ terms pins down $n$ ; then the sum of just the first $n$ terms becomes a building block, and we ask how many copies of that block are needed to exceed the grand total.

Step 1 — Find the common difference.

Since $a_1 = 3$ and $a_2 = 7$ in an arithmetic progression: $d = a_2 - a_1 = 7 - 3 = 4.$

Step 2 — Express the sum of the first $3n$ terms.

Using the AP sum formula $S_k = \dfrac{k}{2}\bigl(2a_1 + (k-1)d\bigr)$ with $k = 3n$ : $S_{3n} = \frac{3n}{2}\bigl(2\cdot 3 + (3n-1)\cdot 4\bigr) = \frac{3n}{2}\bigl(6 + 12n - 4\bigr) = \frac{3n}{2}(12n + 2).$ $S_{3n} = 3n(6n + 1) = 1830.$

Step 3 — Solve for $n$ .

$3n(6n+1) = 1830 \;\Rightarrow\; 18n^2 + 3n - 1830 = 0 \;\Rightarrow\; 6n^2 + n - 610 = 0.$

Using the quadratic formula: $n = \frac{-1 + \sqrt{1 + 4\cdot 6 \cdot 610}}{12} = \frac{-1 + \sqrt{14641}}{12} = \frac{-1 + 121}{12} = \frac{120}{12} = 10.$

So $\boxed{n = 10}$ , meaning the AP has $3n = 30$ terms.

💡 Teacher tip: Always check that $n$ comes out as a positive integer — if it didn't, the problem's setup would be inconsistent. Here $\sqrt{14641} = 121$ is a perfect square, confirming the data is clean.

Step 4 — Compute the sum of the first $n = 10$ terms.

$S_{10} = \frac{10}{2}\bigl(2\cdot 3 + 9 \cdot 4\bigr) = 5\,(6 + 36) = 5 \times 42 = 210.$

Step 5 — Find the smallest $m$ with $m \cdot 210 > 1830$ .

$\frac{1830}{210} = 8.714\ldots$

Since $m$ must be a positive integer and we need strict inequality:

$m$	$m \times 210$	Exceeds 1830?
8	1680	✗
9	1890	✓

Eight copies give $1680 < 1830$ ; nine copies give $1890 > 1830$ . The smallest such $m$ is $9$ .

Answer: B — 9

💡 Closing insight: The sum of the first $n$ terms ( $210$ ) is roughly $\frac{1}{9}$ of the grand total ( $1830$ ), so it takes about 9 copies to surpass it — the slight excess ( $1890 > 1830$ ) is what makes 9 the answer rather than 8.

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Q19PercentagesMCQ

The number of girls appearing for an admission test is twice the number of boys. If 30% of the girls and 45% of the boys get admission, the percentage of candidates who do not get admission is

Show solution

Correct answer: D

The girls outnumber the boys by 2:1, so the overall percentage is a weighted blend of the two groups' admission rates — not a simple average of 30% and 45%.

Step 1 — Pick convenient numbers matching the ratio.

Since only a ratio is given, we may choose any multiple of it. Let the number of boys be 100. Then the number of girls is twice that, i.e. 200. Total candidates = 100 + 200 = 300.

💡 Teacher tip: When a question gives only a ratio and percentages, choose 100 for the smaller group. The percentages then turn into clean whole numbers.

Step 2 — Find how many get admitted.

Girls admitted = 30% of 200 = 60
Boys admitted = 45% of 100 = 45
Total admitted = 60 + 45 = 105

Step 3 — Find how many do not get admitted.

Either subtract from the total, or use the not-admitted percentages directly:

Group	Candidates	Admitted	Not admitted
Girls	200	60	140
Boys	100	45	55
Total	300	105	195

So, candidates not admitted = 300 − 105 = 195.

Step 4 — Convert to a percentage of all candidates.

Not-admitted percentage = (195 / 300) × 100 = 65%.

Answer: D — 65

Alternative approach — weighted average.

Of every 3 candidates, 2 are girls and 1 is a boy. The not-admitted rates are 70% for girls (100% − 30%) and 55% for boys (100% − 45%). So the overall not-admitted percentage is:

(2/3) × 70% + (1/3) × 55% = 140/3 + 55/3 = 195/3 = 65%.

This matches, and it shows why the answer is closer to the girls' 70% than to the boys' 55%: girls form the larger group.

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Q20Profit Loss & DiscountMCQ

If a seller gives a discount of 15% on retail price, she still makes a profit of 2%. Which of the following ensures that she makes a profit of 20%?

Show solution

Correct answer: D

Think of this as: we know how cost, retail price, and one particular discount relate. We need to find the selling price that yields 20% profit, then match it to the right option.

Step 1 — Set up variables. Let the retail price be $R$ and the cost price be $C$ .

Step 2 — Translate the given condition. A 15% discount on retail price means the item sells for $0.85R$ . This still earns a 2% profit, so the selling price equals $1.02C$ : $0.85R = 1.02\,C$ Solving for the cost in terms of the retail price: $C = \frac{0.85R}{1.02}$

💡 Teacher tip: Anchoring cost to the retail price keeps the arithmetic as clean multiplications. Since every option is expressed in terms of $R$ , this is the natural direction to work in.

Step 3 — Find the selling price needed for 20% profit. A 20% profit requires selling at $1.20\,C$ : $\text{Required selling price} = 1.20 \times C = 1.20 \times \frac{0.85R}{1.02}$ Notice the neat cancellation: $1.20 \times 0.85 = 1.02$ So, $1.20 \times \frac{0.85R}{1.02} = \frac{1.02R}{1.02} = R$

Step 4 — Interpret the result. To make a 20% profit, she must sell at exactly $R$ — the retail price. Selling at retail price means giving no discount.

Step 5 — Verify against the options.

Option	Selling price	Profit check
A. 5% discount	$0.95R$	$\frac{0.95R}{C} - 1 = \frac{0.95 \times 1.02}{0.85} - 1 \approx 14\%$ ✗
B. 2% discount	$0.98R$	$\approx 17.6\%$ ✗
C. Increase retail by 2%	$1.02R$	$\approx 22.4\%$ ✗
D. Sell at retail price	$R$	$= 20\%$ ✓

Only option D produces exactly 20% profit.

Answer: D — Sell at retail price.

💡 Why this is elegant: the numbers were chosen so that $1.20 \times 0.85 = 1.02$ . The same 2% that appeared as the profit at a 15% discount reappears as the factor needed to scale cost up to retail — so the 20%-profit price lands exactly on $R$ .

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Q21Time Speed & DistanceTITA

A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train's departure. The distance (in km) from his home to the railway station is

Your answer:

Show solution

Correct answer: 20

Think of this as: the same distance walked at two different speeds swings his arrival from 10 minutes late to 10 minutes early — a total swing of 20 minutes. That 20-minute gap is caused entirely by the difference in walking speeds.

Step 1 — Set up variables.

Let the distance from home to station be $D$ km.

Let $T$ be the scheduled time (in minutes) from the moment he leaves home to the moment the train departs.

Step 2 — Translate each scenario into an equation.

At 12 km/h: He reaches 10 minutes after the train leaves, so his walking time exceeds the scheduled time by 10 minutes.

Walking time at 12 km/h $= \dfrac{D}{12}$ hours $= \dfrac{D}{12} \times 60 = 5D$ minutes.

$5D = T + 10 \quad \cdots (1)$

At 15 km/h: He reaches 10 minutes before the train leaves, so his walking time is 10 minutes less than the scheduled time.

Walking time at 15 km/h $= \dfrac{D}{15}$ hours $= \dfrac{D}{15} \times 60 = 4D$ minutes.

$4D = T - 10 \quad \cdots (2)$

💡 Teacher tip: The key move is expressing both scenarios in terms of the same scheduled time $T$ . Once you do that, subtracting the equations eliminates $T$ entirely — you never need to know when the train actually leaves.

Step 3 — Subtract equation (2) from equation (1).

$5D - 4D = (T + 10) - (T - 10)$

$D = 20$

Step 4 — Verify.

Scenario	Speed	Walking time	Scheduled time $T$	On time?
Slow	12 km/h	$5 \times 20 = 100$ min	$T = 100 - 10 = 90$ min	10 min late ✅
Fast	15 km/h	$4 \times 20 = 80$ min	$T = 80 + 10 = 90$ min	10 min early ✅

Both scenarios give the same scheduled time $T = 90$ minutes, confirming consistency.

Answer: $\boxed{20}$

Alternative approach — using the time difference directly.

The faster speed saves him 20 minutes (from 10 late to 10 early) compared to the slower speed. So the difference in walking times between the two speeds is 20 minutes $= \dfrac{1}{3}$ hour.

$\frac{D}{12} - \frac{D}{15} = \frac{1}{3}$

$D\left(\frac{1}{12} - \frac{1}{15}\right) = \frac{1}{3}$

$D\left(\frac{5 - 4}{60}\right) = \frac{1}{3}$

$\frac{D}{60} = \frac{1}{3} \implies D = 20$

This is faster under exam conditions — no need to introduce $T$ at all.

💡 Closing insight: Whenever two scenarios differ only in speed, the entire difference in arrival times comes from the speed gap. Write "time = scheduled ± offset" for each case and subtract — the scheduled time cancels out cleanly.

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Q22Quadratic & Linear EquationsMCQ

If a and b are integers of opposite signs such that (a + 3)² = 9b² and (a − 1)² = 4(b − 1)², then the ratio a² : b² is

Show solution

Correct answer: D

Think of this as: two "square = square" equations, each splitting into a plus-or-minus branch. We get four possible pairings, and the opposite-signs + integer conditions filter us down to exactly one.

Step 1 — Split the first equation.

Given $(a+3)^2 = 9b^2$ . Since $9b^2 = (3b)^2$ , we have a square equal to a square, so: $a + 3 = \pm 3b$ This gives two linear branches:

$a = 3b - 3$ (taking the $+$ sign)
$a = -3b - 3$ (taking the $-$ sign)

Step 2 — Split the second equation.

Given $(a-1)^2 = 4(b-1)^2$ . Since $4(b-1)^2 = (2(b-1))^2$ , again a square equals a square: $a - 1 = \pm 2(b-1)$ This gives two linear branches:

$a = 2b - 1$ (taking the $+$ sign)
$a = 3 - 2b$ (taking the $-$ sign)

💡 Teacher tip: Whenever you see $X^2 = Y^2$ , immediately rewrite it as $X = \pm Y$ . Each such equation doubles the number of cases — here two equations give $2 \times 2 = 4$ candidate pairings.

Step 3 — Match each pair of branches and check validity.

We equate each branch from Step 1 with each branch from Step 2:

Branch from Eq. 1	Branch from Eq. 2	Solving	Result	Valid?
$a = 3b - 3$	$a = 2b - 1$	$3b - 3 = 2b - 1 \Rightarrow b = 2$	$a = 3,\; b = 2$	Same sign — reject
$a = 3b - 3$	$a = 3 - 2b$	$3b - 3 = 3 - 2b \Rightarrow 5b = 6$	$b = \tfrac{6}{5}$	Not integer — reject
$a = -3b - 3$	$a = 2b - 1$	$-3b - 3 = 2b - 1 \Rightarrow -5b = 2$	$b = -\tfrac{2}{5}$	Not integer — reject
$a = -3b - 3$	$a = 3 - 2b$	$-3b - 3 = 3 - 2b \Rightarrow -b = 6$	$b = -6,\; a = 15$	Opposite signs ✓

Only the last pairing survives both filters: $a$ and $b$ are integers and of opposite signs ( $a = 15 > 0$ , $b = -6 < 0$ ).

Step 4 — Compute the ratio.

$a^2: b^2 = 15^2: (-6)^2 = 225: 36 = 25: 4$

Answer: D — 25: 4

💡 Closing insight: The opposite-signs condition is doing real work here — without it, the pairing $(a, b) = (3, 2)$ would also be valid and the answer wouldn't be unique. Always note which constraint eliminates which candidate.

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Q23Time & WorkTITA

A person can complete a job in 120 days. He works alone on Day 1. On Day 2, he is joined by another person who also can complete the job in exactly 120 days. On Day 3, they are joined by another person of equal efficiency. Like this, everyday a new person with the same efficiency joins the work. How many days are required to complete the job?

Your answer:

Show solution

Correct answer: 15

Think of this as: each day a fresh worker of equal strength joins, so the crew size grows as $1, 2, 3, \dots$ The total work done is the sum of these growing daily contributions, and we need that sum to equal one full job.

Step 1 — Find one worker's daily contribution.

One person completes the entire job in $120$ days, so in a single day he does $\frac{1}{120}$ of the job. Every worker has the same efficiency, so each worker contributes $\dfrac{1}{120}$ per day.

Step 2 — Write the work done on each day.

The number of workers present grows by one each day:

Day	Workers present	Work done that day
$1$	$1$	$\dfrac{1}{120}$
$2$	$2$	$\dfrac{2}{120}$
$3$	$3$	$\dfrac{3}{120}$
$\vdots$	$\vdots$	$\vdots$
$n$	$n$	$\dfrac{n}{120}$

Step 3 — Sum the work over $n$ days and set it equal to one whole job.

$\frac{1}{120} + \frac{2}{120} + \frac{3}{120} + \cdots + \frac{n}{120} = 1$

Factor out $\dfrac{1}{120}$ :

$\frac{1 + 2 + 3 + \cdots + n}{120} = 1$

💡 Teacher tip: Whenever you see $1 + 2 + 3 + \cdots + n$ , reach for the triangular-number formula $\dfrac{n(n+1)}{2}$ . It converts a long wordy sum into a single algebraic expression.

Using $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}$ :

$\frac{n(n+1)}{2 \times 120} = 1$

Step 4 — Solve for $n$ .

$n(n+1) = 240$

We need two consecutive integers whose product is $240$ . Testing:

$10 \times 11 = 110$ — too small
$15 \times 16 = 240$ ✓

So $n = 15$ .

Step 5 — Verify.

After $15$ days, total work done $= \dfrac{1 + 2 + \cdots + 15}{120} = \dfrac{\frac{15 \times 16}{2}}{120} = \dfrac{120}{120} = 1$ whole job. ✓

The job is completed exactly at the end of Day 15.

Answer: $\boxed{15}$

💡 Why this works so cleanly: the crew grows linearly ( $1, 2, 3, \dots$ ), so the total work is a triangular number. Setting $\dfrac{n(n+1)}{2} = 120$ turns a wordy work problem into a one-line equation — and $120$ happens to be $\dfrac{15 \times 16}{2}$ , giving a clean integer answer.

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Q24LogarithmsMCQ

Suppose, log₃ x = a and log₁₂ y = a, where x, y are positive numbers. If G is the geometric mean of x and y, and log₆ G is equal to

Show solution

Correct answer: D

We are given $\log_3 x = a$ and $\log_{12} y = a$ , with $x, y > 0$ . We need to find $\log_6 G$ , where $G$ is the geometric mean of $x$ and $y$ .

Step 1 — Convert the logarithms to powers.

Using the definition $\log_b m = n \iff m = b^n$ :

From $\log_3 x = a$ , we get $x = 3^a$ .
From $\log_{12} y = a$ , we get $y = 12^a$ .

Step 2 — Write the geometric mean.

The geometric mean of $x$ and $y$ is

G = \sqrt{xy}.

Substitute $x = 3^a$ and $y = 12^a$ :

G = \sqrt{3^a \cdot 12^a}.

Step 3 — Combine the powers.

Since the exponents are equal,

3^a \cdot 12^a = (3 \cdot 12)^a = 36^a.

G = \sqrt{36^a} = (36^a)^{1/2} = 36^{a/2}.

Step 4 — Rewrite as a power of 6.

Since $36 = 6^2$ ,

36^{a/2} = (6^2)^{a/2} = 6^a.

Thus,

G = 6^a.

Step 5 — Find $\log_6 G$ .

Now take log base 6:

\log_6 G = \log_6 (6^a) = a.

Answer: D — $a$

💡 Why it works: The key coincidence is that $3 \times 12 = 36 = 6^2$ . Because the geometric mean takes the square root, this square cancels perfectly and leaves $G$ as a clean power of $6$ .

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Q25TrianglesTITA

Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively. The minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour is

Your answer:

Show solution

Correct answer: 24

Think of this as: we have a right-angled triangle, and we need the shortest path from the right-angle vertex to the hypotenuse — then convert that distance into time.

Step 1 — Identify the right angle.

Since $BC$ is the hypotenuse, the right angle is at $A$ . So $AB$ and $AC$ are the two legs, with $AB = 15$ km and $AC = 20$ km.

Step 2 — Find the hypotenuse $BC$ .

Using the Pythagorean theorem: $BC = \sqrt{AB^2 + AC^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25 \text{ km.}$

Step 3 — Find the shortest distance from $A$ to $BC$ .

💡 Teacher tip: The shortest path from a point to a line is always along the perpendicular. So the minimum distance from $A$ to the hypotenuse $BC$ is the altitude from $A$ .

We can find this altitude by equating two expressions for the area of the triangle.

Using the two legs as base and height: $\text{Area} = \tfrac{1}{2} \times AB \times AC = \tfrac{1}{2} \times 15 \times 20 = 150 \text{ km}^2.$

Using $BC$ as the base and $h$ as the altitude from $A$ : $\text{Area} = \tfrac{1}{2} \times BC \times h = \tfrac{1}{2} \times 25 \times h.$

Equating the two: $\tfrac{1}{2} \times 25 \times h = 150 \implies h = \frac{300}{25} = 12 \text{ km.}$

So the shortest distance from $A$ to the hypotenuse is 12 km.

Step 4 — Convert distance to time.

Speed $= 30$ km/h. Time $= \frac{\text{distance}}{\text{speed}}$ : $\text{Time} = \frac{12}{30} \text{ hours} = 0.4 \text{ hours}.$

Converting to minutes: $0.4 \times 60 = 24 \text{ minutes.}$

Answer: 24

💡 Closing insight: This is a classic 15-20-25 right triangle (a scaled-up 3-4-5). In any 3-4-5 family triangle, the altitude to the hypotenuse always works out to $\frac{\text{leg}_1 \times \text{leg}_2}{\text{hypotenuse}} = \frac{15 \times 20}{25} = 12$ — a handy shortcut to remember.

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Q26MensurationMCQ

Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC is

Show solution

Correct answer: B

Think of this as: two curved fences (arcs) both bow outward from the same base $BC$ , one bulging farther than the other. We want the area trapped between them.

Step 1 — Set up the triangle.

$ABC$ is right-angled and isosceles with hypotenuse $BC$ , so the right angle is at $A$ and the two legs are equal: $AB = AC = 6\text{ cm},\quad \angle BAC = 90°.$ By Pythagoras: $BC = \sqrt{6^2 + 6^2} = 6\sqrt{2}\text{ cm}.$

Step 2 — Identify the two arcs.

Arc $BPC$ is part of the circle centered at $A$ with radius $AB = 6$ , sweeping across $\angle BAC = 90°$ . It sits just outside $BC$ , on the side away from $A$ .
Arc $BQC$ is a semicircle with diameter $BC = 6\sqrt{2}$ , also on the side away from $A$ , bulging even farther out.

The region we want is the lens-shaped area between these two arcs.

💡 Teacher tip: The semicircle is the outer boundary and arc $BPC$ is the inner boundary. So the desired area = (semicircle area) − (area between chord $BC$ and arc $BPC$ ).

Step 3 — Area between chord $BC$ and arc $BPC$ (the inner segment).

This is a circular segment of the circle centered at $A$ with radius $6$ and central angle $90°$ : $\text{Sector } BAC = \frac{90°}{360°}\times \pi (6)^2 = \frac{1}{4}\times 36\pi = 9\pi.$ $\text{Triangle } BAC = \frac{1}{2}\times 6\times 6 = 18.$ $\text{Segment} = \text{Sector} - \text{Triangle} = 9\pi - 18.$

Step 4 — Area of the semicircle $BQC$ .

Diameter $= 6\sqrt{2}$ , so radius $= 3\sqrt{2}$ : $\text{Semicircle area} = \frac{1}{2}\pi(3\sqrt{2})^2 = \frac{1}{2}\pi \times 18 = 9\pi.$

Step 5 — Subtract to get the enclosed region.

$\text{Required area} = \text{Semicircle} - \text{Segment} = 9\pi - (9\pi - 18) = 18.$

Notice how the $\pi$ -terms cancel completely — the answer is a clean integer.

Answer: B — $\boxed{18}$

💡 Why this is elegant: The two curved areas each contain $9\pi$ , so subtracting wipes out the irrational part entirely. What remains is just the triangle's area — a neat geometric coincidence that comes from the isosceles right-triangle setup.

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Q27Averages & Weighted AveragesMCQ

A class consists of 20 boys and 30 girls. In the mid-semester examination, the average score of the girls was 5 higher than that of the boys. In the final exam, however, the average score of the girls dropped by 3 while the average score of the entire class increased by 2. The increase in the average score of the boys is

Show solution

Correct answer: A

Think of this as a weighted-average problem where only the change in the boys' average matters — so we anchor on the boys' mid-sem average and let everything shift around it.

Step 1 — Set up the mid-semester averages.

Let the boys' mid-sem average be $B$ .

Since the girls' average was 5 higher, the girls' mid-sem average is $B + 5$ .

The class has 20 boys and 30 girls, so the mid-sem class average is the weighted mean: $\text{Class average}_{\text{mid}} = \frac{20B + 30(B+5)}{50} = \frac{20B + 30B + 150}{50} = \frac{50B + 150}{50} = B + 3.$

So the mid-sem class average is $B + 3$ .

Step 2 — Translate the final-exam changes.

Two things happen in the final exam:

The girls' average drops by 3: new girls' average $= (B+5) - 3 = B + 2$ .
The whole-class average rises by 2: new class average $= (B+3) + 2 = B + 5$ .

Let the boys' final-exam average be $B + x$ , where $x$ is the increase we need to find.

Step 3 — Use the final-exam class total.

The final-exam class average must equal the weighted mean of the boys' and girls' final averages: $\frac{20(B+x) + 30(B+2)}{50} = B + 5.$

Multiply both sides by 50: $20(B+x) + 30(B+2) = 50(B+5).$

Expand: $20B + 20x + 30B + 60 = 50B + 250.$

Combine like terms on the left: $50B + 20x + 60 = 50B + 250.$

The $50B$ cancels from both sides: $20x + 60 = 250 \implies 20x = 190 \implies x = 9.5.$

Step 4 — Verify the numbers make sense.

Group	Mid-sem avg	Final avg	Change
Boys (20)	$B$	$B + 9.5$	$+9.5$
Girls (30)	$B+5$	$B+2$	$-3$
Class (50)	$B+3$	$B+5$	$+2$

Check: $\frac{20(B+9.5) + 30(B+2)}{50} = \frac{50B + 190 + 60}{50} = \frac{50B + 250}{50} = B + 5.$ ✅

The boys' average must rise sharply ( $+9.5$ ) to offset the girls' drop ( $-3$ ) and still lift the whole class by $+2$ — which makes sense because the boys are the smaller group carrying the entire net increase.

Answer: A — 9.5

💡 Why this works cleanly: We never needed the actual value of $B$ . Because only changes were asked, $B$ cancelled out — a common shortcut in weighted-average problems involving shifts rather than absolute values.

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Q28Quadratic & Linear EquationsTITA

If f₁(x) = x² + 11x + n and f₂(x) = x, then the largest positive integer n for which the equation f₁(x) = f₂(x) has two distinct real roots is

Your answer:

Show solution

Correct answer: 24

We need the largest positive integer $n$ for which $f_1(x) = f_2(x)$ has two distinct real roots.

Step 1 — Form a single quadratic equation.

Set $f_1(x) = f_2(x)$ : $x^2 + 11x + n = x$

Bring every term to one side: $x^2 + 11x + n - x = 0 \;\Rightarrow\; x^2 + 10x + n = 0.$

This is now a standard quadratic in $x$ with coefficients $a = 1$ , $b = 10$ , $c = n$ .

Step 2 — Translate "two distinct real roots" into a discriminant condition.

A quadratic $ax^2 + bx + c = 0$ has:

two distinct real roots when the discriminant $\Delta = b^2 - 4ac$ is strictly positive ( $\Delta > 0$ ),
one repeated real root when $\Delta = 0$ ,
no real roots when $\Delta < 0$ .

Since the question asks for two distinct real roots, we need: $\Delta > 0.$

Step 3 — Compute the discriminant and solve the inequality.

$\Delta = 10^2 - 4(1)(n) = 100 - 4n.$

Requiring this to be strictly positive: $100 - 4n > 0 \;\Rightarrow\; 4n < 100 \;\Rightarrow\; n < 25.$

💡 Teacher tip: The word "distinct" is doing critical work here. It forces a strict inequality ( $>$ , not $\ge$ ). If the question had said "real roots" (allowing a repeated root), the bound would be $n \le 25$ and the answer would jump to 25. Always check which inequality the wording buys you.

Step 4 — Pick the largest positive integer strictly below 25.

The integers satisfying $n < 25$ are $\dots, 23, 24$ . The largest positive one is $24$ .

A quick check at the boundary:

$n$	Discriminant $100 - 4n$	Roots	Two distinct real roots?
$24$	$100 - 96 = 4 > 0$	two distinct real	✅
$25$	$100 - 100 = 0$	one repeated root	❌

So $n = 25$ gives a repeated root (not distinct), and $n = 24$ is the largest integer that still gives two distinct real roots.

Answer: $\boxed{24}$

💡 Takeaway: Whenever a question says "two distinct real roots," immediately write $\Delta > 0$ — not $\ge 0$ . The entire question usually reduces to reading off the largest integer below the resulting bound.

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Q29Ratio & ProportionMCQ

A stall sells popcorn and chips in packets of three sizes: large, super, and jumbo. The numbers of large, super, and jumbo packets in its stock are in the ratio 7 : 17 : 16 for popcorn and 6 : 15 : 14 for chips. If the total number of popcorn packets in its stock is the same as that of chips packets, then the numbers of jumbo popcorn packets and jumbo chips packets are in the ratio

Show solution

Correct answer: A

We have two separate ratios for popcorn and chips packet sizes, and one linking condition — the total number of packets of each snack is the same. That single equality is enough to pin down the ratio of the jumbo packets.

Step 1 — Add up the parts in each ratio.

For popcorn, the parts are $7: 17: 16$ , so the total number of parts is $7 + 17 + 16 = 40.$

For chips, the parts are $6: 15: 14$ , so the total number of parts is $6 + 15 + 14 = 35.$

Step 2 — Use the equal-totals condition.

Let the common multiplier for the popcorn ratio be $a$ and for the chips ratio be $b$ . Then the actual total number of popcorn packets is $40a$ and the actual total number of chips packets is $35b$ .

We are told these totals are equal: $40a = 35b \implies \frac{a}{b} = \frac{35}{40} = \frac{7}{8}.$

So we can write $a = 7k$ and $b = 8k$ for some positive number $k$ .

💡 Teacher tip: When two ratios have different total parts but their real-world totals are equal, the scaling factors are inversely proportional to the part-totals. The snack with fewer parts per ratio gets scaled up more.

Step 3 — Compute the jumbo-packet ratio.

Jumbo popcorn packets correspond to $16$ parts, so their count is $16a = 16(7k) = 112k.$

Jumbo chips packets correspond to $14$ parts, so their count is $14b = 14(8k) = 112k.$

Therefore, the required ratio is $112k: 112k = 1: 1.$

Answer: A — 1: 1

The numbers were arranged so that $16 \times 7$ exactly equals $14 \times 8$ . The equal-totals condition forces the multipliers into a $7:8$ ratio, which precisely cancels the $16:14$ base ratio of the jumbo parts.

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Q30Polynomials & ProgressionsMCQ

If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is

Show solution

Correct answer: A

Think of this as: three specific terms of an AP — the 3rd, 7th, and 17th — happen to be in geometric progression. What does that force the first term and common difference to look like?

Step 1 — Write the three terms.

With first term $a$ and common difference $d$ (where $d > 0$ ), the $k$ -th term of an AP is: $T_k = a + (k-1)d$

So the three terms we need are: $T_3 = a + 2d, \quad T_7 = a + 6d, \quad T_{17} = a + 16d.$

Step 2 — Translate the condition.

We're told the square of the 7th term equals the product of the 3rd and 17th terms: $(T_7)^2 = T_3 \cdot T_{17}$

This is exactly the condition for $T_3, T_7, T_{17}$ to be in geometric progression (the middle term squared equals the product of its neighbours).

Substituting: $(a + 6d)^2 = (a + 2d)(a + 16d)$

Step 3 — Expand both sides.

Left side: $(a + 6d)^2 = a^2 + 12ad + 36d^2$

Right side: $(a + 2d)(a + 16d) = a^2 + 16ad + 2ad + 32d^2 = a^2 + 18ad + 32d^2$

Setting them equal: $a^2 + 12ad + 36d^2 = a^2 + 18ad + 32d^2$

Step 4 — Simplify.

Cancel $a^2$ from both sides: $12ad + 36d^2 = 18ad + 32d^2$

Gather the $d^2$ terms on the left and the $ad$ terms on the right: $36d^2 - 32d^2 = 18ad - 12ad$ $4d^2 = 6ad$

Since $d > 0$ , we can safely divide both sides by $2d$ : $2d = 3a$

Step 5 — Find the ratio $a: d$ .

From $2d = 3a$ : $\frac{a}{d} = \frac{2}{3}$

So the ratio of the first term to the common difference is $a: d = 2: 3$ .

Step 6 — Verify with actual numbers.

Let $a = 2,\; d = 3$ :

Term	Formula	Value
$T_3$	$2 + 2(3)$	$8$
$T_7$	$2 + 6(3)$	$20$
$T_{17}$	$2 + 16(3)$	$50$

Check: $T_7^2 = 20^2 = 400$ and $T_3 \cdot T_{17} = 8 \times 50 = 400$ . ✓

Answer: A — $2: 3$

💡 Why this works: The condition $(T_7)^2 = T_3 \cdot T_{17}$ forces three AP terms to also lie in GP. That double constraint is strong enough to pin down a single ratio $a/d$ , regardless of the actual values of $a$ and $d$ . Whenever a problem says "the square of one term equals the product of two others," think: geometric mean condition — and expand directly.

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Q31MensurationMCQ

A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8 : 27 : 27. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to

Show solution

Correct answer: B

Think of this as: when you melt and recast a cube, the total volume stays the same, but the surface area can change dramatically — because volume scales with side³ while surface area scales with side². So we need to compare the squares of the new sides against the square of the original side.

Step 1 — Convert the volume ratio into a side-length ratio.

The five new cubes have volumes in the ratio $1: 1: 8: 27: 27$ .

Since volume of a cube $= (\text{side})^3$ , the side lengths are the cube roots of these numbers:

Cube	Volume (ratio)	Side (ratio)
1	$1$	$1$
2	$1$	$1$
3	$8$	$2$
4	$27$	$3$
5	$27$	$3$

So the five side lengths are in the ratio $1: 1: 2: 3: 3$ .

Step 2 — Find the original cube's side (in the same ratio units).

Let the common scale factor be $k$ , so the five actual side lengths are $k,\, k,\, 2k,\, 3k,\, 3k$ .

The total volume of the five new cubes is: $k^3 + k^3 + (2k)^3 + (3k)^3 + (3k)^3 = k^3(1 + 1 + 8 + 27 + 27) = 64k^3.$

Since the metal is just melted and recast, volume is conserved — the original cube's volume is also $64k^3$ .

Therefore the original side length is: $\sqrt[3]{64k^3} = 4k.$

Step 3 — Compute the original surface area.

A cube has 6 faces, each of area $\text{side}^2$ : $SA_{\text{original}} = 6 \times (4k)^2 = 6 \times 16k^2 = 96k^2.$

Step 4 — Compute the sum of the five new surface areas.

$SA_{\text{new}} = 6k^2 + 6k^2 + 6(2k)^2 + 6(3k)^2 + 6(3k)^2$ $= 6k^2(1^2 + 1^2 + 2^2 + 3^2 + 3^2)$ $= 6k^2(1 + 1 + 4 + 9 + 9) = 6k^2 \times 24 = 144k^2.$

Step 5 — Find the percentage excess.

$\text{Excess} = 144k^2 - 96k^2 = 48k^2.$

$\text{Percentage excess} = \frac{48k^2}{96k^2} \times 100\% = 50\%.$

💡 Teacher tip: The scale factor $k$ cancels out completely — as it must, since the answer is a percentage and should not depend on the actual size of the original cube. You could have set $k = 1$ from the start.

Answer: B — 50

💡 Why this happens: Breaking one large cube into several smaller ones increases total surface area because you're exposing new faces. The more uneven the split (here, several small cubes alongside larger ones), the bigger the surface-area penalty. Volume is conserved; surface area is not.

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Q32Averages & Weighted AveragesTITA

An elevator has a weight limit of 630 kg. It is carrying a group of people of whom the heaviest weighs 57 kg and the lightest weighs 53 kg. What is the maximum possible number of people in the group?

Your answer:

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Correct answer: 11

Think of this as: we have a fixed weight budget of 630 kg, and we want to pack in as many people as possible. Two people's weights are already locked in — so how light can everyone else be?

Step 1 — Identify what's fixed and what's free.

We're told the heaviest person weighs 57 kg and the lightest weighs 53 kg. These two weights are pinned — they cannot change. Every other person in the elevator must weigh somewhere between 53 kg and 57 kg (inclusive), because 53 is the lightest and 57 is the heaviest in the group.

Step 2 — Minimise each free person's weight.

To fit the maximum number of people under a 630 kg ceiling, we want the total weight to be as small as possible. The heaviest person is stuck at 57 kg. Every other person can be made as light as the lightest — 53 kg. So the lightest possible group of $n$ people is:

1 person at 57 kg (the heaviest)
$(n - 1)$ people at 53 kg (the lightest plus everyone else)

Step 3 — Set up the inequality.

The total weight of this lightest possible group must not exceed 630 kg:

$57 + 53(n - 1) \le 630$

Step 4 — Solve for $n$ .

Expand and simplify:

$57 + 53n - 53 \le 630$

$53n + 4 \le 630$

$53n \le 626$

$n \le \frac{626}{53} \approx 11.81$

Since $n$ must be a whole number, the largest possible value is $n = 11$ .

💡 Teacher tip: The key move is recognising that the heaviest person's extra weight above the minimum is what eats into the budget. Here, one person costs 57 instead of 53 — that's only 4 extra kg — so the effective constraint is $53n + 4 \le 630$ .

Step 5 — Verify the boundary.

$n$	Total weight (lightest config)	Fits in 630 kg?
11	$57 + 53 \times 10 = 587$ kg	✅
12	$57 + 53 \times 11 = 640$ kg	❌

At $n = 11$ , the total is 587 kg — comfortably under 630 kg. At $n = 12$ , even with everyone as light as possible, the total is 640 kg — over the limit. So 11 is indeed the maximum.

Answer: 11

💡 Closing insight: In any "maximise the count" problem with a fixed total budget, push every free variable to its smallest allowed value, then check the boundary by hand. The one pinned-heavy value acts like a small surcharge on top of the per-person minimum.

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Q33Profit Loss & DiscountMCQ

In a market, the price of medium quality mangoes is half that of good mangoes. A shopkeeper buys 80 kg good mangoes and 40 kg medium quality mangoes from the market and then sells all these at a common price which is 10% less than the price at which he bought the good ones. His overall profit is

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Correct answer: B

Think of this as: two types of mangoes bought at different rates are blended and sold at one common rate. Profit comes from the gap between total revenue and total cost.

Step 1 — Assign clean prices.

The problem only gives ratios of prices, so we can pick any convenient number. Let the market price of good mangoes be $100$ Rs/kg. Since medium-quality mangoes cost half as much, their price is $50$ Rs/kg.

💡 Teacher tip: When a question gives only ratios ("half", "10% less"), fixing one base value to $100$ keeps every calculation in whole numbers. The final percentage profit is unaffected by the choice.

Step 2 — Compute total cost.

The shopkeeper buys $80$ kg of good mangoes and $40$ kg of medium-quality mangoes:

$\text{Cost of good mangoes} = 80 \times 100 = 8000$

$\text{Cost of medium mangoes} = 40 \times 50 = 2000$

$\text{Total cost} = 8000 + 2000 = \text{Rs }10000.$

Step 3 — Find the common selling price.

All mangoes are sold at one price that is 10% less than the price he paid for the good mangoes.

$\text{Selling price per kg} = 100 - 10\%\text{ of }100 = 90 \text{ Rs/kg}.$

Total quantity sold $= 80 + 40 = 120$ kg.

$\text{Total revenue} = 120 \times 90 = \text{Rs }10800.$

Step 4 — Compute profit percentage.

Profit is measured on the cost (the money the shopkeeper actually put in):

$\text{Profit} = 10800 - 10000 = \text{Rs }800.$

$\text{Profit }\% = \frac{800}{10000} \times 100 = 8\%.$

Answer: B — 8%

💡 Why this works: The good mangoes are sold at a $10\%$ loss (bought at $100$ , sold at $90$ ), but the medium mangoes are sold at a big profit (bought at $50$ , sold at $90$ ). Because the good mangoes form the larger share of the cost ( $8000$ of $10000$ ), they pull the blend toward a loss; the medium mangoes rescue it, landing the overall result at a modest $8\%$ profit.

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Q34Maxima & MinimaTITA

If a, b, c, and d are integers such that a + b + c + d = 30, then the minimum possible value of (a − b)² + (a − c)² + (a − d)² is

Your answer:

Show solution

Correct answer: 2

Think of this as: we're measuring how far $b$ , $c$ , and $d$ each sit from $a$ , and we want all four integers to hug the overall average ( $30/4 = 7.5$ ) as tightly as possible.

Step 1 — Understand what we're minimising.

The expression $(a-b)^2 + (a-c)^2 + (a-d)^2$ is a sum of three squared differences. Since $a, b, c, d$ are integers, each difference $a-b$ , $a-c$ , $a-d$ is an integer, so each squared term is a perfect square (0, 1, 4, 9, …).

To make this sum small, we want $b$ , $c$ , and $d$ to be as close to $a$ as possible — ideally equal to $a$ .

Step 2 — Can the value be 0?

For the sum to be $0$ , every term must be $0$ , so $b = c = d = a$ . Then the total sum is $4a = 30$ , giving $a = 7.5$ , which is not an integer. ❌

Step 3 — Can the value be 1?

For three perfect squares to sum to $1$ , exactly one difference must be $\pm 1$ and the other two must be $0$ . So three of the four variables are equal, and one differs by $1$ .

If three equal $a$ and one is $a+1$ : sum $= 4a + 1 = 30 \Rightarrow 4a = 29$ . Not divisible by $4$ . ❌
If three equal $a$ and one is $a-1$ : sum $= 4a - 1 = 30 \Rightarrow 4a = 31$ . Not divisible by $4$ . ❌

So a value of $1$ is structurally impossible.

Step 4 — Can the value be 2?

For three perfect squares to sum to $2$ , we need two differences equal to $\pm 1$ and one equal to $0$ (since $1 + 1 + 0 = 2$ is the only way).

Try $a = 8$ . Then $b + c + d = 22$ . Choose $b, c, d$ as close to $a = 8$ as possible: $b = 7,\ c = 7,\ d = 8$ (sum $= 22$ ✓).

$\text{Value} = (8-7)^2 + (8-7)^2 + (8-8)^2 = 1 + 1 + 0 = 2.\ \checkmark$

Step 5 — Verification table.

Target	Configuration needed	Sum constraint	Achievable?
$0$	$b = c = d = a$	$4a = 30$	❌ ( $a = 7.5$ )
$1$	three equal, one differs by $\pm 1$	$4a \pm 1 = 30$	❌ (29 or 31, not divisible by 4)
$2$	two differ by $\pm 1$ , one equal	e.g. $8, 7, 7, 8$	✅

Since $0$ and $1$ are both impossible and $2$ is achievable, the minimum is $2$ .

Answer: $\boxed{2}$

💡 Teacher tip: When minimising a sum of squared deviations with an integer constraint, first identify the ideal (all-equal) case, then check which small targets are ruled out by divisibility — this is faster than trial-and-error.

Alternative approach — Substitution with modular arithmetic.

Let $x = a - b$ , $y = a - c$ , $z = a - d$ . Then $b = a - x$ , $c = a - y$ , $d = a - z$ , and the sum becomes: $a + (a-x) + (a-y) + (a-z) = 4a - (x+y+z) = 30$ So $x + y + z = 4a - 30 \equiv 2 \pmod{4}$ .

We need to minimise $x^2 + y^2 + z^2$ subject to $x + y + z \equiv 2 \pmod{4}$ .

Sum of squares $= 0$ : $x = y = z = 0$ , sum $\equiv 0 \pmod{4}$ . ❌
Sum of squares $= 1$ : one of $x, y, z$ is $\pm 1$ , sum $\equiv \pm 1 \pmod{4}$ . ❌
Sum of squares $= 2$ : e.g. $x = 1, y = 1, z = 0$ , sum $= 2 \equiv 2 \pmod{4}$ . ✅

This confirms $2$ is the minimum, reached at $a = 8, b = 7, c = 7, d = 8$ .

💡 Why this works: The modular approach converts the problem into "what's the smallest sum of three squares whose total is $\equiv 2 \pmod 4$ ?" — a clean number-theory question that rules out $0$ and $1$ instantly.

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