CAT 2018 Quant Questions — Slot 1

The actual Quantitative Ability section from CAT 2018, 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 questions22 MCQ · 12 TITA60 min section0/34 attempted

Q1Set TheoryMCQ

If among 200 students, 105 like pizza and 134 like burger, then the number of students who like only burger can possibly be

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Correct answer: D

Think of this as: we know the total number of students and how many like each food, but we don't know the overlap. The question asks what the "burger-only" count can possibly be — so we need to find the full range of valid values, then check which option lands inside it.

Step 1 — Set up the four regions.

With 200 students split by their pizza/burger preferences, there are four disjoint groups:

$a$ = pizza only
$b$ = burger only
$c$ = both pizza and burger
$d$ = neither

From the data: $a + c = 105 \quad \text{(pizza lovers)}, \qquad b + c = 134 \quad \text{(burger lovers)}$ $a + b + c + d = 200 \quad \text{(everyone)}$

So we can express $a$ and $b$ in terms of $c$ : $a = 105 - c, \qquad b = 134 - c.$

Step 2 — Find the allowable range for $c$ (the overlap).

Every region count must be non-negative.

From $a \ge 0$ : $\;105 - c \ge 0 \;\Rightarrow\; c \le 105$ .
From $b \ge 0$ : $\;134 - c \ge 0 \;\Rightarrow\; c \le 134$ (already weaker than $c \le 105$ ).
From $c \ge 0$ : trivially $c \ge 0$ .

Now use the "neither" group. Substitute $a$ and $b$ : $d = 200 - a - b - c = 200 - (105 - c) - (134 - c) - c = 200 - 105 + c - 134 + c - c = c - 39.$

Since $d \ge 0$ : $c - 39 \ge 0 \;\Rightarrow\; c \ge 39.$

Combining both bounds: $\boxed{39 \le c \le 105}.$

💡 Teacher tip: The "neither" region is the hidden constraint here. Even though the question never mentions students who like neither food, that group must still exist as a non-negative count — and that's what pins down the lower bound on the overlap $c$ .

Step 3 — Convert the range of $c$ into a range for burger-only $b$ .

Since $b = 134 - c$ , burger-only is largest when the overlap $c$ is smallest, and vice versa:

Overlap $c$	Burger-only $b = 134 - c$
$c = 39$ (minimum)	$b = 134 - 39 = 95$ (maximum)
$c = 105$ (maximum)	$b = 134 - 105 = 29$ (minimum)

So the number of burger-only students can be any integer in: $29 \le b \le 95.$

Step 4 — Check which option falls inside the range.

Option	Value	In $[29, 95]$ ?
A	$23$	✗ (too low)
B	$26$	✗ (too low)
C	$96$	✗ (too high)
D	$93$	✅

Only 93 lies inside the valid range.

Answer: D — $\boxed{93}$

💡 Why this works: "Can possibly be" is the key phrase — it signals a range question, not a unique-value question. Whenever you see it, derive the minimum and maximum of the requested quantity using non-negativity of every region, then test the options against that range.

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Q2LogarithmsMCQ

If log₂ (5 + log₃ a) = 3 and log₅ (4a + 12 + log₂ b) = 3, then a + b is equal to

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Correct answer: A

Think of this as: two nested-log equations, each with a clean cube on the outside. Undo the outer log first, then the inner one — and the first equation hands you $a$ , which feeds straight into the second.

Step 1 — Translate the first equation.

We are given: $\log_2(5 + \log_3 a) = 3$

The key move: if $\log_x(\text{stuff}) = k$ , then $\text{stuff} = x^k$ . Applying that here with base $2$ : $5 + \log_3 a = 2^3 = 8$

Now subtract $5$ : $\log_3 a = 3$

Apply the same move again with base $3$ : $a = 3^3 = 27$

So $\boxed{a = 27}$ .

Step 2 — Translate the second equation.

We are given: $\log_5(4a + 12 + \log_2 b) = 3$

Again, undo the outer log first — the argument equals $5^3$ : $4a + 12 + \log_2 b = 5^3 = 125$

Now substitute $a = 27$ from Step 1: $4(27) + 12 + \log_2 b = 125$ $108 + 12 + \log_2 b = 125$ $120 + \log_2 b = 125$

Subtract $120$ : $\log_2 b = 5$

Undo this inner log with base $2$ : $b = 2^5 = 32$

So $\boxed{b = 32}$ .

Step 3 — Add them. $a + b = 27 + 32 = 59$

💡 Teacher tip: For nested logarithms, always work outside-in: first rewrite $\log_x(\text{stuff}) = k$ as $\text{stuff} = x^k$ , then deal with whatever log is left inside. Trying to handle both layers at once is where sign and base errors creep in.

Quick check.

Quantity	Value	Verify
$\log_3 a = \log_3 27$	$3$	$3^3 = 27$ ✓
$5 + \log_3 a$	$8$	$\log_2 8 = 3$ ✓
$\log_2 b = \log_2 32$	$5$	$2^5 = 32$ ✓
$4a + 12 + \log_2 b$	$120 + 5 = 125$	$\log_5 125 = 3$ ✓

Everything closes cleanly, so the values are consistent.

Answer: A — $\boxed{59}$

The structure of the question is a chain: the first equation isolates $a$ , and $a$ is then planted inside the second equation to isolate $b$ . Once you see that dependency, each equation is just a single logarithm to undo at a time.

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Q3Mixtures & AlligationsMCQ

Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio

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Correct answer: C

Think of this as: the selling price and profit percentage together tell us the cost price of each blend; each blend's cost is a weighted average of the two ingredient costs — so two different blends give us two equations in two unknowns.

Step 1 — Recover each blend's cost price from the selling price.

The mixture sells at Rs. 40/kg in both cases. Profit percentage links selling price to cost price:

$\text{Cost Price} = \frac{\text{Selling Price}}{1 + \text{profit fraction}}$

10% profit (ratio 3: 2): Cost $= \dfrac{40}{1.10} = \dfrac{40}{11/10} = \dfrac{400}{11}$ Rs/kg
5% profit (ratio 2: 3): Cost $= \dfrac{40}{1.05} = \dfrac{40}{21/20} = \dfrac{800}{21}$ Rs/kg

Step 2 — Write the weighted-cost equations.

When A and B are mixed in ratio $3:2$ , the mixture cost is the weighted average of the ingredient costs:

$\frac{3C_A + 2C_B}{5} = \frac{400}{11}$

Multiplying both sides by 5:

$3C_A + 2C_B = \frac{2000}{11} \quad \cdots (1)$

Similarly, for the ratio $2:3$ mixture:

$\frac{2C_A + 3C_B}{5} = \frac{800}{21}$

$2C_A + 3C_B = \frac{4000}{21} \quad \cdots (2)$

Step 3 — Eliminate $C_B$ and solve for $C_A$ .

Multiply equation (1) by 3 and equation (2) by 2 so the $C_B$ coefficients match:

$9C_A + 6C_B = \frac{6000}{11} \quad \cdots (3)$ $4C_A + 6C_B = \frac{8000}{21} \quad \cdots (4)$

Subtract (4) from (3):

$5C_A = \frac{6000}{11} - \frac{8000}{21}$

Common denominator is $11 \times 21 = 231$ :

$5C_A = \frac{6000 \times 21 - 8000 \times 11}{231} = \frac{126000 - 88000}{231} = \frac{38000}{231}$

$C_A = \frac{7600}{231}$

Step 4 — Back-substitute to find $C_B$ .

From equation (1):

$2C_B = \frac{2000}{11} - 3C_A = \frac{2000}{11} - \frac{22800}{231}$

Convert $\frac{2000}{11}$ to denominator 231 (multiply by $\frac{21}{21}$ ):

$2C_B = \frac{42000 - 22800}{231} = \frac{19200}{231}$

$C_B = \frac{9600}{231}$

Step 5 — Take the ratio.

$C_A: C_B = \frac{7600}{231}: \frac{9600}{231} = 7600: 9600 = 19: 24$

💡 Teacher tip: The denominators (231) cancel the moment you take the ratio — you never need the actual cost prices, only their relative sizes. This is why keeping $C_A$ and $C_B$ symbolic until the final ratio step avoids unnecessary arithmetic noise.

Answer: C — $\boxed{19: 24}$

Alternative approach — Divide the equations directly (faster).

Instead of solving for $C_A$ and $C_B$ individually, let $r = \dfrac{C_A}{C_B}$ and divide the two equations to eliminate the common scale factor $C_B$ .

From (1): $C_B(3r + 2) = \dfrac{2000}{11}$

From (2): $C_B(2r + 3) = \dfrac{4000}{21}$

Dividing the first by the second:

$\frac{3r + 2}{2r + 3} = \frac{2000/11}{4000/21} = \frac{2000 \times 21}{11 \times 4000} = \frac{21}{22}$

Cross-multiply:

$22(3r + 2) = 21(2r + 3)$ $66r + 44 = 42r + 63$ $24r = 19$ $r = \frac{19}{24}$

So $C_A: C_B = 19: 24$ — confirmed.

💡 Why this is faster: by dividing the equations, $C_B$ cancels immediately and you solve one equation in one unknown ( $r$ ) — no common denominators, no back-substitution. Under exam time, this is the method to reach for.

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Q4CirclesMCQ

In a circle with center O and radius 1 cm, an arc AB makes an angle 60 degrees at O. Let R be the region bounded by the radii OA, OB and the arc AB. If C and D are two points on OA and OB, respectively, such that OC = OD and the area of triangle OCD is half that of R, then the length of OC, in cm, is

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Correct answer: C

Think of this as: a sector is cut out of a circle, and inside that sector we place a small isosceles triangle whose area is exactly half the sector's. We need the side length of that triangle.

Step 1 — Find the area of the sector $R$ .

The region $R$ is a sector of radius $1$ cm with central angle $60^\circ$ . The area of a sector is the fraction of the full circle given by $\dfrac{\theta}{360^\circ}$ :

$\text{Area}(R) = \frac{60^\circ}{360^\circ}\cdot \pi (1)^2 = \frac{1}{6}\pi = \frac{\pi}{6}\text{ cm}^2.$

Since the triangle $OCD$ must have half this area:

$\text{Area}(\triangle OCD) = \frac12 \cdot \frac{\pi}{6} = \frac{\pi}{12}\text{ cm}^2.$

Step 2 — Express the area of $\triangle OCD$ in terms of $OC$ .

Let $OC = OD = x$ (we are told the two are equal). The angle between these two sides is the sector's central angle, $\angle COD = 60^\circ$ .

Using the formula for the area of a triangle given two sides and the included angle, $\text{Area} = \tfrac12 ab\sin C$ :

$\text{Area}(\triangle OCD) = \frac12 \cdot x \cdot x \cdot \sin 60^\circ = \frac12 x^2 \cdot \frac{\sqrt3}{2} = \frac{\sqrt3}{4}x^2.$

💡 Teacher tip: Whenever a triangle sits inside a sector with its vertex at the centre, the two radii become two sides of the triangle and the sector angle becomes the included angle — so $\tfrac12 ab\sin C$ is usually the fastest route.

Step 3 — Equate and solve for $x$ .

$\frac{\sqrt3}{4}x^2 = \frac{\pi}{12}.$

Multiply both sides by $4$ :

$\sqrt3\, x^2 = \frac{\pi}{3}.$

Now divide by $\sqrt3$ :

$x^2 = \frac{\pi}{3\sqrt3}.$

Taking the positive square root (since $x$ is a length):

$x = \sqrt{\frac{\pi}{3\sqrt3}}.$

Step 4 — Match with the options.

This is exactly option C.

Answer: C — $\boxed{\sqrt{\dfrac{\pi}{3\sqrt3}}}$

💡 Why this works out neatly: the sector area brings in a factor of $\pi$ (from the curved arc), while the triangle area brings in $\sqrt3$ (from $\sin 60^\circ$ ). Equating them forces $\pi$ and $\sqrt3$ into the same expression, which is why the answer mixes both.

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Q5Mixtures & AlligationsMCQ

A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is Rs. 8 more than that of paint B. If the trader sells the entire mixture for Rs. 264 and makes a profit of 10%, then the highest possible cost of paint B, in Rs. per litre, is

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Correct answer: C

Think of this as: we know the mixture's total cost price and its per-litre cost. Paint A is Rs. 8/L dearer than paint B, and B can be at most half the mixture. We want to push paint B's price as high as possible — so we need to find the extreme blend that still satisfies every condition.

Step 1 — Find the cost price of the mixture.

The trader sells 10 litres for Rs. 264 at a 10% profit. So the selling price is 110% of the cost price:

$\text{Cost price of 10 L} = \frac{264}{1.10} = 240 \text{ Rs.}$

$\text{Cost price per litre} = \frac{240}{10} = 24 \text{ Rs/L}$

This Rs. 24/L is the weighted average of the costs of paint A and paint B.

Step 2 — Set up variables.

Let the cost of paint B be $b$ Rs/L. Then the cost of paint A is $b + 8$ Rs/L.

Let the amount of paint A be $a$ litres, so paint B is $10 - a$ litres.

The condition "amount of B does not exceed amount of A" gives:

$10 - a \le a \implies a \ge 5$

So paint A is at least 5 litres (i.e., A is at least half the mixture).

Step 3 — Write the cost equation.

The total cost of the mixture equals the cost of A plus the cost of B:

$a(b+8) + (10-a)b = 240$

Expand and simplify:

$ab + 8a + 10b - ab = 240$

$8a + 10b = 240$

Divide by 2:

$4a + 5b = 120$

Solve for $b$ :

$b = \frac{120 - 4a}{5} = 24 - \frac{4a}{5}$

Step 4 — Maximise $b$ by minimising $a$ .

Since $b = 24 - \frac{4a}{5}$ , the cost of B decreases as $a$ increases. To make $b$ as large as possible, we need $a$ as small as possible.

The smallest allowed value is $a = 5$ (from Step 2):

$b = 24 - \frac{4 \times 5}{5} = 24 - 4 = 20$

💡 Teacher tip: "Highest possible cost of B" combined with "B ≤ A" pushes the mixture to its most B-heavy legal blend — exactly half A and half B, i.e., 5 litres each. Plug that extreme in directly and you get the answer in one line.

Verification check:

$a$ (litres of A)	$b = 24 - \frac{4a}{5}$	Valid? ( $a \ge 5$ )
5	20	✅ (minimum $a$ , maximum $b$ )
7.5	18	✅
10	16	✅ (all A, minimum $b$ )

As $a$ rises from 5 to 10, $b$ falls from 20 to 16 — confirming the highest possible cost of B is Rs. 20/L.

Answer: C — $\boxed{20}$

💡 Why this works: The per-litre cost of the mixture (Rs. 24) is fixed. Since paint A is costlier, using more A would drag the average up — meaning B's price would have to drop to compensate. So the most B-friendly (and thus highest- $b$ ) scenario is the one with the least A allowed, which is $a = 5$ .

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Q6CirclesMCQ

In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is

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Correct answer: A

Think of this as: two parallel chords sit on the same side of the centre, 1 cm apart. The chord–centre-distance formula links each chord's length to the radius — and since both chords share the same radius, we get two equations we can solve together.

Key property (chord–centre-distance relation). A chord of length $\ell$ whose perpendicular distance from the centre is $d$ satisfies $r^2 = d^2 + \left(\frac{\ell}{2}\right)^2.$ The half-chord, the perpendicular distance to the centre, and the radius form a right triangle.

Step 1 — Set up the two chords.

Chord	Length	Half-length	Distance from centre
$C_1$ (longer)	$6$	$3$	$d_1$
$C_2$ (shorter)	$4$	$2$	$d_2$

Both chords live in the same circle, so each gives an equation for the same $r^2$ : $r^2 = d_1^2 + 3^2 = d_1^2 + 9, \qquad r^2 = d_2^2 + 2^2 = d_2^2 + 4.$

Step 2 — Decide which chord is closer to the centre. A longer chord is always closer to the centre (think of the diameter, the longest chord, passing through the centre). So the 6-cm chord is nearer the centre: $d_1 < d_2$ .

Since the chords are on the same side of the diameter and 1 cm apart, $d_2 - d_1 = 1.$

💡 Teacher tip: "Same side" means the distances subtract; "opposite sides" would mean they add. Getting this sign right is the make-or-break step in chord-gap problems.

Step 3 — Subtract the two radius equations. From $d_1^2 + 9 = d_2^2 + 4$ : $d_2^2 - d_1^2 = 5.$ Factor the left side as a difference of squares: $(d_2 - d_1)(d_2 + d_1) = 5.$ Using $d_2 - d_1 = 1$ : $1 \cdot (d_2 + d_1) = 5 \;\Rightarrow\; d_2 + d_1 = 5.$

Step 4 — Solve for $d_1$ and $d_2$ . $d_2 - d_1 = 1, \qquad d_2 + d_1 = 5.$ Adding: $2d_2 = 6 \Rightarrow d_2 = 3$ . Subtracting: $2d_1 = 4 \Rightarrow d_1 = 2$ .

Step 5 — Compute the radius. $r^2 = d_1^2 + 3^2 = 2^2 + 9 = 4 + 9 = 13 \;\Rightarrow\; r = \sqrt{13}.$

Quick cross-check with the other chord: $r^2 = d_2^2 + 2^2 = 9 + 4 = 13$ . ✓

Answer: A — $\boxed{\sqrt{13}\text{ cm}}$

💡 Why this works so cleanly: subtracting the two $r^2$ -equations turns the problem into a difference of squares, $(d_2-d_1)(d_2+d_1)$ , which immediately yields the sum of the distances from the known gap. That single identity is the engine of every "two parallel chords with a given separation" problem.

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Q7MensurationTITA

A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft. The tip of the cone is cut off with a plane which is parallel to the base and 9 ft from the base. With π = 22/7, the volume, in cubic ft, of the remaining part of the cone is

Your answer:

Show solution

Correct answer: 198

Think of this as: a small cone is sliced off the top of a bigger cone; the leftover solid (a frustum) has volume = big cone volume − small cone volume.

Step 1 — Understand the geometry.

The full cone has height $12$ ft and base diameter $8$ ft, so its base radius is $R = 4$ ft.

The cutting plane is parallel to the base and $9$ ft above the base. Since the total height is $12$ ft, the plane sits $12 - 9 = 3$ ft below the apex. The piece removed — the tip — is itself a small cone of height $h = 3$ ft, and it is similar to the original cone.

Step 2 — Find the radius of the small cone.

When a cone is cut by a plane parallel to its base, the removed tip is similar to the original cone. The linear scale factor is the ratio of their heights:

$\text{scale factor} = \frac{3}{12} = \frac{1}{4}.$

So the small cone's base radius is

$r = 4 \times \frac{1}{4} = 1 \text{ ft}.$

💡 Teacher tip: "Slice a cone parallel to its base" always produces a similar small cone — find its scale factor from the heights, and every linear dimension (radius, slant height, etc.) scales by the same factor.

Step 3 — Volume of the full cone.

$V_{\text{big}} = \frac{1}{3}\pi R^{2}H = \frac{1}{3}\pi (4)^{2}(12) = \frac{1}{3}\pi \cdot 16 \cdot 12 = 64\pi.$

Step 4 — Volume of the small cone (the tip).

$V_{\text{small}} = \frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi (1)^{2}(3) = \frac{1}{3}\pi \cdot 1 \cdot 3 = \pi.$

Step 5 — Subtract to get the remaining frustum.

$V = V_{\text{big}} - V_{\text{small}} = 64\pi - \pi = 63\pi.$

Now substitute $\pi = \frac{22}{7}$ :

$V = 63 \times \frac{22}{7} = 9 \times 22 = 198.$

Answer: $\boxed{198 \text{ cubic ft}}$

Alternative approach — Frustum formula.

The remaining part is a frustum of height $H_f = 9$ ft, lower radius $R = 4$ ft, and upper radius $r = 1$ ft. Using the frustum volume formula:

$V = \frac{1}{3}\pi H_f\left(R^{2} + Rr + r^{2}\right) = \frac{1}{3}\pi (9)\left(16 + 4 + 1\right) = \frac{1}{3}\pi \cdot 9 \cdot 21 = 3\pi \cdot 21 = 63\pi.$

With $\pi = \frac{22}{7}$ , this again gives $63 \times \frac{22}{7} = 198$ . The subtraction method is faster here because the small cone's volume works out to a clean $\pi$ .

💡 Why this is elegant: the small cone is a $\frac{1}{4}$ -scale copy of the big one, so its volume is $\left(\frac{1}{4}\right)^{3} = \frac{1}{64}$ of the big cone's volume — which is exactly why $V_{\text{small}} = \pi$ when $V_{\text{big}} = 64\pi$ . Volume scales with the cube of the linear scale factor.

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Q8Boats Trains & Relative SpeedTITA

Train T leaves station X for station Y at 3 pm. Train S, traveling at three quarters of the speed of T, leaves Y for X at 4 pm. The two trains pass each other at a station Z, where the distance between X and Z is three-fifths of that between X and Y. How many hours does train T take for its journey from X to Y?

Your answer:

Show solution

Correct answer: 15

Think of this as: two trains head toward each other from opposite ends, one starts an hour earlier and is faster. They meet at a point that splits the total distance in a known ratio. We need the faster train's total journey time.

Step 1 — Set up distances and speeds.

Let the total distance $XY = D$ .

Since $Z$ is the meeting point and $XZ = \tfrac{3}{5}$ of $XY$ : $XZ = \frac{3D}{5}, \qquad ZY = D - \frac{3D}{5} = \frac{2D}{5}.$

Let train $T$ 's speed be $v$ . Then train $S$ 's speed is $\tfrac{3}{4}v$ .

Step 2 — Write each train's travel time to reach $Z$ .

Train $T$ travels from $X$ to $Z$ , a distance of $\tfrac{3D}{5}$ , at speed $v$ : $\text{Time of } T = \frac{3D/5}{v} = \frac{3D}{5v}.$ This time is measured from 3 pm, when $T$ started.

Train $S$ travels from $Y$ to $Z$ , a distance of $\tfrac{2D}{5}$ , at speed $\tfrac{3v}{4}$ : $\text{Time of } S = \frac{2D/5}{3v/4} = \frac{2D}{5} \times \frac{4}{3v} = \frac{8D}{15v}.$ This time is measured from 4 pm, when $S$ started.

Step 3 — Use the fact that they reach $Z$ at the same instant.

💡 Teacher tip: "They pass at the same place at the same time" is your single equation. Write both travel times to that place and equate them — but keep careful track of who started when, because $T$ had a one-hour head start.

Since $T$ left one hour earlier than $S$ , $T$ 's travel time is one hour more than $S$ 's: $\frac{3D}{5v} = \frac{8D}{15v} + 1.$

Multiply through by $15v$ to clear denominators: $9D = 8D + 15v.$

$\boxed{D = 15v}.$

Step 4 — Find $T$ 's total journey time from $X$ to $Y$ .

Train $T$ must cover the full distance $D$ at speed $v$ : $\text{Total time} = \frac{D}{v} = \frac{15v}{v} = 15 \text{ hours}.$

Quick sanity check. If $T$ takes 15 hours total, its speed is $v = \tfrac{D}{15}$ . Then $S$ 's speed is $\tfrac{3v}{4} = \tfrac{D}{20}$ . $T$ reaches $Z$ in $\tfrac{3D/5}{D/15} = 9$ hours (from 3 pm → midnight). $S$ reaches $Z$ in $\tfrac{2D/5}{D/20} = 8$ hours (from 4 pm → midnight). Same instant. ✓

Answer: $\boxed{15 \text{ hours}}$

💡 Why this works cleanly: the distance $D$ and speed $v$ always appear together as $\tfrac{D}{v}$ , which is exactly the total time we want — so the equation solves itself without needing numerical values for either.

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Q9Polynomials & ProgressionsMCQ

Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is

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Correct answer: C

Think of this as: we have a GP whose three terms we can write using a common ratio $r$ , and an AP condition that will hand us a quadratic equation in $r$ .

Step 1 — Parametrise the geometric progression.

Since $x, y, z$ are in GP with $x < y < z$ , let the common ratio be $r > 1$ and write

$x = a,\quad y = ar,\quad z = ar^{2}$

for some positive $a$ . The condition $x < y < z$ is exactly what forces $r > 1$ .

Step 2 — Translate the AP condition.

We are told $5x,\ 16y,\ 12z$ are in arithmetic progression. Three numbers are in AP when the middle one equals the average of the other two, i.e.

$2 \times (\text{middle}) = (\text{first}) + (\text{last}).$

$2(16y) = 5x + 12z.$

Substituting the GP parametrisation:

$32(ar) = 5a + 12(ar^{2}).$

Step 3 — Simplify to a quadratic in $r$ .

Divide both sides by $a$ (which is positive, so safe):

$32r = 5 + 12r^{2}$

$\Rightarrow 12r^{2} - 32r + 5 = 0.$

Step 4 — Solve the quadratic.

Using the quadratic formula $r = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ with $a=12,\ b=-32,\ c=5$ :

$r = \frac{32 \pm \sqrt{(-32)^{2} - 4(12)(5)}}{2(12)} = \frac{32 \pm \sqrt{1024 - 240}}{24} = \frac{32 \pm \sqrt{784}}{24} = \frac{32 \pm 28}{24}.$

This gives two candidates:

Candidate	Value	$> 1$ ?
$r = \dfrac{32 + 28}{24} = \dfrac{60}{24}$	$\dfrac{5}{2}$	✅
$r = \dfrac{32 - 28}{24} = \dfrac{4}{24}$	$\dfrac{1}{6}$	❌

Step 5 — Pick the valid root.

Because $x < y < z$ , the common ratio must satisfy $r > 1$ . Only $r = \dfrac{5}{2}$ qualifies.

💡 Teacher tip: "Three terms in AP" $\Rightarrow$ "twice the middle $=$ sum of the ends." Write that single equation, substitute the GP parametrisation, and you almost always land on a clean quadratic in $r$ . Then use the ordering ( $r>1$ or $0<r<1$ ) to pick the right root.

Answer: C — $\boxed{\dfrac{5}{2}}$

The second root $\dfrac{1}{6}$ is not a distractor to discard blindly — it is a genuine solution of the AP equation, but it corresponds to a decreasing GP ( $r<1$ ), which contradicts $x < y < z$ . The ordering condition is what selects the answer.

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Q10Quadratic & Linear EquationsMCQ

If U² + (U − 2V − 1)² = −4V(U + V), then what is the value of U + 3V?

Show solution

Correct answer: C

Think of this as: we have one equation in two unknowns ( $U$ and $V$ ). Normally that's impossible to solve uniquely — unless the equation is secretly a sum of squares equal to zero. Let's find out if it is.

Step 1 — Expand everything and bring all terms to one side.

Start with the given equation: $U^2 + (U - 2V - 1)^2 = -4V(U + V)$

First, expand $(U - 2V - 1)^2$ using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$ : $(U - 2V - 1)^2 = U^2 + 4V^2 + 1 - 4UV - 2U + 4V$

Also expand the right side: $-4V(U + V) = -4UV - 4V^2$

Now substitute both back into the equation: $U^2 + U^2 + 4V^2 + 1 - 4UV - 2U + 4V = -4UV - 4V^2$

Bring every term to the left side: $U^2 + U^2 + 4V^2 + 1 - 4UV - 2U + 4V + 4UV + 4V^2 = 0$

The $-4UV$ and $+4UV$ cancel each other out, and we combine like terms: $2U^2 + 8V^2 - 2U + 4V + 1 = 0$

Step 2 — Complete the square for both $U$ and $V$ .

Factor out the leading coefficients from each variable's terms: $2(U^2 - U) + 8\!\left(V^2 + \tfrac{V}{2}\right) + 1 = 0$

Now complete the square inside each parenthesis:

For $U$ : $U^2 - U = \left(U - \tfrac{1}{2}\right)^2 - \tfrac{1}{4}$
For $V$ : $V^2 + \tfrac{V}{2} = \left(V + \tfrac{1}{4}\right)^2 - \tfrac{1}{16}$

Substitute these in: $2\!\left[\left(U - \tfrac{1}{2}\right)^2 - \tfrac{1}{4}\right] + 8\!\left[\left(V + \tfrac{1}{4}\right)^2 - \tfrac{1}{16}\right] + 1 = 0$

Distribute the coefficients: $2\!\left(U - \tfrac{1}{2}\right)^2 - \tfrac{1}{2} + 8\!\left(V + \tfrac{1}{4}\right)^2 - \tfrac{1}{2} + 1 = 0$

The constants combine neatly: $-\tfrac{1}{2} - \tfrac{1}{2} + 1 = 0$ , leaving: $2\!\left(U - \tfrac{1}{2}\right)^2 + 8\!\left(V + \tfrac{1}{4}\right)^2 = 0$

💡 Teacher tip: A sum of non-negative terms (squares) can equal zero only if each term is individually zero. This is the key principle that lets us solve for two unknowns from one equation.

Step 3 — Read off the values.

Since both squared terms must be zero: $U - \tfrac{1}{2} = 0 \implies U = \tfrac{1}{2}$ $V + \tfrac{1}{4} = 0 \implies V = -\tfrac{1}{4}$

Step 4 — Compute $U + 3V$ . $U + 3V = \tfrac{1}{2} + 3\!\left(-\tfrac{1}{4}\right) = \tfrac{1}{2} - \tfrac{3}{4} = \tfrac{2}{4} - \tfrac{3}{4} = -\tfrac{1}{4}$

Answer: C — $\boxed{-\tfrac{1}{4}}$

💡 Why this works: One equation in two unknowns can pin down both values only when it collapses into a sum of squares $= 0$ . The telltale sign is that all the constant terms cancel perfectly after completing the square — that's not a coincidence, it's the structure of the problem.

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Q11TrianglesMCQ

Given an equilateral triangle T1 with side 24 cm, a second triangle T2 is formed by joining the midpoints of the sides of T1. Then a third triangle T3 is formed by joining the midpoints of the sides of T2. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T1, T2, T3,... will be

Show solution

Correct answer: D

Think of this as: one equilateral triangle spawns a smaller one inside it, that one spawns an even smaller one, and so on forever — what is the total area of all of them put together?

Step 1 — Area of the first triangle $T_1$ .

$T_1$ is equilateral with side $24$ cm. Using the standard area formula for an equilateral triangle,

$\text{Area}(T_1) = \frac{\sqrt{3}}{4}\times(\text{side})^2 = \frac{\sqrt{3}}{4}\times 24^2 = \frac{\sqrt{3}}{4}\times 576 = 144\sqrt{3}\text{ sq cm}.$

Step 2 — What happens when we join the midpoints?

When you join the midpoints of the sides of any triangle, the small triangle formed in the middle is similar to the original, and each of its sides is exactly half the corresponding side of the original.

Since area scales as the square of the side length, halving the side multiplies the area by $\left(\tfrac12\right)^2 = \tfrac14$ .

So each new triangle has one-fourth the area of the one before it:

$\text{Area}(T_{k+1}) = \tfrac14\,\text{Area}(T_k).$

💡 Teacher tip: "Join the midpoints" is a fixed signal in geometry problems — the inner triangle is always a $\tfrac14$ -area copy of the outer one. Spotting this instantly turns the problem into a geometric series.

Step 3 — Write the areas as a geometric series.

The areas form the sequence

$144\sqrt{3},\;\; 36\sqrt{3},\;\; 9\sqrt{3},\;\; \tfrac{9}{4}\sqrt{3},\;\; \dots$

This is a geometric progression with

first term $a = 144\sqrt{3}$ ,
common ratio $r = \tfrac14$ .

Step 4 — Sum the infinite series.

For an infinite geometric series with $|r| < 1$ , the sum is $\dfrac{a}{1-r}$ . Here $r = \tfrac14 < 1$ , so

$S = \frac{144\sqrt{3}}{1-\tfrac14} = \frac{144\sqrt{3}}{\tfrac34} = 144\sqrt{3}\times\frac{4}{3} = 192\sqrt{3}.$

Quick check — partial sums should climb toward $192\sqrt{3}$ :

Triangles added	Partial sum (in units of $\sqrt{3}$ )
$T_1$	$144$
$T_1+T_2$	$144+36=180$
$T_1+T_2+T_3$	$180+9=189$
$T_1+T_2+T_3+T_4$	$189+2.25=191.25$
$\dots$	$\to 192$

The partial sums approach $192\sqrt{3}$ , confirming the infinite-sum result.

Answer: D — $\boxed{192\sqrt{3}\text{ sq cm}}$

💡 Why this is elegant: the side lengths halve each time ( $24, 12, 6, \dots$ ), so the areas quarter each time. The total area is just $\tfrac{4}{3}$ of the very first triangle — a neat reminder that an infinite process can still land on a clean, finite number.

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Q12Profit Loss & DiscountMCQ

A wholesaler bought walnuts and peanuts, the price of walnut per kg being thrice that of peanut per kg. He then sold 8 kg of peanuts at a profit of 10% and 16 kg of walnuts at a profit of 20% to a shopkeeper. However, the shopkeeper lost 5 kg of walnuts and 3 kg of peanuts in transit. He then mixed the remaining nuts and sold the mixture at Rs. 166 per kg, thus making an overall profit of 25%. At what price, in Rs. per kg, did the wholesaler buy the walnuts?

Show solution

Correct answer: A

Think of this as a chain of transactions: wholesaler → shopkeeper → final customer. The shopkeeper's cost is what he paid the wholesaler, and his revenue is what he earned after the transit loss. The 25% profit links those two — and every rupee in the problem traces back to one variable: the peanut's cost price.

Step 1 — Set up the cost prices.

Let the wholesaler's cost price of peanuts be $p$ Rs/kg.

Since walnuts cost thrice as much per kg: $\text{Walnut cost} = 3p \text{ Rs/kg}.$

💡 Teacher tip: In a multi-stage problem, anchor on the cheapest item's price and express everything else as a multiple of it. Here, both nut prices and both selling prices will be multiples of $p$ — so one equation will close the whole problem.

Step 2 — Find the shopkeeper's cost (what he paid the wholesaler).

The wholesaler sells to the shopkeeper at a profit:

Peanuts: $10\%$ profit → selling price $= 1.1p$ Rs/kg.
Walnuts: $20\%$ profit → selling price $= 1.2 \times 3p = 3.6p$ Rs/kg.

The shopkeeper buys $8$ kg of peanuts and $16$ kg of walnuts, so his total cost is: $\text{Shopkeeper's cost} = 8(1.1p) + 16(3.6p) = 8.8p + 57.6p = 66.4p.$

Step 3 — Find the shopkeeper's revenue.

In transit, the shopkeeper loses $5$ kg of walnuts and $3$ kg of peanuts. What remains:

Peanuts: $8 - 3 = 5$ kg
Walnuts: $16 - 5 = 11$ kg
Total mixture sold $= 5 + 11 = 16$ kg

He sells this entire mixture at Rs. $166$ /kg: $\text{Revenue} = 16 \times 166 = 2656 \text{ Rs}.$

Step 4 — Use the 25% overall profit to solve for $p$ .

A $25\%$ profit means revenue is $1.25$ times the cost: $2656 = 1.25 \times 66.4p.$

Divide both sides by $1.25$ : $66.4p = \frac{2656}{1.25} = 2124.8.$

Now solve for $p$ : $p = \frac{2124.8}{66.4} = 32.$

Step 5 — Find the walnut cost price.

$\text{Walnut cost} = 3p = 3 \times 32 = 96 \text{ Rs/kg}.$

Quick verification:

Quantity	Value	Check
Peanut cost $p$	$32$ Rs/kg	—
Walnut cost $3p$	$96$ Rs/kg	matches option A
Shopkeeper's cost	$66.4 \times 32 = 2124.8$ Rs	—
Shopkeeper's revenue	$16 \times 166 = 2656$ Rs	—
Profit check	$\frac{2656 - 2124.8}{2124.8} = 25\%$	✅

Answer: A — $\boxed{\text{Rs. } 96 \text{ per kg}}$

💡 Why this works cleanly: The transit loss doesn't change the shopkeeper's cost — it only reduces how much he can sell. That's why his cost ( $66.4p$ ) uses the full $8+16$ kg he bought, while his revenue uses only the $16$ kg that survived transit. Keeping cost and revenue on their correct sides of the equation is the key to not slipping here.

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Q13Divisibility & FactorsTITA

The number of integers x such that 0.25 ≤ $2^x$ ≤ 200 and $2^x$ + 2 is perfectly divisible by either 3 or 4, is

Your answer:

Show solution

Correct answer: 5

We need to count how many integers $x$ satisfy both conditions: the size bound $0.25 \le 2^x \le 200$ , and the divisibility condition that $2^x + 2$ is divisible by $3$ or $4$ .

Step 1 — Translate the size bound into a range for $x$ .

Write $0.25$ as a power of $2$ : $0.25 = \frac{1}{4} = 2^{-2}$ .

For the upper end, compute consecutive powers of $2$ : $2^7 = 128 \le 200 < 256 = 2^8.$

So the inequality $2^{-2} \le 2^x \le 2^7$ gives: $-2 \le x \le 7.$

That's the candidate set $x \in \{-2, -1, 0, 1, 2, 3, 4, 5, 6, 7\}$ .

Step 2 — Recognise the hidden integer requirement.

The phrase "perfectly divisible by $3$ or $4$ " only makes sense when $2^x + 2$ is an integer. Since $2$ is already an integer, we need $2^x$ to be an integer too — which forces $x \ge 0$ .

💡 Teacher tip: "Perfectly divisible" silently demands an integer dividend. That knocks out $x = -2$ and $x = -1$ before you even start testing divisibility.

So the effective candidate set is: $x \in \{0, 1, 2, 3, 4, 5, 6, 7\}.$

Step 3 — Test each candidate.

We compute $2^x + 2$ for each value and check divisibility by $3$ and by $4$ :

$x$	$2^x + 2$	Div by $3$ ?	Div by $4$ ?	Valid?
$0$	$3$	✅ ( $3 = 3{\times}1$ )	—	✅
$1$	$4$	✗	✅ ( $4 = 4{\times}1$ )	✅
$2$	$6$	✅ ( $6 = 3{\times}2$ )	✗	✅
$3$	$10$	✗	✗	✗
$4$	$18$	✅ ( $18 = 3{\times}6$ )	✗	✅
$5$	$34$	✗	✗	✗
$6$	$66$	✅ ( $66 = 3{\times}22$ )	✗	✅
$7$	$130$	✗	✗	✗

Step 4 — Count the valid values.

The valid $x$ -values are $\{0, 1, 2, 4, 6\}$ — that's $5$ integers.

Answer: $\boxed{5}$

Alternative approach — use modular arithmetic to avoid the table.

Instead of testing all eight values, we can find the pattern directly.

Divisibility by $3$ : We need $2^x + 2 \equiv 0 \pmod{3}$ , i.e. $2^x \equiv 1 \pmod{3}$ . Since $2 \equiv -1 \pmod{3}$ , this becomes $(-1)^x \equiv 1 \pmod{3}$ , which holds when $x$ is even.

Divisibility by $4$ : We need $2^x + 2 \equiv 0 \pmod{4}$ , i.e. $2^x \equiv 2 \pmod{4}$ . This holds only for $x = 1$ (since $2^1 = 2$ ; for $x \ge 2$ , $2^x$ is a multiple of $4$ ).

So the condition is satisfied when $x$ is even or $x = 1$ .

From $\{0,1,2,3,4,5,6,7\}$ :

Even values: $\{0, 2, 4, 6\}$ — four values.
Plus $x = 1$ — one more value.
Total: $4 + 1 = 5$ .

This confirms the answer without a full table. The modular approach is faster under exam conditions once you spot the $(-1)^x$ pattern for divisibility by $3$ .

💡 Closing insight: The divisibility-by- $3$ condition reduces to " $x$ is even" because $2$ and $3$ are coprime — so powers of $2$ cycle through all non-zero residues mod $3$ with period $2$ . This kind of cycling is the backbone of most modular-arithmetic shortcuts.

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Q14Time & WorkMCQ

When they work alone, B needs 25% more time to finish a job than A does. They two finish the job in 13 days in the following manner: A works alone till half the job is done, then A and B work together for four days, and finally B works alone to complete the remaining 5% of the job. In how many days can B alone finish the entire job?

Show solution

Correct answer: A

Think of this as: we know exactly how the total job is split among three phases, so we can write a "work = 1" equation and solve for A's solo time — then convert to B's.

Step 1 — Translate "25% more time" into rates.

Let A alone finish the job in $a$ days. Since B needs 25% more time than A:

$\text{B's solo time} = 1.25\,a = \frac{5a}{4} \text{ days.}$

So their daily work rates are:

A's rate $= \dfrac{1}{a}$ of the job per day,
B's rate $= \dfrac{1}{1.25\,a} = \dfrac{0.8}{a} = \dfrac{4}{5a}$ of the job per day.

💡 Teacher tip: "25% more time" means multiply the time by $1.25$ . The rate goes the other way — divide by $1.25$ , i.e. multiply by $0.8$ . Never add 25% to the rate.

Step 2 — Write the work equation for the three phases.

The entire job equals 1, and it breaks into three pieces:

Phase	Work done
A alone until half the job is done	$0.50$
A and B together for 4 days	$4\!\left(\dfrac{1}{a}+\dfrac{0.8}{a}\right) = \dfrac{4 \times 1.8}{a} = \dfrac{7.2}{a}$
B alone finishes the last 5%	$0.05$

Adding them up and setting the total equal to 1:

$0.5 + \frac{7.2}{a} + 0.05 = 1$

$\frac{7.2}{a} = 1 - 0.55 = 0.45$

$a = \frac{7.2}{0.45} = 16 \text{ days.}$

So A alone finishes the job in 16 days.

Step 3 — Find B's solo time.

$\text{B's solo time} = 1.25 \times a = 1.25 \times 16 = 20 \text{ days.}$

Step 4 — Verify against the 13-day total (consistency check).

Phase	Time taken
A does 50% at rate $1/16$ per day	$0.5 \times 16 = 8$ days
A and B together	$4$ days
B does 5% at rate $1/20$ per day	$0.05 \times 20 = 1$ day
Total	$8 + 4 + 1 = 13$ days ✓

This matches the 13 days stated in the question, confirming our answer.

Answer: A — $\boxed{20 \text{ days}}$

💡 Why this works cleanly: the three phases give us the work split directly (50% + 4 days of joint work + 5%), so we never need the 13-day total to solve — it simply falls out as a built-in check. Whenever a problem hands you both a work breakdown and a total time, use the breakdown to solve and the total to verify.

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Q15Polygons & QuadrilateralsMCQ

Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?

Show solution

Correct answer: A

Think of this as: a rectangle drawn inside a circle always has its diagonal passing through the center, so the diagonal is fixed by the circle's radius. Once the diagonal is fixed, Pythagoras decides which side-pairs are allowed.

Step 1 — Use the rectangle-in-a-circle property.

A rectangle has four right angles. When a rectangle is inscribed in a circle, each of its vertices lies on the circle, so each side of the rectangle is a chord subtending a $90^\circ$ angle at the opposite vertex.

By Thales' theorem, an angle inscribed in a semicircle is a right angle. Conversely, a right angle inscribed in a circle must subtend a diameter. Therefore, the diagonal of the rectangle is a diameter of the circle.

Given the radius $r = 13$ cm, the diameter is: $d = 2r = 2 \times 13 = 26 \text{ cm}.$

Step 2 — Translate into a Pythagoras condition.

For a rectangle with length $l$ and breadth $b$ , the diagonal satisfies: $l^2 + b^2 = d^2 = 26^2 = 676.$

So we need the pair $(l, b)$ whose squares add up to $676$ .

Step 3 — Test each option.

Option	Sides $(l, b)$	$l^2 + b^2$	Equals $676$ ?
A	$24,\ 10$	$576 + 100 = 676$	✅
B	$25,\ 9$	$625 + 81 = 706$	✗
C	$25,\ 10$	$625 + 100 = 725$	✗
D	$24,\ 12$	$576 + 144 = 720$	✗

Only option A satisfies $l^2 + b^2 = 676$ .

💡 Teacher tip: The phrase "rectangle inscribed in a circle" is a standard signal to immediately convert the problem into "diagonal = diameter". From there, it is no longer a geometry problem — it is just a Pythagoras check on the options.

Answer: A — $\boxed{24,\ 10}$

💡 Why this works so cleanly: $24$ and $10$ are part of the scaled Pythagorean triple $(5, 12, 13)$ — multiplied by $2$ they become $(10, 24, 26)$ , which fits the diameter of $26$ exactly.

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Q16Permutation & CombinationTITA

How many numbers with two or more digits can be formed with the digits 1,2,3,4,5,6,7,8,9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?

Your answer:

Show solution

Correct answer: 502

Think of this as: the digits must rise left-to-right, so once we choose which digits to use, their arrangement is completely forced. The real question is just — how many ways can we pick a set of two or more digits?

Step 1 — Recognise that order is forced.

We are forming numbers from the digits $\{1,2,3,4,5,6,7,8,9\}$ , each used at most once, with the digits appearing in ascending order.

Because the digits must rise from left to right, there is exactly one way to arrange any chosen set of digits. For example:

If we choose $\{3, 7, 2\}$ , the only valid number is $237$ .
If we choose $\{5, 1, 9, 4\}$ , the only valid number is $1459$ .

So the number of valid numbers equals the number of ways to choose a subset of digits of size $\ge 2$ .

💡 Teacher tip: A "fixed order" constraint is a gift — it collapses an ordering/permutation problem into a plain combination (subset) problem. Whenever an arrangement is forced by a rule, count the choices, not the arrangements.

Step 2 — Count subsets of size 2 through 9.

The total number of subsets of a 9-element set is $2^9 = 512.$

From this total, we must remove the subsets that are too small to form a two-or-more-digit number:

The empty subset: $\binom{9}{0} = 1$ way.
Subsets of exactly one digit: $\binom{9}{1} = 9$ ways.

So the count of valid subsets is $2^9 - \binom{9}{0} - \binom{9}{1} = 512 - 1 - 9 = 502.$

Equivalently, summing directly: $\sum_{k=2}^{9} \binom{9}{k} = \binom{9}{2} + \binom{9}{3} + \cdots + \binom{9}{9} = 502.$

Step 3 — Quick sanity check on the boundaries.

Subset size	Number of such subsets	Forms a valid number?
0	$\binom{9}{0}=1$	✗ (no digits)
1	$\binom{9}{1}=9$	✗ (only one digit)
2 to 9	$512 - 1 - 9 = 502$	✓

Everything checks out.

Answer: $\boxed{502}$

💡 Why this is elegant: the ascending-order rule removes all the usual permutation complexity. What looks like a counting-of-arrangements problem is secretly just a counting-of-subsets problem — and the answer is the full power set minus the two trivial cases.

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Q17Time & WorkMCQ

Humans and robots can both perform a job but at different efficiencies. Fifteen humans and five robots working together take thirty days to finish the job, whereas five humans and fifteen robots working together take sixty days to finish it. How many days will fifteen humans working together (without any robot) take to finish it?

Show solution

Correct answer: C

Think of this as: two different human-robot teams finish the same job in two different times. From those two facts, we can solve for one human's daily work rate — and then find how long 15 humans alone need.

Step 1 — Define rates and write the work equations.

Let one human complete $h$ of the job per day, and one robot complete $r$ of the job per day (so the whole job $= 1$ ).

Using Work $=$ Rate $\times$ Time, the combined rate of a team equals (work done) $\div$ (time taken):

Team 1: 15 humans + 5 robots finish in 30 days, so their combined daily rate is $\frac{1}{30}$ : $15h + 5r = \frac{1}{30}$
Team 2: 5 humans + 15 robots finish in 60 days, so their combined daily rate is $\frac{1}{60}$ : $5h + 15r = \frac{1}{60}$

Step 2 — Simplify both equations.

Divide the first equation by 5 and the second by 5: $3h + r = \frac{1}{150}, \qquad h + 3r = \frac{1}{300}$

💡 Teacher tip: In a two-unknown work problem, divide each equation by its common factor first. The numbers shrink dramatically before you ever start eliminating — this is what keeps the arithmetic manageable.

Step 3 — Eliminate $r$ and solve for $h$ .

Multiply the first simplified equation by 3 so the $r$ -coefficients match: $9h + 3r = \frac{3}{150} = \frac{1}{50}$

Now subtract the second simplified equation ( $h + 3r = \frac{1}{300}$ ): $9h + 3r - (h + 3r) = \frac{1}{50} - \frac{1}{300}$ $8h = \frac{6 - 1}{300} = \frac{5}{300} = \frac{1}{60}$ $h = \frac{1}{60 \times 8} = \frac{1}{480}$

So one human does $\frac{1}{480}$ of the job per day.

Step 4 — Find the time for 15 humans alone.

Combined rate of 15 humans $= 15h = \frac{15}{480} = \frac{1}{32}$ of the job per day.

Time $= \frac{1}{\text{rate}} = \frac{1}{1/32} = 32$ days.

Back-substitution check.

From $3h + r = \frac{1}{150}$ : $r = \frac{1}{150} - \frac{3}{480} = \frac{1}{150} - \frac{1}{160} = \frac{160 - 150}{24000} = \frac{1}{2400}$

Verify with Team 2's equation $5h + 15r = \frac{1}{60}$ : $5\left(\frac{1}{480}\right) + 15\left(\frac{1}{2400}\right) = \frac{1}{96} + \frac{1}{160} = \frac{5 + 3}{480} = \frac{8}{480} = \frac{1}{60} \checkmark$

The rates are consistent with both given conditions.

Answer: C — $\boxed{32\text{ days}}$

💡 Why this works so cleanly: the two teams are mirror images (15 humans + 5 robots vs. 5 humans + 15 robots). That symmetry is what lets elimination take out the robot rate in one move and isolate the human rate — no quadratic, no trial and error.

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Q18Time & WorkTITA

A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?

Your answer:

Show solution

Correct answer: 10

Think of this as: two unknown pipe rates, two scenarios that pin them down, and one new combination to evaluate.

Step 1 — Translate each scenario into an equation.

Let each filling pipe fill at rate $f$ tanks/hour, and each draining pipe drain at rate $d$ tanks/hour. The net fill rate is (number of filling pipes) $\times f -$ (number of draining pipes) $\times d$ .

Scenario 1: 6 filling + 5 draining fills the tank in 6 hours, so the net rate is $\frac{1}{6}$ tank/hour: $6f - 5d = \frac{1}{6}$
Scenario 2: 5 filling + 6 draining fills the tank in 60 hours, so the net rate is $\frac{1}{60}$ tank/hour: $5f - 6d = \frac{1}{60}$

Step 2 — Add and subtract to isolate $f+d$ and $f-d$ .

Adding the two equations: $11f - 11d = \frac{1}{6} + \frac{1}{60} = \frac{10}{60} + \frac{1}{60} = \frac{11}{60}$ $\Rightarrow f - d = \frac{1}{60}$

Subtracting the second equation from the first: $(6f - 5d) - (5f - 6d) = \frac{1}{6} - \frac{1}{60}$ $f + d = \frac{10}{60} - \frac{1}{60} = \frac{9}{60} = \frac{3}{20}$

💡 Teacher tip: Adding and subtracting the two equations is far cleaner than standard elimination here. The symmetric coefficients $(6, 5)$ and $(5, 6)$ mean the sum gives $11(f-d)$ and the difference gives $f+d$ — two simple expressions instead of messy fractions.

Step 3 — Solve for $f$ and $d$ individually.

From $f - d = \frac{1}{60}$ and $f + d = \frac{3}{20}$ :

$2f = (f - d) + (f + d) = \frac{1}{60} + \frac{3}{20} = \frac{1}{60} + \frac{9}{60} = \frac{10}{60} = \frac{1}{6}$ $\Rightarrow f = \frac{1}{12}$

$2d = (f + d) - (f - d) = \frac{3}{20} - \frac{1}{60} = \frac{9}{60} - \frac{1}{60} = \frac{8}{60} = \frac{2}{15}$ $\Rightarrow d = \frac{1}{15}$

So one filling pipe fills $\frac{1}{12}$ of the tank per hour (12 hours solo), and one draining pipe empties $\frac{1}{15}$ of the tank per hour (15 hours solo).

Step 4 — Verify against the original scenarios.

Scenario	Net rate with $f = \frac{1}{12},\; d = \frac{1}{15}$	Given rate	Match?
6 filling, 5 draining	$6 \cdot \frac{1}{12} - 5 \cdot \frac{1}{15} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$	$\frac{1}{6}$	✅
5 filling, 6 draining	$5 \cdot \frac{1}{12} - 6 \cdot \frac{1}{15} = \frac{5}{12} - \frac{2}{5} = \frac{25}{60} - \frac{24}{60} = \frac{1}{60}$	$\frac{1}{60}$	✅

Both check out.

Step 5 — Compute the target: 2 filling + 1 draining.

$\text{Net rate} = 2f - d = 2 \cdot \frac{1}{12} - \frac{1}{15} = \frac{1}{6} - \frac{1}{15}$

$= \frac{5}{30} - \frac{2}{30} = \frac{3}{30} = \frac{1}{10} \text{ tank/hour}$

A net rate of $\frac{1}{10}$ tank/hour means the tank fills in:

$\frac{1}{\frac{1}{10}} = 10 \text{ hours}$

Answer: $\boxed{10 \text{ hours}}$

💡 Why the times differ so dramatically: A single filling pipe fills in 12 hours and a single drain empties in 15 hours — they are close in strength. So when many pipes run together, the net rate is small and sensitive: swapping one filling pipe for a drain (6F/5D → 5F/6D) cuts the net rate tenfold, from $\frac{1}{6}$ to $\frac{1}{60}$ . That is why the fill time jumps from 6 hours to 60 hours.

Alternative approach — Linear combination (skip solving for $f$ and $d$ ).

Instead of finding $f$ and $d$ separately, express the target $2f - d$ directly as a combination of the two given equations. Suppose:

$2f - d = a(6f - 5d) + b(5f - 6d)$

Matching coefficients of $f$ and $d$ : $6a + 5b = 2, \qquad 5a + 6b = 1$

Solving: multiply the first by 6 and the second by 5, then subtract: $36a + 30b = 12, \qquad 25a + 30b = 5$ $11a = 7 \;\Rightarrow\; a = \frac{7}{11}, \qquad b = -\frac{4}{11}$

Now substitute the known net rates:

$2f - d = \frac{7}{11} \cdot \frac{1}{6} + \left(-\frac{4}{11}\right) \cdot \frac{1}{60} = \frac{7}{66} - \frac{4}{660} = \frac{70}{660} - \frac{4}{660} = \frac{66}{660} = \frac{1}{10}$

Same answer: 10 hours. This method is elegant but the fraction arithmetic is trickier; the add/subtract method in Steps 2–3 is faster under exam conditions.

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Q19Polygons & QuadrilateralsMCQ

In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is

Show solution

Correct answer: A

Think of this as: we know the parallelogram's area and two side lengths, so we can find its height — and that height becomes one leg of a right triangle whose hypotenuse is the other side.

Step 1 — Find the height of the parallelogram using CD as the base.

The area of a parallelogram is $\text{base} \times \text{height}$ . Taking $CD = 9$ cm as the base:

$9 \times h = 72 \quad\Rightarrow\quad h = 8\text{ cm}.$

Since $AB \parallel CD$ and $A$ lies on $AB$ , the perpendicular dropped from $A$ to $CD$ is exactly this height. Therefore:

$AP = 8\text{ cm}.$

Step 2 — Use the right triangle $APD$ to find $PD$ .

Because $AP \perp CD$ and both $P$ and $D$ lie on $CD$ , the angle $\angle APD = 90^\circ$ . So $\triangle APD$ is a right triangle with:

$AP = 8$ (one leg),
$AD = 16$ (hypotenuse),
$PD =?$ (the other leg).

Applying Pythagoras:

$AP^2 + PD^2 = AD^2$ $8^2 + PD^2 = 16^2$ $64 + PD^2 = 256$ $PD^2 = 192 = 64 \times 3$ $PD = 8\sqrt{3}\text{ cm}.$

💡 Teacher tip: In a parallelogram, the height from a vertex to the opposite side is simultaneously a leg of a right triangle whose hypotenuse is the adjacent side. That dual role is what connects the area formula to Pythagoras.

Step 3 — Compute the area of $\triangle APD$ .

With $\angle APD = 90^\circ$ , we can use $AP$ and $PD$ as the two legs:

$\text{Area of } \triangle APD = \tfrac{1}{2} \times AP \times PD = \tfrac{1}{2} \times 8 \times 8\sqrt{3} = 32\sqrt{3}\text{ sq cm}.$

Answer: A — $\boxed{32\sqrt{3}\text{ sq cm}}$

Alternative approach — via the 30-60-90 triangle.

From Step 1, $AP = 8$ and $AD = 16$ . In right triangle $APD$ :

$\sin(\angle ADP) = \frac{AP}{AD} = \frac{8}{16} = \frac{1}{2} \quad\Rightarrow\quad \angle ADP = 30^\circ.$

So $\triangle APD$ is a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle. The side opposite $30^\circ$ is $AP = 8$ , so the side opposite $60^\circ$ is $PD = 8\sqrt{3}$ , and the area is again $\frac{1}{2} \times 8 \times 8\sqrt{3} = 32\sqrt{3}$ . This route is a quick cross-check once you spot the $1:2$ ratio between $AP$ and $AD$ .

💡 Closing insight: The key move was recognising that the parallelogram's height (found from the area) becomes a leg of the right triangle $APD$ . Whenever a parallelogram problem gives you an area and a side, compute the height first — it usually unlocks everything else.

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Q20Time Speed & DistanceMCQ

The distance from A to B is 60 km. Partha and Narayan start from A at the same time and move towards B. Partha takes four hours more than Narayan to reach B. Moreover, Partha reaches the mid-point of A and B two hours before Narayan reaches B. The speed of Partha, in km per hour, is

Show solution

Correct answer: D

Think of this as: two walkers leave A together for B (60 km away). We're given two facts about their arrival times, and we need Partha's speed.

Step 1 — Translate each fact into a time equation.

Let Partha's speed be $p$ km/h and Narayan's speed be $q$ km/h. The midpoint of AB is at $30$ km.

Fact 1: Partha takes 4 hours more than Narayan to cover the full 60 km. $\frac{60}{p} - \frac{60}{q} = 4$
Fact 2: Partha reaches the midpoint (30 km) two hours before Narayan reaches B (60 km). So Narayan's full time minus Partha's midpoint time equals 2. $\frac{60}{q} - \frac{30}{p} = 2$

💡 Teacher tip: "Reaches the midpoint $X$ hours before the other reaches the end" means (other's full time) $-$ (my midpoint time) $= X$ . Keep both times measured on the same clock from the common start.

Step 2 — Eliminate Narayan's time.

From Fact 2: $\frac{60}{q} = \frac{30}{p} + 2$

Substitute this into Fact 1: $\frac{60}{p} - \left(\frac{30}{p} + 2\right) = 4$ $\frac{30}{p} - 2 = 4$ $\frac{30}{p} = 6$ $p = 5$

Step 3 — Verify both conditions hold.

With $p = 5$ : Partha covers 60 km in $60/5 = 12$ h, and the midpoint in $30/5 = 6$ h.

From Fact 2: $\frac{60}{q} = 6 + 2 = 8$ , so $q = 7.5$ km/h, and Narayan takes $\frac{60}{7.5} = 8$ h.

Condition	Partha	Narayan	Difference	Satisfied?
Full 60 km time	12 h	8 h	4 h	✅
Midpoint vs full	6 h (midpoint)	8 h (full)	2 h	✅

Both conditions check out.

Answer: D — $\boxed{5 \text{ km/h}}$

Alternative approach — work with times directly.

Let Narayan take $t$ hours for the full 60 km. Then Partha takes $t + 4$ hours (Fact 1).

Partha's speed is $\frac{60}{t+4}$ , so her midpoint time is: $\frac{30}{60/(t+4)} = \frac{t+4}{2} \text{ hours}$

Fact 2 says this midpoint time is 2 hours less than Narayan's full time $t$ : $t - \frac{t+4}{2} = 2$ $\frac{2t - (t+4)}{2} = 2$ $\frac{t-4}{2} = 2$ $t = 8$

So Narayan takes 8 h and Partha takes $8 + 4 = 12$ h. Partha's speed: $p = \frac{60}{12} = 5 \text{ km/h}$

This route avoids solving two equations in two unknowns — handy under exam time.

💡 Closing insight: When every condition is about times rather than speeds, letting the unknown be a time (not a speed) often collapses the algebra. The speed falls out at the end by a single division.

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Q21Boats Trains & Relative SpeedTITA

Point P lies between points A and B such that the length of BP is thrice that of AP. Car 1 starts from A and moves towards B. Simultaneously, car 2 starts from B and moves towards A. Car 2 reaches P one hour after car 1 reaches P. If the speed of car 2 is half that of car 1, then the time, in minutes, taken by car 1 in reaching P from A is

Your answer:

Show solution

Correct answer: 12

Think of this as: two cars start at the same instant from opposite ends and drive toward each other's starting point. We know how their distances to the meeting point P compare, and how their speeds compare. We're told who reaches P later and by how much — and we need the earlier car's travel time to P.

Step 1 — Assign variables to distances and speeds.

Since $BP = 3 \times AP$ , let's write

$AP = d, \qquad BP = 3d.$

Let car 1's speed be $v$ . Since car 2's speed is half of car 1's,

$\text{speed of car 1} = v, \qquad \text{speed of car 2} = \frac{v}{2}.$

Step 2 — Write each car's time to reach P.

Car 1 travels from $A$ to $P$ , a distance of $d$ , at speed $v$ :

$t_1 = \frac{AP}{v} = \frac{d}{v}.$

Car 2 travels from $B$ to $P$ , a distance of $3d$ , at speed $\frac{v}{2}$ :

$t_2 = \frac{BP}{v/2} = \frac{3d}{v/2} = 3d \times \frac{2}{v} = \frac{6d}{v} = 6 \cdot \frac{d}{v} = 6t_1.$

So car 2 takes six times as long as car 1 to reach P.

💡 Teacher tip: When both a distance ratio and a speed ratio are given, the time ratio falls out immediately: $\text{time} = \frac{\text{distance}}{\text{speed}}$ . Car 2 travels $3\times$ the distance and moves at half the speed, so its time is $3 \times 2 = 6$ times car 1's time.

Step 3 — Use the "one hour later" condition.

Both cars start simultaneously. Car 2 reaches P one hour after car 1 reaches P, so

$t_2 = t_1 + 1.$

Substituting $t_2 = 6t_1$ :

$6t_1 = t_1 + 1 \implies 5t_1 = 1 \implies t_1 = \frac{1}{5}\text{ hour}.$

Step 4 — Convert to minutes.

$t_1 = \frac{1}{5}\text{ hr} \times 60\text{ min/hr} = 12\text{ minutes}.$

Quick check. If $t_1 = 12$ min $= \frac{1}{5}$ hr, then $t_2 = 6t_1 = \frac{6}{5}$ hr $= 72$ min. The gap is $72 - 12 = 60$ min $= 1$ hour. ✅ This matches the given condition.

Answer: $\boxed{12\text{ minutes}}$

Alternative approach — pure ratios.

Distance ratio $AP: BP = 1: 3$ . Speed ratio $v_1: v_2 = 2: 1$ . Since time $= \frac{\text{distance}}{\text{speed}}$ ,

$t_1: t_2 = \frac{1}{2}: \frac{3}{1} = \frac{1}{2}: 3 = 1: 6.$

So $t_2 - t_1 = 5$ parts $= 1$ hour, giving $1$ part $= \frac{1}{5}$ hr $= 12$ min. This is the same algebra, just packaged as a ratio — faster under exam pressure once you spot it.

💡 Why this is clean: the distance ratio ( $1:3$ ) and the speed ratio ( $2:1$ ) combine into a single time ratio ( $1:6$ ), and the one-hour gap then pins down each part exactly. No need to ever find $d$ or $v$ individually.

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Q22LogarithmsMCQ

If log₁₂ 81 = p, then 3((4 − p)/(4 + p)) is equal to

Show solution

Correct answer: D

What we want: Simplify $3\!\left(\dfrac{4-p}{4+p}\right)$ where $p = \log_{12}81$ , and match it to one of the four options.

Step 1 — Express $p$ using a power. Since $81 = 3^4$ : $p = \log_{12}81 = \log_{12}3^4 = 4\log_{12}3$

Step 2 — Simplify the numerator $4 - p$ . $4 - p = 4 - 4\log_{12}3 = 4\bigl(1 - \log_{12}3\bigr)$

Now use the fact that $1 = \log_{12}12$ (since $12^1 = 12$ ): $= 4\bigl(\log_{12}12 - \log_{12}3\bigr) = 4\log_{12}\!\left(\frac{12}{3}\right) = 4\log_{12}4$

Step 3 — Simplify the denominator $4 + p$ . $4 + p = 4 + 4\log_{12}3 = 4\bigl(1 + \log_{12}3\bigr)$

Again using $1 = \log_{12}12$ : $= 4\bigl(\log_{12}12 + \log_{12}3\bigr) = 4\log_{12}(12 \times 3) = 4\log_{12}36$

Step 4 — Form the fraction. $\frac{4-p}{4+p} = \frac{4\log_{12}4}{4\log_{12}36} = \frac{\log_{12}4}{\log_{12}36}$

Step 5 — Multiply by 3 and simplify using powers. Note that $4 = 2^2$ and $36 = 6^2$ : $3 \cdot \frac{\log_{12}4}{\log_{12}36} = \frac{3 \cdot \log_{12}2^2}{\log_{12}6^2} = \frac{3 \cdot 2\log_{12}2}{2\log_{12}6} = \frac{3\log_{12}2}{\log_{12}6}$

Step 6 — Apply change of base. The ratio $\dfrac{\log_{12}2}{\log_{12}6}$ is exactly $\log_6 2$ by the change-of-base formula: $\frac{\log_{12}2}{\log_{12}6} = \log_6 2$

Therefore: $3 \cdot \frac{4-p}{4+p} = 3\log_6 2 = \log_6 2^3 = \log_6 8$

💡 Teacher tip: The change-of-base identity $\dfrac{\log_a x}{\log_a y} = \log_y x$ is the key move that collapses a ratio of same-base logs into a single logarithm. Recognising it saves several lines of working.

Answer: D — $\boxed{\log_6 8}$

Alternative approach. Instead of breaking $4$ and $36$ into $2^2$ and $6^2$ , we can push the factor of 3 into the logarithm first: $3 \cdot \frac{\log_{12}4}{\log_{12}36} = \frac{\log_{12}4^3}{\log_{12}36} = \frac{\log_{12}64}{\log_{12}36}$

Now write $64 = 8^2$ and $36 = 6^2$ : $= \frac{\log_{12}8^2}{\log_{12}6^2} = \frac{2\log_{12}8}{2\log_{12}6} = \frac{\log_{12}8}{\log_{12}6} = \log_6 8$

Both routes arrive at the same answer; the first is slightly faster since it avoids dealing with $64$ .

💡 Why this works: The entire expression is built from $\log_{12}3$ , and the numbers $4$ , $12$ , $3$ , $36$ are all powers or products of $2$ , $3$ , and $6$ . The change-of-base formula lets us "switch" the base from $12$ to $6$ , which is exactly what we need to land on $\log_6 8$ .

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Q23PercentagesMCQ

In an examination, the maximum possible score is N while the pass mark is 45% of N. A candidate obtains 36 marks, but falls short of the pass mark by 68%. Which one of the following is then correct?

Show solution

Correct answer: B

Think of this as: we're given a candidate's score and told how far below the pass mark it lands — we need to work backwards through two percentages to find the maximum score $N$ .

Step 1 — Decode "falls short of the pass mark by $68\%$ ".

The candidate scored $36$ marks. Saying this score "falls short of the pass mark by $68\%$ " means $36$ is $68\%$ below the pass mark. So the candidate's score is only the remaining $32\%$ of the pass mark:

$36 = (100\% - 68\%) \times \text{pass mark} = 0.32 \times \text{pass mark}.$

Solving for the pass mark:

$\text{pass mark} = \frac{36}{0.32} = 112.5.$

💡 Teacher tip: "Falls short of $X$ by $p\%$ " means the score equals $(100-p)\%$ of $X$ — the percentage is anchored on the target (the pass mark), not on the candidate's own score. A common mistake is to take $68\%$ of $36$ ; that reverses the reference.

Step 2 — Convert the pass mark to $N$ .

The pass mark is $45\%$ of the maximum score $N$ :

$0.45\,N = 112.5.$

Solving:

$N = \frac{112.5}{0.45} = 250.$

Step 3 — Match against the options.

Option	Range	Does $N=250$ fit?
A	$N \le 200$	✗
B	$243 \le N \le 252$	✓
C	$201 \le N \le 242$	✗
D	$N \ge 253$	✗

$250$ lies in the interval $[243, 252]$ .

Answer: B — $\boxed{243 \le N \le 252}$

💡 Why this works cleanly: the two percentages chain together — $36 \xrightarrow{\div 0.32}$ pass mark $\xrightarrow{\div 0.45}$ $N$ . Each percentage is anchored on the next quantity in the chain, not on $36$ .

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Q24Ratio & ProportionMCQ

Raju and Lalitha originally had marbles in the ratio 4:9. Then Lalitha gave some of her marbles to Raju. As a result, the ratio of the number of marbles with Raju to that with Lalitha became 5:6. What fraction of her original number of marbles was given by Lalitha to Raju?

Show solution

Correct answer: D

Think of this as: two people start in a known ratio, some marbles move from one to the other, and we land in a new ratio. We need the transferred amount as a fraction of Lalitha's original count.

Step 1 — Anchor on a common multiplier.

Since Raju and Lalitha start in the ratio $4:9$ , write their original counts as:

Raju $= 4k$
Lalitha $= 9k$

Let Lalitha give $x$ marbles to Raju.

💡 Teacher tip: Introducing a single multiplier $k$ is the cleanest way to handle ratios when a transfer happens. The $k$ will cancel at the end when we form a fraction, so its actual value never matters.

Step 2 — Write the new counts and use the new ratio.

After the transfer:

Raju has $4k + x$
Lalitha has $9k - x$

The new ratio is $5:6$ , so: $\frac{4k + x}{9k - x} = \frac{5}{6}$

Cross-multiply: $6(4k + x) = 5(9k - x)$ $24k + 6x = 45k - 5x$

Step 3 — Solve for $x$ in terms of $k$ .

Collect the $x$ -terms on one side and the $k$ -terms on the other: $6x + 5x = 45k - 24k$ $11x = 21k$ $x = \frac{21k}{11}$

So Lalitha gave away $\dfrac{21k}{11}$ marbles.

Step 4 — Form the required fraction.

The question asks for the fraction of Lalitha's original marbles that she gave away: $\frac{x}{\text{Lalitha's original}} = \frac{\,21k/11\,}{9k} = \frac{21k}{11 \times 9k} = \frac{21}{99} = \frac{7}{33}$

Notice how $k$ cancels cleanly — confirming that the answer is a fixed fraction regardless of how many marbles they actually started with.

Quick sanity check. Lalitha gave away $\tfrac{7}{33}$ of her marbles, which is roughly $21\%$ — a reasonable, less-than-half transfer. The new ratio $5:6$ is also close to balanced, consistent with a modest transfer from the person who had more.

Answer: D — $\boxed{\dfrac{7}{33}}$

💡 Closing insight: Whenever a ratio problem involves a transfer and asks for a fraction of an original amount, anchor both people on a common multiplier $k$ . The transfer amount comes out proportional to $k$ , and the final fraction is independent of $k$ — so you never need the actual counts.

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Q25Polygons & QuadrilateralsMCQ

Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD and CG is longer than EB, then the ratio of length of EB to that of CG is

Show solution

Correct answer: D

Think of this as: a smaller square is tucked inside a bigger square, touching each side once. The four corner triangles that get "cut off" must all be identical — and that symmetry is what lets us solve the whole problem with one variable ratio.

Step 1 — Why the four corner triangles are congruent.

Since $EFGH$ is a square, every interior angle is $90°$ . Look at vertex $E$ on side $AB$ : the angles $\angle AEH$ , $\angle HEF$ , and $\angle FEB$ lie on a straight line, so they sum to $180°$ . With $\angle HEF = 90°$ , we get $\angle AEH + \angle FEB = 90°$ .

But in right triangle $AEH$ , we have $\angle AEH + \angle AHE = 90°$ . Comparing: $\angle FEB = \angle AHE$ . By the same logic at every corner, all four corner triangles ( $\triangle AEH$ , $\triangle BFE$ , $\triangle CGF$ , $\triangle DHG$ ) are similar — and since the sides of the inner square are equal ( $EF = FG = GH = HE$ ), they are congruent.

Step 2 — Label the corner cuts.

Let the two legs of each congruent corner triangle be $a$ and $b$ , going around consistently:

$AE = BF = CG = DH = a, \qquad EB = FC = GD = HA = b.$

Each side of $ABCD$ then has length $s = a + b$ .

Step 3 — Express the area of the inner square.

Each side of $EFGH$ is the hypotenuse of a right triangle with legs $a$ and $b$ :

$\text{side of } EFGH = \sqrt{a^2 + b^2}.$

So the area of the inner square is:

$\text{Area}(EFGH) = a^2 + b^2.$

Step 4 — Apply the area condition.

We're told $\text{Area}(EFGH) = 62.5\% = \tfrac{5}{8}$ of $\text{Area}(ABCD) = (a+b)^2$ :

$\frac{a^2 + b^2}{(a+b)^2} = \frac{5}{8}.$

Cross-multiply and expand:

$8(a^2 + b^2) = 5(a^2 + 2ab + b^2)$ $8a^2 + 8b^2 = 5a^2 + 10ab + 5b^2$ $3a^2 - 10ab + 3b^2 = 0.$

Step 5 — Solve for the ratio $a: b$ .

Divide through by $b^2$ and set $t = a/b$ :

$3t^2 - 10t + 3 = 0.$

Using the quadratic formula:

$t = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm 8}{6}.$

So $t = 3$ or $t = \tfrac{1}{3}$ .

💡 Teacher tip: Both roots are geometrically valid — they simply swap which leg ( $a$ or $b$ ) is the longer one. The problem's inequality is what picks the correct root.

Step 6 — Select the right root using $CG > EB$ .

From our labeling: $CG = a$ and $EB = b$ . The condition $CG > EB$ means $a > b$ , so $\frac{a}{b} = 3$ .

Therefore:

$EB: CG = b: a = 1: 3.$

Verification check:

Root	$a/b$	$a > b$ ?	$CG > EB$ ?	Use?
$t = 3$	$3$	✅	✅	✅
$t = \tfrac{1}{3}$	$\tfrac{1}{3}$	❌	❌	❌

Answer: D — $\boxed{1: 3}$

💡 Why this is elegant: The quadratic yields two reciprocal roots ( $3$ and $\tfrac{1}{3}$ ), reflecting a rotational symmetry — swapping the roles of $a$ and $b$ gives an equally valid inscribed square. The inequality $CG > EB$ is the only thing that breaks the symmetry and pins down a unique answer.

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Q26Surds & IndicesMCQ

Given that $x^{2018}$ $y^{2017}$ = 1/2, and $x^{2016}$ $y^{2019}$ = 8, then value of x² + y³ is

Show solution

Correct answer: D

Think of this as: two equations with enormous exponents in two unknowns. The exponents look scary, but if we divide the equations, most of those powers cancel — leaving a simple ratio between $x$ and $y$ .

What we want: the value of $x^2 + y^3$ .

Step 1 — Divide the two equations to eliminate the common powers.

Write the two given equations: $x^{2018}\,y^{2017} = \tfrac12 \qquad\text{(1)}$ $x^{2016}\,y^{2019} = 8 \qquad\text{(2)}$

Divide (1) by (2): $\frac{x^{2018}\,y^{2017}}{x^{2016}\,y^{2019}} = \frac{\,1/2\,}{8}$

Subtract exponents on the left: $x^{2018-2016}\,y^{2017-2019} = \frac{1}{16} \;\Rightarrow\; x^{2}\,y^{-2} = \frac{1}{16} \;\Rightarrow\; \left(\frac{x}{y}\right)^{2} = \frac{1}{16}.$

So $\dfrac{x}{y} = \pm\dfrac14$ , i.e. $x = \pm\dfrac{y}{4}$ .

💡 Teacher tip: When two equations share the same variables with huge exponents, dividing them is almost always the key move — the exponents subtract, and the giant powers collapse to a small ratio.

Step 2 — Substitute back to find $y$ , then $x$ .

Take $x = \dfrac{y}{4}$ (we will check the sign at the end; since we ultimately need $x^2$ , the sign of $x$ will not matter). Substitute into equation (1):

$\left(\frac{y}{4}\right)^{2018} y^{2017} = \frac12.$

Expand $\left(\frac{y}{4}\right)^{2018} = \dfrac{y^{2018}}{4^{2018}}$ :

$\frac{y^{2018}\cdot y^{2017}}{4^{2018}} = \frac12 \;\Rightarrow\; \frac{y^{4035}}{4^{2018}} = \frac12.$

Now solve for $y^{4035}$ :

$y^{4035} = \frac12 \cdot 4^{2018}.$

Express everything as powers of $2$ . Since $4^{2018} = (2^2)^{2018} = 2^{4036}$ :

$y^{4035} = \frac{1}{2}\cdot 2^{4036} = 2^{4035}.$

Since $4035$ is odd, the equation $y^{4035} = 2^{4035}$ has the unique real solution: $y = 2.$

Then $x = \dfrac{y}{4} = \dfrac{2}{4} = \dfrac12$ .

Step 3 — Verify the sign doesn't matter.

If we had taken $x = -\dfrac{y}{4}$ , then $x^{2018} = \left(-\dfrac{y}{4}\right)^{2018} = \left(\dfrac{y}{4}\right)^{2018}$ because $2018$ is even. So equation (1) gives the same $y = 2$ , and $x = -\dfrac12$ . Either way, $x^2 = \dfrac14$ . The answer is unaffected.

Step 4 — Compute $x^2 + y^3$ .

$x^2 + y^3 = \left(\frac12\right)^2 + (2)^3 = \frac14 + 8 = \frac14 + \frac{32}{4} = \frac{33}{4}.$

Quick check with the second equation. With $x = \frac12,\; y = 2$ : $x^{2016}\,y^{2019} = \left(\tfrac12\right)^{2016}\cdot 2^{2019} = 2^{-2016}\cdot 2^{2019} = 2^{3} = 8.\;\checkmark$ Both equations are satisfied.

Answer: D — $\boxed{\dfrac{33}{4}}$

💡 Why this works so cleanly: the exponents $2018, 2017, 2016, 2019$ were chosen so that dividing cancels $2016$ from $x$ and $2017$ from $y$ , leaving tiny powers $x^2$ and $y^2$ . The giant numbers are a smokescreen — only the base values $x$ and $y$ matter.

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Q27Set TheoryTITA

Each of 74 students in a class studies at least one of the three subjects H, E and P. Ten students study all three subjects, while twenty study H and E, but not P. Every student who studies P also studies H or E or both. If the number of students studying H equals that studying E, then the number of students studying H is

Your answer:

Show solution

Correct answer: 52

Think of this as a three-circle Venn diagram where one region is forced to be empty, and the key is that two of the circles must have the same total.

Step 1 — Identify the regions and what we're told.

A three-set Venn has seven regions. Let's label the ones we don't yet know:

Region	Count
$H$ only	$c$
$E$ only	$d$
$H \cap E$ only (not $P$ )	$20$
$H \cap P$ only (not $E$ )	$a$
$E \cap P$ only (not $H$ )	$b$
All three $H \cap E \cap P$	$10$
$P$ only	$0$

The last row is zero because the question says every student who studies $P$ also studies $H$ or $E$ or both — so nobody takes $P$ alone.

Step 2 — Use the grand total of 74 students.

Adding every region: $c + d + 20 + a + b + 10 + 0 = 74$ $c + d + a + b = 44.$

Step 3 — Translate " $\#H = \#E$ " into an equation.

The number studying $H$ is the sum of all regions inside the $H$ circle: $\#H = c + 20 + a + 10 = (c + a) + 30.$

Similarly, the number studying $E$ is: $\#E = d + 20 + b + 10 = (d + b) + 30.$

Since these are equal: $(c + a) + 30 = (d + b) + 30 \quad\Longrightarrow\quad c + a = d + b.$

💡 Teacher tip: When two category-counts are equal, the shared regions (the $20$ and the $10$ ) appear in both counts and cancel out. The equality then lives entirely in the "private" regions — here, $c+a$ versus $d+b$ .

Step 4 — Solve for $c + a$ .

From Step 2, $(c + a) + (d + b) = 44$ . From Step 3, $c + a = d + b$ . Substituting: $(c + a) + (c + a) = 44 \quad\Longrightarrow\quad 2(c + a) = 44 \quad\Longrightarrow\quad c + a = 22.$

Step 5 — Compute $\#H$ .

$\#H = (c + a) + 30 = 22 + 30 = 52.$

Quick check. With $c + a = d + b = 22$ , both $\#H$ and $\#E$ equal $52$ , and the total is $22 + 22 + 20 + 10 = 74$ . ✓ Everything is consistent.

Answer: $\boxed{52}$

💡 Why this works so cleanly: the shared regions ( $20 + 10 = 30$ ) contribute equally to both $\#H$ and $\#E$ , so they drop out of the comparison. The remaining private regions must split the leftover $44$ students evenly — $22$ each — giving $\#H = 22 + 30 = 52$ .

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Q28Polynomials & ProgressionsTITA

If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals

Your answer:

Show solution

Correct answer: 54

Framing. The relation $f(x+2) = f(x) + f(x+1)$ is the Fibonacci rule wearing a costume: each value is the sum of the two values just before it. We are handed two values four steps apart — $f(11)=91$ and $f(15)=617$ — and asked for the value sitting immediately before $f(11)$ .

Method 1 — March forward from the unknown

Step 1. Introduce a letter for what we want. Let $f(10) = a, \qquad f(11) = 91 \text{ (given)}.$

Now climb up the recurrence one rung at a time, keeping every term expressed in terms of $a$ .

Step 2. Apply $f(x+2) = f(x) + f(x+1)$ repeatedly.

$f(12) = f(10) + f(11) = a + 91$
$f(13) = f(11) + f(12) = 91 + (a + 91) = a + 182$
$f(14) = f(12) + f(13) = (a + 91) + (a + 182) = 2a + 273$
$f(15) = f(13) + f(14) = (a + 182) + (2a + 273) = 3a + 455$

Step 3. Pin down $a$ using the second given value. We are told $f(15) = 617$ , so $3a + 455 = 617 \implies 3a = 162 \implies a = 54.$

Step 4. Verify by listing the actual sequence.

$x$	$f(x)$	How it's built
10	54	(the unknown)
11	91	given
12	145	$54 + 91$ ✓
13	236	$91 + 145$ ✓
14	381	$145 + 236$ ✓
15	617	$236 + 381$ ✓

Every term is the sum of the two before it, and the final value matches the given $f(15)=617$ .

💡 Teacher tip: When a recurrence links two given values that are several steps apart, express each intermediate term as a linear combination of the unknown and the known value. The coefficients grow in a predictable Fibonacci-like pattern, so the algebra stays tidy and the equation practically solves itself.

Alternative approach — March backward from $f(15)$

Rewrite the recurrence as $f(x) = f(x+2) - f(x+1)$ : each term is the difference of the two terms after it. This lets us step downward.

Let $f(14) = b$ and descend toward the known $f(11)$ :

$f(13) = f(15) - f(14) = 617 - b$
$f(12) = f(14) - f(13) = b - (617 - b) = 2b - 617$
$f(11) = f(13) - f(12) = (617 - b) - (2b - 617) = 1234 - 3b$

Set this equal to the given $f(11) = 91$ : $1234 - 3b = 91 \implies 3b = 1143 \implies b = 381.$

So $f(14) = 381$ . One more step back gives the target: $f(10) = f(12) - f(11) = (2b - 617) - 91 = 2(381) - 708 = 762 - 708 = 54.$

Both directions agree. The forward method is marginally quicker here because $f(10)$ sits right next to the given $f(11)$ .

Answer: $\boxed{54}$

Closing insight. Notice from Method 1 that $f(15) = 3\,f(10) + 5\,f(11)$ — the coefficients $3$ and $5$ are themselves consecutive Fibonacci numbers. That is no coincidence: in any Fibonacci-style recurrence, a term $n$ steps ahead is always a fixed linear combination of two consecutive terms, with Fibonacci-number coefficients. Spotting this lets you leap across gaps without listing every intermediate value.

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Q29Simple & Compound InterestTITA

John borrowed Rs. 2,10,000 from a bank at an interest rate of 10% per annum, compounded annually. The loan was repaid in two equal instalments, the first after one year and the second after another year. The first instalment was interest of one year plus part of the principal amount, while the second was the rest of the principal amount plus due interest thereon. Then each instalment, in Rs., is

Your answer:

Show solution

Correct answer: 121000

Think of this as: the debt grows at 10% each year, and we knock it down with two equal payments — one at the end of year 1 and one at the end of year 2. We need to find that equal payment amount.

What we know:

Principal borrowed = Rs. 2,10,000
Rate = 10% per annum, compounded annually
Two equal instalments, paid at the end of year 1 and end of year 2
Let each instalment be $x$

Step 1 — Grow the debt through Year 1.

After one year, the bank adds 10% interest to the principal:

$2{,}10{,}000 \times 1.10 = 2{,}31{,}000.$

The first instalment $x$ is paid now, so the outstanding balance after the first payment is:

$2{,}31{,}000 - x.$

Step 2 — Grow the remaining debt through Year 2.

This balance sits for another year and earns 10% interest:

$(2{,}31{,}000 - x) \times 1.10.$

At the end of Year 2, the second instalment (also $x$ ) must clear this balance entirely. So we set them equal:

$(2{,}31{,}000 - x) \times 1.10 = x.$

Step 3 — Solve for $x$ .

Expand the left side:

$2{,}54{,}100 - 1.10x = x.$

Bring the $x$ -terms together:

$2{,}54{,}100 = x + 1.10x = 2.10x.$

$x = \frac{2{,}54{,}100}{2.10} = 1{,}21{,}000.$

Step 4 — Back-substitution check.

Year	Debt at start	After 10% growth	Instalment paid	Balance left
1	2,10,000	2,31,000	1,21,000	1,10,000
2	1,10,000	1,21,000	1,21,000	0 ✓

The balance hits exactly zero after the second instalment. ✓

💡 Teacher tip: "Two equal instalments" problems are just a rolling balance — grow the debt by the rate, subtract the instalment, repeat. The final balance after the last instalment must be zero. That single condition gives you the equation.

Alternative approach — Present Value method.

Each instalment's present value (discounted back to today) must add up to the principal:

$\frac{x}{1.10} + \frac{x}{1.10^2} = 2{,}10{,}000.$

Factor out $x$ :

$x \left(\frac{1}{1.10} + \frac{1}{1.21}\right) = 2{,}10{,}000.$

$x \left(\frac{1.21 + 1.10}{1.331}\right) = 2{,}10{,}000 \implies x \cdot \frac{2.31}{1.331} = 2{,}10{,}000.$

$x = 2{,}10{,}000 \times \frac{1.331}{2.31} = 2{,}10{,}000 \times \frac{121}{210} = 1{,}000 \times 121 = 1{,}21{,}000.$

Both methods agree. The rolling-balance method is more intuitive; the present-value method is faster once you're comfortable discounting.

Answer: Rs. 1,21,000

💡 Why this works out neatly: the principal 2,10,000 is divisible by 210, and the fraction $\frac{1.331}{2.31} = \frac{121}{210}$ cancels beautifully — a sign the numbers were chosen to give a clean answer.

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Q30Maxima & MinimaTITA

Let f(x) = min(2x², 52 − 5x) where x is any positive real number. Then the maximum possible value of f(x) is

Your answer:

Show solution

Correct answer: 32

Think of this as: we have two curves — a rising parabola and a falling line — and $f(x)$ always takes the smaller of the two. Where is that smaller value at its biggest?

Step 1 — Understand what $f(x)$ looks like.

We have two pieces:

$g(x) = 2x^2$ — a parabola opening upward, starting at $0$ and rising as $x$ grows.
$h(x) = 52 - 5x$ — a straight line, falling as $x$ grows.

Since $f(x) = \min(g(x), h(x))$ , the function always picks the lower of the two curves.

For small $x$ : $2x^2$ is tiny while $52 - 5x$ is large, so $f(x) = 2x^2$ — the function rises along the parabola.
For large $x$ : $2x^2$ is huge while $52 - 5x$ is small (eventually negative), so $f(x) = 52 - 5x$ — the function falls along the line.

So $f(x)$ rises, then falls. Its peak must sit exactly where the two curves cross.

💡 Teacher tip: When one curve rises and the other falls, the maximum of their minimum always occurs at the intersection point — no calculus required.

Step 2 — Find where the two curves meet.

Set them equal: $2x^2 = 52 - 5x$ $2x^2 + 5x - 52 = 0$

Solve with the quadratic formula: $x = \frac{-5 \pm \sqrt{25 + 416}}{4} = \frac{-5 \pm \sqrt{441}}{4} = \frac{-5 \pm 21}{4}$

This gives two roots:

$x = \frac{-5 + 21}{4} = \frac{16}{4} = 4$
$x = \frac{-5 - 21}{4} = \frac{-26}{4} = -6.5$

Since $x$ must be positive, we take $x = 4$ .

Step 3 — Evaluate $f$ at the crossing point.

$f(4) = 2(4)^2 = 32$

Check the other curve agrees: $52 - 5(4) = 52 - 20 = 32 \checkmark$

Both pieces equal $32$ at $x = 4$ , which is exactly what should happen at the intersection.

Step 4 — Confirm this is a maximum.

$x$	$2x^2$	$52 - 5x$	$f(x) = \min$	Compared to $32$
$3$	$18$	$37$	$18$	below ✗
$4$	$32$	$32$	$32$	peak ✓
$5$	$50$	$27$	$27$	below ✗

Just below $x = 4$ , $f$ rides the rising parabola (heading up toward $32$ ). Just above $x = 4$ , $f$ rides the falling line (heading down from $32$ ). So $f(x) \le 32$ for every positive $x$ .

Answer: $\boxed{32}$

💡 Why this works: The minimum of two functions can never exceed either one. At the crossing point, both functions equal $32$ , and on either side one of them drops below $32$ — dragging the minimum down. That crossing point is therefore the highest the minimum can ever climb.

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Q31Averages & Weighted AveragesTITA

A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him is

Your answer:

Show solution

Correct answer: 60

Think of this as: when we drop a group of tests, the total sum drops by a known amount (since we know the group's average and size), and the average shifts by a known amount. Two such conditions give us two equations — enough to solve for the total number of tests.

Step 1 — Name the variables and convert averages to sums.

Let the total number of tests be $n$ and the total sum of all scores be $S$ . The overall average is then $\dfrac{S}{n}$ .

The first 10 tests average 20, so their combined score is: $10 \times 20 = 200.$

The last 10 tests average 30, so their combined score is: $10 \times 30 = 300.$

💡 Teacher tip: Average-change problems hinge on the sum removed, not the average removed. Always convert "first 10 averaged 20" into "removed sum = 200" before setting up any equation.

Step 2 — Translate the two conditions into equations.

Condition 1: If the first 10 tests are dropped, the average rises by 1. $\frac{S - 200}{n - 10} = \frac{S}{n} + 1$

Condition 2: If the last 10 tests are dropped, the average falls by 1. $\frac{S - 300}{n - 10} = \frac{S}{n} - 1$

Step 3 — Clear denominators in each equation.

Multiply both sides of each equation by $n(n-10)$ :

From Condition 1: $n(S - 200) = (n-10)(S + n)$ $nS - 200n = nS + n^2 - 10S - 10n$ $-200n = n^2 - 10S - 10n$ $10S = n^2 + 190n \quad \cdots (1)$

From Condition 2: $n(S - 300) = (n-10)(S - n)$ $nS - 300n = nS - n^2 - 10S + 10n$ $-300n = -n^2 - 10S + 10n$ $10S = -n^2 + 310n \quad \cdots (2)$

Step 4 — Equate the two expressions for $10S$ .

Both (1) and (2) equal $10S$ , so set them equal to each other: $n^2 + 190n = -n^2 + 310n$ $2n^2 = 120n$ $2n^2 - 120n = 0$ $2n(n - 60) = 0$

Since $n \neq 0$ (there are tests), we get $n = 60$ .

Step 5 — Verify by back-substitution.

With $n = 60$ : from equation (1), $10S = 3600 + 11400 = 15000$ , so $S = 1500$ and the overall average is $\dfrac{1500}{60} = 25$ .

Scenario	Remaining sum	Remaining tests	New average	Expected	Match?
Drop first 10 (sum 200)	$1500 - 200 = 1300$	$50$	$\frac{1300}{50} = 26$	$25 + 1 = 26$	✅
Drop last 10 (sum 300)	$1500 - 300 = 1200$	$50$	$\frac{1200}{50} = 24$	$25 - 1 = 24$	✅

Both conditions check out perfectly.

Answer: $\boxed{60}$

Alternative approach — Work with the average directly (avoids the quadratic).

Let the overall average be $A$ and the total number of tests be $n$ , so the total sum is $nA$ .

From Condition 1 (drop first 10, average rises by 1): $\frac{nA - 200}{n - 10} = A + 1 \implies nA - 200 = (n-10)(A+1) = nA + n - 10A - 10$ $\implies 10A - n = 190 \quad \cdots (1')$

From Condition 2 (drop last 10, average falls by 1): $\frac{nA - 300}{n - 10} = A - 1 \implies nA - 300 = (n-10)(A-1) = nA - n - 10A + 10$ $\implies 10A + n = 310 \quad \cdots (2')$

Adding $(1')$ and $(2')$ : $20A = 500 \implies A = 25$ .

Substituting into $(2')$ : $250 + n = 310 \implies n = 60$ .

This method trades the quadratic for a clean linear system — faster under exam time.

💡 Why this works so neatly: The two conditions are mirror images — one raises the average by 1, the other lowers it by 1 — so their equations combine symmetrically, eliminating one variable and leaving a single clean equation for $n$ .

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Q32Averages & Weighted AveragesMCQ

In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?

Show solution

Correct answer: D

Think of this as: the overall average is locked at 38. We want to push the under-51 group's average as high as possible — so we need to figure out what makes that group's average rise or fall.

Step 1 — Name the two groups and set up the average equation.

Let:

The ≥ 51 group have 30 people with average age $M$ (where $M \ge 51$ ).
The under-51 group have $n$ people with average age $m$ (where $m < 51$ ), and $n \le 39$ .

The overall average is 38, so the total age of everyone divided by the total number of people equals 38:

$\frac{30M + nm}{30 + n} = 38$

Step 2 — Solve for $m$ (the quantity we want to maximise).

Multiply both sides by $(30 + n)$ :

$30M + nm = 38(30 + n)$

Rearrange to isolate $nm$ :

$nm = 38(30 + n) - 30M$

$m = \frac{38(30 + n) - 30M}{n} = \frac{1140 + 38n - 30M}{n}$

We can also write this as:

$m = 38 + \frac{1140 - 30M}{n}$

Step 3 — Decide what makes $m$ large.

Look at the term $\frac{1140 - 30M}{n}$ added to 38.

Since $M \ge 51$ , the numerator $1140 - 30M \le 1140 - 1530 = -390$ , which is negative.
A negative number divided by $n$ becomes less negative (closer to zero) when $n$ is larger.
So to maximise $m$ , we want $n$ as large as possible: $n = 39$ .
We also want $M$ as small as possible (since a smaller $M$ makes the numerator less negative): $M = 51$ .

💡 Teacher tip: When the overall average is fixed, to maximise one group's average, push the other group to its least-contributing extreme (smallest allowed average) and make your own group as large as allowed. The heavy, low-average group drags the overall average down, letting your group sit higher.

Step 4 — Plug in the extreme values.

$m = \frac{1140 + 38(39) - 30(51)}{39}$

$m = \frac{1140 + 1482 - 1530}{39} = \frac{1092}{39} = 28$

Step 5 — Verify the answer is consistent.

Check	Value	Valid?
Under-51 group average	$m = 28 < 51$	✅
≥ 51 group average	$M = 51 \ge 51$	✅
Under-51 group size	$n = 39 \le 39$	✅
Overall average	$\frac{30(51) + 39(28)}{69} = \frac{1530 + 1092}{69} = \frac{2622}{69} = 38$	✅

Every constraint is satisfied, so $m = 28$ is achievable.

Answer: D — $\boxed{28}$

Closing insight: The under-51 group can average as high as 28 because the 30 people at exactly age 51 pull the overall average up just enough. Making the under-51 group as large as possible (39 people) dilutes the drag from its own below-38 average, letting each member sit closer to 38 — and 28 is the ceiling.

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Q33Maxima & MinimaTITA

While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is

Your answer:

Show solution

Correct answer: 40

Think of this as: a slip in one factor pins down the product of the other two factors, and then we minimise a sum of squares subject to that fixed product.

Step 1 — Translate the error into an equation.

Let the three real numbers be $37$ , $a$ , and $b$ , where $a$ and $b$ are the two numbers Ashok read correctly. Instead of $37$ , he used $73$ .

Correct product: $37ab$
Ashok's product: $73ab$

The product went up by $720$ :

$73ab - 37ab = 720$

$36ab = 720$

$ab = 20$

So the product of the other two numbers is fixed at $20$ .

Step 2 — Minimise $a^2 + b^2$ given $ab = 20$ .

We use the identity

$a^2 + b^2 = (a - b)^2 + 2ab.$

Since $(a-b)^2 \ge 0$ for all real $a, b$ , we get

$a^2 + b^2 \ge 2ab = 2(20) = 40.$

💡 Teacher tip: Whenever a problem fixes a product and asks about a sum of squares, reach for $a^2 + b^2 \ge 2ab$ (equivalently, $(a-b)^2 \ge 0$ ). It converts a product-constraint into a lower bound on the sum of squares instantly.

Step 3 — Check that 40 is actually attainable.

Equality in $a^2 + b^2 \ge 2ab$ holds when $(a-b)^2 = 0$ , i.e., $a = b$ .

If $a = b$ , then $ab = a^2 = 20$ , so $a = b = \sqrt{20}$ (or both $-\sqrt{20}$ ).

Check: $ab = \sqrt{20} \cdot \sqrt{20} = 20$ ✓, and $a^2 + b^2 = 20 + 20 = 40$ ✓.

Candidate	$ab$	$a^2 + b^2$	Attains minimum?
$a = b = \sqrt{20}$	$20$ ✓	$40$ ✓	✅
$a = 4,\; b = 5$	$20$ ✓	$16 + 25 = 41$	✗ (larger)
$a = 2,\; b = 10$	$20$ ✓	$4 + 100 = 104$	✗ (larger)

The closer $a$ and $b$ are to each other, the smaller the sum of squares — and the minimum is reached when they are equal.

Answer: $\boxed{40}$

Why this works: The error $73 - 37 = 36$ cleanly divides $720$ to give $ab = 20$ , and the inequality $a^2 + b^2 \ge 2ab$ turns that product directly into the minimum sum of squares. The two ideas chain together in one line each.

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Q34LogarithmsMCQ

If x is a positive quantity such that $2^x$ = $3^{(\log₅ 2)}$ , then x is equal to

Show solution

Correct answer: D

We're given $2^x = 3^{\log_5 2}$ and need a clean closed form for $x$ . The obstacle is that $x$ sits in an exponent on the left, while $\log_5 2$ sits in an exponent on the right — taking a single logarithm of both sides will flatten both powers into products at once.

Step 1 — Take $\log_5$ of both sides.

A logarithm converts a power into a product: $\log_b(a^c) = c\,\log_b a$ . Applying this to both sides,

$\log_5\!\left(2^x\right) = \log_5\!\left(3^{\log_5 2}\right)$

$x\,\log_5 2 = \left(\log_5 2\right)\!\left(\log_5 3\right).$

Step 2 — Cancel the common factor $\log_5 2$ .

Since $2 \neq 1$ and $2 \neq 5$ , we have $\log_5 2 \neq 0$ , so it is safe to divide both sides by it:

$x = \log_5 3.$

Step 3 — Match this against the options.

The answer $\log_5 3$ does not appear verbatim, so we rewrite each option using $\log_b\!\left(\tfrac{p}{q}\right) = \log_b p - \log_b q$ and $\log_b b = 1$ :

Option	Simplified form	Equals $\log_5 3$ ?
A: $\log_5 8$	$\log_5 2^3 = 3\log_5 2$	✗
B: $1 + \log_3\!\left(\tfrac{5}{3}\right)$	$1 + \log_3 5 - \log_3 3 = \log_3 5$	✗
C: $\log_5 9$	$\log_5 3^2 = 2\log_5 3$	✗
D: $1 + \log_5\!\left(\tfrac{3}{5}\right)$	$1 + \log_5 3 - \log_5 5 = 1 + \log_5 3 - 1 = \log_5 3$	✓

Only Option D simplifies to $\log_5 3$ .

💡 Teacher tip: Whenever an equation has the shape $a^{(\text{stuff})} = b^{(\text{stuff})}$ , take one logarithm of both sides — the powers immediately flatten into multiplications, and the common factor usually cancels.

Answer: D — $\boxed{\log_5 3}$

Alternative approach (natural logs). Take $\ln$ of both sides instead: $x\ln 2 = (\log_5 2)(\ln 3) = \frac{\ln 2}{\ln 5}\cdot\ln 3.$ Dividing by $\ln 2 \neq 0$ gives $x = \dfrac{\ln 3}{\ln 5} = \log_5 3$ , the same result. Either base works; $\log_5$ is slightly faster here because it cancels the $\log_5 2$ factor directly.

Why the answer hides itself: $\log_5 3$ is less than $1$ (since $3 < 5$ ), so it is naturally written as $1 + \log_5(3/5)$ — the $+1$ and $-1$ from $\log_5 5$ cancel, disguising a simple logarithm as a "shifted" expression.

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