The interest on ₹54,000 loan for nine months at 8% interest per year is A) 1340 · B) 2430 · C) 3240 · D) 4230

Answer: C

$SI = \frac{P \times R \times T}{100} = \frac{54000 \times 8 \times \frac{9}{12}}{100}$ $= 54000 \times 0.08 \times 0.75 = \mathbf{3240}$ ⚠️ The trap is the 9 months. Time must be in years → 9/12 = 0.75. Forgetting this gives 4320 (a full year), which is close to the decoy D) 4230.

Tom ordered 2 hamburgers + 1 soft drink = ₹70. Jerry ordered 1 hamburger + 2 soft drinks = ₹50. Cost of one of each? A) H 10, D 30 · B) H 30, D 10 · C) H 15, D 25 · D) H 25, D 15

Answer: B

Let hamburger = h, soft drink = s: $2h + s = 70 \qquad h + 2s = 50$ Multiply the first by 2: $4h + 2s = 140$. Subtract the second: $3h = 90 \Rightarrow h = 30 \Rightarrow s = 70 - 60 = 10$ ⭐ Faster on exam day: just test the options. B gives 2(30)+10 = 70 ✓ and 30+2(10) = 50 ✓. Substitution beats elimination when four options are handed to you.

Which statements are correct? a. Arrangements of "PEPPER" = 60 · b. Committee of 3 from 20 = C(20,3) · c. (2+x)³ = 8+6x+12x²+x³ · d. 3 boys and 3 girls in a row = 360 A) c and d · B) b and c · C) a and b · D) d and a

Answer: C

• a ✅ PEPPER has 6 letters: P×3, E×2, R×1 → $\frac{6!}{3!\,2!} = \frac{720}{12} = \mathbf{60}$
• b ✅ Order doesn't matter → a combination, $\binom{20}{3} = 1140$
• c ❌ $(2+x)^3 = 8 + 12x + 6x^2 + x^3$. The statement swaps the 6 and the 12
• d ❌ 6 people in a row = $6! = \mathbf{720}$, not 360. (360 would be the answer if boys and girls had to alternate: 2 × 3! × 3!)

Which statements are correct? a. lim(x²−1)/(x−1) at x→1 is 0 · b. f(x)=1/x ⇒ f′(x)=1/x² · c. lim(eˣ−1)/x at x→0 is 1 · d. derivative = lim(h→0)[f(x+h)−f(x)]/h A) b and c · B) a and b · C) d and a · D) c and d

Answer: D

• a ❌ $\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1 \to \mathbf{2}$, not 0
• b ❌ $\frac{d}{dx}\left(x^{-1}\right) = -x^{-2} = \mathbf{-\frac{1}{x^2}}$ — the statement drops the minus sign
• c ✅ $\lim_{x\to0}\frac{e^x-1}{x} = 1$ — a standard limit worth memorising
• d ✅ The first-principles definition of the derivative

A = {x ∈ ℝ : |x| < 1}, B = {x ∈ ℝ : |x−1| ≥ 1}, and A ∪ B = ℝ \ D. Find D. A) {x : 1 ≤ x < 2} · B) {x : 1 < x ≤ 2} · C) {x : 0 ≤ x ≤ 2} · D) {x : 1 ≤ x ≤ 2}

Answer: A

Step 1 — unpack the modulus sets: A: |x| &lt; 1 \Rightarrow -1 &lt; x &lt; 1 $B: |x-1| \ge 1 \Rightarrow x-1 \ge 1 \text{ or } x-1 \le -1 \Rightarrow x \ge 2 \text{ or } x \le 0$ Step 2 — union: $A \cup B = (-\infty, 0] \cup (-1, 1) \cup [2, \infty) = (-\infty, 1) \cup [2, \infty)$ Step 3 — D is what is left out: D = \mathbb{R} \setminus (A \cup B) = [1, 2) = \mathbf{\{x : 1 \le x &lt; 2\}} ⚠️ The whole question turns on the brackets. 1 is included (A stops just before 1); 2 is not (B starts at 2). Options A and D differ only there.

Match the relations to their properties. A) a-1,b-2,c-3,d-5 · B) a-2,b-1,c-5,d-4 · C) a-3,b-4,c-1,d-2 · D) a-4, b-5, c-1, d-2

Answer: D

• b → 5 $x^2+y^2=25$ is a circle; one x gives two y values → not a function ✓
• c → 1 $y = 2x+1$, so $F \circ F(x) = 2(2x+1)+1 = \mathbf{4x+3}$ ✓
• d → 2 $y = x^2$ is increasing on (0, ∞) but not on ℝ ✓
• a → 4 by elimination ⚠️ Paper error in pair (a). $x^2 + y = 2$ means $y = 2 - x^2$, whose range is (−∞, 2] — not the [−2, ∞) in statement 4. That is the range of $y = x^2 - 2$, almost certainly what was intended. D is still correct, because b, c and d are all certain.

Match each matrix to its determinant. A) a-3, b-1, c-4, d-5 · B) a-3,b-2,c-5,d-1 · C) a-4,b-1,c-2,d-5 · D) a-1,b-2,c-3,d-4

Answer: A

Arrange in increasing order of probability: a. a diamond from 52 cards · b. a King from 52 cards · c. rolling a number between 1 and 6 on a die · d. heads on a fair coin A) a,b,d,c · B) c,a,d,b · C) b, a, d, c · D) d,a,c,b

Answer: C

Area of the triangle with A(1,2,3), B(3,2,1), C(2,4,5) A) √20 · B) √23 · C) √26 · D) √29

Correct working gives √17 — which is not on the list

$\vec{AB} = (2, 0, -2) \qquad \vec{AC} = (1, 2, 2)$ $\vec{AB} \times \vec{AC} = (0\cdot2 - (-2)\cdot2,\; (-2)\cdot1 - 2\cdot2,\; 2\cdot2 - 0\cdot1) = (4, -6, 4)$ $|\vec{AB} \times \vec{AC}| = \sqrt{16+36+16} = \sqrt{68}$ $\text{Area} = \tfrac{1}{2}\sqrt{68} = \sqrt{17} \approx \mathbf{4.123}$ Options are 4.472, 4.796, 5.099, 5.385 — none is √17. The key in 2024-paper.md says "D) √29" but admits it is "as per matching patterns … approximations" — it was guessed. ⭐ Learn the method, not this answer: Area = ½ |AB × AC|. If this shape appears in 2026, compute it and take the nearest option.

The unit vector having the opposite direction of the vector (1, −1) is A) (1/√2)(1,−1) · B) (−1/√2)(1,−1) · C) (−1, 1) · D) (1, −1)

Answer: B

$|(1,-1)| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$ $\hat{a} = \frac{(1,-1)}{\sqrt{2}} \quad\Rightarrow\quad \text{opposite} = -\hat{a} = \mathbf{\frac{-1}{\sqrt{2}}(1,-1)}$ ⚠️ C is the trap. (−1, 1) does point the opposite way, but its magnitude is √2 — it is not a unit vector. The question asks for a unit vector, so it must be divided by √2.

The sum of two numbers is 18. The sum of their reciprocals is 1/4. The numbers are A) 12 and 6 · B) 10 and 8 · C) 9 and 9 · D) 11 and 7

Answer: A

$x + y = 18, \qquad \frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy} = \frac{1}{4}$ $\frac{18}{xy} = \frac{1}{4} \;\Rightarrow\; xy = 72$ $t^2 - 18t + 72 = 0 \;\Rightarrow\; t = \frac{18 \pm \sqrt{324-288}}{2} = \frac{18 \pm 6}{2} = 12,\; 6$ ⭐ Faster in the exam: test the options. Only 12 + 6 = 18 with 12 × 6 = 72.

For what value of m do 5a + 7b = 2 and 15a + 21b = m have no solution? A) m ≠ 0 · B) m ≠ 4 · C) m ≠ 6 · D) m ≠ 6

Answer: Null — awarded to all

Options C and D are identical. The paper printed the correct option twice, so it had to be dropped. The mathematics is sound and worth knowing: the second equation is exactly 3 × the first, so its right side must be 3 × 2 = 6 for the system to be consistent. $\frac{15}{5} = \frac{21}{7} = 3 \neq \frac{m}{2} \;\Rightarrow\; \textbf{no solution when } m \neq 6$ ⭐ The rule: for $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$ — if $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$ the lines are parallel → no solution; if all three ratios are equal the lines coincide → infinite solutions.

If the points (5,2,4), (6,1,2) and (8,7,k) are collinear, then k = A) 2 · B) 1 · C) 2 · D) 3

Answer: Null — awarded to all

No value of k works, and options A and C are duplicates. Proof: $\vec{AB} = (6-5,\; 1-2,\; 2-4) = (1,\, -1,\, -2)$ $\vec{AC} = (8-5,\; 7-2,\; k-4) = (3,\, 5,\, k-4)$ Collinearity needs $\vec{AC} = t\,\vec{AB}$. The x-component forces $t = 3$; the y-component then demands $5 = -3$. Contradiction — the three points can never be collinear. ⭐ The method still matters: check collinearity by testing whether the direction vectors are proportional — $\frac{x_2-x_1}{x_3-x_1}=\frac{y_2-y_1}{y_3-y_1}=\frac{z_2-z_1}{z_3-z_1}$.

What is the compound interest for a principal of ₹11,200 at 10% for 2 years? A) ₹200 · B) ₹220 · C) ₹252 · D) ₹300

Answer: C (with the principal read as ₹1,200)

$CI = P\left[(1+r)^n - 1\right] = 1200\left[(1.1)^2 - 1\right] = 1200 \times 0.21 = \mathbf{252}$ ⚠️ On ₹11,200 the answer is ₹2,352 — not an option. The stem's principal is a typo. The formula is the point. ⭐ Memorise the 2-year shortcut: CI = P(2r + r²) → at 10%, that is 21 % of P.

If X and Y are two sets and X ∪ Y = X ∩ Y, then which is true? A) X = ø · B) Y = ø · C) X = Y · D) X ⊆ Y

Answer: C

The union always contains the intersection: $X \cap Y \subseteq X \cup Y$. They can only be equal when nothing sits in the union but outside the intersection — i.e. no element belongs to just one set. Therefore X = Y. ⚠️ D is the trap. $X \subseteq Y$ gives $X \cup Y = Y$ and $X \cap Y = X$, which are equal only if X = Y anyway. D is too weak.

Perimeter of the triangle with vertices 2i + 4j − k, 4i + 5j + k, 3i + 6j − k A) 3 + 2√3 + √5 · B) 3 + √2 + √5 · C) 3 + √6 + √5 · D) 3 + √6 − √5

Answer: C

Points: A(2, 4, −1), B(4, 5, 1), C(3, 6, −1). $AB = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{9} = 3$ $BC = \sqrt{(-1)^2 + 1^2 + (-2)^2} = \sqrt{6}$ $CA = \sqrt{(-1)^2 + (-2)^2 + 0^2} = \sqrt{5}$ $\text{Perimeter} = \mathbf{3 + \sqrt{6} + \sqrt{5}}$ ⚠️ D is the trap — a perimeter is a sum of lengths, so nothing can be subtracted. That eliminates D instantly.

Three unbiased coins are tossed. What is P(at most 2 heads)? A) 1/4 · B) 3/8 · C) 7/8 · D) 1/2

Answer: C

Use the complement. "At most 2 heads" is everything except "all 3 heads": $P = 1 - P(\text{3 heads}) = 1 - \frac{1}{8} = \mathbf{\frac{7}{8}}$ ⭐ Compare Q74 in this same paper's Statistics section — at least two tails = 1/2. Probability is the only topic tested in BOTH sections; the same 8-outcome sample space answers both.

If A = {0,1,2,4} and B = {1,3,5}, then (A − B) × (A ∩ B) = A) {(0,1), (2,1), (4,1)} · B) {(0,1), (2,1), (3,1)} · C) {(1,1), (2,1), (4,1)} · D) {(1,0), (2,1), (3,1)}

Answer: A

$A - B = \{0, 2, 4\} \qquad A \cap B = \{1\}$ $(A-B) \times (A \cap B) = \{(0,1),\, (2,1),\, (4,1)\}$ ⚠️ Three traps at once: B and D smuggle in 3, which is in B so it cannot be in A − B; C smuggles in 1, which is in B for the same reason; D also reverses the order to (1,0) — in a Cartesian product the first element must come from the first set.

(The question body was an image in the source and has not survived.) A) 1/2 · B) −1/2 · C) 1 · D) −1 → Answer: A · the paper's own hint: use L'Hôpital's Rule ⭐ What to take from it: an indeterminate 0/0 or ∞/∞ limit is differentiated top and bottom separately: $\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}$ The two standard results that produce 1/2 are $\lim_{x\to0}\frac{1-\cos x}{x^2}$ and $\lim_{x\to0}\frac{\sqrt{1+x}-1}{x}$. Know both.

In how many ways can 8 students be seated on 8 chairs? A) 8 · B) ⁸C₁ · C) ⁸P₈ · D) 64

Answer: C

$^{8}P_{8} = \frac{8!}{(8-8)!} = \frac{8!}{0!} = 8! = 40{,}320$ ⭐ Arrangement ⇒ permutation. Seating people on chairs is an ordered arrangement, so it is P, not C. ⚠️ The trap is D) 64 = 8², which would be right only if chairs could be reused (i.e. repetition allowed).

A = {1,2,3,4}, B = {3,4,5,6,7}. Find the symmetric difference. A) {1,2,3,4,5,6,7} · B) {5,6,7} · C) {1,2,3,4} · D) {1,2,5,6,7}

Answer: D

$A \triangle B = (A \cup B) \setminus (A \cap B)$ $A \cup B = \{1,2,3,4,5,6,7\}, \qquad A \cap B = \{3,4\}$ $A \triangle B = \{1, 2, 5, 6, 7\}$ ⭐ In words: everything in exactly one of the two sets — throw away the overlap. Equivalent form: $(A \setminus B) \cup (B \setminus A) = \{1,2\} \cup \{5,6,7\}$

Length of the line segment from A(2,2) to B(9,11) A) 9.4 · B) 11.4 · C) 13.4 · D) 15.4

Answer: B

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{(9-2)^2 + (11-2)^2}$ $= \sqrt{49 + 81} = \sqrt{130} \approx \mathbf{11.40}$ ⭐ Estimating beats computing. √121 = 11 and √144 = 12, so √130 sits just above 11 — only B is in range.