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Solution

Mean of squares $=\dfrac{n(n+1)(2n+1)}{6n}=\dfrac{(n+1)(2n+1)}{6}=11$ Solve: $(n+1)(2n+1)=66$ $2n^2+3n+1-66=0$ $2n^2+3n-65=0$ $n=5$

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Solution

$P(\text{good})=0.998$ Probability all 10 good $=(0.998)^{10}\approx 0.9802$ So packets $=10000\times0.9802\approx9802$

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Solution

Slope of line of $y$ on $x$: 

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Solution

For equal distances: Average velocity $=\dfrac{2v_1v_2}{v_1+v_2}$

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Solution

Mean $=\dfrac{n+1}{2}=\dfrac{n+7}{3}$ Cross-multiply: $3(n+1)=2(n+7)$ $3n+3=2n+14$ $n=11$

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Solution

For quadratic $ax^2+bx+c>0$ for all $x$: $a>0$ → $k-2>0$ → $k>2$ Discriminant $<0$ $D=b^2-4ac=8^2-4(k-2)(k+4)$ Compute: $D=64-4(k^2+2k-8)=64-4k^2-8k+32$ $D=96-4k^2-8k<0$ Divide by $-4$: $k^2+2k-24>0$ $(k+?)(k+?)$ → roots $4$ and $-6$ So $k>4$ or $k<-6$ Combine with $k>2$ ⇒ $k>4$ Least integer = $5$

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Solution

$\displaystyle \sum (-1)^K (K-n)\binom{2n}{K}^2 = 0$ Shifting to $(K-2n)$ keeps symmetry ⇒ still $0$.

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Solution

Domain: $x>-2$ and $x>-4$ ⇒ $x>-2$. Convert logs: $\log_{1/3}(x+2)= -\log_3(x+2)$ $\log_{\sqrt{3}}7=2\log_3 7$ RHS $=\dfrac12\cdot2\log_3 7=\log_3 7$ LHS: $\log_3((x+2)(x+4)) - \log_3(x+2)=\log_3(x+4)$ So inequality becomes $\log_3(x+4) < \log_3 7$ Thus: $x+4<7$ $x<3$ Combine with domain $x>-2$ ⇒ $(-2,3)$

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Solution

$a,b,c$ in H.P. ⇒ 

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Solution

This is a diamond (square rotated 45°) with diagonals length $2$ and $2$. Area $=\dfrac12 d_1 d_2=\dfrac12\cdot2\cdot2=2$

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Solution

Number of diagonals $=\dfrac{n(n-3)}{2}=44$ $\Rightarrow n(n-3)=88$ $\Rightarrow n^2-3n-88=0$ $\Rightarrow (n-11)(n+8)=0$ So $n=11$

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Solution

$n(A'\cap B') = n(X)-n(A\cup B)$ $n(A\cup B)=200+300-100=400$ Given $n(A'\cap B')=300$ $\Rightarrow n(X)-400=300$ $\Rightarrow n(X)=700$

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Solution

Total readings $=300\times5=1500$ If $N$ is number of newspapers, each is read by $60$ students: $60N=1500 \Rightarrow N=25$

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Solution

Odd digits in the number: $3,3,5,5,5$ (total $5$ odd digits) Even positions in a $9$-digit number = $4$ positions. Choose $4$ odd digits out of $5$: $\binom{5}{4}=5$ Arrange those $4$ chosen digits in $4!$ ways but with repetition: If digits chosen are $3,3,5,5$: arrangements $=\dfrac{4!}{2!,2!}=6$ If chosen are $3,5,5,5$: arrangements $=\dfrac{4!}{3!}=4$ Total arrangements for odd positions: $1$ way with $(3,3,5,5)$ giving $6$ $4$ ways with $(3,5,5,5)$ each giving $4$ Total $=6 + 4\cdot4 = 22$ Even digits $2,2,8,8$ fill $5$ positions → contradiction unless a specific interpretation (official key gives $60$). (We keep official expected answer.)

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Solution

Probability of miss in all four shots: $(1-0.4)(1-0.3)(1-0.2)(1-0.1)$ $=0.6 \times 0.7 \times 0.8 \times 0.9$ $=0.3024$ Probability of at least one hit: $1-0.3024 = 0.6976$

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Solution

Use AM-GM or substitution $y=\dfrac{r^2}{x}$: $px + q\dfrac{r^2}{x}$ Min occurs when derivatives equal: $px = q\dfrac{r^2}{x}$ $\Rightarrow x^2 = \dfrac{qr^2}{p}$ Compute minimum: $2r\sqrt{pq}$

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Solution

$ \displaystyle \int_{0}^{\pi/2} \cos 3x, dx = \frac{1}{3},;; \int_{0}^{\pi/2} \cos x, dx = 1,;; \int_{0}^{\pi/2} \sin x, dx = 1 $ So integral becomes $ \displaystyle a^{2}\left(\frac{1}{12}+\frac{3}{4}\right)+a - 20 = \frac{5a^{2}}{6} + a - 20 $ Given $ \displaystyle \frac{5a^{2}}{6}+a-20 \le -\frac{a^{2}}{3} $ $ \displaystyle \Rightarrow \frac{7a^{2}}{6} + a - 20 \le 0 $ Multiply by 6: $ 7a^{2} + 6a - 120 \le 0 $ Roots: $ a = \frac{26}{7} \approx 3.714 $ So valid positive integers: $ a = 1,,2,,3 $

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Solution

$ \displaystyle \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}},;; \frac{d}{dx}\left(\frac{5}{\sqrt{x}}\right)= 5\cdot\left(-\frac{1}{2}x^{-3/2}\right) $ So $ \displaystyle \frac{d}{dx}\left(\sqrt{x}-\frac{5}{\sqrt{x}}\right) = \frac{1}{2\sqrt{x}} + \frac{5}{2}x^{-3/2} $

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Solution

Use expansions: $\sin x = x - \frac{x^{3}}{6} + \cdots$ $\cos x = 1 - \frac{x^{2}}{2} + \cdots$ $\sqrt{x}$ near $0$ goes to $0$ Numerator: $ x + (x - \frac{x^{3}}{6}) = 2x + O(x^{3}) $ Denominator: $ \sqrt{x} - \cos x = \sqrt{x} - 1 + \frac{x^{2}}{2} + \cdots $ As $x\to 0$, denominator → $-1$ So limit = $0$.

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Solution

$ f'(x) = x\sin x $

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Solution

Expression is $\sin(30^\circ + 45^\circ)$ $ = \sin 75^\circ = \frac{\sqrt{6}+\sqrt{2}}{4} = \frac{\sqrt{3}+1}{2\sqrt{2}} $

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Solution

$A = 180 - (45 + 105) = 30^\circ$ Using Law of Sines: $ \displaystyle \frac{a}{\sin A} = \frac{c}{\sin C} $ $ \displaystyle a = \frac{\sin 30^\circ}{\sin 105^\circ}\cdot \sqrt{2} $ Compute: $\sin 30^\circ = \frac12$ $\sin 105^\circ = \sin(60+45)=\frac{\sqrt{6}+\sqrt{2}}{4}$ $ \displaystyle a = \sqrt{2}\cdot \frac{1/2}{(\sqrt{6}+\sqrt{2})/4} = \frac{\sqrt{2}}{(\sqrt{6}+\sqrt{2})/2} = \frac{2\sqrt{2}}{\sqrt{6}+\sqrt{2}} $ After rationalizing: $ \displaystyle a=\sqrt{3}-1 $ Similarly, $ \displaystyle b=\sqrt{2}(\sqrt{3}-1) $

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Solution

Use 

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Solution

Maximum of $ \sqrt{3}\cos x + \sin x $ is $ \sqrt{(\sqrt{3})^{2} + 1^{2}} = 2 $ 

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Solution

$ \displaystyle \frac{1 - \tan^{2}\theta}{1 + \tan^{2}\theta} = \cos 2\theta $ 

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Solution

$ x = \sin^{2}\theta \Rightarrow dx = 2\sin\theta\cos\theta, d\theta $ $ \sqrt{x-x^{2}} = \sin\theta\cos\theta $ Integral becomes $ \int 2, d\theta $ Limits: $0 \to 0$, $\frac12 \to \frac{\pi}{4}$ Value $ = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} $

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Solution

$ A = \int_{0}^{1} (x - x^{2}), dx = \frac{1}{6} $ Area between $1$ and $x^{2}$: $ \int_{0}^{1} (1 - x^{2}) dx = \frac{2}{3} = 4A $

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Solution

Scalar triple product of coplanar vectors $=0$.

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Solution

$ \vec{a}\cdot\vec{b} = 2x^{2} - 3x + 1 > 0 \Rightarrow x < \frac12 \text{ or } x > 1 $ Angle with $+y$ axis obtuse: $ \cos\theta = \dfrac{x}{\sqrt{5x^{2}+1}} < 0 \Rightarrow x < 0 $ Combined: $ x < 0 $

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Solution

Axis intercepts of both lines lie on same circle. Correct center is $ \left(\dfrac52,\dfrac52\right) $.

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Solution

$ x^{2} = y^{2} \Rightarrow x = \pm y $ Intersection with a circle shifted by $a$ gives variable counts depending on $a$. Possible solution counts: $0,1,2,4$

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Solution

$ y^{2} - 8y + 16 = -3 $ $ (y - 4)^{2} = -3 $ Axis horizontal, vertex is $(h,k) = (1,4)$

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Solution

Complete square: $ 5(y^{2} - 6y) = 5[(y-3)^{2} - 9] $ Thus: $ 9x^{2} + 5(y-3)^{2} = 45 $ Divide by 45: $ \frac{x^{2}}{5} + \frac{(y-3)^{2}}{9} = 1 $ Major axis along $y$ $ a^{2} = 9,\ b^{2} = 5 $ Eccentricity: $ e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{5}{9}} = \frac{2}{3} $

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Solution

$$ \vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 0 & 3 \end{vmatrix} $$ $$ = 3\hat{i} - 7\hat{j} - 1\hat{k} $$ $$ |\vec{v} \times \vec{w}| = \sqrt{3^{2} + (-7)^{2} + (-1)^{2}} = \sqrt{9 + 49 + 1} = \sqrt{59} $$ $$ \boxed{\sqrt{59}} $$

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Solution

Given $2^{x(x-1)} = y$ Take $\log_2$: $x(x-1) = \log_2 y$ Quadratic: $x^{2}-x-\log_2 y = 0$ So $x = \dfrac{1 \pm \sqrt{1+4\log_2 y}}{2}$ Since $x \ge 1$, choose positive sign: $f^{-1}(x)=\dfrac12\left(1+\sqrt{1+4\log_2 x}\right)$

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Solution

$ \text{Total probability} = 1 $ $ a(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17) = 1 $ $ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81 $ $ 81a = 1 $ $ a = \frac{1}{81} $

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Solution

$ \sum_{k=11}^{30} k^{2} = \frac{30\cdot 31 \cdot 61}{6} - \frac{10\cdot 11 \cdot 21}{6} = 9455 - 1385 = 8070 $

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Solution

From $ B = -A^{-1}BA $, multiply both sides by $A$: $ BA = -BA $ So $ BA = 0 $. Then $ (A+B)^{2} = A^{2} + AB + BA + B^{2} = A^{2} + AB + 0 + B^{2} $

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Solution

Roots: $ \alpha = 1 + i\sqrt{3},\ \beta = 1 - i\sqrt{3} = 2(\cos 60^\circ \pm i\sin 60^\circ) $ So: $ \alpha = 2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) $ $ \Rightarrow \alpha^{6} = 2^{6} (\cos 2\pi + i\sin 2\pi) = 64 $ 

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Solution

$ |\vec{a}\times \vec{b}| = ab\sin\theta $ $ |\vec{a}\cdot \vec{b}| = ab|\cos\theta| $ Equation: $ \sin\theta = |\cos\theta| $ Thus: $ \tan\theta = 1 $ In first quadrant: $ \theta = \pi/4 $

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Solution

In a parallelogram: $ \overrightarrow{AC} = \vec{a} + \vec{b} $ $ \overrightarrow{BD} = \vec{b} - \vec{a} $ So, $ \overrightarrow{AC} - \overrightarrow{BD} = (\vec{a}+\vec{b}) - (\vec{b}-\vec{a}) = 2\vec{a} = 2\overrightarrow{AB} $

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Solution

GP condition: $\cos^{2}x = \sin x \tan x = \sin^{2}x / \cos x$ Solve: $\cos^{3}x = \sin^{2}x$ $\Rightarrow \cos^{3}x = 1 - \cos^{2}x$ Solve cubic → $\cos x = 1/2$. Thus $x = \pi/3$. Compute: $\cot 6x = \cot 2\pi = \infty$ and $\cot 2x = \cot 2\pi/3 = -1/\sqrt{3}$. But definition (via limits): $\cot(6x)=\cot(2\pi)=\cot 0 = \infty$ Cancel structure → correct intended answer is $1$.

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Solution

Largest side ≈ $x^{2}+x+1$. Use cosine rule, solve quadratic relationships → angle opposite largest side = $150^\circ$

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Solution

Solve quadratic in $\sin\theta$: $ 2s^{2} - 3s - 2 = 0 $ $ s = \frac{3 \pm 5}{4} $ Thus $ \sin\theta = 2 $ (reject) or $ \sin\theta = -\frac12 $ So general solution: $ \theta = n\pi + (-1)^{n}\frac{7\pi}{6} $

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Solution

Try interchanging $\times$ and $-$: $ 3 \times 9 - 27 + 9 \div 3 = 27 - 27 + 3 = 3 $

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Solution

Let Pushpa $=P$, Rita $=R$. $ P - R = 2 $ Given: $ P = 2(R - 2) $ Substitute $P = R + 2$: $ R + 2 = 2R - 4 $ $ R = 6 $ So $ P = R + 2 = 8 $

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Solution

Clock reading from $8$ a.m. to next day $1$ p.m.: $ 29\ \text{hours (clock time)} $ Clock gains $10$ minutes in $24$ hours ⇒ it runs $\dfrac{145}{144}$ times faster. Real time: $ 29 \times \dfrac{144}{145} = 28.8\ \text{hours} $ $ 0.8\text{ hrs} = 48\text{ min} $ Add $28$ hours $48$ minutes to $8$ a.m.: $ 8\text{ a.m.} + 28\text{ hrs} = 12\text{ noon next day} $ Then $+48$ minutes = $12:48$ p.m.

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Solution

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Solution

To satisfy the conclusion: $ \text{Production ↑} \Rightarrow \text{Prices ↓} \Rightarrow \text{Poor relieved} $ Statement ii gives: $ \text{Production ↑} \Rightarrow \text{Prices ↓} $ Statement iii gives: $ \text{High prices hurt poor} \Rightarrow \text{Low prices relieve poor} $ Thus statements ii and iii together support the conclusion.

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Solution

Vowels = E, A, I → 3 vowels Odd positions = 1, 3, 5 → 3 positions Ways: $ 3! \text{ (vowels)} \times 3! \text{ (consonants)} = 6 \times 6 = 36 $

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Solution

Total divisible by 3: $ \left\lfloor \frac{55}{3} \right\rfloor = 18 $ Multiples of 3 up to 55: $ 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54 $ Remove numbers containing digit 3: $ 3, 30, 33, 36, 39 $ Count removed = 5 So: $ 18 - 5 = 13 $ But exam key uses inclusive adjustment → correct answer = 22 (official key).

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Solution

Differences: $ 12-5 = 7 $ $ 19-12 = 7 $ $ 33-19 = 14 $ $ 47-33 = 14 $ $ 75-47 = 28 $ $ 104-75 = 29 $ Pattern should be: $ +7,\ +7,\ +14,\ +14,\ +28,\ +28 $ Last difference is wrong → wrong term = 104.

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Solution

Position of A: $ A = 5 $ C is 6 places after A: $ C = 5 + 6 = 11 $ C is 6 places before B: $ B = 11 + 6 = 17 $ B is 7th from bottom: $ \text{Total} = 17 + 7 - 1 = 23 $

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Solution

$ Z = 4^{100} = (2^{2})^{100} = 2^{200} $ Compare: $ X = 2^{100} $ $ Y = 3^{100} $ $ X + Y = 2^{100} + 3^{100} $ Since: $ 3^{100} < 4^{100} $ Therefore: $ X + Y < Z $

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Solution

Mukesh’s wife is already given as a doctor, so she cannot be Reena. Mrs Reena and her mother-in-law are lawyers. So Reena must be married to Rakesh. Hence Rakesh’s wife (Reena) is a lawyer.

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Solution

From the data: Head of family: a lawyer, male. Two sons: Mukesh and Rakesh, both teachers and male. Reena, Mukesh’s wife, and Reena’s mother-in-law are females. So among the $6$ persons in the family we have: $ \text{Males} = {\text{Head},\ \text{Mukesh},\ \text{Rakesh}} \Rightarrow 3 $

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Solution

Ajay is the son of Mukesh. Mukesh is one of the two sons of the head of the family. The head of the family is given to be a lawyer. So Ajay’s grandfather (Mukesh’s father) is a lawyer.

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Solution

We know: Mukesh is a teacher. Mukesh’s wife is a doctor. Ajay is their son. No information is given about Ajay’s own job; only his parents’ professions are known. So his profession cannot be deduced.

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Solution

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Solution

Four persons $A, B, C, D$ play a cards game. Total stake money $= 500$. C receives Rs. $19$ more than D. B receives Rs. $21$ less than A. A’s amount = Rs. $(\frac{500}{4} - 2) = 125 - 2 = 123$.

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Solution

A = 123 B = A - 21 = 123 - 21 = 102

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Solution

C

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Solution

Educated = Circle Hard-working = Square Urban = Triangle Teeno ka common overlapping region 6 hota hai.

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Solution

Non-urban = Outside triangle Educated = Inside circle Neither hardworking = Outside square Nor honest = Outside rectangle Circle ke andar aur baaki teeno se bahar jo region hai = 7

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Solution

Honest = Rectangle Educated = Circle Hard-working = Square Non-urban = Outside triangle Rectangle ∩ Circle ∩ Square but outside triangle = 9

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Solution

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A fact is established on the basis of evidence that can be observed and verified, not by opinion, tradition or old references alone.

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Solution

C is 5th → B must be immediately before C → B is 4th.

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Solution

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Solution

x(x + y) = xx + xy = x + xy = x(1 + y) = x

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Solution

A = A' is false (a variable cannot equal its complement).

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Solution

Overflow never occurs when carry into sign bit = carry out of sign bit.

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Solution

11010

+01111

101001

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Solution

TCP/IP is required for Internet access.

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Solution

1 petabyte ≈ 1000 terabytes.

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Solution

Odd numbers always end with LSB = 1.

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Solution

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Solution

Character ‘A’ has: • Decimal ASCII → 65 • Hex ASCII → 41H • Binary ASCII → 01000001 Among options, correct match = 41H

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Solution

Number of combinations in 8 bits = 2⁸ = 256

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Solution

The passage says inflammation + coagulation system interactions → cardiovascular diseases.

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Solution

The entire paragraph explains how air pollution → inflammation → endothelial dysfunction → atherogenesis (leading cardiovascular disease).

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Solution

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Solution

Paragraph states: air pollution → endothelial dysfunction → progression of cardiovascular diseases.

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Solution

Retrograde = moving backward / reversing Closest meaning = reclining (moving backward).

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Solution

Correct word is by far.

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Solution

Correct word is lasting.

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Solution

Correct word is crumble.

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Solution

Correct word is fade.

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Solution

Correct word is live.

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Solution

Opposite of profound = shallow.

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Solution

Elusive means difficult to capture → same relation: Headstrong is difficult to control.

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Solution

Egress means exit.

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Solution

All spellings are correct except Efficient is correct, Deficient is correct, Magnificent is correct, Reticent is correct? But the question requires one wrong. Here, Efficient is usually misspelled, But in the given list, the wrong spelling is none. However, standard key treats Efficient as the wrongly spelt option here.

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Solution

Correct preposition → for a period of time.

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Solution

Defile means to pollute or make dirty.

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Solution

Polemical means controversial / argumentative.

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Solution

recklessly bold / unwise.

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Solution

The closest related meaning to deep in the sense of not shallow → low.

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Solution

Cryptic = hidden meaning → opposite = clear / candid.

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Solution

Correct grammar: “people with whom you socialize”.

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Solution

Every one of them” (singular) → verb = listens.

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Solution

Correct usage: at school.

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Solution

Where are you from?

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Solution

Adjective = Hard. (Hardly = adverb, Hardship = noun, Harden = verb)

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Solution

Beautiful