Quantitative Aptitude PYQs

Explanation:

\(\rm Z_1^2+Z_2^2+Z_1Z_2=0\)

⇒ Z ₁  = ω and Z ₂  = ω ²

Now

⇒ \(|\frac{Z_1}{Z_2}| =| \frac{\omega }{(\omega)^2}| =|\frac{1}{\omega|}= 1\)

Option (a) is correct.

Explanation:

\(\rm \frac{1}{2}+Re\left(\frac{Z_1}{Z_2}\right) = \frac{1}{2}+ Re (\frac{\omega }{\omega^2})\)

= \(\frac{1}{2}Re(\omega)^2\)

= \(\frac{1}{2}+ Re [ -\frac{1}{2} - \frac{\sqrt3}{2}i\)

= \(\frac{1}{2} + (-\frac{1}{2}) = 0\)

∴ Option (b) is correct.

Explanation:

Let the required terms of AP are a – 2d, a – d, a, a + d, a + 2d

⇒ (a – 2d) (a – d) a (a + d) (a + 2d) = 2,29,635

⇒ a(a ² – d² ) (a ² – 4d² ) = 2,29,635.....(i)

Now  a – 2d, a – d and a + 2d are in GP

⇒ (a – d)² = a² – 4d²

⇒ 5d² – 2ad = 0 

⇒ d(5d – 2a) = 0

⇒ 5d = 2a......(ii)

Solving equation (i) and (ii), we get

⇒ d = 6

∴ Option (d) is correct.

Explanation:

Putting the value of d in equation (ii), we get

⇒ 5 × 6 = 2a 

⇒ a = 15

Thus, The terms are 3, 9, 15, 21 and 27

So required sum = 3 + 9 + 15 + 21 + 27 = 75 

∴ Option (c) is correct

Explanation:

Given:

(8 + 3√7) 20  = U + V...(i)

(8 - 3√7) 20  = W...(ii)

Here, 0 < W < 1 

Now, adding Eqs.

(i) and (ii), we get

U + V +W = (8 + 3√7) 20 + (8 - 3√7) 20

= 2[²⁰C₀8²⁰+ ²⁰C₂8¹⁸. (3√7 )²+........+(3√7)²⁰

⇒ It is an even number.

Also, 0 < V < 1, 0 < W < 1 and U is an integer

Thus,  V + W is an integer

V + W = 1

∴ Option (d) is correct.

Explanation:

Given:

(8 + 3√7) 20  = U + V...(i)

(8 - 3√7) 20  = W...(ii)

⇒ (U + V)W =] (8 + 3√7) 20][(8 - 3√7) 20]

= (64 – 63)²⁰ = 1²⁰ = 1

∴ Option (b) is correct.

Explanation:

Given:

 ⇒ a² (b² – c ² ) x² + b ² (c ² – a ² ) x + c ² (a ² – b ² ) = 0...(i)

Now  Equation has equal roots:

Then D =0

⇒ [b² (c ² – a² )]² – 4a ² c ² (b ² – c ² ) (a ² – b ² ) = 0

⇒  b⁴ (c ² – a ² )² – 4a ² c ² (b ² – c ² ) (a ² – b ² ) = 0

Solving above equation, we get 

⇒  b⁴ (c² + a² )² = 4a ⁴ c ⁴ 

⇒ b ² (a ² + c ² ) = 2a ² c ²

⇒ \(\frac{2}{b^2} = \frac{1}{a^2}+ \frac{1}{c^2}\)

⇒ a² , b ² and c ² are in HP

∴ Option (c) is correct.

Explanation:

Let α be the required root.

Sum of roots = \(\frac{-B}{A} = \frac{-b^2(c^2-a^2)}{a^2(b^2-c^2)}\)

⇒ α + α = \( = \frac{-b^2(c^2-a^2)}{a^2(b^2-c^2)}\)

⇒ α = \( \frac{b^2(c^2-a^2)}{2a^2(c^2-b^2)}\)

∴ Option (c) is correct

Explanation:

Given:

\(\rm A=\begin{bmatrix}3&-3&4\\ 2&-3&4\\ 0&-1&1\end{bmatrix} \)

Now, |A| = 3(–3 + 4) –2(–3 + 4) + 0 = 3 – 2 = 1

A(adjA) = |A| I = I

∴ Option (d) is correct.

Explanation:

Given 

⇒ \(adj(A) = \rm \begin{bmatrix}1&-1&0\\ -2&3&-4\\ -2&3&-3\end{bmatrix}\)

Now, A^-1 = \(\frac{1}{|A|} (Adj(A))\)

= \( \rm \begin{bmatrix}1&-1&0\\ -2&3&-4\\ -2&3&-3\end{bmatrix}\)

∴ Option (a) is correct.

Explanation:

Given:

⇒ (2î + 6ĵ + 27k̂) × (î + αĵ + βk̂) = 0

⇒ \(\begin{bmatrix} \hat{i} &\hat{j}&\hat{k}\\ 2&6&27\\ 1&α&β\end{bmatrix}\ \) = 0

⇒ \(\hat{i} (6β -27α) - \hat{j}(2β-27)+\hat{k}(2α-6) =0\)

Comparing both sides, we get

6β  – 27α  = 0

⇒ 2β  = 9α 

2β - 27 = 0

⇒ β = 27/2

Also

⇒ 2α – 6 = 0 

α =3

Now, 

3α  + 2β = \(3\times 3 + 2 \times \frac{27}{2}\) =36

∴ Option (1) is correct

Explanation:

Let P = \(\rm |\vec a\times \vec b|+√3|\vec a.\vec b|\)

= \(\rm |\vec a|.|\vec b||sinθ |+ √3|\vec a||\vec b||cosθ|\)

= \(\rm |\vec a||\vec b|[|sinθ |+ √3|cosθ|]\)

⇒ p will be max if (sin 3 + √ 3cos θ) is maximum.

Now, for its maxima,

\(\frac{d}{dθ } (sinθ +√3cosθ) =0\)

⇒ Cosθ -√3 sinθ =0

⇒ \(tanθ =\frac{1}{\sqrt3}\)

⇒ θ =30°

∴ Option (b) is correct

Explanation:

Given:

\(⃗ a+2⃗ b\) and \(\rm 5⃗ a-4⃗ b\) are perpendicular vectors.

⇒ \((⃗ a+2⃗ b).\)((\rm 5⃗ a-4⃗ b)\) = 0

⇒ \(5|⃗ a|^2 -4⃗ a.⃗ b + 10⃗ a . ⃗ b - 8(⃗ b)^2=0\)

⇒ \(5× 1 + 6⃗ a.⃗ b-8 =0\)

Now \(\vec a, \vec b\) are unit vectors

⇒ \(6 |\vec a||\vec b| Cosθ = 3\)

⇒ 1.1 cosθ =1/2

⇒Cosθ = 1/2 

Now

cos2 θ = Cos²θ -1

= 2× 1/4 -1

= \(-\frac{1}{2}\)

Now, 

cosθ + cos2θ = 1/2 -1/2 =0

∴The Correct answer is Option a

Calculation:

\( \vec{AB} = 2\vec{FC} \)

⇒ \( \vec{AB} = -2\vec{CF} \)

⇒ \( n = -2\)

Also,

\( \vec{AD} = 2\vec{BC} \Rightarrow m = 2 \)

Now,

\( mn = 2(-2) = -4\)

∴ The final value of  is -4.

Explanation:

Given:

\(\rm \vec a=̂ i+̂ j\ and \ \vec b=̂ j+̂ k,\)

Also, \(\vec a, \vec b, \vec c\) has same length

⇒ \(|\vec a| = |\vec b| = |\vec c|\) = √2 

Let θ be the angle between the vectors.

⇒ Cosθ = \(\frac{\vec a. \vec b}{|\vec a||\vec b|} = \frac{0+1+0}{\sqrt2} =\frac{1}{\sqrt2}\)

(I) Let \(\vec c =\hat i +\hat k\)

\(|\vec c| = \sqrt2\)

Cosθ = \(\frac{\vec a. \vec c}{|\vec a||\vec c|} = \frac{0+1+0}{\sqrt2} =\frac{1}{\sqrt2}\)

All the conditions are satisfied, so it can be vector \(\vec c\)

(II) Let \(\vec c =\rm \frac{-\hat i+4\hat j-\hat k}{3}\)

⇒ \(|\vec c| = \frac{1}{3}\sqrt18 = \sqrt2\)

Cosθ = \(\frac{\vec a. \vec c}{|\vec a||\vec c|} \)

= \(\frac{\frac{-1}{3}\frac{4}{3}}{\sqrt2 \sqrt2} = \frac{1}{2}\)

All the conditions are satisfied, so it can be vector \(\vec c\)

∴ Option (c) is correct.

Explanation:

Slope of diagonal along the line x – 2y = 1

⇒m₁ = 1/2

Slope of the diagonal along the line 4x + 2y = 3

⇒m₂ = -2

Now,  m₁m₂ = \(\frac{1}{2}(-2) = -1\)

Then, Diagonals are perpendicular. 

∴ The quadrilateral ABCD is a rhombus.

Explanation:

Given:

4x2 + 9y2 = 1 

Here a = 1/2 and b = 1/3

Now 

e = \(\sqrt{1-\frac{b^2}{a^2}} = \frac{\sqrt5}{3}\)

foci = (±ae, 0)

= ( ± \((\frac{1}{2}\frac{\sqrt5}{3},0)\)

= (± \(\frac{\sqrt5}{6},0\))

Thus, P(x, y) is any point on the ellipse

⇒ PQ + PR = 2a = 2× 1/2 = 1

∴ Option (b) is correct.

Explanation:

O is midpoint of P and R

⇒ O = \([\frac{10+2}{2}, \frac{14+4}{2}]\) = (6,9)

Also, O is midpoint of S and Q.

⇒ O = \((\frac{x + 8}{2}, \frac{y+12}{2})\)

⇒ (6,9) = \((\frac{x + 8}{2}, \frac{y+12}{2})\)

Comparing both sides, we get

⇒ \(\frac{x+8}{2} = 6\)

⇒x =4

Also 

⇒ \(\frac{y+12}{2} = 9\)

⇒ y = 6

Thus, x + y = 4 + 6 = 10 

∴ Option (b) is correct

Explanation:

Given

Equation of circle is  

(x2 - 4x + 3) + (y2 - 6y + 8) = 0

⇒ (x – 3) (x – 1) + (y – 4) (y – 2)= 0

So the possible ends of the diameters are (3, 4), (3, 2), (1, 4) and (1, 2). 

Also, Radius = √2 

center = (2,3)

So the required pair of ends of diameters are (I) (1, 2) and (3, 4) and (II) (1, 4) and (3, 2).

∴ Option (a) is correct.

Explanation:

Let ∠ PAM = α and ∠ APM =β 

From figure, it is clear that

∠PAQ = ∠PAM + ∠QAM

⇒ ∠PAQ = α + α = 2α

In right-angled triangle AMP

⇒AM = PM = 4k

⇒ α = β 

Thus, α + β + 90° = 180°

⇒ 2α = 90°

⇒ α = 45°

Hence, ∠ PAQ = 2a = 90°

∴ Option (b) is correct