Quantitative Aptitude PYQs

Calculation:

Convert \( (10110110)_2 \) to decimal:

\( (10110110)_2 = 1 \times 2^7 + 0 \times 2^6 + 1 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 \)

\( = 128 + 0 + 32 + 16 + 0 + 4 + 2 + 0 = 182 \)

Convert \( (100011)_2 \) to decimal:

\( (100011)_2 = 1 \times 2^5 + 0 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 \)

\( = 32 + 0 + 0 + 0 + 2 + 1 = 35 \)

Add the two decimal numbers

\( 182 + 35 = 217 \)

Convert the sum back to binary

To convert \( 217 \) into binary, divide by 2 and note down the remainders:

\( 217 \div 2 = 108 , \text{remainder} , 1 \)

\( 108 \div 2 = 54 , \text{remainder} , 0 \)

\( 54 \div 2 = 27 , \text{remainder} , 0 \)

\( 27 \div 2 = 13 , \text{remainder} , 1 \)

\( 13 \div 2 = 6 , \text{remainder} , 1 \)

\( 6 \div 2 = 3 , \text{remainder} , 0 \)

\( 3 \div 2 = 1 , \text{remainder} , 1 \)

\( 1 \div 2 = 0 , \text{remainder} , 1 \)

Reading the remainders from bottom to top, \( 217_{10} = (11011001)_2 \)

The sum of \( (10110110)_2 \) and \( (100011)_2 \) is \( (11011001)_2 \).

Therefore, the correct option is 1.

Calculation:

Given,

Cardinality of set X: |X| = 3n

Cardinality of set Y: |Y| = 2n

Cardinality of intersection: |X ∩ Y| = n

Calculate cardinality of set difference (X - Y):

|X - Y| = |X| - |X ∩ Y|

⇒ |X - Y| = 3n - n

⇒ |X - Y| = 2n

Calculate cardinality of set difference (Y - X):

|Y - X| = |Y| - |X ∩ Y|

⇒ |Y - X| = 2n - n

⇒ |Y - X| = n

The number of elements in Cartesian product (X - Y) × (Y - X) is:

|(X - Y) × (Y - X)| = |X - Y| · |Y - X|

⇒ |(X - Y) × (Y - X)| = (2n) · (n)

⇒ |(X - Y) × (Y - X)| = 2n²

∴ The number of elements is 2n², which is option (d).

Calculation:

Given sets:

\( A = {-3, -2, -1, 0, 1, 2, 3} \)

\( B = {0, 1, 4, 9} \)

We are asked to find the number of elements in the subset of \( A \times B \) corresponding to the relation:

\( R = {(x, y) : |x| < y} \)

We evaluate the condition \( |x| < y \) for each \( y \in B \):

⇒ \( y = 0 \Rightarrow |x| < 0 \) No values of x 

⇒ \( y = 1 \Rightarrow |x| < 1\) Only x = 0 

⇒ \( y = 4 \Rightarrow |x| < 4\)  All \(x \in A \)

⇒ \( y = 9 \Rightarrow |x| < 9 \) All \(x \in A \)

Count of valid pairs:

From y = 1: 1 pair → (0, 1)

From y = 4 : 7 pairs → all \( x \in A\)

From y = 9 : \(7 pairs → all ( x \in A) \)

Total \(= 1 + 7 + 7 = 15\)

Hence, the correct answer is Option 3.

Calculation:

We know that for any square matrix \(A\) of order \(n\):

\( |kA| = k^{n}|A| \)

Hence, for a matrix \(X\) of order \(n\):

\( |mX| = m^{n}|X| \)

So, Statement I is correct.

Let

\( X= \begin{bmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{bmatrix} \)

and

\( Y= \begin{bmatrix} ma_{11} & ma_{12} & ma_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{bmatrix} \)

Then,

\( |Y|= \begin{vmatrix} ma_{11} & ma_{12} & ma_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{vmatrix} \)

Taking \(m\) common from the first row:

\( |Y|=m \begin{vmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{vmatrix} \)

\( |Y| = m|X| \)

So, Statement II is also correct and Statement II is the correct explanation of Statement I.

Hence, the correct answer is option 1.

Calculation:

Given the matrix \( M = \begin{pmatrix} 71 & 23 & 48 \ 57 & 28 & 29 \ 65 & 17 & 48 \end{pmatrix} \)

Determinant of Matrix M

The inverse of a matrix exists if the determinant of the matrix is non-zero. So, we calculate the determinant of matrix M using the formula for the determinant of a 3x3 matrix:

\( \text{det}(M) = 71 \times \begin{vmatrix} 28 & 29 \ 17 & 48 \end{vmatrix} - 23 \times \begin{vmatrix} 57 & 29 \ 65 & 48 \end{vmatrix} + 48 \times \begin{vmatrix} 57 & 28 \ 65 & 17 \end{vmatrix} \)

After calculating the individual 2x2 determinants:

\( \text{det}(M) = 71 \times 851 - 23 \times 851 + 48 \times (-851) \)

\( \text{det}(M) = 60421 - 19573 - 40848 = 0 \)

Since the determinant of the matrix \( M \) is 0, this means the matrix \( M \) is singular, and as a result, the inverse of \( M \) does not exist.

Statement-I: "The inverse of M does not exist" is correct.

Statement-II: "M is non-singular" is incorrect because \( M \) is singular (det(M) = 0).

Therefore, the correct answer is:

Statement-I is correct but Statement-II is not correct.

Hence, the correct answer is Option 3.

Calculation:

We are asked to evaluate the expression:

\( \cot^{-1}(9) + \csc^{-1}\left(\frac{\sqrt{41}}{4}\right) \)

Convert cotangent inverse to tangent inverse

⇒ \( \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \)

⇒ \( \cot^{-1}(9) = \tan^{-1}\left(\frac{1}{9}\right) \)

Convert cosecant inverse to tangent inverse

Let \( \theta = \csc^{-1}\left(\frac{\sqrt{41}}{4}\right) \), then:

\( \csc(\theta) = \frac{\sqrt{41}}{4} \). We have Hypotenuse= \( \sqrt{41} \) and Opposite= 4.

Adjacent side \( a = \sqrt{(\sqrt{41})^2 - 4^2} = \sqrt{41 - 16} = \sqrt{25} = 5 \).

⇒ \( \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{4}{5} \)

⇒ \( \csc^{-1}\left(\frac{\sqrt{41}}{4}\right) = \tan^{-1}\left(\frac{4}{5}\right) \)

Add the two inverse tangents

Use the identity (since \( \frac{1}{9} \cdot \frac{4}{5} < 1 \)):

⇒ \( \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \)

Let \( a = \frac{1}{9}, b = \frac{4}{5} \)

⇒ \( \frac{a + b}{1 - ab} = \frac{\frac{1}{9} + \frac{4}{5}}{1 - \frac{1}{9} \cdot \frac{4}{5}} \)

\( = \frac{\frac{5 + 36}{45}}{1 - \frac{4}{45}} = \frac{\frac{41}{45}}{\frac{45 - 4}{45}} = \frac{\frac{41}{45}}{\frac{41}{45}} = 1 \)

The expression equals: \( \tan^{-1}(1) \)

⇒ \( \tan^{-1}(1) = \frac{\pi}{4} \)

∴ The correct answer is \(\frac{\pi}{4}\)

Calculation:

We are given the conditions:

\( \cot \theta = -\sqrt{3} \)

\( \csc \theta = -2 \)

Use identity for cotangent

\( \cot \theta = \frac{\cos \theta}{\sin \theta} = -\sqrt{3} \Rightarrow \cos \theta = -\sqrt{3} \sin \theta \)

Use identity for cosecant

\( \csc \theta = \frac{1}{\sin \theta} = -2 \Rightarrow \sin \theta = -\frac{1}{2} \)

Substitute into cotangent identity

\( \cos \theta = -\sqrt{3} \cdot \left(-\frac{1}{2}\right) = \frac{\sqrt{3}}{2} \)

So we now have:

\( \sin \theta = -\frac{1}{2} \)

\( \cos \theta = \frac{\sqrt{3}}{2} \)

Determine angle

These values correspond to the reference angle \( \theta = \frac{\pi}{6} \), but we need the quadrant where:

sine is negative

cosine is positive

This occurs in Quadrant IV, so:

\( \theta = -\frac{\pi}{6} \)

The correct answer is 1 value of \( \theta\) in the interval \( -\pi < \theta < \pi \)

Hence, the correct answer is option 3.

Calculation:

We are given that:

\( x + \frac{1}{x} = 2 \cos\theta \)

We need to find the value of:

\( x^3 + \frac{1}{x^3} \)

Cube both sides of the given equation:

⇒ \( \left( x + \frac{1}{x} \right)^3 = \left( 2\cos\theta \right)^3 \)

Expanding the left-hand side:

⇒\( \left( x + \frac{1}{x} \right)^3 = x^3 + \frac{1}{x^3} + 3 \left( x + \frac{1}{x} \right) \)

Thus:

⇒ \( x^3 + \frac{1}{x^3} + 3 \left( x + \frac{1}{x} \right) = 8 \cos^3\theta \)

Substitute  \(x + \frac{1}{x} = 2 \cos\theta \) into the equation:

⇒ \( x^3 + \frac{1}{x^3} + 3 \times 2 \cos\theta = 8 \cos^3\theta \)

Simplifying:

⇒ \( x^3 + \frac{1}{x^3} + 6 \cos\theta = 8 \cos^3\theta \)

isolate \(x^3 + \frac{1}{x^3} \)

⇒ \( x^3 + \frac{1}{x^3} = 8 \cos^3\theta - 6 \cos\theta \)

Factor out\( 2 \cos\theta \)

⇒ \( x^3 + \frac{1}{x^3} = 2 \cos\theta (4 \cos^2\theta - 3) \)

⇒ \( x^3 + \frac{1}{x^3} = 2 \cos 3\theta \)

Hence the correct answer is Option 3.

Calculation:

The problem is to find the number of values of \( x \) in the interval \( 0 \le x \le \frac{\pi}{2} \) that satisfy the equation:

\( \tan x + \sec x = 2\cos x \)

Rewrite the equation in terms of \( \sin x \) and \( \cos x \)

⇒ \( \frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x \)

⇒ \( \frac{\sin x + 1}{\cos x} = 2\cos x \)

⇒ \( \sin x + 1 = 2\cos^2 x \)

Convert to a quadratic equation in \( \sin x \)

Using the identity \( \cos^2 x = 1 - \sin^2 x \):

⇒ \( \sin x + 1 = 2(1 - \sin^2 x) \)

⇒ \( \sin x + 1 = 2 - 2\sin^2 x \)

⇒ \( 2\sin^2 x + \sin x - 1 = 0 \)

Solve the quadratic equation for \( \sin x \)

Factor the equation:

⇒ \( (2\sin x - 1)(\sin x + 1) = 0 \)

Possible solutions for \( \sin x \):

⇒ \( \sin x = \frac{1}{2} \) or \( \sin x = -1 \)

Find values of \( x \) in the interval \( [0, \frac{\pi}{2}] \)

Case 1: \( \sin x = \frac{1}{2} \)

In \( [0, \frac{\pi}{2}] \), the solution is \( x = \frac{\pi}{6} \). (1 value)

Case 2: \( \sin x = -1 \)

The solution is \( x = \frac{3\pi}{2} \), which is outside \( [0, \frac{\pi}{2}] \). (0 values)

The only valid value is \( x = \frac{\pi}{6} \).

∴ The number of values of \( x \) satisfying the equation is 1, which is option (b).

Calculation:

We are asked to evaluate the expression:

\( \tan\left[\frac{1}{2} \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\right] \)

Let \(\theta = \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) \)

Then \( \sec \theta = \frac{2}{\sqrt{3}} \Rightarrow \cos \theta = \frac{\sqrt{3}}{2} \)

This corresponds to \( \theta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \)

Evaluate the half-angle

\( \frac{1}{2} \theta = \frac{\pi}{12} \)

So we need to compute \( \tan\left(\frac{\pi}{12}\right) \)

Use known value

\( \tan\left(\frac{\pi}{12}\right) = \tan(15^\circ) = 2 - \sqrt{3} \)

Hence, the correct answer is Option 1

Calculation:

Given data:

Direction ratios of the line parallel to the plane P are (1, 3, 2) .

The equation of the two planes are:

Plane 1: 6x + 4y - 5z = 2 , normal vector \( \mathbf{n_1} = (6, 4, -5) \)

Plane 2:  x - 2y + 3z = 0 , normal vector \( \mathbf{n_2} = (1, -2, 3) \)

The direction ratios of the line of intersection of two planes can be found using the cross product of the normal vectors of the planes:

\( \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \)

\( \mathbf{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 6 & 4 & -5 \ 1 & -2 & 3 \end{vmatrix} \)

Expanding the determinant:

\( \mathbf{d} = \hat{i} \begin{vmatrix} 4 & -5 \ -2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 6 & -5 \ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 6 & 4 \ 1 & -2 \end{vmatrix} \)

\( \mathbf{d} = \hat{i} \left( (4)(3) - (-5)(-2) \right) - \hat{j} \left( (6)(3) - (-5)(1) \right) + \hat{k} \left( (6)(-2) - (4)(1) \right) \)

\( \mathbf{d} = \hat{i} (12 - 10) - \hat{j} (18 + 5) + \hat{k} (-12 - 4) \)

\( \mathbf{d} = \hat{i} (2) - \hat{j} (23) + \hat{k} (-16) \)

Therefore, the direction ratios of the line of intersection are  (2, -23, -16) .

Hence, the Correct answer is Option 2.

Calculation:

Family of planes through the intersection

The general form is:

\( (6x + 4y - 5z - 2) + \lambda(x - 2y + 3z) = 0 \)

Expanding:

\( (6 + \lambda)x + (4 - 2\lambda)y + (-5 + 3\lambda)z - 2 = 0 \)

Apply parallel condition

Let normal vector \( \vec{n} = \langle 6 + \lambda, 4 - 2\lambda, -5 + 3\lambda \rangle \)

Direction vector of the line: \( \vec{d} = \langle 1, 3, 2 \rangle \)

Since the plane is parallel to the line, \( \vec{n} \cdot \vec{d} = 0 \)

Compute dot product:

\( (6 + \lambda)(1) + (4 - 2\lambda)(3) + (-5 + 3\lambda)(2) = 0 \)

\( 6 + \lambda + 12 - 6\lambda - 10 + 6\lambda = 0 \Rightarrow \lambda + 8 = 0 \Rightarrow \lambda = -8 \)

Substitute back to get plane equation

\( (6 - 8)x + (4 + 16)y + (-5 - 24)z - 2 = 0 \Rightarrow -2x + 20y - 29z - 2 = 0 \)

Multiply by -1 to simplify:

\( 2x - 20y + 29z + 2 = 0 \)

Hence, the required plane equation is:

\(2x - 20y + 29z + 2 = 0\)

Calculation:

Let the centre of the sphere be \(O(x,y,z)\).

Distances of the centre from the given points:

⇒ \( OA=\sqrt{(x-1)^2+y^2+z^2} \)

⇒ \( OB=\sqrt{x^2+(y-1)^2+z^2} \)

⇒ \( OC=\sqrt{x^2+y^2+(z-1)^2} \)

Since the sphere passes through all three points: \( OA=OB \)

⇒ \( (x-1)^2+y^2+z^2=x^2+(y-1)^2+z^2 \)

⇒ \( x^2-2x+1+y^2+z^2=x^2+y^2-2y+1+z^2 \)

\( \Rightarrow x=y \)

Also, \( OB=OC \)

⇒ \( x^2+(y-1)^2+z^2=x^2+y^2+(z-1)^2 \)

⇒ \( x^2+y^2-2y+1+z^2=x^2+y^2+z^2-2z+1 \)

\( \Rightarrow y=z \)

Hence, \( x=y=z \)

For the smallest radius:

⇒ \( x=y=z=\frac{1}{3} \)

Radius of the sphere:

⇒ \( r=\sqrt{\left(1-\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2} \)

\( =\sqrt{\frac{4}{9}+\frac{1}{9}+\frac{1}{9}} =\sqrt{\frac{6}{9}} =\sqrt{\frac{2}{3}} \)

Hence, the correct answer is Option 2

Calculation:

We are given three points: A(1, 0, 0),  B(0, 1, 0) , and  C(0, 0, 1) , and we are asked to find the equation of the plane passing through these points and the center of the sphere.

\( \overrightarrow{AB} = B - A = (0 - 1, 1 - 0, 0 - 0) = (-1, 1, 0) \)

\( \overrightarrow{AC} = C - A = (0 - 1, 0 - 0, 1 - 0) = (-1, 0, 1) \)

\( \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & 0 \ -1 & 0 & 1 \end{vmatrix} = \hat{i}(1) - \hat{j}(-1) + \hat{k}(1) = \hat{i} + \hat{j} + \hat{k} \)

Thus, the normal vector to the plane is (1, 1, 1).

The equation of the plane is \(x + y + z - 1 = 0\)

The center of the sphere lies on this plane. Therefore, the center of the sphere lies on the plane with the equation x + y + z - 1 = 0 

Hence, the correct answer is Option 1.

Calculation:

\( A(1, -1, 0), B(-2, 1, 8), C(-1, 2, 7) \) are the three consecutive vertices of the parallelogram. Let the fourth vertex be \( D(x, y, z) \).

In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of diagonal \( AC \) must be equal to the midpoint of diagonal \( BD \).

Midpoint of AC\( = \left( \frac{1 + (-1)}{2}, \frac{-1 + 2}{2}, \frac{0 + 7}{2} \right) = (0, \frac{1}{2}, \frac{7}{2}) \)

Midpoint of BD\(= \left( \frac{-2 + x}{2}, \frac{1 + y}{2}, \frac{8 + z}{2} \right) \)

Equating the midpoints, we get

\( \frac{-2 + x}{2} = 0, \quad \frac{1 + y}{2} = \frac{1}{2}, \quad \frac{8 + z}{2} = \frac{7}{2} \)

Solving for \( x, y, z \):

\( x = 2, \quad y = 0, \quad z = -1 \)

Therefore, the fourth vertex D is \( (2, 0, -1) \).

Hence, the correct answer is Option 2.

Calculation:

\( \vec{BC} = \hat{i} + \hat{j} - \hat{k} \)

\( \vec{CD} = 3\hat{i} - 2\hat{j} - 8\hat{k} \)

\( \cos \theta = \frac{\vec{BC} \cdot \vec{CD}}{|\vec{BC}| \cdot |\vec{CD}|} \)

Compute dot product:

\( \vec{BC} \cdot \vec{CD} = 1 \cdot 3 + 1 \cdot (-2) + (-1) \cdot (-8) = 3 - 2 + 8 = 9 \)

Compute magnitudes:

\( |\vec{BC}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3} \)

\( |\vec{CD}| = \sqrt{3^2 + (-2)^2 + (-8)^2} = \sqrt{9 + 4 + 64} = \sqrt{77} \)

\( \cos \theta = \frac{9}{\sqrt{3} \cdot \sqrt{77}} \)

\( \cos^2 \theta = \left( \frac{9}{\sqrt{3} \cdot \sqrt{77}} \right)^2 = \frac{81}{3 \cdot 77} = \frac{27}{77} \)

Hence, the correct answer is Option 2.

Calculation:

4y = mx-m+2

4y = m(x-1)+2

\(y - \frac{1}{2} = \frac{m}{4}(x -1)\)

for any values of m the family of lines always posses through (1, 1/2) so, all lines are concurrent.

Calculation:

Let locus of point be \( (x, y) \)

⇒ \( (x - a)^2 + (y - b)^2 = (x - c)^2 + (y - d)^2 \)

⇒ \( x^2 - 2ax + a^2 + y^2 - 2by + b^2 = x^2 - 2cx + c^2 + y^2 - 2dy + d^2 \)

⇒ \( 2(a - c)x + 2(b - d)y + c^2 + d^2 - a^2 - b^2 = 0 \)

⇒ \( (a - c)x + (b - d)y + \frac{1}{2}(c^2 + d^2 - a^2 - b^2) = 0 \)

\( \therefore \quad k = \frac{(c^2 + d^2 - a^2 - b^2)}{2} \)

Hence, the correct answer is Option 4.

Calculation:

Given: The equation is \( x^2 + 3y = 0 \)

Rewriting: \( y = -\frac{x^2}{3} \), which is the standard form of a parabola.

Statement I: The equation represents a parabola that opens upwards.

Since \( a = -\frac{1}{3} \) is negative, the parabola opens downwards. Therefore, Statement I is incorrect

Statement II: The axis of the parabola is \( x = 0 \)

In the equation \(x^2 + 3y = 0 \), the variable x  is squared, and there is no linear term in x . Hence, the axis of symmetry is along the y -axis, i.e., x = 0 . Therefore, Statement II is correct

Statement III: The equation of the latus rectum is \( 4y - 3 = 0 \)

The latus rectum's length for a parabola \(y = ax^2\) is\(\frac{4}{|a|} \), where \(a = -\frac{1}{3} \) so the length is \( \frac{4}{\left| -\frac{1}{3} \right|} = 12 \). The equation  4y - 3 = 0  does not represent the latus rectum equation. Therefore, Statement III is incorrect.

Only Statement II is correct.

Therefore, the correct answer is Option 2

Calculation:

Put \( y = 0 \), we get x-intercepts:

\( x = \frac{2a^2}{a^2 + b^2} \)

Put \( x = 0 \), we get y-intercepts:

\( y = \frac{2b^2}{a^2 + b^2} \)

\( \text{Sum} = \frac{2a^2}{a^2 + b^2} + \frac{2b^2}{a^2 + b^2} = 2 \)

Hence, the correct answer is Option 1.