Quantitative Aptitude PYQs

Concept:

Sum of Binomial Coefficients and Greatest Binomial Coefficient:

• The sum of binomial coefficients in the expansion of \( (x + y)^n \) is calculated by substituting x = 1 and y = 1. The result is \(2^n\).
• To find the greatest binomial coefficient, we analyze the coefficients \(C(n, r)\) where r is the term index in the expansion. The greatest coefficient occurs near the middle term(s).
• Key Formulae:
• Sum of binomial coefficients: \( \text{Sum} = 2^n \)
• Binomial coefficient: \( C(n, r) = \frac{n!}{r!(n-r)!} \)
• Greatest binomial coefficient: For even n, it occurs at r = n/2. For odd n, it occurs at r = (n-1)/2 and r = (n+1)/2.

 

Calculation:

Given,

Sum of binomial coefficients = \(2^n = 256\)

We calculate n:

\( 2^n = 256 \)

⇒ \(2^8 = 256\)

Greatest Binomial Coefficient:

For \( n = 8 \) (even), the greatest binomial coefficient occurs at \( r = n/2 = 8/2 = 4 \).

⇒ The term index is r = 4, which corresponds to the 5^th term (since indexing starts from 0).

∴ The greatest binomial coefficient occurs in the 5^th term.

Hence, the correct answer is Option 3.

Calculation:

Given,

The inequality is: \( k < (√{2} + 1)^3 < k + 2 \), where k is a natural number.

We need to compute \((√{2} + 1)^3\)

\( (√{2} + 1)^3 = 2√{2} + 6 + 3√{2} + 1 = 5√{2} + 7 \)

Approximating √ 2≈ 1.414 , we find:

\( 5√{2} + 7 ≈ 5 \times 1.414 + 7 = 7.07 + 7 = 14.07 \)

The inequality becomes:

\( k < 14.07 < k + 2 \)

This implies that k > 12.07 \), so the smallest integer value of k  is:

\( k = 13 \)

∴ The value of k  is 13.

Hence, the correct answer is Option 2.

Calculation:

Multiply Matrix 1 and Matrix 2:

\(\begin{bmatrix} x & 1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix} = \begin{bmatrix} x+4+7 & 2x+5+8 & 3x+6+9 \end{bmatrix}\)

⇒ \(\begin{bmatrix} x+11 & 2x+13 & 3x+15 \end{bmatrix}\)

Multiply the resulting matrix with Matrix 3:

\(\begin{bmatrix} x+11 & 2x+13 & 3x+15 \end{bmatrix} \times \begin{bmatrix} 1 \ 1 \ x \end{bmatrix} \)

\(= (x+11) \cdot 1 + (2x+13) \cdot 1 + (3x+15) \cdot x\)

⇒ \((x+11) + (2x+13) + (3x^2+15x)\)

⇒ \((3x^2 + 18x + 24)\)

Equate the result to 45:

\((3x^2 + 18x + 24 = 45)\)

⇒ \((3x^2 + 18x - 21 = 0)\)

\((x^2 + 6x - 7 = 0)\)

\((x^2 + 7x - x - 7 = 0)\)

⇒ \((x = 1 \text{ or } x = -7)\)

Step 5: Verify:

For \(x = 1\), substitute back:

\((3(1)^2 + 18(1) + 24 = 45)\)

45 =45

∴ The correct value of x is 1.

Hence, the correct answer is Option 4.

Calculation:

Given,

The matrix A is:

\( A = \begin{bmatrix} y & x & x \ z & x & y \ x & y & z \end{bmatrix} \)

Since A  is an orthogonal matrix, we know that:

\( A^T = A^{-1} \quad \Rightarrow \quad A^T A = I \)

This property tells us that A  is orthogonal, and it implies that \(A^T A \) (the product of A's transpose and A is equal to the identity matrix I , which is:

\( A^T A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \)

Now, let’s calculate \(A^T A \) step by step. The transpose of matrix A , denoted \(A^T \) is:

\( A^T = \begin{bmatrix} y & z & x \ x & x & y \ x & y & z \end{bmatrix} \)

Now, we perform matrix multiplication between \(A^T\) and A:

\( A^T A = \begin{bmatrix} y & z & x \ x & x & y \ x & y & z \end{bmatrix} \begin{bmatrix} y & x & x \ z & x & y \ x & y & z \end{bmatrix} \)

Performing this multiplication, we get the following matrix:

\( A^T A = \begin{bmatrix} y^2 + z^2 + x^2 & xy + zx + xy & xz + yz + x^2 \ xy + zx + xy & x^2 + x^2 + y^2 & xy + xz + yz \ xz + yz + x^2 & xy + xz + yz & x^2 + y^2 + z^2 \end{bmatrix} \)

This matrix must be equal to the identity matrix I , which is:

\( \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \)

By comparing the elements of the matrices, we get the following system of equations:

1. \( y^2 + z^2 + x^2 = 1 \) 2. \(xy + zx + xy = 0 \) 3. \(xz + yz + x^2 = 1 \)

Thus, the key result from the orthogonality condition is:

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

Hence, the correct answer is Option 2.

Calculation:

Statement I

\( |M^2| = |M \times M| = |M| \cdot |M| = |M|^2 \)

⇒ Statement I is correct.

Statement II

For a non-singular matrix, \( M \times M^{-1} = I \), where \( I \) is the identity matrix.

\( |M| \cdot |M^{-1}| = |I| = 1 \)

⇒ Statement II is incorrect unless \( |M| = \pm 1 \).

Statement III

The determinant of a matrix is equal to the determinant of its transpose:

\( |M| = |M^T| \)

⇒ Statement III is correct.

Out of the three statements, two are correct: I and III.

Hence, the correct answer is Option 3.

Concept:

Rotation Matrix:

• A rotation matrix is used to perform a rotation in a Euclidean space. It is a square matrix that describes the rotation of a vector space.
• For a 2D rotation, the matrix is given by: \(f(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{bmatrix}\)
• Here, θ is the angle of rotation in radians.
• cos θ: Represents the cosine of the rotation angle.
• sin θ: Represents the sine of the rotation angle.
• Key property of a rotation matrix:
• The transpose of the matrix is equal to its inverse.
• The determinant of the matrix is always equal to 1.
• When θ = π, the rotation matrix becomes: \(f(\pi) = \begin{bmatrix} \cos \pi & \sin \pi \ -\sin \pi & \cos \pi \end{bmatrix} = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}\)

 

Calculation:

Given,

Rotation matrix at θ = π:

\(f(\pi) = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}\)

To find (f(π))², multiply the matrix by itself:

\(f(\pi) × f(\pi) = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix} × \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}\)

Using matrix multiplication:

Top-left element: (-1)(-1) + (0)(0) = 1

Top-right element: (-1)(0) + (0)(-1) = 0

Bottom-left element: (0)(-1) + (-1)(0) = 0

Bottom-right element: (0)(0) + (-1)(-1) = 1

Resulting matrix:

\(f(\pi)^2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\)

∴ (f(π))² is equal to the identity matrix, which is \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\).

Hence, the correct answer is Option 4.

Calculation:

Given,

\(A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}\)

\(\Rightarrow A^2 = A \times A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}\)

 

\(\Rightarrow A^2 = \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix}\)

Now 4A

\(\Rightarrow 4A = 4 \times \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}\)

\(\Rightarrow 4A = \begin{bmatrix} 4 & 8 & 8 \ 8 & 4 & 8 \ 8 & 8 & 4 \end{bmatrix}\)

Also A² - 4A

\(\Rightarrow A^2 - 4A = \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \ 8 & 4 & 8 \ 8 & 8 & 4 \end{bmatrix}\)

\(\Rightarrow A^2 - 4A = \begin{bmatrix} 9-4 & 8-8 & 8-8 \ 8-8 & 9-4 & 8-8 \ 8-8 & 8-8 & 9-4 \end{bmatrix}\)

\(\Rightarrow A^2 - 4A = \begin{bmatrix} 5 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 5 \end{bmatrix}\)

Relate the result to the identity matrix I3I3I3

\(\Rightarrow A^2 - 4A = 5I_3\)

Hence, the correct answer is option 4.

Calculation:

Given,

The number of selections of r things from 5n things is equal to the number of selections of n+r things from 5n things.

This gives the equation:

\( \binom{5n}{r} = \binom{5n}{n+r} \)

Using the symmetry property of binomial coefficients, we know that:

\( \binom{5n}{r} = \binom{5n}{5n-r} \)

Thus, comparing the two sides:

\( r = 5n - (n + r) \)

Simplifying the equation:

\( r = 5n - n - r \)

\( 2r = 4n \)

\( r = 2n \)

∴ The value of r is 2n.

Hence, the correct answer is Option 2.

Calculation:

Given:

Total number of items, n = 6

We need to calculate the total number of selections for r = 0, 1, 2, and 3:

For r = 0

\( ^6C_0 = \frac{6!}{0!(6-0)!} = \frac{1}{1} = 1 \)

⇒ Number of ways to select 0 items = 1

For r = 1

\( ^6C_1 = \frac{6!}{1!(6-1)!} = \frac{6}{1} = 6 \)

⇒ Number of ways to select 1 item = 6

For r = 2

\( ^6C_2 = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \)

⇒ Number of ways to select 2 items = 15

For r = 3

\( ^6C_3 = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \)

⇒ Number of ways to select 3 items = 20

Total number of selections:

\( ^6C_0 + ^6C_1 + ^6C_2 + ^6C_3 = 1 + 6 + 15 + 20 = 42 \)

∴ The total number of selections of at most 3 things from 6 different things is 42.

Hence, the correct answer is Option 4.

Calculation:

Given,

Matrix A is defined as:

\(A = \begin{bmatrix} x & y & z \ y & z & x \ z & x & y \end{bmatrix}\)

It is mentioned that A is an orthogonal matrix. Therefore,

\(A^T A = I\)

Now, we calculate A²:

\(A^2 = A \cdot A\)

\(A^2 = \begin{bmatrix} x & y & z \ y & z & x \ z & x & y \end{bmatrix} \cdot \begin{bmatrix} x & y & z \ y & z & x \ z & x & y \end{bmatrix}\)

Since A is orthogonal, \(A^T A = I\), and hence:

\(A^2 = I\)

∴ A² = Identity matrix (I).

Hence, the Correct answer is Option 2.

Calculation:

Given,

Let\(p = \sin 35^\circ \),\(q = \sin 25^\circ \) and \(r = \sin(-95^\circ) \)

We need to find the value of p + q + r .

First, we know that:

\( \sin(-95^\circ) = -\sin(95^\circ) \).

Thus, the expression p + q + r becomes:

\( p + q + r = \sin 35^\circ + \sin 25^\circ - \sin 95^\circ \).

Using the identity for the sum of two sine functions:

\( \sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \).

Substitute \(A = 35^\circ \) and \(B = 25^\circ \)

\( \sin 35^\circ + \sin 25^\circ = 2 \sin 30^\circ \cos 5^\circ \).

Since \(\sin 30^\circ = \frac{1}{2} \), we get:

\( \sin 35^\circ + \sin 25^\circ = \cos 5^\circ \).

Now, we know that:

\( \sin 95^\circ = \cos 5^\circ \).

Therefore:

\( p + q + r = \cos 5^\circ - \cos 5^\circ = 0 \).

∴ The value of p + q + r  is 0.

Hence, the correct answer is Option 0.

Calculation:

Given,

The expression to simplify is pq + qr + rp, where:

\(p = \sin(35^\circ) \)

\(q = \sin(25^\circ) \)

\(r = \sin(-95^\circ) = -\cos(5^\circ) \)

Thus, we need to evaluate:

\( pq + qr + rp = \sin(35^\circ) \sin(25^\circ) + \sin(25^\circ) (-\cos(5^\circ)) + (-\cos(5^\circ)) \sin(35^\circ) \)

Using trigonometric identities, we simplify the expression:

\( pq + qr + rp = -\sin^2(25^\circ) - \sin(95^\circ) \sin(35^\circ) \)

Using the sum-to-product identity and known values for sine and cosine:

\( = \frac{1}{2} [-2 \sin(25^\circ) - 2 \sin(95^\circ) \cdot \sin(35^\circ)] \)

\( = \frac{1}{2} [-1 + \cos(50^\circ) + \cos(130^\circ) - \cos(60^\circ)] \)

\( = \frac{1}{2} [-1 + \cos(50^\circ) - \cos(50^\circ) - \frac{1}{2}] \)

\( = \frac{1}{2} \left[ -1 - \frac{1}{2} \right] = -\frac{3}{4} \)

∴ The value of pq + qr + rp  is \( -\frac{3}{4} .\)

Hence, the correct answer is Option 1.

Calculation:

Given,

The expression to simplify is \(p^2 + q^2 + r^2\), where:

\(p^2 + q^2 + r^2\)

\(q = \sin(25^\circ) \)

\(r = \sin(-95^\circ) = -\cos(5^\circ) \)

We are given that p + q + r = 0 . Using the identity for squaring a sum:

\( (p + q + r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) \)

Since p + q + r = 0, we substitute this into the identity:

\( 0 = p^2 + q^2 + r^2 + 2(pq + qr + rp) \)

From previous calculations, we know that:

\( pq + qr + rp = -\frac{3}{4} \).

Substitute this value into the equation:

\( 0 = p^2 + q^2 + r^2 + 2\left( -\frac{3}{4} \right) \)

Simplifying:

\( 0 = p^2 + q^2 + r^2 - \frac{3}{2} \)

\( p^2 + q^2 + r^2 = \frac{3}{2} \).

Hence, the correct answer is Option 3.

Calculation:

Given,

We are given the expression:

\( p = | \sin \alpha - \sin(\alpha - 90^\circ) | \)

Using the identity for sine of a difference:

\( \sin(\alpha - 90^\circ) = -\cos \alpha \)

Substituting this identity into the expression for p :

\( p = | \sin \alpha + \cos \alpha | \)

The minimum value of \( | \sin \alpha + \cos \alpha | \) occurs when \(\sin \alpha = -\cos \alpha \), which happens when\(\alpha = 135^\circ \). At this point:

\( \sin 135^\circ = \frac{1}{\sqrt{2}}, \cos 135^\circ = -\frac{1}{\sqrt{2}} \)

Therefore, at \(\alpha = 135^\circ \), the sum of \(\sin \alpha \) and \(\cos \alpha \) is 0, so:

\( p = 0 \)

∴ The minimum value of p  is 0.

Hence, the correct answer is option (a) 0.

Calculation:

Given,

The expression for p  is:

\( p = |\sin \alpha - \sin(\alpha - 90^\circ)| \)

Using the identity for \(\sin(\alpha - 90^\circ) \)

\( \sin(\alpha - 90^\circ) = \cos \alpha \)

Substituting this into the expression for p :

\( p = |\sin \alpha - \cos \alpha| \)

The expression  \(|\sin \alpha - \cos \alpha|\)  achieves its maximum value when the difference between \(\sin \alpha \) and \(\cos \alpha \) is largest.

We can rewrite \(\sin \alpha - \cos \alpha \) as:

\( \sin \alpha - \cos \alpha = \sqrt{2} \left( \sin \left( \alpha - 45^\circ \right) \right) \)

Since \(\sin(\alpha - 45^\circ) \) has a maximum value of 1, the maximum value of \(\sqrt{2} \left( \sin(\alpha - 45^\circ) \right) \) is:

\( \sqrt{2} \)

 

Hence, the correct answer is Option 2

Calculation:

Given,

The sides of the triangle are:

a = 5 cm

b = 7 cm

c = 3 cm

Statement I: The triangle is obtuse-angled.

Using the Cosine Rule:

\( \cos B = \frac{a^2 + c^2 - b^2}{2ac} \)

Substitute the values for a, b = 7 and c

\( \cos B = \frac{5^2 + 3^2 - 7^2}{2 \times 5 \times 3} = \frac{25 + 9 - 49}{30} = \frac{-15}{30} = -\frac{1}{2} \)

Now, calculate angle B:

\( B = \cos^{-1}\left(-\frac{1}{2}\right) = 120^\circ \)

Thus, angle B  is \(120^\circ \), which confirms that the triangle is an obtuse-angled triangle, 

Statement II: The sum of the acute angles of the triangle is also acute.

Since the sum of the angles in a triangle is \(180^\circ \), the sum of the acute angles is:

\( A + C = 180^\circ - 120^\circ = 60^\circ \)

This confirms that the sum of the acute angles is acute,

∴ Both statements are correct.

Hence, the correct answer is Option 3.

Calculation:

Given,

The sides of the triangle are: a = 5 Cm,  b = 7 cm,  c = 3 cm.

 

Apply the Cosine Rule to find angle B:

Using the Cosine Rule:

\( \cos B = \frac{a^2 + c^2 - b^2}{2ac} \)

Substitute the values of a, b and c

\( \cos B = \frac{5^2 + 3^2 - 7^2}{2 \times 5 \times 3} = \frac{25 + 9 - 49}{30} = \frac{-15}{30} = -\frac{1}{2} \)

Now, calculate angle B :

\( B = \cos^{-1}\left(-\frac{1}{2}\right) = 120^\circ \)

Thus, angle B  is \(120^\circ \)

Hence, the correct answer is Option 3.

Calculation:

Given,

The sides of the triangle are: a = 5 cm, c = 3 cm, and the angle B = \(120^\circ \) between sides a and c.

The area of the triangle is given by the formula:

\( \text{Area} = \frac{1}{2} \times a \times c \times \sin B \)

\( \text{Area} = \frac{1}{2} \times 5 \times 3 \times \sin(120^\circ) \)

We know that\(\sin(120^\circ) = \frac{\sqrt{3}}{2} \), so:

\( \text{Area} = \frac{1}{2} \times 5 \times 3 \times \frac{\sqrt{3}}{2} \)

\( \text{Area} = \frac{1}{2} \times 15 \times \frac{\sqrt{3}}{2} = \frac{15 \times \sqrt{3}}{4} \)

∴ The area of the triangle is \(\frac{15\sqrt{3}}{4} \) square cm.

Hence, the Correct answer is Option 1.

Calculation:

The angles of elevation are:

 

From point P,\(\theta = 30^\circ \); from point Q, \(\theta = 45^\circ \); and from point R, \(\theta = 60^\circ \)

Using the tangent formula for each angle:

\( \tan(60^\circ) = \frac{h}{x} \implies h = \sqrt{3}x \)

\( \tan(45^\circ) = \frac{h}{QN} \implies h = QN\)

\( \tan(30^\circ) = \frac{h}{PN} \implies PN = h{\sqrt{3}} \)

from the figure PN = h + a 

\(h+a = h\sqrt3\)

\(h = \frac{a}{\sqrt3 -1} \implies \frac{a \sqrt3 +1}{2}\)

= \((\frac{3 + \sqrt3}{2}) a\)

Hence, the correct answer is Option 2.

Calculation

 

\( \tan(60^\circ) = \frac{h}{x} \Rightarrow h = \sqrt{3}x \)

\( \tan(45^\circ) = \frac{h}{b + x} \Rightarrow h = b + x \)

\( h = b + x \Rightarrow \sqrt{3}x = b + x \)

\( x = \frac{b(\sqrt{3} + 1)}{2} \)

\( h = MN = \sqrt{3}x = \left( \frac{3 + \sqrt{3}}{2} \right) b \)

Hence, the correct answer is Option 1.