Quantitative Aptitude PYQs

Explanation -

Given:

|A| = \(\frac{1}{2\sqrt{2}} \)  and |B| =\( \frac{1}{729}\) 

1. Property of Adjugate:

\(\text{adj}(A) = |A|^{n-1} \cdot A^{-1}\)

For a 3 x 3 matrix A , \(\text{adj}(A) = |A|^2 \cdot A^{-1} \)

2. Determine \(\text{adj}(3A):\)

\(\text{adj}(3A) = (3A)^{*} = 3^{n-1} \cdot \text{adj}(A)\)

For a 3 x 3 matrix, \(\text{adj}(3A) = 3^2 \cdot \text{adj}(A) = 9 \cdot \text{adj}(A) \)

3. Determine |adj(A)|:

|adj(A)| = |A|^{n - 1}

For a 3 x 3 matrix, |adj(A)| = |A|²

Given \(|A| = \frac{1}{2\sqrt{2}} \),

\(|\text{adj}(A)| = \left( \frac{1}{2\sqrt{2}} \right)^2 = \frac{1}{8} \)

4. Determine |adj(3A)|:

\(|\text{adj}(3A)| = 9^3 \cdot |\text{adj}(A)| = 9^3 \cdot \frac{1}{8} = \frac{9^3}{8} \)

5. Combine the Determinants: 

\(|2B(\text{adj}(3A))| = 2^3 \cdot |B| \cdot |\text{adj}(3A)| = 8 \cdot \frac{1}{729} \cdot \frac{9^3}{8}\)  

\(|2B(\text{adj}(3A))| = \frac{8 \cdot 9^3}{729 \cdot 8} = \frac{729}{729} = 1\)

So, the value of |2B(adj(3A))| is 1.

Explanation:

z is any complex number and iz3 + z2 - z + i = 0, where \(i=\sqrt{-1}\)

If z = i then 

iz3 + z2 - z + i = i(i)3 + (i)2 - i + i = 1 - 1 - i + i = 0

So, z = i satisfying the given equation.

Now, |z| = 1

So, (|z| + 1)2 = (1 + 1)² = 2² = 4.

Hence option (2) is true.

Concept -

If number of digit is n then the sum of all n digit numbers formed by using n digits without repetition of digits is given by 

(n-1)! × sum of digits × (11111111.... n times)

Explanation -

Now we have the digits 0, 1, 4, 5

So total digits = n = 4

Now Using the formula -

The sum of all four digit numbers formed by using all digits 0, 1, 4, 5 is

= (4-1)! × (0 + 1 + 4 + 5) × (1111)

= 3! × 10 × 1111

= 66660

But in this sum, there are the numbers which starts from 0 but those are not 4 digit number so we have to subtract the sum of those number from this sum.

Now for digits 1, 4, 5

We have n = 3

So total sum = 2! × 10 × 111 = 20 × 111 = 2220

Now the sum of all four digit numbers = 66660 - 2220 =64440

Hence Option (4) is correct.

Explanation -

The problem of finding the value of xy + yz + zx when x, y and z  are the cube roots of unity, we need to use some properties of the cube roots of unity.

Properties of Cube Roots of Unity:

The cube roots of unity are solutions to the equation x³ = 1

These roots are 1, ω and ω² where ω is a complex number satisfying:

ω = \(e^{2\pi i / 3}\)

ω³ = 1

1 + ω + ω² = 0

Let x = 1 , y = ω , and z = ω² .

xy + yz + zx = \((1 \cdot ω) + (ω \cdot ω^2) + (ω^2 \cdot 1)\)

Simplify each term:

xy = 1 . ω = ω

yz = ω . ω² = ω³ = 1  (since ω³ = 1)

zx = ω² . 1 = ω²

4. Sum the Terms:

xy + yz + zx = ω + 1 + ω²

We know from the properties of the cube roots of unity that:

1 + ω + ω² = 0

Therefore:

xy + yz + zx = 0

So the correct answer is 0.

Explanation -

There are 4 possibilities

(i) 3 ladies from husband's side and 3 gentleman from wife's side .

No. of ways in this case = \(4_{C_3} \times 4_{C_3} = 4 \times 4 =1 6\)

ii) 3 gentleman from husband's side and 3 ladies from wife's side.

No. of ways in this case = \(3_{C_3} \times 3_{C_3} = 1 \times 1 =1 \)

iii) Two ladies & one gentleman from husband's side and one lady and two gentleman from wife's side.

No. of ways in this case = \((4_{C_2} \times 3_{C_1}) ( 3_{C_1} \times 4_{C_2} ) = (3 \times 6)^2 = 324\)

iv) One lady and two gentlemen from husband's side and 2 ladies and one gentleman from wife's side.

No. of ways in this case = \((4_{C_1} \times 3_{C_2}) ( 3_{C_2} \times 4_{C_1} ) = (4 \times 3)^2 =144\)

Total no. of ways are = 16 + 1 + 324 + 144 = 485

Explanation -

The number of triangles that can be formed using points on the sides of a triangle PQR, we need to consider the given points and use combinatorial counting methods.

Given:

 3 points on side PQ

 4 points on side QR

 5 points on side RP

Total points on the sides: 3 + 4 + 5 = 12 points.

Combinatorial Approach:

To form a triangle, we need to select any 3 points out of these 12 points such that no three points lie on the same line. 

Total Ways to Select 3 Points Out of 12:

\(\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220\)

We need to subtract the number of ways we can select 3 collinear points (i.e., all points from the same side).

Invalid Selections for Each Side:

For PQ: Select 3 out of 3.

\(\binom{3}{3} = 1\)

For QR: Select 3 out of 4.

\(\binom{4}{3} = 4\)

For RP: Select 3 out of 5.

\(\binom{5}{3} = 10\)

Total Invalid Selections:

1 + 4 + 10 = 15

Total Valid Selections = \(\binom{12}{3}\) - Total Invalid Selections

= 220 - 15 = 205

So, the number of triangles that can be formed using the points as vertices is 205.

Explanation -

Given:

\(\log_b a = p\) , \( \log_d c = 2p\) and  \(\log_f e = 3p\)  

Convert logarithmic forms to exponential forms:

\(\log_b a = p \implies a = b^p\)  

⇒\( \log_d c = 2p \implies c = d^{2p}\)

⇒ \(\log_f e = 3p \implies e = f^{3p}\)

Substitute the exponential forms into ace :

\(ace = (b^p)(d^{2p})(f^{3p})\)

\(ace = b^p \cdot d^{2p} \cdot f^{3p}\)

\((ace)^{\frac{1}{p}} = (b^p \cdot d^{2p} \cdot f^{3p})^{\frac{1}{p}}\)

⇒ \((ace)^{\frac{1}{p}} = (b^p)^{\frac{1}{p}} \cdot (d^{2p})^{\frac{1}{p}} \cdot (f^{3p})^{\frac{1}{p}} \)

= \(b^{\frac{p}{p}} \cdot d^{\frac{2p}{p}} \cdot f^{\frac{3p}{p}}\)

=\( b^1 \cdot d^2 \cdot f^3\)

= \(bd^2 f^3\)

So, the value of \((ace)^{\frac{1}{p}}\) is \(bd^2 f^3\).

Concept -

The given roots are -√2 and √3, and we know that if a polynomial has real coefficients and a non-real root, its conjugate must also be a root.

Explanation -

Given Roots:

Roots: -√2, √3

Conjugate Roots (since coefficients are real):

Conjugate roots:  √2, -√3

The polynomial with these roots can be expressed as:

(x + √2)(x - √2)(x + √3)(x - √3)

⇒ (x² - 2)(x² - 3)

= \(x^4 - 3x^2 - 2x^2 + 6 = x^4 - 5x^2 + 6\)

The given polynomial is:

\(a_0 + a_1 x + a_2 x^2 + a_3 x^3 + x^4 = 0\)

By comparing coefficients:

The coefficient of x⁴ is 1.

The coefficient of x³ is 0.

The coefficient of x² is -5.

The coefficient of x is 0.

The constant term is 6.

Therefore:

\(a_0 = 6, \quad a_1 = 0, \quad a_2 = -5, \quad a_3 = 0\)

Hence Option(3) is correct.

Explanation:

Let z1 = 1 and z2 = i then

\(\rm \left|\frac{z_1+z_2}{z_1-z_2}\right|\) = \(\rm \left|\frac{1+i}{1-i}\right|\) = \({\sqrt 2\over \sqrt2}\) = 1

So, \(\rm \left|\frac{z_1+z_2}{z_1-z_2}\right|=1\) satisfying.

Now, \(\operatorname{Re}\left(\frac{z_1}{z_2}\right)+1 \) = \(\operatorname{Re}\left(\frac{1}{i}\right)+1 \) = Re(-i) + 1 = 0 + 1 = 1

Hence option (3) is true.

Explanation:

\([{26\over2}]+[{26\over2^2}]+[{26\over2^3}]+[{26\over2^4}]\)

= 13 + 6 + 3 + 1 ([.] is greatest integer function)

= 23

then

26! = a2²³

⇒ 26! = a(2³)⁷× 2²

⇒ 26! = 4a × 8⁷

Comparing with 26! = n8k we get maximum value of k is 7.

Hence option (2) is true.

Concept:

(i) Matrix multiplication is not communicative i.e., AB ≠ BA for two square non-singular matrix A and B.

(ii) For any non-singular matrix A of order n, A(adj A) = |A|× I_n.

(iii) adj(AB) = (adj B)(adj A)

Explanation:

we know that, 

adj(AB) = (adj B)(adj A)

and adj(BA) = (adj A)(adj B)

So, 1. and 2. are false.

A and B are non-singular matrices so, AB is non-singular.

So, using the formula A(adj A) = |A|× In we get

(AB) adj (AB) = |AB|In 

So, (AB) adj (AB) - |AB|In = 0 

i.e., (AB) adj (AB) - |AB|In is a null matrix of order n.

3. is true.

Hence option (2) is correct.

Concept:

Let A be a square matrix then (adj A)^T = adj(A^T)

Explanation:

1. A(adj AT) = A(adj A)T (as (adj A)T = adj(AT))

1. is true.

2. A2 = A

⇒ A2 - A = 0

⇒ A(A - I) = 0

⇒ A = 0 or A = I

Now, given A is non-singular matrix so A ≠ 0, A = I i.e., A is identity matrix of order n.

2. is true.

3. A3 = A

⇒ A3 - A = 0

⇒ A(A² - I) = 0

⇒ A(A + I)(A - I) = 0

⇒ A = 0, A = I, A = - I

Since A is non-singular matrix so A ≠ 0, 

So, its not necessary that A is identity matrix of order n.

3. is false.

Hence option (1) is true.

Concept -

1. The first digit (thousands place) cannot be 0 (since it wouldn't be a four-digit number).
2. The remaining three digits (hundreds, tens, and units place) can be any of the even digits.

Explanation -

1. First digit (thousands place):

   - Possible digits: 2, 4, 6, 8 (4 choices) 2. Remaining three digits (hundreds, tens, and units places):

   - Possible digits: 0, 2, 4, 6, 8 (5 choices for each)

Total number of four-digit natural numbers:

4 × 5 × 5 × 5 = 4 × 125 = 500

Thus, the total number of four-digit natural numbers such that all of the digits are even is 500.

Explanation -

The given equation \((z - 100)^3 + 1000 = 0\),

⇒ \((z - 100)^3 = -1000\)

Next, we recognize that -1000 can be written as -10³.

Thus, we have:

\((z - 100)^3 = -(10)^3\)

Taking the cube root of both sides, we get:

\(z - 100 = ω \cdot (-10)\)

where ω is a cube root of unity (ω³ = 1) and ω ≠ 1. The cube roots of unity are 1, ω, and ω², where ω and ω² are complex numbers and ω² = ω^-1.

So, the possible values for z - 100 are:

z - 100 = -10, -10ω, -10ω²

Solving for z, we get:

z = 100 - 10

z = 100 - 10ω

z = 100 - 10ω²

Therefore, the solutions are:

z = 90, 100 - 10ω, 100 - 10ω²

⇒ z = 90, 10(10 - ω), 10(10 - ω2)

Hence Option (2) is correct.

Concept -

(1 + x)2 = 1 + x² + 2x and i² = -1 

Explanation -

(1 + i)2 = 1 + i² + 2i = 1 - 1 + 2i = 2i

Now (1 + i)4 = 2i x 2i = 4 i² = - 4

and (1 - i)4 = -4

Put these values in given equation we get, 

(1 + i)4 + (1 - i)4  = - 4 - 4 = - 8

Hence Option (4) is correct.

Concept:

(i) A square matrix A is said to be skew-symmetric matrix if A^T = -A.

(ii) A matrix A is said to be orthogonal matrix if AA^T = I i.e., A^T = A^-1, where I is the identity matrix.

Explanation:

A is a skew-symmetric matrix A of order 3.

So, AT = - A

From the property of skew-symmetric matrix we know that,

All the elements of main diagonal elements of skew symmetric matrix are zero.

So, sum of the elements of all the diagonal elements of the matrix is zero. 

Therefore, 1. and 2. are true.

A can't be orthogonal matrix as it is possible only when AT = A-1.

So, 3. is false.

Hence option (1) is correct.

Concept Used:

A number is divisible by 4 if the last two digits form a number that is divisible by 4.

Solution:

The digits available are 1, 2, 3, and 5. Let's check the possible pairs:

• 12 - Divisible by 4
• 21 - Not divisible by 4
• 13 - Not divisible by 4
• 31 - Not divisible by 4
• 15 - Not divisible by 4
• 51 - Not divisible by 4
• 23 - Not divisible by 4
• 32 - Divisible by 4
• 25 - Not divisible by 4
• 52 - Divisible by 4
• 35 - Not divisible by 4
• 53 - Not divisible by 4

 

So, the valid pairs for the last two digits are 12, 32 and 52.

Now, let's calculate the number of four-digit numbers that can be formed with these pairs:

1. For the pair 32:

• We have two remaining digits: 1 and 5.

• The number of permutations of these two digits in the first two places is 2!=22!=22!=2

  .

• So, there are 2 numbers: 1532, 5132.

2. For the pair 52:

• We have two remaining digits: 1 and 3.

• The number of permutations of these two digits in the first two places is 2!=22!=22!=2

  .

• So, there are 2 numbers: 1352, 3152.

3. For the pair 12:

• We have two remaining digits: 3 and 5.
• The number of permutations of these two digits in the first two places is 2!=22!=22!=2

.

• So, there are 2 numbers: 3512, 5312.

 

Therefore, the total number of four-digit numbers that can be formed and are divisible by 4 is 2 + 2 + 2 = 6

Hence Option (4) is correct.

Concept -

Fermat's Little Theorem states that if p is a prime number and a is an integer such that a is not divisible by p, then:

\(a^{p-1} \equiv 1 \pmod{p}\)

Explanation -

For our problem: a = 2 and p = 7

According to Fermat's Little Theorem:

\(2^{7-1} = 2^6 \equiv 1 \pmod{7}\)

This implies that:

\(2^6 \equiv 1 \pmod{7}\)

We need to find the remainder when 2¹²⁰ is divided by 7.

Notice that 120 can be expressed as a multiple of 6:

120 = 20 ×  6

Using this, we can write:

\(2^{120} = (2^6)^{20}\)

Since \(2^6 \equiv 1 \pmod{7}\):

\((2^6)^{20} \equiv 1^{20} \pmod{7}\)

And since 1²⁰ = 1:

\(2^{120} \equiv 1 \pmod{7}\)

Thus, the remainder when 2¹²⁰ is divided by 7 is 1.

Concept:

Combination formula:

C(n, r) + C(n, r - 1) = C(n+1, r)

Explanation:

\(\left|\begin{array}{ccc} C(9,4) & C(9,3) & C(10, n-2) \ C(11,6) & C(11,5) & C(12, n) \ C(m, 7) & C(m, 6) & C(m+1, n+1) \end{array}\right|\) = 0

Using C₁ = C₁ + C₂ and using the formula

\(\left|\begin{array}{ccc} C(10,4) & C(9,3) & C(10, n-2) \ C(12,6) & C(11,5) & C(12, n) \ C(m+1, 7) & C(m, 6) & C(m+1, n+1) \end{array}\right|\) = 0

This determinant will be zero if 1^st and 3^rd column are same,

which is possible only when

n - 2 = 4 i.e., n = 6

Option (3) is correct

Explanation:

\(\left|\begin{array}{ccc} \cos C & \sin B & 0 \ \tan A & 0 & \sin B \ 0 & \tan (B+C) & \cos C \end{array}\right|\)

Let A = 30^∘, B = 90∘, C = 60∘ then the determinant is

\(\left|\begin{array}{ccc} \frac12 &1 & 0 \ \frac1{\sqrt3} & 0 & 1 \ 0 & -\frac1{\sqrt3} &\frac12 \end{array}\right|\)

= \(\frac12(0+\frac1{\sqrt3})\) + 1\((0-\frac1{2\sqrt3})\)

= 0 

Hence option (2) is true.