Quantitative Aptitude PYQs

Concept:

If ω is a non-real cube root of 1, that is, ω³ = 1, then

1 + ω + ω² = 0  ------(1)

and \(\omega = {-1 + i √{3} \over 2}\)

If Z = x + iy then |Z| = √(x² + y²)

Calculation:

Given, ω is a non-real cube root of 1, then

⇒ \(\left|\frac{1-\omega}{\omega+\omega^2}\right| = \left|\frac{1-\omega}{-1}\right| \) ....{From equation (1)}

⇒ \(\left|\frac{1-\omega}{\omega+\omega^2}\right| = \left|{\omega - 1}\right| \)

⇒ \(\left|\frac{1-\omega}{\omega+\omega^2}\right| = \left|{{-1 + i √{3} \over 2} - 1}\right| \)

⇒ \(\left|\frac{1-\omega}{\omega+\omega^2}\right| = \left|{{-3 + i √{3} \over 2} }\right| \)

⇒ \(\left|\frac{1-\omega}{\omega+\omega^2}\right| = \sqrt{[{9 \over 4} + { 3 \over 4}]} = \sqrt 3 \)

∴ The correct option is (1).

Explanation:

If a 6-digit number is formed only by using 0, 1, 2, 3, 4 and 5 (each once), then every digit is used once as six digits are given.

Since the number is 6-digit so 0 should not be at starting place.

and the number is divisible by 6, so it is divisible by 2 and 3.

So only even digit {0,2,4} occur at last place for divisible by 2.

And the number always contains the digits 0,1,2,3,4,5 once, so sum of digits = 15, so such a number will always be divisible by 3.

Case 1: The number ends with 0.

Number of choices for first place = 5

And for second, third, fourth and fifth number of choices will be 4, 3, 2, and 1 respectively.

So, the number of 6 digit numbers which are ending with 0 with given conditions = 5 × 4 × 3 × 2 × 1 = 120

Case 2: The number ends with 2.

Number of choices for first place = 4 {excluding 2 and 0}

And for second, third, fourth and fifth number of choices will be 4, 3, 2, and 1 respectively.

So, the number of 6 digit numbers which are ending with 0 with given conditions = 4 × 4 × 3 × 2 × 1 = 96

Case 3: The number ends with 4.

Number of choices for first place = 4 {excluding 4 and 0}

And for second, third, fourth and fifth number of choices will be 4, 3, 2, and 1 respectively.

So, the number of 6 digit numbers which are ending with 0 with given conditions = 4 × 4 × 3 × 2 × 1 = 96

So total number of ways = 120 + 96 + 96 = 312

∴ The correct option is (4).

Concept:

Decimal number is with base 10 and binary number is with base 2.

Calculation:

Divide the number repeatedly by 2 until the quotient becomes 0.

Write the remainder from bottom to top.

(1011)₁₀ = (1111110011)₂

∴ The correct option is (4).

Concept:

If A is a matrix of order n × n, then

• |k A| = kⁿ |A|
• |adj(A)| = |A|^n-1

 

Calculation:

Given, A is the matrix of order 3 × 3

then, |2adj(3A)| = 2³ |adj(3A)|   {∵ order of adj(3A) is 3 × 3}

⇒ |2adj(3A)| = 23 |3A|^3-1 =  23 |3A|2 

⇒ |2adj(3A)| = 23 (3³|A|)²  = 23 36 4²

⇒ |2adj(3A)| = 2⁷3⁶ = 2α 3β, 

then α = 7 and β = 6

⇒ α + β = 7 + 6 = 13

∴ The correct option is (2).

Concept:

1) The roots of unity (ω):

ω = \({-1 + i√{3} \over 2}\), ω² = \({-1 - i√{3} \over 2}\)

2)  1 + ω + ω² = 0

3)  |x + iy| = √(x² + y²)

Calculation:

Given equation x2 - x + 1 = 0,

⇒ \(x = {1 \pm √{1-4} \over 2}\)

⇒ \(x = {1 \pm i√{3} \over 2}\)

⇒ \(x =- ω, -ω^2\)

where \(ω\) is the cube root of the unity, that is \(ω\)³ = 1

Let α = -\(ω\) and β = -\(ω\)²

⇒  α¹⁰⁰ = \(ω^{100} = ω^{99}.ω = (ω^3)^{33}.ω = ω\) 

and β¹⁰⁰ = \(ω^{200} = ω^{198}.ω^2 = (ω^3)^{66}.ω^2 = ω^2\)

So, \(\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|\) = \(\left|\frac{ω+ω^{2}}{ω-ω^{2}}\right|\)

So, \(\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|\) = \(\left|\frac{ω(1+ω)}{ω(1-ω)}\right|\)

⇒ \(\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|\) = \(\left|\frac{1+ω}{1-ω}\right|\)

⇒ \(\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|\) = \(\left|\frac{1+{-1 + i√{3} \over 2}}{1-{-1 + i√{3} \over 2}}\right|\)

⇒ \(\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|\) = \(\left|\frac{1 + i√{3} }{3 - i√{3} }\right|\)

Rationalizing the denominator,

⇒ \(\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|\) = \(\left|\frac{1 + i√{3} }{3 - i√{3} }\right|\left|\frac{3+ i√{3} }{3+ i√{3} }\right|\)

⇒ \(\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|\) = \(\left|\frac{3 + i√{3}+3i\sqrt {3}-3 }{9+3 }\right|\)

⇒ \(\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|\) = \(\left|\frac{4i\sqrt 3}{12}\times \frac{i}{i}\right|=\left|\frac{i}{\sqrt 3}\right|\)

⇒ \(\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|\) = \(\sqrt{[0 + (\frac{1}{\sqrt 3})^2]} = \frac{1}{√ 3}\)

∴ The correct option is (4).

Concept:

• If A is a symmetric matrix then, A = A', where A' is the transpose of A
• If A is a skew-symmetric matrix then, A = -A', where A' is the transpose of A
• (AB)' = B'A'
• The diagonal of a skew-symmetric matrix consists of zero elements and therefore the sum of elements in the main diagonals is equal to zero.
• The sum of non-diagonal entries of the skew-symmetric matrix is zero.

 

Calculation:

Given, A and B be symmetric matrices of the same order,

⇒ A = A' and B = B' __(1)

Now (AB - BA)' = (AB)' - (BA)' 

⇒ (AB - BA)' = B'A' - A'B'

⇒ (AB - BA)' = BA - AB  (From (1))

⇒ (AB - BA)' = - (AB - BA) 

So (AB - BA) is a skew-symmetric matrix

By using the properties discussed above, 

Diagonal entries of (AB - BA) are zero.

The sum of non-diagonal entries of the skew-symmetric matrix is zero.

∴ The correct statement is (2).

Explanation:

**Statement

1. AB = AC ⇒ B = C if A is non-singular**

If AB = AC, and A is non singular,

Multiply both sides by A^-1 on left,

⇒ A^-1AB = A^-1AC

⇒ B = C

So statement 1 is true.

Statement 2. If BX = CX for every column matrix X having n rows then B = C

If BX = CX for every column matrix X having n rows 

let X be from the Euclidean basis B = {e₁,e₂,....e_n}

then Ae_i = Be_i 

⇒ i^th column of A = i^th column of B for 1 ≤ i ≤ n

Hence A = B

So statement 2 is true.

∴ The correct option is (3).

Given  - 

x + 2y + z = 4, 2x + 4y + 2z = 8 and 3x + 6y + 3z = 10

Concept - 

ρ (A) =  number of no zero row of the matrix A.

if \(ρ(A) \neq ρ(A:B)\) then no solution exist.

Explanation - 

matrix representation of the given equations -

\(\left[ \begin{array}{ccc|c} 1 & 2 & 1 & 4 \ 2 & 4 & 2 & 8 \ 3 & 6 & 3 & 10 \end{array} \right]\)

Now use \(R_2 \to R_2-2R_1 \ \ and \ \ R_3 \to R_3-3R_1 \)

\(\left[ \begin{array}{ccc|c} 1 & 2 & 1 & 4 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & -2 \end{array} \right]\)

So ρ (A) = 1 because in the first part of the matrix only one non zero row exist.

and ρ(A:B) = 3 because in the  matrix three non zero row exist.

Hence \(ρ(A) \neq ρ(A:B)\)

So no solution exist.

Important Points Crammer's rule is applied only when determinant of the matrix is not Zero.

Concept:

In a system of linear equations, if we have two distinct solutions, say X₁ and

X₂, then any linear combination of these solutions, such as aX₁ + bX₂, will

also, be a solution to the system if certain conditions are met.

Calculation:

Given that X₁ and X₂ are distinct solutions of the system AX = B, we can

write the following equations:

AX1 = B        ---(1)

AX2 = B        ---(2)

Now, let's consider the combination aX₁ + bX₂.

We want to check if this combination satisfies the equation AX = B.

Substituting X = aX₁ + bX₂ into AX = B, we have:

A × (aX₁ + bX₂) = B

Using the linearity property of matrix multiplication, we can distribute the

matrix A over the linear combination:

a × (AX₁) + b(AX₂) = B

Substituting equations (1) and (2) into the above equation, we get:

aB + bB = B

a + b = 1

Therefore, the relation between a and b is that their sum is equal to 1.

Calculation:

Given, \(\left|\begin{array}{ccc} 0 & x-a & x-b \ 0 & 0 & x-c \ x+b & x+c & 1 \end{array}\right|=0 \)

Expanding along column 1,

0 + 0 + (x + b)[(x - a)(x - c) - 0(x - b)] = 0

⇒ (x + b)(x - a)(x - c) = 0

So, the roots of the equation are -b, a, c.

⇒ The sum of the roots = a - b + c

∴ The correct option is (2).

Concept:

The roots of a quadratic equation with real coefficients are either real or occur in conjugate pairs(if complex).

For the quadratic equation ax² + bx + c = 0, the sum of roots = -b/a and product of roots = c/a

Calculation:

Given, 2 - i√3 where i = √-1 is a root of the equation x2 + ax + b = 0,

Since complex roots occur in conjugate pairs, 

⇒ 2 + i√3 is also the root of the equation.

So the roots of the equation are 2 + i√3 and 2 - i√3.

⇒ The sum of roots = 4 = -a/1

⇒ a = -4

And the product of roots = (2 + i√3)(2 - i√3) = (4 + 3) = b/1

⇒ b = 7

⇒ a + b = -4 + 7 = 3

∴ The correct option is (4).

Concept:

If z = a + ib then arg(z) = tan^-1(b/a)

Calculation:

Given, \(z=\frac{1+i √{3}}{1-i √{3}}\)

Multiplying numerator and denominator by (1 + i√3)

⇒ \(z=\frac{1+i √{3}}{1-i √{3}} \times \frac{1+i √{3}}{1+i √{3}}\)

⇒ \(z=\frac{(1+i √{3})^2}{1-(i √{3})^2}\)

⇒ \(z=\frac{1+(\sqrt 3 i)^2+ 2 i √{3}}{1+ 3}\)

⇒ \(z=\frac{1 - 3+ 2 i √{3}}{4} \)

⇒ \(z={-1 \over 2} + i {\sqrt 3 \over 2}\) 

So, arg(z) = \(\tan^{-1} {{\sqrt 3 \over 2} \over { 1\over 2} } = \tan^{-1} \sqrt 3\) in second quadrant

(where x is negative and y is positive, that is second quadrant.)

⇒ arg(z) = π - \(\frac{\pi}{3}\) = \(\frac{2\pi}{3}\)

∴ The correct option is (2).

Concept:

If three terms a, b, c are in AP, then 2b = a + c

Calculation:

Given, a, b, c are in AP, then 2b = a + c

Given determinant,

\(\left|\begin{array}{lll} x+1 & x+2 & x+3 \ x+2 & x+3 & x+4 \ x+a & x+b & x+c \end{array}\right|\)

Apply C₁ → C₁ + C₃ - 2C₂,

⇒ \(\left|\begin{array}{lll} 0 & x+2 & x+3 \ 0 & x+3 & x+4 \ a+ c- 2b & x+b & x+c \end{array}\right|\)

Since a + c = 2b

⇒ \(\left|\begin{array}{lll} 0 & x+2 & x+3 \ 0 & x+3 & x+4 \0 & x+b & x+c \end{array}\right|\)

If any row or column of matrix is 0 then, detreminant equals to 0.

∴ The correct option is (2).

Concept:

If three terms p, q, r are in GP, then pr = q2.

 Property of log- \(\log_q p = \frac{\log p}{\log q} \) for usual log

Calculation:

Given, logxa, ax and logbx are in GP, 

⇒ \((a^x)^2=\log_x a . \log_b x\)

⇒ \(a^{2x}={\log a \over \log x}.{ \log x \over \log b}\) 

⇒ \(a^{2x}={\log a \over \log b}\)

⇒ \(a^{2x}={\log_b a}\)

⇒ \({2x}=\log_a({\log_b a})\)

⇒ x = \(\frac{\log _a\left(\log _b a\right)}{2}\)

∴ The correct option is (3).

Concept:

If three terms p, q, r are in GP, then pr = q².

If three terms a, b, c are in AP, then 2b = a + c

Calculation:

Given, \(2^{\frac{1}{c}}, 2^{\frac{b}{a c}}, 2^{\frac{1}{a}}\) are in GP, 

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

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

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

⇒ 2b = a + c

⇒ a, b, c are in AP

∴ The correct option is (1).

Concept:

If a is the first term of AP and d is the common difference between terms then, 

n^th term of the AP is given by, a_n = a + (n - 1)d

Calculation:

Given,

The first and the second terms of an AP are \(\frac{5}{2}\) and \(\frac{23}{12}\) respectively

⇒ a = \(\frac{5}{2}\) and d =  \(\frac{23}{12}\) - \(\frac{5}{2}\) = \(\frac{-7}{12}\)

The largest negative term will be the first negative term.

T_n < 0

⇒ a + (n - 1) < 0

⇒ \(\frac{5}{2}\) + (n - 1)\(\frac{-7}{12}\) < 0

⇒ \(\frac{5}{2}\) + \(\frac{7}{12}\) - \(\frac{7n}{12}\) < 0

⇒ (30 + 7)/12 - 7n/12 < 0

⇒ 37/12 - 7n/12 < 0

⇒ 37 - 7n < 0

⇒ 7n - 37 > 0

⇒ n > 37/7 or n > 5.2

But n will be the integer (as it is the term of AP)

⇒ n = 6 

∴ The correct option is (2).

Concept:

For ax² + bx + c = 0 to have real roots, 

Δ = b² - 4ac ≥ 0

Calculation:

Given, x2 - 4x + k = 0

For roots to exist, (- 4)² - 4k ≥ 0

⇒ 16 - 4k ≥ 0

⇒ k ≤ 4 ...(i)

Equation given is, x2 - 4x + k = 0

⇒ x = \({4 ± \sqrt{16-4k} \over 2}\)

According to the question, the roots lie between (0, 5)

⇒ 0 < \({4 ± \sqrt{16-4k} \over 2}\) < 5

⇒ 0 < \(4 ± \sqrt{16-4k}\) < 10

⇒ - 4 < \(± \sqrt{16-4k}\) < 6

⇒ 0 < 16 - 4k < 36

⇒ - 16 < - 4k < 20

⇒ - 4 < - k < 5

⇒ - 5 < k < 4 ...(ii)

Using (i) and (ii), the possible values of k are  1, 2, 3.

∴ Number of possible values of k is 3

Concept:

For an AP with first term a and common difference d,

n^th term is a_n = a + (n - 1)d

And sum of first n terms = S_n = \({n \over 2} [2a + (n-1)d]\)

Calculation:

Given, first term = x and S_n = 0

⇒ \({n \over 2} [2x + (n-1)d]\) = 0

⇒ 2x + (n - 1)d = 0 __(1)

Sum of next m terms = S_m+n - S_n = S_m

⇒ Sum of next m terms = Sm+n = \({m+n \over 2} [2x + (m+n-1)d]\)

⇒ Sum of next m terms = \({m+n \over 2} [2x + (n-1)d + md]\)

⇒ Sum of next m terms = \({m+n \over 2} [md]\)   { From equation (1)}

Putting the value of d from (1),

⇒ Sum of next m terms = \({m+n \over 2}m {(-2x) \over n-1}\) 

⇒ Sum of next m terms = \(\frac{m x(m+n)}{1-n}\)

∴ The correct option is (2).

Calculation:

Statement:1 

25! = 25 × 24 × 23 × .....× 13 .....× 2 × 1

Since 13 & 2 are the factor of 25!, it is divisible by 26.

Now,

Remainder of (25! + 1)/26 = Rem (25!/26) + Rem(1/26)

Remainder of (25! + 1)/26 = 0 + 1 = 1

∴ 25! + 1 is not divisible by 26.

Statement:2

6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

6! + 1 = 720 + 1 = 721 

Which is divisible by 7.

∴ Only statement 2 is correct.

Concept:

If a number ω = a + ib is purely imaginary, then

• a = 0
• and ω = -\(\overline {\omega}\)

Calculation:

Given, z is a complex number such that \(\frac{z-1}{z+1}\) is purely imaginary,

Let \(\frac{z-1}{z+1}\) = a + ib, then a = 0

And \(\frac{z-1}{z+1} = - \overline{(\frac{z-1}{z+1})}\) ⇒ \(\frac{z-1}{z+1} = -(\frac{ \overline{z}-1}{ \overline{z}+1})\)

⇒ \(({z-1})({ \overline{z}+1}) = -({ \overline{z}-1})({z+1})\) ⇒ \(({z \overline{z}+z }- { \overline{z}-1}) = -({z \overline{z}-z }+ { \overline{z}-1})\) 

⇒ \(|z|^2 -1 = -|z|^2 + 1\) ⇒ |z|2 = 1 ⇒ |z| = 1

∴ The correct option is (3).