Quantitative Aptitude PYQs

Given:

z z̅ = |z + z̅ |,  where z = x + iy, i = \(√{-1}\), 

Formula Used: 

(a + b) (a - b) = a² - b²

|z| = \(√ {{x^2}+{y^2}}\)

zz̅ = |z|²

Calculation:

z = x + iy, z̅ = x - iy

|z| = √(x² + y²)

By using the above formula, 

⇒ z z̅ = (√x² + y²)² 

⇒ z z̅ = x2 + y2          -----(1)

|z + z̅ | = |x + iy + x - iy|

⇒ |z + z̅ | = |2x|         -----(2)

As, z z̅ = |z + z̅ |

x2 + y2 = 2|x|  

Where x is any constant.

Hence, the correct answer is circle.

Given:

Expression is 1! + 3! + 5! + 7! + ... + 199!

Calculation:

Here, given expression is 1! + 3! + 5! + 7! + ... + 199!

1! = 1

3! = 3 × 2 × 1 = 6

5! = 5 × 4! = 120

7! = 7 × 6 × 5 × 4! = 5040

So, 4! It is divisible by 24.

Similarly, every term greater than 4! in the given expression will be divisible by 24.

So, the sum of terms that are not divisible by 24 is

1!  + 3! = 1 + 6 = 7

∴ The remainder when 1! + 3! + 5! + 7! + ... + 199! is divided by 24 is 7.

Formula Used: 

(a + b) (a - b) = a² - b²

i² = -1

Calculation:

Let x =\(\sqrt{12+5 i}+\sqrt{12-5 i}\), where \(i=\sqrt{-1}\)

Squaring both sides,

x² = \({12+5 i}+{12-5 i} +2\sqrt{12+5 i}\sqrt{12-5 i}\)

x² = \({24} +2\sqrt{144+25}\)

x2 = \({24} +2\sqrt{169}\)

x2 = \(24 +26\)

x2 = \(50\)

x = \(5\sqrt 2\)

Given: 

\(A=\left[\begin{array}{l}1 \ 2 \ 3\end{array}\right]\)and \(I=\left[\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right]\)

Calculation:

AA' = \(\left[\begin{array}{l}1 \ 2 \ 3\end{array}\right] \times \left[\begin{array}{lll}1 & 2 & 3 \end{array}\right]\)

AA' = \(\left[\begin{array}{lll}1 & 2 & 3 \ 2 & 4 & 6 \ 3 & 6 & 9\end{array}\right]\)

I + AA' = \(\left[\begin{array}{lll}1 & 2 & 3 \ 2 & 4 & 6 \ 3 & 6 & 9\end{array}\right]\)+ \(\left[\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right]\)

I + AA' = \(\left[\begin{array}{lll}2 & 2 & 3 \ 2 & 5 & 6 \ 3 & 6 & 10\end{array}\right]\)

det(I + AA') = [2(50 - 36) - 2 (20 - 18) + 3(12 - 15)]

det(I + AA') = [2(14) - 2 (2) + 3(-3)]

det(I + AA') = [28 - 4 - 9]

det(I + AA') = 15

Concept:

• If A is a square matrix of order n, then |kA| = kⁿ|A|
• det (AB) = det (A) det (B)

 

Calculation:

Given that 

det (BC) = 2 det (A)          ---(1)

Let, D = det (2A-1BC) 

By using the above properties.

⇒ D = 2³ det(A-1BC)  

Again, by using the above concept,

⇒ D = 23 det(A-1) det (BC)

⇒ D = 23 det(A-1) .2det (A)     [From equation (1)]  

⇒ D = 23.2

⇒ D = 16

Hence, the value of det (2A-1BC) is 16.

Given: 

The nth term of a sequence is \(\frac{2 n+5}{7}\)

Formula Used:

n^th of AP is given by 

T_n = a + (n - 1)d

Sum of n term is given by

\(S_n = {n \over2}{[2a +(n-1)d]}\)

Where a & d are the 1^st term & common difference of AP

Calculation:

We have

\(T_n = \frac{2 n+5}{7}\)

\(T_1 = \frac{2 (1)+5}{7} = {7 \over 7}\)

\(T_2 = \frac{2 (2)+5}{7} = {9 \over 7}\)

\(T_3 = \frac{2 (3)+5}{7} = {11 \over 7}\)

\(T_4 = \frac{2 (4)+5}{7} = {13 \over 7}\)

Hence, the sequence is \({7 \over 7}, {9 \over 7}, {11 \over 7}, {13 \over 7}.........\)

First term, a = 1

Common Difference, d = \({9 \over 7} - {7 \over 7} = {2 \over 7}\)

Now, by using the above formula,

\(S_n = {n \over2}{[2a +(n-1)d]}\)

\(⇒ S_n = {140 \over2}{[2(1) +(140-1)({2 \over 7})]}\)

\(⇒ S_n = {140 \over2}{(2 +{278 \over 7})}\)

\(⇒ S_n = {70}\times{(14 +278) \over 7}\)

\(⇒ S_n = {10}{(292)}\)

⇒ S_n = 2920

∴  Sum of first 140 terms of the sequence is 2920.

Given: 

A is a skew-symmetric matrix of order 3.

Concept:

• The determinant of a skew-symmetric matrix of order 3 is zero.
• If A is a skew-symmetric matrix, then A² is a symmetric matrix.
• |kA| = kⁿ|A|, n is the order of the matrix A.

 

Calculation:

|A| = 0, |-I| = -1

|A²| = |A4| = |A3| = 0

det(4A4) - det(3A3) + det(2A2) - det(A) + det(-I) 

= 4³det(A4) - 3³det(A3) + 2³det(A2) - det(A) + det(-I) 

= 0 - 0 + 0 - 0 + (-1)

= -1

Concept:

If A is symmetric, then

aij =  aji for all i, j.

If A is skew-symmetric, then

aij = - aji for all i, j

Multiplication of Matrices:

Here, C₁ = a₁b₁ + a₂b₄ + a₃b₆

Calculation:

Given that,

\(A=\left[\begin{array}{rrr} 0 & 3 & 4 \ -3 & 0 & 5 \ -4 & -5 & 0 \end{array}\right]\)

\(A^2=\left[\begin{array}{rrr} 0 & 3 & 4 \ -3 & 0 & 5 \ -4 & -5 & 0 \end{array}\right] × \left[\begin{array}{rrr} 0 & 3 & 4 \ -3 & 0 & 5 \ -4 & -5 & 0 \end{array}\right]\)

Multiple first row with first column to get 11 value. Similarly, mul

\(A^2=\left[\begin{array}{rrr} -25 & -20 & 15 \ -20 & -34 &-12 \ 15 & -12 & -41 \end{array}\right] \)

A² is symmetric as aij =  aji

Now, we have to find the determinant of A²

\(|A^2|=\left|\begin{array}{rrr} -25 & -20 & 15 \ -20 & -34 &-12 \ 15 & -12 & -41 \end{array}\right|\)

|A²| = -25[-34 × (-41) - (-12)(-12)

         - (-20)[-20 × (-41) - (-12)(15)]

         + (15)[(-20)(-12) - (-34) × 15]

|A²| = -25(1394 - 144) + 20 (820 + 180) + 15(240 + 510)

|A²| = -25(1250) + 20(1000) + 15(750)

|A²| = - 31250 + 20000 + 11250 = 0

Hence, A2 is symmetric matrix with det(A2) = 0.

Hence Option 1 is the correct answer.

Given: 

\(A=\left[\begin{array}{lll} 2 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 4 \end{array}\right]\)

Concept:

• A singular matrix is a square matrix whose determinant is zero.
• A diagonal matrix is a square matrix in which all of the elements except the principal diagonal elements are zeroes.
• It is both upper and lower triangular, as all the elements except the main diagonal elements are zeros.
• A matrix is said to be symmetric if and only if, aij = aji for all i, j such that 1 ≤ n, j ≤ n
• For symmetric matrix A,  A = A^T

 

Calculation:

To evaluate the statement for Aⁿ, Let us take n = 2, and find A²

\(A^2=\left[\begin{array}{lll} 2 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 4 \end{array}\right] × \left[\begin{array}{lll} 2 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 4 \end{array}\right]\)

\(A^2=\left[\begin{array}{lll} 2 × 2 & 0 × 3 & 0 × 4 \ 0 × 0 & 3 × 3 & 0 × 0 \ 0 × 0 & 0 ×0 & 4 × 4 \end{array}\right] \)

\(A^2=\left[\begin{array}{lll} 4 & 0 & 0 \ 0 & 9 & 0 \ 0 & 0 & 16 \end{array}\right] \)

Now, Let us evaluate each statement

Statement 1: An will always be singular for any positive integer n.

A singular matrix is a square matrix whose determinant is zero.

det (A) = \(\left|\begin{array}{lll} 2 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 4 \end{array}\right|\)

det (A) = 2 × 3 × 4 = 24

Hence, An is not a singular for any positive integer n.

So this, statement 1 is incorrect.

Statement 2: An will always be a diagonal matrix for any positive integer n.

By using the above concept

A square matrix A = [a_ij] is said to be a diagonal matrix if

a_ij = 0 when i ≠ j

Here, An will always be a diagonal matrix for any positive integer n.

So, statement 2 is correct.

Statement 3: 

Aⁿ will always be a symmetric matrix for any positive integer n.

\(A^2 = (A^2)^T=\left[\begin{array}{lll} 4 & 0 & 0 \ 0 & 9 & 0 \ 0 & 0 & 16 \end{array}\right] \)

Similarly, |A| will be symmetric for all the values of n.

So, statement 3 is correct.

∴ Correct answer is option (2).

Given:

(a + b), 2b, (b + c) are in HP

Concept:

If a, b, c are in AP then 1/a, 1/b, 1/c in HP

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

If a, b, c are in GP then \({a\over b} = {b\over c}\)

Calculation:

(a + b), 2b, (b + c) are in HP, then \({1 \over {a + b}}, {1 \over {2b}}, {1 \over {b + c}}\) are in AP

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

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

⇒ \({1 \over {b}} = {{b+c+a+b} \over {(a + b)(b + c)}}\)

⇒ \( (a + b)(b + c) = b(2b+c+a)\)

⇒\(ab+ac+b^2+bc = 2b^2+bc+ab\)

\(ac= b^2\)

\({a\over b} = {b\over c}\)

Hence, a, b and c are in GP.

Concept:

If a, b, c are in GP, then \({a\over b} = {b\over c}\)

t_n = ar^n-1

The term of GP are "a, ar, ar²....arⁿ"

Sum of n consecutive numbers = n(n + 1)/2

Where n is the number of terms,

a is the first term, r is the common ratio

Calculation:

Let 

P = \(\rm \left(t_1 t_3 \ldots t_{21}\right)^{\frac{1}{11}}\)

Given that 

t1, t2, t3 ... be a, ar, ar²  

Therefore

⇒ P = \(\rm \left(a. ar^2 \ldots ar^{20}\right)^{\frac{1}{11}}\)

There are 11 terms in the series,

⇒ P = \(\rm \left(a^{11}. r^{2+4+6+...20}\right)^{\frac{1}{11}}\)

⇒ P =\(\rm \left(a^{11}. r^{2(1+2+3+...10)}\right)^{\frac{1}{11}}\)

Sum of 1 + 2 + 3 ...... 10 = (10 × 11)/2 = 55

⇒ P = \(\rm \left(a^{11}. r^{110}\right)^{\frac{1}{11}}\)

⇒ P = \(\rm a. r^{10}\)

⇒ P = t11

∴ Correct answer is option (3).

Calculation:

Let x = \(-\sqrt{-1} \)

As, \(i =\sqrt{-1} \)

x = - i

Taking Square root,

\(\sqrt{x} = \sqrt{- i}\)

By squaring the options, 

Option 1: 1 + i

(1 + i)² = 1 - 1 + 2i = 2i

Option 2: \(\frac{1-i}{\sqrt{2}}\)

\((\frac{1-i}{\sqrt{2}})^2 = \frac{1-1-2i}{{2}} = - i\)

Option 3: \(\frac{1+i}{\sqrt{2}}\)

\((\frac{1+i}{\sqrt{2}})^2 = \frac{1-1+2i}{{2}} = i\)

Option 4:\(\frac{1}{\sqrt{2}} i\)

\((\frac{1}{\sqrt{2}} i)^2 =-\frac{1}{{2}}\)

Hence, option 2 is correct.

Given: Number of circles = 10

Calculation:

Two circles can intersect at a maximum of 2 points

Maximum number of points of intersection of 10 circles,

ⁿC_r = \({{n!}\over {{(n-r)!}r!}} \)

In this case, n = 10 (total circles) and r = 2 (pairs of circles).

So, the number of unique pairs of circles is

¹⁰C₂ = \({{10!}\over {{(10-2)!}2!}} \) = 45.

Since each pair can intersect at most twice, the maximum number of intersection points is 45 × 2 = 90.

∴ Correect answer is option (C).

Formula Used:

\(: ^{2n+1}C_0 + : ^{2n+1}C_1 + \dots : ^{2n+1}C_{n+1} + \dots+ : ^{2n+1}C_{2n+1} = 2^{2n+1}\)

\(: ^{n}C_r = : ^{n}C_{n-r}\)

 

Calculation:

There are 4096 subsets of S which contain at most n elements.

By using the above formula,

\(^{2n+1}C_{n+1} = ^{2n+1}C_{n}\) &  \(^{2n+1}C_{n+2} = ^{2n+1}C_{n-1}\)

According to the question,

\(: ^{2n+1}C_0 + : ^{2n+1}C_1 + : ^{2n+1}C_2 .......+: ^{2n+1}C_n\) = 4096    ---(1)

\(: ^{2n+1}C_0 + : ^{2n+1}C_1 + \dots : ^{2n+1}C_{n+1} + \dots+ : ^{2n+1}C_{2n+1} = 2^{2n+1}\)

\(2(: ^{2n+1}C_0 + : ^{2n+1}C_1 + .......+: ^{2n+1}C_n + : ^{2n+1}C_{n}) = 2^{2n+1}\)

\(: ^{2n+1}C_0 + : ^{2n+1}C_1 + .......+: ^{2n+1}C_n + : ^{2n+1}C_{n} = 2^{2n}\)    ----(2)

From equation (1) & (2),

2²ⁿ = 4096

\(2^{2n} = 2^{12}\)

∴ n = 6

Calculation:

Given that,

\(\left|\begin{array}{ccc} x^2+3 x & x-1 & x+3 \ x+1 & -2 x & x-4 \ x-3 & x+4 & 3 x \end{array}\right|\) = ax4 + bx3 + cx2 + dx + e

Put x = 0,

 \(\left|\begin{array}{ccc} 0 & -1 & 3 \ 1 & 0 & -4 \ -3 & 4 & 0 \end{array}\right|\) = e

0[0 + 16] - 1 [0 - 12] + (-3) [4 - 0] = e

0[16] - 1 [-12] -3 [4] = e

0 + 12 - 12 = e

∴ e = 0

Calculation:

If all elements of a third order determinant are equal to 1 or -1, then the value of the determinant is an even number but not necessarily 0.

By setting the first row of the matrix to 1 1 -1, the second row to 1 -1 1, and the third row to -1 1 1.

The determinant of this matrix is 4.

Concept: 

The property related to the adjugate (or adjoint) of a square matrix A.

• A(adj A) = |A| . I

We know that, for a square matrix of order n,

• adj(adjA) = |A|^n-2A, |A| ≠ 0
• det kA = kⁿdet A

Calculation:

Given that,

\(A=\left[\begin{array}{rrr} 2 & -1 & 0 \ -1 & 3 & 0 \ 1 & 0 & 1 \end{array}\right]\)      -----(1)

det A = 2[3 × 1 - (-1) × 0] -(-1)[-1 × 1 - 1 × 0]     

⇒ det (A) = 2(3) + 1 × (-1)

⇒ det (A) = 6 - 1

⇒ det (A) = 5             -----(2) 

We know that, for a square matrix of order n,

adj(adjA) = |A|n-2A, |A| ≠ 0

⇒ |adj (adj A)| = ||A|^n-2A|

⇒ |adj (adj A)| = (|A|^n-2)ⁿ|A|

⇒ |adj (adj A)| = |A|n²-2n|A|   

⇒ |adj (adj A)| = |A|n²-2n+1    

At, n = 3 & |A| = 5

⇒ |adj (adj A)| = 5^9-6+1 = 5⁴ = 625

∴ |adj (adj A)| = 625

Given: 

\(A=\left[\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right]\)

Concept:

If A = I, then Aⁿ = I

det |I| = 1

Calculation:

23A3 - 19A2 - 4A 

⇒ 23I3 - 19I2 - 4I

⇒ 23 I - 19 I - 4 I

⇒ 23 - 19 - 4 = 0

Hence, 23A3 - 19A2 - 4A equal to Null matrix of order 3.

Given:

|A|_3×3 = 3

C = cofactors of the matrix A

Formula Used:

(Cofactors of the matrix A)^T = adj A

|adj A| = |A|^n-1 , where n is the order of matrix

Calculation:

C = cofactors of the matrix A

C = (adj A)T = adj AT

|C2| = |adj AT|²

As, |adj A| = |A|n-1

|C2| = (3^3-1)²

∴ |C2| = 81

Formula Used: 

Sum of an AP is given by,

\(S_n = {n \over 2}{[2a + (n-1)d}]\)

Where, a = First term,

d = common difference, n = no. of terms

Calculation:

\(A_k=\left[\begin{array}{cc} k-1 & k \ k-2 & k+1 \end{array}\right] \)

⇒ |A_k| = (k + 1)(k - 1) - k (k - 2)

⇒ |Ak| = k² + k - k - 1 - k² + 2k

⇒ |Ak| = 2k - 1

⇒ |A1| = 2(1) - 1 = 1

⇒ |A2| = 2(2) - 1 = 3

⇒ |A3| = 2(3) - 1 = 5

⇒ |A4| = 2(4) - 1 = 7

⇒ |A100| = 2(100) - 1 = 199

So, 

det(A1) + det(A2) + det(A3) + ... + det(A100) 

⇒ 1 + 3 + 5 + 7........... 199

This is an A.P. series, with first term = 1, last term = 199, Common Difference = 2,

By using the above formula

⇒ \(S_n = {n \over 2}{[2a + (n-1)d}]\)

⇒ \(S_n = {100 \over 2}{[2(1) + (100-1)2}]\)

⇒ \(S_n = {50}{[2 + 198}]\)

∴ S_n = 10000