Quantitative Aptitude PYQs

Concept:

If a₁, a₂, …., a_n is a GP then the general term is given by: a_n = a × r^{n - 1} where a is the 1^st term and r is the common ratio.

Calculation:

Given: 25, -125, 625, -3125, …….

Here, first term a = 25 and common ratio r = -5.

As we know that, if a₁, a₂, …., a_n is a GP then the general term is given by: a_n = a × r^{n - 1} where a is the first term and r is the common ratio.

⇒ The general term is : a_n = 25 × (-5)^{n – 1} = (-1)^{n – 1} × 5^{n + 1}

Concept:

For any two non-empty sets A and B, a relation R from A to B is a subset of the cartesian product A X B.

Reflexive:

Let R be a relation on a non-empty set A, if every element of A is related to itself then R is said to be a reflexive relation.

Thus, R is reflexive if and only if (a, a) ∈ R, ∀ a ∈ A.

Symmetric:

Let R be a relation on a non-empty set A, then the relation R is said to be symmetric relation if and only if (a, b) ∈ R ⇒ (b, a) ∈ R

Transitive:

Let R be a relation on a non-empty set A, then the relation R is said to be transitive relation if and only if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R

Equivalence:

Let R be a relation on a non-empty set A, then the relation R is said to be equivalence relation if R is reflexive, symmetric and transitive.

Calculation:

Given: X = {1, 2, 3, 4} and R is a relation on X given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)}

Reflexive:

The given relation R on X is not reflexive ∵ (a, a) ∉ R, ∀ a ∈ X i.e (4, 4) ∉ R.

Symmetric:

The given relation R on X is symmetric ∵ for all (a, b) ∈ R ⇒ (b, a) ∈ R 

Transitive:

The given relation R on X is transitive ∵ for all (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R

Concept:

For any two non-empty sets A and B, a relation R from A to B is a subset of the cartesian product A X B.

Reflexive:

Let R be a relation on a non-empty set A, if every element of A is related to itself then R is said to be a reflexive relation.

Thus, R is reflexive if and only if (a, a) ∈ R, ∀ a ∈ A.

Symmetric:

Let R be a relation on a non-empty set A, then the relation R is said to be symmetric relation if and only if (a, b) ∈ R ⇒ (b, a) ∈ R ∀ a, b ∈ A.

Transitive:

Let R be a relation on a non-empty set A, then the relation R is said to be transitive relation if and only if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ A.

Equivalence:

Let R be a relation on a non-empty set A, then the relation R is said to be equivalence relation if R is reflexive, symmetric and transitive.

Calculation:

Given: R is a relation on N and is defined as xRy ⇒ x² – 4xy + 3y² = 0.

⇒ x² – 4xy + 3y² = 0

⇒ (x - y) (x – 3y) = 0     ...1)

Reflexive:

The given relation R is reflexive ∵ if x = a and y = a then (x = a, y = a) satisfies the equation (1) ⇒ (a, a) ∈ R, ∀ a ∈ N.

Hence, R is reflexive.

Symmetric:

∵ (3, 1) satisfies the equation (1) ⇒ (3, 1) ∈ R.

Let’s check whether (1, 3) ∈ R or not. If we substitute x = 1 and y = 3 in equation (1), we get

⇒ (x - y) (x – 3y) = 16 ≠ 0.

⇒ (1, 3) ∉ R.

Hence, the given relation R is not symmetric ∵ (3, 1) ∈ R but (1, 3) ∉ R.

Transitive:

As we can see that (9, 3) and (3, 1) satisfy the equation (1).

⇒ (9, 3) ∈ R and (3, 1) ∈ R.

Let’s check whether (9, 1) satisfies the equation (1) or not.

⇒ (x - y) (x – 3y) = (9 - 1) (9 - 3) = 48 ≠ 0.

⇒ (9, 1) ∉ R.

Hence, the given relation R is not transitive ∵ (9, 3) ∈ R and (3, 1) ∈ R but (9, 1) ∉ R.

Concept:

1, ω and ω² are called cube roots of unity where \(\omega = \frac{{ - 1 + i\sqrt 3 }}{2};and;{\omega ^2} = \frac{{ - 1 - i\sqrt 3 }}{2}\)

1 + ω + ω² = 0 and ω³ = 1.

Calculation:

Given: A = {x ϵ Z : x³ – 1 = 0} and B = {x ϵ Z: x² + x + 1= 0}

As we know that, the roots of the x³ – 1 = 0 are 1, ω and ω².

⇒ A = {1, ω, ω²}

As we know that, the roots of x² + x + 1= 0 are ω and ω².

⇒ B = {ω, ω²}

∴ A ∩ B = B

Alternate MethodThe roots of equation x3 – 1 = 0 are

\(\left{ {1, \frac{{-1 + \sqrt 3 i}}{2},\frac{{-1 - \sqrt 3 i}}{2}} \right}\)

And the roos of the equation x2 + x + 1 = 0 are

\(\left{ {\frac{{-1 + \sqrt 3 i}}{2},\frac{{-1 - \sqrt 3 i}}{2}} \right}\)

So, the common roots between A and B are, \(\left{ {\frac{{-1 + \sqrt 3 i}}{2},\frac{{-1 - \sqrt 3 i}}{2}} \right}\)

Therefore, A ∩ B = \(\left{ {\frac{{-1 + \sqrt 3 i}}{2},\frac{{-1 - \sqrt 3 i}}{2}} \right}\)

Concept:

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Calculation:

Let, us suppose the universal set U = A ∪ B.

Statement 1:

⇒ (A ∩ B) ∪ (A ∩ B̅) ∪ (A̅ ∩ B) = {A ∪ (A ∩ B̅)} ∩ {B ∪ (A ∩ B̅)} ∪ (A̅ ∩ B)

⇒ (A ∩ B) ∪ (A ∩ B̅) ∪ (A̅ ∩ B) = {(A ∪ A) ∩ (A ∪ B̅)} ∩ {(A ∪ B) ∩ (B ∪ B̅)} ∪ (A̅ ∩ B)

⇒ (A ∩ B) ∪ (A ∩ B̅) ∪ (A̅ ∩ B) = {A ∩ U} ∩ {U ∩ U} ∪ (A̅ ∩ B)

⇒ (A ∩ B) ∪ (A ∩ B̅) ∪ (A̅ ∩ B) = A ∪ (A̅ ∩ B) = (A ∪ A̅) ∩ (A ∪ B) = A ∪ B

Hence statement 1 is Correct.

Statement 2:

⇒ (A ∪ (A̅ ∩ B̅)) = (A ∪ A̅) ∩ (A ∪ B̅)

⇒ (A ∪ (A̅ ∩ B̅)) = U ∩ (B ∩ A̅) = A̅ ∩ B

Hence, statement 2 is wrong.

Concept:

1. A ∆ B is the symmetric difference that represents the objects that belong to A or B but not to their intersection.

• A Δ B = (A ∪ B) - (A ∩ B)

2. A ⊆ B subset A is a subset of B. set A is included in set B.
3. A ⋃ B, ⋃ is a union, that represents the objects that belong to set A or set B
4. A ⋂ B, ⋂ is an intersection, that represents the objects that belong to set A and set B
5. a ∈ A means x is an element of A, i.e. belongs to set membership

Calculation:

Given: A, B, C be subsets of X

Statement 1: A ⊂ C ⇒ (A ∩ B) ⊂ (C ∩ B), (A ∪ B) ⊂ (C ∪ B)

We know that,

If A ∪ B = A ∪ C and A ∩ B=A ∩ C, then B = C which gives 

⇒ A ⊂ C 

⇒ A ⊂ C ⇒ (A ∩ B) ⊂ (C ∩ B), (A ∪ B) ⊂ (C ∪ B)

Hence statement 1 is true.

Statement 2: (A ∩ B) ⊂ (C ∩ B) for all sets B ⇒ A ⊂ C

Let A = {1, 2, 3}, B = {3, 4, 5} and C = {1, 3, 6, 7, 8}

⇒ A ∩ B = {3} ⊂ (C ∩  B) but A ⊂ C is not true.

Hence, statement 2 is false.

Statement 3: (A ∪ B) ⊂ (C ∪ B) for all sets B ⇒ A ⊂ C

This is true as (A ∪ B) ⊂ (C ∪ B) for all sets B ⇒ A ⊂ C

Concept:

If A = [a_ij] is a square matrix of order n and C = [c_ij] is cofactor matrix of A. Then C^T = [c_ji] is called the adjoint of matrix A and is denoted as: adj (A) = C^T = [c_ji], 1 ≤ i, j ≤ n.

Calculation:

Given: \(B = \left[ {\begin{array}{*{20}{c}} 3&2&0\ 2&4&0\ 1&1&0 \end{array}} \right]\)

As we know that, the adj (B) = C^T where C is cofactor matrix of B and it is given by:

\(C = \left[ {\begin{array}{*{20}{c}} {{B_{11}}}&{{B_{12}}}&{{B_{13}}}\ {{B_{21}}}&{{B_{22}}}&{{B_{23}}}\ {{B_{31}}}&{{B_{32}}}&{{B_{33}}} \end{array}} \right];where;{B_{ij}} = {\left( { - 1} \right)^{i + j}} \times {M_{ij}};and;{M_{ij}} = Minor;of;{b_{ij}}\)

⇒ B₁₁ = 0, B₁₂ = 0, B₁₃ = - 2, B₂₁ = 0, B₂₂ = 0, B₂₃ = - 1, B₃₁ = 0, B₃₂ = 0 and B₃₃ = 8.

\(\Rightarrow ;C = \left[ {\begin{array}{*{20}{c}} 0&0&{ - 2}\ 0&0&{ - 1}\ 0&0&8 \end{array}} \right];and;adj;\left( B \right) = {C^T} = ;\left[ {\begin{array}{*{20}{c}} 0&0&0\ 0&0&0\ { - 2}&{ - 1}&8 \end{array}} \right]\)

Concept:

If \(f\left( x \right) = \left| x \right|,;then;f\left( x \right) = \left{ {\begin{array}{*{20}{c}} { - x,;x < 0}\ {x,;x \ge 0} \end{array}} \right.\)

Calculation:

Given: |x² – x - 6|= x + 2

We can factorize x² – x – 6 as (x - 3) (x + 2)

\(\Rightarrow ;\left| {\left( {x - 3} \right)\left( {x + 2} \right)} \right| = ;\left{ {\begin{array}{*{20}{c}} {\left( {x - 3} \right)\left( {x + 2} \right),;if;\left( {x - 3} \right)\left( {x + 2} \right) \ge 0}\ { - ;\left( {x - 3} \right)\left( {x + 2} \right),;if;\left( {x - 3} \right)\left( {x + 2} \right) < 0} \end{array}} \right.\)

Case – 1: If (x - 3) (x + 2) ≥ 0

⇒ x ∈ (- ∞, - 2] ∪ [3, ∞)

⇒ x² – x – 6 = x + 2

⇒ x² – 2x – 8 = 0

⇒ (x - 4) (x + 2) = 0

⇒ x = 4 or – 2 ∈ (- ∞, - 2] ∪ [3, ∞)

So, the roots of the given quadratic equation are 4 and – 2.

Case – 2: If (x - 3) (x + 2) < 0

⇒ x ∈ [- 2, 3]

⇒ - (x² – x – 6) = x + 2

⇒ x² – 4 = 0

⇒ x = 2 or – 2 ∈ [- 2, 3]

So, the roots of the given quadratic equation are 2 and – 2

Hence, the roots of the given quadratic equation are – 2, 2 and 4.

Concept:

Singular Matrix: Any square matrix of order n is said to be singular if |A| = 0.

Involuntary Matrix: Any square matrix of order n is said to be an involuntary matrix if A² = I, where I is the identity matrix of order n.

Nilpotent Matrix: Any square matrix of order n is said to be nilpotent matrix if there exist least positive integer m such that A^m = O, where O is the null matrix of order n.

Idempotent Matrix: Any square matrix of order n is said to be an idempotent matrix if A² = A.

Calculation:

Given: \(A = \left[ {\begin{array}{*{20}{c}} 0&1\ 1&0 \end{array}} \right],\)

\(\Rightarrow ;{A^2} = ;\left[ {\begin{array}{*{20}{c}} 0&1\ 1&0 \end{array}} \right] \times ;\left[ {\begin{array}{*{20}{c}} 0&1\ 1&0 \end{array}} \right] \= ;\left[ {\begin{array}{*{20}{c}} 1&0\ 0&1 \end{array}} \right] = I\)

Hence, A is an involuntary matrix as A² = I.

Calculation:

Given: \(\left| {\begin{array}{*{20}{c}} x&-3i&1\ y&1&{i}\ 0&2i&-i \end{array}} \right|=6+11i\)

Let \(A=\left| {\begin{array}{*{20}{c}} x&-3i&1\ y&1&{i}\ 0&2i&-i \end{array}} \right|\)

⇒ det A = x[-i - 2i²] - (-3i)[-yi - 0] + 1[2yi - 0]

As we know that, i2 = -1 

det A = x[-i + 2] + 3i[-yi] + [2yi]

det A = -ix + 2x + 3y + 2yi

det A = (2x + 3y) + (2y - x)i

According to the question, 

det A =  6 + 11i

(2x + 3y) + (2y - x)i = 6 + 11i

⇒ 2x + 3y = 6 and 2y – x = 11

Solving the above equations we get

⇒ x = - 3, y = 4

Concept:

1, ω and ω² are called cube roots of unity where \(\omega = \frac{{ - ;1 + i\sqrt 3 }}{2};and;{\omega ^2} = \frac{{ - ;1 - i\sqrt 3 }}{2}\)

Properties of the cube roots of unity:

1 + ω + ω² = 0 and ω³ = 1.

Calculation:

Given: The equations z³ + 2z² + 2z + 1 = 0 and z²⁰¹⁷ + z²⁰¹⁸ + 1 = 0.

⇒ z³ + 2z² + 2z + 1 = 0

⇒ (z + 1) × (z² + z + 1) = 0

⇒ z = - 1, ω and ω².

Hence, the roots of the equation z³ + 2z² + 2z + 1 = 0 are – 1, ω and ω².

It is given that the equations z³ + 2z² + 2z + 1 = 0 and z²⁰¹⁷ + z²⁰¹⁸ + 1 = 0 has common roots.

Let f(z) = z²⁰¹⁷ + z²⁰¹⁸ + 1 = z²⁰¹⁷(z + 1) + 1

If z = -1, then f(- 1) = 1 ≠ 0.

Hence, z = - 1 is not a root of f(z).

If z = ω, then f(ω) = ω²⁰¹⁷ (ω + 1) + 1

⇒ f(ω) = (ω³)⁶⁷² × ω × (- ω²) + 1 = 0     ...(∵ 1 + ω + ω² = 0 and ω³ = 1)

Hence, z = ω is a root of f(z).

If z = ω², then f(ω²) = (ω²)²⁰¹⁷ × (ω² + 1) + 1

⇒ f(ω²) = ((ω³)⁶⁷² × ω)² × (- ω) + 1 = 0     ...(∵ 1 + ω + ω² = 0 and ω³ = 1)

Hence, z = ω² is a root of f(z).

So, the common roots of the equations z³ + 2z² + 2z + 1 = 0 and z²⁰¹⁷ + x²⁰¹⁸ + 1 = 0 are ω and ω²

Concept:

The notation C(n, r) is the number of combinations/groups of n different things taking r at a time and is given by: \(C;\left( {n,;r} \right) = \frac{{{\rm{n}}!}}{{{\rm{r}}!\left( {{\rm{n;}} - {\rm{;r}}} \right)!}}\)

If C(n, x) = C(n, y), then x + y = n.

Calculation:

Given C(20, n + 2) = C(20, n – 2)

As we know that, if C(n, x) = C(n, y), then x + y = n.

⇒ (n + 2) + (n – 2) = 20

⇒ 2n = 20

⇒ n = 10

Concept

If there are n points in a plane and no three points lie on a straight line, then the total number of straight line segments that can be formed is given by: \(C\left( {n,;2} \right) = \frac{{{\rm{n}}!}}{{2!\left( {{\rm{n;}} - {\rm{;}}2} \right)!}}\)

Calculation

Given: n = 10.

\(\Rightarrow ;C\left( {10,;2} \right) = \frac{{10!}}{{2!\left( {10 - {\rm{;}}2} \right)!}} = \frac{{10 \times 9}}{2} = 45\)

Hence, the number of straight lines that can be formed is 45.

Concept:

For any quadratic equation, ax² + bx + c = 0. We have discriminant, D = b² – 4ac then the given quadratic equation has:

I. Distinct and real roots if D > 0.

II. Real and repeated roots, if D = 0.

III. Complex roots and conjugate of each other, if D < 0.

Calculation:

In the given quadratic equation px² + qx + r = 0, coefficients p, q, r are positive with roots a and b.

⇒ a + b = - q/p and a × b = r/p.

From the condition: a × b = r/p, we can say that there are two possibilities which are:

Case 1: Both roots are negative so that the product is positive i.e a < 0 and b < 0.

Case 2: Both roots are positive so that the product is positive i.e a > 0 and b > 0.

From the relation a + b = - q/p, we can say that the sum of roots is negative.

Out of the two cases, Case 1 is a valid case satisfying both relations a + b = - q/p and a × b = r/p.

Concept:

Power set: Let A be a non-empty set. Then, \(P\left( A \right) = \left{ {B;|;B; \subseteq A} \right}\) is called as the power set of A.

For any non-empty set A with n(A) = x. Then the total no. of subsets of set A is given by: 2^x.

i.e n(P(A)) = 2^x.

Calculation:

Given: A = {λ, {λ, μ}}

Subsets of A are: {λ}, {{λ, μ}}, {λ, {λ, μ}}, φ

Power set of A, P(A) = {φ, {λ}, {{λ, μ}}, {λ, {λ, μ}}}

Calculation:

In the above Venn diagram, we have represented chess, carom and table tennis with three circles.

Given: 60 play chess, 50 play table tennis, 48 play carom, 12 play chess and carom, 15 play carom and table tennis, 20 play table tennis and chess.

⇒ a + b + d + e = 60     ...1)

⇒ d + e + f + g = 50     ...2)

⇒ b + c + e + f = 48     ...3)

⇒ b + e = 12     ...4)             

⇒ e + f = 15     ...5)

⇒ d + e = 20     ...6)

By using (1) + (2) + (3) – (4) – (5) – (6)

⇒ a + b + c+ d + f + g = 60 + 50 + 48 – 12 – 15 - 20 = 111

We know that, total number of students in school is given by: a + b + c + d + f + g + e = 111 + e

Total number of students in school will be minimum, when e = 0.

⇒ Minimum number of students in school = 111.

Calculation:

⇒ a + b + d + e = 60     ...1)

⇒ d + e + f + g = 50     ...2)

⇒ b + c + e + f = 48     ...3)

⇒ b + e = 12     ...4)

⇒ e + f = 15     ...5)

⇒ d + e = 20     ...6)

By using (1) + (2) + (3) – (4) – (5) – (6)

⇒ a + b + c + d + f + g = 60 + 50 + 48 – 12 – 15 - 20 = 111

Total number of students in school = a + b + c + d + f + g + e = 111 + e.

Maximum number of students will depend on the maximum value of e.

From the equations (4), (5) and (6) we have the following three cases:

Case 1: Let b = 0

⇒ e = 12.

By substituting the value of e in (5), we get

⇒ f = 15 – e = 3

By substituting the value of e in (5), we get

⇒ d = 20 – e = 8

Case 2: Let f = 0

⇒ e = 15.

By substituting the value of e in (4), we get

⇒ b = 12 – e = -3

We got b as negative, so we need to discard this case ∵ b ≥ 0.

Case 3: Let d = 0

⇒ e = 20.

By substituting the value of e in (4), we get

⇒ b = 12 – e = -8

We got b as negative, so we need to discard this case ∵ b ≥ 0.

So the maximum possible value of e is from case 1 i.e. e = 12.

∴ Maximum possible number of students in school = 111 + e = 111 + 12 = 123.

Concept

If A is any matrix of order n and it’s inverse exists, then we can write

AA^-1 = A^-1A = I, where I = Identity matrix of order n

Calculation

Given: A is an identity matrix of order 3 i.e. A = I

Multiplying both sides by A^-1 we get

⇒ AA^-1 = IA^-1

⇒ I = A^-1 [∵ A matrix multiplied by the identity matrix is the matrix itself i.e. AI = A]

⇒ A = A^-1

Concept:

For any square matrix A of order n, |A| = |A^T|

Calculation:

Given: A is a square matrix of order 3 such that its determinate is 4

⇒ |A| = 4

As we know that, for any square matrix A of order n, |A| = |A^T|

⇒ |A^T| = 4

Concept:

The notation C(n, r) is the number of combinations/groups of n different things taking r at a time and is given by: \(C\left( {n,;r} \right) = \frac{{{\rm{n}}!}}{{{\rm{r}}!\left( {{\rm{n;}} - {\rm{;r}}} \right)!}}\)

Calculation:

Given: 6 programmers and 4 typists, an office wants to recruit 5 people.

Here, n = 10 and r = 5.

Case 1: Group of 5 with 1 typist

1 typist can be chosen from 4 in C(4, 1) = 4 ways

4 programmers can be chosen from 6 in C(6, 4) = 15 ways

⇒ No. of ways to form a group of 5 with 1 typist = 4 × 15 = 60.

Case 2: Group of 5 with 2 typists

2 typist can be chosen from 4 in C(4, 2) = 6 ways

3 programmers can be chosen from 6 in C(6, 3) = 20 ways

⇒ No. of ways to form a group of 5 with 2 typist = 6 × 20 = 120.

Case 3: Group of 5 with 3 typists

3 typist can be chosen from 4 in C(4, 3) = 4 ways

2 programmers can be chosen from 6 in C(6, 2) = 15 ways

⇒ No. of ways to form a group of 5 with 2 typist = 4 × 15 = 60.

Case 4: Group of 5 with 4 typists

4 typist can be chosen from 4 in C(4, 4) = 1 way

1 programmers can be chosen from 6 in C(6, 1) = 6 ways

⇒ No. of ways to form a group of 5 with 2 typist = 1 × 6 = 6.

Total number of ways in forming a group of 5 with at least one typist is = 60 + 120 + 60 + 6 = 246.