Quantitative Aptitude PYQs

Concept:

Let A and B be the subsets of X, then then Venn Diagram to represent that is:

• Union of sets (A ∪ B):

​

 

• Intersection of subsets (A ∩ B):

 

 

• Complement of subset (A’):

 

• Difference of subsets (A – B):

​

 

Calculation:

Given:

Only Hindi = 54; Only Mathematics = 63; Only English = 41; Both Mathematics & Hindi = 18; English, Hindi & Mathematics = 10.

The Venn diagram below shows the number of students good in either Hindi or Mathematics but not in English.

= Only Hindi + Only Mathematics + Both Hindi & Mathematics – (Hindi, English & Mathematics)

= {H – E – M} + {M – E – H} + {H ∩ B} – {H ∩ E ∩ M}

= 54 + 63 + 18 – 10

= 125

Concept:

Let A and B be the subsets of X, then then Venn Diagram to represent that is:

• Union of sets (A ∪ B):

​

 

• Intersection of subsets (A ∩ B):

 

 

• Complement of subset (A’):

 

• Difference of subsets (A – B):

​

 

Calculation:

Given:

Only Hindi = 54; Only Mathematics = 63; Only English = 41; Both Mathematics & Hindi = 18; English, Hindi & Mathematics = 10.

The Venn diagram below shows the number of students good in Hindi and Mathematics but not in English.

= Both Hindi & Mathematics – (Hindi, English & Mathematics)

= {H ∩ B} – {H ∩ E ∩ M}

= 18 – 10

= 8

Concept:

Let ax² + bx + c = 0 be a quadratic equation then,

\({\rm{x}} = \frac{{ - {\rm{b}} \pm \sqrt {{{\rm{b}}^2} - 4{\rm{ac}}} }}{{2{\rm{a}}}}\)

Modulus: If f(x) = |x – a| then f(x) is defined as,

\({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {x - a,\;\;x \ge a}\\ { - \left( {x - a} \right),\;\;x \le a} \end{array}} \right.\)

Calculation:

For 1 – x > 0 i.e., x < 1

1 – x + x² = 5

⇒ x² – x – 4 = 0

\(\Rightarrow x = \frac{{1 \pm \sqrt {1 - \;4\left( { - 4} \right)} }}{2}\)

\(\Rightarrow x = \frac{{1 \pm \sqrt {17} }}{2}\)

But, x < 1

So, \(x = \frac{{1 - \sqrt {17} }}{2}\)

For 1 – x < 0 i.e., x > 1

-(1 – x) + x² = 5

⇒ x² + x – 6 = 0

\(\Rightarrow x = \frac{{ - 1 \pm \sqrt {1 - 4\left( { - 6} \right)} }}{2}\)

\(\Rightarrow x = \frac{{ - 1 \pm \sqrt {25} }}{2}\)

\(\Rightarrow x = \frac{{ - 1 \pm 5}}{2}\)

⇒ x = -3 and x = 2

But, x > 1

So, x = 2

Hence, the solution of given equation are \(x = \frac{{1 - \sqrt {17} }}{2}\) and x = 2.

∴ Equation has a rational root and an irrational root

Concept:

Formulas used:

• i = √(-1)
• i² = -1
• i³ = -i × i² = -i
• i⁴ = (i²)² = (-1)² = 1

 

Calculation:

To find: i¹⁰⁰⁰ + i¹⁰⁰¹ + i¹⁰⁰² + i¹⁰⁰³

We know that i⁴ = 1. So, 1000/4 = 250.

= (i⁴)²⁵⁰ + i × (i⁴)²⁵⁰ + i² × (i⁴)²⁵⁰ + i³ × (i⁴)²⁵⁰

= 1 + i + i² + i³

= 1 + i + (-1) + (-i)

= 0

Concept:

Formula used:

• \({\log _a}b = \frac{1}{{{{\log }_b}a}}\)
• Log_a M + log_a N = log_a (MN)

Factorial:

• n! = 1 × 2 × 3 × ⋯ × (n – 1) × n

 

Calculation:

Using \({\log _a}b = \frac{1}{{{{\log }_b}a}}\)

\(\frac{1}{{{{\log }_2}{\rm{N}}}} + \frac{1}{{{{\log }_3}{\rm{N}}}} + \ldots + \frac{1}{{{{\log }_{100}}{\rm{N\;}}}} = {\log _{\rm{N}}}2 + {\log _{\rm{N}}}3 + \ldots + {\log _{\rm{N}}}100\)

= log_N (2 × 3 × ⋯ × 100)

= log_N (100!)

\(= \frac{1}{{{{\log }_{100!}}N}}\)

Concept:

Formula used:

• aⁿ + bⁿ = {a + b} × {a^n-1 - a^n-2b + a^n-1b² – ⋯ – ab^n-2 + b^n-1}

Calculation:

Using aⁿ + bⁿ = {a + b} × {a^n-1 - a^n-2b + a^n-1b² – ⋯ – ab^n-2 + b^n-1}

5⁵ + 7⁵ = {5 + 7} × {5⁴ – 5³ × 7 + ⋯ +7⁴}

= 12 × {5⁴ – 5³ × 7 + ⋯ +7⁴}

Hence, 5⁵ + 7⁵ is divisible 12.

Concept:

Let a, ar, ar², … up to infinite terms, be an infinite G.P. with first term as a & r as common ratio.

So, sum \(= \frac{a}{{1 - r}}\) for |r| < 1

Note:

• N^th term of GP = ar^{(r – 1)}
• Sum of first n terms, \({{\rm{S}}_{\rm{n}}} = \left\{ {\begin{array}{*{20}{c}} {\frac{{{\rm{a}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - 1}},\;\;\;r \ne 1}\\ {na\;\;\;\;\;\;\;\;,\;\;\;r = 1} \end{array}} \right.\)

Calculation:

Given:

x = 1 – y + y² – y³ + ⋯ up to infinite terms.

|y| < 1

So, y ∈ (-1,1) is given.

x = 1 + y² + y⁴ + ⋯ up to infinite terms – {y + y³ + y⁵ + ⋯ up to infinite terms}

Here, we can see two infinite G.P.s are formed with r = y² and |y| < 1.

Sum of an infinite G.P. is given as \(= \frac{a}{{1 - r}}\) for |r| < 1

\(\Rightarrow x = \frac{1}{{1 - {y^2}}} - \frac{y}{{1 - {y^2}}}\)

\(\Rightarrow x = \frac{{1 - y}}{{\left( {1 - y} \right)\left( {1 + y} \right)}}\)

\(\Rightarrow x = \frac{1}{{1 + y}}\)

Alternate method:

x = 1 – y + y² – y³ + ⋯

⇒ x = 1 – y + y²{1 – y + y² – y³ ⋯}

⇒ x = 1 – y + y²x

⇒ x – y²x = 1 – y

⇒ x(1 – y²) = 1 – y

⇒ x(1 – y)(1 + y) = 1 – y

⇒ x(1 + y) = 1      [∵y ≠ 1, as y ∈ (-1,1)]

\(\Rightarrow x = \frac{1}{{1 + y}}\)

Concept:

Let A be a non-singular 3 × 3 matrix, then inverse of A is given as:

\({{\rm{A}}^{ - 1}} = \frac{1}{{\left| {\rm{A}} \right|}} \times {\rm{adj\;}}\left( {\rm{A}} \right)\)

Steps for calculating inverse of a matrix:

• Find the determinant of given matrix. If |A| ≠  0, then inverse  exists.
• Find the minors of the matrix:
• Find the co-factor of matrix formed:

           C = [c_ij]_{3 × 3} = (-1)^{(i + j)} × m_ij

• Take the transpose of the matrix i.e., C^T: it gives Adj of matrix
• Swapping the (ij)^th element with (ji)^th element.

 

Calculation:

Given: \(A = \left( {\begin{array}{*{20}{c}} {\cos \theta }&{\sin \theta }&0\\ { - \sin \theta }&{\cos \theta }&0\\ 0&0&1 \end{array}} \right)\)

Determinant of A:

|A| = cos θ {cos θ × 1 – 0} – sin θ {-sin θ × 1 – 0} + 0

= cos² θ + sin² θ

= 1

Finding minors of the matrix:

m₁₁ = cos θ × 1 – 0 = cos θ

m₁₂ = -sin θ × 1 – 0 = -sin θ

m₁₃ = 0

m₂₁ = sin θ × 1 – 0 = sin θ

m₂₂ = cos θ × 1 – 0 = cos θ

m₂₃ = 0

m₃₁ = sin θ × 0 – 0 × cos θ = 0

m₃₂ = sin θ × 0 – 0 × cos θ = 0

m₃₃ = cos θ × cos θ – (-sin θ) × sin θ = 1

Finding co-factors:

\(C = \left[ {\begin{array}{*{20}{c}} {\cos \theta }&{\sin \theta }&0\\ { - \sin \theta }&{\cos \theta }&0\\ 0&0&1 \end{array}} \right]\)

Now, adj (A) = C^T

\( \Rightarrow adj\;\left( A \right) = \left[ {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta }&0\\ {\sin \theta }&{\cos \theta }&0\\ 0&0&1 \end{array}} \right]\)

∴ \({A^{ - 1}} = \frac{1}{{\left| A \right|}} \times \left[ {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta }&0\\ {\sin \theta }&{\cos \theta }&0\\ 0&0&1 \end{array}} \right]\)

\(\Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta }&0\\ {\sin \theta }&{\cos \theta }&0\\ 0&0&1 \end{array}} \right]\)

Concept:

Multiplication of matrices:

Let A be a matrix of m × n, B be a matrix of p × q then AB is a matrix of m × q and for multiplication to exist n = p.

Calculation:

Let B be a p × q matrix.

Now, we know that if A be a matrix of m × n, B be a matrix of p × q then AB is a matrix of m × q and for multiplication to exist n = p.

So, A_{2 × 3} × B_{p × q} = (AB)_{2 × 5}

Hence, p = 3 and q = 5.

∴ B should be a 3 × 5 matrix.

Concept:

Matrix Multiplication:

• First make sure the number of columns of first matrix is equal to number of rows of second matrix.
• Multiply elements of each row of first matrix to elements of each column of second matrix.

Let A = [a_ij]_{m × p} & B = [b_ij]_{p × n} then AB = [c_ij]_{m × n} where \({{\rm{c}}_{{\rm{ij}}}} = \mathop \sum \limits_{{\rm{k}} = 1}^{\rm{p}} \left( {{{\rm{a}}_{{\rm{ik}}}}{{\rm{b}}_{{\rm{kj}}}}} \right)\).

Formulas used:

• Let A = [a_ij]_{m × n} be a matrix then kA = [b_ij]_{m × n} where b_ij = ka_ij.
• I_n is an identity matrix of n × n.
• O is a null matrix, with every element zero.

 

Calculation:

Given:

\(A = \left( {\begin{array}{*{20}{c}} 1&2\\ 2&3 \end{array}} \right)\)

A² – kA – I₂ = 0

To find: Value of k

\({{\rm{A}}^2} = \left( {\begin{array}{*{20}{c}} 1&2\\ 2&3 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1&2\\ 2&3 \end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}} {1 \times 1 + 2 \times 2}&{1 \times 2 + 2 \times 3}\\ {2 \times 1 + 3 \times 2}&{2 \times 2 + 3 \times 3} \end{array}} \right)\)

\(= \left( {\begin{array}{*{20}{c}} 5&8\\ 8&{13} \end{array}} \right)\)

\({\rm{kA}} = {\rm{k}}\left( {\begin{array}{*{20}{c}} 1&2\\ 2&3 \end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}} {1{\rm{k}}}&{2{\rm{k}}}\\ {2{\rm{k}}}&{3{\rm{k}}} \end{array}} \right)\)

\({{\rm{I}}_2} = \left( {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right)\)

Now, \({{\rm{A}}^2} - {\rm{kA}} - {{\rm{I}}_2} = 0\)

\(\Rightarrow \left( {\begin{array}{*{20}{c}} 5&8\\ 8&{13} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} {1{\rm{k}}}&{2{\rm{k}}}\\ {2{\rm{k}}}&{3{\rm{k}}} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right) = 0\)

So, 5 – k – 1 = 0

⇒ k = 4

Concept:

If n points are there in a plane and none of the 3 are collinear then,

Number of triangles formed = ⁿC₃

Formulas used:

• ⁿC_r \(= \frac{{n!}}{{\left( {n - r} \right)!\left( r \right)!}}\)
• n! = 1 × 2 × 3 × ⋯ × n

Calculation:

Number of triangles formed by 12 points = ¹²C₃

But, 7 points are collinear, so triangles cannot be formed. And hence, the number of triangles which could not be formed by these 7 points mutually should be removed.

∴ Number of triangles formed will be = ¹²C₃ – ⁷C₃

\(= \frac{{12!}}{{\left( {12 - 3} \right)!\left( 3 \right)!}} - \frac{{7!}}{{\left( {7 - 3} \right)!\left( 3 \right)!}}\)

\(= \frac{{12!}}{{\left( 9 \right)!\left( 3 \right)!}} - \frac{{7!}}{{\left( 4 \right)!\left( 3 \right)!}}\)

\(= \frac{{12 \times 11 \times 10 \times 9!}}{{\left( 9 \right)!\left( 3 \right)!}} - \frac{{7 \times 6 \times 5 \times 4!}}{{\left( 4 \right)!\left( 3 \right)!}}\)

\(= \frac{{12 \times 11 \times 10}}{{3 \times 2}} - \frac{{7 \times 6 \times 5}}{{3 \times 2}}\)

= 220 – 35

= 185

Concept:

Formula used:

• ⁿC_r + ⁿC_{(r – 1)} = ^{(n + 1)}C_r

Calculation:

ⁿC_r + 2ⁿC_{(r – 1)} + ⁿC_r-2­ = ⁿC_r + ⁿC_{(r – 1)} + ⁿC_{(r – 1)} + ⁿC_{(r – 2)}

Using ⁿC_r + ⁿC_{(r – 1)} = ^{(n + 1)}C_r

= ^{(n + 1)}C_r + ^{(n + 1)}C_{(r – 1)}

Again using ⁿC_r + ⁿC_{(r – 1)} = ^{(n + 1)}C_r

= ^{(n + 2)}C_r

Concept:

Let A and B be the subsets of X, then then Venn diagram to represent that is:

Union of sets (A ∪ B):

 

Intersection of subsets (A ∩ B):

 

Complement of subset (A’):

 

Difference of subsets (A - B):

 

Formulas used:

• (A ∩ B) = (A) + (B) - (A ∪ B)

 

Calculation:

Given:

Total students (X) = 850

Students like music (M) = 680

Students like dance (D) = 215

 

We need to find number of students who like both music and dance = (M ∩ D)

We know that, (A ∩ B) = (A) + (B) - (A ∪ B)

So, (M ∩ D) = (M) + (D) - (M ∪ D)

= 680 + 215 - 850

= 895 - 850

= 45

Concept:

Arithmetic Progression (A.P.): Let a be the first term with common difference d of an A.P. then:

• N^th term, a_n = a + (n - 1)d
• Sum \(= \frac{{\rm{n}}}{2}\left[ {2{\rm{a}} + \left( {{\rm{n}} - 1} \right){\rm{d}}} \right] = \frac{{\rm{n}}}{2}\left[ {{\rm{a}} + {{\rm{a}}_{\rm{n}}}} \right]\)

 

Calculation:

Two-digit numbers that leave remainder 2 when divided by 3 are 11, 14, 17, …, 98

Sum = 11 + 14 + 17 + … + 98

So, it is sum of an A.P.

Let’s find the number of terms

a_n = a + (n - 1)d

⇒ 98 = 11 + (n - 1)3

⇒ 87 = (n - 1)3

⇒ 29 = n - 1

⇒ n = 30

Sum \(= \frac{{\rm{n}}}{2}\left[ {{\rm{a}} + {{\rm{a}}_{\rm{n}}}} \right]\)

\(= \frac{{30}}{2}\left[ {11 + 98} \right]\)

= 15 × 109

= 1635

Concept:

Formulas used:

• i = √(-1)
• i² = -1
• i³ = -i × i² = -i
• i⁴ = (i²)² = (-1)² = 1

 

So, iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³ = 0

Calculation:

\(\mathop \sum \limits_{{\rm{n}} = 2}^{11} \left( {{{\rm{i}}^{\rm{n}}} + {{\rm{i}}^{{\rm{n}} + 1}}} \right) = {{\rm{i}}^2} + {{\rm{i}}^3} + {{\rm{i}}^3} + \cdots + {{\rm{i}}^{10}} + {{\rm{i}}^{11}} + {{\rm{i}}^{11}} + {{\rm{i}}^{12}}\)

= i² + 2{i³ + i⁴ + ⋯ + i¹¹} + i¹²

We know that iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³ = 0

So, i³ + i⁴ + ⋯ + i¹⁰ = 0

\(\mathop \sum \limits_{{\rm{n}} = 2}^{11} \left( {{{\rm{i}}^{\rm{n}}} + {{\rm{i}}^{{\rm{n}} + 1}}} \right) = {{\rm{i}}^2} + 2{{\rm{i}}^{11}} + {{\rm{i}}^{12}}\)

= -1 + 2i × (i²)⁵ + (i²)⁶

= -1 + 2i × (-1) + (-1)⁶

= -1 - 2i + 1

= -2i

Concept:

Trigonometric formulas used:

• sin (x + y) = sin x × cos y + cos x × sin y
• \(\cos {\rm{\theta }} = \sqrt {1 - {{\sin }^2}{\rm{\theta }}} \)

 

Calculation:

Given: \(\sin {\rm{x}} = \frac{1}{{\sqrt 5 }}\) & \(\sin {\rm{y}} = \frac{1}{{\sqrt {10} }}\)

We know that, sin (x + y) = sin x × cos y + cos x × sin y

Now, \(\cos {\rm{\theta }} = \sqrt {1 - {{\sin }^2}{\rm{\theta }}} \)

So, \(\cos {\rm{x}} = {\rm{\;}}\sqrt {1 - {{\sin }^2}{\rm{x}}} \)

⇒ cos x = √(1 - 1/5)

⇒ cos x = 2/√5

And, \(\cos {\rm{y}} = {\rm{\;}}\sqrt {1 - {{\sin }^2}{\rm{y}}} \)

⇒ cos y = √(1 - 1/10)

⇒ cos y = 3/√10

So, sin (x + y) = sin x × cos y + cos x × sin y

\(= \frac{1}{{\sqrt 5 }} \times \frac{3}{{\sqrt {10} }} + \frac{2}{{\sqrt 5 }} \times \frac{1}{{\sqrt {10} }}\)

\( = \frac{3}{{\sqrt {50} }} + \frac{2}{{\sqrt {50} }}\)

= 5/√50

= 1/√2

So, sin (x + y) = 1/√2

⇒ sin (x + y) = sin π/4

⇒ x + y = π/4

Concept:

Trigonometric formulas used:

sin C - sin D = 2[cos (C + D)/2] × [sin (C - D)/2]

cos C + cos D = 2[cos (C + D)/2] × [cos (C - D)/2]

Calculation:

Given trigonometric expression is

\(\frac{{\sin 5{\rm{x}} - \sin 3{\rm{x}}}}{{\cos 5{\rm{x}} + \cos 3{\rm{x}}}}\)

According to the formula used

⇒ (2sinx.cos4x)/(2cosx.cos4x)

⇒ tanx

∴ The value of the expression \(\frac{{\sin 5{\rm{x}} - \sin 3{\rm{x}}}}{{\cos 5{\rm{x}} + \cos 3{\rm{x}}}}\) is tanx.

Concept:

Trigonometric formulas used:

• cos (90° + θ) = -sin θ
• sin (90° + θ) = cos θ
• sin C - sin D = 2cos (C + D)/2 × sin (C - D)/2

 

Calculation:

sin 105° + cos 105° = sin 105° + cos (90° + 15°)

We know that, cos (90° + θ) = -sin θ

= sin 105° - sin (15°)

Now, sin C + sin D = 2cos (C + D)/2 × sin (C - D)/2

= 2 × cos (105° + 15°)/2 × sin (105° - 15°)/2

= 2 × cos 60° × sin 45°

= 2 × (1/2) × (1/√2)

= 1/√2

Concept:

Formulas used:

sin (π - θ) = sin θ

sin^-1 (sin θ) = θ if \(- \frac{{\rm{\pi }}}{2} \le {\rm{\theta }} \le \frac{{\rm{\pi }}}{2}\)

Calculation:

\({\sin ^{ - 1}}\left( {\sin \frac{{2{\rm{\pi }}}}{3}} \right) = {\sin ^{ - 1}}\left( {\sin \left( {{\rm{\pi }} - \frac{{\rm{\pi }}}{3}} \right)} \right)\)

We know that, sin (π - θ) = sin θ

⇒ sin^-1 (sin π/3)

We know that, sin^-1 (sin θ) = θ if \(- \frac{{\rm{\pi }}}{2} \le {\rm{\theta }} \le \frac{{\rm{\pi }}}{2}\)

∴ The value of sin^-1 (sin π/3) is π/3.

Concept:

Sum of internal angles of n sided polygon = (n - 2) × 180°

 

Calculation:

Given:

x, x - y, x + y are the angles of triangles.

Sum of angles of triangles = (n - 2) × 180°

= (3 - 2) × 180°

= 180°

Sum of angles of triangles = 180°

⇒ x + (x - y) + (x + y) = 180°

⇒ 3x = 180°

⇒ x = 60°