Quantitative Aptitude PYQs

Concept:

Logarithm properties

1. Product rule: The log of a product equals the sum of two logs.

\({\log _a}\left( {mn} \right) = ;{\log _a}m + ;{\log _a}n\)

1. Quotient rule: The log of a quotient equals the difference of two logs.

\({\log _a}\frac{m}{n} = ;{\log _a}m - ;{\log _a}n\)

1. Power rule: In the log of a power the exponent becomes a coefficient.

\({\log _a}{m^n} = n{\log _a}m\)

1. Change of base rule

\({\log _m}n = \frac{{{{\log }_a}n}}{{{{\log }_a}m}}\)

If m = n;

⇒ \({\log _m}m = \frac{{{{\log }_a}m}}{{{{\log }_a}m}} = 1\)

1. \({\log _m}n = \frac{1}{{{{\log }_n}m}}\)

 

Calculation:

Here, we have to find the value of \({\log _7}{\rm{;}}{\log _7}\sqrt {7\sqrt {7\sqrt 7 } } \)

\({\log _7}{\rm{;}}{\log _7}\sqrt {7\sqrt {7\sqrt 7 } } \)

= log₇ log₇ (7^1/2 × 7^1/4 × 7^1/8)

= log₇ log₇ (7(^{1/2 + 1/4 + 1/8)})

= log₇ log₇ (7^{(4 + 2 + 1)/8})

= log₇ log₇ (7^7/8)

From power rule;

= log₇ (7/8) log₇7

= log₇ (7/8) × 1 = log₇ (7/8) = log₇ 7 – log₇ 8

= 1 – log₇ 8 = 1 – log₇ 2³

= 1 – 3 log₇ 2

Concept:

If a, ar, ar², …. is an infinite GP, then the sum of infinite geometric series is given by.

Sum of infinite GP = \({{\rm{s}}_\infty } = {\rm{;}}\frac{{\rm{a}}}{{1{\rm{;}} - {\rm{;r}}}}{\rm{;}}\); |r| < 1

Calculation:

Given:

First term of GP = x

Sum of infinite GP = 5

We know that;

S_∞ = a/ (1 - r)

⇒ 5 = x/ (1 - r)

⇒ 5 × (1 - r) = x

⇒ 5 – 5r = x

⇒ 5 – x = 5r

∴ r = (5 – x)/5

For infinite GP |r| < 1

|r| < 1

⇒ -1 < r < 1

⇒ -1 < (5 – x)/5 < 1

⇒ -5 < 5 - x < 5

⇒ -10 < -x < 0

⇒ 0 < x < 10

Concept:

Rational expressions: A rational expression is nothing more than a fraction in which the numerator and/or the denominator are polynomials.

A rational expression is considered simplified if the numerator and denominator have no factors in common.

Steps for obtaining a rational expression.

1. Factor the numerator and the denominator.
2. List restricted values.
3. Cancel common factors.
4. Simplify and note any restricted values not implied by the expression.

 

Calculation:

1. \(x + {x^2} - \frac{1}{x} = {\rm{;}}\frac{{{x^2} + {\rm{;}}{x^3}{\rm{;}} - 1{\rm{;}}}}{x}\)
2. By definition of rational function, it is a rational expression.
3. \(\sqrt {a{x^2} + bx + x - c + \frac{d}{x} - \frac{e}{{{x^2}}}} \)
4. It is not a rational expressions because it is not a polynomials.
5. 3x² - 5x + ab
6. It is in form of a fraction of denominator 1 so it is a rational expression.
7. \(\frac{2}{{{x^{2{\rm{;}}}} - {\rm{;}}ax{\rm{;}} + {\rm{;}}{b^2}}}\)
8. By definition of rational function it is a rational expressions.
9. 1/x - 1/x + 5 = (x + 5 - x) / x(x + 5) = 5 / x(x + 5) = 5 / x² + 5x
10. By definition of rational function it is a rational expressions.

 

∴ Option 1^st is correct.

Concept:

Orthogonal matrix: When the product of a matrix to its transpose gives identity matrix.

Suppose A is a square matrix with real elements and of n x n order and A^T or A’ is the transpose of A.

AA^T = I

Calculation:

Suppose A is a square matrix with real elements and of n x n order and A^T or A’ is the transpose of A.

Then according to the definition;

AA^T = I

Pre multiplication by A^-1

A^-1 AA^T = A^-1 I

IA^T = A^-1

A^T = A^-1 or A’ = A^-1

then A is orthogonal matrix.

∴ Option 2 is correct

Concept:

If A, B and C are subsets of a set X. Then

Commutative property: A ⋃ B = B ⋃ A

Associative property: (A ⋃ B) ∪ C = A ⋃ (B ⋃ C)

Distributive property: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) or A ⋂ (B ⋃ C) = (A ⋂ B) ⋃ (A ⋂ C)

Some others property

I. A ∪ A = A, and A ∩ A = A

II. A ∩ (A ∪ B) = A, A ∪ (A ∩ B) = A

III. (A ∩ B) ∪ C = (A∩C) ∪ (B∩C)

IV. (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

V. (A ⋃ B)’ = A’ ⋂ B’ and (A ⋂ B)’ = A’ ⋃ B’

VI. (A’)’ = A

Calculation:

1. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

According to Distributive property option 1 is true. 2. L.H.S = A’ ∪ (A ∪ B) = A ∪ B

 R.H.S = (B’ ∩ B)’ ∪ A’ = ((B’)’ ⋃ B’) ⋃ A’ = (B ⋃    B’) ⋃ A’ = (A ∪ B) ⋃ A’ = A ∪ B

 So, option 2 is correct. 3. L.H.S = A’ ∪ (B ∪ C)

 R.H.S = (C’ ∩ B)’ ∩ A’ = ((C’)’ ⋃ B) ⋂ A’ = (C ⋂ B) ⋂ A’

 L.H.S ≠ R.H.S

 So, option 3 is incorrect.

Calculation:

Total number of integers lying between 2999 and 8001 = 8001 - 2999 - 1 = 5001

These are four digits numbers.

For distinct digits, starting digit can be filled by 5 ways (3, 4, 5, 6 or 7)

Total number of all distinct digits = 5 × 9 × 8 × 7 = 2520

Now, Total number which have at least two digits equal = 5001 - 2520 = 2481

∴ x = 2481

Concept:

If a, ar, ar², …. is an infinite GP, then the sum of infinite geometric series is given by.

Sum of infinite GP = \({s_\infty } = ;\frac{a}{{1; - ;r}}\); |r| < 1

Calculation:

Here, we have to find the sum of the series \(3 - 1 + \frac{1}{3} - \frac{1}{9} + \ldots \)

As, we can see that sequence is an GP with a = 3 and r = - 1/3

Sum of the series = S_∞ = a/ (1 - r)

Sum of the series = \(\frac{3}{{1; - ;\frac{{ - 1}}{3}}} = ;\frac{3}{{1 + ;\frac{1}{3}}} = ;\frac{3}{{\left( {\frac{4}{3}} \right)}} = ;\frac{9}{4}\)

Concept:

Let A, B and C denote the three set of elements.

n(A), n(B) and n(C) are the number of elements present in sets A, B and C.

n(A ⋃ B) is the number of elements present in either of the sets A or B.

n(A ⋂ B), n(B ⋂ C) and n(A ⋂ C) are the number of elements present in both the sets (A and B) ,

(B and C) and (A and C).

1. No. of elements present in at least one set

n(A ⋃ B ⋃ C) = n(A) + n(B) + n(C) - n(A ⋂ B) - n(B ⋂ C) - n(A ⋂ C) + n(A ⋂ B ⋂ C)

2. Number of elements present in exactly only one set

n(A) + n(B) + n(C) - 2 × [n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C)] + 3 × n(A ⋂ B ⋂ C)]

3. Number of elements present in exactly two of the sets

n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C)] - 3 × n(A ⋂ B ⋂ C)

Calculation:

Let A, B, and C be the set of persons who can like to play cricket, football and tennis

Given: n(A ⋃ B ⋃ C) = 300, n(A) = 125, n(B) = 145, n(C) = 90

n(A ⋃ B ⋃ C) = n(A) + n(B) + n(C) - n(A ⋂ B) - n(B ⋂ C) - n(A ⋂ C) + n(A ⋂ B ⋂ C)

⇒ 300 = 125 + 145 + 90 - n(A ⋂ B) - n(B ⋂ C) - n(A ⋂ C) + n(A ⋂ B ⋂ C)

⇒ n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C) = 360 - 300 + n(A ⋂ B ⋂ C)

⇒ n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C) = 60 + n(A ⋂ B ⋂ C)     ….(1)

Again;

32 students like to play exactly two games out of the three games.

n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C) - 3 × n(A ⋂ B ⋂ C) = 32

⇒ n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C) = 32 + 3 × n(A ⋂ B ⋂ C)     ….(2)

Put this value in equation 1^st

n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C) = 60 + n(A ⋂ B ⋂ C)

⇒ 32 + 3 × n(A ⋂ B ⋂ C) = 60 + n(A ⋂ B ⋂ C)

⇒ 2 × n(A ⋂ B ⋂ C) = 60 - 32

⇒ n(A ⋂ B ⋂ C) = 28/2

∴ n(A ⋂ B ⋂ C) = 14

Calculation:

Number of students like to play exactly only one game = n(A) + n(B) + n(C) - 2 × [n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C)] + 3 × n(A ⋂ B ⋂ C)

From equation 2

= 125 + 145 + 90 - 2 × [32 + 3 × n(A ⋂ B ⋂ C)] + 3 × 14

= 360 - 2 × [32 + 3 × 14] + 42

= 360 - 2 × 74 + 42

= 402 - 148

= 254

Concept:

Let us consider the standard form of a quadratic equation,

ax² + bx + c = 0

Where a, b and c are constants with a ≠ 0

Let α and β be the two roots of the above quadratic equation. 

The sum of the roots of a quadratic equation are:

\({\rm{\alpha }} + {\rm{\beta }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient;of;x}}}}{{{\rm{coefficient;of;}}{{\rm{x}}^2}}}\)The product of the roots is given by:

\({\rm{\alpha \beta }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant;term}}}}{{{\rm{coefficient;of;}}{{\rm{x}}^2}}}\)

Greatest value of a quadratic equation = \(\frac{{ - D}}{{4a}} = \frac{{ - ;\left( {{b^2} - 4ac} \right)}}{{4a}}\)

Calculation:

Given: x² + αx - β = 0

The sum of the roots = α + β = - α     ….

(1) The product of the roots = α × β = - β

⇒ α β + β = 0

⇒ (α + 1) β = 0

∴ (α + 1) = 0 and β ≠ 0

∴ α = - 1 and β ≠ 0

Now put the value of α in equation 1^st

α + β = - α

⇒ - 1 + β = - ( - 1)

∴ β = 2

Again; - x² + α x + β = - x² - x + 2

Greatest value of - x² + α x + β = - D/4a

Greatest value = - (1 + 8) / - 4 = 9/4

Concept:

General term: General term in the expansion of (x + y)ⁿ is given by

\({T_{\left( {r; + ;1} \right)}} = ;{;^n}{C_r} \times {x^{n - r}} \times {y^r}\)

Middle terms: The middle terms is the expansion of (x + y) ⁿ depends upon the value of n.

• If n is even, then the total number of terms in the expansion of (x + y) ⁿ is n +1. So there is only one middle term i.e. \(\left( {\frac{n}{2} + 1} \right){{\rm{;}}^{th}}\) term is the middle term.

\({T_{\left( {\frac{n}{2}; + ;1} \right)}} = ;{;^n}{C_{\frac{n}{2}}} \times {x^{\frac{n}{2}}} \times {y^{\frac{n}{2}}}\)

• If n is odd, then the total number of terms in the expansion of (x + y) ⁿ is n + 1. So there are two middle terms i.e. \({\left( {\frac{{n; + ;1}}{2}} \right)^{th}};\)and \({\left( {\frac{{n; + ;3}}{2}} \right)^{th}}\) are two middle terms.

 

Calculation:

Here, we have to find the coefficient of the middle term in the binomial expansion of (2 + 3x) ⁴

Here n = 4 (n is even number)

∴ Middle term = \(\left( {\frac{n}{2} + 1} \right) = \left( {\frac{4}{2} + 1} \right) = 3^rd;term\)

T₃ = T _{(2 + 1)} = ⁴C₂ × (2) ^{(4 - 2)} × (3x) ²

T₃ = 6 × 4 × 9x² = 216 x²

∴ Coefficient of the middle term = 216

Concept:

If A is a square matrix;

1. (AB) ^{- 1} = B ^{- 1} A ^{- 1}
2. (A ^{- 1}) ^{- 1} = A
3. (A^T) ^{- 1} = (A ^{- 1})^T
4. (KA ^{- 1}) = \(\frac{1}{k};{A^{ - 1}}\) for any K ≠ 0
5. (Aⁿ) ^{- 1} = (A ^{- 1})ⁿ
6. AA ^{- 1} = A ^{- 1}A = I

Calculation:

1. If matrix A is square matrix and invertible matrix. Such that AA ^{- 1} = A ^{- 1}A = I also (AB) ^{- 1} = B ^{- 1} A ^{- 1}

AA ^{- 1} = I

(AA ^{- 1}) ^{- 1} = I ^{- 1}

We know that (AB) ^{- 1} = B ^{- 1} A ^{- 1}

∴ (A ^{- 1}) ^{- 1} A ^{- 1} = I ^{- 1} = I

Post multiple by A

⇒ (A ^{- 1}) ^{- 1} A ^{- 1} A = IA

⇒ (A ^{- 1}) ^{- 1} I = A

∴ (A ^{- 1}) ^{- 1} = A

∴ Option 1 is correct. 2. We know that det (AB) = det (A) det (B)

If matrix A is square matrix and invertible matrix. Such that AA ^{- 1} = A ^{- 1}A = I

det (AA ^{- 1}) = det (I)

det (A) det (A ^{- 1}) = 1

∴ \(\det \left( {{A^{ - 1}}} \right) = \frac{1}{{detA}}\)

∴ Option 2 is correct. 3. Given: (λA) ^{- 1} = λA ^{- 1} where λ is a scalar

We know that (KA ^{- 1}) = \(\frac{1}{k};{A^{ - 1}}\) for any K ≠ 0

∴ Option 3 is wrong.

Concept:

Elementary row or column transformations do not change the value of the determinant of a matrix.

Calculation:

\(\left| {\begin{array}{*{20}{c}} x&y&3\ {{x^2}}&{5{y^3}}&9\ {{x^3}}&{10{y^3}}&{27} \end{array}} \right|\)

Apply C₁ → C₁ - C₃

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {x - 3}&y&3\ {{x^2} - 9}&{5{y^3}}&9\ {{x^3} - 27}&{10{y^3}}&{27} \end{array}} \right|\)

\(\Rightarrow ;\left| {\begin{array}{*{20}{c}} {x - 3}&y&3\ {\left( {{\rm{x}} - 3} \right)\left( {x + 3} \right)}&{5{y^3}}&9\ {\left( {{\rm{x;}} - 3} \right)\left( {{x^2} + 9 + 3x} \right)}&{10{y^3}}&{27} \end{array}} \right| = \left( {x - 3} \right)\left| {\begin{array}{*{20}{c}} 1&y&3\ {\left( {x + 3} \right)}&{5{y^3}}&9\ {\left( {{x^2} + 9 + 3x} \right)}&{10{y^3}}&{27} \end{array}} \right|\)

∴ (x - 3) is a factors of the expansion of the determinant.

Concept:

Let \(\rm A; = ;\left[ {{a_{ij}}} \right]\) be a square matrix.

Steps for obtaining adjoint of the matrix

1. We need to calculate cofactors of all elements

⇒ \(\rm {C_{ij}};or;{A_{ij}} = {\left( { - 1} \right)^{i + j}};{M_{ij}}\)

Here, \(\rm {M_{ij}}\) = Minor of matrix A

2. Transpose of cofactor is called adjoint of the matrix

⇒ adj (A) = C^T

Calculation:

⇒ cos ( - θ) = cos θ

⇒ sin ( - θ) = sin θ

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

Cofactors of matrix A,

A₁₁ = ( - 1)^{1 + 1} cos θ = cos θ

A12 = ( - 1)^{1 + 2} sin θ = - sin θ

A21 = ( - 1)^{2 + 1} sin θ = - sin θ

A22 = ( - 1)^{2 + 2} cos θ = cos θ

Cofactors of matrix A = \(= {A_{ij}} = ;\left( {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta }\ { - \sin \theta }&{\cos \theta } \end{array}} \right)\)

adj (A) = \({\left[ {{A_{ij}}} \right]^T} = \left( {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta }\ { - \sin \theta }&{\cos \theta } \end{array}} \right)\)

∴ Option 1 is correct.

Shortcut Method

Let A = \(\left( {\begin{array}{*{20}{c}} {\rm{a}}&b\ {\rm{c}}&{\rm{d}} \end{array}} \right)\)

adj A = \(\left( {\begin{array}{*{20}{c}} {\rm{d}}&{ - b}\ { - {\rm{c}}}&{\rm{a}} \end{array}} \right)\) 

Note: Interchange the diagonal elements and change the sign of remaining elements**.**

Concept:

Cube Roots of unity are 1, ω and ω²

Here, ω = \(\frac{{ - ;1; + ;i\sqrt 3 }}{2}\) and ω² = \(\frac{{ - ;1; - ;i\sqrt 3 }}{2}\)

Property of cube roots of unity

1. ω³ = 1
2. 1 + ω + ω² = 0
3. ω = 1 / ω ² and ω² = 1 / ω
4. ω³ⁿ = 1

Calculation:

We know that ω = \(\frac{{ - ;1; + ;i\sqrt 3 }}{2}\) and ω² = \(\frac{{ - ;1; - ;i\sqrt 3 }}{2}\)

\({\left( {\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}} \right)^{3{\rm{n}}}} + {\left( {\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}} \right)^{3{\rm{n}}}} = {\rm{;}}{{\rm{\omega }}^{3{\rm{n}}}} + {({{\rm{\omega }}^2})^{3{\rm{n}}}} = ;{\left( {{\omega ^3}} \right)^n} + ;{\left( {{\omega ^3}} \right)^{2n}}\) = 1 + 1 = 2

Concept:

Combination: Selecting r objects from given n objects.

The number of selections of r objects from the given n objects is denoted by

\({^n}{C_r}; = ;\frac{{n!}}{{r!\left( {n; - ;r} \right)!}}\)

Note: Use combinations if a problem calls for the number of ways of selecting objects.

Calculation:

There are 17 cricket players, out of which 5 players can bowl.

We have to make a team of 11 players so to include 3 bowlers

So, we select 3 bowlers out of 5 players and remaining 8 players must be selected from 12 players

Total ways = C (5, 3) × C (12, 8)

∴ Option 4 is correct.

Concept:

Logarithm properties

Product rule:

The log of a product equals the sum of two logs.

\({\log _a}\left( {mn} \right) = ;{\log _a}m + ;{\log _a}n\)

Quotient rule:

The log of a quotient equals the difference of two logs.

\({\log _a}\frac{m}{n} = ;{\log _a}m - ;{\log _a}n\)

Power rule:

In the log of power, the exponent becomes a coefficient.

\({\log _a}{m^n} = n{\log _a}m\)

Change of base rule

\({\log _m}n = ;\frac{{{{\log }_a}n}}{{{{\log }_a}m}}\)

If m = n;

⇒ \({\log _m}m = ;\frac{{{{\log }_a}m}}{{{{\log }_a}m}} = 1\)

 

Calculation:

log₉ 27 + log₈ 32

= \(\frac{{\log 27}}{{\log 9}} + ;\frac{{\log 32}}{{\log 8}}\)

= \(\frac{{\log {3^3}}}{{\log {3^2}}} + ;\frac{{\log {2^5}}}{{\log {2^3}}} \)

\(= ;\frac{{3;\log 3}}{{2;\log 3}} + ;\frac{{5;\log 2}}{{3;\log 2}}\)

\(= ;\frac{3}{2}; + ;\frac{5}{3}\)

\( = ;\frac{{9 + 10}}{6} = ;\frac{{19}}{6}\)

Concept:

If A is a square matrix;

(AB) ^{- 1} = B ^{- 1} A ^{- 1}

Calculation:

We know that AA ^{- 1} = I

So, (AB) (AB) ^{- 1} = I

Pre multiply by A ^{- 1},

⇒ (A ^{- 1} AB) (AB) ^{- 1} = A ^{- 1} I

⇒ (I B) (AB) ^{- 1} = A ^{- 1}

⇒ B (AB) ^{- 1} = A ^{- 1}

Pre multiply by B ^{- 1},

⇒ B ^{- 1} B (AB) ^{- 1} = B ^{- 1} A ^{- 1}

⇒ I (AB) ^{- 1} = B ^{- 1} A ^{- 1}

∴ (AB) ^{- 1} = B ^{- 1} A ^{- 1}

Note: (AB) ^{- 1} ≠ A ^{- 1} B ^{- 1}

Concept:

Elementary row or column transformations do not change the value of the determinant of a matrix.

Calculation:

\(\left| {\begin{array}{*{20}{c}} {a - x}&c&b\ c&{b - x}&a\ b&a&{c - x} \end{array}} \right| = 0\)

Apply R₁ → R₁ + R₂ + R₃

\(\left| {\begin{array}{*{20}{c}} {a + b + c - x}&{a + b + c - x}&{a + b + c - x}\ c&{b - x}&a\ b&a&{c - x} \end{array}} \right| = 0\)

Take common (a + b + c - x) from row R₁

⇒ (a + b + c - x) \(\left| {\begin{array}{*{20}{c}} 1&1&1\ c&{b - x}&a\ b&a&{c - x} \end{array}} \right| = 0\)

Now; (a + b + c - x) = 0 and \(\left| {\begin{array}{*{20}{c}} 1&1&1\ c&{b - x}&a\ b&a&{c - x} \end{array}} \right| = 0\)

(a + b + c - x) = 0

⇒ x = a + b + c

Given: a + b + c = 0

∴ x = 0

Concept:

Consider a matrix A and let its inverse be A ^{- 1}

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

Here; adj (A) is adjoint of matrix A and det (A) is determinant of matrix A.

⇒ If det (A) ≠ 0, so inverse of matrix is exist.

⇒ If det (A) = 0, so inverse of matrix is not exist.

If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\) then determinant of A is given by: |A| = (a­₁₁ × a₂₂) - (a₁₂ - a₂₁).

Calculation:

Let A = \(\left( {\begin{array}{*{20}{c}} 2&4\ { - 8}&{\rm{x}} \end{array}} \right)\)

Given Matrix A does not have an inverse.

So, det (A) = 0

|A| = (2x) - (4 × - 8) = 0

⇒ 2x - ( - 32) = 0

⇒ 2x = - 32

∴ x = - 16