Quantitative Aptitude PYQs

Concept:

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

1. Distinct and real roots if D > 0.
2. Real and repeated roots, if D = 0.
3. Complex roots and conjugate of each other, D < 0.

Calculation:

Given:

Both p and q belong to the set {1, 2, 3, 4} and the quadratic equation of the form, px² + qx + 1 = 0 has real roots.

By comparing the quadratic equation of the form, px² + qx + 1 = 0 with the standard quadratic equation ax² + bx + c = 0. We get, a = p, b = q and c = 1.

D = q² - 4p

∵ The roots are real.

⇒ D ≥ 0 ⇒ q2 - 4p ≥ 0

⇒ q² ≥ 4p

∵ p and q belong to the set {1, 2, 3, 4}

Case -1:

• If p = 1 then q² ≥ 4 ⇒ q ∈ {2, 3, 4}
• Hence, for p = 1, q can take 3 values
• So, 3 quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 1.

Case -2:

• If p = 2 then q² ≥ 8 ⇒ q ∈ {3, 4}
• Hence, for p = 2, q can take 2 values
• So, 2 quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 2.

Case -3:

• If p = 3 then q² ≥ 12 ⇒ q ∈ {4}
• Hence, for p = 1, q can take 1 value
• So, only 1 quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 3.

Case -4:

• If p = 4 then q² ≥ 16 ⇒ q ∈ {4}
• Hence, for p = 4 , q can take 1 value
• So, only 1 quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 4.

So, in total we get, 3 + 2 + 1 + 1 = 7 possible quadratic equations of the form px² + qx + 1 = 0 that can be formed such that both p and q belong to the set {1, 2, 3, 4} and have real roots.

Concept:

If a₁, a₂, …., a_n be an AP then \({S_n} = {a_1} + {a_2} + \ldots + {a_n} = \frac{n}{2} \times \left( {2a + \left( {n - 1} \right)d} \right)\) where a is the 1^st term and d is the common difference.

If a₁, a₂, …, a_n be an AP then the general term is given by: a_n = a + (n - 1) × d where a is the 1^st term and d is the common difference.

Calculation:

Here, we have to find the value of 1 - 2 + 3 - 4 + 5 - ______ + 101.

⇒ 1 - 2 + 3 - 4 + 5 - ______ + 101 = (1 + 3 + …… + 101) - (2 + 4 + ….. + 100)

As, we can see that (1, 3, ……, 101) is an AP with a = 1 and d = 2.

⇒ a_n = 101 = 1 + (n - 1) × 2

⇒ n = 51

\(\Rightarrow {S_{51}} = 1 + 3 + \ldots + 101 = \frac{{51}}{2} \times \left( {2 + 50 \times 2} \right) = 2601\)

Similarly, (2, 4, …, 100) is an AP with a = 2 and d = 2

⇒ a_n = 100 = 2 + (n - 1) × 2

⇒ n = 50

\(\Rightarrow {S_{50}} = 2 + 4 + \ldots + 100 = \frac{{50}}{2} \times \left( {4 + 49 \times 2} \right) = 2550\)

⇒ 1 - 2 + 3 - 4 + 5 - ______ + 101 = (1 + 3 + …… + 101) - (2 + 4 + ….. + 100) = 2601 - 2550 = 51.

Concept:

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

I. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

II. A ∪ A = A, A ∩ (A ∪ B) = A, A ∪ (A ∩ B) = A and A ∩ A = A

III. (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)

IV. (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

Calculation:

⇒ A ∪ (A ∩ B) = (A ∪ A) ∩ (A ∪ B) = A ∩ (A ∪ B) = A       --- (Using property I and II)

So, option 1 is not correct.

⇒ A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B) = A ∪ (A ∩ B) = A       --- (Using property I and II)

So, option 2 is correct.

⇒ (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)       --- (Using property III)

So, option 3 is correct.

⇒ (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)       --- (Using property IV)

So, option 4 is correct.

Concept:

S_n denotes the sum of the first n terms of the sequence.

T_n = a_n = S_n - S_n-1

Calculation:

Given: S_n = n + 12

By substituting n = 2 in the given equation, we get

⇒ S₂ = 2 + 12 = 14

Similarly by substituting n = 3 in the given equation, we get

⇒ S₃ = 3 + 12 = 15

⇒ a₃ = S₃ - S₂ = 15 - 14 = 1.

Concept:

If α and β are the roots of the quadratic equation, ax² + bx + c = 0. Then

\(\alpha + \beta = - \frac{b}{a};and;\alpha × \beta = \frac{c}{a}\)

Calculation:

Given: 2x² - 2(k - 2) x - (k + 1) = 0

By comparing the given equation with the quadratic equation, ax² + bx + c = 0.

We get, a = 2, b = - 2(k - 2) and c = - (k + 1).

Let α and β be the roots of the equation 2x² - 2(k - 2) x - (k + 1) = 0.

\(\Rightarrow \alpha + \beta = - \frac{{ - ;2\left( {k - 2} \right)}}{2} = k - 2\)

\(\Rightarrow \alpha × \beta = \frac{{ - \left( {k + 1} \right)}}{2}\)

As we know that, α² + β² = (α + β) ² - 2αβ

⇒ α² + β² = (k - 2) ² + (1 + k)

⇒ α2 + β2 = k² + 4 - 4k + k + 1

⇒ α2 + β2 = k2 - 3k + 5

⇒ α2 + β2 = k2 - (3 × 2/2)k + 9/4 - 9/4 + 5

⇒ α² + β² = (k - 3/2)² + 11/4

So, the minimum value of α² + β² is found when k - 3/2 = 0 ⇒ k = 3/2.

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, D < 0.

If a, b and c are in HP, then the harmonic mean, \(b = \frac{{2ac}}{{a + c}}\)

Calculation:

Given: a (b - c) x² + b (c - a) x + c (a - b) = 0 has equal roots ⇒ Discriminant, D = 0.

By comparing the given equation with the quadratic equation, ax² + bx + c = 0. We get, a’ = a (b - c) , b’ = b (c - a) and c’ = c (a - b)

As, Discriminant, D = 0.

⇒ D = b² - 4ac = b² × (c - a) ² - 4 × a (b - c) × c (a - b) = 0

⇒ D = (bc + ab - 2ac) ² = 0

⇒ bc + ab - 2ac = 0

⇒ b × (a + c) = 2ac

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

Hence a, b and c are in HP.

Concept:

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

Calculation:

Given: |x² - 3x + 2| > x² - 3x + 2

Case-1: If x ≥ 0

⇒ x² - 3x + 2 > x² - 3x + 2

∴ No real value of x satisfies the above equation.

Case-2:- If x < 0

⇒ - (x² - 3x + 2) > x² - 3x + 2

⇒ x² - 3x + 2 < 0

⇒ (x - 1) (x - 2) < 0

⇒ 1 < x < 2.

Concept:

If a₁, a₂, ….., a_n are in GP then common ratio is given by: r = a_{i + 1} / a_i ∀ i =1, 2, …., n - 1.

The n^th term in a GP is given by: a_n = ar^{n - 1}, where a is the first term and r is the common ratio.

Calculation:

Given: The sum of odd terms of the GP is m and sum of even terms of the GP is n.

\(\Rightarrow \frac{{Sum;of;even;terms;of;GP}}{{Sum;of;odd;terms;of;GP}} = \frac{{\left( {ar + a{r^3} + \ldots + a{r^{199}}} \right)}}{{\left( {a + a{r^2} + \ldots + a{r^{198}}} \right)}} = \frac{n}{m}\)

\(\Rightarrow \frac{{Sum;of;even;terms;of;GP}}{{Sum;of;odd;terms;of;GP}} = \frac{{ar \times \left( {1 + {r^2} + \ldots + {r^{198}}} \right)}}{{a \times \left( {1 + {r^2} + \ldots + {r^{198}}} \right)}} = r = \frac{n}{m}\)

⇒ r = n / m.

Concept:

If A and B are two sets then, n (A ∪ B) = n (A) + n (B) - n (A ∩ B)

Calculation:

Given: n (A) = 3 and n (B) = 6.

As we know that, if A and B are two sets then, n (A ∪ B) = n (A) + n (B) - n (A ∩ B)

⇒ n (A ∪ B) = 3 + 6 - n (A ∩ B)

In order to minimize n (A ∪ B) we have to maximize n (A ∩ B) .

If A is a subset of B, then A ∩ B = A ⇒ n (A ∩ B) = n (A) = 3

⇒ n (A ∪ B) = 3 + 6 - 3 = 6.

Hence, the minimum number of elements that (A∪B) can have is 6.

Concept:

The no. of diagonals that can be drawn by joining the angular points of a polygon having n sides is given by: \({;^n}{C_2} - n = \frac{{n \times \left( {n - 3} \right)}}{2}\)

Calculation:

Given: The polygon is an octagon ⇒ The no. of sides of the given polygon n = 8.

As we know that, no. of diagonals that can be drawn by joining the angular points of a polygon having n sides is given by: \({;^n}{C_2} - n = \frac{{n \times \left( {n - 3} \right)}}{2}\)

⇒ \({;^8}{C_2} - 8 = \frac{{8 \times 5}}{2} = 20\)

Concept:

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) .

 

If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\) then determinant of A is given by:

|A| = a₁₁ × { (a₂₂ × a₃₃) - (a₂₃ × a₃₂) } - a₁₂ × { (a₂₁ × a₃₃) - (a₂₃ × a₃₁) } + a₁₃ × { (a₂₁ × a₃₂) - (a₂₂ × a₃₁) }

Calculation:

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {1!}&{2!}&{3!}\ {2!}&{3!}&{4!}\ {3!}&{4!}&{5!} \end{array}} \right| \)

\(= 1! \times \left( {3! \times 5! - 4! \times 4!} \right) - 2! \times \left( {2! \times 5! - 4! \times 3!} \right) + 3! \times \left( {2! \times 4! - 3! \times 3!} \right);\)

 

= 144 - 192 + 72 = 24

Concept:

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) .

If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\) then determinant of A is given by:

|A| = a₁₁ × { (a₂₂ × a₃₃) - (a₂₃ × a₃₂) } - a₁₂ × { (a₂₁ × a₃₃) - (a₂₃ × a₃₁) } + a₁₃ × { (a₂₁ × a₃₂) - (a₂₂ × a₃₁) }

Calculation:

\(\Rightarrow \left| {\begin{array}{*{20}{c}} x&0&2\ {2x}&2&1\ 1&1&1 \end{array}} \right| = x \times \left( {2 - 1} \right) - 0 \times \left( {2x - 1} \right) + 2 \times \left( {2x - 2} \right) = 5x - 4\)

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {3x}&0&2\ {{x^2}}&2&1\ 0&1&1 \end{array}} \right| = 3x \times \left( {2 - 1} \right) - 0 \times \left( {{x^2} - 0} \right) + 2 \times \left( {{x^2} - 0} \right) = 2{x^2} + 3x\)

\(\Rightarrow \left| {\begin{array}{*{20}{c}} x&0&2\ {2x}&2&1\ 1&1&1 \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {3x}&0&2\ {{x^2}}&2&1\ 0&1&1 \end{array}} \right| = \left( {5x - 4} \right) + \left( {2{x^2} + 3x} \right) = 2{x^2} + 8x - 4 = 0\)

⇒ 2x² + 8x - 4 = 0

By comparing the equation 2x² + 8x - 4 = 0 with ax² + bx + c = 0, we get a = 2, b = 8 and c = - 4.

\(\Rightarrow x = \frac{{ - b \pm \sqrt {{b^2} - 4ac;} }}{{2a}} = \frac{{ - 8 \pm \sqrt {64 + 32} }}{4} = ; - 2 \pm \sqrt 6 ;\)

Concept:

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

Calculation:

Given: x + a + b + c = 0

By applying R₁ ↔ R₁ - R₂ we get

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

By applying R₂ ↔ R₂ - R₃ we get

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

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

\(\Rightarrow \left| {\begin{array}{*{20}{c}} x&{ - x}&0\ 0&x&{ - x}\ a&b&{x + c} \end{array}} \right| = x \times \left( {{x^2} + cx + bx} \right) + x \times \left( {ax} \right) = x \times \left( {{x^2} + cx + bx + ax} \right)\)

As we know it is given that, x + a + b + c = 0

\(\Rightarrow \left| {\begin{array}{*{20}{c}} x&{ - x}&0\ 0&x&{ - x}\ a&b&{x + c} \end{array}} \right| = x \times \left( {{x^2} + cx + bx + ax} \right) = {x^2} \times \left( {x + a + b + c} \right) = 0\)

Concept:

Let A and B be any two matrices of order m × n and p × q respectively. Then the product AB exists if and only if n = p. Similarly, the product BA will exist if and only if q = m.

Zero Matrix or Null Matrix: 

• Any matrix in which all the elements are zero is called zero matrix. It is also known as null matrix and is denoted by O.

i.e \(O = ;{\left[ {\begin{array}{*{20}{c}} 0& \cdots &0\ \vdots & \ddots & \vdots \ 0& \cdots &0 \end{array}} \right]_{m \times ;n}}\) 

Identity Matrix:

• A diagonal matrix in which all the principal diagonal elements are equal to 1 is called an identity matrix. It is also known as unit matrix whereas an identity matrix of order n is denoted by I or In

If A and B are two matrices such that the no. of columns of A is equal to the no. of rows of B. If A = [a_ij] is a m × n matrix and B = [b_ij] be a n × p matrix, then the product AB is the resultant matrix of order m × p and is defined as:

\({\left( {AB} \right)_{ij}} = \mathop \sum \limits_{k = 1}^n {a_{ik}} \times {b_{kj}} = {a_{i1}};{b_{1j}} + ;{a_{i2}};{b_{2j}} + \ldots + {a_{in}};{b_{nj}};\forall ;i = 1,;2, \ldots ,m;and;j = 1,;2,; \ldots .,;p\)

Calculation:

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

\(\Rightarrow {A^3} = \left[ {\begin{array}{*{20}{c}} 4&{ - 4}\ { - 4}&4\end{array}} \right];and;2{A^2} = 2 \times \left[ {\begin{array}{*{20}{c}} 2&{ - 2}\ { - 2}&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 4&{ - 4}\ { - 4}&4 \end{array}} \right]\)

\(\Rightarrow ;{A^3} - 2{A^2} = ;\left[ {\begin{array}{*{20}{c}} 4&{ - 4}\ { - 4}&4 \end{array}} \right] - ;\left[ {\begin{array}{*{20}{c}} 4&{ - 4}\ { - 4}&4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { 0}&0\ 0&{ 0} \end{array}} \right] \)

Calculation:

Given:

m and n (m < n) be the roots of the equation x² - 16x + 39 = 0

⇒ x² - 13x - 3x + 39 = 0

⇒ x (x - 13) - 3 (x - 13) = (x - 13) × (x - 3) = 0

⇒ x = 13 or 3

⇒ m = 3 and n = 13

∵ m, p, q, r, s and n are in AP

⇒ p + q + r + s = (m + d) + (m + 2d) + (m + 3d) + (m + 4d)

= 4m + 10d = 2 × (2m + 5d)

= 2 × [m + (m + 5d)]

= 2 × (m + n) ----------∵ (n = m + 5d)

⇒ p + q + r + s = 2 × (m + n)

∵ m + n = 16

⇒ p + q + r + s = 2 × (16) = 32

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, D < 0.

Calculation:

Given: x² + mx + 2 = 0

By comparing the given equation with the standard quadratic equation ax² + bx + c = 0, we get a = 1, b = m and c = 2.

⇒ D = b² - 4ac = m² - 8

For real roots, D ≥ 0

⇒ D = m² - 8 ≥ 0 ⇒ m² ≥ 8

∴ m2 ≥ 8.

Concept:

• i² = -1, i³ = -i, i⁴ⁿ = 1, where n = natural number
• The cube roots of unity are: 1, , ω and ω2
• \(ω = ;\frac{{ - 1 + i\sqrt 3 ;}}{2};and;{ω ^2} = ;\frac{{ - 1 - i\sqrt 3 ;}}{2};\)
• ω³ = 1

Calculation:

We know that,​

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

and i2 = -1

⇒ \(ω = ;\frac{{{i^2} + i\sqrt 3 ;}}{2};and;{ω ^2} = ;\frac{{{i^2} - i\sqrt 3 ;}}{2};\)

⇒ \(- ;iω = \frac{{i + \sqrt 3 }}{2};and - i{ω ^2} = \frac{{i - \sqrt 3 }}{2}\)

\(\Rightarrow {\left[ {\frac{{i + \sqrt 3 }}{2}} \right]^{2019}} + {\left[ {\frac{{i - \sqrt 3 }}{2}} \right]^{2019}} = {\left( { - ;iω } \right)^{2019}} + {\left( { - i{ω ^2}} \right)^{2019}} \)

⇒ -(i)^{2016 +3} (ω)²⁰¹⁹ - (i)^{2016 +3} (ω)4038

⇒ -(i)2016. (i)³ × 1 - (i)2016. (i)3 × 1

⇒ - 1 × (-i) × 1 - 1 × (-i) × 1 

⇒ (i) + (i) = 2i

∴ The required value is 2i.

Concept:

The cube roots of unity are: 1, ω, ω²

1 + ω + ω² = 0 and ω³ = 1

Calculation:

Given: x² + x + 1 = 0, here a = 1, b = 1 and c = 1.

\(\Rightarrow x = \frac{{ - 1 \pm \sqrt {1 - 4} }}{2} = \frac{{ - 1 \pm \sqrt 3 i}}{2}\)

⇒ The roots of the given equation are ω and ω² .

Let α = ω and β = ω²

\(\Rightarrow \mathop \sum \limits_{j = 0}^3 \left( {{α ^j} + {β ^j}} \right) = 2 + ω + {ω ^2} + {ω ^2} + {\left( {{ω ^2}} \right)^2} + {ω ^3} + {\left( {{ω ^2}} \right)^3} = 2\)

Calculation:

Let A = Students who play cricket, B = Students who play football

⇒ n (A) = 50% and n (B) = 40%

⇒ n (A ∩ B) = 10%

⇒ n (A ∪ B) = n (A) + n (B) - n (A ∩ B) = 50 + 40 - 10 = 80%

As we know that (A ∪ B) ’ = A’ ∩ B’

⇒ n ( (A ∪ B) ’) =n (U) - n (A ∪ B) = 100 - 80 = 20% = n (A’ ∩ B’)

Concept:

The prime numbers = A number that can be divided only by itself and 1

Calculation:

Given: A = (x : 0 ≤ x ≤ 2) and B = {y; y is a prime number}

A = (0, 1, 2) and B = (2, 3, 5, 7, 11, 13, 17, 19, .....)

So, from the above, The common number between A and B is 2.

⇒ A ∩ B = {2}