Quantitative Aptitude PYQs

Concept:

1. The null set is a subset of every set. (ϕ ⊆ A)
2. Every set is a subset of itself. (A ⊆ A)
3. The number of subsets of a set with n elements is 2ⁿ.

 

Explanation:

1. The null set is a subset of every set - The intersection of two sets is a subset of each of the original sets.

So if {} is the empty set and A is any set then {} intersect A is {} which means {} is a subset of A and {} is a subset of {}.

You can prove it by contradiction. Let's say that you have the empty set {} and a set A. 2. Every set is a subset of itself. (A ⊆ A) 3. If n = 10 then  2¹⁰ = 1024

So, all three statements are true.

Concept:

If x ∈  R then we express it by writing xRy and say that " x is related to y with relation R"

Thus, (x, y) ∈ R ⇔ xRy

Calculation:

Given

2x + 3y = 20

The relation R can be written as

R = {(1,6), (4, 4), (7, 2)}

There are 3 elements in the form (x, y) are there in R.

Concept:

Onto Functions - 

A function f from X to Y is onto (or surjective), if and only if for every element y ∈ Y there is an element x ∈ X with f(x) = y.

In words: “Each element in the co-domain of f has a pre-image” 

Mathematical Description: f : X →Y is onto ⇔ \(\rm \forall \) y \(\rm \exists \)x, f(x) = y 

One-to-one Correspondence

A function f is a one-to-one correspondence (or bijection), if and only if it is both one-to-one and onto

In words: “No element in the co-domain of f has two (or more) pre images” (one-to-one) and “Each element in the co-domain of f has a pre-image” (onto).

Calculation:

1. A function f : Z → Z, defined by f(x) = x + 1, is one-one as well as onto.

f(x) = x + 1,

calculate f(x1) :

 f(x1) = x1 + 1

calculate f(x2) :

 f(x2) = x2 + 1

Now, f(x1) = f(x2) 

⇒ x1 + 1 =  x2 + 1

⇒ x1 =  x2 

So, f is one-one function.

Consider f(x) = y

y = x + 1

x = y - 1

f(y-1) = y - 1 + 1 = y

f is onto. 2. A function f : N → N, defined by f(x) = x + 1, is one-one but not onto.

f(x) = x + 1,

calculate f(x₁) :

 f(x1) = x₁ + 1

calculate f(x₂) :

 f(x₂) = x₂ + 1

Now, f(x1) = f(x2) 

⇒ x1 + 1 =  x2 + 1

⇒ x1 =  x2 

So, f is one-one function.

Clearly, f(x) = x + 1 ≥ 2 for all x ∈ N

So, f(x) does not assume values 1.

 f is not an onto function.

So, Both 1 and 2 are correct.

Concept:

Consider z = a + ib

Multiplicative identity z.z^-1 = 1

zz̅ = a² + b² = (a + ib) (a - ib)

Explanation

Given:

1. \(\rm {\overline{\left(z^{-1}\right)}}=(\bar{z})^{-1}\)

consider z = a - ib

⇒ z^-1 = \(\rm \frac 1 z = \frac 1 {a - ib}\) =\(\rm \frac{a + ib}{a^{2} + b^{2}}\)

⇒ \(\rm \overline{z^{-1}}\) = \(\rm \frac{a - ib}{a^{2} + b^{2}}\)      ....(1)

⇒ \(\rm \overline{z}\) = a + ib

⇒ (\(\rm \overline{z}\))^-1 = \(\rm \frac 1 {\bar{z}} = \frac 1 {a +ib}\)= \(\rm \frac{a - ib}{a^{2} + b^{2}}\)      ....(2)

from eq(1) and (2)

Statement 1 is correct. 

2. zz-1 = |z|2

consider z = a - ib

⇒ z^-1 = \(\rm \frac{a + ib}{a^{2} + b^{2}}\)

⇒ z.z-1 = (a - ib)\(\rm \frac{a + ib}{a^{2} + b^{2}}\) = \(\rm \frac{a^{2} + b^{2}}{a^{2} + b^{2}} =1\)

Statement 2 is not correct

Concept:

Let z = x + iy be a complex number,

Where x is called the real part of the complex number or Re (z) and y is called the Imaginary part of the complex number or Im (z)

Conjugate of z = z̅ = x - iy  

Calculation:

1. The difference of Z and its conjugate is an imaginary number.

Consider z = a + ib           ....(i)

conjugate of z = z̅ = a - ib      ....(ii)

eq(i) - eq (ii)

z - z̅ = a + ib - a + ib

⇒ 2ib

Thus it is clear that the difference of z and its conjugate is an imaginary number. 2. The sum of Z and its conjugate is a real number.

eq (i) + eq(ii)

z + z̅ = a + ib + a - ib

⇒ 2a

Thus it is clear that the sum of Z and its conjugate is a real number.

So, Both 1 and 2 are correct.

Concept:

Iota power:

• i² = -1, i³ = -i, i⁴ = 1, i⁴ⁿ = 1
• Number of the form 2n is always even, n ∈ N
• Numebr of the form 2n +1 or 2n - 1 is always odd, n ∈ N
• (-a)^{2n -1} = -(a)^{2n -1}

Calculation:

Let,

Z = i2n + 1(-i)2n - 1

⇒ Z = i2n+1 × (i)2n-1 × (-1)^2n-1

⇒ Z =  i2n +1+2n-1 ×(-1)^2n-1

⇒ Z = i⁴n × (-1)          [∵ (-1)2n-1 = -1]

⇒ Z = (-1)⁴ⁿ ×

⇒ Z = i4n × (-1)

⇒ Z = -(i)4n × (-1)

⇒  Z = 1 × (-1)         [∵ i⁴ⁿ = 1] 

⇒ Z = -1

⇒ |Z| = |-1| = 1

Hence, the modulus of the complex number i2n + 1(-i)^{2n - 1} is 1.

Concept:

General Quadratic Equation

ax2 + bx + c = 0

• Product of roots (αβ) = c/a
• Sum of roots (α + β) = - b/a
• Formula to find Roots,  \(\displaystyle \rm x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
• sin 3θ = 3 sinθ - 4 sin³θ

Calculation:

Given: 4x2 + 2x - 1 = 0

a = 4, b = 2, c = -1

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

⇒ \(\displaystyle x = {-2 \pm \sqrt{2^2-4(4)(-1)} \over 2(4)}\)

⇒ \(\displaystyleα = {-1 + \sqrt{5} \over 4}\) and \(\displaystyle β = {-1 - \sqrt{5} \over 4}\)

As we know that \(\displaystyle sin\ 18° = {-1 + \sqrt{5} \over 4}\) and \(\displaystyle sin\ 54° = {1 + \sqrt{5} \over 4}\)

So, We can say that α = sin18° and β = - sin54°      ------(i)

Now, using the formula, sin 3θ = 3 sinθ - 4 sin3θ 

On putting θ = 18° in the above trigonometric formula, we get 

⇒ sin 54° = 3 sin18° - 4 sin318°      ------(ii)

From (i) and (ii), we get 

⇒ - β = 3α - 4α³

⇒ β = 4α3 - 3α

∴ The correct relation is β = 4α3 - 3α.

Concept: 

Let us consider the standard form of a quadratic equation, ax2 + bx + c =0

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

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

The product of the roots is given by:  \({\rm{α β }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant;term}}}}{{{\rm{coefficient;of;}}{{\rm{x}}^2}}}\)

 

Calculation:

Let the roots of 5x2 + 26x + k = 0 are α and β

Given one root is reciprocal of the other So, α = 1/β 

⇒ α.β = 1    ----(i)

Compare with general equation ax² + bx + c = 0

a = 5, b = 26, c = k

According to the concept used

⇒ α.β = k/5    ----(ii)

From (i) and (ii), we get 

k/5 = 1

∴The value of k is 5.

Concept:

Combination Formula

ⁿC_r = \(\rm \frac{n!}{r! (n - r)!}\)

Calculation:

A particular player should not be included 

We have to select 5 players from (8 - 1) = 7 players.

⇒  required number of ways =  ⁷C₅

\(⇒ \frac{7!}{5! (7-5)!}\) = 21

∴ The required number of ways is 21.

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.

(1−x)n=∑k=0n(nk)1n−k(−x)k(1−x)n=∑k=0n(nk)1n−k(−x)k(1−x)n=∑k=0n(nk)1n−k(−x)

Calculation:

Given:

(1 + 4x + 4x2)5

⇒ [(1 + 2x)²]⁵

⇒ (1+ 2x)¹⁰ 

Here n = 10 (n is even number)

∴ Middle term = \(\left( {\frac{n}{2} + 1} \right) = \left( {\frac{10}{2} + 1} \right) \) = 6^th term 

Middle Term, T₆ =  T_{5 + 1} = ¹⁰C₅ (1)⁵ (2x)⁵

⇒ \(\frac {10!}{5!5!}\)× 32x⁵

⇒ 8064 x⁵

∴ The coefficient of the middle term in the expansion of (1 + 4x + 4x2)5  is 8064.

Concept:

tan x = \(\rm \frac{Perpendicular}{Base}\) = \(\rm \frac{P}{B}\) 

sin x = \(\rm \frac{Perpendicular}{Hypotenuse}\) = \(\rm \frac{P}{H}\) 

cos x = \(\rm \frac{Base}{Hypotenuse}\) = \(\rm \frac{B}{H}\) 

By Pythagoras theorem

Hypotenuse2 = Perpendicular2 + Base2 

 

Calculation:

Given:

\(\rm tan:x=-\dfrac{3}{4}\)

P= 3, B = - 4

By Pythagoras theorem

H² = P² + B²

=  3²  + (-4)²

Hypotenuse(H) = 5

sin x  = \(\rm \frac{3}{5}\)

cos x = \(\rm \frac{-4}{5}\)

sin x . cos x = \(\rm \frac{3}{5} . \rm \frac{-4}{5}\) = \(\rm \frac{-12}{25}\)

Concept:

sec (360° -  θ) = sec θ;   

cosec (180° + θ) = - cosec θ

Calculation:

\(\rm cosec\left(\dfrac{7\pi}{6}\right)sec\left(\dfrac{5\pi}{3}\right)\)

= cosec (210^∘) sec (300^∘)

=  cosec(180^∘ + 30^∘) sec(360^∘ - 60^∘)

= - cosec 30^∘ sec 60∘

= - 2 (2)

= - 4

 

Additional Information

sin (360° -  θ) = - sin θ            

cos (360° -  θ) = cos θ      

tan (360° -  θ*)* = - tan θ      

cosec (360° -  θ) = - cosec θ        

sec (360° -  θ) = sec θ   

cot (360° -  θ*)* = - cot θ

sin (180° + θ) = - sin θ

 

cos (180° + θ) = - cos θ

tan (180° + θ) = tan θ

cosec (180° + θ) = -cosec θ

sec (180° + θ) = - sec θ

cot (180° + θ) = cot θ

 

sin (90° - θ) = cos θ

cos (90° - θ) = sin θ

tan (90° - θ) = cot θ

cosec (90° - θ) = sec θ

sec (90° - θ) = cosec θ

cot (90° - θ) = tan θ

Concept:

Let A = \(\rm \begin{vmatrix} a_{1} & a_{2} & a_{3}\ b_{1} & b_{2} & b_{3}\ c_{1} & c_{2} & c_{3} \end{vmatrix}\)

Then determinant is given by:

det(A) = a₁ (b₂ c₃ - b₃ c₂) - a₂ (b₁ c₃ - b₃ c₁) + a₃ (b₁ c₂ - b₂ c₁)

Calculation:

Given:

\(\left| {\begin{array}{*{20}{c}} x&1&3\ 0&0&1\ 1&x&4 \end{array}} \right| = 0\)

⇒ x(0 - x) - 1(0 - 1) + 3(0 - 0) = 0

⇒ -x² + 1  = 0

⇒ x² = 1

∴ x = 1 or -1

Concept:

tan A = cot (90- A) 

cot A = \(\rm \frac{1}{tan A}\) --- (2) or tan A × cot A = 1

Calculation:

Given,

tan 31° tan 33° ..... tan 47°tan49° ...... tan 57° tan 59°

By using the above formula

tan 31° tan 33° .....tan(90° - 47°) tan(90° - 49°) ......tan (90° - 31°) 

tan 31° tan 33° .....tan(90° - 47°) tan(90° - 49°) ......tan (90° - 31°) 

tan 31° tan 33°....cot 43°cot 45°....cot 33°cot 31° 

Since, tan θ × cot θ = 1 

Therefore,

tan 31° tan 33° ..... tan 47°tan49° ...... tan 57° tan 59° = 1

Concept:

Let A = \(\rm \begin{vmatrix} a_{1} & a_{2} & a_{3}\ b_{1} & b_{2} & b_{3}\ c_{1} & c_{2} & c_{3} \end{vmatrix}\)

det(A) = a1 (b2 c3 - b3 c2) - a2 (b1 c3 - b3 c1) + a3 (b1 c2 - b2 c1)

Calculation:

Given:

\(f(-1) = \rm \left| {\begin{array}{*{20}{c}} 1&-1&{0}\ {-2}&{{2}}&{0}\ {6}&{12}&0 \end{array}} \right|= 1(0 - 0) + 1(0 - 0) + 0(-24 - 12) = 0\)

\(f(0) = \rm \left| {\begin{array}{*{20}{c}} 1&0&{1}\ {0}&{{0}}&{0}\ {0}&{4}&0 \end{array}} \right|= 1(0 - 0) - 0(0 - 0) + 1(0 - 0) = 0\) 

\(f(1) = \rm \left| {\begin{array}{*{20}{c}} 1&1&{2}\ {2}&{{0}}&{2}\ {0}&{0}&0 \end{array}} \right| = 1(0 - 0) -1(0 - 0) + 2(0 - 0) = 0\) 

Now;

f(-1) + f(0) + f(1) = 0

Concept:

sin-1x + cos-1x = \(\rm \frac{\pi}{2} \)

cos (30^∘ ) = \(\rm \frac{\sqrt{3}}{2}\)

Calculation:

Given

⇒ \(sin^{-1}x-cos^{-1}x=\frac{\pi}{6}\)

⇒ sin^-1x + cos^-1x - 2 cos^-1x = \(\rm \frac{\pi}{6}\)

⇒ \(\rm \frac{\pi}{2} \)- 2 cos-1x = \(\rm \frac{\pi}{6}\)

⇒ 2 cos-1x  = \(\rm \frac{\pi}{2} - \rm \frac{\pi}{6}\)

⇒ cos-1x = \(\rm \frac{\pi}{6}\)

⇒ x = \(\rm \frac{\sqrt{3}}{2}\)

∴ The equation has unique solution.

Concept:

cos (90^∘ - θ ) = sin θ 

sin (90∘ - θ ) = cos θ 

Calculation:

= (sin 24° + cos 66°)(sin 24° - cos 66°)

= (sin 24^∘ + sin (90^∘ - 66^∘)) (sin 24∘ - sin (90∘ - 66∘)) 

= (sin 24∘ + sin 24∘) (sin 24∘ - sin 24∘) 

= (sin 24∘ + sin 24∘) × 0

= 0

Concept:

Length of chord = 2r sin(\(\rm \frac{θ }{2}\)​)

Where r = radius of the circle

And θ = Angle subtended by the chord at the centre.

Calculation:

Given: r = 1 and angle = 120°

Now, Length of chord = 2 (1) sin(\(\rm \frac{120^{∘} }{2}\)​)

⇒ 2 sin (60^∘)

⇒ 2 x \(\rm \frac{√{3}}{2}\) 

⇒ √3 units

∴ The length of chord is √3 units.

Concept:

Trigonometry Formula

cot θ = \(\rm \frac{cos θ }{sin θ }\)

tan θ = \(\rm \frac{sin θ }{cos θ }\)

cosec θ = \(\rm \frac{1 }{sin θ }\)

sec θ = \(\rm \frac{1}{cos θ }\)

sin² θ + cos² θ = 1

(sin θ + cos θ)² = sin² θ + cos² θ + 2 sin θ cos θ 

 

Calculation:

To Find: Value of (1 + cot θ - cosec θ)(1 + tan θ + sec θ)

= (1 + \(\rm \frac{cos θ }{sin θ }\) - \(\rm \frac{1 }{sin θ }\)) (1 + \(\rm \frac{sin θ }{cos θ }\) + \(\rm \frac{1}{cos θ }\))

= \(\rm \frac{(sin θ + cos θ - 1) ( sin θ + cos θ + 1)}{sin θ : cosθ }\)

= \(\rm \frac{( sin θ + cos θ )^{2} - 1}{ sin θ cos θ }\)

= \(\rm \frac{sin^{2}θ + cos^{2}θ + 2 sin θ cos θ - 1}{sinθ : cos θ }\)

= \(\rm \frac{1 + 2 sin θ cos θ - 1}{sinθ : cos θ }\)

= \(\rm \frac{2 sin θ cos θ }{sinθ : cos θ }\)

= 2

Concept:

Trigonometry Formula

sec² θ = 1 + tan² θ

cosec2 θ = 1 + cot2 θ

cot θ = \(\rm \frac{1}{tan \theta }\)

sec θ = \(\rm \frac{1}{cos \theta }\)

cosec θ = \(\rm \frac{1}{sin \theta }\)

 

Calculation:

\(\rm \frac{1+tan^2\theta}{1+cot^2\theta}-\left(\frac{1-tan\theta}{1-cot\theta}\right)^2\)

\(= \frac{sec^2θ}{cosec^2θ}-\left(\frac{1-tanθ}{1-\frac{1}{tan θ }}\right)^2\)

\(= \rm \frac{\frac{1}{cos^{2}θ}}{\frac{1}{sin^{2}θ}} - \left (\frac{1 - tanθ}{\frac{tan θ - 1}{tan θ }} \right )^{2}\)

\(\rm = \frac{sin^{2}θ }{cos^{2} θ } - \left (\frac{1 - tanθ}{\frac{- (1 - tan θ )}{tan θ }} \right )^{2}\)

= tan² θ - (-tan θ)²

= tan2 θ - tan2 θ 

= 0