Quantitative Aptitude PYQs

Concept:

Consider three linear eqaution in two variable:

a₁x + b₁y + c_{1 = 0}

a₂x + b₂y + c₂ = 0

a₃x + b₃y + c₃ = 0

Condition for the consistency of three simultaneous linear equations in 2 variables:

​​​​\( \left| {\begin{array}{*{20}{c}} a_1&b_1&c_1\ a_2&b_2&c_2\ a_3&b_3&c_3 \end{array}} \right|=0\)

Formula for Quadratic equation:

ax² + bx + c = 0

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

Calculation:

2k2x + 3y - 1 = 0      ....(1)

7x - 2y + 3 = 0      ....(2)

6kx + y + 1 = 0      ....(3)

For consistency of given simultaneous equation,

\( \left| {\begin{array}{*{20}{c}} 2k^2&3&-1\ 7&-2&3\ 6k&1&1 \end{array}} \right|=0\)

⇒ 2k²(-2- 3) - 3(7 - 18k) - 1(7 + 12k) = 0

⇒ -10k² - 21 + 54k - 7 - 12k = 0

⇒ -10k² + 42k - 28 =  0

⇒ 5k2 - 21k + 14 =  0

By using the formula,

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

\(\Rightarrow k=\rm \frac{-(-21) \pm \sqrt{(-21)^{2} - 4(5)(14)}}{2\times 5}\)

\(\therefore k=\rm \frac{21 \pm \sqrt{161}}{10}\)

Concept:

AA^-1 = I

Calculation:

A^-1 = \(\rm \begin{bmatrix} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}\)

Let A = \(\rm \begin{bmatrix} a & b \c & d \end{bmatrix}\)

We know that AA^-1 = I

∴ \(\rm \begin{bmatrix} a & b \c & d \end{bmatrix}\)(\rm \begin{bmatrix} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}\) = \(\rm \begin{bmatrix} 1 & 0\0 & 1 \end{bmatrix}\)

⇒ \(\rm \begin{bmatrix} -2a +\frac{3}{2}b & a -\frac{b}{2} \-2c +\frac{3d}{2}& c-\frac{d}{2} \end{bmatrix}\)= \(\rm \begin{bmatrix} 1 & 0\0 & 1 \end{bmatrix}\)

Two matrices are equal if their order and corresponding elements are equal.

Therefore, - 2a + \(\frac{3}{2}b = 1\)  ---(1)

and a - b/2 = 0 ⇒ a = b/2 --- (2)

From equations (1) and (2),

⇒ \(-2(\frac{b}{2})+\frac{3}{2}b = 1\)

⇒ \(-b +\frac{3b}{2} =1\)

⇒ \(\frac{b}{2} = 1\)

∴ b = 2 

From equation (2), a = 1

Similarly, \(-2c +\frac{3d}{2} = 0\)  --- (3)

and \(c -\frac{d}{2} = 1\) --- (4)

Solving equation (3) and (4), we get c = 3 and d = 4

∴ Matrix A = \(\rm \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}\)

Formula used:

cos2θ = 1 - 2sin²θ 

Calculation:

f(x) = ln(2 + sin2x)

⇒ f(x) = ln[2 + (1 - cos2x)/2]

⇒ f(x) = ln[(5 - cos2x)]/2      ----(i)

If we put (x + π) in place of x, we get

⇒ f(x + π) = ln[(5 - cos2(x + π)/2]

⇒ f(x + π) = ln[5 - cos(2π + 2x)]/2 

⇒ f(x + π) = ln[(5 - cos2x)]/2      ----(ii) 

From (i) and (ii), we get 

f(x + π) = f(x)

∴ The period of the function f(x) = ln(2 + sin2x) is π.

Formula used:

sin \(\rm \frac{\pi}{2}\) = 1

sin \(\rm \frac{\pi}{6}\) = \(\rm \frac{1}{2}\)

Calculation:

Given

sin(A + B) = 1 

⇒ sin(A + B) = sin \(\rm \frac{\pi}{2}\)

⇒ A + B = \(\rm \frac{\pi}{2}\)     ----(1)

2 sin(A - B) = 1

⇒ sin(A - B) = \(\rm \frac{1}{2}\)

⇒ sin(A - B) = sin \(\rm \frac{\pi}{6}\) 

⇒ A - B = \(\rm \frac{\pi}{6}\)     ----(2)

On solving (1) and (2), we get

A = \(\rm \frac{\pi}{3}\), B = \(\rm \frac{\pi}{6}\)

Now, We have to find the value of "tan A ∶ tan B"

⇒ \(\rm \frac{tan \frac{\pi}{3}}{tan \frac{\pi}{6}}\) = \(\rm \frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}\) = 3 : 1

∴ The value of "tan A ∶ tan B" equal to 3 : 1.

Formula used:

Triangle having no side common with those of the polygon = 

\(\rm \frac{n(n-4)(n-5)}{6}\)

n is the sides of the polygon.

Calculation:

Here n = 10

Triangle having no side common with those of the polygon

= \(\rm \frac{10(10-4)(10-5)}{6}\)

= \(\rm \frac{10\times 6\times 5}{6}\)

= 50

∴ The number of triangles is 50.

Formula used:

Quadratic Equation formula: 

A quadratic equation can be written in a standard form as

ax² + bx + c = 0

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

Concept Used:

Algebraically, the absolute value of a number equals the nonnegative square root of its square

i.e Modulus of that number.

Calculation:

Given equation:

x4 - 10x2 + 9 = 0.

⇒ x² = \(\rm \frac{10 ± \sqrt{100 - 4\times 9}}{2}\)

⇒ x2 = \(\rm \frac{10 ± \sqrt{64}}{2}\)

⇒ x² = \(\rm \frac{10 ± 8}{2}\)

Case I. Take positive sign

x² = 9

⇒ x = ± 3

Case II. Take negative sign

x² = 1

⇒ x = ± 1

∴ The sum of the absolute values of the roots = 3 + 1 + |-3| + |-1| = 8.

Formula used:

This binomial expansion formula gives the expansion of (x + y)ⁿ where 'n' is a natural number. The expansion of (x + y)ⁿ has (n + 1) terms. 

(x + y)ⁿ = ⁿC ₀ xⁿ y⁰ + ⁿC ₁ x^{n - 1} y¹ + ⁿC ₂ x^n-2 y² + ^_nC ₃ x^{n - 3} y³ + ... + ⁿC _n−1 x y^{n - 1} + ⁿC _n x⁰yⁿ

Calculation:

(1 + x)ⁿ = 1 + n x + ...  + n.x^n-1+ xⁿ

From above equation we can write 

p = 1, q = n

r = n, s = 1

Now, We have to find the value of (ps + qr)

⇒  1 + n²

∴ (ps + qr) equal to (1 + n2).

Formula used:

sin^-1 x = \(\rm \frac{\pi}{2}\)

x = sin (\(\rm \frac{\pi}{2}\))

sin (\(\rm \frac{\pi}{2}\)) = 1

Calculation:

sin-1 x + sin-1 y + sin-1 z = \(\frac{3\pi}{2}\) 

This is possible only when

sin-1 x = sin-1 y = sin-1 z = \(\rm \frac{\pi}{2}\) 

⇒ sin-1 x = \(\rm \frac{\pi}{2}\)

⇒ x = 1    ----(i)

⇒ sin-1 y = \(\rm \frac{\pi}{2}\)

⇒ y = 1      ----(ii)

⇒ sin-1 z = \(\rm \frac{\pi}{2}\)

⇒ z = 1      ----(iii)

Now, we have to find the value of 

x1000 + y1001 + z1002

Now, from (i), (ii), and (iii), we get

⇒ (1)1000 + (1)1001 + (1)1002

⇒ 1 + 1 + 1

⇒ 3

∴ The value of x1000 + y1001 + z1002 is 3.

Formula used:

sin x + sin y \(=2\ sin\ {\left(\frac{x+y}{2}\right)}\ cos\ {\left(\frac{x-y}{2}\right)}\)

cos x + cos y \(=2\ cos\ {\left(\frac{x+y}{2}\right)}\ cos\ {\left(\frac{x-y}{2}\right)}\)

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

 

Calculation:

Given, sin x + sin y = cos x + cos y 

We know that

sin x + sin y \(=2\ sin\ {\left(\frac{x+y}{2}\right)}\ cos\ {\left(\frac{x-y}{2}\right)}\)

cos x + cos y \(=2\ cos\ {\left(\frac{x+y}{2}\right)}\ cos\ {\left(\frac{x-y}{2}\right)}\)

sin x + sin y = cos x + cos y

\(\Rightarrow 2\ sin\ {\left(\frac{x+y}{2}\right)}\ cos\ {\left(\frac{x-y}{2}\right)}=2\ cos\ {\left(\frac{x+y}{2}\right)}\ cos\ {\left(\frac{x-y}{2}\right)}\)

\(\Rightarrow \ sin\ {\left(\frac{x+y}{2}\right)}=\ cos\ {\left(\frac{x+y}{2}\right)}\)

\(\frac{\ sin\ \frac{x+y}{2}}{ cos\frac{x+y}{2}}=1\)................(\(\frac{sinx}{cosx}=tanx\))

\(tan (\frac{x+y}{2})=1\)

\(tan \left(\frac{x}{2}+\frac{y}{2}\right)=1\)

∴ The value of \(\rm tan \left(\frac{x}{2}+\frac{y}{2}\right)\) is equal to 1.

Formula used:

Multiplication matrix: 

\(\rm \begin{bmatrix} a & b\ c & d \end{bmatrix} \times \begin{bmatrix} w & x\ y & z \end{bmatrix} = \begin{bmatrix} aw + by & ax + bz\ cw + dy & cx + dz \end{bmatrix}\)

Identity matrix: 

An identity matrix is a given square matrix of any order which contains on its main diagonal elements with a value of one, while the rest of the matrix elements are equal to zero.

I = \(\rm \begin{bmatrix} 1 & 0\ 0 & 1 \end{bmatrix} \)

Calculation:

\(I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\)

⇒  \(I^2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\)

\(A = \begin{bmatrix} 0 & 2 \ -2 & 0 \end{bmatrix}\)

⇒  A² = \( \begin{bmatrix} -4 & 0 \ 0 & -4 \end{bmatrix}\)

According to question,

(mI + nA)2 = A

⇒ m²I² + n²A² + 2mnA = A   

⇒  m2I2 + n2A2 = A(1 - 2mn)

⇒  m² \(\rm \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\) + n² \( \rm \begin{bmatrix} -4 & 0 \ 0 & -4 \end{bmatrix}\) = (1 - 2mn)\(\rm \begin{bmatrix} 0 & 2 \ -2 & 0 \end{bmatrix}\) \(\rm \begin{bmatrix} m^{2} & 0\ 0 & m^{2} \end{bmatrix} + \begin{bmatrix} -4n^{2} & 0\ 0 & -4n^{2} \end{bmatrix} = \begin{bmatrix} 0 & 2(1-2mn)\ -2(1-2mn) & 0 \end{bmatrix} \)

On equating on both sides 

⇒ m² - 4n² = 0

⇒ m = 2n     ----(1)

⇒ 2(1 - 2mn) = 0

⇒ 1 - 2mn = 0

⇒  mn = \(\rm \frac{1}{2}\)

From equation (1)

(2n)n = \(\rm \frac{1}{2}\)

⇒ n = \(\rm ± \frac{1}{2}\)

Taking positive sign 

When n =  \(\rm \frac{1}{2}\) then, m = 1

∴ m + n = 1 + \(\rm \frac{1}{2}\) = \(\rm \frac{3}{2}\)

Formula used:

cot (cot^–1 x) = x, x ϵ R

sin^–1 x + sin^–1  y = sin^–1 [x √(1 - y²) + y √(1 - x2)]

where x > 0, y > 0 & (x² + y²) ≤ 1

cot^-1x = tan–1 (1/x) 

Calculation:

\(\rm cot \left[sin^{-1} \frac{3}{5}+cot^{-1}\frac{3}{2} \right]\)    ----(i)

Let a = cot^-1(3/2)

⇒ cot a = 3/2

⇒ sin a = 2/√13

⇒ a = sin^-1(2/√13)

From (i)

\(\Rightarrow \rm cot \left[sin^{-1} \frac{3}{5}+sin^{-1}\frac{2}{√13} \right]\)

⇒ cot [sin^-1{(3/5) √{1 - (4/13)} + (2/√13) √{1 - (9/25)}]

⇒ cot [sin-1(17/5√13)]

⇒ cot [cot-1(6/17)]

⇒ 6/17

∴ The value of \(\rm cot \left[sin^{-1} \frac{3}{5}+cot^{-1}\frac{3}{2} \right]\)is 6/17.

Formula used:

sin 60^∘ = \(\rm \frac{\sqrt{3}}{2}\), tan 180∘ = 0

If sin a = sin b then a = b

Calculation:

Given, 4sin2 x = 3

⇒ sin x = \(\rm \frac{\sqrt{3}}{2}\)

⇒ sin x = sin60° 

⇒ x = 60^∘ or 120°   [∵ 0 ≤ x ≤ π]

We have to find the value of tan 3x

⇒ tan 3(60^∘) or tan 3 × 120° 

⇒ tan 180^∘ or tan360° 

⇒ 0 or 0

∴ The value of tan3x is equal to 0.

Formula used:

The formula for Arithmetic Sequence:

an = a1 + (n – 1)d

Where,

• an = nth term that has to be found
• a1 = 1^st term in the sequence
• n = Number of terms
• d = Common difference
• Sn = Sum of n terms

Calculation:

According to the question

q = p + (3 - 1)d   

⇒ q = p + 2d    ----(1)

And 3 = p + (5 - 1)d 

⇒ 3 = p + 4d    ----(2)

Solving (1) and (2), we get 

q = 3 - 2d

and p = 3 - 4d      [from (2)]

According to the question pq is minimum 

⇒ (3 - 4d) (3 - 2d) is minimum  

⇒ (9 - 18d + 8d²) is minimum 

Differentiation the above equation with respect to d is 0

⇒ -18 + 16d = 0 

⇒ d = 9/8 

∴ The value of d is 9/8.

Concept:

Non-Collinear Points: The set of points that do not lie on the same line are called non-collinear points. We cannot draw a single straight line through these points.

Formula Used:

a³ - b³ = (a - b) (a² + ab + b²)

Calculation:

Given equation x³ - 8 = 0

⇒ x3 - 2³ = 0

⇒ (x - 2) (x² + 2x + 4) = 0

⇒ x - 2 = 0 or x2 + 2x + 4 = 0 

⇒ x = 2 or x = -1 + i√3 or x = -1 - i√3

We can take these three points on complex plane 

So, we can write 1^st point as 2 + i.0, -1 + i√3, -1 - i√3

The three points on complex plain is (2, 0), (-1,√3), (-1, -√3)

Case I. 

Equation of line passes through (-1,√3), (-1, -√3) is

(y - √3) = (2√3/0) (x + 1)

⇒ x + 1 = 0 

And this line does not passes through (2, 0).

∴ These points (roots) are not collinear.

Case 2. 

We know that these points are non-collinear it means they will definitely pass through a circle 

Let the equation of circle be (x - a)² + (y - b)² = r²

Now, According to our assumption 

(2 - a)2 + (0 - b)2 = r2     ----(i)

(-1 - a)2 + (√3 - b)2 = r2     ----(ii)

(-1 - a)2 + (-√3 - b)2 = r2     ----(iii)

On solving (i), (ii) and (iii), we get 

a = 0, b = 0 and r = 2

Hence the equation of circle is (x - 0)2 + (y - 0)2 = 22

∴ The roots represents a circle but radius of circle is 2.

Formula used:

sec x = \(\rm \frac{1}{cos :x}\), cosec x = \(\rm \frac{1}{sin :x}\)

(sin x - cos x)2 = sin2 x + cos2 x - 2 sin x cos x

sin2 x + cos2 x = 1, tan x = \(\rm \frac{sin : x}{cos : x}\)

Calculation:

We have sec x × cosec x = p

⇒ \(\rm \frac{1}{cos x}× \frac{1}{sin x}\) = p

⇒ sinx.cos x = \(\rm \frac{1}{p}\)     ----(i)

We know that, 

(sin x - cos x)2 = sin2 x + cos2 x - 2 sin x cos x

From (i), we get 

⇒ (sin x - cos x)2 = 1 - 2 × \(\rm \frac{1}{p}\)    ----(ii)

We know tha the square of any number is ≥ 0

(sin x - cos x)2 ≥ 0

From (ii), we get

\(1 - 2 \times \rm \frac{1}{p} ≥ 0\)

⇒ 1 ≥ 2/p

⇒ 1/1 ≤ p/2

⇒ p ≥ 2

∴ The smallest value of p is 2.

Formula used:

sin 2θ = 2 sinθ cosθ 

cos 2θ = 2cos²θ - 1

Concept:

f'(x) = 0 at a point x = a is 

• If f'(x) changes the sign from positive to negative while the graph of the function passes through x = a then x = a is said to be a point local maxima.
• If f''(a) < 0 then x = a is a point of local maxima.

Calculation:

Let f(θ) = sinθ + sinθ cosθ

⇒ f(θ) = sinθ + \(\rm \frac{1}{2}\)​ sin 2θ

⇒ f'(θ) = cosθ + \(\rm \frac{1}{2}\)​ cos 2θ × 2

⇒ f'(θ) =  cosθ + cos 2θ 

Put f'(θ) = 0

⇒ cosθ + 2cos2θ - 1 = 0

⇒ 2cos2θ + 2cosθ - cosθ  - 1 = 0

⇒ 2cosθ (cosθ + 1) - 1(cosθ + 1) = 0 

⇒ (cosθ + 1) (2cosθ - 1) = 0

⇒ cosθ = 1/2, cosθ = -1

⇒ θ = π/3, π 

According to the given range 0 < θ < \(\frac{π}{2}\)

θ = π/3

At this value of θ = π/3, the function can be maximum and minimum 

So, we have to check second derivatives

f"(θ) = - sinθ - 2sin2θ 

at θ = π/3 

f"(π/3 ) ≤ 0

So, we can say that θ = π/3 is a point of local maxima.

Concept:

Definition of De Morgan’s law: 

The complement of the union of two sets is equal to the intersection of

their complements and the complement of the intersection of two sets

is equal to the union of their complements.

Explanation:

**Statement:**1

The union over the intersection of sets is distributive.

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Statement 1 is correct

**Statement:**2 

According to De Morgan law,

(A U B)' = A' ∩ B' 

Hence, statement 2 is correct.

Statement:3 

If A and B are two sets, then their difference is given by

A - B or B - A. 

If A = {2, 3, 4} and B = {2, 3, 4, 5, 6} 

A - B means elements of A that are not the elements of B i.e,

in the above example A - B = ϕ

But here, A and B are not equal sets so statement 3 is incorrect

Also, B - A = {5, 6} ≠ ϕ 

Additional Information 

In general, B - A = {x : x ∈ B, and x ∉ A} 

• If A - B is the empty set, then we can show that A is a subset of B
• If B - A is also the empty set, then we can also show that B is a subset of A

Concept:

If A and B are two sets, then their difference is given by A - B or B - A. 

If A = {2, 3, 4} and B = {4, 5, 6} 

A - B means elements of A that are not the elements of B i.e,

in the above example A - B = {2, 3} 

In general, B - A = {x : x ∈ B, and x ∉ A} 

Proper subset = 2ⁿ - 1,

n is the number of elements

Calculation:

Given:

Number of elements in X = 6

Number of eleelementsent in Y = 5

Number of elements in Z = 4

All elements are different.

Since, X, Y,  have distnict elements. 

Therefore,

n(X - Y) = 6, n(Z) = 4

⇒ S = (X - Y) ∪ Z

⇒ S = 6 + 4 = 10

Hence,

Number of proper subsets 

N = 2¹⁰ - 1

⇒ N = 1024 - 1

⇒ N = 1023

S has 1023 proper subsets.

Explanation:

Statement:1. All relations are functions but all functions are not relations.

All functions are relations, but not all relations are functions. A function

is a relation that for each input, there is only one output.

So, statement 1 is incorrect.

Statement: 2.  A relation from A to B is a subset of Cartesian product A × B.

If A and B are set, the Cartesian product of A and B is the set

A × B = {(a, b) : (a ∈ A) and (b ∈ B)}.

Every relation is a subset of the Cartesian product

So, statement 2 is correct.

Statement: 3. A relation in A is a subset of Cartesian product A × A.

A relation from a non-empty set A to a non-empty set A is a subset of

cartesian product A × A. The first element is called the preimage of the

second and the second element is called the image of the first.

So, statement 3 is correct.

Formula used:

log_b(b^x) = x

b = logarithmic base

log_b b = 1

Calculation:

We have, 

log10 2 log₂ 10 + log10(10x) = 2

\(\rm \frac{log 2}{log 10} \times \frac{log 10}{log 2} + log_{10}({10})^x= 2\)

1 + x log10(10) = 2

1 + x × 1 = 2

⇒1 +  x = 2

⇒ x = 2 - 1

⇒ x = 1

∴ The value of x is 1.