Quantitative Aptitude PYQs

Concept:

A proper subset is one that contains few elements of the original set.

A superset is one that contains all the elements including elements of the original set.

Calculation:

The superset of {1, 2, 3, 4} and supersets of set {4} are

{4}, {1, 4}, {2, 4}, {3, 4}, {1, 2, 4}, {2, 3, 4), (1, 3, 4}, {1, 2, 3, 4}

Hence, there are 8 supersets of the set {4}.

∴ The required number of set is 8.

Concept:

• Complement of set A or A' contains elements other than elements of set A.
• x ∉ A means x is not an element of A, i.e. no set membership
• x ∈ A means x is an element of A, i.e. belongs to set membership

 

Calculation:

Given that, A and B are two non-empty sets.

Statement I: x ∉ (A ∪ B) ⇒ x ∉ A or x ∉ B

As x ∉ (A ∪ B)

⇒ x ∈ (A ∪ B)'

⇒ x ∈ A' and x ∈ B'

⇒ x ∉ A and x ∉ B

Statement I is incorrect.

Statement II: x ∉ (A ∩ B) ⇒ x ∉ A and x ∉ B

As x ∉ (A ∩ B)

x ∈ (A ∩ B)'

x ∈ A' ∪ B'

⇒ x ∈ A' or x ∈ B'

⇒ x ∉ A or x ∉ B

Statement II is incorrect.

∴ Neither statement I nor II is correct.

Concept:

1). A ∆ B is the symmetric difference that represents the objects that belong to A or B but not to their intersection.

• A Δ B = (A ∪ B) - (A ∩ B)

2). A ⊆ B subset A is a subset of B. set A is included in set B.

3). A ⋃ B, ⋃ is a union, that represents the objects that belong to set A or set B

4). A ⋂ B, ⋂ is an intersection, that represents the objects that belong to set A and set B

5). x ∈ A means x is an element of A, i.e. belongs to set membership

Calculation:

Given that, A and B  are two non-empty sets.

Statement I: A ∪ B = A ∩ B if A = B

Let x ∈ A ⇒ x ∈ A ⋃ B, [Because A ⊆ A ⋃ B]

⇒ x ∈ A ⋂ B, [As A ⋃ B = A ⋂ B]

⇒ x ∈ A and x ∈ B ⇒ x ∈ A,

⇒ B ⊆ A

Now, A ⊆ B, B ⊆ A ⇒ A = B

Statement I is correct.

Statement II: A Δ B = φ if A = B

⇒ A Δ B = φ

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

⇒ (A ∪ B) - (A ∩ B) = ϕ 

⇒ (A ∪ B) = (A ∩ B)

⇒ A = B

Statement II is correct.

∴ Statement I and II both are correct.

Concept:

1. Reflexive: Each element is related to itself.

• R is reflexive if for all x ∈ A, xRx.

2. Symmetric: If any one element is related to any other element, then the second element is related to the first.

• R is Symmetric if for all x, y ∈ A, if xRy, then yRx.

3. Transitive: If any one element is related to a second and that second element is related to a third, then the first element is related to the third.

• R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

4. R is an equivalence relation if A is nonempty and R is reflexive, symmetric, and transitive.

Calculation:

We have R = {(x,y): x2 - 5xy + 4y2 = 0, x,y ∈ N}.

Statement I: Reflexive

If y = x, then x2 - 5x² + 4x2 = 0

⇒ (x,x) ∈ R.

⇒ R is reflexive

Statement II: Symmetric

Take (4, 1) i.e. x = 4 and y = 1

We have (4)² − 5(4)(1) + 4(1)² = 16 − 20 + 4 = 0

⇒ (4,1) ∈ R.

Also (1)²− 5(1)(4) + 4(4)² = 1 − 20 + 64 = 45 ≠ 0

⇒ (1,4) ∉ R.

⇒ R is not symmetric.

Statement III: Transitive

(16,4) ∈ R because

(16)² − 5(16)(4) + 4(4)² = 256 − 320 + 64 = 0

Also (4,1) ∈ R because

⇒ (4)² − 5(4)(1) + 4(1)² = 16 − 20 + 4 = 0

Now, (16,1) ∈ R if (16)² − 5(16)(1) + 4(1)² = 0

⇒ 256 − 80 + 4 = 180 ≠ 0, which is not so.

⇒ (16,4), (4,1) ∈ R and (16,1) ∉ R

⇒ R is not transitive.

∴ R is only Reflexive.

Concept:

1). Reflexive: Each element is related to itself.

• R is reflexive if for all x ∈ A, xRx.

2). Symmetric: If any one element is related to any other element, then the second element is related to the first.

• R is Symmetric if for all x, y ∈ A, if xRy, then yRx.

3). Transitive: If any one element is related to a second and that second element is related to a third, then the first

    element is related to the third.

• R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

4). R is an equivalence relation if A is nonempty and R is reflexive, symmetric, and transitive.

5). Let R be a relation from a set A to another set B. Then R is of the form {(x, y): x ∈ A and y ∈ B}. The inverse

    relationship of R is denoted by R^-1 and its formula is R^-1 = {(y, x): y ∈ B and x ∈ A}.

Calculation:

Statement I: If R is reflexive, then R-1 is also reflexive

R is reflexive

⇒ (a,a) ∈ R,  a ∈ A

⇒ (a,a) ∈ R⁻¹      [by the definition of R^-1]

⇒ R⁻¹ is also reflexive relation.

Statement II: If R is symmetric, then R-1 is also symmetric

Let (b,a) ∈ R−1

⇒ (a,b) ∈ R, a,b ∈ A      [by the definition of R]

⇒ (b,a) ∈ R         [R is symmetric]

⇒ (a,b) ∈ R⁻¹        [by the definition of R-1]

If (b,a) ∈ R−1 then (a,b) ∈ R−1

⇒ R⁻¹ is also symmetric relation.

Statement III: If R is transitive, then R-1 is also transitive

Let (b,a), (a,c) ∈ R−1

⇒ (a,b), (c,a) ∈ R       [by the definition of R-1]

⇒ (c,a),(a,b) ∈ R

⇒ (c,b) ∈ R        [R is transitive]

⇒ (b,c) ∈ R^-1       [by the definition of R-1]

If (b,a), (a,c) ∈ R-1 then (b,c) ∈ R-1

⇒ R⁻¹ is also transitive relation.

∴ R⁻¹ is reflexive, symmetric and transitive.

Concept:

Let z = x + iy be a complex number.

 

Calculation:

Let, x + iy = \(\frac{1}{1 + i}\)

⇒ x + iy = \(\frac{1}{1 + i}\times \frac{1-i}{1-i}\)

We know that, (a - b)(a + b) = a² - b²

⇒ x + iy = \(\frac{1-i}{1 - i^2}\)

⇒ x + iy = \(\frac{1-i}{2} = \frac{1}{2}-\frac{i}{2}\)     [∵ i = √-1]

Compare the real and imaginary part

⇒ x = 1/2 & y = -1/2

So, x > 0 and y < 0

Therefore, principal argument

 \(-\rm \tan^{-1} \left|\frac{y}{x}\right|\)

⇒ \(-\rm \tan^{-1} \left|\frac{-1/2}{1/2}\right|=-tan^{-1}1\)

⇒ - π/4 

∴ The principal argument of \(\frac{1}{1 + i}\) is \(-\frac{\pi}{4}\).

Formula used:

For a complex number z = x + iy

|zⁿ| = |z|ⁿ

Where i = √-1 and

|z| = √(x² + y²)

Calculation:

Let, z = \(\left(\frac{\sqrt{-3}}{2}-\frac{1}{2}\right) \)

This can be written as:

\(z=\left(\frac{{i\sqrt 3}}{2}-\frac{1}{2}\right) \)

⇒ |z²⁰⁰| = \(|\left(\frac{{i\sqrt 3}}{2}-\frac{1}{2}\right)^{200}| \)

Using the property of modulus:

⇒ |z|²⁰⁰ = \(|\left(\frac{{i\sqrt 3}}{2}-\frac{1}{2}\right)^{200}| \)

We know that |z| = √(x2 + y2)

⇒ |z|200 = \(\left(\sqrt{(\frac{{\sqrt3}}{2})^2+(\frac{-1}{2})^2}\right)^{200} \)

⇒ |z|200 = \((\sqrt {\frac{3}{4}+\frac{1}{4}})^{200} = 1^{200} \)

∴ \(\left(\frac{\sqrt{-3}}{2}-\frac{1}{2}\right)^{200} =1\)

Concept:

Factorial of a non-negative integer is the multiplication of all integers smaller than or equal to n.

For example factorial of 6 (6!) is 6 × 5 × 4 × 3 × 2 × 1

Calculation:

Statement I: \(\frac{n!}{3!}\) is divisible by 6, where n > 3

Let n = 4

⇒ \(\displaystyle \frac{n!}{3!}=\frac{4!}{3!}=4\)

4 is not divisible by 6.

Statement I is incorrect.

Statement I: \(\frac{n!}{3!}+3 \) is divisible by 7, where n > 3

Let n = 5

⇒ \(\displaystyle \frac{n!}{3!}+3=\frac{5!}{3!}+3=\frac{5×4×3!}{3!}+3=20+3=23\)

23 is not divisible by 7.

Statement II is incorrect.

∴ Neither statement I nor II is correct.

Given:

5 members be selected out of 9 members

Two particular members must always be excluded

Formula used:

nCr = n!/(n – r)! r!

Calculation:

According to the question, 

Two members have always been excluded then actually we are choosing

7 members and the number of members chosen in 5 ways

⁷C5 = 7!/(7 – 5)! 5!

⇒(7 × 6 × 5!)/(5! × 2!)

⇒ (7 × 6)/2

⇒ 7 × 3

⇒ 21

∴ The number of ways is 21.

Concept:

1). We know that the rth term from the end in the expansion (a + b)n = (n − r + 2)th term from the beginning of the expansion (a + b)n

2). rth term for the binomial expression, (x + y)n, Tr = nCr-1 xn-r+1 yr-1

 

Calculation:

In the expansion \(\left(x + \frac{1}{x}\right)^{2n} \),

The (n + 1)^th term from the end = (2n − n − 1 + 2) = (n + 1)^th term from the beginning.

We know that,  Tr = nCr-1 xn-r+1 yr-1

Now, the t_n+1 term is given by 

⇒ Tn+1 = ²ⁿC_n (x)^2n-n (1/x)ⁿ

⇒ Tn+1 = 2nCn (x)2n-n-n

⇒ Tn+1 = 2nCn

∴ Tn+1 = 2nCn or C(2n, n)

Given:

Sum of the first 9 terms of an AP = sum of the first 11 terms

Formula used:

Sum of n terms of AP:

S_n = \(\frac{n}{2}[2a+ (n-1)d]\)

Where a and d are the first terms and common differences respectively.

Calculation:

Let, a and d be first terms and common differences respectively

According to the question,  S₉ = S11

By using the formula of the sum of AP

\(⇒ \frac{9}{2}[2a+8d]=\frac{11}{2}[2a+10d]\)

⇒ 18a + 72d = 22a + 110d

⇒ 4a = - 38d

⇒ a = \(\frac{-19d}{2}\)         -----(1)

Now, sum of the first 20 terms

\({{S}_{20}}=\frac{20}{2}\left[ 2a+19d \right]\)

From equation (1)

S₂₀ = 10 [2 × (\(\frac{-19d}{2}\)) + 19d]

⇒ S20 = 10[-19d + 19d] = 0

∴ The sum of the first 20 terms is 0.

Given:

5^th term of an AP is \(\frac{1}{10}\) and its 10^th term is \(\frac{1}{5},\)

Formula used:

n^th term of AP:

T_n = a + (n - 1)d

Sum of n terms of AP:

S_n = \(\frac{n}{2}\)[2a + (n - 1)d]

Where a and d are the first term and common differences of AP.

Calculation:

According to the question,

T₅ = \(\frac{1}{10}\)

Using the formula of n^th term of AP

⇒ T₅ = a + 4d 

⇒  a + 4d = \(\frac{1}{10}\)      ------(1)

Similarly, we have

a + 9d =  \(\frac{1}{5}\)          ------(2)

Doing equation (2) - (1)

⇒ (a + 9d) - (a + 4d) = \(\frac{1}{5}\) - \(\frac{1}{10}\)

⇒ 5d = \(\frac{1}{10}\)

⇒ d = \(\frac{1}{50}\)

Put this value in equation (1)

a  + \(4×\frac{1}{50}\) = \(\frac{1}{10}\)

⇒ a = \(\frac{1}{10}- \frac{2}{25} = \frac{1}{50}\)

Now using the formula, Sn = \(\frac{n}{2}\)[2a + (n - 1)d]

Sum of 50 terms of AP = S₅₀ = \(\frac{50}{2}\)[2×\(\frac{1}{50}\)  + (50 - 1)\(\frac{1}{50}\)]

⇒ S₅₀ = 25\((\frac{2}{50}+ \frac{49}{50} )= \frac{51}{2}\) = 25.5

∴  The sum of the first 50 terms is 25.5.

Concept:

Binary to decimal conversion formula

N = bn-1 × 2n-1 + ......... + b₄ × 24 + b3 × 23 + b₂ × 22 + b1 × 21 + b0 × 20 

Where N is decimal equivalent, b is the binary digit, 2 is the base value

Decimal to Binary conversion

Step 1: Divide the decimal number n by 2. Use the integer quotient obtained in this step as the dividend for the next step.

Step 2: Write the remainder from bottom to top i.e. in the reverse chronological order.

Calculation:

We know that, the binary to decimal conversion formula is

N = bn-1 × 2n-1 + ............ + b4 × 24 + b3 × 23 + b2 × 22 + b1 × 21 + b0 × 20 

Now, converting (1110011)2 to decimal form, we get,

(1110011)2 = 1 × 26 + 1 × 25 + 1 × 24 + 0 × 23 + 0 × 22 + 1 × 21 + 1 × 20 

⇒ (1110011)2 = 64 + 32 + 16 + 2 + 1

⇒ (1110011)2 = 115      ------(i)

Again converting (10111)2 to decimal form, we get,

(10111)2 = 1 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 1 × 20

⇒ (10111)2 =  16 + 4 + 2 + 1

⇒(10111)2  =  23      ------(ii)

Putting the value of equations (i) and (ii), we get,

We have to find (1110011)2 ÷ (10111)2

Putting the value of equations (i) and (ii), we get,

⇒ 115 ÷ 23 = 5

Now, convert the decimal number 5 into binary form, 5 = (101)2

∴ (1110011)2  ÷ (10111)2 = (101)2

Concept:

Formulae

• x3 + y3 = (x + y)(x2 - xy + y2)
• (x - y)2 = x2 - 2xy + y2

Binary to decimal conversion formula

N = b_n-1 × 2^n-1 + ...... + b4 × 24 + b3 × 23 + b2 × 22 + b₁ × 21 + b₀ × 20 

Where N is decimal equivalent, b is the binary digit, 2 is the base value

Decimal to Binary conversion

Step 1: Divide the decimal number n by 2. Use the integer quotient obtained in this step as the dividend for the next step.

Step 2: Write the remainder from bottom to top i.e. in the reverse chronological order.

Calculation:

Given:

x3 + y3 = (100010111)2 and x + y = (11111)2

Converting (100010111)2 to decimal form, we get,

x3 + y3 = (100010111)2 = 1 × 2⁸ + 0 × 2⁷ + 0 × 2⁶ + 0 × 2⁵ + 1 × 2⁴ + 0 × 2³ + 1 × 2² + 1 × 2¹ + 1 × 2⁰ 

⇒ x3 + y3 = 256 + 16 + 4 + 2 + 1 

⇒ x3 + y3 = 279      ------(i)

Converting (11111)2 to decimal form, we get,

x + y = (11111)2 = 1 × 2⁴ + 1 × 2³ + 1 × 2² + 1 × 2¹ + 1 × 2⁰

⇒ x + y =  16 + 8 + 4 + 2 + 1

⇒ x + y =  31      ------(ii)

Also, we know that, 

x3 + y3 = (x + y) (x² - xy + y²)

⇒ x3 + y3 = (x + y) (x2 - 2xy + y2 + xy)

⇒ x3 + y3 = (x + y) [(x - y)² + xy]

⇒ (x - y)2 + xy = \(\displaystyle \frac{x^3+y^3}{x+y}\)

Putting the value of equation (i) and (ii), we get,

⇒ (x - y)2 + xy = \(\displaystyle \frac{279}{31}=9\)

Now, convert this decimal number 9 into binary form, 9 = (1001)₂

The value of (x - y)2 + xy = (1001)2

Calculation:

Converting the inequations to equations, we obtain:

5x - 4y + 12 = 0, x + y = 2, x = 0, y = 0.

For 5x - 4y + 12 = 0

• Putting x = 0, 0 - 4y = - 12 ⇒ y = 3
• Putting y = 0, 5x - 0 = - 12 ⇒ x = - 2.4
• This line meets the x-axis at (-2.4, 0) and the y-axis at (0, 3). Draw a line through these points.
• We see that the origin (0, 0) does not satisfy the inequation 5x - 4y + 12 < 0
• Therefore, the region that does not contain the origin is the solution of the inequality 5x - 4y + 12 < 0

For x + y = 2

• Putting x = 0, y = 2
• Putting y = 0, x = 2
• This line meets the x-axis at (2, 0) and the y-axis at (0, 2). Draw a line through these points.
• We see that the origin (0, 0) does satisfy the inequation x + y < 2
• Therefore, the region contains the origin is the solution of the inequality x + y < 2

The solution to the inequalities is the intersection of the 5x - 4y + 12 < 0, x + y < 2, x < 0 and y > 0. Thus, the shaded region represents the solution set of the given set of inequalities.

Now solving 5x - 4y + 12 = 0 and x + y = 2, we get the point P(-0.444, 2.444).

Now, putting (-1, 2) in 5x - 4y + 12 < 0, 

• 5 - 8 + 12 < 0 ⇒ -1 < 0 which is true.

Again, putting (-1, 2) in x + y < 2, 

-1 + 2 < 2 ⇒ 1 < 2 which is true.

⇒ Point (-1, 2) satisfies 5x - 4y + 12 < 0 and x + y < 2

∴ (-1, 2) lies in the common region.

Mistake Points 

Now, putting (0, 0) in 5x - 4y + 12 < 0, 

0 - 0 + 12 < 0  ⇒ 12 < 0 which is False.

⇒ Point (0, 0) not satisfies 5x - 4y + 12 = 0 and x + y = 2

Concept:

Greatest Integer Function: (Floor function)

The function f (x) = [x] is called the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x.

The domain of [x] is R and the range is I.

Note:

Any function is differentiable only if it is continuous.

The floor function f(x) = ⌊x⌋ is differentiable in every open interval between integers, (n, n + 1) for any integer n.

Calculation:

Given that,

y = [x]

Statement:1  Its derivative is 0 at x = 0.5

We know that the floor function is differentiable at all points except integer points.

Hence, y = [x] is differentiable at x = 0.5

⇒ y = [0.5] = 0

⇒ dy/dx = 0

Statement:2  It is continuous at x = 0

We know that y = [x] is only continuous in the open interval between integers and discontinuous at all integer values.

∴ Only statement 1 is correct.

Concept:

Order of a differential equation: The order of a differential equation is the order of the highest order derivative appearing in the equation.

Degree of a differential equation: The degree of a differential equation is the degree of the highest order derivative when differential coefficients are made free from radicals and fractions.

Formula used:

(a + b)3 = a3 + b3 + 3a2b + 3ab2

(a^m)ⁿ = a^mn

Calculation:

We have, \(1+\left(\frac{dy}{dx}\right)^2 =\left(\frac{d^2y}{dx^2}\right)^{\frac{4}{3}}\)

Since power 4/3 is a fraction, hence we need to remove it first.

Doing cube of both sides

\([1+\left(\frac{dy}{dx}\right)^2]^3 =[\left(\frac{d^2y}{dx^2}\right)^{\frac{4}{3}}]^3\)

⇒ \(1+[\left(\frac{dy}{dx}\right)^2]^3 + 3.1^2.\left(\frac{dy}{dx}\right)^2+3.1.[\left(\frac{dy}{dx}\right)^2]^2 =\left(\frac{d^2y}{dx^2}\right)^{\frac{4\times 3}{3}}\)

⇒ \(1+\left(\frac{dy}{dx}\right)^6 + 3\left(\frac{dy}{dx}\right)^2+3\left(\frac{dy}{dx}\right)^4=\left(\frac{d^2y}{dx^2}\right)^{4}\)

Clearly, the highest order differential coefficient in this equation is \(\frac{d^2y}{dx^2}\) and its power is 4. ∴ The degree is 4.

Concept

Half-life:  \(\displaystyle t_{\frac{1}{2}}=\frac{log_e2}{k}\)

Where k is the decay constant.

Calculation:

We know that  \(\displaystyle t_{\frac{1}{2}}=\frac{log_e2}{k}\)

Putting the value of t_1/2 = 100 years,

⇒ 100 = \(\displaystyle \frac{log_e2}{k}\)

⇒ k = \(\displaystyle \frac{log_e2}{100}\)

∴ Decay constant is \(\displaystyle \frac{log_e2}{100}\).

Concept:

A function y = √f(x)  is define for f(x) > 0.

If (x - a)(x - b) ≤ 0 ⇒ x ∈ [a, b]

Calculation:

Given  that, \(f(x) =\sqrt{1 - (x-1)^2} \ \)

For a function to be defined,

1 - (x - 1)² ≥ 0

⇒ (x - 1)2 - 1 ≤ 0

Since, (a - b)² = a² - 2ab + b²

⇒ x² + 1  - 2x - 1 ≤ 0

⇒ x² - 2x ≤ 0

⇒ x(x - 2) ≤ 0

∴ x ∈ [0, 2]

Calculation:

The equation of parabola is y² = 4kx      ------(1)

Let O be the vertex, S be the focus, and LL' be the latus rectum of the parabola.

The equation of latus rectum is x = k.

Also, here the parabola is symmetric about x−axis.

Required area, A = 2(area of OSL)

⇒ Required area, A = 2 \(\displaystyle \int_0^ky\ dx\)

⇒ A = 2 \(\displaystyle \int_0^k2√ k\ √ x\ dx\)

⇒ A = 2.2√k \(\displaystyle \int_0^k \ x^{\frac{1}{2}} dx\)

⇒ A = 4√k \(\displaystyle \ \left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^k \)

⇒ A = \(\displaystyle \frac{8}{3}\ \sqrt k\ [k^\frac{3}{2}-0^\frac{3}{2}]\)

⇒ A = \(\displaystyle \frac{8}{3}\ k^2\)

Since the area of the given parabola is 24.

⇒ 24 = \(\displaystyle \frac{8}{3}\ k^2\)

⇒ k = ± 3

As the value of  k > 0, k = 3.

∴ The value of k = 3.