Quantitative Aptitude PYQs

Given:

p = \(\sqrt[3]{\left(a+\sqrt{a^2+b^3}\right)}+\sqrt[3]{\left(a-\sqrt{a^2+b^3}\right)}\)

Calculation:

We can easily solve this by putting values of a and b,

Lets assume a = 1 and b = 0

so,

p = \(\sqrt[3]{\left(1+\sqrt{1^2+0^3}\right)}+\sqrt[3]{\left(1-\sqrt{1^2+0^3}\right)}\)

⇒ p = \(\sqrt[3]{\left(1+1\right)}+\sqrt[3]{\left(1-1\right)}\)

⇒ p = \(\sqrt[3]{\left(2\right)}\)

Putting value of p and b,

⇒ p3 + 3bp = \(\sqrt[3]{\left(2\right)}^3\) + 3 × 0 × \(\sqrt[3]{\left(2\right)}\)

⇒ p3 + 3bp = 2 ----- (i)

Putting value of a = 1 in options,

Option 1 = -2a = -2 × 1 = -2

Option 2 = a = 1 = 1

Option 3 = 2a = 2 × 1 = 2

Option 4 = 3a = 3 × 1 = 3

Option 3 matches with equation (i).

∴ The correct answer is option 3.

Given:

Length of plank = 4.25 m

Width of plank = 3.4 m

Concept:

To find the largest size of square pieces that can be cut without

wastage, we need to find the greatest common divisor (GCD) of the

length and width of the plank.

Formula Used:

GCD of two numbers a and b is the largest number that divides both a

and b without leaving a remainder.

Calculation:

Length = 4.25 m = 425 cm

Width = 3.4 m = 340 cm

GCD(425, 340) = 85 cm

⇒ Number of pieces along the length = 425 / 85 = 5

⇒ Number of pieces along the width = 340 / 85 = 4

⇒ Total number of pieces = 5 × 4 = 20

Hence, the number of square pieces of the largest size that can be cut from the plank is 20.

Given:

x4 - 13x2y2 - 300y4,

x3 - 4x2y - 4xy2 - 5y3,

x3 - 125y3

Calculation:

x3 - 125y3 = (x)² - (5y)² = (x - 5y)(x² + 25y² + 5xy) .....

(i) x4 - 13x2y2 - 300y4 = x⁴ - 25x²y² + 12x²y² - 300y²

x²(x² - 25y²) + 12y²(x² - 25y²)

(x² + 12y²)(x2 - 25y2) 

(x2 + 12y2)(x - 5y)(x + 5y) .....(ii)

Since (x - 5y) is zero of both the polynomial (i) and (ii)

Put x - 5y = 0, x = 5y in third polynomial

to confirm (x - 5y) as HCF,

x3 - 4x2y - 4xy2 - 5y3 = 125y³ - 100y³ -20y³ -5y³ = 0

Hence HCF is (x - 5y).

∴ The correct answer is x - 5y.

Given:

HCF of 768 and x⁶y² is 32xy for natural numbers x ≥ 2, y ≥ 2.

Concept Used:

1. Highest Common Factor (HCF)
2. Prime factorization

Calculation:

Prime factorization of 768:

768 = 2⁸ × 3

Prime factorization of 32xy:

32xy = 2⁵ × x × y

Prime factorization of x⁶y²:

x⁶y² = x⁶ × y²

For HCF to be 32xy, the minimum power of each prime factor in both numbers should be considered:

2⁵ × x × y = 32xy

Here, x and y must be such that:

2⁵ × x × y is a factor of both 768 and x⁶y².

Since 32 = 2⁵,

Therefore, x and y must be:

x = 2

y = 3

Sum of x and y:

⇒ x + y = 2 + 3 = 5

∴ The value of (x + y) is 5.

Given:

We need to find the smallest natural number n such that (n + 1) × n × (n - 1) × (n - 2) × ... × 3 × 2 × 1 is divisible by 910.

Concept Used:

To determine the smallest n such that the product is divisible by 910, we

need to factorize 910 and ensure that the factorial (n!) contains these

factors.

Formula Used:

Factorize the number 910.

Calculation:

Factorize 910:

910 = 2 × 5 × 7 × 13

We need n! to include all these prime factors.

⇒ The smallest n must be at least 13 to include the factor 13.

Check n = 13:

13! = 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

13! contains the factors 2, 5, 7, and 13.

∴ 13! is divisible by 910.

Hence, the smallest natural number n is 12.

Calculation:

555777 + 777555

⇒ (5 × 111)^{7 × 111} + (7 × 111)^{5 × 111}

⇒ 111⁵⁵⁵ (5⁷⁷⁷ × 111²²² + 7⁵⁵⁵)

111 is divisible by 3 and 37 both as 111 = 3 × 37,

Also, both the terms in bracket are odd so their sum will be even and hence divisible by 2.

∴ The correct answer is option 4.

Given:

Consider the following

Statements:

1. If (3m³ + 2m² + 5m + n) / m is not an integer, where m and n are integers, then n is not divisible by m.
2. 5(8m) + 23m is divisible by 48 for all whole numbers m.

Formula Used:

For statement 1: Verify if the expression (3m³ + 2m² + 5m + n) / m is an integer or not.

For statement 2: Check divisibility of 5(8m) + 23m by 48.

Calculation:

Statement 1:

Expression: (3m³ + 2m² + 5m + n) / m

⇒ 3m² + 2m + 5 + n/m

If (3m² + 2m + 5 + n/m) is not an integer, then n/m is not an integer.

⇒ n is not divisible by m.

Statement 1 is correct.

Statement 2:

5(8^m) + (2³)^m

⇒ 8^m(5 + 1)

⇒ 6 × 8^m

For m = 0, 6 × 1 = 6 which is not divisible by 48.

Statement 2 is incorrect.

∴ The correct answer is option 1.

Given:

Sum of two positive numbers = 40

The GM of these two numbers is 20% lower than their AM

Concept:

AM (Arithmetic Mean) = (a + b) / 2

GM (Geometric Mean) = √(a × b)

GM = 0.8 × AM

Calculation:

Let the two numbers be a and b.

a + b = 40

AM = (a + b) / 2 = 40 / 2 = 20

GM = √(a × b)

Given GM = 0.8 × AM

⇒ √(a × b) = 0.8 × 20

⇒ √(a × b) = 16

⇒ a × b = 16²

⇒ a × b = 256

We have the equations:

1. a + b = 40
2. a × b = 256

Solving these two equations:

Let a and b be the roots of the quadratic equation x² - (a+b)x + ab = 0

⇒ x² - 40x + 256 = 0

Solving for x using the quadratic formula x = [ -b ± √(b² - 4ac) ] / 2a

⇒ x = [ 40 ± √(1600 - 1024) ] / 2

⇒ x = [ 40 ± √576 ] / 2

⇒ x = [ 40 ± 24 ] / 2

⇒ x = 32 or x = 8

So, the two numbers are 32 and 8.

Difference between the two numbers = 32 - 8 = 24

Hence, the difference between the two numbers is 24.

Shortcut Trick

Total Work = 50 × 40 = 2000 units.

Work done in 10-day intervals: 10 × (50 + 45 + 40 + 35 + 30) = 10 × 200 = 2000 units.

The total work is completed in exactly 5 such intervals.

Total time = 5 × 10 = 50 days.

∴ The correct answer is 50 days.

Alternate Method

Given: Initial Men = 50, Target Days = 40, Reduction = 5 men every 10 days.

Formula Used: Total Work = Men × Days

⇒ Total Work to be done = 50 × 40 = 2000 units.

⇒ Work done in first 10 days = 50 × 10 = 500 units.

⇒ Work done in next 10 days = 45 × 10 = 450 units.

⇒ Work done in third 10 days = 40 × 10 = 400 units.

⇒ Work done in fourth 10 days = 35 × 10 = 350 units.

⇒ Work done in fifth 10 days = 30 × 10 = 300 units.

⇒ Total Work completed = 500 + 450 + 400 + 350 + 300 = 2000 units.

⇒ Total Time taken = 10 + 10 + 10 + 10 + 10 = 50 days.

∴ The correct answer is 50 days.

Additional Information

Man-Days Concept

The total amount of work is constant and is expressed as the product of the number of men and the time taken: W = M × D.

Variable Manpower Rule

When manpower changes in stages, the total work is the sum of products of men and duration for each stage: W = M₁D₁ + M₂D₂ + ... + M_nD_n.

Arithmetic Progression in Work

If the number of men decreases by a constant value 'd' at regular intervals 't', the work done forms an AP: Work = t × [n/2 × {2M + (n−1)(−d)}], where n is the number of intervals.

Calculation:

x = \(\frac{1}{2+\frac{3}{4+\frac{5}{6+\frac{7}{8+\frac{9}{10}}}}}\)

We know that the value of will decrease if we increase the denominator,

Let a = \(\frac{3}{4+\frac{5}{6+\frac{7}{8+\frac{9}{10}}}}\)

So, x = \(\frac{1}{2 + a}\) 

Since a is a positive number,

The value of \(\frac{1}{2 + a}\) will always be less then \(\frac{1}{2}\) = 0.5

∴ The correct answer is option 1.

Given:

A bottle contains spirit and water in the ratio 1 ∶ 4

Another identical bottle contains spirit and water in the ratio 4 ∶ 1

Calculation:

Let the quantity of the first mixture be x and the quantity of the second mixture be y.

In the first bottle, the ratio of spirit to water is 1 ∶ 4, so the spirit content is x/5 and the water content is 4x/5.

In the second bottle, the ratio of spirit to water is 4 ∶ 1, so the spirit content is 4y/5 and the water content is y/5.

In the new mixture, the spirit content should be (x/5 + 4y/5) and the water content should be (4x/5 + y/5).

The required ratio is 1 ∶ 3, so:

⇒ (x/5 + 4y/5) / (4x/5 + y/5) = 1/3

⇒ (x + 4y) / (4x + y) = 1/3

⇒ 3(x + 4y) = 4x + y

⇒ 3x + 12y = 4x + y

⇒ 12y - y = 4x - 3x

⇒ 11y = x

⇒ x/y = 11/1

∴ Ratio in which mixtures in the two bottles should be mixed is 11 ∶ 1

Given:

3 sin θ + 5 cos θ = 5

Concept Used:

We can use the trigonometric identities and algebraic manipulation to solve the given equation.

Calculation:

Let 3 sin θ + 5 cos θ = R

Given R = 5

We need to find the value of 5 sin θ - 3 cos θ

Square both sides of the given equation:

⇒ (3 sin θ + 5 cos θ)² = 5²

⇒ 9 sin² θ + 30 sin θ cos θ + 25 cos² θ = 25

Using the identity sin² θ + cos² θ = 1:

⇒ 9 (1 - cos² θ) + 30 sin θ cos θ + 25 cos² θ = 25

⇒ 9 - 9 cos² θ + 30 sin θ cos θ + 25 cos² θ = 25

⇒ 9 + 16 cos² θ + 30 sin θ cos θ = 25

⇒ 16 cos² θ + 30 sin θ cos θ = 16

Now, consider the expression we need to find:

Let 5 sin θ - 3 cos θ = K

Square both sides of this equation:

⇒ (5 sin θ - 3 cos θ)² = K²

⇒ 25 sin² θ - 30 sin θ cos θ + 9 cos² θ = K²

Using the identity sin² θ + cos² θ = 1:

⇒ 25 (1 - cos² θ) - 30 sin θ cos θ + 9 cos² θ = K²

⇒ 25 - 25 cos² θ - 30 sin θ cos θ + 9 cos² θ = K²

⇒ 25 - 16 cos² θ - 30 sin θ cos θ = K²

Since we know from earlier:

16 cos² θ + 30 sin θ cos θ = 16

Substitute this into the equation:

⇒ 25 - 16 = K²

⇒ 9 = K²

⇒ K = ±3

Since the options provided are negative, the correct answer is:

⇒ K = -3

∴ The value of 5 sin θ - 3 cos θ is -3.

Given:

x4k + x4k+2 + x4k+4 + x4k+6

Calculation:

We have,

x4k + x4k+2 + x4k+4 + x4k+6

⇒ x^4k (x⁰ + x²) + x^4k+4 (x⁰ + x²)

⇒ (x4k + x4k+4) (x0 + x2)

⇒ x4k (x⁰ + x4) (x0 + x2)

⇒ x4k (1 + x4) (1+ x2)

After factorizing the polynomial, we can see the given polynomial is a multiple of (1 + x²) and (1 + x⁴).

Thus, the given polynomial is divisible by both (1 + x2) and (1 + x4).

Hence, both statements are correct.

∴ The correct answer is option (3).

Given:

\(\frac{\sin ^2 A+5 \sin A+1}{\sin A}\), where 0 < A ≤ \(\frac{\pi}{2}\)

Concept Used:

The minimum Value of \(x + \frac{1}{x}\)  is always 2.

Calculation:

\(\frac{\sin ^2 A+5 \sin A+1}{\sin A}\)

⇒ Sin A + 5 + \(\frac{1}{sin A}\)

So, the minimum value of Sin A + \(\frac{1}{sin A}\) = 2

Minimum Value of the given expression,

⇒ 2 + 5 = 7

∴ The correct answer is 7.

Given:

\(\frac{3}{1^2 \times 2^2}\) + \(\frac{5}{2^2 \times 3^2}\) + \(\frac{7}{3^2 \times 4^2}\) + ... 

Calculation:

\(\frac{3}{1^2 \times 2^2}\) + \(\frac{5}{2^2 \times 3^2}\) + \(\frac{7}{3^2 \times 4^2}\) + ... 

⇒ \(\frac{2^2 - 1^2}{1^2 \times 2^2}\) + \(\frac{3^2 - 2^2}{2^2 \times 3^2}\) + \(\frac{4^2 - 3^2}{3^2 \times 4^2}\) + ... 

⇒ \(\frac{1}{1^2} - \frac{1}{2^2}\) + \(\frac{1}{2^2} - \frac{1}{3^2}\) + \(\frac{1}{3^2} - \frac{1}{4^2}\) + ...... - \(\frac{1}{\infty^2 }\)

⇒ 1 - \(\frac{1}{\infty^2 }\)

⇒ 1

∴ The correct answer is 1

Given:

\(\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}}\) = \(\frac{421}{972}\)

Calculation:

For this type of question, we solve the reciprocal of the value and the integers will be values of the variables,

\(\frac{972}{421}\) = 2 + \(\frac{130}{421}\) ⇒ a = 2

\(\frac{421}{130}\) = 3 + \(\frac{31}{130}\) ⇒ b = 3

\(\frac{130}{31}\) = 4 + \(\frac{6}{31}\) ⇒ c = 4

\(\frac{31}{6}\) = 5 + \(\frac{1}{6}\) ⇒ d = 5

\(\frac{6}{1}\) = 6 ⇒ e = 6

⇒ a × b × c × d × e = 2 × 3 × 4 × 5 × 6

⇒ a × b × c × d × e = 720

∴ The correct answer is 720.

Given:

Edge of the cube = 14 cm

Radius of each circle = 7 cm

π = 22/7

Concept:

To find the unpainted surface area of the cube, we need to calculate the total surface area of the cube and then subtract the total area of the painted circles.

Formula Used:

Total surface area of the cube = 6a²

Area of one circle = πr²

Total area of painted circles = 6 × Area of one circle

Unpainted surface area = Total surface area of the cube - Total area of painted circles

Calculation:

⇒ Total surface area of the cube = 6 × (14 cm)²

⇒ Total surface area of the cube = 6 × 196 cm²

⇒ Total surface area of the cube = 1176 cm²

⇒ Area of one circle = π × (7 cm)²

⇒ Area of one circle = (22/7) × 49 cm²

⇒ Area of one circle = 154 cm²

⇒ Total area of painted circles = 6 × 154 cm²

⇒ Total area of painted circles = 924 cm²

⇒ Unpainted surface area = 1176 cm² - 924 cm²

⇒ Unpainted surface area = 252 cm²

Hence, the total area of the unpainted surface is 252 cm².

Given:

Radius of circular plate (R) = 7 cm

Thickness of plate (h) = 0.16 mm = 0.016 cm

Angle of the sector (θ) = 150°

Concept:

The volume of the remaining piece after cutting the sector is equal to the volume of the spherical bead formed.

Formula used:

Volume of the sector of a cylinder = (θ/360) × π × R² × h

Volume of the sphere = (4/3) × π × r³

Calculation:

Volume of the sector cut off:

⇒ (150/360) × π × (7)² × 0.016

⇒ (5/12) × π × 49 × 0.016

⇒ (5/12) × π × 0.784

⇒ 3.14 × 0.3267

⇒ 1.026 cm³

Volume of the remaining piece:

⇒ Total volume of the plate - Volume of the sector cut off

⇒ π × R² × h - 1.026

⇒ π × 49 × 0.016 - 1.026

⇒ 3.14 × 0.784 - 1.026

⇒ 2.46176 - 1.026

⇒ 1.43576 cm³

Volume of the spherical bead:

⇒ (4/3) × π × r³ = 1.43576

⇒ r³ = (1.43576 × 3)/(4 × π)

⇒ r³ = 0.3427 cm³ ≈ 0.343 cm³

⇒ r = ³√0.343

⇒ r ≈ 0.7 cm

∴ The value of r is 0.7 cm.

Shortcut Trick

Let the central angle of the sector be 2θ. Chord length = 2r × sin(θ) and Height = r(1 − cos(θ)).

Given 2r × sin(θ) = 2√3 × r(1 − cos(θ)) ⇒ tan(θ/2) = 1/√3 ⇒ θ = 60° ⇒ Central angle = 120°.

Sector area P = πr² / 3. Segment area Q = πr² / 3 − √3r² / 4.

Ratio P / Q = 4π / (4π − 3√3) = (4 × 3.14) / (4 × 3.14 − 3 × 1.7).

P / Q = 12.56 / 7.46 ≈ 1.68 ≈ 1.7.

∴ The correct answer is 1.7.

Alternate Method

Given:

Chord AB of a circle is 2√3 times the height of the minor segment.

π = 3.14, √3 = 1.7

Formula Used:

For a circle with radius r and central angle 2θ:

Length of chord = 2r × sin(θ)

Height of minor segment = r(1 − cos(θ))

O
A
B
120°
h
r
r

Calculations:

⇒ 2r × sin(θ) = 2√3 × r(1 − cos(θ))

⇒ 2 × sin(θ/2) × cos(θ/2) = √3 × 2 × sin²(θ/2)

⇒ cos(θ/2) = √3 × sin(θ/2)

⇒ tan(θ/2) = 1 / √3

⇒ θ/2 = 30° ⇒ θ = 60°

⇒ Central angle 2θ = 120°

⇒ P (Area of sector) = (120° / 360°) × πr² = πr² / 3

⇒ Area of triangle OAB = (1/2) × r² × sin(120°) = √3r² / 4

⇒ Q (Area of minor segment) = P − Area of triangle OAB = (πr² / 3) − (√3r² / 4)

⇒ P / Q = (π / 3) ÷ [(π / 3) − (√3 / 4)]

⇒ P / Q = 4π / (4π − 3√3)

⇒ P / Q = (4 × 3.14) / (4 × 3.14 − 3 × 1.7)

⇒ P / Q = 12.56 / (12.56 − 5.10) = 12.56 / 7.46

⇒ P / Q ≈ 1.683 ≈ 1.7

∴ The correct answer is 1.7.

Additional Information

Geometry Circle Formulas

Length of Chord = 2r × sin(θ/2), where θ is the central angle. Area of Sector = (θ / 360°) × πr².

Area of Segment

The area of a minor segment is found by subtracting the area of the triangle formed by the radii and chord from the area of the corresponding sector.

Trigonometric Half-Angle Identities

sin(θ) = 2 × sin(θ/2) × cos(θ/2) and 1 − cos(θ) = 2 × sin²(θ/2).

Calculation:

Let the radius of the bigger circle = R 

The radius of a smaller circle = r

AB = 14 cm.

AM = 14/2 = 7 cm

In triangle OAM 

OA2 = OM2 + AM2

R2 = r2 + 72

R2 - r2 = 49

Area of between two triangle = π(R2 - r2)

⇒ (22/7) × 49 = 154 square cm

∴ The correct answer is 154 square cm