Quantitative Aptitude PYQs

Given:

A pendulum swings through an angle = 9° 

Length of arc = 14.3 cm

Formula used:

Length of acr of sector of a angle = (θ/360°) × 2πr

Calculation:

Angle of AOB = 9° 

14.3 = (9°/360°) × 2 × (22/7) × r

\(r = \frac{14.3\times 70}{11}\)

r = 91 cm

∴ The length of the pendulum is 91 cm

Given:

The arch of a bridge is in the form of an arc of a circle. Span of the bridge is 40m & height in the middle is 8m.

Formula used:

Pythagoras theorem:

(Hypotenuse)² = (Base)² + (Perpendicular)² 

Calculation:

According to the question,

In Δ OAP

According to the formula,

⇒ R² =  20² + (R - 8)² 

⇒ R² = 400 + R² + 64 - 16R

⇒ 400 + 64 = 16R

⇒ 464 = 16R

⇒ R = 29

∴ The radius of the curvature of the bridge is 29m.

Given:

If a, b, c are sides of a right angle triangle, where a> b> c

Calculation:

Let's assume the sides of right angles triangle be 3, 4, and 5

Therefore,

⇒ (a + b + c) (a + b - c) ( a - b + c) (a - b - c)

⇒ (5 + 4 + 3) (5 + 4 - 3) (5 - 4 + 3) (5 - 4 - 3)

⇒ (12) (6) (4) (-2) = - 576

From the options,

Option1 → 4b²c² = 4(4²)(3²) = 576

Option2 → -4b²c² = -4(4²)(3²) = - 576

Option3 → -2a²b² = -2(5²)(4²) = - 800

Option4 → -4a²b²  = -4(5²)(4²) = - 1600    

∴ (a + b + c) (a + b - c) (a - b + c) ( a - b - c) = -4b²c²

Given:

The cube rot of x varies inversely as the square root of y.

Calculation:

According to the question,

∛x ∝ \(\frac{1}{√{y}}\) 

⇒ x^1/3 = k/y^1/2

Now,

⇒ x = 8, y = 3

⇒ (8)^1/3 = k/3^1/2

⇒ k = 2√3

Now, 

⇒ x^1/3 = k/√y

⇒ x^1/3 = 2√3 /√∛3 

⇒ x^1/3 = 2 × 3^1/2 /3^1/6

⇒ x^1/3 = 2 × 3^1/3

⇒ x = (2 × 3^1/3)³ = 8 × 3

∴ x = 24.

Shortcut Trick

Radius of new sphere R = ³√(r₁³ + r₂³ + r₃³) = ³√(6³ + 8³ + 10³) = ³√(216 + 512 + 1000) = 12 cm.

Sum of surface areas = 4π(6² + 8² + 10²) = 4π(36 + 64 + 100) = 4π(200) = 800π.

New surface area = 4π(12²) = 4π(144) = 576π.

Diminution % = [(800π − 576π) ÷ 800π] × 100 = (224 ÷ 800) × 100 = 28%.

∴ The correct answer is 28%.

Alternate Method

Given: Radii of three spheres r₁ = 6 cm, r₂ = 8 cm, r₃ = 10 cm.

Formula Used: Volume of sphere = (4/3)πr³; Surface area = 4πr².

⇒ Total Volume = (4/3)π(6³ + 8³ + 10³) = (4/3)π(216 + 512 + 1000) = (4/3)π(1728) cm³.

⇒ Let R be the radius of the new sphere: (4/3)πR³ = (4/3)π(1728) ⇒ R³ = 1728 ⇒ R = 12 cm.

⇒ Sum of initial surface areas (S₁) = 4π(6² + 8² + 10²) = 4π(36 + 64 + 100) = 800π cm².

⇒ Surface area of new sphere (S₂) = 4π(12²) = 4π(144) = 576π cm².

⇒ Diminution in area = S₁ − S₂ = 800π − 576π = 224π cm².

⇒ Percentage diminution = (224π ÷ 800π) × 100 = 28%.

∴ The correct answer is 28%.

Additional Information

Volume Conservation

When objects are melted and recast, the total volume remains constant, while the surface area usually decreases when multiple shapes are combined into one.

Sphere Properties

For a sphere of radius r, Volume V = (4/3)πr³ and Surface Area A = 4πr². The ratio V/A = r/3.

Pythagorean Triplets in Cubes

A common mathematical property used in exams is 6³ + 8³ + 10³ = 12³, which simplifies calculations for melting spheres.

Given:

A solid sphere of radius 3 cm is melted to form a hollow cylinder of height 4cm & external diameter 10cm.

Formula used:

Volume of sphere = 4/3π r³

Volume of hollow cylinder = π(R² - r²)h

Thickness of cylinder = External radius - Internal radius

Calculation:

According to the question,

External diameter = 10cm

External radius = 10/2= 5cm

Volume of sphere = Volume of hollow cylinder

⇒ 4/3π (3³) = π(5²-r²)4

⇒ 9 = 25 - r²

⇒ r² = 16

⇒ r = 4

Thickness of cylinder = 5 - 4 = 1 cm

∴ The thickness of the cylinder is 1 cm.

Given:

Three glasses P, Q, and R have capacities in the ratio 1 ∶ 2 ∶ 3

The ratio of milk to water in P, Q & R is 1 ∶ 2, 2 ∶ 3, 3 ∶ 1 respectively

Calculation:

Let the quantity of mixture in P, Q & R be 1L, 2L & 3L respectively

In glass P

⇒ Quantity of Milk = 1 × (1/3) = (1/3) L

⇒ Quantity of Water = 1 × (2/3) = (2/3) L

In glass Q

⇒ Quantity of Milk = 2 × (2/5) = (4/5) L

⇒ Quantity of Water = 2 × (3/5) = (6/5) L

In glass R

⇒ Quantity of Milk = 3 × (3/4) = (9/4) L

⇒ Quantity of Water = 3 × (1/4) = (3/4) L

In Bigger Container

⇒ Quantity of Milk = \(\frac{1}{3}\) + \(\frac{4}{5}\) + \(\frac{9}{4}\) = \(\frac{20+48+135}{60}\) = \(\frac{203}{60}\)

⇒ Quantity of Water = \(\frac{2}{3}\) + \(\frac{6}{5}\) + \(\frac{3}{4}\) = \(\frac{40+72+45}{60}\) = \(\frac{157}{60}\)

∴ The ratio of milk to water in the container is 203 : 157.

Given:

 x4 + x2y2 + y4, x3 + y3, x3 - y3

Concept:

LCM: It is the product of the greatest power of each prime factor,

involved in the numbers.

Formula Used:

(a⁴ + a²b² + b⁴) = (a² + ab + b²)(a² - ab + b²)

(a2 - b2) = (a + b)(a - b)

(a³ - b³) = (a - b)(a² + ab + b²)

(a3 + b3) = (a + b)(a2 - ab + b2)

Calculation:

x⁴  + x²y² + y⁴ = (x2 + y2 - xy)(x2 + y2 + xy)

x3 + y3 = (x + y)(x² - xy + y²) 

x3 - y3 = (x - y)(x² + xy + y²)

LCM = (x + y)(x - y)(x² - xy + y²)(x²+ xy + y²)

LCM = (x² - y²)(x⁴ + x²y² + y⁴)

Again, we know that 

a³ - b³ = (a - b)(a² + ab + b²)

LCM = (x⁶ - y⁶)

∴ The correct answer is (x⁶ - y⁶).

Given:

 \( \frac{x^2-y^2-z^2-2 y z}{x^2+y^2-z^2+2 x y}\) + \(\frac{x^2-y^2-z^2-2 y z}{x^2-y^2+z^2-2 x z}\) 

Formula Used:

1. (a + b)² = a² + b² + 2ab
2. (a - b)² = a² + b² - 2ab
3. (a² - b²) = (a - b)(a + b)

Calculation:

 \( \frac{x^2-y^2-z^2-2 y z}{x^2+y^2-z^2+2 x y}\) + \(\frac{x^2-y^2-z^2-2 y z}{x^2-y^2+z^2-2 x z}\) 

⇒ \( \frac{x^2-(y^2+z^2+2 y z)}{(x^2+y^2+2 x y)-z^2}\) + \(\frac{x^2-(y^2+z^2+2 y z)}{(x^2+z^2-2 x z)-y^2}\)

By using the identity (1) & (2)

⇒  \( \frac{x^2-(y+z)^2}{(x+y)^2-(z)^2}\) +  \(\frac{x^2-(y+z)^2}{(x-z)^2- (y)^2}\) 

By using the identity 3) 

⇒\(\frac{(x + y + z)(x-y-z)}{(x+y-z)(x+y+z)}\) + \(\frac{(x-y-z)(x+y+z)}{(x-z+y)(x-y-z)}\)

⇒ \(\frac{(x-y-z)}{(x+y-z)}\) + \(\frac{(x+y+z)}{(x-z+y)}\)

⇒ \(\frac{x-y-z+x+y+z}{x+y-z}\)= \(\frac{2x}{x+y-z}\)

∴ The correct answer is \(\frac{2x}{x+y-z}\).

Given:

If tan A + cot A = 2

Concept used: 

tan 45^° = 1

Calculation:

Putting A = 45^° 

tan 45^°  + cot 45^°  = 2 

So, tan 45^° = 1

Thus, given condition is satisfied. 

According to the question,

tan²A + tan³A + tan⁴A + .....+ tanⁿA 

⇒ 1 + 1 + 1 + .... (n - 1)times

∴  tan2 A + tan3 A + tan4 A + .... + tann A =  n-1

Formula used:

a³ + b³ = (a + b)(a² + b² - ab)

a3 - b3 = (a - b)(a2 + b2 + ab)

Calculation:

Statement: 1 \(\frac{\sin ^3 \theta+\cos ^3 \theta}{\sin \theta+\cos \theta}+\sin \theta \cos \theta\) = 1

Put θ = 45° [0< θ < π/2]

Sin 45° = 1/√2, Cos 45° = 1/√2

\(\frac{1}{2√2}\ +\ \frac{1}{2√2}\over\frac{1}{√2}\ +\ \frac{1}{√2}\) +\(\frac{1}{√2} \times \frac{1}{√2}\) = \(\frac{2}{2√2}\over\frac{2}{√2}\) + 1/2 = 1/2 + 1/2 = 1

Hence, Statement  1 is correct

Statement:

2 1 - Sin6 θ = Cos2 θ(Cos4 θ + 3 Sin2 θ)

LHS: 1³ - (Sin² θ)³

⇒ (1 -  Sin2 θ)[1² + (Sin² θ)² + Sin2 θ]

⇒ (1 -  Sin2 θ)[12 + Sin4 θ+ Sin2 θ]

RH:  (1 - Sin2 θ)[(1 - Sin2 θ)² + 3 Sin2 θ]

⇒ (1 - Sin2 θ)[1² - 2Sin2 θ + (Sin² θ)² + 3 Sin2 θ]

⇒ (1 - Sin2 θ)[12 - 2Sin2 θ + Sin4 θ + 3 Sin2 θ]

⇒ (1 - Sin2 θ)[12 + Sin4 θ + Sin2 θ]

Hence statement 2 is correct

∴ Both the statement is correct

Shortcut Trick

Putting θ = 30°,45°, 60° values one by one in given equation

θ = 30°

7 sin4θ + 9 cos4θ + 42 sin2θ

= 7 (1/2)4  + 9 (√3/2)4  + 42 (1/2)2 

7/16 + 81/16 + 42/4 = 256/16 = 16 (satisfied)

So, tanθ = tan30° = 1/√3

Alternate Method 

Given:

7 sin⁴ θ + 9 cos⁴ θ + 42 sin² θ = 16 

Formula used:

1) cos² θ = 1 - sin² θ

2) \(x = {-b ± √{b^2-4ac} \over 2a}\)

Calculation:

According to the question,

7 sin⁴ θ + 9 cos⁴ θ + 42 sin² θ = 16

⇒ 7 sin⁴ θ + 9(cos² θ)² + 42 sin² θ = 16

⇒ 7 sin⁴ θ + 9(1 - sin² θ)² + 42 sin² θ = 16

Here, using the identity (a - b)².

⇒ 7 sin⁴ θ + 9(1 - 2 sin² θ + sin⁴ θ) + 42 sin² θ = 16

⇒ 7 sin⁴ θ + 9 - 18 sin² θ + 9 sin⁴ θ + 42 sin² θ = 16

⇒ 16 sin⁴ θ + 24 sin² θ - 7 = 0

Putting  sin² θ = x

⇒ 16x² + 24x - 7 = 0

Using Quadratic formula 

\(x = {-b ± √{b^2-4ac} \over 2a}\)

Where, a = 16, b = 24, c = -7

⇒ \(x = {-24 ± √{24^2-[4\times 16 \times (-7)]} \over 2 \times 16 }\) 

⇒  \(x = {-24 ± √{576 + 448} \over 32}\)

⇒ - 24 ± √1024

⇒ \(-24 \pm 32 \over 32\) 

⇒ \(-24 - 32 \over 32\) or \(-24 + 32 \over 32 \) 

⇒ \(-56 \over 32\) or  \(8 \over 32\)

⇒ \(-7 \over4\) or \(1\over4\)

Solving above equation

⇒ x = \(-7 \over4\) or  x =  \(1\over4\)

Since  0< θ < \(\pi \over2\), θ lies in 1^st quadrant

Thus, sin θ is positive

⇒ sin² θ = \(1\over4\) 

⇒ sin θ \(√{\frac{1}{4}}\) = \(1\over2\) 

⇒ θ = 30^° 

⇒ tan θ = tan 30^° = \({\frac{1}{√ 3}} \) 

∴ tan θ equal to \({\frac{1}{√ 3}} \)

Given:

An isosceles triangle has its base length 2a & its height is h.

Formula used:

Area of Triangle = 1/2 × base × height

Area of Square = side²

Calculation:

According to the figure,

Area of isosceles triangle = 1/2 × b × h 

⇒ 1/2 × 2a × h  ( base = 2a)     .....(1)

Again, according to the figure,

By Pythagoras theorem

⇒ AC = √DC² + AD² 

⇒ AC = √a² + h²

Then, Area of square drawn on AB & AC 

Side² = (√a² + h²)² = (a² + h²)

Now, both side will be equal then, 

Area = 2(a² + h²)     .....(2)

Again, according to the figure,

Area of square drawn on BC which is (2a)²

Area of square = (2a)² = (4a²)    .....(3)

By adding above equation (1), (2) & (3) we get,

Area of the figure,

(1/2 × 2a × h) + 2(a² + h²) + (4a²)

⇒ ah + 2a² + 2h² + 4a²

⇒(2a² +4a²) + 2h² + ah

⇒ 6a² + 2h² + ah

∴ The area of the figure is  6a2 + 2h2 + ah

Given:

p = \(\frac{a^2}{(b-a)(c-a)}\), q = \(\frac{b^2}{(c-b)(a-b)}\), r = \(\frac{c^2}{(a-c)(b-c)}\)

Calculation:

Let a = 1, b = 2, c = 3

According to the question,

⇒ p = \(\frac{1^2}{(2-1)(3-1)}\) = \(\frac{1}{2}\)

⇒ q = \(\frac{2^2}{(3-2)(1-2)}\) = -4

⇒ r = \(\frac{3^2}{(1-3)(2-3)}\) = \(\frac{9}{2}\)

Now, (p + q + r)² = (\(\frac{1}{2}\) - 4 + \(\frac{9}{2}\))² = (1)² = 1

∴ The value of (p + q + r)2 is 1.

Formula Used:

1. (a - b)² = a² + b² - 2ab

2. a² - b² = (a + b)(a - b)

Calculation:

a2 - b2 - c2 + 2bc + a + b - c

⇒ a2 - (b2 + c2 - 2bc) + a + b - c

⇒ a2 - (b - c)² + a + b - c

⇒ (a + b - c) (a - b + c) + (a + b - c)

⇒ (a + b - c) [(a - b + c) + 1]

∴ Factors of the given equation are (a + b - c) and (a - b + c + 1).

Given:

Roots of the equation \(\rm \frac{1}{x+a+b}=\frac{1}{x}+\frac{1}{a}+\frac{1}{b}\) are α and β

Where, a ≠ 0, b ≠ 0, x ​≠ 0

Concept Used:

If the roots of a quadratic equation are p and q

Quadratic equation is →  x² - (p + q)x + pq = 0

Calculation:

According to the question,

\(\frac{1}{x+a+b}\) = \(\frac{1}{x}+\frac{1}{a}+\frac{1}{b}\)

⇒ \(\frac{1}{x+a+b}\) - \(\frac{1}{x}\) = \(\frac{1}{a}+\frac{1}{b}\)

⇒ \(\frac{x-(x+a+b)}{x(x+a+b)}\) = \(\frac{a+b}{ab}\)

⇒ \(\frac{-(a+b)}{x(x+a+b)}\) = \(\frac{a+b}{ab}\)

⇒ x(x + a + b) = -ab

⇒ x² + ax + bx + ab = 0

⇒ x(x + a) + b(x + a)  = 0

⇒ (x + a)(x + b) = 0 

⇒ x = -a and x = -b

Thus, roots of the equation are α = -a and β = -b

⇒ α² = a² and β² = b²

Quadratic equation whose roots are α² and β²

⇒ x² - (a² + b²)x + a²b² = 0

∴ The required quadratic equation is x2 - (a2 + b2)x + a2b2 = 0.

Shortcut Trick

\(\rm x=\frac{6}{7-\frac{6}{7-\frac{6}{7-\frac{6}{7-x}}}}\)

Clearly we can see, at x = 6,

LHS = RHS

Therefore,

x2 - 3x + 2 = 6² - 3 × 6 + 2

x2 - 3x + 2 = 36 - 18 + 2

∴ x2 - 3x + 2 = 20

Alternate MethodGiven:

\( x = \frac{6}{7 - \frac{6}{7 - \frac{6}{7 - \frac{6}{7 - x}}}}, x > 1 \)

Calculate \( x^2 - 3x + 2 \)

Concept Used:

Recurrence and simplification of nested fractions followed by algebraic simplification.

Calculation:

Let's solve the recurrence:

⇒ Let \( y = 7 - x \).

⇒ Then \( x = \frac{6}{7 - \frac{6}{7 - \frac{6}{7 - \frac{6}{7 - x}}}} \)

⇒ Substitute \( y \) back in, \( x = \frac{6}{y} \) where \( y = 7 - x \)

⇒ \( x = \frac{6}{7 - x} \)

⇒ \( x(7 - x) = 6 \)

⇒ \( 7x - x^2 = 6 \)

⇒ \( x^2 - 7x + 6 = 0 \)

⇒ \( (x - 6)(x - 1) = 0 \)

⇒ \( x = 6 \) or \( x = 1 \) (Since \( x > 1 \), \( x = 6 \))

Now, calculate \( x^2 - 3x + 2 \):

⇒ Substitute \( x = 6 \) into \( x^2 - 3x + 2 \),

⇒ \( 6^2 - 3 \times 6 + 2 \)

⇒ \( 36 - 18 + 2 \)

⇒ \( 20 \)

∴ \( x^2 - 3x + 2 = 20 \).

Formula used:

Distance = Speed × Time

Concept used:

1. While a train crosses a person, it covers a distance equal to its length.

2. When two objects move in the same direction,

Relative speed = Speed₁ - Speed₂

3. Km/h = 5/18 m/sec

Calculation:

Let the speed of the train be 'x' km/h and the length of the train be 'L'

According to the question, overtakes 2 persons, walking in the same direction with speeds of 3 km/h & 4 km/h in 9 sec & 75/8 sec

Here, the distance covered is the length of the train

⇒ (x - 3) × 9/3600 = (x - 4) × 75/3600

⇒ (x - 3) × 9 = (x - 4)× 75/8

⇒ 24x - 72 = 25x - 100

⇒ x = 28 km/h

Length of train = [(x - 3) × 9/3600] km

⇒ Length of train = (28 - 3) × 1/400 km

⇒ Length of train = 25/400 = 1/16 km = 1/16 × 1000 = 62.5 m

∴ The length of the train is 62.5 m.

Concept Used:

SP = CP + Profit

Profit% = (Profit/CP) × 100

CP = Cost Price, SP = Selling Price

Calculation:

A person bought an article and sold it at a profit of 20%:

Let's consider CP₁ is 100.

⇒ SP = 100 + 100 × (20/100) = 120

Had he bought it at 20% less:

⇒ CP₂ = 100 - 100 × (20/100) = 80

Since he still sells it at the same selling price of 120,

⇒ Profit = 120 - 80 = 40

⇒ Profit% = (40/80) × 100 = 50%

∴ The Profit Percentage is 50%.

Shortcut Trick ⇒ SP / CP = (100 + 20) / (100 - 20)

⇒  SP / CP = 120 / 80

⇒ Profit% = [(120 - 80) × 100] / 80 

⇒ 400 / 80 = 50% 

∴ The profit percentage is 50%.

Shortcut Trick

Given that, 

2s = a + b + c

's' is not present in any option.

Therefore, Put s = 0.

⇒ 02 + (0 - a)(0 - b) + (0 - b)(0 - c) + (0 - c)(0 - a)

⇒ (-a)(-b) + (-b)(-c) + (-c)(-a)

⇒ ab + bc + ca

∴ The correct option is 2 which is ab + bc + ca.