Quantitative Aptitude PYQs

Given:

Formula:

Curved surface area of cone = πrl

Calculation:

Let length of cloth be a m.

Area of cloth = a × 3 = 3a m²

Diameter of tent = 12 m

Radius of tent = 12/2 = 6 m

Curved surface area of cloth = 22/7 × 6 × 7 = 132 m²

Then,

⇒ 3a = 132

⇒ a = 44 m

∴ Length of cloth is 44 m.

Given:

Diameter of sphere = 6cm

Radius of sphere = 3cm

Radius of  vessel = 6cm

Formula:

Volume of sphere = 4/3 × πr³

Volume of cylinder = πr²h

Calculation:

⇒ Volume of sphere = 4/3 × π × 3³ = 36π cm³

⇒ Volume of vessel = π × 6² × h = 36πh cm³

⇒ 36π = 36πh

⇒ h = 1cm

∴ 1cm water level will be raised.

Given:

Radius of circle = 21cm

Formula:

Let l = length of arc, A = area of arc and θ = sector angle and r = radius of circle

⇒ l = θ/360 × 2πr

⇒ A = θ/360 × πr²

Calculation:

⇒ 55 = θ/360 × 2 × 22/7 × 21

⇒ θ = 150º

⇒ Area of sector = 150/360 × 22/7 × 21²

∴ Area of sector is 577.5 cm²

Given:

Radius of wire = 70 cm

Formula:

⇒ Circumference of circle = 2πr

⇒ Perimeter of rhombus = 4 × side

Calculation:

⇒ Circumference of wire = 2 × 22/7 × 70 = 440 cm

⇒ Circumference of wire = perimeter of rhombus

∴ Side of rhombus = 440/4 = 110cm

Given:

The perimeter of semicircular park = 360 m

Formula:

Circumference of semicircular park = πr + 2r

Area of semicircular park = πr²/2

Calculation:

360 = r(22/7 + 2)

⇒ r = 70 m

⇒ Area of park = 22/7 × 1/2 × 70² = 7700 m²

∴ Area of park is 7700 m².

Given:

Concept:

The diagonals of a Trapezium divide each other proportionally.

Calculation:

⇒ AP : PC = BP : PD

⇒ 4/(4x - 4) = (2x - 1) /(2x + 4)

⇒ 8x + 16 = 8x² - 4x - 8x + 4

⇒ 8x² - 20x - 12 = 0

⇒ 4(2x² – 5x – 3) = 0 

⇒ 2x² – 6x + x – 3 = 0 

⇒ 2x(x – 3) + 1(x – 3) = 0 

⇒ (2x + 1) (x – 3) = 0 

⇒ x = -1/2, 3

∴ The value of x is 3.

Concept used:

Ratio of sides, perimeter, median, circumradius, inradius and square root of area of two similar triangle will be equal always.

Calculation:

ΔABC ~ ΔDEF

⇒ AB/DE = BC/EF = AC/DF = (AB + BC + AC)/(DE + EF + DF)

⇒ AB + BC + AC = 40

⇒ DE + EF + DF = 30

⇒ AB/DE = BC/EF = AC/DF = 40/30 = 4/3

Then,

Let BC and EF be 4a and 3a respectively.

Let AC and DF be 4b and 3b respectively.

∴ (BC + CA) : (EF + FD) = (4a + 4b) : (3a + 3b) = 4 : 3

Given:

The vertical angle of an isosceles triangle is equal.

The ratio of the area of both triangles = 4.84 ∶ 5.29

Concept:

Properties of similar triangles:

Angle – Angle similarity

Side – Angle – Side similarity

Side – Side – Side similarity

If two triangles are similar, the ratio of areas is equal to the

ratio of squares of corresponding sides.

So, \(\rm\frac{\Delta_1}{\Delta_2}=\frac{\left(S_1\right)^2}{\left(S_2\right)^2}\)

Calculation:

Let Δ ABC and Δ DEF are isosceles triangle.

⇒ AB/BC = 1

⇒ DE/EF = 1

⇒ AB/AC = DE/EF

⇒ AB/DE = AC/EF

⇒ ∠ A = ∠ C

ΔABC and ΔDEF are similar triangles.

⇒ A(ΔABC) : A(ΔDEF) = (height of ΔABC)² : (height of ΔDEF)²

⇒ 4.84 : 5.29 = (height of ΔABC)² : (height of ΔDEF)²

∴ Ratio of their corresponding heights = 22 : 23

Given:

Δ ABC is right-aa ngle triangle and  AD ⊥ BC

BD = 8 cm, DC = 12.5 cm

Concept:

In right ∆ ABC, ∠ A = 90˚ and AD is perpendicular to BC then

 AD2 = BD × CD

Calculation:

Using the above concept

AD² = BD × CD

⇒ AD² = 8 × 12.5

∴ AD = 10cm

Given:

Let r = radius of sphere and l = side of cube

Formula:

⇒ Surface area of cube = 6l²

⇒ Surface area of sphere = 4πr²

Calculation:

⇒ 6l² = 4πr²

⇒ l/r = √(4π/6)

Then,

⇒ x² : y² = [l³]² : [(4/3 × π × r³)]²

⇒ x² : y² = (3/4π)² × (l/r)⁶

⇒ x² : y² = (3/4π)² × (4π/6)^{(6 × 1/2 = 3)}

∴ x² : y² = π  : 6

Given:

⇒ Volume of cone = volume of hemisphere

⇒ Radius of cone = radius of hemisphere

Formula:

⇒ Volume of cone = 1/3 × πr²h

⇒ Volume of hemisphere = 2/3 × πr³

Calculation:

⇒ 1/3 × πr²h = 2/3 × πr³

∴ h : r = 2 : 1

Given:

Radius of sphere = 60/2 = 30mm = 3cm

Length of wire = 144 cm.

Formula:

Volume of sphere = 4/3 × πr³

Volume of cylinder = πr²L

Calculation:

The volume of solid sphere

= 4/3 × π × 3³ = 36π cm³

⇒ 36π = π × r² × 144

⇒ r = 0.5cm

⇒ Radius of wire = 0.5cm

∴ Diameter of wire = 2 × 0.5 = 1cm

Given:

r : h = 2 : 3

Volume of cylinder = 1617 cm³

Formula:

Let radius & height of the cylinder are r & h respectively. 

1. Volume of cylinder = πr²h

2. Curved surface area of cylinder = 2πrh

Shortcut Trick

Let the radius be 2x and height be 3x

Volume of cylinder = π(2x)².3x

⇒ 1617 = 22/7 × 4x² × 3x

⇒ x³ = 343/8

⇒ x = 7/2

So, r = 2x= 7 cm

h = 3x = 21/2 cm

Now, Curved surface area 

⇒ 2 × 22/7 × 7 × 21/2

⇒ 463 cm2

∴ The correct answer is option (3)

Alternate Method

Calculation:

We know that,  

Volume of cylinder = πr2h

⇒ 1617 = π × r² × h

⇒ 1617 = (22/7)  × r2 × h

⇒ 1617 × (7/22) =  r2 × h

⇒ r²h = 514.5     ----(1)

But according to question,

r/h = 2/3

⇒ h = 3r/2        ----(2)

Put this value in equation (1)

r2 × (3r/2)  = 514.5

⇒ r³ = 343

⇒ r = 7 cm.

Hence, 

h = 3 × 7/2 = 21/2

⇒ h = 10.5 cm.

Hence, curved surface area 

A = 2πrh

⇒ A = 2 × (22/7) × 7 × (21/2)

⇒ A = 463 cm².

∴ The curved surface area of cylinder 462 cm².

Given:

⇒ Height of cylindrical pipe = 14cm

⇒ Volume of metal used = 99 cm

Formula:

Let r = radius of cylinder and h = height of cylinder

⇒ Volume of cylinder = πr²h

⇒ Curved surface area of cylinder = 2πrh

Calculation:

⇒ 2πRh - 2πrh = 44

⇒ R - r = 1/2       ----(1)

Then,

⇒ πR²h - πr²h = 99

⇒ πh(R2 - r2) = 99

⇒ (22/7) × 14 ×(R2 - r2) = 99

⇒ (R² - r²) = 9/4

⇒ (R + r)(R - r) = 9/4

⇒ (R + r) × (1/2) = 9/4

⇒ (R + r) = 9/2 = 4.5

∴ (R + r) = 4.5

Given:

⇒ Edge of cube = 24cm

⇒ Edge of small cube = 12cm

⇒ Edge of small cube = 16cm

Let edge of other third cube be m cm.

Formula:

Let a = edge of cube

⇒ Surface area of cube = 6a²

⇒ Volume of cube = a³

Calculation:

⇒ Volume of large cube = 24³ = 13824 cm³

⇒ 13824 = 12³ + 16³ + m³

⇒ m = 20cm

∴ Surface area of third cube = 6 × 20² = 2400 cm²

Given:

Internal radius of vessel = 5cm

Height of vessel = 24cm

Formula:

Let Inner radius = r, Outer radius = R and h = height and l = slant = height

Volume of cone = 1/3 × πr²h

Volume of cylinder = πr²h

Calculation:

Volume of conical vessel = (1/3 × π × 5² × 24) cm³

Volume of cylindrical vessel = (π × 10² × height) cm³

Then,

(1/3 × π × 5² × 24) = (π × 10² × height)

Height of cylindrical vessel = 2 cm

∴ water rises by 2cm.

Given:

⇒ Side of cube = 22cm

⇒ Height of cone = 21cm

Formula:

⇒ Volume of cube = l³

⇒ Volume of cone = 1/3 × πr²h

Calculation:

⇒ Volume of cone formed = 22³ = 10648 cm³

⇒ 10648 = 1/3 × π × r² × 21

∴ Radius of cone = 22cm

Given:

Height of cone = 24

The curved surface area of cone = 550 cm²

Concept:

Pythagoras theorem:

It states that in a right-angled triangle, the square of the hypotenuse is

equal to the sum of the squares of the other two sides

The curved surface area of cone = πrl

Where,

r = radius of the cone, h = Height and  l = slant height

Calculation:

Let, r, l, and h are radius, height and slant height of cone respectively.

According to question,

Curved surface area of cone = 550 cm2

⇒ 550 = πrl      

⇒ rl × 3.14 = 550       (∵ π  = 3.14)

⇒ rl  ≈ 175 

⇒ r = 175/l   ----(1)

Using pythagoras theorem,

⇒ l² = h² + r²

⇒ l² - r² = 576     (∵ h = 24 cm)

⇒ l2 - (175/l)2 = 576      [From equation (1)]

⇒ l⁴ - 576 l² - 30625 = 0

⇒ l4 - 625 l² + 49 l2 - 625 × 49 = 0

⇒ l²(l² - 625) + 49(l² - 625) = 0

⇒ (l2 - 625)(l2 + 49) = 0

⇒ l² = 625 and l² = -49 (not possible)

⇒ l² = 625

⇒ l = 25 cm.

Hence,

r = 175/l

⇒ r = 7 cm

∴ Required ratio = r : l = 7 : 25

Shortcut Trick Using Pythagoras triplet 7, 24 and 25.

⇒ Height = 24 cm

​⇒ Slant height = 25 cm and radius = 7 cm  [using triplet]

⇒ Radius : Slant height = 7 : 25

∴ The correct answer is option (3).

Shortcut Trick 

2πR₁ = 44 (rolling along length)

⇒ R₁ = 7 cm

2πR₂ = 22 (rolling along width)

⇒ R₂ = 7/2 cm

⇒ x : y = 7² × 22 : (7/2)² × 44 = 2 : 1

∴ x : y = 2 : 1

Alternate Method 

Given:

Length of paper = 44cm

Width of paper = 22cm

Formula:

Volume of cylinder = πr²h

Calculation:

Case: 1

When x = volume of cylinder then

Height (h₁) = 22 cm and

Circumference = 44 = 2π(r₁)

r₁ = 7 cm

Case: 2

When y = volume of cylinder then

Height (h₂) = 44cm

Circumference = 22 = 2π(r₂)

r₂ = 7/2

We know that,

Volume of cylinder = πr2h

⇒ x = π × 7² × 22

⇒ y = π × (7/2)² × 44 =  π × 72 × 22

∴ x : y = 2 : 1

Shortcut Trick 

Mass of the shell = (4/3) × (22/7) × (6³ - 5³) × 3

⇒ Mass of the shell = (88/7) × 91 = 1144 g

∴ Mass of shell = 1144/3 × 3 = 1144 g

Alternate Method 

Given:

⇒ Density of shell = 3 g/cm³

⇒ Inner radius = 5cm

⇒ Outer radius = 6cm

Formula:

⇒ Volume of sphere = 4/3 × πr³

Calculation:

⇒ Volume of shell = 4/3 × 22/7 × (6³ - 5³) = 1144/3 cm³

∴ Mass of shell = 1144/3 × 3 = 1144 g