Quantitative Aptitude PYQs

Given:

In quadrilateral ABCD, 

∠B = 90°

and AB2 + BC2 + CD2 - AD2 = 0.

Concept Used:

In ΔABC, ∠B = 90° Using Pythgoras Theorem 

AB2 + BC2 = AC2

Calculation:

Observe the diagram given below carefully,

According to the question,

AB2 + BC2 + CD2 - AD2 = 0

⇒ (AB2 + BC2) + CD2 - AD2 = 0

⇒ (AC²) + CD2 - AD2 = 0      ----(∵ From Pythgoras Theorem, AB² + BC² = AC²)

⇒ AC2 + CD2 = AD2      ----(i)

Here, Equation (i) is the Equation form of Pythagoras theorem applied in Δ ACD.

For Equation (i) to hold true, the side AD must be the Hypotenuse of the Δ ACD.

So, the angle opposite to the Hypotenuse i.e. AD must be 90°.

∴ ∠ ACD = 90°.

Given:

In a Δ ABC,

AC = 12 cm,

AB = 16 cm,

AD is biesctor of ∠A. 

& BD = 4 cm.

Concept Used:

Angle Bisector Theorem: An angle bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.

Calculation:

Observe the diagram below carefully,

From the Angle Bisector Theorem,

We know that,

AB/AC = BD/DC

⇒ 16/12 = 4/DC

⇒ DC = 4 × (12/16)

⇒ DC = 4 × (3/4)

⇒ DC = 3 cm

∴ The length of DC = 3 cm.

 Shortcut Trick

When an angle bisector of an angle of a triangle divides the opposite side in two segments in  a and b  then other two sides of the triangle will be in ratio a:b

Given:

ABCD is a cyclic quadrilateral. 

& the bisectors of the angles A, B, C and D cut the circle at P, Q, R and S respectively,

Calculation:

Observe the diagram given below carefully,

According to the question,

Bisectors of A, B, C and D cut the circle at P, Q, R and S respectively. 

PQRS is also a cyclic quadrilateral, Sum of opposite angles of the cyclic quadrilateral is 180° 

∠PQR is opposite to ∠RSP 

∴ ∠PQR + ∠RSP = 180°.

Given: 

In equilateral Δ ABC, BC is trisected at D such that BC = 3 × BD. 

Calculation:

Observe the diagram given below carefully,

According to the question,

BC = 3 × BD

⇒ BD = BC/3

Also, We know that,

BE = BC/2      ----(∵ AE is the median of the Δ ABC which bisects BC)

Now, 

In Δ AEB, 

DE = BE - BD

⇒ (BC/2) - (BC/3)

⇒ (3 × BC - 2 × BC)/6

⇒ BC/6

Now, Applying Pythagoras Theorem in Δ ADE,

(AD)² = (AE)² + (DE)²      ----(i)

Again, Applying Pythagoras Theorem in Δ ABE,

(AB)² = (AE)² + (BE)²      ----(ii)

Subtracting equation (i) from equation (ii),

(AB)² - (AD)² = {(AE)² - (AE)²}  + {(BE)²  - (DE)²}

⇒ (AB)² - (AD)² = 0  + {(BC/2)²  - (BC/6)²}

⇒ (AB)² - (AD)² = {(BC)²/4}  - {(BC)²/36}

⇒ (AB)² - (AD)² = (8/36) × (BC)²

⇒ (AB)² - (AD)² = (2/9) × (AB)²      ----(∵ AB = BC = CA, as Δ ABC is an equilateral triangle)

⇒ (AD)² = (AB)² - {(2/9) × (AB)²}

⇒ (AD)² = (7/9) × (AB)²

⇒ (AD)²/(AB)² = 7/9

⇒ (AD)² ∶ (AB)² = 7 ∶ 9

∴ The ratio of (AD)2 to (AB)2 is 7 ∶ 9.

Calculations :

Statement 1

 

There is a trapezium WXYZ in which ZX and WY are two diagonals and PQ || WX || ZY 

In ∆ WZY, PQ || ZY so

WP/PZ = WQ/QY      ....(1) 

In ∆ WZX, PQ || WX, so

ZP/PW = ZQ/QX or PW/ZP = QX/ZQ     ....(2)

From equation (1) and (2)

WQ/QY = QX/ZQ

So proved that diagonals divide each other proportionally. 

Statement 2

 

In trapezium WXYZ there is a diagonal ZX and PR || WX || ZY and PQ || ZY || WX

In ∆ ZWX PR || WX 

ZP/PW = RZ/XR      ....(1) 

In ∆ XYZ, RQ || ZY 

XQ/QY= XR/RZ or QY/XQ = RZ/XR      ....(2) 

From equation (1) and (2) 

ZP/PW = YQ/QX 

So proved that any line drawn parallel to the parallel lines divides non-parallel line proportionally.

∴ Option 3 is the correct choice.

Given:

H is the height of the cone

C is the curved surface area of the cone

V is the volume of the cone

Formula used:

The volume of cone = (πr²H/3)

The curved surface area of cone = πrl

Calculation:

Now we have to find the value of 3πVH3 + 9V2 

⇒ 3V(πH³ + 3V)

Now, we substitute the value of V

⇒ 3 × (1/3) × πr² × H(πH³ +  3 × (1/3) × πr²× H)

⇒ πr2H(πH3 + πr2H)

⇒ π²r2H²(H² + r2)

⇒ π²r²l²H²

⇒ C²H²

∴ The value of 3πVH3 + 9V2 is C2H2

Given:

The diameter of small lead ball = 2mm

The radius of big lead ball = 8 cm

Formula used:

The volume of sphere = (4πr³)/3

Required number of the balls = Volume of big lead ball/ Volume of small lead ball

Calculation:

The radius of a small lead ball = 1mm 

We know, 1cm = 10 mm

⇒ Radius of small lead ball = 0.1 cm

The volume of small lead ball = (4π ×  0.13)/3

The volume of big lead ball = (4π ×  83)/3

Required number of the balls = (4π ×  83)/3/(4π ×  0.13)/3

⇒ 8³/0.1³ = 512000

∴ The number of small lead ball made from the big lead ball is 512000.

Shortcut TrickLength of 3^rd side = 90 - (40 + 41) = 9 cm

We know that (9, 40, 41) is a Pythagorean triple,

hence, the triangle is a right-angle triangle with hypotenuse = 41 cm

Area of a right angle triangle = (Base × Height)/2

⇒ Area of Δ = (40 × 9)/2 = 180 cm²

∴ The correct answer is 180 cm².

Given:

The sides of a triangle are 40 cm and 41 cm

The perimeter of triangle = 90 cm

Formula used:

Area of triangle = √\(√{s(s - a)(s - b)(s - c)}\)

Where a, d and c sides of triangle 

s = ( a+ b +c)/2

Calculation:

Perimeter of triangle is 90 

⇒ 90 = a + b + c

⇒ 90 = 81 + c

⇒ 9 = c

Semi perimeter of triangle (s) = 90/2 = 45

Area of the triangle = \(√{45(45 - 40)(45 - 41)(45 - 9)}\)

⇒ √32400

⇒ 180 cm²

∴ Area of the triangle is 180 cm²

Given:

The difference between diagonal = 2

Perimeter is 6 more to the sum of diagonal

Formula used:

Area of rhombus = (d₁ × d₂)/2

Side of rhombus = √{(d₁/2)² + (d₂/2)²}

The perimeter of rhombus = 4 × Side of the rhombus

Calculation:

Let the diagonal of rhombus d₁ be 2x

According to the question 

The diagonal d₂ be 2x + 2

As we know all sides of rhombus are equal and diagonal of rhombus bisect each other at 90° 

⇒ Side of rhombus = √{x² + (x + 1)²}

Now again according to the question

Perimeter of rhombus = sum of diagonal + 6

⇒ 4√{x2 + (x + 1)2} = 4x + 8 

⇒ √{x2 + (x + 1)2} = x + 2

On squaring both side, we get

x2 + (x + 1)² = (x +2)²

⇒ x² - 3 - 2x = 0 

⇒ (x + 1)(x - 3) = 0

⇒ x = 3

So, the length of d₁ = 6 and d₂ = 8 

⇒ Area of the rhombus = (6 × 8)/2 = 24 cm²

∴ The are of rhombus is 24 cm²

Given:

Radius of circumcircle = 5 cm

Altitude to the hypotenuse = 4 cm 

Concept used:

The hypotenuse is the diameter of a circle if an angle of a triangle is 90° which is inscribed in a circle, that triangle formed by the diameter of the circle.

Formula used:

Area of triangle = (B × H)/2

Where

B is the base of the triangle 

H is the height of the triangle 

Calculation:

The hypotenuse of tringle is BC =  2 × of the radius of circles(OC)

⇒ BC = 2 × 5 = 10 cm

The altitude drawn to hypoteneus is AD = 4 cm

The area of ΔABC = (10 × 4)/2 = 20 cm²

∴ The area of the right-angle triangle is 20 cm²

Concept:

Sum of interior angle of triangle = 180° 

Calculation:

1)1 : 2 : 3

So, x + 2x + 3x = 6x

Then 6x = 180°, x = 30° 

Angles 30°, 60° and 90° 

2. 3 : 4 : 5

So, 3x + 4x + 5x = 180° 

Then 12x = 180°, x = 15

Angles 45°, 60° and 75° 

3. 5 : 6 : 7

So, 5x + 6x + 7x = 180° 

Then 18x = 180°, x = 10° 

Angles 50°, 60° and 70° 

4. 6 : 7 : 8

So, 6x + 7x + 8x = 180° 

Then 21x = 180°, x = 8.57° 

Angles 51.42°, 59.99° and 68.57° 

∴ Option 4 is not integar angle in triangle.

Shortcut Trick 

 

⇒ Area is decreased by = (1/100) × 100

⇒ Area is decreased by = 1%

∴ The correct answer is Option 3).

Given:

Increment in length = 10%

decrement in breadth = 10%

formula used:

Percentage = (Final value - Initial value)/Initial value

Area of rectangle = ab

where a and b is length and breadth respectively

Calculation:

Let length and breadth be a and b respectively

⇒ The area before change = ab 

⇒ The area after change = (a + 10a/100)(b - 10b/100)

⇒ The area after change = (11a/10)(9b/10)

⇒ The area after change = 99ab/100

⇒ Percentage change = [{(99ab/100) - ab}/ab] × 100%

⇒ Percentage change = (-1/100) × 100%

⇒ Percentage change = -1%

Here (-) shows decrement in area 

∴ The area of the new rectangle is decreased by 1%.

Alternate Method

Given:

Increment in length = 10%

decrement in breadth = 10%

Formula used: 

Using AB rule

Percentage change = (A + B + AB/100)%

Where A and B are successive percentage changes

Calculation:

A = 10% and B = -10%

put values in the formula

Percentage change = [10 - 10 + 10(-10)/100]%

Percentage change = -1%

∴ The area of the new rectangle is decreased by 1%

Given:

The ratio of the surface areas of two spheres = 1 ∶ 4

Formula used:

The surface area of the sphere = 4πr²

The volume of the sphere = (4/3)πr³

where r is the radius of the sphere

Calculation:

let r₁ and r₂ be the radius of sphere

(surface area of the sphere)₁/(surface area of the sphere)₂ = 1/4

⇒ 4πr₁²/4πr₂2 = 1/4

⇒ r₁/r₂ = 1/2

⇒ (The volume of the sphere)₁/(The volume of the sphere)₂ = (4/3)πr₁3/(4/3)πr₂3

⇒ (The volume of the sphere)1/(The volume of the sphere)2 = (r₁/r₂)³

Put value r1/r2 = 1/2 in above equation

⇒ (The volume of the sphere)1/(The volume of the sphere)2 = (1/2)3

⇒ (The volume of the sphere)1/(The volume of the sphere)2 = 1 : 8

∴ The ratio of the volume of the sphere is 1 : 8

Given:

Dimensions of brick

length = 20 cm = 0.20 m

breadth = 15 cm = 0.15 m

height = 10 cm = 0.10 m

Dimentions of wall 

length = 45 m 

breadth = 0.15 m

height = 3 m 

Formula used: 

Volume of cuboid = length × breadth × height

Number of brick = Volume of the wall/Volume of brick 

Calculation:

According to the formula,

⇒ The volume of the wall = 45 × 0.15 × 3 m³

⇒ Volume of brick = 0.20 × 0.15 × 0.10 m³

⇒ Number of brick = (45 × 0.15 × 3)/( 0.20 × 0.15 × 0.10)

⇒ Number of brick = 6750

∴ The number of bricks required to construct a wall is 6750.

Given:

The sum of all interior angles of a regular polygon is twice the sum of all its exterior angles

Concept: 

Sum of all interior angles of a regular polygon = (n−2) × 180° 

where n= number of sides of polygon

Sum of all exterior angle of a regular polygon = 360° 

Calculation:

According to question 

= 180° (n − 2) = 2 × 360° 

= n − 2 = 4

= n = 6

∴ Number of sides of the polygon 6(Hexagon)

Given:

The number of revolutions of wheel = 5000

Covered distance = 11 km =11000 m

Formula used: 

The perimeter of wheel = 2πr

where r is the radius of the wheel

(2πr) × N = D

2πr = Perimeter of wheel 

N = Number of revolutions

D = Covered distance

Calculation:

Let r be a radius of the wheel 

According to the formula,

⇒ 2πr × 5000 = 11000

⇒ 2r × (22/7) × 5000 = 11000 

⇒ r = (11 × 7)/(22 × 2 × 5) m

Multiply by 100 to convert meter to centimeter

⇒ r = (7/20) × 100 cm

⇒ r = 35 cm

∴ The radius of the wheel is 35 cm.

Given:

The ratio of the volume of cones = 1 : 4

The ratio of the diameter of cones = 4 : 5

Formula used:

The volume of cone = (1/3)πr²h

Calculation:

Let the radius of cones be r₁ and r_2.

Let the height of cones be h1 and h2.

According to the question,

The ratio of the volume of cones

⇒ (1/3)πr₁2h₁/ (1/3)πr₂2h₂ = 1/4

⇒ r₁²h₁/r22h2 = 1/4      ----(1)

The ratio of the diameter of cones 

⇒ d₁/d₂ = r₁/r₂ = 4/5      ----(2)

where d1, d2 are diameters of cones

Put value in equation 1 from equation 2

⇒ (4/5)²(h₁/h₂) = 1/4

⇒ h1/h2 = (1/4) × (25/16)

⇒ h1/h2 = 25/64

∴ The ratio of heights of cones is 25 : 64. 

Shortcut Trick ⇒ r12h1/r22h2 = 1/4 

​Put r₁ = 4 and r₂ = 5 

​⇒ 4²×h₁/5²×h2 = 1/4 

​⇒ h1/h2 = (1/4) × (25/16)

⇒ h1/h2 = 25/64

∴ The ratio of heights of cones is 25 : 64.

Shortcut Trick Let us find the ratio of ∠A, ∠B, and ∠C

LCM of 2, 3, and 6 = 6

Dividing 2 ∠A = 3 ∠B = 6 ∠C by 6 we get,

⇒ ∠A/3 = ∠B/2 = ∠C/1

⇒ ∠A : ∠B : ∠C = 3 : 2 : 1

⇒ ∠A = 180^∘ × (3/6) = 90^∘ 

⇒ ∠C = 180^∘ × (1/6) = 30^∘ 

⇒ ∠A + ∠C = 90^∘ + 30^∘ = 120^∘ 

∴ The value of ∠A  + ∠C is 120°.

Given:

If ∠A, ∠B, and ∠C are the angles of a triangle then

2∠A = 3∠B = 6∠C.

Concept Used:

The sum of the angles of a triangle is 180°

Calculation:

According to the question,

2∠A = 3∠B = 6∠C.

⇒ ∠A = 3∠C and ∠B = 2∠C

Now, according to the concept used,

∠A + ∠B + ∠C = 180°

⇒ 3∠C + 2∠C + ∠C = 180°

⇒ 6∠C = 180°

⇒ ∠C = 30°

⇒ ∠A = 3∠C = 3 × 30° = 90°

⇒ ∠B = 2∠C = 2 × 30° = 60°

⇒ ∠A  + ∠C = 90° + 30° = 120°

∴ The value of ∠A  + ∠C is 120°.

Given: 

The perimeter of the circle is equal to that of the square.

Formula:

If “r” is the radius of the circle and a” is the side of the square,

1) Perimeter of a circle = 2πr

2. Area of a circle = πr²

3) Perimeter of a square = 4a

4. Area of a square = a²

Calculation:

Let “r” be the radius of the circle and “a” be the side of the square,

According to the question,

The perimeter of the circle is equal to that of the square.

2πr = 4a

⇒ r/a = 4/(2π) = 2/π      ----(1)

Now, the ratio of the area of circle and Area of square:

⇒ πr²/a²      ----(2)

From equation (1) and (2),

πr²/a² = π(2/π)²

⇒ πr²/a² = 4/π

∴ The ratio of the area of a circle to that of a square is 4 ∶ π.

Shortcut Trick

We know that the Pythagorean triple 6, 8, and 10 follows the conditions of the question,

The product of these integers = 6 × 8 × 10 = 480

∴ The correct answer is 480.

Additional InformationSome other important Pythagorean triples are:

 

Alternate MethodConcept:

In ΔABC, ∠B is 90° so we use the Pythagoras theorem 

(Hypotenuse)² = (base)² + (Perpendicular)²

(AC)² = (AB)² + (BC)²

Calculation:

Let sides of right angle triangle as p, p + 2 and p + 4

As in triangle biggest side is hypotenuse 

(p + 4)² = (p)² + (p + 2)²

= p² + 16 + 8p = p² + p² + 4p + 4

= p² - 4p - 12 = 0

= p² - 6p + 2p - 12 = 0

= (p - 6)(p + 2) = 0

p = - 2 and 6

• 2  is canceled because the side is always positive

p = 6 the n side of right-angle triangle 6, 8 and 10

Product of these integer = 6 × 8 × 10 = 480

∴ The product of these integers is 480.