Quantitative Aptitude PYQs

Given:

two parallel line segments PQ = 5

RS = 3 cm

Concept Used:

If three lines are parallel as shown below,

then, \(\frac{1}{c} = \frac{1}{a} + \frac{1}{b}\)

Calculation:

Using the above formula,

\(\frac{1}{MN} = \frac{1}{PQ} + \frac{1}{RS}\)

\(\frac{1}{MN} = \frac{1}{5} + \frac{1}{3}\)

\(\frac{1}{MN} = \frac{8}{15}\)

MN  = 15/8 cm

∴ The correct answer is 15/8 cm.

Given:

two parallel line segments PQ = 5

RS = 3 cm

Concept Used:

If three lines are parallel as shown below,

then, \(\frac{1}{c} = \frac{1}{a} + \frac{1}{b}\)

Area of Trapezium = \(\frac{1}{2}\times (sum\ of\ parallel\ sides) \times Height\)

Calculation:

Using the above formula,

\(\frac{1}{MN} = \frac{1}{PQ} + \frac{1}{RS}\)

\(\frac{1}{MN} = \frac{1}{5} + \frac{1}{3}\)

\(\frac{1}{MN} = \frac{8}{15}\)​

MN  = 15/8 cm

ΔPQS and ΔMNS are similar,

So,

\(\frac{h2}{h1\ +\ h2} = \frac{\frac{15}{8}}{5} = 3/8\)

h2 = 3 and h1 = 5

Area of trapezium PQNM =  \(\frac{1}{2} \times [5 + \frac{15}{8}]\times 5\)

Area of trapezium RSNM = \(\frac{1}{2} \times [3 + \frac{15}{8}]\times 3\)

Ratio of area of both trapezium = \(\frac{\frac{1}{2} \times [5 + \frac{15}{8}]\times 5}{\frac{1}{2} \times [3 + \frac{15}{8}]\times 3}\) = 275/117

∴ The correct answer is 275/117.

Given:

If the people coming from a particular State belonging to S are 15% and 24,000 in number

Calculation:

According to the question,

Total People × 20% × 15% = 24000

⇒ Total People = 24000 × 10000/300

⇒ 800000

The total number of migrating people belonging to the age group B 15%

⇒ 800000 × 15/100

⇒ 1.2 lac

∴ The total number of migrating people belonging to the age group B is 1.2 Lac

Solution:

Total People = 800000

According to the question,

The maximum of differences between the number of people coming from different groups P, Q, R and S is 40% - 10%

⇒ 30%

⇒ 30% of 800000

⇒ 2.4 lac

∴ The maximum of differences between the number of people coming from different groups P, Q, R and S is 2.4 lac

Solution:

Total number of People = 800000

The number of people coming from R having age group A 

⇒ 800000 × 10/100 × 50/100

⇒ 40000

The number of people coming from Q having age group D

⇒ 800000 × 30/100 × 20/100

⇒ 48000

The difference between number of people coming from R having age group A and those coming from Q having age group D 

⇒ 48000 - 40000

⇒ 8000

∴ The difference between number of people coming from R having age group A and those coming from Q having age group D is 8000.

Calculation:

Let the DG = a = GB

In Δ AGB

⇒ a² = (2 - a)² + 1²

⇒ a² = 4 + a² – 4a + 1

⇒ 4a = 5 

⇒ a = 5/4 cm.

So, required area = Area of rectangle – 2 (area of Δ AGB)

⇒ (1 × 2) - 2 × [1/2 × 1 × (2 - 5/4)]

⇒ (2 - 3/4) = 5/4 cm².

Concept :

 Area of non-overlapping region = 2 [Area of rectangle] – 2 [area of overlapping region]

Calculation:

We have to find first Area of overlapping region

Let the DG = a = GB

In Δ AGB

⇒ a2 = (2 - a)2 + 12

⇒ a2 = 4 + a2 – 4a + 1

⇒ 4a = 5 

⇒ a = 5/4 cm.

So,

Area of overlapping region = Area of rectangle – 2 (area of Δ AGB)

⇒ (1 × 2) - 2 × [1/2 × 1 × (2 - 5/4)]

⇒ (2 - 3/4) = 5/4 cm2.

Now, we have  to find Area of non-overlapping region 

Area of non-overlapping region = 2 [Area of rectangle] – 2 [area of overlapping region]

⇒ 2 × 1 × 2 - 2 ×  5/4

⇒ 4 - 5/2 = 3/2 cm²

∴ Area of non-overlapping region is 3/2 cm 2

Given:

ABC is a right-angled triangle with ∠ABC = 90°.

The radius of circle having radius O is 2 cm.

MA ∶ MC = 2 ∶ 3.

Concept:

1. In a right-angled triangle, Hypotenuse² = Base² + Perpendicular²
2. Tangents drawn from an external point to a circle are equal.

Calculation:

In an ΔABC, using the Pythagoras theorem

⇒ (5x)² = (2x + 2)² + (3x + 2)²    

⇒ 25x² = 4x² + 4 + 8x + 9x² + 4 + 12x

⇒ 25x² = 13x² + 20x + 8

⇒ 12x² - 20x - 8 = 0

⇒ 3x² - 5x - 2 = 0

⇒ 3x² - 6x + x - 2 = 0

⇒ 3x(x - 2) + 1(x - 2) = 0

⇒ (3x + 1)(x - 2) = 0

x = 2 & x = -1/3 (not possible) 

Now, 

AB + BC = 2x + 2 + 3x + 2

⇒ AB + BC = 5x + 4

⇒ AB + BC = 5 × 2 + 4 

⇒ AB + BC = 14

∴ AB + BC equal to 14.

Given:

ABC is a right-angled triangle with ∠ABC = 90°.

The radius of circle having radius O is 2 cm.

MA ∶ MC = 2 ∶ 3.

Concept:

1. In a right-angled triangle, Hypotenuse2 = Base2 + Perpendicular2
2. Tangents drawn from an external point to a circle are equal.
3. \(\frac{r_1}{r_2} = \frac{1-sin θ }{1 + sin θ}​​\)

Calculation:

Let the radius of circle of center O₁ is r₁

In an ΔABC, using the Pythagoras theorem

⇒ (5x)2 = (2x + 2)2 + (3x + 2)2    

⇒ 25x2 = 4x2 + 4 + 8x + 9x2 + 4 + 12x

⇒ 25x2 = 13x2 + 20x + 8

⇒ 12x2 - 20x - 8 = 0

⇒ 3x2 - 5x - 2 = 0

⇒ 3x2 - 6x + x - 2 = 0

⇒ 3x(x - 2) + 1(x - 2) = 0

⇒ (3x + 1)(x - 2) = 0

x = 2 & x = -1/3 (not possible) 

Now, 

2x = 4

Using the Pythagoras theorem 

AO² = 2² + 4²

AO = 2√5

Let, ∠ OAM = θ 

sin θ = 2/2√5 = 1/√5

We know that 

\(\frac{r_1}{r} = \frac{1-sin θ }{1 + sin θ}\)

\(\frac{r_1}{2} = \frac{1-1/√ 5 }{1 +1/√ 5}\)

\(\frac{r_1}{2} = \frac{√5-1 }{√5 +1}\)

\(\frac{r_1}{2} = \frac{√5-1 }{√5 +1}\times \frac{√5-1 }{√5 -1}\)

\(\frac{r_1}{2} = \frac{6-2√5 }{4}\)

r₁ = 3 - √5

Given:

ABC is a right-angled triangle with ∠ABC = 90°.

The radius of circle having radius O is 2 cm.

MA ∶ MC = 2 ∶ 3.

Concept:

1. In a right-angled triangle, Hypotenuse2 = Base2 + Perpendicular2
2. Tangents drawn from an external point to a circle are equal.
3. \(\frac{r_1}{r_2} = \frac{1-sin θ }{1 + sin θ}​​\)

Calculation:

Let the radius of circle of center O1 is r2

In an ΔABC, using the Pythagoras theorem

⇒ (5x)2 = (2x + 2)2 + (3x + 2)2    

⇒ 25x2 = 4x2 + 4 + 8x + 9x2 + 4 + 12x

⇒ 25x2 = 13x2 + 20x + 8

⇒ 12x2 - 20x - 8 = 0

⇒ 3x2 - 5x - 2 = 0

⇒ 3x2 - 6x + x - 2 = 0

⇒ 3x(x - 2) + 1(x - 2) = 0

⇒ (3x + 1)(x - 2) = 0

x = 2 & x = -1/3 (not possible) 

Now, 

3x = 6

Using the Pythagoras theorem 

CO2 = 22 + 62

CO = 2√10

Let, ∠ MCO = ϕ  

sin ϕ  = 2/(2√10) = 1/√10

We know that 

\(\frac{r_1}{r} = \frac{1-sin θ }{1 + sin θ}\)

\(\frac{r_2}{2} = \frac{1-{1\over\sqrt{10}} }{1+{1\over\sqrt{10}}}\)

\(\frac{r_2}{2} = \frac{\sqrt{10}-1 }{\sqrt{10} +1}\)

\(\frac{r_2}{2} = \frac{\sqrt{10}-1 }{\sqrt{10} +1} \times \frac{\sqrt{10}-1 }{\sqrt{10} -1}\)

\(\frac{r_2}{2} = \frac{(\sqrt{10}-1)^2 }{(\sqrt{10})^2-1^2}\)

\(\frac{r_2}{2} = \frac{10 + 1- 2\sqrt{10} }{(10-1)}\)

\(\frac{r_2}{2} = \frac{11- 2\sqrt{10} }{(9)}\)

r₂ = \(\frac{22-4 \sqrt{10}}{9}\)

Given:

Two identical semicircles and one circle inscribed in a rectangle of length 10 cm

Calculation:

DC is the diameter of semi circle,

DO = OC = EO = FO all radius of semicircle,

Since, EF is the diameter of a circle angle between EO and FO = 90° 

Area of a right angle triangle EOF = (EO × FO)/2 = 5 × 5/2 = 12.5 square cm

∴ The correct answer is 12.5 square cm.

Given:

Two identical semicircles and one circle inscribed in a rectangle of length 10 cm

Calculation:

DC is the diameter of semi circle,

DO = OC = EO = FO all radius of semicircle,

∠OEF = ∠ AOQ = 45° 

Using pythagorous in ΔEOF,

EF = 5√2

AB = 10 (given)

Height of the trapezium AEFB = EF/2 =  5/√2

Area of trapezium = Height × (Sum of parallel sides)/2

Area of trapezium = \(\frac{5}{\sqrt2} × (\frac{10 + 5\sqrt2}{2}) = \frac{5× (10\sqrt2 + 10)}{\sqrt2× 2\sqrt2}\)

⇒ 120 × 4 = 30

∴ The correct answer is 30 square cm.

Given:

Two identical semicircles and one circle inscribed in a rectangle of length 10 cm

Calculation:

DC is the diameter of semi circle,

DO = OC = EO = FO all radius of semicircle,

Area of shaded region = 2 ×  area of segment EQF,

∠EOF = 90°

Area of segment = π r² ×  (90° /360° ) - Area of ΔEOF,

Area of segment =  [3.14 × 25 × (1/4) - 25/2]

Area of segment = 7.125 cm²

The area of the shaded region = 2 × 7/125 = 14.25 square cm.

∴ The correct answer is 14.25 square cm.

Given:

Circle of area = 9π square unit,

Equilateral triangle ABC

Formula Used:

Height of equilateral triangle = √3a/2 

Calculation:

Let side of the equilateral triangle = a

Radius of the circle = r

Height of the equilateral triangle = 2 × r

⇒ √3a/2 = 2r

⇒ a = 4r/√ 3

Area of circle = πr²

9π = π r²

r = 3

a = (4 × 3)/√3 = 4√3 

∴ The correct answer is 4√3.

Given:

Circle of area = 9π square unit,

Equilateral triangle ABC

Formula Used:

Height of equilateral triangle = √3a/2 

Calculation:

Let side of the equilateral triangle = a

Radius of the circle = r

Height of the equilateral triangle = 2 × r

⇒ √3a/2 = 2r

⇒ a = 4r/√ 3

Area of circle = πr2

9π = π r2

r = 3

a = (4 × 3)/√3 = 4√3 

Area of equilateral triangle = √3a²/4

Area of equilateral = √3 × 48/4 = 12√3 cm².

Area of circle = 9π 

Area of circle + area of equilateral = 12√3 + 9π 

∠A = 60° 

Area of unshaded region = [2 × (1/2 × 3 × 3 × sin 120° + 9π (120/360))]

Area of unshaded Region = (9√3/2) + 3π 

Total area of shaded region = 12√3 + 9π - 2 × (9√3/2) + 3π)

Total area of shaded region = 12√3 + 9π - 9√3 - 6π 

Total area of shaded region = 3π  + 3√3

∴ The correct answer is 3π  + 3√3.

Given:

Two circles with centers at O1 and O2 touching each other are placed inside a rectangle of sides 10 cm and 8 cm 

Calculation:

Radius of bigger circle = 4 cm

Radius of smaller circle = r cm

Area of the bigger circle = π R² = 16π cm²

O₁O₂ = 4 + r

O₁E = 4 - r

EO₂ = 5 - r

Using pythagorous Theorm in ΔO₁EO₂, 

(4 + r)² = (4 - r)² + (5 - r)²

r² + 16 + 8r = 16 + r2 -8r + 25 + r2 -  10r

r2 + 26r -25 = 0

⇒ r = 25, 1

Since r can not be 25,

r = 1

Area of smaller circle = π × 1

The sum of the areas of the two circles = 16π + π = 17π

∴ The correct answer is 17π square unit.

Given:

Two circles with centers at O1 and O2 touching each other are placed inside a rectangle of sides 9 cm and 8 cm 

Calculation:

Radius of bigger circle = 4 cm

Radius of smaller circle = r cm

Area of the bigger circle = π R2 = 16π cm2

O1O2 = 4 + r

O1E = 4 - r

EO2 = 5 - r

Using pythagorous Theorm in ΔO1EO2, 

(4 + r)2 = (4 - r)2 + (5 - r)2

r2 + 16 + 8r = 16 + r2 -8r + 25 + r2 -  10r

r2 + 26r -25 = 0

⇒ r = 25, 1

Since r can not be 25,

r = 1

O1O2 = 4 + 1 = 5

O1E = 4 - 1 = 3

EO2 = 5 - 1 = 4

tan θ = EO2/O1E = 4/3 = 1.333

tan 45° = 1 and tan 60° = 1.73

Since tanθ lies between both values,

θ lie between 45°  and 60° 

∴ The correct answer is 45° < θ < 60°.

Given:

Two circles with centers at O1 and O2 touching each other are placed inside a rectangle of sides 9 cm and 8 cm 

Calculation:

Radius of bigger circle = 4 cm

Radius of smaller circle = r cm

Area of the bigger circle = π R2 = 16π cm2

O1O2 = 4 + r

O1E = 4 - r

EO2 = 5 - r

Using pythagorous Theorm in ΔO1EO2, 

(4 + r)2 = (4 - r)2 + (5 - r)2

r2 + 16 + 8r = 16 + r2 -8r + 25 + r2 -  10r

r2 + 26r -25 = 0

⇒ r = 25, 1

Since r can not be 25,

r = 1

O1O2 = 4 + 1 = 5

O1P = 4

PF = 5 - 1 = 4

O₂B = 1

Trapezium O₁O₂FE,

Area of trapezium = \(\frac{1}{2} × (Sum\ of \ parallel\ sides) × Height\)

= \(\frac{1}{2}× 5× 4 = 10\)

Area of sector O₁GP = 16π (θ/360°) 

∠O1O2E = 180 - θ (Parallel line property)

Area of Sector GO₂F = π × 1 × (180 - θ)/360 

Area of the shaded region = Area of trapezium - (Area of big circle sector + Area of small circle sector)

⇒ 10 - [16π (θ/360°) + π (180 - θ)/360]

⇒ 10 - [16πθ/360 + π/2 - πθ/360]

⇒ 10 - 15πθ/350 + π/2

⇒ \(\frac{240-12 \pi-\pi \theta}{24}\)

∴ The correct answer is option 4.

Given:

ABCD be the diameter of a circle of radius 6 cm

The lengths AB, BC and CD are equal

Calculation:

Area of Shaded area = Area of semi circle - area of 2 small semi circles

⇒  π × 36/2 - [2π + 8π] = 8π

Unshaded Area = area of semi circle + area of other small semi circles

⇒ 18π + 2π + 8π = 28π 

Required Ratio = 8π/28π = 2/7

∴ The correct answer is 2:7.

Given:

ABCD be the diameter of a circle of radius 6 cm

The lengths AB, BC and CD are equal

Calculation:

Perimeter of semi circle without Diameter = πR

Required perimeter = sum of perimeter of all three semi circle.

Required Perimeter = 6π + 2π + 4π = 12π cm

∴ The correct answer is 12π.