Quantitative Aptitude PYQs

Formula used:

Mode = L + \(\frac{D_1}{D_1+D_2}\times i\)

L = lower limit,

D₁ = ​​The difference between the frequency of the modal class and the

frequency of the class preceding the modal class,

D₂ = The difference between the frequency of the modal class and the

frequency of the class following the modal class.

i = The class width

Calculation:

Since class 60 - 90 has the highest frequency. Therefore,

Model class = 60 - 90   

L = 60, D₁ = 7 - 5 = 2

D₂ = 7 - 4 = 3, i = 90 - 60 = 30

Mode = 60 +  \(\frac{2}{2+3}\times 30\) = 72

∴ Mode of the distribution is 72.

Formula used:

Mode = L + \(\frac{D_1}{D_1+D_2}× i\)

L = lower limit,

D1 = ​​The difference between the frequency of the modal class and the

frequency of the class preceding the modal class,

D2 = The difference between the frequency of the modal class and the

frequency of the class following the modal class.

i = The class width,

Median \( = l+ \left[\frac{\left(\frac{n}{2} – cf\right)}{f}\right] × i\)

Where l = lower limit of median class,

n = number of observations,

i = class size, f = frequency of median class,

cf = cumulative frequency of class preceding the median class.

Calculation:

 

Since class 60 - 90 has the highest frequency. Therefore,

Model class = 60 - 90   

L = 60, D1 = 7 - 5 = 2

D2 = 7 - 4 = 3, i = 90 - 60 = 30

Mode = 60 +  \(\frac{2}{2+3}× 30\) = 72

Let's calculate the median 

Here, n = 20. (n/2) = 20/2 = 10 

So, the 10^th observation will be in class 60 - 90.

Median = \( 60+ \left[\frac{\left(\frac{20}{2} – 9\right)}{7}\right] × 30 \)

Median = 60 + (30/7) = 450/7

Median = P = 450/7

Mode = Q = 72

According to the question, 7(Q - P) = 9R

⇒ 7 (72 - 450/7) = 9R

⇒ 7 (504 - 450)/7 = 9R

⇒  54 = 9R

∴ R = 6

Formula used:

Mean = ∑  (f.x̅) / ∑f

Calculation:

 

Therefore,

Mean = (560/10)

∴ Mean = 56

Formula Used: 

Median \( = l+ \left[\frac{\left(\frac{n}{2} – cf\right)}{f}\right] \times h \)

Where l = lower limit of median class,

n = number of observations,

h = class size, f = frequency of median class,

cf = cumulative frequency of class preceding the median class.

Calculation: 

 

n = 10. Therefore n/2 = 10/2 = 5

Therefore, 50 - 60 will be the median class.

Median (M) = \( = 50+ \left[\frac{\left(\frac{10}{2} – 4\right)}{3}\right] \times 10\)

Median (M) = 50 + 10/3 = 160/3

∴ Required value 3M  = 160.

Formula used:

Area of rectangle = length× breadth

Calculation:

Let the length be 2X and the breadth be X. Then, 

Area of plinth = length x breadth

⇒  (2X)*(X)

⇒ 2x² = 200 sq. m.

⇒ X = √(100)

⇒ X = 10 m.

The dimensions of the plinth are 2X = 20 m (length) & X = 10 m (breadth).

Now, the owner extends the terrace by 1 m on each side. So, the new

dimensions of the terrace are (20 + 2) m (length) and (10 + 2) m (breadth).

Area of new terrace = (20 + 2) x (10+2)

Area of new terrace = 264 sq. m

Increase in area = 264 - 200

Increase in area = 64 sq. m

Percentage increase = (increase in area/area of plinth) x 100

Percentage increase = (64/200) x 100 = 32%

∴ The percentage increase in the area is 32%.

Formula used:

Perimeter = Base of cylinder = 2π × radius

Volume of cylinder = π(radius)² × h

Calculation:

When the square sheet is rolled along one of its sides 

The circumference of the base of the cylinder = Length of the side of the square sheet

Therefore, the circumference of the base of the cylinder = 44 cm

⇒ 2πr = 44

⇒ 2 × (22/7) × r = 44 

⇒ r = 7 cm

The height of the cylinder is equal to the length of the side of the square sheet, which is 44 cm.

Volume = π × radius² × height

⇒ V = (22/7) × 7² × 44

⇒ V = 22 × 7 × 44

⇒ V = 6776 cm³

Therefore, the volume of the cylinder is 6776 cm³.

Shortcut Trick

Relation between Volume (V) and areas of three adjacent faces (A₁, A₂, A₃) is V² = A₁ × A₂ × A₃.

Given: V = 3600, A₁ = 225, A₂ = 144.

3600² = 225 × 144 × A₃

A₃ = (3600 × 3600) ÷ (225 × 144) = (3600 ÷ 180)² = 20² = 400.

∴ The correct answer is 400 square cm.

Alternate Method

Given:

Volume of cuboid (V) = 3600 cm³

Area of first face (A₁ = l × w) = 225 cm²

Area of second face (A₂ = w × h) = 144 cm²

Formula Used:

Volume of cuboid = l × w × h

Relationship: V² = (l × w) × (w × h) × (h × l)

Calculations:

⇒ Let the three adjacent faces be A₁, A₂, and A₃.

⇒ V² = A₁ × A₂ × A₃

⇒ 3600 × 3600 = 225 × 144 × A₃

⇒ 12,960,000 = 32,400 × A₃

⇒ A₃ = 12,960,000 ÷ 32,400

⇒ A₃ = 400

∴ The correct answer is 400 square cm.

Additional Information

Total Surface Area of Cuboid

The total surface area is the sum of the areas of all six faces: TSA = 2(lw + wh + hl).

Diagonal of a Cuboid

The longest rod that can be placed inside a cuboid is its diagonal: d = √(l² + w² + h²).

Lateral Surface Area

The area of the four side walls of a cuboid is LSA = 2h(l + w).

Given:

The perimeter and area of a right-angled triangle are 36 cm and 54 cm2

Formula used:

The perimeter of a right-angled triangle = Base + Perpendicular + Hypotenuse

Area of a right-angled triangle = \(\frac{1}{2} \) × Base × Perpendicular 

Hypotenuse2 = Base2 + Perpendicular2

Calculation:

Let B , P, and H be the base, Perpendicular, and Hypotenuse of the right-angled triangle

B + P + H = 36

B + P = 36 - H      -----(1)

\(\frac{1}{2} × B× P\) = 54

B × P = 108         -----(2)

H2 = B2 + P2

We know that, (a² + b²) = (a + b)² - 2ab

⇒ H² = (B + P)2 - 2BP

Substituting (1) and (2)

⇒ H² = (36 - H)2 - 216

⇒ H2 = 1296 + H2 - 2(36)(H) - 216

⇒ 72 H = 1296 - 216

⇒ 72 H = 1080

⇒ H = 1080/72 = 15 cm

∴ The correct answer is 15 cm.

Calculation:

The set X is given by

{2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, 98}.

We want to find the maximum size of a subset S of X such that no two elements sum to 100.

The pairs in X that sum to 100 are

(2, 98), (6, 94), (10, 90), (14, 86), (18, 82), (22, 78), (26, 74), (30, 70), (34, 66), (38, 62), (42, 58), (46, 54), (50, 50){

note:

50 appears only once in X }

Therefore, 

To maximize the number of elements in S while ensuring no two elements sum to 100:

• Choose one element from each of the 12 pairs (but not both)
• Additionally, include the element 50

 

The maximum possible number of elements in S = 13

∴ The maximum possible number of elements in S be 13.

Given:

The perimeter of a sector of a circle of radius = 16.4 cm

The radius of a sector of a circle of radius 5.2 cm.

Formula used:

The perimeter (P) of a sector of a circle with radius (r) and central angle (θ) is 

P = θr + 2r

Area of sector = (θ/2π) πr²

Calculation:

Using the given values, we have:

16.4 = θ(5.2) + 2(5.2)

16.4 = 5.2θ + 10.4

θ = 6/5.2 ≈ 1.153 radians (rounded to three decimal places)

Area of sector = (θ/2π) πr2

A = (θ/2π)πr²

A = (1.153/2π) × π(5.2)²

A = (1.153 × 5.2 × 5.2)/2 

A = 15.6 cm² (rounded to one decimal place)

Therefore, the area of the sector is approximately 15.6 cm².

Given:

AB = 6 cm, BC = 8 cm and AD = 10 cm

Concept used:

By Pythagoras theorem

AC² = AB² + BC²

Angle bisector theorem: An angle bisector of an angle of a triangle divides the opposite side into two parts that are proportional to the other two sides of the triangle.

Calculation:

Now, according to the above concept,

⇒ AB × BC = AC × BD

⇒ 6 × 8 = 10 × r

⇒ 48 = 10r

⇒ r = 4.8 cm

So, BD = 4.8 cm

Now, QC = BC - BQ

⇒ QC = (8 - 4.8) cm = 3.2 cm

∴ The required value of QC is 3.2 cm

Given:

AB = 6 cm, BC = 8 cm and AC = 10 cm

Concept used:

If a perpendicular is drawn from the vertex of the right angle of a right triangle

to the hypotenuse then, AB² = BC.BD

Calculation:

According to the above concept,

⇒ AB² = AD × Ac

⇒ 6² = AD ×10

⇒ AD = 3.6 cm

Now, according to the question

⇒ Sinθ = \(\frac{AD}{{AB}}\)

⇒ Sinθ = \(\frac{3.6}{{6}}\) = 0.6

∴ The required value of sinθ is 0.6

Given:

AB = 6 cm, BC = 8 cm and AD = 10 cm

Concept used:

By Pythagoras theorem

AC2 = AB2 + BC2

Angle bisector theorem: An angle bisector of an angle of a triangle divides the opposite side into two parts that are proportional to the other two sides of the triangle.

Calculation:

Now, according to the above concept,

⇒ AB × BC = AC × BD

⇒ 6 × 8 = 10 × r

⇒ 48 = 10r

⇒ r = 4.8 cm

Given:

A circle is inscribed in a square PQRS

A rectangle at the corner P that measures 4 cm × 2 cm and a square at the corner R are drawn.

Concept used:

By Pythagoras theorem,

AC² = AB² + BC²

Area of circle = πr²

Calculation:

According to the above figure,

⇒ OD = (r - 2) cm, CD = (r - 4) cm, OC = r cm

Now, according to the above concept,

In ΔOCD,

⇒ OC² = (OD)² + (CD)²

⇒ r² = (r - 2)² + (r - 4)²

⇒ r² = r² + 4 - 4r + r² + 16 - 8r

⇒ r² - 12r + 20 = 0

⇒ r² - 10r - 2r + 20 = 0

⇒ r(r - 10) - 2(r - 10) = 0

⇒ (r - 10)(r - 2) = 0

⇒ r = 10, 2   [Here we take the radius = 10]

Now, area of circle is

⇒ π × (10)² = 100π square cm

∴ The required value is 100π square cm.

Given:

A circle is inscribed in a square PQRS

A rectangle at the corner P that measures 4 cm × 2 cm and a square at the corner R are drawn.

Concept used:

By Pythagoras theorem,

AC2 = AB2 + BC2

Area of circle = πr²

Diagonal of square = √2a

Area of the square = a2

Calculation:

Z

According to the above figure,

⇒ OD = (r - 2) cm, CD = (r - 4) cm, OC = r cm

Now, according to the above concept,

In ΔOCD,

⇒ OC2 = (OD)2 + (CD)2

⇒ r2 = (r - 2)2 + (r - 4)2

⇒ r2 = r2 + 4 - 4r + r2 + 16 - 8r

⇒ r2 - 12r + 20 = 0

⇒ r2 - 10r - 2r + 20 = 0

⇒ r(r - 10) - 2(r - 10) = 0

⇒ (r - 10)(r - 2) = 0

⇒ r = 10, 2   [Here we take the radius = 10]

So, r = 10 cm

Now, In ΔOAR,

Here the radius i.e. OA = r = 10 cm

Here, OR is diagonal of square i.e. 10√2 cm

⇒ (r + √2a) = 10√2

⇒ 10 + √2a = 10√2

⇒ √2a = 10√2 - 10

⇒ a = \(\frac{10(√2 - 1)}{{√ 2}} \)

Now, area of the small square = a²

⇒ (\(\frac{10(√2 - 1)}{{√ 2}} \))² 

⇒ 50(2 + 1 - 2√2) = 50(3 - 2√2) cm²

∴ The required area is 50(3 - √2) square cm

Given:

A circle is inscribed in a square PQRS

A rectangle at the corner P that measures 4 cm × 2 cm and a square at the corner R are drawn

Concept used:

Area of square = a²

Area of rectangle = length × breadth

Area of sector = (\(\frac{θ}{{360°}}\))πr²

Calculation:

Now, according to the question,

Area of shaded region = Area of square - Area of sector - Area of rectangle

⇒ Area of shaded region = [(10)² - \(\frac{90°}{{360°}}\) × π(10)² - (2 × 4)] cm²

⇒ (100 - \(\frac{1}{{4}}\) × 100π - 8) cm²

⇒ (100 - 25π - 8) = (92 - 25π) cm²

∴ The required area of the shaded region is (92 - 25π) square cm.

Concept:

If in a right-angle triangle,

The ratio of angle are 30° : 60° : 90°  

The ratio of sides length opposite to the angle = 1 : √3 : 2 

Formula used:

Area of a circle = πr²

Area of a rectangle = Lenght × Breadth

Calculation:

The ratio of sides length opposite to the angle = 1 : √3 : 2 

In an ΔADC, side AC is opposite to the ∠D which is 90° 

We know that, if in the right-angle triangle 

The ratio of angle are 30° : 60° : 90° 

⇒ AD : DC : AC = r : r√3 : 2r

Therefore, the ratio of the area of the circle to the area of the rectangle

(πr²)/(AD × DC) = πr² / (r × r√3) = \(\frac{\pi}{\sqrt 3}\)

∴ Required ratio is \(\frac{\pi}{\sqrt 3}\)

Concept:

If in a right-angle triangle,

The ratio of angle are 30° : 60° : 90°  

The ratio of sides length opposite to the angle = 1 : √3 : 2 

Formula used:

Area of a circle = πr2

Area of a rectangle = Lenght × Breadth

Calculation:

If in a right-angle triangle,

The ratio of angle are 30° : 60° : 90°  

The ratio of sides length opposite to the angle = 1 : √3 : 2 

In an ΔADC, side AC is opposite to the ∠D which is 90° 

We know that, if in the right-angle triangle 

The ratio of angle are 30° : 60° : 90° 

⇒ AD : DC : AC = r : r√3 : 2r

Area of ΔAEC = Area of ΔADC - Area of ΔADE

Area of ΔAEC = \(\frac{1}{2}\) (AD × DC) - \(\frac{1}{2}\) (AD × DE)

Area of ΔAEC = \(\frac{1}{2}\) (r × r√3) - \(\frac{1}{2}\) (r × DE)

But, in the ΔADE,

tan 30°  = DE/AD = DE/r

= 1/√3 = AD/r

⇒ AD = r/√3

Area of ΔAEC = \(\frac{1}{2}\) (r × r√3) - \(\frac{1}{2}\) (r × r/√3)

Area of ΔAEC = \(\frac{\sqrt 3}{2}\)r² - \(\frac{1}{2\sqrt3}\)r² = \(\frac{3r^2-r^2}{2\sqrt 3}\)

∴ The required area is \(\frac{\pi}{√ 3}\).

Shortcut Trick 

in ∠OAQ

∠AOP = ∠OAQ + ∠OQA

x° = 2y° + y°  = 3y° 

Alternate Method 

Given:

∠POA = x° and ∠OQB = y°

OB = BQ

Concept used:

The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles of the triangle.

The angle which forms a straight line is called the 180-degree angle

Calculation:

∠OBA = ∠BQO + ∠BOQ = 2y°

AO = OB = Radius of the circle

So, ∠OAB = ∠OBA = 2y°

So, ∠AOB = 180° - 4y°

Again,

∠AOB = 180° - x° - y°       [As PQ is a straight line]

So,

180° - 4y° = 180° - x° - y°

⇒ x = 4y - y

⇒ x = 3y

∴ The relation between x and y is x = 3y.

Given:

∠POA = x° and ∠OQB = y°

OB = BQ, y = 15

Concept used:

The measure of an exterior angle of a triangle is equal to the sum of the

measures of the two non-adjacent interior angles of the triangle.

Central angle theorem: It states that for any arc of a circle, the angle formed

by the radii at the center of the circle is twice the angle formed by the arc at

any point on the circumference. 

Calculation:

∠OBA = ∠BQO + ∠BOQ = 30°

AO = OB = Radius of the circle

So, ∠OAB = ∠OBA = 30°

So, ∠AOB = 180° - 30° - 30° = 120°

So, according to the concept,

∠ACB = 120/2 = 60°

∴ ∠ACB is equal to 60°.