Quantitative Aptitude PYQs

Given:

x is the harmonic mean between y and z.

Concept:

Harmonic mean (HM):  If a, b & c are in AP then,

1/a, 1/b, 1/c are in harmonic progression (HP). 

If H is the harmonic mean of a & b, then

H = \(\frac{2}{\frac{1}{a} \ + \ \frac{1}{b} }\) or H = \(\frac{2ab}{a\ + \ b}\)

Calculation:

We have,

x is the harmonic mean of y and z

⇒ x = \(\frac{2yz}{y \ + \ z}\)

⇒ xy + xz = 2yz

⇒ xy + xz - 2yz = 0

∴ If x is the harmonic mean between y and z, then xy + xz - 2yz = 0.

Concept:

Median:

Case 1: If the number of observations (n) is even

\({\rm{Median;}} = {\rm{;}}\frac{{{\rm{value;of;}}{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{;observation;}} + {\rm{;;value;of;}}{{\left( {\frac{{\rm{n}}}{2}{\rm{;}} + 1} \right)}^{{\rm{th}}}}{\rm{;observation}}}}{2}\)

Case 2: If the number of observations (n) is odd

\({\rm{Median;}} = {\rm{value;of;}}{\left( {\frac{{{\rm{n}} + 1}}{2}} \right)^{{\rm{th}}}}{\rm{;observation}}\)

Calculation:

Possible factors of 120:

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

The middle values of the set are 10 and 12

⇒ \(\frac{10 \ + \ 12}{2}\) = \(\frac{22}{2}\) = 11

∴ The median of all possible factors of 120 is 11.

Given:

The sum of deviations of a set of n values measured from 50 = -10

The sum of deviations of the values measured from 46 = 70

Formula used:

Sum of deviation:

(x₁ - d) + (x₂ - d) + .......+ (x_n - d) = \(\displaystyle\sum_{i = 1}^{n}\)x_i

Mean = \(\frac{1}{{n}}\)(\displaystyle\sum_{i = 1}^{n}\) xi 

Calculation:

We have,

⇒ \(\displaystyle\sum_{i = 1}^{n}\) xi − 50n = -10          -----(1)

⇒ \(\displaystyle\sum_{i = 1}^{n}\) xi − 46n = 70           ------(2)

Subtracting equation (2) from (1), we get

⇒ -4n = -80

⇒ n = 20

Now, from equation (1)

⇒ \(\displaystyle\sum_{i = 1}^{n}\) x_i − 50 × 20 = -10

⇒ \(\displaystyle\sum_{i = 1}^{n}\) x_i = 990

Now, the value of  mean

⇒ \(\frac{1}{{n}}\)(\displaystyle\sum_{i = 1}^{n}\) x_i = \(\frac{990}{{20}}\)

⇒ Mean = 49.5

∴ The require mean is 49.5

Given:

The average marks of the students of all four schools = 58

Formula used:

Average = Sum of observations/Number of observations

Calculation:

According to the question

⇒ (60 × 60 + 50 × 80 + 50 × 40 + x × 50)/(60 + 50 + 50 + x) = 58

⇒ (3600 + 4000 + 2000 + 50x)/(160 + x) = 58

⇒ (9600 + 50x)/160 + x) = 58

⇒ 9600 + 50x = 9280 + 58x

⇒ (58x − 50x) = (9600 − 9280)

⇒ 8x = 320

⇒ x = 40

∴ The required students appeared from school - IV is 40.

Given:

The data of different natural numbers 4, 7, 10, 14, 2x + 3, 2x + 5, 22, 23, 30, 50 are in ascending order.

Concept used:

Natural number: Positive integers from 1 and end at infinity

When n is even,

Median = (n + 1)/2 term

When n is even,

Median = [(n/2) + (n/2 + 1)]/2 term

Shortcut TrickWe need to fill only integer value the observation b/w 14 & 22

 

∴ There are five possible values.

Alternate Method

Calculation:

The given numbers = 4, 7, 10, 14, 2x + 3, 2x + 5, 22, 23, 30, 50 

Here, n = 10, which is even

So, Median =  [(n/2) + (n/2 + 1)]/2 term

⇒ [(10/2) + (10/2 + 1)]/2 term

⇒ (5^th term + 6^th term)/2

⇒ (2x + 3 + 2x + 5)/2

⇒ (4x + 8)/2

⇒ Median = 2x + 4

We need to fill only integer values the observation b/w 14 & 22

∴ There are five possible values.

Given:

M is the mean of the first 100 even natural numbers

Concept used:

Mean of first n even numbers = (n + 1)

Mean of first n odd numbers = n

Calculation:

According to the question

Let the mean of the first 100 even numbers = M

⇒ (n + 1) = M        -----(1)

Mean of first 100 odd numbers = n

⇒ (n + 1) = M

From equation (1)

⇒ n = (M − 1) 

∴ The required value is M − 1.

Given: 

Mean marks of 50 students = 78.4

95 was misread as 59, 25 was misread as 52

Concept Used:

Mean of observations = sum of observations/Number of observations  

Calculation

The sum of observations = X

⇒ Initial average = X/50 = 78.4

After correction,

Final average = [X + (95 - 59) + (25 - 52)] / 50

⇒ Final average = (X + 9) / 50

Therefore, required difference in average

⇒ (X + 9 - X) / 50 = 9/50 = 0.18

∴ The required difference is 0.18

Given:

In a triangle ABC, ∠A = 2θ, ∠B = ∠C = 4θ and θ satisfies 4 sin2θ + 2 sin θ - 1 = 0.

Concept used:

Isosceles triangle:

The altitude from vertex A to the base BC is the perpendicular bisector of the base BC.

Quadratic formula

\({-b \pm \sqrt{b^2-4ac} \over 2a}\)

sin θ = Perpendicular / Hypotenuse

Calculation:

4sin2θ + 2sin θ - 1 = 0

⇒ sinθ = \( {-2 \pm \sqrt{2^2-4\times (-1) \times (4)} \over 8}\)

⇒ sinθ = \( {-2 \pm 2\sqrt{5} \over 8}\)

⇒ sinθ = \( {-2 + 2\sqrt{5} \over 8}\)

⇒ sinθ = \( {\sqrt{5} - 1 \over 4}\)      ----(1)

In ΔABD,

⇒ sinθ = \(\frac{BD}{AB}\)           (sin θ = P/H)

Using equation (1)

⇒\(\frac{b}{{a}}\) = \( {\sqrt{5} - 1 \over 4}\)             -----(2)

Now, in Δ ABD

⇒ \(\frac{BC}{{AB}}\) = \(\frac{2b}{{a}}\) = \(2\times \frac{b}{{a}}\)

⇒ \(\frac{BC}{{AB}}\) = (2 × \( {\sqrt{5} - 1 \over 4}\))       [From equation (2)]

⇒ \(\frac{BC}{{AB}}\) =\(\frac{(\sqrt5-1)}{2}\)

∴ The required value is \(\frac{(\sqrt5-1)}{2}\).

Shortcut Trick

In an Inscribed Circle, the angle formed by the Contact Triangle vertices (like ∠EDF) is directly related to the opposite vertex angle.

The segments from vertices to contact points are equal tangents, creating Isosceles Triangles at each corner: ▵AFE, ▵BDF, and ▵CDE.

Property: The angle subtended by the chord joining two contact points at the third contact point is 90° − (Opposite Angle/2).

For ∠EDF, the opposite vertex is A. ⇒ ∠EDF = 90° − (A/2).

Similarly, ∠DEF = 90° − (B/2) and ∠EFD = 90° − (C/2).

∴ The correct answer is 90° − (A/2).

Alternate Method

Given:

A circle is inscribed in triangle ABC.

It touches the sides BC, CA and AB at points D, E and F respectively.

Concept Used:

• The angle subtended by a chord at any point on the circumference is half the angle subtended by the same chord at the centre of the circle.
• The sum of all angles of a quadrilateral is 360°.

Calculation:

We know that the radius is perpendicular to the tangent.

⇒ ∠AEO = ∠AFO = 90°

In quadrilateral AEOF, using the above concept:

⇒ 90° + 90° + A + ∠O = 360°

⇒ ∠O = 180° − A

We know that ∠EDF = 1/2 × ∠EOF

⇒ ∠EDF = (180° − A) ÷ 2

⇒ ∠EDF = 90° − A/2

∴ The value of ∠EDF is 90° − A/2.

Additional Information

Incenter and Inradius

The center of the inscribed circle (Incenter) is the intersection of the angle bisectors of the triangle. The inradius r = Area / s, where s is the semi-perimeter.

Tangent Properties

The distance from a vertex to the points of contact is s − a, s − b, and s − c for vertices A, B, and C respectively.

Contact Triangle

The triangle formed by joining the points of contact D, E, and F is called the Gergonne triangle or the Contact triangle of ▵ABC.

Given:

The length of a room = \(\frac{21}{16}\)  times it's breadth.

The breadth is \(\frac{4}{3}\) times its height.

Formula used:

Longest diagonal of the room = √(l² + b² + h²)

Where, l = length, b = breadth and h = height

Calculation:

Let the value of Height = 12

According to the question

Then, Breadth = (\(\frac{4}{3}\) × 12) = 16

Length = (\(\frac{21}{16}\) × 16) = 21

Now, longest diagonal of the room = √(l2 + b2 + h2)

⇒ L = √(21² + 16² + 12²)

⇒ L = √(441 + 256 + 144)

⇒ L = √841

⇒ L = 29

Now,

⇒ \(\frac{H}{{L}}\) = \(\frac{12}{{29}}\)

⇒ 12L = 29H

∴ The required value is 12L = 29H.

Given:

The radius of cone = 12 cm

The height of the cone = 16 cm

Concept used:

For similar triangles, the ratio of corresponding sides will be equal

Formula used:

Slant height of cone, l² = √(r² + h²)

Where, i = length, r = radius and h = height

Calculation:

According to the figure

The slant height of the cone,

(AO)= √(12² + 16²)

⇒ √(144 + 256) = √400

⇒ AO = 20 cm

In ΔOBA ~ ΔOCD

⇒ \(\frac{AB}{{CD}}\) = \(\frac{OA}{{OD}}\)

⇒ \(\frac{12}{{r}}\)= \(\frac{20}{{16-r}}\)           [∵ OD = OB – BD]

⇒ \(\frac{3}{{r}}\) = \(\frac{5}{{16-r}}\)

⇒ 48 – 3r = 5r

⇒ 8r = 48

⇒ r = 6

∴ The radius of a sphere is 6 cm.

Given:

The radius of cone = 12 cm

The height of the cone = 16 cm

Concept used:

For similar triangles, the ratio of corresponding sides will be equal

Formula used:

Slant height of cone, l2 = √(r2 + h2)

Where, i = length, r = radius and h = height

Calculation:

According to the figure

The slant height of the cone,

(AO)= √(122 + 162)

⇒ √(144 + 256) = √400

⇒ AO = 20 cm

In ΔOBA ~ ΔOCD

⇒ \(\frac{AB}{{CD}}\) = \(\frac{OA}{{OD}}\)

⇒ \(\frac{12}{{r}}\)= \(\frac{20}{{16-r}}\)           [∵ OD = OB – BD]

⇒ \(\frac{3}{{r}}\) = \(\frac{5}{{16-r}}\)

⇒ 48 – 3r = 5r

⇒ 8r = 48

⇒ r = 6

The remaining water = Volume of cone – Volume of sphere

⇒ [\(\frac{1}{{3}}\)π × (12)² × 16] – [\(\frac{4}{{3}}\)π × (6)³]

⇒ (\(\frac{1}{{3}}\)π × 144 × 16) – (\(\frac{4}{{3}}\)π × 216)

⇒ π(768 – 288) = 480π 

∴ The water remain in the vessel after the overflow is 480π ml.

Given:

The radius of cone = 12 cm

The height of the cone = 16 cm

Concept used:

For similar triangles, the ratio of corresponding sides will be equal

Formula used:

Slant height of cone, l2 = √(r2 + h2)

Where, i = length, r = radius and h = height

Calculation:

According to the figure

The slant height of the cone,

(AO)= √(122 + 162)

⇒ √(144 + 256) = √400

⇒ AO = 20 cm

In ΔOBA ~ ΔOCD

⇒ \(\frac{AB}{{CD}}\) = \(\frac{OA}{{OD}}\)

⇒ \(\frac{12}{{r}}\)= \(\frac{20}{{16-r}}\)           [∵ OD = OB – BD]

⇒ \(\frac{3}{{r}}\) = \(\frac{5}{{16-r}}\)

⇒ 48 – 3r = 5r

⇒ 8r = 48

⇒ r = 6

⇒ Lateral surface area of the vessel: surface area of the sphere is

⇒ \(\frac{πrl}{{4πr^2}} \) = \(\frac{\pi \times 12 \times 20}{{4 \times \pi \times 6 \times 6}}\) = \(\frac{5}{3}\)

∴ The required value is \(\frac{5}{3}\).

Formula used:

Area of circle = π(radius)2

Pythagoras theorem: In a right-angled triangle 

Base2 + Perpendicular2 = Hypotenuse2

Calculation:

Now in ΔOAB, by Pythagoras theorem,

l2 = r2 + r2

⇒ l2 = 2r2

⇒ r = \(\frac{l}{{\sqrt2}}\)

We know that, the area of segment

A1 = \(\frac{1}{2}(\frac{π}{2}-sin\ \theta)r^2\)

⇒ A1 = \(\frac{1}{2}(\frac{π}{2}-sin\ 90^{\circ})(\frac{l}{\sqrt2})^2\)

⇒ A1 = \(\frac{l^2}{4}\left(\frac{π}{2}-1\right) \)

∴ The required value is \(\frac{l^2}{4}\left(\frac{\pi}{2}-1\right) \).

Formula used:

Area of circle = π(radius)²

Pythagoras theorem: In a right-angled triangle 

Base2 + Perpendicular2 = Hypotenuse2

Calculation:

Now in ΔOAB, by Pythagoras theorem,

l2 = r2 + r2

⇒ l2 = 2r²

⇒ r = \(\frac{l}{{\sqrt2}}\)

We know that, the area of segment

A₁ = \(\frac{1}{2}(\frac{π}{2}-sin\ \theta)r^2\)

⇒ A₁ = \(\frac{1}{2}(\frac{π}{2}-sin\ 90^{\circ})(\frac{l}{\sqrt2})^2\)

⇒ A₁ = \(\frac{l^2}{4}\left(\frac{π}{2}-1\right) \)

Area of major segment = Area of circle – Area of minor segment

⇒ A₂ = (π × \(\frac{l^2}{{2}}\)) – \(\frac{l^2}{4}\left(\frac{π}{2}-1\right) \)

⇒ A2​ = \(\frac{π l^2}{{2}}\) – \(\frac{π l^2}{{8}}\) + \(\frac{ l^2}{{4}}\)

⇒ A2​ = \(\frac{3π l^2}{{8}}\) + \(\frac{ l^2}{{4}}\)

⇒ A2​ = \(\frac{l^2}{4}\left(\frac{3π}{2} +1\right)\)

∴ The required value is \(\frac{l^2}{4}\left(\frac{3π}{2} +1\right)\).

Shortcut Trick

In a right-angled triangle, the square of the altitude (perpendicular) drawn to the hypotenuse is equal to the product of the segments of the hypotenuse it creates.

Property: AD² = BD × DC

⇒ AD² = 7.5 × 10 = 75

⇒ AD = √75 = √(25 × 3) = 5√3 cm

∴ The correct answer is 5√3 cm.

Alternate Method

Given:

Triangle ABC is right-angled at A (∠BAC = 90°).

AD ⊥ BC, BD = 7.5 cm, and DC = 10 cm.

Formula Used:

Geometric Mean Theorem: In a right triangle, the altitude to the hypotenuse divides the triangle into two similar triangles.

Calculations:

In ΔABC, ∠BAC = 90° and AD ⊥ BC.

⇒ ΔABD ∼ ΔCAD (by AA similarity criterion)

⇒ AD / CD = BD / AD

⇒ AD² = BD × CD

⇒ AD² = 7.5 × 10

⇒ AD² = 75

⇒ AD = √75

⇒ AD = √(25 × 3)

⇒ AD = 5√3 cm

∴ The correct answer is 5√3 cm.

Additional Information

Right Triangle Similarity Theorem

If an altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other.

Geometric Mean (Leg) Theorem

Each leg of the right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg (e.g., AB² = BD × BC).

Pythagoras Theorem

In any right-angled triangle, the sum of the squares of the two legs is equal to the square of the hypotenuse: a² + b² = c².

Concept used:

Pythagoras theorem: It states that In a right-angled triangle, the

square of the hypotenuse side is equal to the sum of the squares of

the other two sides.

Calculation:

Let, BD = 3, AD = 4 AB = 5

⇒ CD = 1         (∵ DB = 3 CD)

⇒ AC = √(16 + 1) = √17     (From Pythagoras theorem) 

(AB + AC) (AB - AC) = AB² - AC²

⇒ (AB + AC) (AB - AC) = 5² - (√17)² = 8

From option (1)

2 (AB + AC) (AB - AC) = BC2

⇒ 2 × 8 = (3 + 1)²

⇒ 16 = 16

∴ The required value is 2 (AB + AC) (AB –  AC) = BC2

Formula used:

Cosine rule: If a, b, and c are the sides opposite the angle A, B & C of ΔABC respectively then,

\(cos\ A=\frac{b^2+c^2-a^2}{2bc}\)

Calculation:

Applying the cosine formula in ΔABC

⇒ \(cos\ 60^{\circ}=\frac{AB^2+AC^2-BC^2}{2\ AB \times AC}\)

⇒ \(\frac{1}{2}=\frac{AB^2+AC^2-BC^2}{2\ AB \times AC}\)      (∵ cos 60° = 1/2)

⇒ AB2 + AC2 – BC² = AB.AC

∴ The required value is AB.AC

Shortcut Trick

Volume of Sphere = Volume of Wire (Cylinder)

(4/3) × R_s³ = r_w² × L

(4/3) × (1.5)³ = r_w² × 50

4.5 = 50 × r_w² ⇒ r_w² = 0.09 ⇒ r_w = 0.3 cm

Diameter of wire = 2 × 0.3 = 0.6 cm

∴ The correct answer is 0.6 cm.

Alternate Method

Given:

Diameter of Copper Sphere = 3 cm ⇒ Radius (R) = 1.5 cm

Length of Wire (h) = 0.5 m = 50 cm

Formula Used:

Volume of Sphere = (4/3) × π × R³

Volume of Cylinder = π × r² × h

Calculations:

⇒ Volume of Sphere = (4/3) × π × (1.5)³

⇒ Volume of Wire = π × r² × 50

⇒ (4/3) × π × 3.375 = π × r² × 50

⇒ (4/3) × 3.375 = 50r²

⇒ 4 × 1.125 = 50r²

⇒ 4.5 = 50r²

⇒ r² = 4.5 ÷ 50

⇒ r² = 0.09

⇒ r = √0.09 = 0.3 cm

⇒ Diameter of wire = 2 × r = 2 × 0.3 = 0.6 cm

∴ The correct answer is 0.6 cm.

Additional Information

Volume of Cylinder

The volume of a cylinder with radius 'r' and height 'h' is V = πr²h.

Volume of Sphere

The volume of a sphere with radius 'R' is V = (4/3)πR³.

Conservation of Volume

When a solid is melted and recast into another solid shape, the total volume remains constant (V_initial = V_final).

Unit Conversion

Always ensure all dimensions are in the same unit. 1 metre = 100 cm.