Quantitative Aptitude PYQs

Shortcut Trick

Let the Marked Price (MP) be 100 units.

Discounted price = 100 − 20% = 80 units.

Cost Price (CP) including fee = 80 + (15% of 80) = 92 units.

Profit = 25% of 92 units = 23 units.

Given 23 units = ₹7800 ⇒ 100 units = (7800 ÷ 23) × 100 = ₹33,913.04.

∴ The correct answer is ₹33,913.04.

Alternate Method

Given:

Discount = 20%

Installation and support fee = 15%

Profit made by reseller = 25%

Amount of profit = ₹7800

MP (100x)−20%80x+15%CP (92x)+25%SPProfit = 23x = ₹7800

Calculation:

⇒ Let the Marked Price (MP) of the license = 100x

⇒ Price to the reseller after 20% discount = 100x − (20% of 100x) = 80x

⇒ Final Cost Price (CP) for the reseller = 80x + (15% of 80x) = 80x + 12x = 92x

⇒ Profit earned by the reseller = 25% of CP = 25% of 92x = 23x

⇒ Since the profit earned is ₹7800

⇒ 23x = 7800

⇒ x = 7800 ÷ 23

⇒ Original MP = 100x = 100 × (7800 ÷ 23) = 780000 ÷ 23

⇒ Original MP = 33913.043...

∴ The correct answer is ₹33,913.04.

Additional Information

Marked Price and Discount

Discount is always calculated on the Marked Price (MP). Selling Price (SP) = MP × (100 − Discount%) ÷ 100.

Profit and Loss

Profit or loss is always calculated on the Cost Price (CP) unless stated otherwise. Profit = SP − CP.

Effective Cost Price

Any additional overheads like transportation, installation fees, or repairs are added to the initial purchase value to determine the final effective Cost Price.

Shortcut Trick

Initial Milk = (7/9) × 90 = 70 L, Water = (2/9) × 90 = 20 L.

Since the same amount 'x' is added to both, the difference between their quantities remains unchanged.

Initial difference = 70 − 20 = 50 L.

New ratio is 5 : 2, so the difference is 5 − 2 = 3 parts, which means 3 parts = 50 L.

Water added (x) = New water (2 parts) − Old water = (100/3) − 20 = 40/3 = 13.33 L.

∴ The correct answer is 13.33 litres.

Alternate Method

Given:

Total mixture = 90 litres

Initial ratio of Milk to Water = 7 : 2

New ratio of Milk to Water after adding 'x' litres of both = 5 : 2

Formula Used:

Ratio = Quantity of Component 1 ÷ Quantity of Component 2

Initial (7:2) Final (5:2)

Milk: 70L

W: 20L

Milk: (70 + x)L

W: (20 + x)L

Adding 'x' litres of both changes ratio to 5 : 2

Calculations:

⇒ Quantity of milk initially = (7/9) × 90 = 70 litres

⇒ Quantity of water initially = (2/9) × 90 = 20 litres

⇒ After adding 'x' litres of each, new Milk = (70 + x) and new Water = (20 + x)

⇒ According to the given condition:

⇒ (70 + x) / (20 + x) = 5 / 2

⇒ Cross-multiplying: 2 × (70 + x) = 5 × (20 + x)

⇒ 140 + 2x = 100 + 5x

⇒ 5x − 2x = 140 − 100

⇒ 3x = 40

⇒ x = 40 / 3 = 13.33 litres

∴ The correct answer is 13.33 litres.

Additional Information

Constant Difference Method

Whenever equal amounts of two components are added or removed from a mixture, the absolute difference between their quantities remains strictly constant. This acts as a powerful shortcut by equating the part difference of the ratios.

Cross-Multiplication Principle

For equations of the form (a + x) / (b + x) = c / d, the difference between the terms translates directly to: x(c − d) = ad − bc, allowing faster mental calculation without writing intermediate steps.

Mixtures & Replacement Concept

If 'y' litres of a mixture is removed and replaced by 'y' litres of one of the components (e.g., water), the ratio of the components in the remaining mixture right before replacement stays exactly identical to the initial original ratio.

Shortcut Trick

The volume of a cylinder that completely circumscribes a sphere is exactly (3/2) times the volume of the sphere.

Ratio of Cylinder Volume to Sphere Volume = 3 : 2.

Vacant Space = Cylinder Volume − Sphere Volume = (3/2 − 1) = 1/2 of Sphere Volume.

Vacant Space = (1/2) × 36π = 18π cm³.

∴ The correct answer is 18π cm³.

Alternate Method

Given:

Volume of the sphere = 36π cm³

Height and diameter of the cylinder = Diameter of the sphere

Formula Used:

Volume of Sphere = (4/3)πr³

Volume of Cylinder = πr²h

r = 3h = 2r = 6d = 2r = 6

Calculations:

⇒ (4/3)πr³ = 36π

⇒ r³ = 36 × 3 ÷ 4

⇒ r³ = 27

⇒ Radius of sphere (r) = 3 cm

⇒ Radius of cylinder (R) = 3 cm

⇒ Height of cylinder (h) = diameter of sphere = 2 × 3 = 6 cm

⇒ Volume of cylinder = πR²h = π × 3² × 6 = 54π cm³

⇒ Volume of vacant space = Volume of cylinder − Volume of sphere

⇒ Vacant space = 54π − 36π = 18π cm³

∴ The correct answer is 18π cm³.

Additional Information

Ratio of Volumes (Cone, Sphere, Cylinder)

If a cone, a sphere, and a cylinder have the same radius (r) and the same height (2r), their volumes will always be exactly in the ratio of 1 : 2 : 3.

Archimedes' Hat-Box Theorem

A sphere inscribed perfectly inside a right circular cylinder has exactly 2/3 the volume and 2/3 the surface area of the cylinder.

Key Mensuration Formulas

Curved Surface Area of Cylinder = 2πrh | Surface Area of Sphere = 4πr² | Volume of Cone = (1/3)πr²h

Shortcut Trick

When a solid is melted and recast into another shape, their volumes remain equal.

Volume of Hemisphere = Volume of Cone

⇒ ²/₃ × π × R³ = ¹/₃ × π × R² × h

⇒ 2 × R = h

∴ The correct answer is 2R.

Alternate Method

Given:

Radius of the hemisphere = R

Radius of the recast cone = R

Formula Used:

Volume of a hemisphere = ²/₃ × π × r³

Volume of a cone = ¹/₃ × π × r² × h

R Hemisphere

Melt & Recast

R

h Cone

Calculations:

⇒ Let the height of the newly formed cone be h.

⇒ Since the hemisphere is simply melted to form the cone, the original material volume completely transforms into the new volume.

⇒ Volume of Hemisphere = Volume of Cone

⇒ ²/₃ × π × R³ = ¹/₃ × π × R² × h

⇒ Canceling the common terms (¹/₃ and π) from both sides:

⇒ 2 × R³ = R² × h

⇒ Dividing both sides by R²:

⇒ h = 2R

∴ The correct answer is 2R.

Additional Information

Melting and Recasting Property

When any 3D solid is melted and recast into another shape, the total volume of the material remains strictly constant, even though dimensions and total surface area typically change.

Volume Relation of Cylinder & Cone

For a given identical radius (r) and height (h), the volume of a cylinder (V = πr²h) is exactly 3 times the volume of a cone (V = ¹/₃πr²h).

Hemisphere Surface Areas

The Curved Surface Area (CSA) of a solid hemisphere is 2πr², while its Total Surface Area (TSA) includes the flat circular base, resulting in 3πr².

Shortcut Trick

When 3D shapes are melted and recast, their total volume is conserved.

For hemispheres, the volume constant (2/3)π cancels out, giving: R³ = r₁³ + r₂³

R³ = 6³ + 8³ = 216 + 512 = 728

Since 9³ = 729, the approximate value of R is 9.

∴ The correct answer is 9 cm.

Alternate Method

Given:

Radius of first hemisphere (r₁) = 6 cm

Radius of second hemisphere (r₂) = 8 cm

Formula Used:

Volume of a hemisphere = (2/3)πr³

r=6

r=8

=

R=?

V1 + V2 = Vnew

Calculations:

Let the radius of the newly formed hemisphere be R.

⇒ Total volume of melted hemispheres = Volume of the new single hemisphere

⇒ (2/3)π(6)³ + (2/3)π(8)³ = (2/3)πR³

⇒ (2/3)π × (6³ + 8³) = (2/3)πR³

Canceling out (2/3)π from both sides:

⇒ 6³ + 8³ = R³

⇒ 216 + 512 = R³

⇒ R³ = 728

⇒ R = (728)^1/3

⇒ We know that 9³ = 729. Therefore, (728)^1/3 ≈ 9 cm.

∴ The correct answer is 9 cm.

Additional Information

Volume of a Sphere

The volume of a full sphere is (4/3)πr³, which is exactly double the volume of a hemisphere.

Surface Area of a Hemisphere

The Curved Surface Area (CSA) of a hemisphere is 2πr². The Total Surface Area (TSA) includes the flat circular base and is 3πr².

Concept of Melting and Recasting

When any 3D object is melted and recast into another shape, the Total Volume remains conserved. The sum of the volumes of the smaller parts directly equals the volume of the newly formed object.

Shortcut Trick

For a regular square pyramid, Vertical Height (h) = √(l² − (a/2)²).

⇒ h = √(10² − 6²) = √(100 − 36) = √64 = 8 cm.

Volume = 1/3 × Base Area × h

⇒ Volume = 1/3 × (12 × 12) × 8 = 48 × 8 = 384 cm³.

∴ The correct answer is 384 cm³.

Alternate Method

Given:

⇒ Side of the square base (a) = 12 cm

⇒ Slant height (l) = 10 cm

Formula Used:

⇒ Volume of a right pyramid = 1/3 × Base Area × Vertical Height

⇒ Vertical Height (h) = √(l² − r²) where r is half of the base side length.

Calculations:

⇒ Base Area = a × a = 12 × 12 = 144 cm²

⇒ Half of the base side length (r) = 12 ÷ 2 = 6 cm

⇒ Applying Pythagoras theorem to find the vertical height:

⇒ h² + r² = l²

⇒ h² + 6² = 10²

⇒ h² + 36 = 100

⇒ h² = 64

⇒ h = 8 cm

⇒ Volume = 1/3 × 144 × 8

⇒ Volume = 48 × 8 = 384 cm³

∴ The correct answer is 384 cm³.

Additional Information

Lateral Surface Area of Pyramid

It is the total area of all the triangular lateral faces. Formula: LSA = 1/2 × Perimeter of Base × Slant Height.

Total Surface Area of Pyramid

It comprises both the lateral faces and the base. Formula: TSA = Lateral Surface Area + Base Area.

Slant Height vs Vertical Height

Vertical height is the perpendicular distance from the apex straight down to the center of the base, whereas slant height is measured along the lateral face from the apex to the midpoint of the base edge.

Shortcut Trick

The ratio of the areas of two circles is equal to the square of the ratio of their radii.

Ratio of areas = 1² : (√2)² = 1 : 2

Here, 1 unit → 100 cm²

Larger tyre area = 2 units → 2 × 100 = 200 cm²

∴ The correct answer is 200 cm².

Alternate Method

r Area = 100 cm²

r√2 Area = ?

Given:

Ratio of radii of two tyres = 1 : √2

Area of the smaller tyre (A₁) = 100 cm²

Formula Used:

Area of a circle = πr²

Calculation:

⇒ Let the radii of the smaller and larger tyres be r and r√2 respectively.

⇒ Area of smaller tyre (A₁) = π × (r)² = πr²

⇒ πr² = 100 cm²

⇒ Area of larger tyre (A₂) = π × (r√2)²

⇒ A₂ = π × 2r² = 2πr²

⇒ Substitute the value of πr² into the equation for A₂:

⇒ A₂ = 2 × 100

⇒ A₂ = 200 cm²

∴ The correct answer is 200 cm².

Additional Information

Circumference of Circle

The circumference of a circle is directly proportional to its radius (C = 2πr). The ratio of circumferences of two circles is the same as the ratio of their radii.

Volume of Spheres

For 3D objects like spheres, the volume is proportional to the cube of the radius (V = 4 ÷ 3 × πr³). The ratio of volumes is r₁³ : r₂³.

Area of a Sector

The area of a sector with a central angle θ in a circle of radius r is given by (θ ÷ 360) × πr².

Shortcut Trick

For any triangle, the angle subtended by the incenter at any two vertices is 90° plus half of the third angle.

⇒ ∠BIC = 90° + (∠BAC / 2)

⇒ 125° = 90° + (∠BAC / 2)

⇒ ∠BAC / 2 = 125° − 90°

⇒ ∠BAC = 35° × 2 = 70°

∴ The correct answer is 70°.

Alternate Method

Given:

The angle bisectors of ∠ABC and ∠ACB meet at the incenter I.

Measure of ∠BIC = 125°

Formula Used:

Sum of interior angles of a triangle = 180°

An angle bisector divides an angle into two equal parts.

A B C I 125° ? (70°)

Calculations:

In ΔBIC, the sum of angles is 180°.

⇒ ∠IBC + ∠ICB + ∠BIC = 180°

Since BI and CI are the angle bisectors of ∠ABC and ∠ACB respectively:

⇒ ∠IBC = ∠ABC / 2

⇒ ∠ICB = ∠ACB / 2

Substituting these into the equation for ΔBIC:

⇒ (∠ABC / 2) + (∠ACB / 2) + 125° = 180°

⇒ (∠ABC + ∠ACB) / 2 = 180° − 125°

⇒ (∠ABC + ∠ACB) / 2 = 55°

⇒ ∠ABC + ∠ACB = 55° × 2 = 110°

Now, in ΔABC, the sum of all interior angles is 180°.

⇒ ∠BAC + ∠ABC + ∠ACB = 180°

⇒ ∠BAC + 110° = 180°

⇒ ∠BAC = 180° − 110°

⇒ ∠BAC = 70°

∴ The correct answer is 70°.

Additional Information

Incenter

The incenter is the point where all three internal angle bisectors of a triangle intersect. The central angle formed by two bisectors is 90° + (∠Opposite Vertex) / 2.

Circumcenter

The intersection of the perpendicular bisectors of the sides. The angle subtended by an arc at the circumcenter is double the angle subtended at any point on the circle (∠BOC = 2 × ∠A).

Orthocenter

The point where the altitudes of the triangle meet. The sum of the angle at the orthocenter and the opposite vertex angle is always 180° (∠BHC + ∠A = 180°).

Excenter

The intersection of the external bisectors of two angles and the internal bisector of the third angle. The angle formed by the external bisectors is 90° − (∠Opposite Vertex) / 2.

Shortcut Trick

In quadrilateral ABCD, we are given AB is parallel to CD and AB = CD.

Since the alternate interior angles formed by transversal AC are equal, ∠BAC = ∠DCA.

For ΔABC and ΔCDA, we evaluate the known equal parts:

⇒ AB = CD (Given) | ∠BAC = ∠DCA (Alternate angles) | AC = CA (Common side)

Since two sides and their included angle are equal, the triangles are congruent by the Side-Angle-Side rule.

∴ The correct answer is ΔABC ≅ ΔCDA by SAS.

Alternate Method

Given:

⇒ Quadrilateral ABCD with diagonal AC.

⇒ AB is parallel to CD.

⇒ AB = CD.

Concept Used:

⇒ If two parallel lines are intersected by a transversal, their alternate interior angles are equal.

⇒ SAS Congruence Rule: If two sides and the included angle of one triangle are equal to the corresponding parts of another triangle, they are congruent.

A B C D

AB = x CD = x

Calculations:

⇒ Consider ΔABC and ΔCDA.

⇒ Since AB is parallel to CD and AC acts as the transversal, the alternate interior angles are exactly equal.

⇒ ∠BAC = ∠DCA

⇒ We are already given that side AB = side CD.

⇒ The diagonal AC is a common side to both triangles, so AC = CA.

⇒ We now have two sides and the included angle equal for both triangles.

⇒ Therefore, ΔABC ≅ ΔCDA by the SAS congruence criterion.

∴ The correct answer is ΔABC ≅ ΔCDA by SAS.

Additional Information

SAS (Side-Angle-Side) Congruence

Two triangles are congruent if two sides and the included angle of one are identically equal to the corresponding parts of the other.

SSS (Side-Side-Side) Congruence

Two triangles are completely congruent if all three sides of one triangle are equal to the corresponding three sides of the other.

ASA (Angle-Side-Angle) Congruence

If two angles and the included side of one triangle are equal to the corresponding parts of another triangle, they are completely congruent.

Shortcut Trick

The centroid of a triangle divides the median in the ratio 2 : 1 from the vertex, so the ratio of lengths AG : AM is 2 : 3.

Since line DE is parallel to base BC, ΔADE and ΔABC are similar triangles.

The ratio of the areas of similar triangles is the square of the ratio of their corresponding sides or medians: (2/3)² = 4/9.

This implies that if the area of ΔADE is 4 units, the area of the entire ΔABC is 9 units.

The area of trapezoid DECB = 9 − 4 = 5 units.

∴ The correct answer is 4:5.

Alternate Method

Given:

Line DE is parallel to side BC.

DE passes through the centroid G of ΔABC.

Formula Used:

For two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding heights or medians.

Area(ΔADE) ÷ Area(ΔABC) = (AG ÷ AM)²

2 parts 1 part

A B C D E M G

Area = 4 units Area = 5 units

Calculations:

⇒ Let AM be the median to the side BC. Since G is the centroid, it divides the median AM in the ratio 2 : 1.

⇒ AG : GM = 2 : 1

⇒ AG ÷ AM = AG ÷ (AG + GM) = 2 ÷ (2 + 1) = 2 / 3

⇒ In ΔADE and ΔABC, since DE || BC, their corresponding angles are equal, meaning ΔADE ∼ ΔABC.

⇒ Ratio of Area(ΔADE) to Area(ΔABC) = (AG ÷ AM)² = (2 / 3)² = 4 / 9

⇒ Let the area of ΔADE be 4x and the area of ΔABC be 9x.

⇒ The area of trapezoid DECB = Area(ΔABC) − Area(ΔADE) = 9x − 4x = 5x.

⇒ The required ratio = Area(ΔADE) : Area(DECB) = 4x : 5x = 4 : 5.

∴ The correct answer is 4:5.

Additional Information

Centroid of a Triangle

The centroid is the point of intersection of the three medians of a triangle. It always divides each median into a 2 : 1 ratio, with the longer part being closest to the vertex.

Similarity Area Theorem

If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides, altitudes, or medians.

Basic Proportionality Theorem (Thales Theorem)

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Shortcut Trick

The length of an internal (transverse) common tangent is given directly by the formula: √[d² − (r₁ + r₂)²]

Here, d = 17, r₁ = 9, and r₂ = 5.

Length = √[17² − (9 + 5)²]

Length = √[289 − 196] = √93 cm

∴ The correct answer is √93 cm.

Alternate Method

d = 17 cm

9 cm 5 cm

Internal Common Tangent

Given:

⇒ Radius of the first circle (r₁) = 9 cm

⇒ Radius of the second circle (r₂) = 5 cm

⇒ Distance between their centers (d) = 17 cm

Formula Used:

⇒ Length of internal common tangent = √[d² − (r₁ + r₂)²]

Calculations:

⇒ Length = √[17² − (9 + 5)²]

⇒ Length = √[17² − 14²]

⇒ Length = √[289 − 196]

⇒ Length = √93 cm

∴ The correct answer is √93 cm.

Additional Information

Internal (Transverse) Common Tangent

It is the tangent that intersects the line segment joining the centers of the two circles. Formula: TCT = √[d² − (r₁ + r₂)²]

External (Direct) Common Tangent

It is the tangent that does not intersect the line segment joining the centers. Formula: DCT = √[d² − (r₁ − r₂)²]

Number of Tangents Rule

Two non-intersecting circles can have a maximum of 4 common tangents (2 direct and 2 transverse). If they touch externally, they have exactly 3 common tangents.

Shortcut Trick For two circles with radii r1 and r2, and distance between centers d, the formula is: Transverse Common Tangent = √(d2 − (r1 + r2)2) TCT = √(202 − (7 + 3)2) TCT = √(400 − 100) = √300 cm ∴ The correct answer is √300 cm.

Alternate Method

C1

C2 r₁=7 r₂=3

d = 20 cm T1 T2

TCT

Given: Radius of first circle (r1) = 7 cm Radius of second circle (r2) = 3 cm Distance between centers (d) = 20 cm Formula Used: Length of Transverse Common Tangent (TCT) = √(d2 − (r1 + r2)2) Calculations: ⇒ TCT = √(202 − (7 + 3)2) ⇒ TCT = √(202 − 102) ⇒ TCT = √(400 − 100) ⇒ TCT = √300 cm ∴ The correct answer is √300 cm.

Additional Information Direct Common Tangent (DCT) The formula for the length of a Direct Common Tangent is: DCT = √(d2 − (r1 − r2)2). Number of Common Tangents If two circles are strictly apart (d > r1 + r2), they have exactly 4 common tangents (2 direct and 2 transverse). Intersection Property The Transverse Common Tangent crosses the line joining the centers of the two circles, dividing the distance internally exactly in the ratio of their radii (r1 : r2).

Shortcut Trick According to circle properties, a shorter chord is always farther away from the center compared to a longer chord. The given chord lengths are 9 cm and 7 cm. Since 7 cm < 9 cm, the 7 cm chord must be farther from the center than the 9 cm chord. The 9 cm chord is at a distance of 4 cm. Therefore, the distance of the 7 cm chord from the center must be greater than 4 cm. ∴ The correct answer is greater than 4 cm.

Alternate Method Given: Length of the first chord = 9 cm Distance of the first chord from center (d1) = 4 cm Length of the second chord = 7 cm Formula Used: By Pythagoras theorem in a right-angled triangle formed by the radius, distance to the chord, and half of the chord: r2 = d2 + (L ÷ 2)2

O

A B 9 cm

C D 7 cm

4 cm d = ?

r r

Calculations: ⇒ For the 9 cm chord: r2 = 42 + (9 ÷ 2)2 ⇒ r2 = 16 + 4.52 ⇒ r2 = 16 + 20.25 = 36.25 ⇒ For the 7 cm chord: r2 = d2 + (7 ÷ 2)2 ⇒ 36.25 = d2 + 3.52 ⇒ 36.25 = d2 + 12.25 ⇒ d2 = 36.25 − 12.25 ⇒ d2 = 24 ⇒ d = √24 = 2√6 ≈ 4.89 cm ⇒ Since 4.89 cm is strictly greater than 4 cm, the distance is greater than 4 cm. ∴ The correct answer is greater than 4 cm.

Additional Information Distance of Chords from Center Of any two chords of a circle, the larger chord is nearer to the center, and the shorter chord is farther from the center. Perpendicular from Center The perpendicular drawn from the center of a circle to a chord always bisects the chord into two equal segments. Equal Chords Property Chords that are equal in length are precisely equidistant from the center of the circle.

Shortcut Trick

By the extended Sine Rule for any circumscribed triangle:

Side / sin(θ) = 2 × Circumradius (r)

Let the required side opposite to the 120° angle be x.

⇒ x / sin(120°) = 2r

⇒ x = 2r × (√3 / 2) = r√3

∴ The correct answer is r√3.

Alternate Method

Given:

Angle of the inscribed triangle, ∠A = 120°

Radius of the circumscribed circle = r

Formula Used:

Opposite angles of a cyclic quadrilateral sum to 180°

Angle at the center = 2 × Angle at the circumference

sin(θ) = Perpendicular / Hypotenuse

O

120° B C A M

r 60°

BC = ?

Calculations:

⇒ Let the required side opposite to the 120° angle be chord BC, and A be the vertex containing the 120° angle.

⇒ Let D be an imaginary point on the major arc. ABDC forms a cyclic quadrilateral.

⇒ Opposite angles sum to 180° ⇒ ∠BDC = 180° − 120° = 60°.

⇒ The angle subtended by chord BC at the center O is ∠BOC = 2 × ∠BDC = 2 × 60° = 120°.

⇒ Draw a perpendicular OM from center O to BC. It bisects both ∠BOC and chord BC.

⇒ In right-angled ΔOMB, ∠BOM = 120° ÷ 2 = 60°.

⇒ sin(60°) = BM / OB

⇒ √3 / 2 = BM / r ⇒ BM = r√3 / 2

⇒ Length of side BC = 2 × BM = 2 × (r√3 / 2) = r√3

∴ The correct answer is r√3.

Additional Information

Sine Rule (Extended Law of Sines)

For any triangle inscribed in a circle of radius R, the ratio of any side to the sine of its opposite angle is constant: a/sin(A) = b/sin(B) = c/sin(C) = 2R.

Angle Subtended by Arc

The angle subtended by an arc at the center is double the angle subtended by it at any remaining part of the circle.

Cyclic Quadrilateral Property

In a cyclic quadrilateral, the sum of opposite interior angles is always 180°. This property helps connect major and minor arc angles.

Shortcut Trick

Convert the standard fractions into their known decimal values directly.

1/8 = 0.125

1/16 = 0.0625

Sum = 0.125 + 0.0625 = 0.1875

∴ The correct answer is 0.1875.

Alternate Method

Given:

Expression = 1/8 + 1/16

Calculation:

⇒ Find the Least Common Multiple (LCM) of 8 and 16, which is 16.

⇒ Multiply numerator and denominator of 1/8 by 2 to make denominators equal.

⇒ Expression = (1 × 2)/(8 × 2) + 1/16

⇒ Expression = 2/16 + 1/16

⇒ Expression = 3/16

⇒ Now, convert the fraction into a decimal by dividing 3 by 16.

⇒ 3 ÷ 16 = 0.1875

∴ The correct answer is 0.1875.

Additional Information

Common Fraction to Decimal Conversions

1/2 = 0.5 | 1/4 = 0.25 | 1/8 = 0.125 | 1/16 = 0.0625. Memorising these standard fraction equivalents significantly speeds up calculations.

Addition of Decimals

Always align the decimal points vertically before adding. You can add trailing zeros to make all numbers have the same number of decimal places (e.g., 0.1250 + 0.0625 = 0.1875).

Simplification Rules

While solving arithmetic expressions involving fractions, always equalise the denominators first by taking their LCM before adding or subtracting the numerators.

Shortcut Trick

In surd square roots √(A + B), split B into 2 × a × b.

Here, 6√3 = 2 × 3 × √3. Let a = 3 and b = √3.

Check if a² + b² = A ⇒ 3² + (√3)² = 9 + 3 = 12. This matches perfectly!

So, the expression is exactly (a + b)² = (3 + √3)².

Taking the square root cancels the square, giving 3 + √3.

∴ The correct answer is 3 + √3.

Alternate Method

Given:

√(12 + 6√3)

Formula Used:

(a + b)² = a² + b² + 2ab

Calculations:

⇒ Let us express the term inside the square root in the form of a perfect square.

⇒ The given expression is 12 + 6√3.

⇒ We can rewrite the middle term 6√3 as 2 × 3 × √3.

⇒ This implies a = 3 and b = √3 in the identity (a + b)².

⇒ Now, let us check a² + b²: 3² + (√3)² = 9 + 3 = 12.

⇒ So, 12 + 6√3 can be written as 3² + (√3)² + 2 × 3 × √3.

⇒ This perfectly matches the expansion of (3 + √3)².

⇒ Therefore, √(12 + 6√3) = √((3 + √3)²).

⇒ The square root and square cancel each other out.

⇒ Simplification gives 3 + √3.

∴ The correct answer is 3 + √3.

Additional Information

Square Root of Surds

To find the square root of a mixed surd A + √B, assume it equals √x + √y. Squaring both sides gives A + √B = x + y + 2√(xy). Then equate rational and irrational parts to solve for x and y.

Algebraic Identities

The identities (a + b)² = a² + b² + 2ab and (a − b)² = a² + b² − 2ab are crucial. If the given expression was √(12 − 6√3), the simplified answer would be 3 − √3.

Conjugate Surds

When surds appear in the denominator, multiply both numerator and denominator by the conjugate (changing the middle sign) to rationalise it: 1 ÷ (a + √b) × (a − √b) ÷ (a − √b).

Shortcut Trick

Using the shift method or blank space filling technique:

W : X : Y : Z = (5 × 3 × 8) : (6 × 3 × 8) : (6 × 7 × 8) : (6 × 7 × 9)

W : X : Y : Z = 120 : 144 : 336 : 378

Simplifying by dividing all ratio terms by their HCF, which is 6:

W : X : Y : Z = 20 : 24 : 56 : 63

∴ The correct answer is 20 : 24 : 56 : 63.

Alternate Method

Given:

W : X = 5 : 6

X : Y = 3 : 7

Y : Z = 8 : 9

W X Y Z

5 6

6

6

3 3 7

7

8

8 8 9

120 : 144 : 336 : 378

20 : 24 : 56 : 63

(÷ 6)

Calculations:

⇒ First, equate the common term X in the first two ratios by making them equal to the LCM of 6 and 3.

⇒ W : X = 5 : 6

⇒ X : Y = 3 : 7 = (3 × 2) : (7 × 2) = 6 : 14

⇒ Combining W, X, and Y gives W : X : Y = 5 : 6 : 14

⇒ Now, equate the common term Y with the third ratio using the LCM of 14 and 8, which is 56.

⇒ Y : Z = 8 : 9

⇒ W : X : Y = (5 × 4) : (6 × 4) : (14 × 4) = 20 : 24 : 56

⇒ Y : Z = (8 × 7) : (9 × 7) = 56 : 63

⇒ By combining all four variables, we obtain W : X : Y : Z = 20 : 24 : 56 : 63

∴ The correct answer is 20 : 24 : 56 : 63.

Additional Information

Compound Ratio

When two or more ratios are multiplied together, it is known as a compound ratio. For ratios a : b and c : d, the compound ratio is ac : bd.

Duplicate and Sub-duplicate Ratio

The duplicate ratio of a : b is determined as a² : b², whereas its sub-duplicate ratio is expressed as √a : √b.

Proportionality Rule

If four numbers a, b, c, d are in a perfect proportion (a : b :: c : d), then the product of extremes is always equal to the product of means (ad = bc).

Shortcut Trick

For an infinite series of the form √(a + √(a + √(a + ...))), we find two consecutive integers whose product is a.

Here, a = 6, which can be factored into consecutive integers as 2 × 3.

Since the series has a positive (+) sign throughout, the answer will be the larger integer.

The larger integer between 2 and 3 is 3.

∴ The correct answer is 3.

Alternate Method

Given:

The infinite nested radical expression is √(6 + √(6 + √(6 + ...∞)))

Calculations:

⇒ Let x = √(6 + √(6 + √(6 + ...∞)))

⇒ Squaring both sides of the equation, we get:

⇒ x² = 6 + √(6 + √(6 + ...∞))

⇒ Since the expression goes to infinity, the repeating part is equal to x:

⇒ x² = 6 + x

⇒ x² − x − 6 = 0

⇒ x² − 3x + 2x − 6 = 0

⇒ x(x − 3) + 2(x − 3) = 0

⇒ (x − 3)(x + 2) = 0

⇒ x = 3 or x = −2

⇒ Since the value of a positive square root cannot be negative, we take x = 3.

∴ The correct answer is 3.

Additional Information

Negative Infinite Nested Radical

For a series √(a − √(a − √(a − ...))), if a can be written as the product of two consecutive integers, the answer is the smaller integer. Example: √(6 − √(6 − ...)) = 2.

Multiplicative Infinite Nested Radical

For an infinite series under multiplication like √(a × √(a × √(a × ...))), the final value evaluates exactly to a. Example: √(6 × √(6 × ...)) = 6.

Non-Factorable Infinite Radical

If the term a cannot be factored into two consecutive integers, the exact value of √(a + √(a + ...)) is found using the direct formula (√(4a + 1) + 1) ÷ 2.

Shortcut Trick

Let the Marked Price (MP) of 1 shirt be 100.

Total Cost Price (CP) for 7 shirts = MP of 6 shirts = 600.

Total Selling Price (SP) for 7 shirts after 4% discount = 7 × 96 = 672.

Profit = SP − CP = 672 − 600 = 72.

Profit % = (72 ÷ 600) × 100 = 12%.

∴ The correct answer is 12%.

Alternate Method

Given:

Number of shirts purchased = 7

Cost Price of 7 shirts = Marked Price of 6 shirts

Discount on Marked Price = 4%

Formula Used:

Selling Price (SP) = Marked Price (MP) − Discount

Profit % = (Profit ÷ Cost Price) × 100

Cost, Selling & Marked Price Relationship

CP = 600 (7 shirts)

SP = 672 (7 shirts)

MP = 700 (7 shirts)

• Profit: 72 (12%)

• Discount: 28 (4%)

Calculation:

Let the Marked Price (MP) of 1 shirt be ₹100.

⇒ MP of 7 shirts = 7 × 100 = ₹700

⇒ Total CP of 7 shirts = MP of 6 shirts = 6 × 100 = ₹600

⇒ Discount applied on each shirt = 4% of ₹100 = ₹4

⇒ Selling Price (SP) of 1 shirt = 100 − 4 = ₹96

⇒ Total SP of 7 shirts = 7 × 96 = ₹672

⇒ Profit earned = SP − CP = 672 − 600 = ₹72

⇒ Profit % = (72 ÷ 600) × 100

⇒ Profit % = 12%

∴ The correct answer is 12%.

Additional Information

Marked Price & Discount

Discount is an allowance given to the buyer and is strictly calculated on the Marked Price (MP), not the Cost Price (CP).

Formula: SP = MP × (100 − Discount%) ÷ 100

Profit & Loss Basics

Profit and Loss percentages indicate the financial gain or loss relative to the cost. Both are always calculated on the Cost Price (CP) unless stated otherwise.

Formula: Profit% = (Profit ÷ CP) × 100

Markup Percentage

Markup is the additional percentage by which the Cost Price is increased to set the Marked Price (label price) of an item.

Formula: MP = CP × (100 + Markup%) ÷ 100

Shortcut Trick

Total capacity = LCM(16, 24) = 48 units.

Initial efficiency: A = 3 units/min, B = 2 units/min ⇒ Total = 5 units/min.

Work completed in 4 mins = 5 × 4 = 20 units ⇒ Remaining = 48 − 20 = 28 units.

New efficiency: A = 6 (doubled), B = 1 (halved) ⇒ New total = 7 units/min.

Time for remaining work = 28 ÷ 7 = 4 min.

∴ The correct answer is 4 min.

Alternate Method

Given:

Time taken by Pipe A = 16 minutes

Time taken by Pipe B = 24 minutes

Calculations:

⇒ Initial combined efficiency of A and B = 3 + 2 = 5 units/min

⇒ Work completed by both pipes in the first 4 minutes = 5 × 4 = 20 units

⇒ Remaining work to be done = 48 − 20 = 28 units

⇒ After 4 minutes, the rate of Pipe A doubles ⇒ New efficiency of A = 3 × 2 = 6 units/min

⇒ Rate of Pipe B becomes half ⇒ New efficiency of B = 2 ÷ 2 = 1 unit/min

⇒ New combined efficiency of both pipes = 6 + 1 = 7 units/min

⇒ Time required to complete the remaining work = Remaining work ÷ New combined efficiency

⇒ Time required = 28 ÷ 7 = 4 minutes

∴ The correct answer is 4 min.

Additional Information

Inlet vs Outlet Pipes

An inlet pipe fills the tank (taken as positive efficiency), while an outlet pipe or leak empties it (taken as negative efficiency). Net efficiency is the algebraic sum of all operating pipes.

Proportional Efficiency

The efficiency of a pipe is inversely proportional to the time it takes to fill the tank independently. If a pipe's efficiency is doubled, the time it takes to do the same work is halved.

Varying Efficiencies

Whenever efficiency changes mid-way, always calculate the exact total work completed using the old efficiency first, and then apply the new efficiency strictly to the remaining units of work.