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Shortcut Trick

Let Cost Price (CP) be 100%.

Marked Price (MP) = 180% (since markup is 80% above CP).

Selling Price (SP) after 30% discount = 180% × 0.7 = 126%.

With an additional profit of 12%, the actual SP = CP + 12% profit = 112%.

The difference in SP represents the loyalty discount: 126% − 112% = 14%.

CP (100%) = (₹250 ÷ 14) × 100 ≈ ₹1,785.71.

∴ The correct answer is ₹1,785.71.

Alternate Method

CP

SP

MP

Markup: +80%

Discount: -30% & -₹250

Profit: +12%

Given:

Markup = 80%

Discounts = 30% and an additional loyalty discount = ₹250

Profit = 12%

Formula Used:

MP = CP × (1 + Markup% ÷ 100)

SP = MP × (1 − Discount% ÷ 100) − Loyalty Discount

SP = CP × (1 + Profit% ÷ 100)

Calculations:

⇒ Let the Cost Price (CP) be 100x.

⇒ MP = 100x × 1.80 = 180x

⇒ SP after standard 30% discount = 180x × (1 − 0.30) = 180x × 0.70 = 126x

⇒ Final SP = 126x − 250

⇒ According to the profit margin, SP = 100x × 1.12 = 112x

⇒ Equating the selling prices: 126x − 250 = 112x

⇒ 126x − 112x = 250

⇒ 14x = 250

⇒ x = 250 ÷ 14 = 125 ÷ 7

⇒ CP = 100x = 100 × (125 ÷ 7) = 12500 ÷ 7

⇒ CP ≈ ₹1,785.71

∴ The correct answer is ₹1,785.71.

Additional Information

Marked Price & Discount

Discount is always calculated on the Marked Price (MP) unless specified otherwise. Net Selling Price = MP − Discount.

Successive Discounts

If two successive discounts of a% and b% are given, the equivalent single discount is [a + b − (a × b)÷100]%.

Direct CP to MP Ratio

When Profit% (P) and Discount% (D) are given, their relationship is simply derived as CP ÷ MP = (100 − D) ÷ (100 + P).

Shortcut Trick

By Archimedes' principle, the volume of water displaced by a completely submerged object is exactly equal to its own volume.

The cylinder's dimensions are extra information just to confirm the ball fits entirely inside.

Volume of water displaced = Volume of the sphere = 4/3 × π × r³

⇒ Volume = 4/3 × π × 4³ = 256/3 π cm³.

∴ The correct answer is 256/3 π cm³.

Alternate Method

r = 4 cm

R = 5 cm

h = 30 cm

Displaced Volume = Volume of Submerged Sphere

Given:

Radius of the cylindrical container (R) = 5 cm

Height of the cylindrical container (h) = 30 cm

Radius of the spherical ball (r) = 4 cm

Formula Used:

Volume of a sphere = 4/3 × π × r³

Calculations:

⇒ When an object is completely immersed in water, the volume of water displaced is equal to its own volume.

⇒ The spherical ball has a diameter of 8 cm, which is less than the cylinder's 10 cm diameter. It fits completely inside the container.

⇒ Volume of water displaced = Volume of the spherical ball

⇒ Volume = 4/3 × π × (4)³

⇒ Volume = 4/3 × π × 64

⇒ Volume = 256/3 π cm³

∴ The correct answer is 256/3 π cm³.

Additional Information

Archimedes' Principle

When an object is fully immersed in a liquid, it displaces a volume of liquid exactly equal to its own total volume, regardless of the container's shape or size.

Identifying Redundant Data

In competitive exam mensuration problems, dimensions of the outer container are often provided merely to act as distractors or verify physical constraints. Only extract the required dimensions for your calculation.

Volume of a Cylinder

If required to calculate container capacity, the volume of a cylinder is found using the formula V = π × r² × h.

Shortcut Trick

Radius of the dome (r) = 20 ÷ 2 = 10 m

Curved Surface Area (CSA) of Hemisphere = 2 × π × r²

CSA ≈ 2 × 3.14 × (10)²

CSA ≈ 628 m²

∴ The correct answer is 628 m².

Alternate Method

Given:

Diameter of the circular dome area = 20 m

Radius of the dome (r) = 20 ÷ 2 = 10 m

r = 10 m 10 m

Curved Surface Area = 2 × π × r²

Diameter = 20 m

Formula Used:

Curved Surface Area (CSA) of a Hemisphere = 2 × π × r²

Calculations:

⇒ CSA = 2 × 3.14 × (10)²

⇒ CSA = 2 × 3.14 × 100

⇒ CSA ≈ 628 m²

∴ The correct answer is 628 m².

Additional Information

Total Surface Area of Hemisphere

Includes the curved part and the flat circular base. Formula: TSA = 3 × π × r².

Volume of a Hemisphere

Exactly half the volume of a complete sphere. Formula: Volume = (2 ÷ 3) × π × r³.

Surface Area of a Full Sphere

A hemisphere's curved area is exactly half of a full sphere's area. Formula: SA = 4 × π × r².

Shortcut Trick

The curved surface area of a hemisphere is directly proportional to the square of its radius.

Ratio of outer to inner curved surface area = R² : r²

Given R = 2r, the ratio becomes (2r)² : r²

Ratio = 4r² : r² = 4 : 1

∴ The correct answer is 4 : 1.

Alternate Method

Given:

Inner radius = r

Outer radius = R = 2r

r

R = 2r

Center O Hollow Hemisphere Cross-Section

Formula Used:

Curved Surface Area (CSA) of a hemisphere = 2π × (radius)²

Calculations:

⇒ Outer CSA = 2πR²

⇒ Inner CSA = 2πr²

⇒ Ratio = (Outer CSA) ÷ (Inner CSA)

⇒ Ratio = 2πR² ÷ 2πr²

⇒ Ratio = R² ÷ r²

Substitute R = 2r into the equation:

⇒ Ratio = (2r)² ÷ r²

⇒ Ratio = 4r² ÷ r²

⇒ Ratio = 4 : 1

∴ The correct answer is 4 : 1.

Additional Information

Total Surface Area of a Hollow Hemisphere

TSA = Outer CSA + Inner CSA + Area of base ring = 2πR² + 2πr² + π(R² − r²) = π(3R² + r²).

Volume of a Hollow Hemisphere

Volume = (2 ÷ 3)π(R³ − r³), where R is the outer radius and r is the inner radius.

Solid Hemisphere Properties

For a solid hemisphere of radius r: CSA = 2πr², TSA = 3πr², Volume = (2 ÷ 3)πr³.

Shortcut Trick

For a square pyramid, Lateral Surface Area (LSA) = 2 × base side × slant height = 2al

Base Area = a²

Given LSA = 2 × Base Area

⇒ 2al = 2a² ⇒ l = a

Since the base side (a) is 6 cm, the slant height (l) is also 6 cm.

∴ The correct answer is 6 cm.

Alternate Method

Given:

Base side (a) = 6 cm

Lateral Surface Area = 2 × Base Area

Formula Used:

Area of square base = a²

Lateral Surface Area of square pyramid = 4 × (1/2 × base × slant height) = 2 × a × l

Base side (a) = 6 cm Slant height (l) Height (h) Square Pyramid

Calculations:

⇒ Base Area = 6² = 36 cm²

⇒ Lateral Surface Area = 2 × 36 = 72 cm²

⇒ 2 × a × l = 72

⇒ 2 × 6 × l = 72

⇒ 12l = 72

⇒ l = 72 ÷ 12 = 6 cm

∴ The correct answer is 6 cm.

Additional Information

Total Surface Area (TSA)

The TSA of a square pyramid is the sum of its base area and lateral surface area. Formula: TSA = a² + 2al

Volume of Square Pyramid

The volume is one-third of the product of the base area and the vertical height. Formula: V = (1/3) × a² × h

Relation between Heights

Using the Pythagorean theorem inside the pyramid, the slant height (l), vertical height (h), and base side (a) form a right-angled triangle: l² = h² + (a/2)²

Shortcut Trick

The required fencing length is simply 75% of the total circumference.

Required length = 75% × π × D

⇒ Length = (3/4) × (22/7) × 28

⇒ Length = 3 × 22 = 66 m

∴ The correct answer is 66 m.

Alternate Method

r = 14 m Fenced Path = 75% of Circumference Unfenced (25%)

Given:

Diameter of the circular garden (D) = 28 m

Fraction of the garden to be fenced = 75% = 75/100 = 3/4

Formula Used:

Circumference of a circle (C) = π × D

Calculations:

⇒ Total Circumference (C) = (22/7) × 28

⇒ C = 22 × 4 = 88 m

⇒ The gardener wants to fence 75% of the garden circumference.

⇒ Length of fencing needed = 75% of 88 m

⇒ Length of fencing needed = (3/4) × 88

⇒ Length of fencing needed = 3 × 22 = 66 m

∴ The correct answer is 66 m.

Additional Information

Area of a Circle

The area of a circle is the space enclosed within its boundary, calculated as Area = π × r².

Length of an Arc

An arc is a continuous portion of the circumference. Its length is given by Arc Length = (θ ÷ 360) × 2 × π × r, where θ is the central angle.

Area of a Sector

A sector is a portion of the circle enclosed by two radii and an arc. The formula is Area of Sector = (θ ÷ 360) × π × r².

Shortcut Trick

In an equilateral triangle, the circumcenter, incenter, and centroid all coincide at the exact same point.

The centroid divides the median (which is also the altitude) in the ratio 2:1.

The longer segment from the vertex to the centroid is the circumradius (R).

The shorter segment from the centroid to the base is the inradius (r).

Thus, the ratio of circumradius to inradius (R:r) is directly 2:1 without any calculation.

∴ The correct answer is 2:1.

Alternate Method

Given:

Side length of the equilateral triangle = a

R (Circumradius) r (Inradius) Side = a A B C

Formulas Used:

Circumradius (R) of an equilateral triangle = a ÷ √3

Inradius (r) of an equilateral triangle = a ÷ 2√3

Calculations:

⇒ Ratio = R : r

⇒ R : r = (a ÷ √3) : (a ÷ 2√3)

⇒ R : r = 1 : (1 ÷ 2)

⇒ Multiplying both sides by 2

⇒ R : r = 2 : 1

∴ The correct answer is 2:1.

Additional Information

Area of Equilateral Triangle

The area is given by the formula: (√3 ÷ 4) × a².

Altitude of Equilateral Triangle

The height or altitude (h) is given by: h = (√3 ÷ 2) × a.

Coinciding Centers

In an equilateral triangle, all four key centers — the centroid, orthocenter, circumcenter, and incenter — merge and lie at the exact same point.

Shortcut Trick

By the Tangent-Secant Theorem: AD² = BD × CD

⇒ AD² = (BC + CD) × CD

⇒ 6² = (BC + 4) × 4

⇒ 36 = 4 × BC + 16

⇒ 4 × BC = 20 ⇒ BC = 5

∴ The correct answer is 5 cm.

Alternate Method

Given:

Length of tangent segment AD = 6 cm

Length of outer secant segment CD = 4 cm

Let the length of side BC be x cm.

O A B C D 6 cm 4 cm x

Formula Used:

Tangent-Secant Theorem: If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the length of the tangent segment is equal to the product of the lengths of the external secant segment and the entire secant segment.

⇒ AD² = CD × BD

Calculations:

From the figure, the entire secant segment BD is the sum of the side BC and the external part CD.

⇒ BD = BC + CD

⇒ BD = x + 4

Substitute the given values into the Tangent-Secant theorem equation:

⇒ 6² = 4 × (x + 4)

⇒ 36 = 4 × (x + 4)

⇒ x + 4 = 36 ÷ 4

⇒ x + 4 = 9

⇒ x = 9 − 4

⇒ x = 5 cm

∴ The correct answer is 5 cm.

Additional Information

Intersecting Secants Theorem

When two secants are drawn to a circle from an external point, the product of the entire secant and its external part is equal for both secants: AP × BP = CP × DP.

Intersecting Chords Theorem

When two chords intersect inside a circle, the product of their respective segments is equal: AE × EC = BE × ED.

Alternate Segment Theorem

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment of the circle.

Shortcut Trick

In a trapezoid ABCD with AB || CD, the intersection of diagonals forms two similar triangles along the parallel sides (▵ABE ∼ ▵CDE).

For any two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

⇒ Area(▵ABE) ÷ Area(▵CDE) = (AB ÷ CD)²

∴ The correct answer is The ratio of AB to CD squared.

Alternate Method

ABCDE

Given:

Trapezoid ABCD with side AB parallel to side CD.

Diagonals AC and BD intersect at point E.

Formula Used:

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

Calculations:

Consider ▵ABE and ▵CDE:

⇒ ∠EAB = ∠ECD (Alternate interior angles, AB || CD)

⇒ ∠EBA = ∠EDC (Alternate interior angles, AB || CD)

⇒ ∠AEB = ∠CED (Vertically opposite angles)

⇒ Since all three corresponding angles are equal, ▵ABE ∼ ▵CDE by the AAA similarity criterion.

⇒ According to the area theorem for similar triangles:

⇒ Area(▵ABE) ÷ Area(▵CDE) = (AB ÷ CD)²

∴ The correct answer is The ratio of AB to CD squared.

Additional Information

Area of Similar Triangles

When two triangles are similar, the ratio of their areas is strictly equal to the square of the ratio of their corresponding side lengths: Area(▵1) ÷ Area(▵2) = (s₁ ÷ s₂)².

Area of a Trapezoid

The area of a trapezoid is equal to half the product of the sum of parallel sides and the perpendicular height between them: Area = ¹÷₂ × (AB + CD) × h.

Isosceles Trapezoid Properties

In an isosceles trapezoid, the non-parallel sides are equal, both pairs of base angles are equal, and the diagonals are exactly equal in length.

Mid-Segment of a Trapezoid

The segment connecting the midpoints of the non-parallel sides is parallel to the bases, and its length is the arithmetic average of the two parallel bases.

Shortcut Trick

When two circles with radii r₁ and r₂ touch externally, the length of their Direct Common Tangent (DCT) is given directly by the formula 2√(r₁ × r₂).

Given, r₁ = 2x and r₂ = x.

DCT = 2√(2x × x)

DCT = 2√(2x²) = 2x√2

∴ The correct answer is 2x√2.

Alternate Method

C1C2AB2xx3xDCT

Given:

Radius of the first circle, r₁ = 2x

Radius of the second circle, r₂ = x

Since the circles touch each other externally, the distance between their centers (d) is equal to the sum of their radii.

Formula Used:

Distance between centers, d = r₁ + r₂

Length of Direct Common Tangent (DCT) = √[d² − (r₁ − r₂)²]

Calculations:

⇒ d = 2x + x = 3x

⇒ DCT = √[(3x)² − (2x − x)²]

⇒ DCT = √[9x² − x²]

⇒ DCT = √(8x²)

⇒ DCT = 2x√2

∴ The correct answer is 2x√2.

Additional Information

Transverse Common Tangent (TCT)

The length of the Transverse Common Tangent is calculated using the formula: TCT = √[d² − (r₁ + r₂)²].

Number of Common Tangents

If two circles touch each other externally, they have exactly 3 common tangents (2 direct and 1 transverse). If they intersect, they have 2 direct common tangents.

Direct Common Tangent (DCT)

A tangent that touches two separate circles and does not cross the line segment connecting their centers is known as a Direct Common Tangent.