Search Previous Year Questions

Shortcut Trick

Intermediate Price after 1^st discount =  ₹20,000 × 85/100 = ₹17,000

Value of 2^nd discount = ₹17,000 − ₹15,300 = ₹1,700

Percentage of 2^nd discount = (1700 ÷ 17000) × 100 = 10%

∴ The correct answer is 10%.

Alternate Method

Given:

Marked Price (MP) = ₹20,000

First discount = 15%

Final Selling Price (SP) = ₹15,300

Formula Used:

Price after discount = MP × (100 − D%)/100

Discount % = (Discount Amount ÷ Base Price) × 100

₹20,000 Marked Price

− 15%

₹17,000

2^nd Discount (−₹1,700)

₹15,300 Selling Price

Calculations:

⇒ Price after 1^st discount = 20000 × (100 − 15)/100

⇒ Price after 1^st discount = 20000 × 85/100 = ₹17,000

⇒ 2^nd discount amount = Price after 1^st discount − Final SP

⇒ 2^nd discount amount = 17000 − 15300 = ₹1,700

⇒ 2^nd discount percentage = (1700 ÷ 17000) × 100

⇒ 2^nd discount percentage = 1/10 × 100 = 10%

∴ The correct answer is 10%.

Additional Information

Equivalent Successive Discount

For two successive discounts of a% and b%, the single equivalent discount formula is (a + b − ab/100)%.

Marked Price and Cost Price Relation

MP/CP = (100 + Profit%) ÷ (100 − Discount%). This directly gives the ratio of Marked Price to Cost Price without knowing the Selling Price.

Multiplier Method for Successive Discounts

Final Selling Price = Marked Price × (1 − d₁/100) × (1 − d₂/100) × ...

Shortcut Trick

Let the Cost Price (CP) be 100.

Since the product is marked 30% above CP, Marked Price (MP) = 130.

With a profit of 17%, Selling Price (SP) = 117.

Discount = MP − SP = 130 − 117 = 13.

Discount % = (13 ÷ 130) × 100 = 10%.

∴ The correct answer is 10%.

Alternate Method

Given:

Mark-up percentage = 30%

Profit percentage = 17%

Formula Used:

MP = CP × (1 + Mark-up% ÷ 100)

SP = CP × (1 + Profit% ÷ 100)

Discount % = [(MP − SP) ÷ MP] × 100

CP: 100

SP: 117

MP: 130

Mark Up: +30%

Profit: +17%

Discount: ?

Calculations:

Let the Cost Price (CP) of the product be 100.

⇒ Marked Price (MP) = 100 × (1 + 30 ÷ 100) = 130

⇒ Selling Price (SP) = 100 × (1 + 17 ÷ 100) = 117

⇒ Discount given = MP − SP = 130 − 117 = 13

⇒ Rate of Discount = (Discount ÷ MP) × 100

⇒ Rate of Discount = (13 ÷ 130) × 100 = 10%

∴ The correct answer is 10%.

Additional Information

Direct Relation Between CP and MP

There is a direct ratio relationship connecting Cost Price, Marked Price, Profit% and Discount%: CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%).

Successive Discounts

If two successive discounts of a% and b% are offered, the single equivalent discount is calculated as: (a + b − a × b ÷ 100)%.

Dishonest Dealer Concept

When a dealer cheats by using false weights while selling at CP, the profit percentage is calculated as: (Error ÷ True Value) × 100.

Shortcut Trick

Alcohol from Container A = (4 ÷ 5) × 10 = 8 L and Water from A = 10 − 8 = 2 L

Alcohol from Container B = (5 ÷ 7) × 14 = 10 L and Water from B = 14 − 10 = 4 L

Total Alcohol in new mixture = 8 + 10 = 18 L

Total Water in new mixture = 2 + 4 = 6 L

Required Ratio of Alcohol to Water = 18 : 6 = 3 : 1

∴ The correct answer is 3:1.

Alternate Method

Given:

Ratio of alcohol to water in Container A = 4:1

Ratio of alcohol to water in Container B = 5:2

Mixture drawn from Container A = 10 liters

Mixture drawn from Container B = 14 liters

Formula Used:

Quantity of component = (Ratio of component ÷ Total ratio parts) × Total quantity

Container A 10 L (Ratio 4:1) Alc: 8L | Wat: 2L

Container B 14 L (Ratio 5:2) Alc: 10L | Wat: 4L

New Mixture Alc: 18L | Wat: 6L Ratio → 3:1

Calculations:

⇒ In Container A, the sum of ratio parts = 4 + 1 = 5

⇒ Quantity of alcohol drawn from A = (4 ÷ 5) × 10 = 8 liters

⇒ Quantity of water drawn from A = (1 ÷ 5) × 10 = 2 liters

⇒ In Container B, the sum of ratio parts = 5 + 2 = 7

⇒ Quantity of alcohol drawn from B = (5 ÷ 7) × 14 = 10 liters

⇒ Quantity of water drawn from B = (2 ÷ 7) × 14 = 4 liters

⇒ Total alcohol in the third container = 8 + 10 = 18 liters

⇒ Total water in the third container = 2 + 4 = 6 liters

⇒ Ratio of alcohol to water in the new mixture = 18 : 6

⇒ Simplifying the ratio by dividing both sides by 6, we get 3 : 1

∴ The correct answer is 3:1.

Additional Information

Multiple Mixtures Concept

When components from multiple different mixtures are mixed together, individually calculate the absolute quantities of each component from every source. Sum these respective quantities up to find the final total amount before constructing the new ratio.

Alligation Method

Used to find the ratio in which two mixtures must be mixed to obtain a desired concentration. The fundamental rule relates the quantities: (Quantity of Cheaper) ÷ (Quantity of Dearer) = (Price of Dearer − Mean Price) ÷ (Mean Price − Price of Cheaper).

Replacement Method

If a certain volume 'y' is replaced 'n' times from a total pure volume 'x', the final quantity of the original pure substance left is given by the formula: Final Quantity = Initial Quantity × (1 − y/x)ⁿ.

Shortcut Trick

For any 3D shape, if its linear dimensions (like radius) are multiplied by a factor 'n', its surface area becomes n² times and volume becomes n³ times.

Here, the radius is doubled, meaning the scaling factor n = 2.

Ratio of New Surface Area to Original = 2² : 1 = 4 : 1

Ratio of New Volume to Original = 2³ : 1 = 8 : 1

∴ The correct answer is 4:1 and 8:1.

Alternate Method

r Original Sphere SA = 4×π×r²

2r New Sphere SA = 16×π×r²

Given:

Let the original radius of the sphere be r.

The new radius of the sphere is doubled, so new radius = 2r.

Formula Used:

Surface Area of a Sphere = 4 × π × radius²

Volume of a Sphere = 4/3 × π × radius³

Calculations:

⇒ Original Surface Area = 4 × π × r²

⇒ New Surface Area = 4 × π × (2r)² = 16 × π × r²

⇒ Ratio of New Surface Area to Original = (16 × π × r²) : (4 × π × r²) = 4 : 1

⇒ Original Volume = 4/3 × π × r³

⇒ New Volume = 4/3 × π × (2r)³ = 4/3 × π × 8r³ = 8 × (4/3 × π × r³)

⇒ Ratio of New Volume to Original = 8 × (4/3 × π × r³) : (4/3 × π × r³) = 8 : 1

∴ The correct answer is 4:1 and 8:1.

Additional Information

Scaling Principle

For any 3D geometric solid, if all its linear dimensions are scaled by a factor 'k', its total surface area directly increases by a factor of k², and its volume increases by a factor of k³.

Hemisphere Volume

When a sphere is cut exactly in half, it forms a hemisphere. The volume of a solid hemisphere is given by 2/3 × π × r³.

Hemisphere Surface Area

The Curved Surface Area (CSA) of a hemisphere is 2 × π × r², whereas the Total Surface Area (TSA), which includes the flat circular base, is 3 × π × r².

Shortcut Trick

Since the hemisphere and the cone have equal volumes and share the same base radius R, we directly equate their volume formulas.

⇒ (2/3) × πR³ = (1/3) × πR²h

⇒ 2R = h

∴ The correct answer is 2R.

Alternate Method

Given:

Volume of Hemisphere = Volume of Cone

Radius of Hemisphere = Radius of Cone = R

R Hemisphere V = (2/3)×π×R³

R

h Cone V = (1/3)×π×R²h

=

Formula Used:

Volume of a Hemisphere = (2/3) × πr³

Volume of a Cone = (1/3) × πr²h

Calculation:

Equating the volume of the hemisphere and the cone:

⇒ (2/3) × πR³ = (1/3) × πR²h

Multiplying both sides by 3 to remove the fraction:

⇒ 2 × πR³ = πR²h

Dividing both sides by πR²:

⇒ 2R = h

⇒ h = 2R

∴ The correct answer is 2R.

Additional Information

Cylinder vs Cone Relationship

A cylinder with base radius r and height h has a volume of V = πr²h. Its volume is exactly 3 times that of a cone with the identical base and height.

Sphere vs Hemisphere Volume

A complete sphere has a volume of V = (4/3)πr³, while a hemisphere (which is half a sphere) has a volume of V = (2/3)πr³.

Surface Area of a Cone

The total surface area of a cone is calculated as TSA = πr(r + l), where l is the slant height given by l = √(r² + h²).

Shortcut Trick

When a hemisphere is melted into 2 equal cones, their total volumes remain equal.

Volume of Hemisphere = Volume of 2 Cones

2/3 × π × R³ = 2 × (1/3 × π × r² × h)

⇒ R³ = r² × h

⇒ 10³ = 5² × h ⇒ 1000 = 25 × h ⇒ h = 40

∴ The correct answer is 40 cm.

Alternate Method

Given:

Radius of the solid hemisphere (R) = 10 cm

Radius of each newly formed cone (r) = 5 cm

Number of cones formed = 2

Formula Used:

Volume of a hemisphere = 2/3 × π × R³

Volume of a cone = 1/3 × π × r² × h

R = 10 Hemisphere

Melted

r = 5 h = ?

r = 5 h = ?

Two Equal Cones

Calculations:

⇒ According to the problem, Total Volume of 2 Cones = Volume of the Hemisphere

⇒ 2 × (1/3 × π × r² × h) = 2/3 × π × R³

⇒ 2/3 × π × (5)² × h = 2/3 × π × (10)³

⇒ Canceling (2/3 × π) from both sides:

⇒ 25 × h = 1000

⇒ h = 1000 ÷ 25

⇒ h = 40 cm

∴ The correct answer is 40 cm.

Additional Information

Principle of Recasting

When a solid object is melted and recast into new shapes, the total volume always remains constant regardless of the new shape forms.

Hemisphere Formulas

Volume = 2/3 × π × r³ | Curved Surface Area = 2 × π × r² | Total Surface Area = 3 × π × r²

Cone Formulas

Volume = 1/3 × π × r² × h | Slant height (l) = √(r² + h²) | Curved Surface Area = π × r × l

Shortcut Trick

For any regular pyramid, the lateral surface area can be calculated directly using its base perimeter and slant height.

Lateral Surface Area = ½ × Base Perimeter × Slant Height

⇒ LSA = ½ × 40 × 10

⇒ LSA = 200 cm²

∴ The correct answer is 200 cm².

Alternate Method

Given:

Base perimeter = 40 cm

Slant height (l) = 10 cm

l = 10 cm a = 10 cm Base Perimeter = 40 cm

Formula Used:

Area of one lateral triangle = ½ × Base × Slant height

Side of square base (a) = Perimeter ÷ 4

Calculation:

⇒ Side of the base (a) = 40 ÷ 4 = 10 cm

⇒ A square pyramid has 4 identical lateral triangular faces.

⇒ Area of one triangular face = ½ × 10 × 10

⇒ Area of one triangular face = 50 cm²

⇒ Total Lateral Surface Area = 4 × Area of one face

⇒ Total LSA = 4 × 50 = 200 cm²

∴ The correct answer is 200 cm².

Additional Information

Total Surface Area (TSA)

The TSA of a regular pyramid is the sum of its Lateral Surface Area and Base Area. For a square pyramid of base side 'a': TSA = LSA + a².

Volume of a Pyramid

The volume of any pyramid is one-third of the product of its base area and true vertical height. V = ⅓ × Base Area × h.

Relation between Heights

For a regular square pyramid, the true vertical height (h), slant height (l), and base side (a) form a right-angled triangle. By Pythagoras theorem: l² = h² + (a/2)².

Shortcut Trick

Ratio of areas = (Ratio of radii)²

Area ratio = 2² : 3² = 4 : 9

Given 4 units = 100π cm² ⇒ 1 unit = 25π cm²

Area of larger pizza = 9 units = 9 × 25π = 225π cm²

∴ The correct answer is 225π cm².

Alternate Method

Given:

Ratio of radii of two circular pizzas, r₁ : r₂ = 2 : 3

Area of smaller pizza = 100π cm²

Formula Used:

Area of a circle = πr²

r₁ = 2x A₁ = 100π

r₂ = 3x A₂ = ?

Calculations:

Let the radii of the smaller and larger pizzas be 2x and 3x respectively.

⇒ Area of smaller pizza = π(2x)²

⇒ 4πx² = 100π

⇒ 4x² = 100

⇒ x² = 25

⇒ Area of larger pizza = π(3x)²

⇒ 9πx²

Substituting the value of x²:

⇒ Area of larger pizza = 9π × 25

⇒ Area of larger pizza = 225π cm²

∴ The correct answer is 225π cm².

Additional Information

Area of a Sector

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

Circumference of a Circle

The total boundary length of a circle is its circumference, calculated as C = 2πr.

Similar Figures Property

For any two similar geometric figures, the ratio of their areas is strictly equal to the square of the ratio of their corresponding linear dimensions (e.g., radii, diameters, or sides).

Shortcut Trick

The intersection of medians of a triangle is called its centroid.

The centroid always divides a median in the ratio 2 : 1 (from vertex to base).

Length of AG = (2/3) × Total length of median AD

AG = (2/3) × 12 cm = 8 cm

∴ The correct answer is 8 cm.

Alternate Method

Given:

Length of median AD = 12 cm

Intersection of medians AD and BE is G.

Concept Used:

The point where the medians of a triangle intersect is called the centroid.

The centroid divides each median in the ratio 2 : 1, measuring from the vertex to the midpoint of the opposite side.

 A B C D E G 2 Units 1 Unit

Calculations:

⇒ AG : GD = 2 : 1

⇒ Total parts on median AD = 2 + 1 = 3 parts

⇒ 3 parts = 12 cm

⇒ 1 part = 12 ÷ 3 = 4 cm

⇒ Length of AG = 2 parts = 2 × 4 cm

⇒ AG = 8 cm

∴ The correct answer is 8 cm.

Additional Information

Centroid

The point of intersection of the three medians of a triangle. It divides each median in the ratio 2 : 1. The centroid always lies strictly inside the triangle.

Incenter

The point of intersection of the internal angle bisectors of the triangle. It is equidistant from all three sides and forms the center of the inscribed circle (incircle).

Circumcenter

The point of intersection of the perpendicular bisectors of the sides. It is equidistant from all three vertices and forms the center of the circumscribed circle (circumcircle).

Orthocenter

The point of intersection of the three altitudes (perpendiculars dropped from vertices to opposite sides) of the triangle.

Shortcut Trick

In the given right-angled triangles ABC and DEF, we are given:

∠B = ∠E = 90° (Right angles)

AC = DF (Hypotenuses are equal)

AB = DE (One pair of corresponding sides are equal)

Since we have matching right angles, equal hypotenuses, and one pair of equal sides, the RHS (Right angle − Hypotenuse − Side) congruence criterion is perfectly satisfied.

∴ The correct answer is Yes, by RHS.

Alternate Method

Given:

⇒ Two right-angled triangles, ΔABC and ΔDEF.

⇒ ∠B = ∠E = 90°

⇒ Hypotenuse AC = Hypotenuse DF

⇒ Side AB = Side DE

A B C

D E F

Proof:

⇒ In right-angled ΔABC and right-angled ΔDEF:

⇒ ∠B = ∠E = 90° (Given right angles)

⇒ AC = DF (Given equal hypotenuses)

⇒ AB = DE (Given equal corresponding sides)

⇒ According to the RHS (Right angle − Hypotenuse − Side) congruence rule, if the hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of another right-angled triangle, then the two triangles are congruent.

⇒ ∴ ΔABC ≅ ΔDEF directly by the RHS congruence criterion.

∴ The correct answer is Yes, by RHS.

Additional Information

SSS (Side-Side-Side) Congruence

If all three sides of one triangle are exactly equal to the corresponding three sides of another triangle, the two triangles are congruent.

SAS (Side-Angle-Side) Congruence

If two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of another triangle, they are congruent.

ASA (Angle-Side-Angle) Congruence

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

AAS (Angle-Angle-Side) Congruence

If two angles and a non-included side of one triangle are equal to the corresponding parts of another triangle, the two triangles are congruent.