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Shortcut Trick

Let Cost Price (CP) = 100%.

Marked Price (MP) = 100% + 20% = 120%.

Selling Price (SP) = 120% − (15% of 120%) = 120% − 18% = 102%.

Profit = SP − CP = 102% − 100% = 2%.

Given, 2% = ₹300 ⇒ 100% = ₹15,000.

∴ The correct answer is ₹15,000.

Alternate Method

Given:

Markup = 20%

Discount = 15%

Profit = ₹300

Formula Used:

MP = CP × (100 + Markup%) ÷ 100

SP = MP × (100 − Discount%) ÷ 100

Profit = SP − CP

CP = 100xMP = 120xSP = 102x+20% Markup-15% DiscountProfit = 2x = ₹300

Calculations:

⇒ Let the Cost Price (CP) be 100x.

⇒ Marked Price (MP) = 100x × (100 + 20) ÷ 100 = 120x

⇒ Selling Price (SP) = 120x × (100 − 15) ÷ 100 = 120x × 85 ÷ 100 = 102x

⇒ Profit = 102x − 100x = 2x

⇒ 2x = 300

⇒ x = 150

⇒ CP = 100x = 100 × 150 = 15000

∴ The correct answer is ₹15,000.

Additional Information

Marked Price & Discount

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

Successive Discounts

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

Relation between CP and MP

CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%). This is a direct shortcut formula to find CP or MP without having to calculate SP.

Shortcut Trick Let Cost Price (CP) = 100 units ⇒ Marked Price (MP) = 100 + 25% = 125 units. Net discount = 10% + 20% − (10 × 20) ÷ 100 = 28%. Selling Price (SP) = 125 × (1 − 0.28) = 125 × 0.72 = 90 units. Loss = CP − SP = 100 − 90 = 10 units. 10 units = ₹400 ⇒ 1 unit = ₹40. MP = 125 units × 40 = ₹5000. ∴ The correct answer is ₹5000. Alternate Method Given: Markup = 25% above CP Consecutive discounts = 10% and 20% Loss = ₹400

CP = 100x

MP = 125x

SP = 90x

+25%

-10%, -20%

Loss = 100x − 90x = 10x 10x = ₹400 ⇒ x = 40

Calculations: Let the Cost Price (CP) of the article be 100x. ⇒ Marked Price (MP) = 100x + (25/100) × 100x = 125x Two consecutive discounts are given on the Marked Price: 10% and 20%. ⇒ Selling Price (SP) = MP × (100 − D1)/100 × (100 − D2)/100 ⇒ SP = 125x × (100 − 10)/100 × (100 − 20)/100 ⇒ SP = 125x × 0.90 × 0.80 ⇒ SP = 125x × 0.72 = 90x Loss is incurred when CP > SP. ⇒ Loss = CP − SP ⇒ Loss = 100x − 90x = 10x According to the given data, Loss = ₹400. ⇒ 10x = 400 ⇒ x = 40 Now, substitute the value of x to find the Marked Price: ⇒ MP = 125x = 125 × 40 = ₹5000 ∴ The correct answer is ₹5000. Additional Information Successive Discounts When two discounts d1% and d2% are given consecutively, the effective discount is: Net Discount = d1 + d2 − (d1 × d2) ÷ 100. Marked Price & Profit/Loss Marked price is the price printed on an article. Discount is always calculated on Marked Price (MP), while Profit or Loss is always calculated on Cost Price (CP). Single Equivalent Discount for 3 Discounts For three successive discounts a%, b% and c%, final SP = MP × (100 − a) ÷ 100 × (100 − b) ÷ 100 × (100 − c) ÷ 100.

Shortcut Trick

Let the remaining mixture (after removing 30 L) be 10x (since ratio is 3:5:2).

Quantities of P and Q in the remaining mixture are 3x and 5x respectively.

After adding 12 L of P and 8 L of Q, the difference is given as 20 L: (5x + 8) − (3x + 12) = 20.

⇒ 2x − 4 = 20 ⇒ x = 12.

Remaining mixture = 10 × 12 = 120 L. Initial total quantity = 120 + 30 = 150 L.

∴ The correct answer is 150 litres.

Alternate Method

Initial MixP:Q:R = 3:5:2− 30 LRemainingP=3x, Q=5x+12L P+8L QFinalP = 3x + 12Q = 5x + 8Given: Final Q − Final P = 20 L

Given:

Initial ratio of chemicals P, Q, and R = 3:5:2

Quantity of mixture removed = 30 litres

Quantity of P added = 12 litres

Quantity of Q added = 8 litres

Final amount of Q − Final amount of P = 20 litres

Calculations:

⇒ Let the remaining quantity of the mixture be 10x (since 3 + 5 + 2 = 10).

⇒ In the remaining mixture, quantity of P = 3x and quantity of Q = 5x.

⇒ After addition, resultant P = 3x + 12.

⇒ After addition, resultant Q = 5x + 8.

⇒ According to the given condition: Resultant Q − Resultant P = 20

⇒ (5x + 8) − (3x + 12) = 20

⇒ 2x − 4 = 20

⇒ 2x = 24 ⇒ x = 12

⇒ Remaining quantity of the mixture = 10x = 10 × 12 = 120 litres.

⇒ Initial total quantity = Remaining quantity + Removed quantity

⇒ Initial total quantity = 120 + 30 = 150 litres.

∴ The correct answer is 150 litres.

Additional Information

Mixture Replacement Rule

When a certain volume of a mixture is withdrawn, the ratio of the constituent elements in the remaining mixture remains exactly the same as in the original mixture.

Ratio & Proportion

If elements are present in the ratio a:b:c, their respective quantities can be assumed to be ax, bx, and cx, where the common multiplier 'x' represents a unit part of the total mixture.

Alligation Method

It is a general rule used to find the ratio in which two components of different concentrations are mixed to yield a desired concentration. The ratio of components is inversely proportional to the differences in their concentrations.

Shortcut Trick

Total effective interest rate = (3 × 5%) + (5 × 7%) + (2 × 9%)

⇒ Total rate = 15% + 35% + 18% = 68%

We are given that 68% of the total Principal = ₹6800

⇒ Principal (100%) = (6800 ÷ 68) × 100 = ₹10000

∴ The correct answer is ₹10000.

Alternate Method

Total Time = 10 Years

3 yrs @ 5% 15%

5 yrs @ 7% 35%

2 yrs @ 9% 18%

Total Interest = 15% + 35% + 18% = 68% 68% = ₹6800 → 100% = ₹10000

Given:

Rate for the first 3 years (R₁) = 5% per annum

Rate for the next 5 years (R₂) = 7% per annum

Total time given = 10 years, so remaining time (T₃) = 10 − (3 + 5) = 2 years

Rate for the remaining 2 years (R₃) = 9% per annum

Total Simple Interest (SI) paid at the end = ₹6800

Formula Used:

Simple Interest (SI) = (P × R × T) ÷ 100

Calculations:

⇒ Total SI = SI₁ + SI₂ + SI₃

⇒ 6800 = (P × 5 × 3) ÷ 100 + (P × 7 × 5) ÷ 100 + (P × 9 × 2) ÷ 100

⇒ 6800 = (15P ÷ 100) + (35P ÷ 100) + (18P ÷ 100)

⇒ 6800 = (15P + 35P + 18P) ÷ 100

⇒ 6800 = 68P ÷ 100

⇒ P = (6800 × 100) ÷ 68

⇒ P = 10000

∴ The correct answer is ₹10000.

Additional Information

Effective Rate of Simple Interest

When money is borrowed at different rates for different periods, the total effective percentage interest is simply the sum of the products of each rate and its respective time period: R_effective = R₁T₁ + R₂T₂ + ...

Amount in Simple Interest

The total amount to be repaid at the end of the term is the sum of the Principal and the Simple Interest generated: A = P + SI = P(1 + RT ÷ 100).

Difference Between SI and CI

Unlike Compound Interest where interest is calculated on the accumulated interest at the end of each cycle, Simple Interest is always calculated solely on the initial Principal amount for the entire duration.

Shortcut Trick Base area of a cone = πr2 Lateral surface area = πrl Equating both as per the given condition: πr2 = πrl Canceling π and r from both sides directly gives: l = r ∴ The correct answer is r.

Alternate Method

r l h

Given: Radius of the cone = r Slant height of the cone = l Area of the base = Lateral surface area Formula Used: Area of base of a cone = πr2 Lateral surface area of a cone = πrl Calculations: ⇒ According to the question, Area of base = Lateral surface area ⇒ πr2 = πrl ⇒ Dividing both sides by πr (since r > 0) ⇒ r = l ∴ The correct answer is r.

Additional Information Total Surface Area of a Cone The total surface area includes the area of the circular base and the curved lateral surface: TSA = πr(l + r). Volume of a Cone The volume represents the 3D space enclosed by the cone: V = (1÷3)πr2h, where h is the perpendicular height. Slant Height Relationship The slant height (l), base radius (r), and perpendicular height (h) form a right-angled triangle. Applying Pythagoras theorem: l = √(r2 + h2).

Shortcut Trick

Area of remaining field = Area of Square − Area of Pond

Area of remaining field = (Side)² − 616

Area of remaining field = 28² − 616

Area of remaining field = 784 − 616 = 168 m²

∴ The correct answer is 168 m².

Alternate Method

Pond Area = 616 m2

28 m

28 m Field Field

Given:

Side of the square field (a) = 28 m

Area of the circular pond = 616 m²

Formula Used:

Area of a square = a²

Area of remaining part = Area of square − Area of internal shape

Calculations:

⇒ Area of the square field = 28 × 28

⇒ Area of the square field = 784 m²

⇒ Area of the remaining field = Area of square field − Area of pond

⇒ Area of the remaining field = 784 − 616 = 168 m²

∴ The correct answer is 168 m².

Additional Information

Square Properties

Area = a² | Perimeter = 4a | Diagonal = a√2

Circle Properties

Area = πr² | Circumference = 2πr (where r is the radius)

Largest Circle in a Square

When a circle perfectly fits inside a square of side a, its radius is a/2. It occupies exactly π/4 (approx. 78.5%) of the square's total area.

Shortcut Trick

Volume of any right prism = Base Area × Height.

Since the base area remains constant, Volume ∝ Height.

Therefore, the percentage change in volume will be exactly equal to the percentage change in height.

Given percentage increase in height = 50%.

⇒ Percentage increase in volume = 50%.

∴ The correct answer is 50%.

Alternate Method

Given:

Increase in height of the right prism = 50%

Base area remains constant.

Formula Used:

Volume of a Right Prism = Base Area × Height

h V = Ah Area = A

1.5h V' = 1.5Ah Area = A

+50% Height

Calculations:

⇒ Let the initial height of the prism = h

⇒ Let the constant base area of the prism = A

⇒ Initial Volume (V₁) = A × h = Ah

⇒ New height of the prism = h + (50 ÷ 100) × h = 1.5h

⇒ New Volume (V₂) = A × 1.5h = 1.5Ah

⇒ Increase in Volume = V₂ − V₁ = 1.5Ah − Ah = 0.5Ah

⇒ Percentage Increase in Volume = (Increase in Volume ÷ Initial Volume) × 100

⇒ Percentage Increase = (0.5Ah ÷ Ah) × 100 = 50%

∴ The correct answer is 50%.

Additional Information

Volume of a Prism

Volume = Base Area × Height. If the base area is kept constant, the volume is directly proportional to the height.

Volume of a Pyramid or Cone

Volume = 1/3 × Base Area × Height. It is exactly one-third the volume of a right prism having the same base area and height.

Successive Percentage Change

If the height changes by a% and the base area changes by b%, the overall percentage change in volume can be evaluated using the successive formula: a + b + (a × b)/100.

Shortcut Trick

The percentage area of a minor segment formed by an equilateral triangle at the center is independent of the circle's radius.

Required % = [(Area of Sector − Area of Triangle) ÷ Total Area] × 100

Required % = (1 ÷ 6 − √3 ÷ 4π) × 100 = (0.1667 − 1.732 ÷ 12.56) × 100

Required % = (0.1667 − 0.1379) × 100 = 0.0288 × 100 = 2.88%

∴ The correct answer is 2.88%.

Alternate Method

Given:

Radius of the clock (r) = 20 cm

Central angle (θ) = 60° (since the chord and two radii form an equilateral triangle)

π = 3.14, √3 = 1.732

Formula Used:

Total Area of Clock = πr²

Area of Sector = (θ ÷ 360) × πr²

Area of Equilateral Triangle = (√3 ÷ 4) × r²

O

60° r = 20 r = 20 Chord A B

Painted Minor Segment

Calculations:

⇒ Total Area = 3.14 × 20² = 3.14 × 400 = 1256 cm²

⇒ Area of Sector = (60 ÷ 360) × 1256 = 1256 ÷ 6 ≈ 209.33 cm²

⇒ Area of Triangle = (1.732 ÷ 4) × 400 = 1.732 × 100 = 173.20 cm²

⇒ Area of Painted Segment = Area of Sector − Area of Triangle

⇒ Area of Painted Segment = 209.33 − 173.20 = 36.13 cm²

⇒ Required Percentage = (36.13 ÷ 1256) × 100 = 2.876% ≈ 2.88%

∴ The correct answer is 2.88%.

Additional Information

Area of a Sector

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

Area of a Circular Segment

A segment is the region bounded by a chord and an arc. The area of a minor segment is calculated as the Area of the corresponding Sector − Area of the Triangle formed by the two radii and the chord.

Equilateral Triangle Area

An equilateral triangle has all sides equal and internal angles of exactly 60°. Its total area can be determined using the direct formula (√3 ÷ 4) × side².

Shortcut Trick

Since the hole is bored through the entire height, the ratio of the volume removed depends only on the square of the radii.

Volume Removed % = (r_hole ÷ R_original)² × 100

⇒ Volume Removed % = (2 ÷ 4)² × 100

⇒ (1/2)² × 100 = 25%

∴ The correct answer is 25%.

Alternate Method

Given:

Radius of original cylinder (R) = 4 cm

Height of original cylinder (h) = 10 cm

Radius of bored hole (r) = 2 cm

Height of bored hole = 10 cm (full height)

R = 4 cm r = 2 cm

h = 10 cm

Volume Removed = Inner Hole Volume

Formula Used:

Volume of cylinder = πr²h

Percentage removed = (Volume removed ÷ Original volume) × 100

Calculations:

⇒ Original volume = π × (4)² × 10

⇒ Original volume = 160π cm³

⇒ Volume removed (Hole) = π × (2)² × 10

⇒ Volume removed = 40π cm³

⇒ Percentage removed = (40π ÷ 160π) × 100

⇒ Percentage removed = (1 ÷ 4) × 100 = 25%

∴ The correct answer is 25%.

Additional Information

Volume of a Hollow Cylinder

The volume of the material remaining after boring a hole is determined by calculating the difference in volume: V = πh(R² − r²).

Curved Surface Area of a Hollow Cylinder

The total curved surface area becomes the sum of the inner hole's area and the outer cylinder's area: CSA = 2πRh + 2πrh = 2πh(R + r).

Total Surface Area of a Hollow Cylinder

The total surface area includes the inner and outer curved areas plus the area of the top and bottom solid rings: TSA = 2πh(R + r) + 2π(R² − r²).

Shortcut Trick

For sectors of the same circle, the area is directly proportional to the central angle.

Angle of the first sector = 45°

Angle of the second sector = π/4 radians = (180° ÷ 4) = 45°

Since both central angles are equal, their areas will be exactly the same.

∴ The correct answer is 1:1.

Alternate Method

Given:

Radius of the circle (r) = 15 cm

Central angle of first sector (θ₁) = 45°

Central angle of second sector (θ₂) = π/4 radians

45° r = 15 cm First Sector

π/4 r = 15 cm Second Sector

Ratio = ?

Formula Used:

Area of a sector (when θ is in degrees) = (θ ÷ 360°) × πr²

Area of a sector (when θ is in radians) = (1 ÷ 2) × r²θ

Calculations:

⇒ Area of first sector = (45° ÷ 360°) × π × (15)²

⇒ Area of first sector = (1 ÷ 8) × 225π = 28.125π cm²

⇒ Area of second sector = (1 ÷ 2) × (15)² × (π ÷ 4)

⇒ Area of second sector = (1 ÷ 2) × 225 × (π ÷ 4) = 28.125π cm²

⇒ Ratio of areas = Area of first sector : Area of second sector

⇒ Ratio = 28.125π : 28.125π = 1 : 1

∴ The correct answer is 1:1.

Additional Information

Area of a Sector

The area of a sector of a circle with radius r and central angle θ is given by (θ ÷ 360°) × πr² if θ is measured in degrees, and (1 ÷ 2)r²θ if θ is measured in radians.

Radian to Degree Conversion

To convert an angle from radians into degrees, multiply the radian value by (180° ÷ π). For example, π radians equals 180°.

Length of an Arc

The arc length of a corresponding sector is calculated as (θ ÷ 360°) × 2πr (for degrees) or simply rθ (for radians).