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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.