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Shortcut Trick For two circles, check the relation between the distance of their centers (d) and the sum of their radii. Sum of radii = 6 + 2 = 8 cm. Distance between centers d = 10 cm. Since 10 > 8 (d > r1 + r2), the circles are disjoint and external to each other. A maximum of 4 common tangents can be drawn (2 direct and 2 transverse). ∴ The correct answer is 4 tangents: 2 direct and 2 transverse.

Alternate Method Given: Radius of larger circle, R = 6 cm Radius of smaller circle, r = 2 cm Distance between centers, d = 10 cm

10 cm

6 cm

2 cm

DCT TCT

Calculations: ⇒ Sum of the radii = R + r = 6 + 2 = 8 cm ⇒ Compare distance d with the sum of radii: ⇒ d = 10 cm ⇒ 10 > 8 ⇒ d > R + r ⇒ When the distance between the centers is strictly greater than the sum of their radii, the two circles do not intersect or touch. ⇒ For completely separated circles, we can draw exactly 2 Direct Common Tangents (green lines) and 2 Transverse Common Tangents (orange lines). ⇒ Total common tangents = 2 + 2 = 4 tangents. ∴ The correct answer is 4 tangents: 2 direct and 2 transverse.

Additional Information Tangent Count Rules If d = r1 + r2 (touch externally), there are 3 common tangents. If |r1 − r2| < d < r1 + r2 (intersecting circles), there are 2 common tangents. If d < |r1 − r2|, there are 0 common tangents.

Length of Direct Common Tangent (DCT) Calculated using the formula: DCT = √(d2 − (r1 − r2)2), where d is the distance between centers.

Length of Transverse Common Tangent (TCT) Calculated using the formula: TCT = √(d2 − (r1 + r2)2). Valid only when the circles do not intersect internally or completely overlap.

Shortcut Trick

By the Intersecting Chords Theorem, the product of the segments of each chord is equal.

AP × PB = CP × PD

6 × 8 = 4 × PD

PD = 48 ÷ 4 = 12

∴ The correct answer is 12 cm.

Alternate Method

P

A B C D

6 cm 8 cm 4 cm PD = ?

Given:

Chords AB and CD intersect at point P inside the circle.

AP = 6 cm

PB = 8 cm

CP = 4 cm

Formula Used:

When two chords of a circle intersect internally, the product of the segment lengths on each chord are equal. Thus, AP × PB = CP × PD.

Calculations:

⇒ AP × PB = CP × PD

⇒ 6 × 8 = 4 × PD

⇒ 48 = 4 × PD

⇒ PD = 48 ÷ 4

⇒ PD = 12 cm

∴ The correct answer is 12 cm.

Additional Information

Intersecting Chords Theorem

When two chords intersect inside a circle, the products of their respective segments are equal: AP × PB = CP × PD.

Intersecting Secants Theorem

When two secant lines intersect outside a circle, the product of the external segment and the total secant length is equal for both secants: PA × PB = PC × PD.

Tangent-Secant Theorem

When a tangent and a secant intersect outside a circle, the square of the tangent segment equals the product of the external secant segment and the total secant length: PT² = PA × PB.

Shortcut Trick

The perpendicular drawn from the center of a circle to a chord bisects the chord.

Half of the given chord length = 12 ÷ 2 = 6 cm.

Using Pythagoras theorem: Radius = √(Distance² + Half-chord²)

r = √(4² + 6²) = √(16 + 36) = √52 cm.

∴ The correct answer is √52 cm.

Alternate Method

Given:

⇒ Length of chord AB = 12 cm

⇒ Distance of the chord from the center (OM) = 4 cm

Formula Used:

⇒ Radius r = √(OM² + AM²)

O

A B

M

4 cm 6 cm 6 cm r r = √(4² + 6²) = √52 cm

Calculations:

⇒ A perpendicular drawn from the center of a circle to a chord bisects it.

⇒ AM = MB = AB ÷ 2

⇒ AM = 12 ÷ 2 = 6 cm

⇒ In right-angled ΔOMA, using Pythagoras theorem: OA² = OM² + AM²

⇒ r² = 4² + 6²

⇒ r² = 16 + 36 = 52

⇒ r = √52 cm

∴ The correct answer is √52 cm.

Additional Information

Perpendicular from Center Rule

The perpendicular drawn from the center of a circle to a chord bisects the chord into two equal halves.

Equal Chords Theorem

Chords of equal length in a circle are equidistant from the center of the circle.

Intersecting Chords Theorem

If two chords AB and CD intersect internally at point P, then PA × PB = PC × PD.

Tangent-Secant Theorem

If a tangent segment PT and a secant segment PAB are drawn to a circle from an exterior point P, then PT² = PA × PB.

Shortcut Trick

Observe the relationship between the numerator and denominator terms.

Each term in the numerator is exactly 1/10 of the corresponding term in the denominator (0.02 = 0.2 ÷ 10 and 0.004 = 0.04 ÷ 10).

Since all terms are cubed, the overall multiplying factor becomes (1/10)³ = 1/1000.

The whole expression simplifies directly to 1/1000 = 0.001.

∴ The correct answer is 0.001.

Alternate Method

Given:

(0.02³ + 0.004³) ÷ (0.2³ + 0.04³)

Formula Used:

(x/y)ⁿ = xⁿ ÷ yⁿ

Calculations:

⇒ Let us rewrite the decimal terms in the numerator as fractions of the denominator terms.

⇒ 0.02 = 0.2 ÷ 10

⇒ 0.004 = 0.04 ÷ 10

⇒ Numerator = (0.2 ÷ 10)³ + (0.04 ÷ 10)³

⇒ Numerator = (0.2³ ÷ 1000) + (0.04³ ÷ 1000)

⇒ Numerator = (1 ÷ 1000) × (0.2³ + 0.04³)

⇒ Substitute this back into the original expression.

⇒ Expression = [(1 ÷ 1000) × (0.2³ + 0.04³)] ÷ (0.2³ + 0.04³)

⇒ Cancel out the common term (0.2³ + 0.04³).

⇒ Expression = 1 ÷ 1000 = 0.001

∴ The correct answer is 0.001.

Additional Information

Algebraic Identities

a³ + b³ = (a + b)(a² − ab + b²) and a³ − b³ = (a − b)(a² + ab + b²).

Laws of Exponents

When dividing terms with the same base, powers are subtracted: a^m ÷ aⁿ = a^m−n. Also, (ab)ⁿ = aⁿbⁿ.

BODMAS Rule

Always follow the logical sequence of Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction while simplifying algebraic and decimal expressions.

Shortcut Trick

Observe that each term in the denominator is exactly 3 times the corresponding term in the numerator.

Denominator = (3 × 0.4)³ + (3 × 0.02)³

Denominator = 3³ × [(0.4)³ + (0.02)³]

Denominator = 27 × Numerator

Dividing the numerator by the denominator gives 1/27 directly.

∴ The correct answer is 1/27.

Alternate Method

Given:

Expression = (0.4 × 0.4 × 0.4 + 0.02 × 0.02 × 0.02) ÷ (1.2 × 1.2 × 1.2 + 0.06 × 0.06 × 0.06)

Formula Used:

Laws of exponents: (x × y)ⁿ = xⁿ × yⁿ

Calculations:

⇒ Numerator = (0.4)³ + (0.02)³

⇒ Denominator = (1.2)³ + (0.06)³

⇒ Denominator = (3 × 0.4)³ + (3 × 0.02)³

⇒ Denominator = 3³ × (0.4)³ + 3³ × (0.02)³

⇒ Denominator = 27 × [(0.4)³ + (0.02)³]

⇒ Expression = Numerator ÷ Denominator

⇒ Expression = [(0.4)³ + (0.02)³] ÷ {27 × [(0.4)³ + (0.02)³]}

⇒ Expression = 1 / 27

∴ The correct answer is 1/27.

Additional Information

Algebraic Identities for Cubes

Formulas such as a³ + b³ = (a + b)(a² − ab + b²) and a³ − b³ = (a − b)(a² + ab + b²) are frequently used to expand and simplify complex fractional equations.

Laws of Exponents

When multiplying numbers within a shared exponent, the rule is (x × y)ⁿ = xⁿ × yⁿ. This allows proportional multipliers to be factored out perfectly.

BODMAS Rule

Always follow the correct order of operations: Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction. Solve all terms within brackets before executing outer division.

Shortcut Trick

Plants arranged in the square formation = Total plants − Left out plants.

⇒ Plants in square = 3984 − 15 = 3969

For a perfect square formation, rows = columns, so the number of plants per row is √3969.

Since 60² = 3600 and 70² = 4900, and 3969 ends in 9, the root must end in 3.

⇒ Number of plants per row = 63

∴ The correct answer is 63 plants.

Alternate Method

Given:

Total number of plants = 3984

Number of plants left out = 15

n × n = n2 plants Area = 3969

n plants

n rows

• 15 plants left out

Total = n2 + 15 = 3984

Formula Used:

Total plants = (Number of rows × Number of plants per row) + Plants left out

Calculations:

Let the number of rows be n. Since it is a square formation, the number of plants per row is also n.

⇒ Total plants used in the square formation = n × n = n²

⇒ n² + 15 = 3984

⇒ n² = 3984 − 15

⇒ n² = 3969

⇒ n = √3969

⇒ n = 63

∴ The correct answer is 63 plants.

Additional Information

Square Formation Principle

When objects are organized in a solid square arrangement, the number of columns is equal to the number of rows. Therefore, the total number of objects making up the square will always be a perfect square number (n²).

Unit Digits of Perfect Squares

Perfect squares can only end in the digits 0, 1, 4, 5, 6, or 9. If a perfect square ends in 9, its square root must end in either 3 or 7 (since 3 × 3 = 9 and 7 × 7 = 49). Numbers ending in 2, 3, 7, or 8 are never perfect squares.

Shortcut Trick

Convert the given percentage increases into multiplying fractions:

25% more means 125% = 5/4

20% more means 120% = 6/5

A : C = (A/B) × (B/C) = (5/4) × (6/5) = 30/20 = 3/2

∴ The correct answer is 3 : 2.

Alternate Method

Given:

A is 25% more than B.

B is 20% more than C.

C = 100

+20%

B = 120

+25%

A = 150

Ratio A : C = 150 : 100

Calculations:

⇒ Let the value of C be 100.

⇒ Value of B = 100 + 20% of 100 = 100 + 20 = 120

⇒ Value of A = 120 + 25% of 120

⇒ Value of A = 120 + (25/100) × 120 = 120 + 30 = 150

⇒ Ratio of A to C = 150 : 100

⇒ A : C = 3 : 2

∴ The correct answer is 3 : 2.

Additional Information

Compound Ratio

The compound ratio of two individual ratios a:b and c:d is obtained by multiplying their respective terms: (a/b) × (c/d) = (a × c) : (b × d).

Identifying the Base Value

In percentage comparison statements like "A is x% more than B", the quantity appearing immediately after the word "than" (which is B in this case) is taken as the original base (100%).

Successive Percentage Change

When a value undergoes multiple successive changes of a% and b%, the net percentage change can be calculated using the formula: a + b + (a × b)/100.

Shortcut Trick

For an infinite series of the form √(n + √(n + √(n + ...))), check if n is a product of two consecutive integers.

If n = a × (a + 1), then the answer is the larger integer (a + 1).

Here, the given number is 12.

12 can be factored into consecutive integers as 3 × 4.

Since the series has a positive (+) sign, the answer is the larger factor, which is 4.

∴ The correct answer is 4.

Alternate Method

Given:

The expression is √(12 + √(12 + √(12 + ...∞)))

Method Used:

Assume the given infinite series is equal to x, then substitute x back into the expression to form a quadratic equation.

Calculations:

⇒ Let x = √(12 + √(12 + √(12 + ...∞)))

Since the series continues infinitely, the term inside the first square root is also equal to x:

⇒ x = √(12 + x)

Squaring both sides of the equation:

⇒ x² = 12 + x

⇒ x² − x − 12 = 0

Factorising the quadratic equation:

⇒ x² − 4x + 3x − 12 = 0

⇒ x(x − 4) + 3(x − 4) = 0

⇒ (x − 4)(x + 3) = 0

⇒ x = 4 or x = −3

Since the principal square root is always non-negative, x cannot be −3.

⇒ x = 4

∴ The correct answer is 4.

Additional Information

Infinite Series with Subtraction

For √(n − √(n − √(n ...∞))), if n = a × (a+1), the answer is the smaller integer, a.

Infinite Series with Multiplication

For √(n × √(n × √(n ...∞))), the answer is always n.

Finite Series with Multiplication

For √(n × √(n × ... k times)), the exact formula used is n^{(2^k − 1) ÷ 2^k}.

When Factors are not Consecutive Integers

For √(n + √(n + ...∞)), use the direct root formula: x = (√(4n+1) + 1) ÷ 2.

Shortcut Trick

Selling Price (SP) = 80% of Marked Price (MP) = 0.8 × 600 = ₹480.

Loss = Cost Price (CP) − SP = 500 − 480 = ₹20.

Loss % = (20 ÷ 500) × 100 = 4%.

∴ The correct answer is 4% Loss.

Alternate Method

Given:

Cost Price (CP) = ₹500

Marked Price (MP) = ₹600

Discount = 20%

Formula Used:

Selling Price (SP) = MP × [(100 − Discount %) ÷ 100]

Loss % = (Loss ÷ CP) × 100

CP = ₹500

SP = ₹480

MP = ₹600

Discount = 20%

Loss = ₹20

Calculations:

⇒ SP = 600 × [(100 − 20) ÷ 100]

⇒ SP = 600 × (80 ÷ 100) = ₹480

⇒ Here, Cost Price is greater than Selling Price, so there is a loss.

⇒ Loss = CP − SP = 500 − 480 = ₹20

⇒ Loss % = (20 ÷ 500) × 100

⇒ Loss % = 4%

∴ The correct answer is 4% Loss.

Additional Information

Marked Price & Discount

Discount is always calculated on the Marked Price (MP). So, Selling Price = MP − Discount.

Profit/Loss Percentage

Profit or loss is always calculated on the Cost Price (CP) unless specifically stated otherwise. Loss % = (Loss ÷ CP) × 100.

Successive Discounts

If two successive discounts of x% and y% are given on an article, the equivalent single discount percentage is (x + y − xy ÷ 100)%.

Shortcut Trick

Total capacity of the tank = LCM of 20, 30 and 60 = 60 units.

Efficiency of A = +3, B = +2 and C (drainpipe) = −1 unit/min.

Combined efficiency of all pipes = 3 + 2 − 1 = 4 units/min.

Time taken to fill = 60 ÷ 4 = 15 minutes.

∴ The correct answer is 15 min.

Alternate Method

Given:

Time taken by Pipe A to fill the tank = 20 minutes

Time taken by Pipe B to fill the tank = 30 minutes

Time taken by drainpipe C to empty the tank = 60 minutes

Formula Used:

Efficiency = Total Work ÷ Time Taken

Time = Total Work ÷ Combined Efficiency

Calculations:

⇒ Total efficiency when all pipes are open = Efficiency of A + Efficiency of B + Efficiency of C

⇒ Combined Efficiency = 3 + 2 + (−1)

⇒ Combined Efficiency = 4 units/min

⇒ Time taken to fill the tank = 60 ÷ 4

⇒ Time taken = 15 minutes

∴ The correct answer is 15 min.

Additional Information

Inlet vs Outlet Pipes

An inlet pipe fills the tank, hence its efficiency (work per unit time) is considered positive. An outlet or drainpipe empties the tank, so its efficiency is considered negative.

Combined Work Formula

If pipe A fills a tank in x units of time, pipe B fills it in y units, and pipe C empties it in z units, then the net part filled in 1 unit of time is (1/x + 1/y − 1/z).

Total Time to Fill or Empty

If the net efficiency is positive, the tank gets filled over time. If the net efficiency is negative, a partially or completely full tank will eventually get emptied.