Quantitative Aptitude PYQs

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.

Shortcut Trick

Let A = x − y, B = y − z, and C = z − x.

Sum = A + B + C = (x − y) + (y − z) + (z − x) = 0.

If A + B + C = 0, then directly A³ + B³ + C³ = 3ABC.

Substituting A, B, and C back directly yields 3(x − y)(y − z)(z − x).

∴ The correct answer is 3(x − y)(y − z)(z − x).

Alternate Method

Given:

⇒ x + y + z = 0

⇒ x³ + y³ + z³ = 3xyz

Formula Used:

⇒ If a + b + c = 0, then a³ + b³ + c³ = 3abc

Calculations:

⇒ We need to evaluate (x − y)³ + (y − z)³ + (z − x)³

⇒ Let a = x − y, b = y − z, and c = z − x

⇒ Checking the sum of these variables:

⇒ a + b + c = (x − y) + (y − z) + (z − x)

⇒ a + b + c = x − y + y − z + z − x = 0

⇒ Since the sum of a, b, and c is 0, the special algebraic identity applies directly:

⇒ a³ + b³ + c³ = 3abc

⇒ Substitute the values of a, b, and c back into the equation:

⇒ (x − y)³ + (y − z)³ + (z − x)³ = 3 × (x − y) × (y − z) × (z − x)

⇒ (x − y)³ + (y − z)³ + (z − x)³ = 3(x − y)(y − z)(z − x)

∴ The correct answer is 3(x − y)(y − z)(z − x).

Additional Information

Standard Cubic Identity

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

Modified Cubic Identity

a³ + b³ + c³ − 3abc = (1 ÷ 2) × (a + b + c) × [(a − b)² + (b − c)² + (c − a)²]

Condition for 3abc

If a + b + c = 0 OR if a = b = c, then a³ + b³ + c³ perfectly evaluates to 3abc.

Shortcut Trick

Overall percentage of interest = (390 ÷ 5000) × 100 = 7.8%

By alligation method for 6% and 9% interest rates:

Ratio of parts = (9 − 7.8) : (7.8 − 6) = 1.2 : 1.8 = 2 : 3

Sum lent at 6% = 5000 × [2 ÷ (2 + 3)] = 5000 × (2 ÷ 5) = ₹2000

∴ The correct answer is ₹2000.

Alternate Method

Given:

Total Sum (P) = ₹5000

Total simple interest (I) = ₹390

Rates of interest = 6% and 9%

Time (T) = 1 year

Formula Used:

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

₹ x at 6% ₹ (5000 − x) at 9%

Total Principal = ₹5000 Total Interest = ₹390

Calculations:

⇒ Let the sum lent at 6% be ₹x.

⇒ Then, the sum lent at 9% will be ₹(5000 − x).

⇒ Total Interest = Interest on first part + Interest on second part

⇒ [(x × 6 × 1) ÷ 100] + [((5000 − x) × 9 × 1) ÷ 100] = 390

⇒ 6x + 45000 − 9x = 39000

⇒ 3x = 45000 − 39000

⇒ 3x = 6000

⇒ x = 2000

∴ The correct answer is ₹2000.

Additional Information

Simple Interest (SI)

It is calculated only on the principal amount for the entire tenure. Formula: SI = (P × R × T) ÷ 100.

Compound Interest (CI)

Interest is calculated on the principal and also on the accumulated interest of previous periods. Formula: Amount = P × (1 + R ÷ 100)^T.

Difference between CI and SI

For a period of 2 years, the difference between Compound Interest and Simple Interest on a given principal is calculated as: D = P × (R ÷ 100)².

Alligation Method

A technique used to find the ratio in which two quantities at different rates or percentages are mixed to yield a mixture of a known average rate.

Shortcut Trick

Total interest for 1 year = 4950 ÷ 3 = ₹1650.

Assume the entire ₹15,000 is lent at X's rate of 10%. Expected interest = 10% of 15000 = ₹1500.

Extra interest generated = 1650 − 1500 = ₹150.

This ₹150 extra is due to the additional 2% rate (12% − 10%) on Y's share.

2% of Y = 150 ⇒ Y = 7500. Therefore, X's share = 15000 − 7500 = ₹7500.

∴ The correct answer is ₹7,500.

Alternate Method

Given:

Total principal = ₹15,000

Total Simple Interest (SI) for 3 years = ₹4,950

Rate for X = 10%, Rate for Y = 12%

Formula Used:

SI = (P × R × T) ÷ 100

Total = ₹15,000

Lent to X Rate = 10% p.a. Time = 3 years

Lent to Y Rate = 12% p.a. Time = 3 years

Total SI = ₹4,950

Calculation:

⇒ Let the amount lent to X be P.

⇒ Then, the amount lent to Y = 15000 − P.

⇒ Interest from X = (P × 10 × 3) ÷ 100 = 0.3P

⇒ Interest from Y = ((15000 − P) × 12 × 3) ÷ 100 = (15000 − P) × 0.36

⇒ Total Interest = 0.3P + 5400 − 0.36P = 4950

⇒ 5400 − 0.06P = 4950

⇒ 0.06P = 5400 − 4950

⇒ 0.06P = 450

⇒ P = 450 ÷ 0.06 = 7500

∴ The correct answer is ₹7,500.

Additional Information

Alligation Method for SI

Calculate the overall average rate of interest using the formula: Average Rate = (Total SI × 100) ÷ (Total Principal × Time). Use the alligation cross method to quickly find the ratio in which the total sum is divided.

Shift / Assume Method

Whenever a total amount is split into two parts at two different rates, assuming the entire amount is invested at the lower rate calculates the minimum expected interest. The excess interest is solely due to the higher rate part, directly revealing its value.

Simple Interest Key Fact

Unlike Compound Interest, Simple Interest remains constant every year if the principal, rate, and time do not change.

Shortcut Trick

In a 30°-60°-90° right-angled triangle, the ratio of sides opposite to the angles is 1 : √3 : 2.

The base adjacent to the 60° angle (which is opposite to 30°) corresponds to 1 part.

The length of the wire forms the hypotenuse, corresponding to 2 parts.

Given the base length is 10 m, we can multiply by 2 to get the hypotenuse directly.

Length of the wire = 10 × 2 = 20 m.

∴ The correct answer is 20 m.

Alternate Method

A B C 10 m 60° L = ?

Given:

Distance of the wire fixed on the ground from the base of the pole = 10 m

Angle of elevation of the wire from the ground = 60°

Formula Used:

cos(θ) = Base ÷ Hypotenuse

Calculation:

⇒ Let the length of the taut wire be L.

⇒ From the right-angled triangle formed, cos(60°) = 10 ÷ L

⇒ 1/2 = 10 ÷ L

⇒ L = 10 × 2

⇒ L = 20 m

∴ The correct answer is 20 m.

Additional Information

Trigonometric Ratios

For any right-angled triangle, sine = Perpendicular ÷ Hypotenuse, cosine = Base ÷ Hypotenuse, and tangent = Perpendicular ÷ Base.

Angle of Elevation

The angle formed by the line of sight with the horizontal plane when viewing an object located above the observer's eye level.

Standard 30°-60°-90° Triangle

The sides opposite to the angles 30°, 60°, and 90° respectively are always strictly in the ratio of 1 : √3 : 2.

Shortcut Trick

Volume of Cone = Volume of Sphere ⇒ (1/3)πr_c²h = (4/3)πr_s³

Substitute cone height h = 2r_c directly into the equation:

⇒ r_c² × 2r_c = 4r_s³

⇒ 2r_c³ = 4r_s³ ⇒ r_c³/r_s³ = 2/1

Taking cube roots on both sides gives the direct ratio of their radii.

∴ The correct answer is ∛2 : 1.

Alternate Method

Given:

Volume of the cone = Volume of the sphere

Height of the cone (h) = 2 × Radius of the cone (r_c)

Formula Used:

Volume of a Cone = (1/3)πr²h

Volume of a Sphere = (4/3)πr³

h = 2rc rc Cone

rs Sphere

Volume of Cone = Volume of Sphere

Calculations:

Let the radius of the cone be r_c and the radius of the sphere be r_s.

⇒ (1/3) × π × r_c² × h = (4/3) × π × r_s³

Substitute h = 2r_c in the above equation:

⇒ (1/3) × π × r_c² × (2r_c) = (4/3) × π × r_s³

Cancel out (π/3) from both sides of the equation:

⇒ 2r_c³ = 4r_s³

⇒ r_c³ / r_s³ = 4 / 2

⇒ (r_c / r_s)³ = 2 / 1

Taking the cube root on both sides:

⇒ r_c / r_s = ∛2 / 1

∴ The correct answer is ∛2 : 1.

Additional Information

Volume of a Hemisphere

A hemisphere is exactly half of a sphere. Its volume is calculated using the formula V = (2/3)πr³.

Surface Area of a Sphere

The total surface area of a sphere is equivalent to 4 times the area of its great circle: TSA = 4πr².

Curved Surface Area of a Cone

The lateral or curved surface area is CSA = πrl, where l = √(r² + h²) is the slant height.

Shortcut Trick

Total Surface Area (TSA) of Hemisphere = 3πr²

Total Surface Area (TSA) of Cylinder = 2πr(h + r)

Equating both: 3πr² = 2πr(h + r)

⇒ 3r = 2h + 2r ⇒ r = 2h

⇒ h/r = 1/2

∴ The correct answer is 1:2.

Alternate Method

Given:

Radius of hemisphere = Radius of cylinder = r

Total Surface Area (TSA) of hemisphere = TSA of cylinder

Formula Used:

TSA of a solid hemisphere = 3πr²

TSA of a solid cylinder = 2πrh + 2πr²

r Hemisphere

r

h

Cylinder

Calculation:

⇒ 3πr² = 2πr(r + h)

⇒ 3r = 2(r + h)

⇒ 3r = 2r + 2h

⇒ 3r − 2r = 2h

⇒ r = 2h

⇒ h/r = 1/2

⇒ Ratio of height to radius (h : r) = 1 : 2

∴ The correct answer is 1:2.

Additional Information

Cylinder Formulas

Curved Surface Area (CSA) = 2πrh | Total Surface Area (TSA) = 2πr(h + r) | Volume = πr²h

Hemisphere Formulas

Curved Surface Area (CSA) = 2πr² | Total Surface Area (TSA) = 3πr² | Volume = (2/3)πr³

Sphere Formulas

Total Surface Area (TSA) = 4πr² | Volume = (4/3)πr³

Cone Formulas

Curved Surface Area (CSA) = πrl | Total Surface Area (TSA) = πr(l + r) | Volume = (1/3)πr²h

Shortcut Trick

For a cube to be divided into n³ identical pieces, it must be cut into 'n' equal parts along all 3 dimensions.

Total identical pieces = 64 = 4³

⇒ Number of parts along each dimension (n) = 4

Number of cuts required for 'n' parts = n − 1 cuts per dimension.

⇒ Minimum total cuts = 3 × (4 − 1) = 3 × 3 = 9

∴ The correct answer is 9 cuts.

Alternate Method

Given:

Total number of identical smaller cubical pieces = 64

Formula Used:

Total number of pieces = (x + 1) × (y + 1) × (z + 1)

Where x, y, and z are the number of parallel straight cuts along the length, breadth, and height of the cube respectively.

To strictly minimize the total number of cuts (x + y + z), the individual factors must be as close to each other as possible.

Large Cube cut into 64 pieces (4 × 4 × 4) 3 cuts along Height 3 cuts along Breadth 3 cuts along Length

Calculations:

⇒ 64 = 4 × 4 × 4

⇒ (x + 1) = 4 ⇒ x = 3

⇒ (y + 1) = 4 ⇒ y = 3

⇒ (z + 1) = 4 ⇒ z = 3

⇒ Minimum number of total cuts = x + y + z

⇒ Minimum cuts = 3 + 3 + 3 = 9

∴ The correct answer is 9 cuts.

Additional Information

Maximum Pieces from N Cuts

To get the maximum number of pieces from a given number of cuts (N), distribute the cuts equally among all three axes. Maximum pieces = (N/3 + 1)³.

Cubes with 3 Faces Painted

When a fully painted large cube is cut into n × n × n smaller identical pieces, the number of cubes with exactly 3 faces painted is always 8 (these are situated at the corners).

Cubes with 2 Faces Painted

The number of smaller cubes with exactly 2 faces painted lie precisely along the middle of the edges and is given by the formula 12 × (n − 2).

Cubes with 1 Face Painted

The number of smaller cubes with exactly 1 face painted lie in the center of the outer faces and is given by the formula 6 × (n − 2)².

Shortcut Trick

The volume of a cone is proportional to r² × h, so the percentage change depends on three successive changes: +20%, +20%, and −10%.

First, apply the successive percentage formula (x + y + xy/100) on the radius (+20% and +20%):

⇒ 20 + 20 + (20 × 20) ÷ 100 = 40 + 4 = +44%

Next, apply the formula on the combined radius change (+44%) and height change (−10%):

⇒ 44 − 10 + (44 × −10) ÷ 100 = 34 − 4.4 = +29.6%

Since the result is positive, the overall change is a 29.6% increase.

∴ The correct answer is 29.6% increase.

Alternate Method

Given:

Percentage increase in radius = 20%

Percentage decrease in height = 10%

Formula Used:

Volume of a cone = 1/3 × π × r² × h

r h

Original Cone Volume = 100%

+29.6%

1.2r 0.9h

New Cone Volume = 129.6%

Calculations:

⇒ Let the original radius (r) be 10 units and original height (h) be 10 units.

⇒ Original Volume (V₁) = 1/3 × π × (10)² × 10 = 1000π / 3 cubic units.

⇒ New radius = 10 + (20% of 10) = 10 + 2 = 12 units.

⇒ New height = 10 − (10% of 10) = 10 − 1 = 9 units.

⇒ New Volume (V₂) = 1/3 × π × (12)² × 9 = 1/3 × π × 144 × 9 = 1296π / 3 cubic units.

⇒ Change in volume = V₂ − V₁ = (1296π / 3) − (1000π / 3) = 296π / 3 cubic units.

⇒ Percentage change = (Change in volume ÷ Original Volume) × 100

⇒ Percentage change = [(296π / 3) ÷ (1000π / 3)] × 100

⇒ Percentage change = (296 ÷ 1000) × 100 = 29.6%

∴ The correct answer is 29.6% increase.

Additional Information

Volume of Cone vs Cylinder

The volume of a cone is exactly one-third the volume of a cylinder that shares the same base radius and height (V = 1/3 × πr²h).

Successive Percentage Change

When a value undergoes multiple percentage changes sequentially, the net percentage change for two variables is found using (a + b + ab/100)%. For formulas depending on three variables like r × r × h, apply this successive change formula twice.

Surface Area of a Cone

The curved surface area is given by πrl and the total surface area is πr(r + l), where l = √(r² + h²) is the slant height of the cone.

Shortcut Trick

Since A lies in the 2^nd quadrant, sine is positive and cosine is negative.

Using the 3-4-5 right angle triplet, if sinA = 3/5, then cosA = −4/5.

Expression: (sinA + cosA)²

⇒ (3/5 − 4/5)²

⇒ (−1/5)² = 1/25

∴ The correct answer is 1/25.

Alternate Method

Given:

sinA = 3/5

A lies in the second quadrant.

Formula Used:

Algebraic Identity: (a + b)² = a² + b² + 2ab

Trigonometric Identity: sin²A + cos²A = 1

A Base = −4 Perp = 3 Hyp = 5 y-axis x-axis 2^nd Quadrant

Calculation:

⇒ In the 2^nd quadrant, cosine values are negative.

⇒ cosA = −√(1 − sin²A)

⇒ cosA = −√(1 − 9/25) = −√(16/25) = −4/5

Now, expanding the given expression to solve it directly:

⇒ (sinA + cosA)² = sin²A + cos²A + 2 × sinA × cosA

⇒ (sinA + cosA)² = 1 + 2 × sinA × cosA

⇒ (sinA + cosA)² = 1 + 2 × (3/5) × (−4/5)

⇒ (sinA + cosA)² = 1 − 24/25

⇒ (sinA + cosA)² = (25 − 24) / 25 = 1/25

∴ The correct answer is 1/25.

Additional Information

ASTC Rule (Signs in Quadrants)

1^st Quadrant: All positive. 2^nd Quadrant: sin & csc positive. 3^rd Quadrant: tan & cot positive. 4^th Quadrant: cos & sec positive.

Pythagorean Identity

The foundational identity is sin²θ + cos²θ = 1 for any generic angle θ.

Common Pythagorean Triplets

Sets of integers that perfectly satisfy a² + b² = c². The most frequent sets in exams include (3, 4, 5), (5, 12, 13), and (8, 15, 17).

Shortcut Trick

According to the basic trigonometric identity, 1 + tan² A = sec² A.

We know that sec² A = 1 / cos² A = 1 / (1 − sin² A).

Substitute sin A = m / n into the equation:

⇒ sec² A = 1 / [1 − (m / n)²] = 1 / [(n² − m²) / n²]

⇒ sec² A = n² / (n² − m²)

∴ The correct answer is n² / (n² − m²).

Alternate Method

Given:

sin A = m / n = Perpendicular / Hypotenuse

A m (Perpendicular) n (Hypotenuse) Base = √(n² - m²)

Calculations:

⇒ Perpendicular (P) = m

⇒ Hypotenuse (H) = n

Using Pythagoras theorem:

⇒ Base (B) = √(H² − P²) = √(n² − m²)

⇒ tan A = P / B = m / √(n² − m²)

Now, we need to find the value of (1 + tan² A):

⇒ 1 + [m / √(n² − m²)]²

⇒ 1 + m² / (n² − m²)

⇒ (n² − m² + m²) / (n² − m²)

⇒ n² / (n² − m²)

∴ The correct answer is n² / (n² − m²).

Additional Information

Pythagorean Identities

Fundamental relations between trigonometric functions: sin² θ + cos² θ = 1, sec² θ − tan² θ = 1, and cosec² θ − cot² θ = 1.

Reciprocal Identities

Functions that are inverses of each other: sin θ = 1 / cosec θ, cos θ = 1 / sec θ, and tan θ = 1 / cot θ.

Quotient Identities

Relation of tangent and cotangent with sine and cosine: tan θ = sin θ / cos θ and cot θ = cos θ / sin θ.