Quantitative Aptitude PYQs

Shortcut Trick

Net efficiency = Filling efficiency − Emptying efficiency

LCM of 3 and 6 = 6 units (Total Capacity)

Efficiency: Pipe A = 6/3 = 2, Pipe B = 6/6 = 1

Time to fill half tank (3 units) = 3 ÷ (2 − 1) = 3 minutes

∴ The correct answer is 3 minutes.

Alternate Method

Given: Filling time = 3 min, Emptying time = 6 min.

Formula: Work = Efficiency × Time

⇒ Net efficiency of both pipes = 2 − 1 = 1 unit/min

⇒ Capacity of half the tank = 6 ÷ 2 = 3 units

⇒ Time required = 3 units ÷ 1 unit/min = 3 minutes

∴ The correct answer is 3 minutes.

Additional Information

Net Work in Pipes & Cisterns

The work done by an inlet pipe is positive, while the work done by an outlet pipe (leak) is negative.

Combined Time Formula

If pipe A fills in X and pipe B empties in Y, combined time T = (X × Y) ÷ (Y − X), provided Y > X.

Shortcut Trick

Let the total quantity be 2 units. Half quantity is 1 unit.

Total required gain = 2 units × 20% = 40%

Gain from first half = 1 unit × 10% = 10%

Required gain from remaining half = 40% − 10% = 30%

Since the remaining quantity is 1 unit, the gain percentage = 30% ÷ 1 = 30%.

∴ The correct answer is 30%.

Alternate Method

Given: Total CP = ₹8,000, Gain on first half = 10%, Total desired gain = 20%.

Formulas: Profit = SP − CP | SP = CP × (100 + P%) ÷ 100

⇒ Total Cost Price (CP) = ₹8,000

⇒ Total Selling Price (SP) required for 20% gain = 8000 × (120 ÷ 100) = ₹9,600

⇒ CP of half goods = 8000 ÷ 2 = ₹4,000

⇒ SP of first half at 10% gain = 4000 × (110 ÷ 100) = ₹4,400

⇒ Required SP of remaining half = 9600 − 4400 = ₹5,200

⇒ Profit needed on remaining half = 5200 − 4000 = ₹1,200

⇒ Profit% on remaining = (1200 ÷ 4000) × 100 = 30%

∴ The correct answer is 30%.

Additional Information

Weighted Average Method

When a transaction is split, the overall profit % is the weighted average of individual profit percentages: Total P% = (n₁p₁ + n₂p₂) ÷ (n₁ + n₂).

Independent of Value

In problems involving ratios or fractions of goods, the actual monetary value (like ₹8,000) does not affect the final percentage result.

Shortcut Trick

If a + 1/a = k, then a² + 1/a² = k² − 2.

Let a = √x, then a² = x.

Given √x + 1/√x = 5, we have k = 5.

Required value x + 1/x = 5² − 2 = 25 − 2 = 23.

∴ The correct answer is 23.

Alternate Method

Given:

√x + 1/√x = 5

Formula Used:

(a + b)² = a² + b² + 2ab

Calculations:

Square both sides of the given algebraic equation:

⇒ (√x + 1/√x)² = 5²

⇒ (√x)² + (1/√x)² + 2 × (√x) × (1/√x) = 25

⇒ x + 1/x + 2 = 25

⇒ x + 1/x = 25 − 2

⇒ x + 1/x = 23

∴ The correct answer is 23.

Additional Information

Square of Difference

If a − 1/a = k, then a² + 1/a² = k² + 2.

Cube of Sum

If a + 1/a = k, then a³ + 1/a³ = k³ − 3k.

Relation between Identities

(a + b)² = (a − b)² + 4ab.

Standard Algebraic Identity

(a − b)² = a² + b² − 2ab.

Shortcut Trick

Relative Speed in same direction = 114 − 60 = 54 km/hr.

Convert to m/s: 54 × (5/18) = 15 m/s.

Total distance (2L) = Speed × Time = 15 × 144 = 2160 m.

Length of one train (L) = 2160 ÷ 2 = 1080 m.

∴ The correct answer is 1080 m.

Alternate Method

Given:

Speed of Rajdhani Express (S₁) = 114 km/hr

Speed of Garib Rath Express (S₂) = 60 km/hr

Time taken (t) = 144 seconds

Lengths are equal: L₁ = L₂ = L

Formula Used:

Relative Speed (Same direction) = S₁ − S₂

Total Distance = Sum of lengths of trains = L₁ + L₂

Distance = Speed × Time

Calculations:

⇒ Relative Speed = 114 − 60 = 54 km/hr

⇒ Speed in m/s = 54 × (5/18) = 15 m/s

⇒ Total Distance = 2L

⇒ 2L = 15 × 144

⇒ 2L = 2160 m

⇒ L = 2160 ÷ 2 = 1080 m

∴ The correct answer is 1080 m.

Additional Information

Relative Speed Concepts

When moving in the same direction, relative speed is (S₁ − S₂). When moving in opposite directions, it is (S₁ + S₂).

Unit Conversion

To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.

Train Crossing Distance

When a train crosses another train or a bridge, the total distance covered is the sum of their lengths (L₁ + L₂).

Shortcut Trick

For a Square, the Perimeter is 4 times its Side.

Side = Perimeter ÷ 4

Side = 68 ÷ 4 = 17 m

∴ The correct answer is 17 m.

Alternate Method

Given:

Perimeter of the Square = 68 m

Formula Used:

Perimeter of a Square = 4 × Side

Calculations:

⇒ Let the Side of the Square be 'a' meters.

⇒ 4 × a = 68

⇒ a = 68 ÷ 4

⇒ a = 17 m

∴ The correct answer is 17 m.

Additional Information

Area of a Square

The total space occupied by a square is calculated as Side × Side or a².

Diagonal of a Square

The Diagonal of a square with side 'a' is given by the formula d = a√2.

Properties of a Square

All four sides are equal, all internal angles are 90°, and diagonals bisect each other at 90°.

Shortcut Trick

Directly substitute the standard trigonometric values: cot 45° = 1, sec 30° = 2/√3, and sin 30° = 1/2.

Expression → 2(1)² + 3(2/√3)² − 4(1/2)²

⇒ 2(1) + 3(4/3) − 4(1/4) = 2 + 4 − 1 = 5.

∴ The correct answer is 5.

Alternate Method

Given: 2cot²45° + 3sec²30° − 4sin²30°

Formula Used: Values of trigonometric ratios at specific angles.

⇒ 2 × (1)² + 3 × (2/√3)² − 4 × (1/2)²

⇒ 2 × 1 + 3 × (4/3) − 4 × (1/4)

⇒ 2 + 4 − 1

⇒ 6 − 1 = 5

∴ The correct answer is 5.

Additional Information

Trigonometric Identities

Fundamental relations such as sin²θ + cos²θ = 1 and 1 + tan²θ = sec²θ.

Reciprocal Relations

sec θ = 1/cos θ, cosec θ = 1/sin θ, and cot θ = 1/tan θ.

Complementary Angles

sin(90° − θ) = cos θ, tan(90° − θ) = cot θ, and sec(90° − θ) = cosec θ.

Shortcut Trick

Net change formula: Net % = a + b + (a × b) ÷ 100.

Here, markup a = +80% and discount b = −35%.

Net Profit % = 80 − 35 + [80 × (−35)] ÷ 100.

Net Profit % = 45 − 2800 ÷ 100 = 45 − 28 = 17%.

∴ The correct answer is 17%.

Alternate Method

Given: Markup = 80% over CP, Discount = 35% on MP.

Formulas: MP = CP × (100 + Markup%) ÷ 100 | SP = MP × (100 − Discount%) ÷ 100 | Profit% = (SP − CP) ÷ CP × 100.

⇒ Let Cost Price (CP) = 100.

⇒ Marked Price (MP) = 100 + 80% of 100 = 180.

⇒ Selling Price (SP) = 180 − 35% of 180.

⇒ SP = 180 − (35 ÷ 100 × 180) = 180 − 63 = 117.

⇒ Profit = SP − CP = 117 − 100 = 17.

⇒ Profit % = (17 ÷ 100) × 100 = 17%.

∴ The correct answer is 17%.

Additional Information

Markup and Discount Relation

Profit occurs when the markup percentage is greater than the equivalent discount percentage applied to the marked price.

Successive Percentage Application

Markup and Discount are successive operations on the CP. The final result compared to 100 gives the net profit or loss percentage.

Shortcut Trick

Let total capacity of cistern be LCM(99, 36, 44) = 396 units.

Efficiency of Pipe 1 = 396 ÷ 99 = 4 units/min.

Efficiency of Pipe 2 = 396 ÷ 36 = 11 units/min.

Combined efficiency with Waste Pipe = 396 ÷ 44 = 9 units/min.

Efficiency of Waste Pipe = (4 + 11) − 9 = 6 units/min.

Time taken = 396 ÷ 6 = 66 min.

∴ The correct answer is 66 min.

Alternate Method

Given: Filling times for individual pipes and total time when all pipes (including emptying pipe) are open.

Formula Used: Total Work = Efficiency × Time; Net Efficiency = ∑(Inlet Efficiencies) − ∑(Outlet Efficiencies).

⇒ Let efficiency of emptying pipe be C.

⇒ Efficiency of Pipe A = 396 ÷ 99 = 4

⇒ Efficiency of Pipe B = 396 ÷ 36 = 11

⇒ Net efficiency (A + B − C) = 396 ÷ 44 = 9

⇒ 4 + 11 − C = 9

⇒ 15 − C = 9

⇒ C = 15 − 9 = 6 units/min

⇒ Time taken by pipe C = Total capacity ÷ Efficiency of C = 396 ÷ 6 = 66 min.

∴ The correct answer is 66 min.

Additional Information

Pipes and Cisterns Convention

Inlet pipes contribute positive efficiency (+) while outlet or leak pipes contribute negative efficiency (−).

Net Work Formula

If pipe A fills in x hours and pipe B empties in y hours, then net work per hour = (1/x − 1/y). If the result is negative, the tank is being emptied.

Efficiency-Time Relationship

Efficiency is inversely proportional to the time taken to complete the work, provided total work remains constant.

Shortcut Trick

Let the elder son be x. Then father = 2x and younger son = x − 8.

Total Ratio sum: 2x + x + (x − 8) = 72 ⇒ 4x − 8 = 72.

4x = 80 ⇒ x = 20.

Father's age = 2 × 20 = 40 years.

∴ The correct answer is 40 years.

Alternate Method

Given: Sum of ages = 72 years. Father = 2 × Elder son. Younger son = Elder son − 8.

⇒ Let Elder son's age = E

⇒ Father's age = 2E

⇒ Younger son's age = E − 8

⇒ Sum = Father + Elder son + Younger son = 72

⇒ 2E + E + (E − 8) = 72

⇒ 4E − 8 = 72

⇒ 4E = 80

⇒ E = 20 years

⇒ Father's age = 2 × 20 = 40 years

∴ The correct answer is 40 years.

Additional Information

Ratio of Ages

If age ratio is A:B, then actual ages are Ax and Bx. The sum is (A+B)x.

Ages in Future/Past

If present age is x, then age 'n' years ago was (x − n) and age 'n' years hence will be (x + n).

Age Difference

The difference between the ages of two persons remains constant regardless of the number of years passed.

Shortcut Trick

Check odd numbers between 358 and 368: 359, 361, 363, 365, 367.

359: Not divisible by 2, 3, 5, 7, 11, 13, 17 → Prime.

361: 19 × 19 → Not Prime.

367: Not divisible by 2, 3, 5, 7, 11, 13, 17, 19 → Prime.

∴ The correct answer is 2.

Alternate Method

Given: Range between 358 and 368.

Step 1: List all integers in the range: 359, 360, 361, 362, 363, 364, 365, 366, 367.

Step 2: Eliminate even numbers (360, 362, 364, 366) as they are divisible by 2.

Step 3: Test remaining odd numbers:

⇒ 359: Sum of digits = 17 (not div. by 3), ends in 9 (not div. by 5), 359 ÷ 7 (rem 2), 359 ÷ 11 (rem 7), 359 ÷ 13 (rem 8), 359 ÷ 17 (rem 2) → Prime.

⇒ 361: √361 = 19. Since 19 × 19 = 361 → Composite.

⇒ 363: Sum of digits = 12 (divisible by 3) → Composite.

⇒ 365: Ends in 5 (divisible by 5) → Composite.

⇒ 367: Sum of digits = 16 (not div. by 3), 367 ÷ 7 (rem 3), 367 ÷ 11 (rem 4), 367 ÷ 13 (rem 3), 367 ÷ 17 (rem 10), 367 ÷ 19 (rem 6) → Prime.

⇒ Total prime numbers = 359 and 367.

∴ The correct answer is 2.

Additional Information

Definition of Prime Numbers

A natural number greater than 1 that has no positive divisors other than 1 and itself.

Testing for Primality

To check if N is prime, test divisibility by all prime numbers up to √N.

Divisibility Rule of 3

A number is divisible by 3 if the sum of its digits is a multiple of 3.

Shortcut Trick

For a hollow cylindrical tube open at both ends:

Total Surface Area (TSA) = 2π(R + r)(h + R − r)

Given: External radius (R) = 6 cm, Internal radius (r) = 5.6 cm, Height (h) = 63 cm.

⇒ TSA = 2 × (22/7) × 11.6 × (63 + 0.4) = 2 × (22/7) × 11.6 × 63.4 ≈ 4622.77 cm².

Rounding off to the nearest multiple of 5, we get 4625 cm².

∴ The correct answer is 4625 cm².

Alternate Method

Given:

External Diameter = 12 cm ⇒ External Radius (R) = 12 ÷ 2 = 6 cm

Thickness (t) = 0.4 cm

Internal Radius (r) = R − t = 6 − 0.4 = 5.6 cm

Length (h) = 63 cm

Formula Used:

Total Surface Area (TSA) of a hollow cylinder = Outer CSA + Inner CSA + 2 × Base Area

⇒ TSA = 2πRh + 2πrh + 2π(R² − r²)

Calculations:

⇒ TSA = 2π[Rh + rh + (R + r)(R − r)]

⇒ TSA = 2 × (22/7) × [(6 × 63) + (5.6 × 63) + (6 + 5.6)(6 − 5.6)]

⇒ TSA = (44/7) × [378 + 352.8 + (11.6 × 0.4)]

⇒ TSA = (44/7) × [730.8 + 4.64]

⇒ TSA = (44/7) × 735.44

⇒ TSA = 44 × 105.0628...

⇒ TSA ≈ 4622.76 cm²

⇒ Rounding to the nearest multiple of 5: 4622.76 is between 4620 and 4625.

⇒ Distance to 4620 = 2.76, Distance to 4625 = 2.24.

∴ The correct answer is 4625 cm².

Additional Information

Volume of Hollow Cylinder

The volume represents the material used and is given by V = π(R² − r²)h.

Curved Surface Area (CSA)

For a hollow cylinder, the total CSA is the sum of the external and internal areas: CSA = 2πh(R + r).

Base Area of Ring

The area of the circular cross-section (ring) is given by A = π(R² − r²).

Shortcut Trick

Total distance = 390 + 390 = 780 m

Total time = 65 + 195 = 260 sec

Average speed = 780 ÷ 260 = 3 m/sec

∴ The correct answer is 3 m/sec.

Alternate Method

Given:

⇒ Distance of first lap = 390 m

⇒ Time for first lap = 65 sec

⇒ Distance of second lap = 390 m

⇒ Time for second lap = 195 sec

Formula Used:

⇒ Average Speed = Total Distance ÷ Total Time

Calculations:

⇒ Total Distance = 390 + 390 = 780 m

⇒ Total Time = 65 + 195 = 260 sec

⇒ Average Speed = 780 ÷ 260

⇒ Average Speed = 3 m/sec

∴ The correct answer is 3 m/sec.

Additional Information

Average Speed for Equal Distances

When a body travels equal distances with different speeds u and v, the average speed is given by the harmonic mean: 2uv ÷ (u + v).

Speed-Time-Distance Basics

Speed is directly proportional to distance when time is constant, and inversely proportional to time when distance is constant.

Shortcut Trick
Net time for filling = (Filling Time × Emptying Time) ÷ (Emptying Time − Filling Time)
Time = (3 × 5) ÷ (5 − 3)
Time = 15 ÷ 2 = 7.5 hr
∴ The correct answer is 7.5 hr.
Alternate Method

Component
Time (hr)
Total Work (LCM)
Efficiency

Filling Tap
3
15 Units
+5

Emptying Hole
5
−3

Calculations:
⇒ Net Efficiency when both are open = 5 − 3 = 2 units/hr
⇒ Total Time taken to fill = Total Work ÷ Net Efficiency
⇒ Time = 15 ÷ 2
⇒ Time = 7.5 hr
∴ The correct answer is 7.5 hr.
Additional Information
Pipes and Cisterns Concept
Filling pipes have positive efficiency, while outlet pipes (leaks) have negative efficiency. Total capacity is the LCM of individual times.
Combined Rate Formula
If pipe A fills in x hours and pipe B empties in y hours (where y > x), then the net rate is (1/x − 1/y) per hour.

Shortcut Trick

Let Emma's age be 1 unit.

Sarah's age = 2 units + 5 years.

Sum: 3 units + 5 = 29 ⇒ 3 units = 24 ⇒ 1 unit = 8 years.

Sarah's age = (2 × 8) + 5 = 21 years.

∴ The correct answer is 21 years.

Alternate Method

Given:

Sum of Sarah and Emma's ages = 29 years.

Sarah's age is 5 years more than twice Emma's age.

Let Emma's age = x years.

Then, Sarah's age = 2x + 5.

⇒ x + (2x + 5) = 29

⇒ 3x + 5 = 29

⇒ 3x = 29 − 5

⇒ 3x = 24

⇒ x = 8

⇒ Sarah's age = 2(8) + 5 = 16 + 5 = 21 years.

∴ The correct answer is 21 years.

Additional Information

Linear Equations in Ages

Problems involving ages are typically solved by assigning a variable to the youngest person to simplify multiplication and addition.

Ratio of Ages

If the ratio of ages is given as a:b, the ages can be represented as ax and bx to form an equation based on the given sum or difference.

Verification Tip

Always verify by adding the results: 21 (Sarah) + 8 (Emma) = 29, which matches the total age provided in the question.

Shortcut Trick

Any non-zero number raised to power 0 gives 1

⇒ 5^x+1 = 1 ⇒ exponent must be 0

⇒ x + 1 = 0

⇒ x = −1

∴ Answer = −1

Alternate Method

Given:

5^x+1 = 1

Formula used:

a⁰ = 1 (for a ≠ 0)

Calculation:

⇒ Since base 5 ≠ 1, exponent must be 0

⇒ x + 1 = 0

⇒ x = −1

∴ The correct answer is −1.

Additional Information

Exponent Rule

a⁰ = 1 for any non-zero number a.

Exponential Equations

If base is same, equate exponents; if result is 1, exponent must be 0.

Special Case

Only 1^{any number} = 1, otherwise exponent = 0.

Shortcut Trick

Combined efficiency of P + Q = 1/12 and efficiency of P = 1/21.

Efficiency of Q = 1/12 − 1/21 = (7 − 4)/84 = 3/84 = 1/28.

Time taken by Q to fill the full cistern = 28 hours.

Time taken by Q to fill one-fourth (1/4) cistern = 28 ÷ 4 = 7 hours.

∴ The correct answer is 7 hours.

Alternate Method

Given: Time taken by (P + Q) = 12 hours, Time taken by P alone = 21 hours.

⇒ Efficiency of Q = Efficiency of (P + Q) − Efficiency of P

⇒ Efficiency of Q = 7 − 4 = 3 units/hour

⇒ Required work (1/4 of total) = 84 ÷ 4 = 21 units

⇒ Time taken by Q = Required work ÷ Efficiency of Q

⇒ Time taken by Q = 21 ÷ 3 = 7 hours

∴ The correct answer is 7 hours.

Additional Information

Pipes & Cisterns Rule

Efficiency is inversely proportional to the time taken to complete the work. Total Work = Time × Efficiency.

Negative Work

In cistern problems, an inlet pipe does positive work while an outlet (leak) pipe does negative work.

Work & Wages

Wages are always distributed in the ratio of the efficiency or the actual work done by the individuals.

Shortcut Trick

Put 3x = 0° in the given expression.

Expression ⇒ (3/4)(sin² 0° − tan² 0°) + (3/4)(cos² 0° + sec² 0°)

⇒ (3/4)(0 − 0) + (3/4)(1² + 1²)

⇒ 0 + (3/4) × 2 = 1.5

∴ The correct answer is 1.5.

Alternate Method

Given: (3/4)(sin² 3x − tan² 3x) + (3/4)(cos² 3x + sec² 3x)

Formula Used: sin² θ + cos² θ = 1 and sec² θ − tan² θ = 1

⇒ Factor out 3/4 from the entire expression:

⇒ (3/4) [sin² 3x − tan² 3x + cos² 3x + sec² 3x]

⇒ Rearrange the terms: (3/4) [(sin² 3x + cos² 3x) + (sec² 3x − tan² 3x)]

⇒ Substitute the identities: (3/4) [1 + 1]

⇒ (3/4) × 2 = 1.5

∴ The correct answer is 1.5.

Additional Information

Trigonometric Identities

Standard Pythagorean identities include sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Value Substitution Method

For expressions independent of the angle, substituting simple values like 0°, 30°, 45°, or 90° helps in quick evaluation, provided the function is defined at that point.

Shortcut Trick

For any quadrilateral PQRS, the angle formed by the bisectors of two adjacent angles (∠R and ∠S) is equal to half the sum of the remaining two angles (∠P and ∠Q):

∠STR = (∠P + ∠Q) ÷ 2

⇒ 33° = (∠P + ∠Q) ÷ 2

⇒ ∠P + ∠Q = 66°

Given the ratio of ∠P : ∠Q = 3 : 8, the sum of the ratio terms is 11 units (representing 66°), so 1 unit = 6°.

The difference between ∠Q and ∠P is 8 − 3 = 5 units ⇒ 5 × 6° = 30°.

∴ The correct answer is 30°.

Alternate Method

Given:

• In quadrilateral PQRS, the bisectors of ∠R and ∠S meet at T.
• ∠STR = 33°
• Ratio of ∠P to ∠Q = 3 : 8

Formula Used:

• Angle sum property of a triangle: Sum of all interior angles of a triangle is 180°.
• Angle sum property of a quadrilateral: Sum of all interior angles of a quadrilateral is 360°.

Calculations:

In triangle STR:

⇒ ∠TSR + ∠TRS + ∠STR = 180°

Since ST and RT are the angle bisectors of ∠S and ∠R respectively:

⇒ (1/2)∠S + (1/2)∠R + 33° = 180°

⇒ (1/2)(∠S + ∠R) = 180° − 33°

⇒ (1/2)(∠S + ∠R) = 147°

⇒ ∠S + ∠R = 294°

Now, in quadrilateral PQRS, the sum of all interior angles is 360°:

⇒ ∠P + ∠Q + ∠R + ∠S = 360°

Substitute the value of (∠S + ∠R):

⇒ ∠P + ∠Q + 294° = 360°

⇒ ∠P + ∠Q = 360° − 294°

⇒ ∠P + ∠Q = 66°

Let the measures of ∠P and ∠Q be 3x and 8x respectively:

⇒ 3x + 8x = 66°

⇒ 11x = 66°

⇒ x = 6°

The difference between the measures of ∠Q and ∠P is:

⇒ ∠Q − ∠P = 8x − 3x = 5x

⇒ 5 × 6° = 30°

Explanation of incorrect options:

• Option (1) 26°: If the difference were 26°, then 5x = 26° ⇒ x = 5.2°, which gives ∠P + ∠Q = 11 × 5.2° = 57.2°, which is incorrect as the sum must be 66°.
• Option (3) 34°: If the difference were 34°, then 5x = 34° ⇒ x = 6.8°, which gives ∠P + ∠Q = 11 × 6.8° = 74.8°, which is incorrect as the sum must be 66°.
• Option (4) 32°: If the difference were 32°, then 5x = 32° ⇒ x = 6.4°, which gives ∠P + ∠Q = 11 × 6.4° = 70.4°, which is incorrect as the sum must be 66°.

∴ The correct answer is 30°.

Additional Information

Angle Bisector Theorem in Quadrilaterals

The angle formed by the intersection of the bisectors of two adjacent angles of a quadrilateral is equal to half the sum of the other two angles: ∠STR = (∠P + ∠Q) ÷ 2.

Sum of Interior Angles of a Polygon

For any n-sided polygon, the sum of all interior angles is given by the formula (n − 2) × 180°. For a quadrilateral (n = 4), this sum is always 360°.

Angle Bisectors of Opposite Angles

In a cyclic quadrilateral, the angle bisectors of any two opposite angles intersect on the circumcircle of the quadrilateral.

Shortcut Trick

Triangle ratio 3:4:5 implies a Right-Angled Triangle with Perimeter = 12 units and Area = (1/2 × 3 × 4) = 6 units².

Square perimeter = Triangle perimeter = 12 units ⇒ Side of square = 12/4 = 3 units.

Area of Square = 3² = 9 units².

Difference in area = (9 − 6)x² = 3x² = 48 ⇒ x² = 16 ⇒ x = 4.

Perimeter of triangle = 12x = 12 × 4 = 48 cm.

∴ The correct answer is 48 cm.

Alternate Method

Given:

Ratio of sides of Triangle = 3 : 4 : 5

Difference in Areas = 48 cm²

Formula Used:

Area of Right Triangle = 1/2 × base × height

Perimeter of Triangle = Sum of all sides

Perimeter of Square = 4 × side

Area of Square = (side)²

Calculations:

Let the sides of the triangle be 3x, 4x, and 5x.

⇒ Perimeter of Triangle (P_T) = 3x + 4x + 5x = 12x

⇒ Since (3x)² + (4x)² = (5x)², it is a Right-Angled Triangle.

⇒ Area of Triangle (A_T) = 1/2 × 3x × 4x = 6x²

⇒ Perimeter of Square (P_S) = P_T = 12x

⇒ Side of Square (a) = 12x ÷ 4 = 3x

⇒ Area of Square (A_S) = (3x)² = 9x²

⇒ Difference = A_S − A_T = 9x² − 6x² = 3x²

⇒ 3x² = 48

⇒ x² = 16 ⇒ x = 4

⇒ Perimeter of Triangle = 12x = 12 × 4 = 48 cm

∴ The correct answer is 48 cm.

Additional Information

Pythagorean Triplets

Sets of three positive integers (a, b, c) such that a² + b² = c². Common examples include (3, 4, 5), (5, 12, 13), and (8, 15, 17).

Maximum Area for Fixed Perimeter

For a given perimeter, a Circle has the largest area, followed by a regular polygon (like a square) and then a triangle.

Heron's Formula

Used to find the area of any triangle when all three sides (a, b, c) are known: Area = √[s(s−a)(s−b)(s−c)], where s = (a+b+c)/2.

Shortcut Trick

Voters participated = 86% of 18,000

⇒ 86 × 180 (Cancelling two zeros of 100 with 18,000)

⇒ (80 + 6) × 180 = 14,400 + 1,080 = 15,480

∴ The correct answer is 15,480 voters.

Alternate Method

Given:

Total registered voters = 18,000

Voting percentage = 86%

Calculations:

⇒ Voters participated = (86 ÷ 100) × 18,000

⇒ Voters participated = 86 × 180

⇒ Voters participated = 15,480

∴ The correct answer is 15,480 voters.

Additional Information

Percentage Basics

To find x% of a value y, the formula is (x ÷ 100) × y.

Non-Participating Voters

The percentage of voters who did not vote is (100 − participated percentage). In this case, 14%.

Percentage to Fraction

86% can be expressed as 86/100 or 43/50 in simplified fraction form.