Quantitative Aptitude PYQs

Shortcut Trick

Sum of ages 5 years ago → 67 − (5 × 2) = 57 years.

Age ratio 5 years ago (Elder : Younger) = 2 :

1. Elder sister's age 5 years ago = (2/3) × 57 = 38 years.

Current age of elder sister = 38 + 5 = 43 years.

∴ The correct answer is 43 years.

Alternate Method

Given: Sum of present ages = 67 years. Ratio 5 years ago = 2 :

1. Formula: Age after 'n' years = Age 'n' years ago + (n × 2 for two people).

⇒ Let ages 5 years ago be 2x and x.

⇒ Present ages = (2x + 5) and (x + 5).

⇒ (2x + 5) + (x + 5) = 67

⇒ 3x + 10 = 67

⇒ 3x = 57 ⇒ x = 19.

⇒ Elder sister's current age = 2x + 5 = (2 × 19) + 5 = 38 + 5 = 43.

∴ The correct answer is 43 years.

Additional Information

Age Difference Constant

The difference between the ages of two people always remains constant, regardless of the number of years passed.

Sum of Ages Change

If there are 'n' people, the sum of their ages increases or decreases by (n × number of years).

Ratio Method

When a constant number of years is added to both parts of a ratio, the difference between ratio units should be balanced if the age difference is known.

Shortcut Trick

Given sec θ = Hypotenuse / Base = 17 / 12.

Using Pythagoras Theorem: Perpendicular² = Hypotenuse² − Base².

P² = 17² − 12² = 289 − 144 = 145 ⇒ P = √145.

sin θ = Perpendicular / Hypotenuse = √145 / 17.

∴ The correct answer is √145/17.

Alternate Method

Given: sec θ = 17/12

Formula Used: sin²θ + cos²θ = 1 OR sec θ = 1 / cos θ

⇒ cos θ = 1 / sec θ = 12 / 17

⇒ sin²θ = 1 − cos²θ

⇒ sin²θ = 1 − (12/17)²

⇒ sin²θ = 1 − 144 / 289

⇒ sin²θ = (289 − 144) / 289 = 145 / 289

⇒ sin θ = √(145 / 289) = √145 / 17

∴ The correct answer is √145/17.

Additional Information

Trigonometric Identities

Fundamental relations include sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Right Angled Triangle Ratios

In a right triangle: sin θ = P/H, cos θ = B/H, tan θ = P/B, where P is perpendicular, B is base, and H is hypotenuse.

Pythagoras Theorem

For any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: H² = P² + B².

Shortcut Trick

Average Speed = Total Distance ÷ Total Time

Total Distance = 430 + 470 + 620 = 1520 m

Total Time = 10 + 11 + 17 = 38 minutes

Average Speed = 1520 ÷ 38 = 40 m/minutes

∴ The correct answer is 40 m/minutes.

Alternate Method

Given:

Distance 1 (d₁) = 430 m, Time 1 (t₁) = 10 minutes

Distance 2 (d₂) = 470 m, Time 2 (t₂) = 11 minutes

Distance 3 (d₃) = 620 m, Time 3 (t₃) = 17 minutes

Formula Used:

Average Speed = (d₁ + d₂ + d₃) ÷ (t₁ + t₂ + t₃)

Calculations:

⇒ Total Distance = 430 + 470 + 620 = 1520 m

⇒ Total Time = 10 + 11 + 17 = 38 minutes

⇒ Average Speed = 1520 ÷ 38

⇒ Average Speed = 40 m/minutes

∴ The correct answer is 40 m/minutes.

Additional Information

Average Speed for Equal Distances

If a person covers two equal distances at speeds x and y, the average speed is given by (2xy) ÷ (x + y).

Relative Speed

Speed of two bodies moving in the same direction is (u − v) and in opposite directions is (u + v).

Unit Conversion

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

Shortcut Trick

Deviation = New Score − Old Average = 157 − 90 = 67 runs.

Increase in Average = Deviation ÷ Total Innings = 67 ÷ 7 ≈ 9.57 runs.

Rounding 9.57 to the nearest whole number gives 10 runs.

∴ The correct answer is 10 runs.

Alternate Method

Given:

Average for 6 innings = 90 runs

Runs in 7^th inning = 157 runs

Formula Used:

Average = Total Runs ÷ Number of Innings

Calculations:

⇒ Total runs in 6 innings = 6 × 90 = 540 runs

⇒ Total runs in 7 innings = 540 + 157 = 697 runs

⇒ New average = 697 ÷ 7 = 99.5714... runs

⇒ Increase in average = 99.57... − 90 = 9.57... runs

⇒ Rounding 9.57 to the nearest whole number ≈ 10

∴ The correct answer is 10 runs.

Additional Information

Average Change Formula

When a new value is added, New Average = Old Average + (Extra Runs ÷ New Total Innings).

Weighted Average

Weighted Avg = (n₁ × A₁ + n₂ × A₂) ÷ (n₁ + n₂), where n is the number of items and A is the average.

Batting Average in Cricket

Batting Average = (Total Runs Scored) ÷ (Number of times Out). In simple problems, innings played is often treated as times out.

Shortcut Trick

Check options for Sum = 88 and LCM = 870.

Option (1): 30 + 58 = 88. LCM of 30 and 58: 30 = 2 × 3 × 5; 58 = 2 × 29. LCM = 2 × 3 × 5 × 29 = 870.

Option (2): 28 + 60 = 88. LCM ends in 0 but 28 is not a factor of 870.

Option (3): 23 + 65 = 88. 23 is prime, 870 is not divisible by 23.

∴ The correct answer is 30, 58.

Alternate Method

Given: Sum of numbers (x + y) = 88, LCM(x, y) = 870.

Formula: Product of two numbers = HCF × LCM.

⇒ Let the numbers be x and y.

⇒ x + y = 88

⇒ LCM(x, y) = 870

⇒ Prime factors of 870 = 2 × 3 × 5 × 29

⇒ Since sum is 88, one number must be 2 × 29 = 58 (as 29 is a large prime factor).

⇒ If one number is 58, then the other number = 88 − 58 = 30.

⇒ Verification: LCM(30, 58) = LCM(2 × 3 × 5, 2 × 29) = 2 × 3 × 5 × 29 = 870.

∴ The correct answer is 30, 58.

Additional Information

Product Rule

For any two numbers a and b, a × b = HCF(a, b) × LCM(a, b).

HCF of Sum and LCM

The HCF of two numbers also divides their sum and their LCM.

Prime Factorization Method

LCM is the product of the highest powers of all prime factors involved in the numbers.

Shortcut Trick

Identify square roots by looking at unit digits and nearest tens.

√6889 → ends in 9 (3 or 7), between 80² and 90² → 83.

√4761 → ends in 1 (1 or 9), between 60² and 70² → 69.

8√x = 83 + 69 = 152 ⇒ √x = 19 ⇒ x = 361.

∴ The correct answer is 361.

Alternate Method

Given: 8(x)^1/2 = √6889 + √4761

Step 1: Calculate √6889

⇒ 80² = 6400 and 90² = 8100. Since 6889 ends in 9, the square root must be 83 or 87.

⇒ 85² = 7225. Since 6889 < 7225, √6889 = 83.

Step 2: Calculate √4761

⇒ 60² = 3600 and 70² = 4900. Since 4761 ends in 1, the square root must be 61 or 69.

⇒ 65² = 4225. Since 4761 > 4225, √4761 = 69.

Step 3: Substitute values back into the equation

⇒ 8√x = 83 + 69

⇒ 8√x = 152

⇒ √x = 152 ÷ 8

⇒ √x = 19

Step 4: Squaring both sides to find x

⇒ x = 19²

⇒ x = 361

∴ The correct answer is 361.

Additional Information

Square Root Extraction

For a perfect square ending in 9, the unit digit of the root is 3 or 7; for 1, it is 1 or 9.

Squaring trick for 5

A number (n5)² can be calculated as [n × (n+1)] followed by 25. e.g., 65² = (6×7)25 = 4225.

BODMAS Rule

Always solve operations in the order: Brackets, Orders (powers/roots), Division, Multiplication, Addition, Subtraction.

Shortcut Trick

Let Total Capacity be LCM(9, 12, 18) = 36 units.

Efficiencies: A = 4, B = 3, C = −2. Combined efficiency (A+B−C) = 4 + 3 − 2 = 5.

Work in first 4 hours = 5 × 4 = 20 units. Remaining work = 36 − 20 = 16 units.

Time for A+B to finish = 16 ÷ 7 hours. Total time = 4 + 16/7 = (28 + 16)/7 = 44/7 hours.

∴ The correct answer is 44/7 hr.

Alternate Method

Given: Filling time Pipe A = 9 hr, Pipe B = 12 hr, Emptying time Pipe C = 18 hr.

⇒ Total capacity of tank = 36 units (LCM of 9, 12, 18).

⇒ Efficiency of (A + B + C) = 4 + 3 − 2 = 5 units/hr.

⇒ Water filled in 4 hours = 5 × 4 = 20 units.

⇒ Remaining capacity to be filled = 36 − 20 = 16 units.

⇒ Time taken by (A + B) to fill remaining part = 16 ÷ (4 + 3) = 16/7 hours.

⇒ Total time taken = 4 + 16/7 = 44/7 hours.

∴ The correct answer is 44/7 hr.

Additional Information

Inlet and Outlet Pipes

Inlet pipes add water (positive efficiency) while outlet pipes remove water (negative efficiency). Net work per hour is the algebraic sum of efficiencies.

Time and Work Relation

Total Work = Efficiency × Time. If a pipe can fill a tank in 'x' hours, its 1-hour work is 1/x of the total capacity.

Shortcut Trick

When multiple identical spheres are melted to form a single sphere, the volume remains constant.

Relationship: R³ = n × r³, where R is the new radius, n is the count, and r is the old radius.

R³ = 8 × 10³

R³ = 8 × 1000 = 8000 ⇒ R = 20 cm

∴ The correct answer is 20 cm.

Alternate Method

Given:

Number of spherical balls (n) = 8

Radius of each small ball (r) = 10 cm

Formula Used:

Volume of a Sphere = 4/3 × π × radius³

Calculations:

⇒ Total volume of 8 small balls = 8 × (4/3 × π × 10³)

⇒ Volume of the new single large ball = 4/3 × π × R³

⇒ According to the principle of conservation of volume:

⇒ 4/3 × π × R³ = 8 × 4/3 × π × 1000

⇒ R³ = 8 × 1000

⇒ R³ = 8000

⇒ R = √³8000

⇒ R = 20 cm

∴ The correct answer is 20 cm.

Additional Information

Volume Conservation

When a solid is melted and recast into another shape, the total volume remains identical unless specified otherwise.

Surface Area of Sphere

The Total Surface Area (TSA) of a sphere is given by 4 × π × r². Note that TSA changes during recasting.

Radius Ratio

If n spheres are melted into 1, the ratio of radii is R/r = n^1/3. For n=8, R/r = 8^1/3 = 2.

Shortcut Trick

Assume a base mean of 50 and find deviations: +28, +16, +39, +5, +23, −5, −27, −4, −21.

Calculate Net Deviation: 28 + 16 + 39 + 5 + 23 − 5 − 27 − 4 − 21 = +54.

Average Deviation = 54 ÷ 9 = +6.

Final Mean = Assumed Mean + Average Deviation = 50 + 6 = 56.

∴ The correct answer is 56.

Alternate Method

Given observations: 78, 66, 89, 55, 73, 45, 23, 46, 29

Number of observations (n) = 9

Formula Used: Arithmetic Mean = (Sum of all observations) ÷ (Total number of observations)

⇒ Total Sum = 78 + 66 + 89 + 55 + 73 + 45 + 23 + 46 + 29

⇒ Total Sum = 504

⇒ Arithmetic Mean = 504 ÷ 9

⇒ Arithmetic Mean = 56

∴ The correct answer is 56.

Additional Information

Properties of Arithmetic Mean

The sum of the deviations of all observations from their arithmetic mean is always zero: Σ(x_i − Mean) = 0.

Effect of Operations

If each observation is increased, decreased, multiplied, or divided by a constant k, the mean also changes by the same operation with k.

Combined Mean Formula

For two groups with sizes n₁, n₂ and means x₁, x₂, the combined mean is (n₁x₁ + n₂x₂) ÷ (n₁ + n₂).

Shortcut Trick

Total multiplier for consecutive discounts of 2%, 8%, and 15%.

Effective Multiplier = 0.98 × 0.92 × 0.85 = 0.76636.

Equivalent Discount% = (1 − 0.76636) × 100 = 23.364%.

Rounding to two decimal places, we get 23.36%.

∴ The correct answer is 23.36%.

Alternate Method

Given → Discounts are 2%, 8%, and 15%.
Formulas: Successive Discount = a + b − (a × b) ÷ 100.

⇒ First, combine 2% and 8%: 2 + 8 − (2 × 8) ÷ 100 = 10 − 0.16 = 9.84%.
⇒ Now, combine 9.84% and 15%: 9.84 + 15 − (9.84 × 15) ÷ 100.
⇒ Total = 24.84 − (147.6 ÷ 100) = 24.84 − 1.476.
⇒ Total Discount = 23.364%.

∴ The correct answer is 23.36%.

Additional Information

Single Equivalent Discount

It is the total reduction offered on the Marked Price through multiple stages. For three discounts a, b, and c, use the formula iteratively or the Multiplier Method.

Marked Price and Selling Price

The Selling Price (SP) after discount is: SP = MP × (100 − D₁)/100 × (100 − D₂)/100...

Shortcut Trick

Combine Ratio: A : B : C = (4 × 9) : (15 × 9) : (15 × 3) = 36 : 135 : 45.

Simplified Ratio: 12 : 45 : 15 ⇒ 4 : 15 : 5.

Total units = 4 + 15 + 5 = 24 units ⇒ ₹21,800.

B's share = (15 ÷ 24) × 21,800 = (5 ÷ 8) × 21,800 = 5 × 2725 = ₹13,625.

∴ The correct answer is ₹13,625.

Alternate Method

Given: Total Profit = ₹21,800, A : B = 4 : 15, B : C = 9 : 3.

Formula: Combined Ratio A : B : C = (A × B₁) : (B₁ × B₂) : (B₂ × C).

⇒ To equalize B, multiply A : B by 3 and B : C by 5.

⇒ A : B = 12 : 45.

⇒ B : C = 45 : 15.

⇒ A : B : C = 12 : 45 : 15 = 4 : 15 : 5.

⇒ Sum of ratio parts = 4 + 15 + 5 = 24.

⇒ B's share = (15 ÷ 24) × 21,800.

⇒ B's share = 0.625 × 21,800 = ₹13,625.

∴ The correct answer is ₹13,625.

Additional Information

Compound Ratio

When two or more ratios are multiplied together (a/b × c/d), the result is the compound ratio (ac/bd).

Partnership Profit Distribution

Profit is distributed in the ratio of (Capital × Time). If time is constant, profit ratio equals the investment ratio.

Proportion Rule

If a : b = c : d, then the product of extremes equals the product of means (a × d = b × c).

Shortcut Trick

Use the direct formula for reduction in consumption: [r / (100 + r)] × 100%

Here, increase rate r = 20%.

Reduction = [20 / (100 + 20)] × 100 = (20 / 120) × 100 = 1/6 × 100

Calculation: 100 / 6 = 16(2/3)%

∴ The correct answer is 16(2/3)%.

Alternate Method

Given:

Initial Price of milk = 100 units

Increased Price = 100 + 20% of 100 = 120 units

Formula Used:

Expenditure = Price × Consumption

To keep Expenditure constant, the new consumption must be brought back to the base of the new price.

⇒ Required reduction in consumption = New Price − Old Price = 120 − 100 = 20 units

⇒ Percentage reduction = (Reduction / New Price) × 100

⇒ Percentage reduction = (20 / 120) × 100

⇒ Percentage reduction = (1 / 6) × 100

⇒ Percentage reduction = 16.66% or 16(2/3)%

∴ The correct answer is 16(2/3)%.

Additional Information

Price-Consumption Relationship

When Expenditure is constant, Price is inversely proportional to Consumption (P ∝ 1/C).

Price Decrease Case

If price decreases by r%, the increase in consumption to keep expenditure constant is [r / (100 − r)] × 100%.

Fraction Conversion

20% increase is 1/5. For constant product, if one factor increases by 1/n, the other must decrease by 1/(n+1). Here, 1/(5+1) = 1/6 = 16(2/3)%.

Shortcut Trick

The biggest even number less than 30 is 28.

Check products from options:

(1) 33, (2) 21, (3) 26, (4) 10.

The largest even product less than 28 (or specifically less than 30) among options is 26.

∴ The correct answer is 2 and 13.

Alternate Method

Given: Product must be an even number less than 30.

Property: To get an even product from two prime numbers, one of the numbers must be the only even prime, which is 2 (since Odd × Odd = Odd and Even × Any = Even).

⇒ Target: Largest product P < 30 where P is even.

⇒ Option 1: 3 × 11 = 33 (Odd, > 30) ×

⇒ Option 2: 3 × 7 = 21 (Odd) ×

⇒ Option 3: 2 × 13 = 26 (Even, < 30) ✓

⇒ Option 4: 2 × 5 = 10 (Even, < 30) × [26 is larger than 10]

∴ The correct answer is 2 and 13.

Additional Information

Even Prime Number

The number 2 is the only even prime number. All other prime numbers are odd.

Product Parity Rules

Odd × Odd = Odd; Even × Odd = Even; Even × Even = Even.

Prime Number Definition

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

Shortcut Trick

Calculate the semi-perimeter (s) of the triangle: s = (35 + 44 + 75) ÷ 2 = 77 cm.

Find the Area using Heron's Formula: Area = √(77 × (77−35) × (77−44) × (77−75)) = √(77 × 42 × 33 × 2).

Area = √(7 × 11 × 7 × 6 × 3 × 11 × 2) = 7 × 11 × 6 = 462 cm².

Altitude (h) corresponding to the 44 cm base = (2 × Area) ÷ base = (2 × 462) ÷ 44 = 21 cm.

∴ The correct answer is 21 cm.

Alternate Method

Given: Sides of the triangle are a = 35 cm, b = 44 cm, and c = 75 cm.

Formula Used:

1. Semi-perimeter (s) = (a + b + c) ÷ 2
2. Area = √(s(s − a)(s − b)(s − c))
3. Area = (1 ÷ 2) × base × altitude

Calculations:

⇒ s = (35 + 44 + 75) ÷ 2

⇒ s = 154 ÷ 2 = 77 cm

⇒ Area = √(77 × (77 − 35) × (77 − 44) × (77 − 75))

⇒ Area = √(77 × 42 × 33 × 2)

⇒ Area = √((7 × 11) × (7 × 6) × (11 × 3) × 2)

⇒ Area = √(7² × 11² × 6²) = 7 × 11 × 6 = 462 cm²

⇒ 462 = (1 ÷ 2) × 44 × altitude

⇒ 462 = 22 × altitude

⇒ altitude = 462 ÷ 22 = 21 cm

∴ The correct answer is 21 cm.

Additional Information

Heron's Formula for Area

Used to find the area of a triangle when all three sides are known: Area = √(s(s−a)(s−b)(s−c)), where s is the semi-perimeter.

Relationship between Area and Altitude

For any side taken as the base, the area is always 1/2 × base × corresponding height. Thus, height = (2 × Area) ÷ base.

Triangle Inequality Theorem

A triangle is valid only if the sum of any two sides is greater than the third side (e.g., 35 + 44 = 79, which is > 75).

Shortcut Trick

The given expression follows the algebraic identity (a² − b²) / (a + b) = (a − b).

Here, let a = 100.5 and b = 79.2.

The solution is simply the difference between the two numbers: 100.5 − 79.2 = 21.3.

∴ The correct answer is 21.3.

Alternate Method

Given:

Expression: ((100.5)² − (79.2)²) / (100.5 + 79.2)

Formula Used:

Algebraic Identity: a² − b² = (a + b)(a − b)

Calculations:

⇒ Let a = 100.5 and b = 79.2.

⇒ Numerator = (100.5)² − (79.2)² = (100.5 + 79.2) × (100.5 − 79.2)

⇒ Substituting this into the main expression:

⇒ [(100.5 + 79.2) × (100.5 − 79.2)] / (100.5 + 79.2)

⇒ 100.5 − 79.2 (Cancelling the common term 100.5 + 79.2)

⇒ 21.3

The calculated value corresponds to option (4).

∴ The correct answer is 21.3.

Additional Information

Difference of Squares

The identity a² − b² = (a + b)(a − b) is used to simplify expressions involving squares of two numbers.

Square of a Sum

The formula (a + b)² = a² + 2ab + b² expands the square of the sum of two variables.

Square of a Difference

The formula (a − b)² = a² − 2ab + b² expands the square of the difference between two variables.

Fraction Simplification

If the numerator and denominator share a common factor, they can be cancelled: (x × y) / x = y, provided x ≠ 0.

Shortcut Trick

Convert all into same root form

√8 = 2√2, √32 = 4√2, √128 = 8√2, √50 = 5√2

Expression = 2√2 + 2×4√2 − 8√2 + 2×5√2

= (2 + 8 − 8 + 10)√2 = 12√2

∴ Answer = 12√2

Alternate Method

Given:

√8 + 2√32 − √128 + 2√50

Concept Used:

Simplification of surds

Calculation:

⇒ √8 = √(4×2) = 2√2

⇒ √32 = √(16×2) = 4√2

⇒ √128 = √(64×2) = 8√2

⇒ √50 = √(25×2) = 5√2

⇒ Substitute values

⇒ = 2√2 + 2(4√2) − 8√2 + 2(5√2)

⇒ = 2√2 + 8√2 − 8√2 + 10√2

⇒ = (2 + 8 − 8 + 10)√2

⇒ = 12√2

∴ The correct answer is 12√2.

Additional Information

Surd Simplification

√(a×b) = √a × √b

Like Terms

Only same radicals can be added/subtracted

Perfect Squares

Extract perfect square factors for simplification

Shortcut Trick

Required percentage = (Value ÷ Base Value) × 100

Since both fractions have the same denominator (20), the percentage is simply (Numerator 1 ÷ Numerator 2) × 100.

Required % = (1 ÷ 11) × 100 ≈ 9.09%

∴ The correct answer is 9.09%.

Alternate Method

Given:

First value = 1/20

Second value (Base) = 11/20

Formula Used:

Percentage = (A/B) × 100

Calculations:

⇒ Required Percentage = [(1/20) ÷ (11/20)] × 100

⇒ Required Percentage = (1/20 × 20/11) × 100

⇒ Required Percentage = (1/11) × 100

⇒ Required Percentage = 100 ÷ 11

⇒ Required Percentage = 9.090909...

⇒ Required Percentage ≈ 9.09%

∴ The correct answer is 9.09%.

Additional Information

Fraction to Percentage Conversion

To convert any fraction x/y into a percentage, multiply the fraction by 100 and add the % symbol.

Reciprocal Percentages

Knowing common reciprocals like 1/11 ≈ 9.09% and 1/9 ≈ 11.11% helps in solving Simplification and Percentage questions faster.

Percentage Change

Percentage change = [(New Value − Original Value) ÷ Original Value] × 100.

Shortcut Trick

0.2x + 60 = x

0.8x = 60 ⇒ x = 75

70% of x = 0.7 × 75

= 52.5

∴ Answer = 52.5

Alternate Method

Given:

20% of a number when added to 60 gives the number

Formula used:

Percentage = (value ÷ 100) × number

Calculation:

⇒ Let the number be x

⇒ 20% of x = (20/100) × x = 0.2x

⇒ According to question

⇒ 0.2x + 60 = x

⇒ x − 0.2x = 60

⇒ 0.8x = 60

⇒ x = 60 ÷ 0.8 = 75

⇒ Required percentage = 70% of 75

⇒ = (70/100) × 75

⇒ = 52.5

∴ The correct answer is 52.5.

Additional Information

Percentage Formula

x% of number = (x/100) × number

Equation Formation

Translate statement into linear equation

Decimal Conversion

20% = 0.2, 70% = 0.7

Shortcut Trick

Profit = SP − CP = 335 − 300 = 35.

Profit % = (Profit ÷ CP) × 100.

Profit % = (35 ÷ 300) × 100 = 35 ÷ 3 = 11²/₃%.

∴ The correct answer is 11(2)/(3)%.

Alternate Method

Given → Cost Price (CP) = ₹ 300, Selling Price (SP) = ₹ 335.
Formulas: Profit = SP − CP | Profit% = (Profit ÷ CP) × 100

→ Profit = 335 − 300 = 35
→ Profit % = (35 ÷ 300) × 100
→ Profit % = 35 ÷ 3 = 11.66...%
→ In fraction form, 35/3 = 11²/₃%

∴ The correct answer is 11(2)/(3)%.

Additional Information

Profit and Loss Basics

Profit occurs when Selling Price (SP) is greater than Cost Price (CP). Loss occurs when CP > SP.

Percentage Calculation

Profit or Loss percentage is always calculated on the Cost Price (CP) unless stated otherwise.

Shortcut Trick

The area of a triangle sharing the same base and lying between the same parallel lines as a parallelogram is always half the area of that parallelogram.

Area(Δ APB) = 1/2 × Area(ABCD) (since they share base AB and lie between AB || CD).

Area(Δ BQC) = 1/2 × Area(ABCD) (since they share base BC and lie between BC || AD).

Since both triangles are equal to half of the same parallelogram, their areas must be equal.

∴ The correct answer is the area of Δ BQC.

Alternate Method

Given:ABCD is a parallelogram. P is a point on side DC and Q is a point on side AD.

Formula Used: Area of triangle = 1/2 × base × height; Area of parallelogram = base × height.

⇒ Area(Δ APB) = 1/2 × AB × (height of parallelogram with respect to base AB)

⇒ Area(ABCD) = AB × (height of parallelogram with respect to base AB)

⇒ Area(Δ APB) = 1/2 × Area(ABCD) --- (i)

⇒ Area(Δ BQC) = 1/2 × BC × (height of parallelogram with respect to base BC)

⇒ Area(ABCD) = BC × (height of parallelogram with respect to base BC)

⇒ Area(Δ BQC) = 1/2 × Area(ABCD) --- (ii)

⇒ From equations (i) and (ii), we get: Area(Δ APB) = Area(Δ BQC)

∴ The correct answer is the area of Δ BQC.

Additional Information

Area of Parallelogram

The area of a parallelogram is given by the product of its base and the corresponding altitude (height).

Triangle and Parallelogram Relationship

When a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is half the area of the parallelogram.

Parallelogram Properties

In a parallelogram, opposite sides are parallel and equal, and diagonals bisect each other, dividing the parallelogram into two triangles of equal area.