Quantitative Aptitude PYQs

Shortcut Trick

Effective percentage change = 20 − 24 − (20 × 24)/100 = −8.8%

Final cost as percentage of original = 100% − 8.8% = 91.2%

Given that 91.2% = ₹1,938

Original cost (100%) = (1,938 ÷ 91.2) × 100 = ₹2,125

∴ The correct answer is ₹2,125.

Alternate Method

Given:

First year increase = 20%

Second year decrease = 24%

Final cost of the article = ₹1,938

Calculations:

⇒ Let the original cost of the article be ₹x.

⇒ After 20% increase, cost = x × (1 + 20/100) = 1.2x

⇒ After 24% decrease on the increased cost, final cost = 1.2x × (1 − 24/100)

⇒ 1,938 = 1.2x × 0.76

⇒ 1,938 = 0.912x

⇒ x = 1,938 ÷ 0.912

⇒ x = 2,125

∴ The correct answer is ₹2,125.

Additional Information

Successive Percentage Change Formula

When a value is changed by a% and then by b%, the net change is given by (a + b + ab/100)%. Use positive for increase and negative for decrease.

Multiplying Factor Method

For an increase of r%, multiply by (1 + r/100). For a decrease of r%, multiply by (1 − r/100).

Reverse Percentage Calculation

To find the original value when the final value and percentage change are known: Original Value = Final Value ÷ (Net Multiplying Factor).

Shortcut Trick

Check the sum: 23 + (−11) + (−12) = 23 − 23 = 0.

Using the conditional identity: If a + b + c = 0, then a³ + b³ + c³ = 3abc.

Result = 3 × 23 × (−11) × (−12) = 69 × 132 = 9108.

∴ The correct answer is 9108.

Alternate Method

Given: a = 23, b = −11, c = −12

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

⇒ First, evaluate the sum of the terms: a + b + c

⇒ 23 + (−11) + (−12) = 23 − 11 − 12 = 0

⇒ Since a + b + c = 0, the entire right-hand side of the formula becomes zero.

⇒ a³ + b³ + c³ − 3abc = 0 × (a² + b² + c² − ab − bc − ca) = 0

⇒ Therefore, a³ + b³ + c³ = 3abc

⇒ 23³ + (−11)³ + (−12)³ = 3 × 23 × (−11) × (−12)

⇒ 3 × 23 × 132 = 69 × 132 = 9108

∴ The correct answer is 9108.

Additional Information

Special Case of Cubic Identity

When the sum of three variables is zero, the identity a³ + b³ + c³ = 3abc simplifies complex cubic calculations into simple multiplication.

Alternative Form of Identity

The expression a² + b² + c² − ab − bc − ca can also be written as ½ [(a−b)² + (b−c)² + (c−a)²], which is always non-negative for real numbers.

BODMAS Rule in Simplification

Always resolve brackets and orders (powers) before addition or multiplication unless an identity is applicable.

Shortcut Trick

Difference in the number of notebooks = 7 − 4 = 3 notebooks.

Since the number of pens is the same (6 in both cases), the price difference is solely due to these 3 notebooks.

Price difference = ₹ 333 − ₹ 228 = ₹ 105.

Cost of 1 notebook = 105 ÷ 3 = 35.

∴ The correct answer is ₹ 35.

Alternate Method

Given:

Cost of 4 notebooks and 6 pens = ₹ 228

Cost of 7 notebooks and 6 pens = ₹ 333

Formula Used:

The Elimination Method for solving a System of Linear Equations.

Calculations:

⇒ Let the cost of one notebook be N and the cost of one pen be P.

⇒ 4N + 6P = 228 — (Equation 1)

⇒ 7N + 6P = 333 — (Equation 2)

⇒ Subtract Equation 1 from Equation 2:

⇒ (7N + 6P) − (4N + 6P) = 333 − 228

⇒ (7N − 4N) + (6P − 6P) = 105

⇒ 3N = 105

⇒ N = 105 ÷ 3

⇒ N = 35

∴ The correct answer is ₹ 35.

Additional Information

System of Linear Equations

A set of equations with the same variables. The solution is the set of values that satisfy all equations simultaneously.

Elimination Method

A process of solving a system by adding or subtracting equations to remove one variable, making it easier to find the value of the other.

Unitary Method

A technique where we first find the value of a single unit (like 1 notebook) and then multiply it to find the value of any required number of units.

Shortcut Trick

Direct application of Speed-Time-Distance relation.

Time = Distance ÷ Speed

Time = 12 ÷ 6 = 2 hr.

∴ The correct answer is 2 hr.

Alternate Method

Given:

Walking speed of Rohit = 6 km/hr

Total Distance to be covered = 12 km

Formula Used:

Time = Distance ÷ Speed

Calculations:

⇒ Time = 12 km ÷ 6 km/hr

⇒ Time = 2 hr

∴ The correct answer is 2 hr.

Additional Information

Distance Formula

The total path length covered is given by Distance = Speed × Time.

Unit Conversion

To convert speed from km/hr to m/s, multiply the value by 5/18. To convert m/s to km/hr, multiply by 18/5.

Average Speed

If a person covers different distances at different speeds, Average Speed = Total Distance ÷ Total Time.

Shortcut Trick

Use the Ratio Method: Initial ratio of weights and the profit percentage.

Weight Ratio (CP : SP) = 950 : 1000

Profit Ratio (CP : SP) = 100 : 115

Effective CP : SP = (950 × 100) : (1000 × 115) = 95 : 115 = 19 : 23

Total Profit % = [(23 − 19) ÷ 19] × 100 = (4 ÷ 19) × 100 ≈ 21.05%

∴ The correct answer is 21.05%.

Alternate Method

Given: Profit on price = 15% | False weight = 950 g | True weight = 1000 g (1 kg).

Let the Cost Price (CP) of 1000 g be ₹1000.

⇒ CP of 1 g = ₹1.

⇒ The trader sells 950 g but charges for 1000 g plus a 15% profit.

⇒ Actual CP for the trader (for 950 g) = 950 × ₹1 = ₹950.

⇒ Selling Price (SP) = ₹1000 + 15% of ₹1000 = 1000 + 150 = ₹1150.

⇒ Total Profit = SP − CP = 1150 − 950 = ₹200.

⇒ Total Profit % = (Profit ÷ CP) × 100 = (200 ÷ 950) × 100.

⇒ Profit % = (4 ÷ 19) × 100 = 400 ÷ 19 ≈ 21.05%.

⇒ Therefore, the total profit percentage is 21.05%, which is Option 2.

∴ The correct answer is 21.05%.

Additional Information

Dishonest Dealer Formula

Profit % = [(True Value − False Value + % Profit on True Value) ÷ False Value] × 100. This accounts for both weight cheating and price markup.

Successive Percentage Method

Total Profit can also be viewed as successive gains. Net Profit % = [a + b + (a × b ÷ 100)] where 'a' is profit from price and 'b' is profit from weight manipulation.

Weight Gain Percentage

Profit due to weight cheating alone = [Error ÷ (True Value − Error)] × 100. In this case, (50 ÷ 950) × 100 ≈ 5.26%.

Given:

(26/7)⁵ × (7/26)¹¹ × (26/7)¹⁹ = (7/26)^{9m + 17}

Formula used:

Law of Exponents:

a^x × a^y = a^{x + y}

(a/b)ⁿ = (b/a)⁻ⁿ

Calculations:

(26/7)⁵ × (7/26)¹¹ × (26/7)¹⁹ = (7/26)^{9m + 17}

⇒ (26/7)^{(5 + 19)} × (7/26)¹¹ = (7/26)^{9m + 17}

⇒ (26/7)²⁴ × (7/26)¹¹ = (7/26)^{9m + 17}

Using the property (a/b)ⁿ = (b/a)⁻ⁿ to make the bases identical:

⇒ (7/26)⁻²⁴ × (7/26)¹¹ = (7/26)^{9m + 17}

⇒ (7/26)^{(−24 + 11)} = (7/26)^{9m + 17}

⇒ (7/26)⁻¹³ = (7/26)^{9m + 17}

On comparing the exponents on both sides of the equation:

⇒ −13 = 9m + 17

⇒ 9m = −13 − 17

⇒ 9m = −30

⇒ m = −30/9

⇒ m = −10/3

∴ The correct answer is option (3).

Shortcut Trick

Apply successive discounts as a single multiplying factor: 0.75 (for 25% off) and 0.90 (for 10% off).

Final Selling Price = 2,000 × 0.75 × 0.90

Final Selling Price = 2,000 × 0.675 = 1,350

∴ The correct answer is ₹1350.

Alternate Method

Given:

Marked Price (MP) = ₹2,000

First Discount = 25%

Second Discount = 10%

Formula Used:

Final Selling Price = MP × (100 − d₁)/100 × (100 − d₂)/100

⇒ Price after 25% discount = 2,000 − (25% of 2,000)

⇒ 2,000 − 500 = ₹1,500

⇒ Final Selling Price after 10% discount = 1,500 − (10% of 1,500)

⇒ 1,500 − 150 = ₹1,350

∴ The correct answer is ₹1350.

Additional Information

Successive Discount Formula

For two successive discounts of a% and b%, the net equivalent discount is (a + b − ab/100)%.

Single Equivalent Discount

In this case, 25 + 10 − (25 × 10)/100 = 35 − 2.5 = 32.5% single discount on ₹2,000.

Selling Price Relation

The Selling Price (SP) is always calculated as Marked Price (MP) − Total Discount.

Given:

Expression: 156 ÷ [12 + {7 × (15 − 9) − 28}]

Formula used:

BODMAS Rule (Brackets, Of, Division, Multiplication, Addition, Subtraction)

Calculations:

156 ÷ [12 + {7 × (15 − 9) − 28}]

⇒ 156 ÷ [12 + {7 × 6 − 28}]

⇒ 156 ÷ [12 + {42 − 28}]

⇒ 156 ÷ [12 + 14]

⇒ 156 ÷ 26

⇒ 6

∴ The correct answer is option (3).

Shortcut Trick

By inspection, the value 5/2 can be written as the sum of 2 + 1/2.

Since the equation is in the form y + 1/y = 5/2 (where y = x / (x + 1)), we can equate y = 2 or y = 1/2.

If x / (x + 1) = 2 ⇒ x = 2x + 2 ⇒ x = −2.

If x / (x + 1) = 1/2 ⇒ 2x = x + 1 ⇒ x = 1.

∴ The correct answer is 1 or −2.

Alternate Method

Given: x / (x + 1) + (x + 1) / x = 5/2

Method: Simplify the algebraic expression into a standard Quadratic Equation.

⇒ [x × x + (x + 1) × (x + 1)] / [x × (x + 1)] = 5/2

⇒ [x² + (x² + 2x + 1)] / [x² + x] = 5/2

⇒ [2x² + 2x + 1] / [x² + x] = 5/2

⇒ 2 × (2x² + 2x + 1) = 5 × (x² + x)

⇒ 4x² + 4x + 2 = 5x² + 5x

⇒ 5x² − 4x² + 5x − 4x − 2 = 0

⇒ x² + x − 2 = 0

⇒ x² + 2x − x − 2 = 0

⇒ x(x + 2) − 1(x + 2) = 0

⇒ (x − 1)(x + 2) = 0

⇒ x = 1 or x = −2

∴ The correct answer is 1 or −2.

Additional Information

Quadratic Roots Formula

For any equation of the form ax² + bx + c = 0, the roots are calculated using x = [−b ± √(b² − 4ac)] / 2a.

Factorization Method

This involves splitting the middle term bx into two terms such that their product equals ac and their sum equals b.

Reciprocal Sum Property

If y + 1/y = k, and k is in the form of a fraction a/b + b/a, then the solutions for y are simply a/b and b/a.

Shortcut Trick

The Mode is simply the value that appears most often in the Data Set.

Count the occurrences: 2 (twice), 3 (thrice), 4 (once), 5 (four times).

Since the number 5 has the highest Frequency, it is the mode.

∴ The correct answer is 5.

Alternate Method

Given: The number of books read by students: 3, 5, 2, 3, 4, 5, 3, 2, 5, 5.

Formula Used: Mode = Value with the maximum Frequency in the distribution.

⇒ Arrange observations: 2, 2, 3, 3, 3, 4, 5, 5, 5, 5

⇒ Frequency of 2 = 2

⇒ Frequency of 3 = 3

⇒ Frequency of 4 = 1

⇒ Frequency of 5 = 4

⇒ The value 5 occurs 4 times, which is the Highest Frequency.

∴ The correct answer is 5.

Additional Information

Empirical Formula

For a moderately skewed distribution, the relationship is: Mode = 3 × Median − 2 × Mean.

Bimodal Data

A data set that contains two values with the highest frequency is called Bimodal. If there are more than two, it is Multimodal.

Measures of Central Tendency

Mean is the average, Median is the middle value, and Mode is the most frequent value in a Data Set.

Shortcut Trick

Use the standard trigonometric identity: tan θ + cot θ = 2 cosec 2θ.

Given the value of the expression tan θ + cot θ = b.

Equating both, we get b = 2 cosec 2θ.

Solving for the required term: cosec 2θ = b/2.

∴ The correct answer is b/2.

Alternate Method

Given: sec θ + cosec θ = a and tan θ + cot θ = b

Formula Used: tan θ = sin θ / cos θ, cot θ = cos θ / sin θ, and sin 2θ = 2 sin θ cos θ

Calculations:

⇒ tan θ + cot θ = b

⇒ (sin θ / cos θ) + (cos θ / sin θ) = b

⇒ (sin² θ + cos² θ) / (sin θ cos θ) = b

⇒ 1 / (sin θ cos θ) = b [Since sin² θ + cos² θ = 1]

⇒ sin θ cos θ = 1/b

⇒ sin 2θ = 2 sin θ cos θ = 2 × (1/b) = 2/b

⇒ cosec 2θ = 1 / sin 2θ = 1 / (2/b) = b/2

∴ The correct answer is b/2.

Additional Information

Double Angle Identities

The primary double angle formulas are sin 2θ = 2 sin θ cos θ and cos 2θ = cos² θ − sin² θ = 2 cos² θ − 1 = 1 − 2 sin² θ.

Reciprocal Relations

The sum tan θ + cot θ is always equal to sec θ × cosec θ, which further simplifies to 2 cosec 2θ.

Pythagorean Identities

Fundamental identities include sin² θ + cos² θ = 1, 1 + tan² θ = sec² θ, and 1 + cot² θ = cosec² θ.

Shortcut Trick

Since the frequency (20) is identical for all 5 class intervals, the data is perfectly uniform and symmetric.

In a symmetric distribution with equal frequencies, the Median lies exactly at the midpoint of the entire range.

Total Range = 54 to 64.

Median = (54 + 64) ÷ 2 = 59.

∴ The correct answer is ₹59.

Alternate Method

Given: Total workers (N) = 100. Classes and frequencies (f) are provided.

Formula Used: Median = l + [((N ÷ 2) − cf) ÷ f] × h

Cumulative Frequencies (CF):

⇒ 54-56: 20

⇒ 56-58: 20 + 20 = 40

⇒ 58-60: 40 + 20 = 60 (Median class since 60 ≥ 50)

Median Calculation:

⇒ l (Lower limit) = 58

⇒ N/2 = 100 ÷ 2 = 50

⇒ cf (Previous CF) = 40

⇒ f (Frequency of median class) = 20

⇒ h (Class height) = 2

⇒ Median = 58 + [(50 − 40) ÷ 20] × 2

⇒ Median = 58 + [10 ÷ 20] × 2

⇒ Median = 58 + 0.5 × 2 = 58 + 1

⇒ Median = 59

∴ The correct answer is ₹59.

Additional Information

Empirical Relationship

For a moderately skewed distribution: Mode = 3 × Median − 2 × Mean.

Median for Grouped Data

The median is the value that divides the total frequency into two equal halves (N/2).

Class Interval (h)

Class width or height (h) is the difference between the upper and lower limits of a class (e.g., 56 − 54 = 2).

Shortcut Trick

The given expression follows the algebraic identity: (a³ − b³) ÷ (a² + ab + b²) = (a − b).

In this problem, a = 1.4 and b = 1.2.

Applying the trick: 1.4 − 1.2 = 0.2.

∴ The correct answer is 0.2.

Alternate Method

Given:

Numerator = (1.4 × 1.4 × 1.4 − 1.2 × 1.2 × 1.2)

Denominator = (1.4 × 1.4 + 1.4 × 1.2 + 1.2 × 1.2)

Formula Used:

The difference of two cubes formula is: a³ − b³ = (a − b)(a² + ab + b²)

⇒ Let a = 1.4 and b = 1.2.

⇒ Numerator = a³ − b³ = (a − b)(a² + ab + b²)

⇒ Denominator = (a² + ab + b²)

⇒ Expression = [(a − b)(a² + ab + b²)] ÷ (a² + ab + b²)

⇒ After cancelling the common term (a² + ab + b²), we get: a − b

⇒ Value = 1.4 − 1.2

⇒ Value = 0.2

∴ The correct answer is 0.2.

Additional Information

Sum of Cubes Identity

The corresponding identity for addition is a³ + b³ = (a + b)(a² − ab + b²), where the middle term in the second bracket is negative.

Algebraic Simplification

Always check for standard patterns like (a + b)², (a − b)², and a² − b² to avoid lengthy decimal calculations.

Cubing Decimals

Remember that (0.x)³ will have three decimal places. For example, 1.4³ = 2.744 and 1.2³ = 1.728.

Shortcut Trick

Efficiency of Pipe X = +4 and Pipe Y = −3 (Total work = LCM of 15, 20 = 60).

Combined efficiency for first 10 hours = 4 − 3 = 1 unit/hr.

Work completed in 10 hours = 1 × 10 = 10 units.

Remaining work to be filled by Pipe X alone = 60 − 10 = 50 units.

Time taken by Pipe X for remaining part = 50 ÷ 4 = 12.5 hr.

Total time = 10 + 12.5 = 22.5 hr.

∴ The correct answer is 22.5 hr.

Alternate Method

Given: Filling time of Pipe X = 15 hr, Emptying time of Pipe Y = 20 hr, Opening duration of Pipe Y = 10 hr.

Formula: Total Work = Efficiency × Time; Net Efficiency = Efficiency_Fill − Efficiency_Empty

⇒ Let total capacity of the reservoir = LCM(15, 20) = 60 units.

⇒ Efficiency of Pipe X = 60 ÷ 15 = 4 units/hr.

⇒ Efficiency of Pipe Y = 60 ÷ 20 = 3 units/hr (Negative as it empties).

⇒ Work done in 10 hours with both pipes open = (4 − 3) × 10 = 10 units.

⇒ Remaining capacity to be filled = 60 − 10 = 50 units.

⇒ Time taken by Pipe X to fill the remaining 50 units = 50 ÷ 4 = 12.5 hr.

⇒ Total time required = Initial 10 hr + Subsequent 12.5 hr = 22.5 hr.

∴ The correct answer is 22.5 hr.

Additional Information

Pipes and Cisterns Basics

If a pipe fills a tank in 'a' hours and another empties it in 'b' hours, the net work done in 1 hour is (1/a − 1/b).

Negative Efficiency

An outlet pipe or leak is always treated as having negative efficiency in work calculations.

Time and Efficiency Relation

Efficiency is inversely proportional to the time taken to complete a task (Efficiency = Total Work ÷ Time).

Additional Information

` And the three concepts:

Product of Two Numbers Formula

For any two numbers A and B, the product of the numbers is equal to the product of their HCF and LCM: A × B = HCF × LCM.

LCM of Ratio Numbers

If two numbers are in the ratio a : b (where a and b are co-prime), then their HCF = x, the numbers are ax and bx

Shortcut Trick

Mean proportional of two numbers a and b = √(a × b)

Given: a = 0.16 and b = 0.49

Mean proportional = √(0.16 × 0.49) = 0.4 × 0.7 = 0.28

∴ The correct answer is 0.28.

Alternate Method

Given:

First number (a) = 0.16

Second number (b) = 0.49

Formula Used:

If x is the mean proportional between a and b, then a : x :: x : b

This implies: x² = a × b or x = √(a × b)

Calculations:

⇒ Let the mean proportional be x.

⇒ 0.16/x = x/0.49

⇒ x² = 0.16 × 0.49

⇒ x = √(0.16 × 0.49)

⇒ x = √((16/100) × (49/100))

⇒ x = (4/10) × (7/10)

⇒ x = 28/100

⇒ x = 0.28

∴ The correct answer is 0.28.

Additional Information

Third Proportional

For two numbers a and b, the third proportional c is such that a:b = b:c, which gives c = b²/a.

Fourth Proportional

For three numbers a, b, and c, the fourth proportional d is such that a:b = c:d, which gives d = (b × c)/a.

Continued Proportion

Three quantities a, b, and c are said to be in continued proportion if a:b = b:c.

Shortcut Trick

Since the entire selling price of the first article is reinvested to buy the second article, the overall percentage change is completely independent of the initial cost price of ₹347.

We can directly apply the successive percentage change formula:

Net Percentage Change = a + b + (a × b)/100

Here, a = −6% (loss) and b = +65% (gain).

Overall Gain = −6 + 65 + (−6 × 65)/100 = 59 − 3.9 = 55.1%.

∴ The correct answer is 55.1%.

Alternate Method

Given:

Initial Cost Price (C₁) = ₹347

Loss percentage on the first article = 6%

Gain percentage on the second article = 65%

Formula Used:

Selling Price (SP) = Cost Price (CP) × (1 − Loss%/100) or

Shortcut Trick

Net efficiency = Efficiency of Tap A + Efficiency of Tap B − Efficiency of Tap C.

Since both Tap B and Tap C take 15 hr, their filling and emptying rates are equal and cancel each other out.

Net efficiency = Efficiency of Tap A = 1/4 of the cistern per hour.

Time required to fill 1/4 of the empty cistern = (1/4) ÷ (1/4) = 1 hour.

∴ The correct answer is 1 hr.

Alternate Method

Given: Tap A fills in 4 hr, Tap B fills in 15 hr, and Tap C empties in 15 hr.

Let the Total Capacity (Total Work) of the cistern be the LCM of 4, 15, and 15.

⇒ Total Capacity = 60 units.

⇒ Net Efficiency of all taps together = Efficiency(A) + Efficiency(B) − Efficiency(C)

⇒ Net Efficiency = 15 + 4 − 4 = 15 units/hr.

⇒ Capacity to be filled = One-fourth of the cistern = 1/4 × 60 = 15 units.

⇒ Time taken = Capacity to be filled ÷ Net Efficiency

⇒ Time = 15 ÷ 15 = 1 hour.

∴ The correct answer is 1 hr.

Additional Information

Pipes and Cisterns Concept

Filling pipes are considered as positive work (+), while emptying or outlet pipes (leaks) are considered as negative work (−).

Net Rate Formula

If Pipe A fills in 'x' hours and Pipe B empties in 'y' hours, the net work done in one hour is (1/x − 1/y).

Cistern Capacity

The capacity of the tank is often assumed as the LCM of the individual times taken by the pipes to simplify calculations.

Shortcut Trick

Net multiplier for successive discounts of 20% and 30% = 0.80 × 0.70 = 0.56

Final Selling Price = Marked Price × 0.56

SP = 5000 × 0.56 = 2800

∴ The correct answer is ₹2,800.

Alternate Method

Given:

Marked Price (MP) of the jacket = ₹5,000

First discount (d₁) = 20%

Second discount (d₂) = 30%

Formula Used:

Final SP = MP × (100 − d₁)/100 × (100 − d₂)/100

Calculations:

⇒ Price after 1^st discount = 5000 − (20% of 5000)

⇒ Price after 1^st discount = 5000 − 1000 = 4000

⇒ Price after 2^nd discount = 4000 − (30% of 4000)

⇒ Price after 2^nd discount = 4000 − 1200 = 2800

⇒ Final Selling Price = ₹2,800

∴ The correct answer is ₹2,800.

Additional Information

Single Equivalent Discount

For two successive discounts x% and y%, the single equivalent discount is given by (x + y − xy/100)%.

Relationship between MP and SP

The Selling Price is always calculated on the Marked Price after deducting the discount: SP = MP × (100 − Discount%)/100.

Order of Discounts

The final Selling Price remains the same regardless of the order in which the successive discounts are applied (e.g., 20% then 30% is the same as 30% then 20%).

Shortcut Trick

A diagonal divides a parallelogram into two congruent triangles.

The sides of the triangle are 9 cm, 12 cm, and 15 cm.

Recognize the Pythagorean triplet: 9² + 12² = 81 + 144 = 225 = 15².

Since it is a right-angled triangle, the area of the parallelogram is simply the product of its adjacent sides.

Area = 12 cm × 9 cm = 108 cm².

∴ The correct answer is 108 cm².

Alternate Method

Given: Adjacent sides of the parallelogram are a = 12 cm and b = 9 cm. Length of the diagonal is d = 15 cm.

Formula Used: Area of parallelogram = 2 × Area of the triangle formed by its sides and diagonal.

⇒ Let the triangle be ΔABC with sides 9 cm, 12 cm, and 15 cm.

⇒ Checking for a right-angled triangle: 9² + 12² = 81 + 144 = 225.

⇒ Square of the longest side (diagonal) = 15² = 225.

⇒ Since 9² + 12² = 15², the triangle is right-angled at the vertex between sides 9 cm and 12 cm.

⇒ Area of ΔABC = 1 ÷ 2 × base × height = 1 ÷ 2 × 12 × 9 = 54 cm².

⇒ Area of parallelogram = 2 × Area of ΔABC = 2 × 54 = 108 cm².

∴ The correct answer is 108 cm².

Additional Information

Properties of Parallelogram Diagonals

Each diagonal of a parallelogram divides it into two triangles of equal area.

Heron's Formula

For any triangle with sides a, b, and c, Area = √[s × (s − a) × (s − b) × (s − c)], where s = (a + b + c) ÷ 2.

Parallelogram Law

The sum of the squares of the four sides of a parallelogram is equal to the sum of the squares of its two diagonals: 2(a² + b²) = d₁² + d₂².