Quantitative Aptitude PYQs

Shortcut Trick

Let present ages of Father and Son be F and S.

Condition 9 years ago: (F − 9) = 2 × (S − 9) + 25 ⇒ F − 2S = 16

Let after 'x' years, Father's age becomes twice his Son's age: F + x = 2 × (S + x)

⇒ x = F − 2S

By simply comparing both equations, we directly get x = 16 years.

∴ The correct answer is 16 years.

Alternate Method

Given:

9 years ago, the Father's age was 25 years more than twice his Son's age.

9 Years Ago
Present
After x Years

F - 9
F
F + x

S - 9
S
S + x

F-9 = 2(S-9)+25

F+x = 2(S+x)

Calculation:

Let the present age of the Father be F and the Son be S.

According to the given condition for 9 years ago:

⇒ (F − 9) = 2 × (S − 9) + 25

⇒ F − 9 = 2S − 18 + 25

⇒ F − 2S = 7 + 9

⇒ F − 2S = 16       ... (Equation 1)

Let 'x' be the number of years from now when the Father's age will be twice his Son's age:

⇒ (F + x) = 2 × (S + x)

⇒ F + x = 2S + 2x

⇒ x = F − 2S

Substituting the derived value from Equation 1:

⇒ x = 16

∴ The correct answer is 16 years.

Additional Information

Constant Age Difference

The difference between the ages of two individuals remains constant at any given point in time. If a person is older by 'd' years today, they will remain older by 'd' years forever.

Framing Age Equations

Always assume present ages as the base variables (x, y). For an age 'n' years ago, subtract 'n' from both (x − n, y − n). For an age 'n' years hence, add 'n' to both (x + n, y + n).

Ratio Behavior over Time

While the absolute difference in age stays the same, the ratio of their ages strictly changes and gradually tends towards 1:1 as time progresses to infinity.

Shortcut Trick

The total surface area of a cube is calculated as 6 × (side)².

Given the side of the cube = 74 cm.

First, calculate the square of the side: 74² = 5476.

Now, multiply by the number of faces: 6 × 5476 = 32856 cm².

∴ The correct answer is 32856 cm².

Alternate Method

Given:

Side of the cube (a) = 74 cm

Formula Used:

Total Surface Area of Cube = 6 × a²

a = 74 cm
Area of 1 face = a²
Total faces = 6
TSA = 6a²

Calculations:

⇒ Total Surface Area = 6 × a²

⇒ Total Surface Area = 6 × (74)²

⇒ Total Surface Area = 6 × 5476

⇒ Total Surface Area = 32856 cm²

∴ The correct answer is 32856 cm².

Additional Information

Lateral Surface Area of a Cube

It includes the area of the four vertical faces only, excluding the top and bottom. Formula: LSA = 4 × a²

Volume of a Cube

It measures the total 3D space enclosed by the cube. Formula: Volume = a³

Diagonal of a Cube

The length of the longest space diagonal connecting two opposite corners of the cube is given by a√3.

Shortcut Trick

The average of the first n even numbers is directly calculated using the formula: (n + 1).

Given, the number of terms n = 171.

Average = 171 + 1 = 172

∴ The correct answer is 172.

Alternate Method

Given:

Number of even numbers (n) = 171

Concept Used:

The sequence of the first n even numbers forms an Arithmetic Progression (AP) with the first term 2 and common difference 2.

Average of an AP = (First Term + Last Term) ÷ 2

Average of First 171 Even Numbers

2
1^st Term

342
171^st Term

172
Average
Average = (2 + 342) ÷ 2 = 172

Calculations:

⇒ First even number (a) = 2

⇒ 171^st even number (l) = 171 × 2 = 342

⇒ Average = (a + l) ÷ 2

⇒ Average = (2 + 342) ÷ 2

⇒ Average = 344 ÷ 2

⇒ Average = 172

∴ The correct answer is 172.

Additional Information

Average of first n Odd Numbers

The average of the first n odd numbers is always equal to n. Formula: Average = n.

Sum of first n Even Numbers

The sum of the first n even numbers is calculated by multiplying n with (n + 1). Formula: Sum = n × (n + 1).

Sum of first n Odd Numbers

The sum of the first n odd numbers is equal to the square of n. Formula: Sum = n².

Average of Consecutive Numbers

If a series of numbers are in an Arithmetic Progression, their average is the midpoint. Formula: Average = (First term + Last term) ÷ 2.

Shortcut Trick

Using the deviation method, let's assume the average weight is 50 kg.

Deviations from 50 kg: (40 − 50), (49 − 50), (56 − 50), (76 − 50), (35 − 50)

Net deviation = −10 − 1 + 6 + 26 − 15 = +6

Average deviation = +6 ÷ 5 = +1.2 kg

Actual Average = 50 + 1.2 = 51.2 kg

∴ The correct answer is 51.2 kg.

Alternate Method

Given:

Weights of family members = 40 kg, 49 kg, 56 kg, 76 kg, 35 kg

Total number of members (n) = 5

Formula Used:

Average = Sum of all observations ÷ Total number of observations

Avg = 51.2

40
49
56
76
35

Weights of 5 Family Members (kg)

Calculations:

⇒ Sum of weights = 40 + 49 + 56 + 76 + 35

⇒ Sum of weights = 256 kg

⇒ Average weight = 256 ÷ 5

⇒ Average weight = 51.2 kg

∴ The correct answer is 51.2 kg.

Additional Information

Deviation Method in Averages

This method assumes a mean value to make calculations simpler. The True Average is equal to Assumed Average + (Net Deviation ÷ Total Observations).

Weighted Average

Used when multiple values have different frequencies or weights. Formula: A_w = (w₁x₁ + w₂x₂ + ... + w_nx_n) ÷ (w₁ + w₂ + ... + w_n).

Combined Average

Used to find the combined average of two or more independent groups. Formula: A_c = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Shortcut Trick

Directly expand the brackets using the distributive property:

⇒ x(3x − 7) + 7(x² − 4) + 18

⇒ 3x² − 7x + 7x² − 28 + 18

Combine the like terms (x², x, and constant):

⇒ 10x² − 7x − 10

∴ The correct answer is 10x² − 7x − 10.

Alternate Method

Given:

Expression = x(3x − 7) + 7(x² − 4) + 18

Calculations:

⇒ Multiply x inside the first bracket: x × 3x − x × 7 = 3x² − 7x

⇒ Multiply 7 inside the second bracket: 7 × x² − 7 × 4 = 7x² − 28

⇒ Substitute these expanded values back into the full expression:

⇒ 3x² − 7x + 7x² − 28 + 18

⇒ Group the like terms together:

⇒ (3x² + 7x²) − 7x + (−28 + 18)

⇒ 10x² − 7x − 10

∴ The correct answer is 10x² − 7x − 10.

Additional Information

Distributive Property

The distributive property states that a(b + c) = ab + ac. It is used to multiply a single term with two or more terms inside a set of parentheses.

Like Terms

Like terms are terms that contain the exact same variables raised to the exact same powers.

Only like terms can be combined by addition or subtraction.

BODMAS Rule

Always solve mathematical expressions in the exact order of Brackets, Orders (powers and roots), Division or Multiplication, and lastly Addition or Subtraction.

Shortcut Trick

Let the original number be equal to 100%.

After a 60% increase, the new value becomes 160%.

Given value of 160% = 3580

Original Number (100%) = 3580 × (100 ÷ 160)

Original Number = 3580 × (5 ÷ 8) = 2237.5

∴ The correct answer is 2237.5.

Alternate Method

Given:

A number increased by 60% yields 3580.

Formula Used:

Final Value = Initial Value × (1 + Percentage Increase ÷ 100)

Original Number (100%)

Increase (+60%)

Total Result = 160% = 3580

Calculations:

⇒ Let the required number be x.

⇒ x + 60% of x = 3580

⇒ x + 0.6x = 3580

⇒ 1.6x = 3580

⇒ x = 3580 ÷ 1.6

⇒ x = 35800 ÷ 16

⇒ x = 2237.5

∴ The correct answer is 2237.5.

Additional Information

Percentage Decrease

If a number is decreased by x%, the new number becomes (100 − x)% of the original number. Formula: Final = Initial × (1 − x ÷ 100).

Successive Percentage Change

When a number is changed by a% and then by b%, the net percentage change is given by (a + b + ab ÷ 100)%. Use a negative sign for any decrease in percentage.

Percentage to Fraction Conversion

To convert a percentage directly into a fraction, simply divide the percentage value by 100. For example, 60% becomes 60 ÷ 100, which simplifies to 3 ÷ 5.

Shortcut Trick

When the total amount is constant, the quantity sold is inversely proportional to the selling price percentage.

⇒ Q₁ × (100 − Loss%) = Q₂ × (100 + Gain%)

⇒ 6 × (100 − 45) = Q₂ × (100 + 10)

⇒ 6 × 55 = Q₂ × 110

⇒ Q₂ = 330 ÷ 110 = 3

∴ The correct answer is 3 buttons.

Alternate Method

Given:

Selling price (SP₁) of 6 buttons = 1 Rupee

Loss percentage = 45%

Desired gain percentage = 10%

SP₁ (Loss 45%)
55%

Cost Price
100%

SP₂ (Gain 10%)
110%

−45%
+10%

6 Buttons

? Buttons

For 1 Rupee

Calculations:

⇒ SP₁ of 1 button = 1 ÷ 6 Rupee

⇒ Since the man loses 45%, SP₁ = (100 − 45)% of CP = 55% of CP

⇒ 55% of CP = 1 ÷ 6

⇒ CP = (1 ÷ 6) × (100 ÷ 55) = 10 ÷ 33 Rupee

⇒ To gain 10%, the new selling price (SP₂) = (100 + 10)% of CP = 110% of CP

⇒ SP₂ of 1 button = (110 ÷ 100) × (10 ÷ 33)

⇒ SP₂ of 1 button = (11 ÷ 10) × (10 ÷ 33) = 11 ÷ 33 = 1 ÷ 3 Rupee

⇒ Thus, the selling price is 1 Rupee for 3 buttons.

∴ The correct answer is 3 buttons.

Additional Information

Price and Consumption Relation

When the total amount spent is constant, the quantity of items purchased is inversely proportional to their price. Thus, Q₁ × P₁ = Q₂ × P₂.

Successive Discount

If two successive discounts of a% and b% are offered, the effective net discount is given by [a + b − (a × b) ÷ 100]%.

Dishonest Dealer

If a dealer sells goods at cost price but uses a false weight, his profit percentage is given by [(True Weight − False Weight) ÷ False Weight] × 100%.

Shortcut Trick

Since the time period is constant, Profit Ratio = Investment Ratio = 610 : 630 : 1450 = 61 : 63 : 145.

Given B's profit (63 parts) = ₹591

Difference between A and C's profit = 145 − 61 = 84 parts.

Required difference = (591 ÷ 63) × 84 = 591 × (4 ÷ 3) = 197 × 4 = 788.

∴ The correct answer is ₹788.

Alternate Method

Given:

Investment of A = ₹610

Investment of B = ₹630

Investment of C = ₹1450

Profit of B = ₹591

Formula Used:

When the time period is the same, Profit Ratio = Investment Ratio

Investment & Profit Ratio

A

₹610 → Ratio: 61

B

₹630 → Ratio: 63 (Profit = ₹591)

C

₹1450 → Ratio: 145

Difference (C − A) = 145 − 61 = 84 parts

Calculations:

⇒ Profit ratio of A, B and C = 610 : 630 : 1450 = 61 : 63 : 145

⇒ Let their profits be 61x, 63x and 145x respectively.

⇒ Profit earned by B = 63x = 591

⇒ x = 591 ÷ 63 = 197 ÷ 21

⇒ Difference in profit earned by A and C = 145x − 61x = 84x

⇒ Required difference = 84 × (197 ÷ 21)

⇒ Required difference = 4 × 197 = 788

∴ The correct answer is ₹788.

Additional Information

Partnership Principles

When partners invest for the same time duration, their profit is shared strictly in the ratio of their invested capitals. Profit Ratio = C₁ : C₂ : C₃.

Different Time Periods

If investments are made for different durations, then the effective capital rule is applied. Profit Ratio = (C₁ × T₁) : (C₂ × T₂) : (C₃ × T₃).

Active vs Sleeping Partner

An active partner usually receives a fixed salary or a specific percentage of the total profit for managing the business before the remaining profit is distributed according to capital investment ratios.

Shortcut Trick

When equal distances are covered at speeds x and y, the average speed is given by the direct formula:

Average speed = 2xy ÷ (x + y)

Here, x = 12 km/hr and y = 24 km/hr.

Average speed = (2 × 12 × 24) ÷ (12 + 24)

Average speed = 576 ÷ 36 = 16 km/hr

∴ The correct answer is 16 km/hr.

Alternate Method

Bhopal
Indore

Speed = 12 km/hr

Speed = 24 km/hr

Distance = 24 km (LCM)

Given:

Speed from Bhopal to Indore (S₁) = 12 km/hr

Speed from Indore to Bhopal (S₂) = 24 km/hr

Calculation:

Let the one-way distance between the two cities be the LCM of the given speeds to simplify calculations.

⇒ LCM of 12 and 24 = 24 km

⇒ Time taken to reach Indore = Distance ÷ S₁ = 24 ÷ 12 = 2 hours

⇒ Time taken to return to Bhopal = Distance ÷ S₂ = 24 ÷ 24 = 1 hour

⇒ Total distance covered for entire journey = 24 + 24 = 48 km

⇒ Total time taken = 2 + 1 = 3 hours

⇒ Average Speed = Total Distance ÷ Total Time

⇒ Average Speed = 48 ÷ 3 = 16 km/hr

∴ The correct answer is 16 km/hr.

Additional Information

Average Speed for Equal Time Intervals

If a person travels at speeds x and y for equal amounts of time, the average speed is simply the arithmetic mean: (x + y) ÷ 2.

Average Speed for Three Equal Distances

If equal distances are covered at three different speeds x, y, and z, the average speed formula is: 3xyz ÷ (xy + yz + zx).

Relative Speed

When two objects move in opposite directions, their relative speed is the sum of their individual speeds (x + y). In the same direction, it is the difference (x − y).

Shortcut Trick

In the range 411 to 424, ignore all even numbers and numbers ending in 5.

The remaining odd candidates are 411, 413, 417, 419, 421, and 423.

Eliminate multiples of 3 using the sum of digits trick: 411, 417, and 423 are divisible by 3.

Test the remaining numbers (413, 419, 421) for divisibility by 7: 413 = 7 × 59.

The only remaining numbers are 419 and 421, both of which are prime.

∴ The correct answer is 2 prime numbers.

Alternate Method

Given:

Range of numbers: 411 to 424 (both included)

Method Used:

To check if a number N is prime, we test divisibility by all prime numbers up to √N. Here, √424 ≈ 20, so we check divisibility by 2, 3, 5, 7, 11, 13, 17, and 19.

Calculations:

⇒ All even numbers in the range (412, 414, 416, 418, 420, 422, 424) are divisible by 2 and thus not prime.

⇒ 415 is divisible by 5 and thus not prime.

⇒ Check divisibility of the remaining odd numbers by 3, 7, 11, 13, 17, and 19:

⇒ 411 ÷ 3 = 137 (Not prime)

⇒ 413 ÷ 7 = 59 (Not prime)

⇒ 417 ÷ 3 = 139 (Not prime)

⇒ 419 is not divisible by any prime up to 19 (Prime)

⇒ 421 is not divisible by any prime up to 19 (Prime)

⇒ 423 ÷ 3 = 141 (Not prime)

⇒ The prime numbers in the given range are 419 and 421.

∴ The correct answer is 2 prime numbers.

Additional Information

Prime Number

A natural number greater than 1 that is only divisible by 1 and itself.

Twin Primes

A pair of prime numbers that have a difference of exactly 2. For example, 419 and 421 form a pair of twin primes.

Prime Testing Rule

To determine if a number N is prime, you only need to systematically test for divisibility by prime numbers up to the square root of N (√N).

Shortcut Trick

Solve the first few terms: (1 − 1/2) = 1/2; (1 − 1/3) = 2/3; (1 − 1/4) = 3/4.

Notice the cancellation pattern: The denominator of one term cancels the numerator of the next.

In this telescoping product, only the first numerator and the last denominator remain.

Result = First Numerator / Last Denominator = 1 / m.

∴ The correct answer is 1/m.

Alternate Method

Given: A product of terms in the form (1 − 1/n) from n = 2 to m.

Formula Used: (1 − 1/n) = (n − 1)/n

⇒ Term 1: (1 − 1/2) = 1/2

⇒ Term 2: (1 − 1/3) = 2/3

⇒ Term 3: (1 − 1/4) = 3/4

⇒ Term m: (1 − 1/m) = (m − 1)/m

⇒ Product = (1/2) × (2/3) × (3/4) × (4/5) × ... × (m−1)/m

⇒ After cross-cancellation of intermediate terms: 1/m

∴ The correct answer is 1/m.

Additional Information

Telescoping Series

A sequence where most terms cancel out, leaving only the first and last parts. For addition: ∑(a_n − a_n+1) = a₁ − a_last.

Successive Fractions Product

In products like (1 + 1/n), the pattern is (n+1)/n. Here, denominators cancel numerators differently: result is (m+1)/initial_denominator.

Pattern Recognition

Always simplify the first three terms of an infinite or large finite mathematical series to identify the cancellation behavior.

Shortcut Trick

Use the direct relation between Marked Price (MP) and Cost Price (CP): MP ÷ CP = (100 + Gain%) ÷ (100 − Discount%).

MP ÷ 391 = (100 + 68) ÷ (100 − 31) = 168 ÷ 69.

MP = (168 ÷ 69) × 391.

Dividing 391 by 69: 69 × 5 = 345, 391 − 345 = 46 (not direct). Simplification: 391 ÷ 23 = 17 and 69 ÷ 23 = 3.

MP = (168 ÷ 3) × 17 = 56 × 17 = 952.

∴ The correct answer is ₹952.

Alternate Method

Given → CP = ₹391, Gain% = 68%, Discount% = 31%.
Formulas: SP = CP × (100 + P%) ÷ 100 | MP = SP × 100 ÷ (100 − D%).

⇒ Step 1: Find Selling Price (SP).
⇒ SP = 391 × (168 ÷ 100) = 391 × 1.68 = 656.88.
⇒ Step 2: Find Marked Price (MP).
⇒ MP × (100 − 31) ÷ 100 = 656.88.
⇒ MP × 0.69 = 656.88.
⇒ MP = 656.88 ÷ 0.69 = 952.

∴ The correct answer is ₹952.

Additional Information

Relation between CP and MP

When both profit percentage and discount percentage are known, the ratio of MP to CP is always (100 + Profit%) / (100 − Discount%).

Marked Price & Discount

The price printed on an item is the Marked Price. Discount is always calculated on the MP to arrive at the Selling Price (SP = MP − Discount).

Shortcut Trick

Combined efficiency of A and B = 1/3

Efficiency of B alone = 1/12

Efficiency of A = 1/3 − 1/12 = (4 − 1)/12 = 3/12 = 1/4

Time for A to do single work = 4 days

Time for A to do double the work = 4 × 2 = 8 days

∴ The correct answer is 8 days.

Alternate Method

Given: Time taken by (A + B) = 3 days; Time taken by B = 12 days.

Formula: Total Work = LCM of individual days; Efficiency = Total Work ÷ Days.

⇒ Efficiency of A = Efficiency of (A + B) − Efficiency of B

⇒ Efficiency of A = 4 − 1 = 3 units/day

⇒ Double the work = 12 × 2 = 24 units

⇒ Time taken by A alone = 24 ÷ 3 = 8 days

∴ The correct answer is 8 days.

Additional Information

Efficiency-Time Relationship

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

Work-Time Formula

If A can do a work in x days and B in y days, together they take (xy)/(x+y) days.

Work & Wages

Wages are distributed in the ratio of the efficiency of the workers or the amount of work done.

Shortcut Trick

The age difference between Juli and her daughter is always constant at 24 years.

In 6 years, the ratio of their ages will be 2 : 1.

Difference in ratio parts = 2 − 1 = 1 unit.

1 unit → 24 years. Therefore, Juli's age in 6 years = 2 × 24 = 48 years.

Present age of Juli = 48 − 6 = 42 years.

∴ The correct answer is 42 years.

Alternate Method

Let the present age of the daughter be x years.

Given: Juli is 24 years older than her daughter.

⇒ Juli's present age = x + 24

⇒ In 6 years, Juli's age = (x + 24) + 6 = x + 30

⇒ In 6 years, Daughter's age = x + 6

⇒ According to the problem: x + 30 = 2 × (x + 6)

⇒ x + 30 = 2x + 12

⇒ 2x − x = 30 − 12

⇒ x = 18

⇒ Juli's present age = 18 + 24 = 42

∴ The correct answer is 42 years.

Additional Information

Constant Age Difference

The numerical difference between the ages of two people remains constant throughout their lives, regardless of how many years pass.

Ratio Method for Age

If the ratio of ages at two different times is known, equate the parts of the ratio to the constant age difference to find the multiplier.

Linear Equations in Ages

Always define present age as the base variable (x) to simplify calculations for future (+n years) or past (−n years) conditions.

Shortcut Trick

Apply the successive percentage formula: a + b + (a × b)/100

Here, a = −5% (decrease) and b = +8% (increase).

Effective % = −5 + 8 + (−5 × 8)/100 = 3 − 0.4 = 2.6%

Positive result indicates an increase.

∴ The correct answer is 2.6%.

Alternate Method

Given: First week decrease = 5%, Second week increase = 8%.

Assume initial price = 100 units

⇒ Price after 5% decrease = 100 − 5 = 95 units

⇒ Price after 8% increase = 95 + (8/100 × 95)

⇒ Price after 8% increase = 95 + 7.6 = 102.6 units

⇒ Net change = 102.6 − 100 = 2.6 units

⇒ Effective percentage increase = (2.6/100) × 100 = 2.6%

∴ The correct answer is 2.6%.

Additional Information

Successive Percentage Change

Used when a value is changed multiple times consecutively. Formula: Net % = a + b + (ab/100). Use positive for increase and negative for decrease.

Multiplying Factor Method

Final Value = Initial Value × (1 ± r₁/100) × (1 ± r₂/100). For 5% decrease and 8% increase, factor = 0.95 × 1.08 = 1.026.

Base Change Property

If price increases by x%, then to restore it, consumption must decrease by [x/(100+x)] × 100%.

Shortcut Trick

Assume all 6 fruits are bananas: Cost = 6 × ₹10 = ₹60.

Total difference from actual cost = ₹80 − ₹60 = ₹20.

Extra cost of an apple compared to a banana = ₹20 − ₹10 = ₹10.

Number of apples = 20 ÷ 10 = 2.

Number of bananas = 6 − 2 = 4.

Difference (Bananas − Apples) = 4 − 2 = 2.

∴ The correct answer is 2.

Alternate Method

Given:

Cost of 1 apple = ₹20

Cost of 1 banana = ₹10

Total fruits = 6

Total amount paid = ₹80

Let the number of apples be 'a' and the number of bananas be 'b'.

⇒ a + b = 6 --- (Equation 1)

⇒ 20a + 10b = 80 --- (Equation 2)

Divide Equation 2 by 10:

⇒ 2a + b = 8 --- (Equation 3)

Subtract Equation 1 from Equation 3:

⇒ (2a + b) − (a + b) = 8 − 6

⇒ a = 2

Substitute a = 2 in Equation 1:

⇒ 2 + b = 6

⇒ b = 4

The number of bananas is 4 and the number of apples is 2.

⇒ Difference = b − a = 4 − 2 = 2

∴ The correct answer is 2.

Additional Information

Linear Equations in Two Variables

A system of two equations like ax + by = c can be solved using substitution, elimination, or cross-multiplication methods.

Alligation Method

The ratio of the quantities can be found using (Price_higher − Average) : (Average − Price_lower). Here, Average = 80 ÷ 6 = ₹13.33.

Unitary Method Application

Calculating the cost by assuming an extreme case (e.g., all items are of one type) helps identify the quantity of the other type through the residual cost.

Shortcut Trick

Teacher's Age = New Average + (Initial Count × Increase in Average)

⇒ Teacher's Age = 20 + (30 × (20 − 19))

⇒ Teacher's Age = 20 + 30 = 50

∴ The correct answer is 50 years.

Alternate Method

Given:

Initial number of students = 30

Initial average age = 19 years

New number of people (students + teacher) = 31

New average age = 20 years

Formula Used:

Sum of ages = Average × Number of persons

⇒ Sum of ages of 30 students = 30 × 19 = 570 years

⇒ Sum of ages of 31 persons (including teacher) = 31 × 20 = 620 years

⇒ Age of the teacher = Total sum − Sum of students

⇒ Age of the teacher = 620 − 570 = 50 years

∴ The correct answer is 50 years.

Additional Information

Average Concept

Average is the sum of observations divided by the total number of observations. Average = Σx ÷ n.

Inclusion Rule

When a new person joins and the average increases, the new person's value = Old Average + (New Total Strength × Increase in Average).

Exclusion Rule

When a person leaves and the average decreases, the excluded person's value = Old Average + (Initial Strength × Decrease in Average).

Shortcut Trick

First, calculate the semi-perimeter (s): s = (34 + 18 + 20) ÷ 2 = 36 cm.

Find the Area using Heron's formula: Area = √(36 × (36−34) × (36−18) × (36−20)) = √(36 × 2 × 18 × 16).

Area = √(36 × 36 × 16) = 6 × 6 × 4 = 144 cm².

The altitude (h) corresponding to the 18 cm base: h = (2 × Area) ÷ Base = (2 × 144) ÷ 18 = 16 cm.

∴ The correct answer is 16 cm.

Alternate Method

Given:

Sides of the triangle: a = 34 cm, b = 18 cm, c = 20 cm.

Formula Used:

1. Heron's Formula for Area: Area = √[s(s − a)(s − b)(s − c)], where s = (a + b + c) ÷ 2.
2. General Area formula: Area = (1/2) × Base × Altitude.

Calculations:

⇒ Semi-perimeter (s) = (34 + 18 + 20) ÷ 2

⇒ s = 72 ÷ 2 = 36 cm

⇒ Area = √[36 × (36 − 34) × (36 − 18) × (36 − 20)]

⇒ Area = √[36 × 2 × 18 × 16]

⇒ Area = √[36 × 36 × 16]

⇒ Area = 6 × 6 × 4 = 144 cm²

⇒ Area = (1/2) × 18 × Altitude

⇒ 144 = 9 × Altitude

⇒ Altitude = 144 ÷ 9

⇒ Altitude = 16 cm

∴ The correct answer is 16 cm.

Additional Information

Triangle Inequality Theorem

The sum of the lengths of any two sides of a triangle must be greater than the length of the third side (a + c).

Smallest Side and Altitude

In any triangle, the longest altitude always corresponds to the shortest side, and the shortest altitude corresponds to the longest side.

Area via Sine of Angle

Area can also be calculated if an angle is known: Area = (1/2) × a × b × sin(C), where C is the angle between sides a and b.

Shortcut Trick

Apply the algebraic identity: x² + y² = (x + y)² − 2xy.

Substitute the given values: Sum = 4 and Product = 2.

Calculation: 4² − 2 × 2 = 16 − 4.

Result = 12.

∴ The correct answer is 12.

Alternate Method

Given:

Sum of variables (x + y) = 4

Product of variables (xy) = 2

Formula Used:

(x + y)² = x² + y² + 2xy

Calculations:

⇒ (x + y)² = x² + y² + 2xy

⇒ (4)² = x² + y² + 2 × 2

⇒ 16 = x² + y² + 4

⇒ x² + y² = 16 − 4

⇒ x² + y² = 12

∴ The correct answer is 12.

Additional Information

Square of a Difference

The identity for the square of a subtraction is (x − y)² = x² + y² − 2xy. This can be used to find x² + y² if the difference and product are known.

Difference of Two Squares

The formula x² − y² = (x − y)(x + y) is used to factorize the difference of two squared terms.

Relationship between Identities

The two basic square identities are related by the formula: (x + y)² − (x − y)² = 4xy.

Shortcut Trick

Let the number be x.

20% × 10% × 2/3 × x = 697

1/5 × 1/10 × 2/3 × x = 697

1/75 × x = 697 ⇒ x = 697 × 75 = 52275

∴ The correct answer is 52275.

Alternate Method

Given: Result of the calculation = 697

⇒ 20% of 10% of 2/3 × number = 697

⇒ (20/100) × (10/100) × (2/3) × number = 697

⇒ (1/5) × (1/10) × (2/3) × number = 697

⇒ (2/150) × number = 697

⇒ (1/75) × number = 697

⇒ number = 697 × 75

⇒ number = 52275

∴ The correct answer is 52275.

Additional Information

Percentage to Fraction Conversion

To simplify calculations, convert percentages to fractions: 20% = 1/5 and 10% = 1/10.

Successive Percentage

When applying successive percentages, multiply the fraction equivalents directly: a% of b% of X = (a/100) × (b/100) × X.

Reverse Calculation

If the final value is known, find the initial number by multiplying the final value by the reciprocal of each factor.