Quantitative Aptitude PYQs

Shortcut Trick

Increase in average = (New score − Old average) ÷ Total number of innings

Increase = (193 − 76) ÷ 11

Increase = 117 ÷ 11 = 10.636...

Rounding 10.636 to the nearest whole number, we get 11.

∴ The correct answer is 11.

Alternate Method

Given:

Average of 10 innings = 76 runs

Runs in 11^th inning = 193 runs

Formula Used:

Average = Total Runs ÷ Total Innings

Increase in Average = New Average − Old Average

Calculations:

⇒ Total runs in 10 innings = 10 × 76 = 760 runs

⇒ Total runs in 11 innings = 760 + 193 = 953 runs

⇒ New average = 953 ÷ 11 = 86.6363...

⇒ Increase in average = 86.6363 − 76 = 10.6363...

⇒ Rounding to the nearest whole number ≈ 11

∴ The correct answer is 11.

Additional Information

Effect of New Entry on Average

If a new value is added, New Average = (Old Sum + New Value) ÷ (n + 1). If New Value > Old Average, the average increases.

Weighted Average

When two groups with averages A₁ and A₂ and sizes n₁ and n₂ are combined, Combined Average = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Deviation Method

The sum of deviations of all observations from their mean is always zero. This property helps in finding missing values quickly.

Shortcut Trick

60% of First Number = 12 → 100% of First Number = 20.

Ratio value 10 units = 20 → 1 unit = 2.

Difference of Third and Second (21 − 16) = 5 units = 10.

40% of 10 = 4.

∴ The correct answer is 4.

Alternate Method

Given: Ratio = 10 : 21 : 16 | 60% of 1^st = 12

Formula Used: Number = (Given Value ÷ Percentage) × 100

⇒ Let the numbers be 10x, 21x, and 16x.

⇒ 60% of 10x = 12

⇒ 0.6 × 10x = 12 ⇒ 6x = 12 ⇒ x = 2

⇒ Second number = 21 × 2 = 42

⇒ Third number = 16 × 2 = 32

⇒ Difference = 42 − 32 = 10

⇒ Required value = 40% of 10 = 0.4 × 10 = 4

∴ The correct answer is 4.

Additional Information

Percentage to Fraction Conversion

To simplify calculations, use 60% = 3/5 and 40% = 2/5.

Ratio Scaling

If a : b : c is the ratio, any change in one part according to a multiplier x applies to all parts to maintain the proportion.

Shortcut Trick

Assume an average of 40 to find deviations: −22, +34, −7, −24, −5, +56, 0, +28, +3.

Net deviation = (−22 − 7 − 24 − 5) + (34 + 56 + 28 + 3) = −58 + 121 = +63.

Average deviation = 63 ÷ 9 = +7.

Actual Arithmetic Mean = 40 + 7 = 47.

∴ The correct answer is 47.

Alternate Method

Given: Observations = 18, 74, 33, 16, 35, 96, 40, 68, 43

Total count (n) = 9

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

⇒ Sum = 18 + 74 + 33 + 16 + 35 + 96 + 40 + 68 + 43

⇒ Sum = 423

⇒ Mean = 423 ÷ 9

⇒ Mean = 47

∴ The correct answer is 47.

Additional Information

Properties of Arithmetic Mean

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

Effect of Constant on Mean

If every observation is increased or decreased by a constant k, the mean also increases or decreases by k. If multiplied by k, the mean is multiplied by k.

Weighted Mean Formula

When observations have different weights, the mean is calculated as ∑(w_i × x_i) ÷ ∑w_i.

Shortcut Trick

Use the successive discount formula: Equivalent Discount = a + b − (a × b) ÷ 100.

Here, a = 20% and b = 29%.

Equivalent Discount = 20 + 29 − (20 × 29) ÷ 100.

Equivalent Discount = 49 − (580 ÷ 100) = 49 − 5.8 = 43.2%.

∴ The correct answer is 43.2%.

Alternate Method

Given: First discount = 20%, Second discount = 29%.

Formula: Net SP = MP × (100 − d₁)% × (100 − d₂)%

⇒ Let the Marked Price (MP) be 100.

⇒ Price after 20% discount = 100 − 20 = 80.

⇒ Price after 29% discount on 80 = 80 − (29% of 80).

⇒ 29% of 80 = (29 × 80) ÷ 100 = 23.2.

⇒ Final Selling Price (SP) = 80 − 23.2 = 56.8.

⇒ Total Discount = MP − SP = 100 − 56.8 = 43.2.

∴ The correct answer is 43.2%.

Additional Information

Successive Discounts

When two or more discounts are applied one after another on the reduced price, they are called successive discounts. The total effective discount is always less than the sum of individual discounts.

Marked Price & Selling Price

Marked Price is the price printed on the label. Selling Price is the final amount paid after subtracting the discount from the MP: SP = MP × (1 − d/100).

Shortcut Trick

To find percentage of A with respect to B, use: (A ÷ B) × 100.

Required % = (1/30 ÷ 1/100) × 100 = (100/30) × 100.

Required % = (10/3) × 100 = 1000/3 %.

Required % = 333.33%.

∴ The correct answer is 333.33%.

Alternate Method

Given: First fraction = 1/30, Second fraction = 1/100.

Formula Used: Percentage = (Value / Base Value) × 100

⇒ Required Percentage = [(1/30) / (1/100)] × 100

⇒ Required Percentage = (1/30 × 100/1) × 100

⇒ Required Percentage = (100/30) × 100

⇒ Required Percentage = (10/3) × 100

⇒ Required Percentage = 1000 / 3

⇒ Required Percentage = 333.333...

∴ The correct answer is 333.33%.

Additional Information

Fraction to Percentage Conversion

To convert any fraction to percentage, multiply the fraction by 100 and add the % symbol.

Reciprocal Percentages

Common reciprocals: 1/3 ≈ 33.33%, 1/6 ≈ 16.67%, 1/9 ≈ 11.11%.

Rounding Off Rules

When rounding to two decimal places, if the third decimal digit is 5 or more, increase the second decimal digit by 1.

Shortcut Trick

Combined efficiency of A, B, and C = 8 + 3 + 1 = 12 units/day (taking total work as 24 units).

To complete twice the work (48 units), they will take 48 ÷ 12 = 4 days.

∴ The correct answer is 4 days.

Alternate Method

Given:

Time taken by A = 3 days

Time taken by B = 8 days

Time taken by C = 24 days

Formula Used:

Total Work = LCM of individual times

Efficiency = Total Work ÷ Days

Time taken = Work ÷ Combined Efficiency

Calculations:

⇒ Total original work = 24 units

⇒ Twice the work = 2 × 24 = 48 units

⇒ Combined efficiency of A, B and C = 8 + 3 + 1 = 12 units/day

⇒ Time taken to complete twice the work = 48 ÷ 12 = 4 days

∴ The correct answer is 4 days.

Additional Information

LCM Method in Time and Work

Using the LCM of individual days as the total work helps avoid working with complex fractions, simplifying calculations into whole number efficiencies.

Work and Wages Relation

Wages are always distributed in the proportion of the total work done by each individual, or in the ratio of their efficiencies if they work for the same duration.

Shortcut Trick

For a number to be divisible by 8, its last three digits must be divisible by 8.

The last three digits of 130A1568 are 568.

Check: 568 ÷ 8 = 71 (Exactly divisible).

Since 568 is already divisible by 8, the value of A does not affect the divisibility of the last three digits.

The question asks for the minimum value of A.

∴ The correct answer is 0.

Alternate Method

Given: The 8-digit number is 130A1568.

Divisibility Rule for 8: A number is divisible by 8 if the number formed by its last 3 digits is divisible by 8.

⇒ Last three digits of 130A1568 = 568

⇒ 568 ÷ 8 = 71

⇒ Since 568 is divisible by 8, the entire number 130A1568 is divisible by 8 regardless of the value of A.

⇒ Possible values for digit A are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

⇒ The minimum value among these is 0.

∴ The correct answer is 0.

Additional Information

Divisibility by 2ⁿ

A number is divisible by 2ⁿ if its last n digits are divisible by 2ⁿ. For 2 (2¹), check last 1 digit; for 4 (2²), check last 2 digits; for 8 (2³), check last 3 digits.

Divisibility by 3 and 9

A number is divisible by 3 or 9 if the sum of its digits is divisible by 3 or 9 respectively.

Divisibility by 11

The difference between the sum of digits at odd places and the sum of digits at even places must be 0 or a multiple of 11.

Shortcut Trick

When the distance for two journeys is equal (A to B and B to A), use the harmonic mean formula.

Average Speed = [2 × S₁ × S₂] ÷ [S₁ + S₂]

Average Speed = [2 × 48 × 52] ÷ [48 + 52] = 4992 ÷ 100 = 49.92 km/hr.

∴ The correct answer is 49.92 km/hr.

Alternate Method

Given: Distance (D) = 964 km, Speed A to B (S₁) = 48 km/hr, Speed B to A (S₂) = 52 km/hr.

Formula Used: Average Speed = Total Distance ÷ Total Time

⇒ Total Distance = 964 + 964 = 1928 km

⇒ Time taken (A to B) = 964 ÷ 48 = 241 ÷ 12 hours

⇒ Time taken (B to A) = 964 ÷ 52 = 241 ÷ 13 hours

⇒ Total Time = (241 ÷ 12) + (241 ÷ 13) = [241 × (13 + 12)] ÷ (12 × 13)

⇒ Total Time = (241 × 25) ÷ 156 hours

⇒ Average Speed = 1928 ÷ [(241 × 25) ÷ 156] = (1928 × 156) ÷ (241 × 25)

⇒ Average Speed = (8 × 156) ÷ 25 = 1248 ÷ 25 = 49.92 km/hr

∴ The correct answer is 49.92 km/hr.

Additional Information

Average Speed for Equal Distances

When an object covers two equal distances at speeds x and y, the average speed is given by [2xy] ÷ [x + y]. Note that the actual distance value does not affect the final result.

Three Equal Distances

For three equal distances covered at speeds x, y, and z, the average speed formula is [3xyz] ÷ [xy + yz + zx].

General Average Speed

Always defined as Total Distance ÷ Total Time. If the distances are not equal, individual times must be calculated and summed up.

Shortcut Trick

In any quadrilateral, the angle formed by the bisectors of two adjacent angles is equal to half the sum of the other two angles.

∠DOC = (∠A + ∠B) ÷ 2

∠DOC = (62° + 19°) ÷ 2

∠DOC = 81° ÷ 2 = 40.5°

∴ The correct answer is 40.5°.

Alternate Method

Given: ∠A = 62°, ∠B = 19°, CO and DO are angle bisectors.

Formula Used: Sum of interior angles of a quadrilateral = 360° | Sum of angles in a triangle = 180°.

⇒ In quadrilateral ABCD, ∠A + ∠B + ∠C + ∠D = 360°

⇒ 62° + 19° + ∠C + ∠D = 360°

⇒ ∠C + ∠D = 360° − 81° = 279°

⇒ In ΔDOC, ∠DOC + ∠OCD + ∠ODC = 180°

⇒ Since CO and DO are bisectors, ∠OCD = ∠C ÷ 2 and ∠ODC = ∠D ÷ 2

⇒ ∠DOC + (∠C + ∠D) ÷ 2 = 180°

⇒ ∠DOC + 279° ÷ 2 = 180°

⇒ ∠DOC + 139.5° = 180°

⇒ ∠DOC = 180° − 139.5° = 40.5°

∴ The correct answer is 40.5°.

Additional Information

Angle Sum Property

The sum of the interior angles of an n-sided polygon is given by (n − 2) × 180°. For a quadrilateral (n=4), it is 360°.

Angle Bisector Property

An angle bisector divides an angle into two equal parts. In any triangle, the sum of the angles is 180°, which is used to find the third angle when the other two (or their sum) are known.

1

Shortcut Trick

LCM of 12, 15, 18 = 180 units (Total Capacity).

Efficiencies: A = 15, B = 12, C = 10.

Hour 1 (A+B): 15 + 12 = 27 units | Hour 2 (A+C): 15 + 10 = 25 units.

2-hour cycle = 27 + 25 = 52 units. In 3 cycles (6 hours), work done = 52 × 3 = 156 units.

Remaining 24 units filled by (A+B) in 24/27 = 8/9 hours. Total = 6 + 8/9 = 6 8/9 hours.

∴ The correct answer is 6 8/9 hr.

Alternate Method

⇒ Hour 1 (A and B are open): Efficiency = 15 + 12 = 27 units/hr

⇒ Hour 2 (A and C are open): Efficiency = 15 + 10 = 25 units/hr

⇒ Work done in 2 hours = 27 + 25 = 52 units

⇒ Work done in 6 hours (3 cycles) = 52 × 3 = 156 units

⇒ Remaining work = 180 − 156 = 24 units

⇒ Time taken for remaining work (by A+B) = 24 ÷ 27 = 8/9 hr

⇒ Total time = 6 + 8/9 = 6 8/9 hr

∴ The correct answer is 6 8/9 hr.

Additional Information

Pipes and Cisterns Efficiency

Efficiency is inversely proportional to the time taken. If a pipe fills a tank in 'x' hours, its 1-hour work is 1/x.

Alternate Work Principle

In alternate timing problems, calculate the work done in one complete cycle (sum of individual hours) before multiplying to reach the target capacity.

Inlet and Outlet Pipes

Inlet pipe efficiency is taken as positive, while outlet (leakage) pipe efficiency is taken as negative during net efficiency calculation.

Shortcut Trick

Substitute the options into the equation 25^{(r − 1)} + 100 = 5^{(2r − 1)} to see which one satisfies it.

Checking for r = 2:

⇒ 25^{(2 − 1)} + 100 = 5^{(2 × 2 − 1)}

⇒ 25¹ + 100 = 5³

⇒ 125 = 125 (Satisfied)

∴ The correct answer is 2.

Alternate Method

Given:

25^{(r − 1)} + 100 = 5^{(2r − 1)}

Formula Used:

(a^m)ⁿ = a^{m × n}

Calculation:

⇒ (5²)^{(r − 1)} + 100 = 5^{(2r − 1)}

⇒ 5^{(2r − 2)} + 100 = 5^{(2r − 1)}

Rearranging the terms by taking base 5 values to one side:

⇒ 100 = 5^{(2r − 1)} − 5^{(2r − 2)}

Taking 5^{(2r − 2)} common on the right hand side:

⇒ 100 = 5^{(2r − 2)} × (5¹ − 1)

⇒ 100 = 5^{(2r − 2)} × 4

⇒ 25 = 5^{(2r − 2)}

⇒ 5² = 5^{(2r − 2)}

Equating the exponents since the bases are equal:

⇒ 2r − 2 = 2

⇒ 2r = 4

⇒ r = 2

∴ The correct answer is 2.

Additional Information

Law of Product

When multiplying expressions with the same base, add their exponents: a^m × aⁿ = a^{m + n}.

Law of Quotient

When dividing expressions with the same base, subtract their exponents: a^m ÷ aⁿ = a^{m − n}.

Law of Power of a Power

When raising a power to another power, multiply the exponents: (a^m)ⁿ = a^{m × n}.

Equality of Exponents

If a^m = aⁿ (where a > 0 and a ≠ 1), then m = n.

Shortcut Trick

To find the two numbers quickly using Sum (S) and Difference (D):

Larger number = (S + D) ÷ 2 = (24 + 8) ÷ 2 = 16

Smaller number = (S − D) ÷ 2 = (24 − 8) ÷ 2 = 8

Product = 16 × 8 = 128

∴ The correct answer is 128.

Alternate Method

Given:

Sum of two numbers (x + y) = 24

Difference of two numbers (x − y) = 8

Formula Used:

Algebraic product identity: xy = ((x + y)² − (x − y)²) ÷ 4

Calculations:

⇒ x + y = 24 --- (Equation 1)

⇒ x − y = 8 --- (Equation 2)

⇒ Adding both equations: (x + y) + (x − y) = 24 + 8

⇒ 2x = 32

⇒ x = 32 ÷ 2 = 16

⇒ Substitute x = 16 in Equation 1: 16 + y = 24

⇒ y = 24 − 16 = 8

⇒ Product = x × y = 16 × 8 = 128

∴ The correct answer is 128.

Additional Information

Product-Sum-Difference Identity

The product of two numbers can be directly calculated using the identity: Product = [(Sum)² − (Difference)²] ÷ 4.

Sum of Squares Relationship

The sum of the squares of two numbers is given by: x² + y² = [(Sum)² + (Difference)²] ÷ 2.

Relationship with Arithmetic Mean

The Average of the two numbers is always exactly half of their sum (S ÷ 2). The numbers are (Average + D/2) and (Average − D/2).

Shortcut Trick

To find the larger value (Sister) when the sum and difference are known:

Sister age = (Sum + Difference) ÷ 2

Sister age = (26 + 2) ÷ 2 = 28 ÷ 2 = 14

∴ The correct answer is 14 years.

Alternate Method

Given:

Sum of ages = 26 years

Age difference = 2 years

Formula Used:

Linear Equation solving for two variables.

Let the age of the Sister be S and the age of the Brother be B.

⇒ S + B = 26 --- (Equation 1)

⇒ S − B = 2 --- (Equation 2)

Adding Equation 1 and Equation 2:

⇒ (S + B) + (S − B) = 26 + 2

⇒ 2S = 28

⇒ S = 28 ÷ 2

⇒ S = 14

∴ The correct answer is 14 years.

Additional Information

Sum and Difference Rule

If sum (S) and difference (D) of two numbers are given, Larger = (S+D)/2 and Smaller = (S−D)/2.

Age Constant Property

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

Shortcut Trick

Total Surface Area (TSA) = CSA of cylinder + CSA of hemisphere + CSA of cone + Area of annular base.

TSA = 2πRH + 2πR² + πrl + (πR² − πr²) = π[2RH + 3R² + rl − r²]

Substitute R=14, H=48, r=7, h=24, l=25: TSA = (22/7) × [2(14)(48) + 3(14²) + 7(25) − 7²]

TSA = (22/7) × [1344 + 588 + 175 − 49] = (22/7) × 2058 = 22 × 294 = 6468 cm².

∴ The correct answer is 6468 cm².

Alternate Method

Given:

Height of cylinder (H) = 48 cm, Radius of cylinder (R) = 14 cm

Radius of conical cavity (r) = 7 cm, Height of conical cavity (h) = 24 cm

Radius of hemisphere (R_h) = 14 cm

Formula Used:

Curved Surface Area (CSA) of cylinder = 2πRH

CSA of cone = πrl, where l = √(r² + h²)

CSA of hemisphere = 2πR²

Calculations:

⇒ Slant height of cone (l) = √(7² + 24²) = √(49 + 576) = √625 = 25 cm

⇒ CSA of cylinder = 2 × (22/7) × 14 × 48 = 2 × 22 × 2 × 48 = 4224 cm²

⇒ CSA of cone = (22/7) × 7 × 25 = 550 cm²

⇒ CSA of hemisphere = 2 × (22/7) × 14 × 14 = 1232 cm²

⇒ Area of top annular region = πR² − πr² = (22/7) × (14² − 7²) = (22/7) × (196 − 49) = (22/7) × 147 = 462 cm²

⇒ Total Surface Area = 4224 + 550 + 1232 + 462

⇒ Total Surface Area = 6468 cm²

∴ The correct answer is 6468 cm².

Additional Information

Volume of Remaining Solid

The volume is calculated by subtracting the volumes of the cavities: V = πR²H − (1/3)πr²h − (2/3)πR³.

Surface Area Change

When a cavity is made, the Total Surface Area usually increases as the internal curved surface of the cavity is added, while only the flat base area of the cavity is removed.

Hemisphere Surface Area

For a solid hemisphere, TSA = 3πR² (Curved + Base), but for a hemispherical cavity, only the curved part 2πR² is added to the internal surface.

Shortcut Trick

The given expression follows the algebraic identity (a² − b²) ÷ c.

Notice that (75983 − 45983) = 30000, which is exactly the denominator.

Since (a² − b²) = (a + b)(a − b), the expression reduces to (a + b) when (a − b) is the divisor.

Result = 75983 + 45983 = 121966.

∴ The correct answer is 121966.

Alternate Method

Given: Expression = (75983 × 75983 − 45983 × 45983) ÷ 30000

Formula Used: a² − b² = (a + b)(a − b)

⇒ Let a = 75983 and b = 45983.

⇒ a² − b² = (75983 + 45983) × (75983 − 45983)

⇒ a² − b² = 121966 × 30000

⇒ Expression = (121966 × 30000) ÷ 30000

⇒ Expression = 121966

∴ The correct answer is 121966.

Additional Information

Difference of Squares

The identity a² − b² = (a + b)(a − b) is used to simplify the multiplication and division of large square terms.

Sum and Difference Cubes

Similarly, a³ − b³ = (a − b)(a² + ab + b²) and a³ + b³ = (a + b)(a² − ab + b²).

BODMAS Rule

Always follow the order of operations: Brackets, Orders, Division, Multiplication, Addition, and Subtraction for correct results.

Shortcut Trick

Total Cost Price (CP) = Purchase Price + Repair Cost = 5,600 + 1,400 = 7,000.

Selling Price (SP) = 7,700.

Profit = 7,700 − 7,000 = 700.

Profit% = (Profit ÷ CP) × 100 = (700 ÷ 7,000) × 100 = 10%.

∴ The correct answer is 10%.

Alternate Method

Given: Purchase Price = ₹5,600, Repair cost = ₹1,400, Selling Price (SP) = ₹7,700.
Formulas: Total CP = Purchase Price + Overhead Expenses | Profit = SP − CP | Profit% = (Profit ÷ CP) × 100

⇒ Total Cost Price (CP) = 5,600 + 1,400 = ₹7,000.
⇒ Selling Price (SP) = ₹7,700.
⇒ Profit = SP − CP = 7,700 − 7,000 = ₹700.
⇒ Profit Percentage = (700 ÷ 7,000) × 100.
⇒ Profit Percentage = 10%.

∴ The correct answer is 10%.

Additional Information

Effective Cost Price

Any additional expenditure incurred on an item like repairs, transportation, or labor after purchase is added to the original price to find the effective CP.

Profit and Loss Calculation

Profit or loss is always calculated on the Cost Price unless specified otherwise. Formula: Profit% = (Profit ÷ CP) × 100.

Shortcut Trick

Start from inner bracket: 2² = 4

Expression becomes: 3 × (10 × 4) ÷ 20 + 13 − 3

10 × 4 = 40 ⇒ 3 × 40 ÷ 20

120 ÷ 20 = 6 ⇒ 6 + 13 − 3

∴ Answer = 16

Alternate Method

Given:

3 × (10 × (2²)) ÷ 20 + 13 − 3

Concept Used:

BODMAS rule

Calculation:

⇒ 3 × (10 × 4) ÷ 20 + 13 − 3

⇒ 3 × 40 ÷ 20 + 13 − 3

⇒ 120 ÷ 20 + 13 − 3

⇒ 6 + 13 − 3

⇒ 19 − 3

⇒ 16

∴ The correct answer is 16.

Additional Information

BODMAS Rule

Brackets → Orders → Division → Multiplication → Addition → Subtraction

Order of Operations

Solve brackets first, then powers, then × and ÷, then + and −

Simplification Tip

Always simplify inner brackets first to reduce steps

Exponent Rule

a² = a × a

Shortcut Trick

Total valid votes = 88% of 25,000 = 22,000.

Percentage of votes for Candidate C = 100% − (40% + 30%) = 30%.

Votes for Candidate C = 30% of 22,000 = 6,600.

∴ The correct answer is 6600.

Alternate Method

Given:

Total voters = 25,000

Invalid votes = 12%

Candidate A votes = 40% of valid votes

Candidate B votes = 30% of valid votes

⇒ Total Valid Votes = 100% − 12% = 88%

⇒ Total Valid Votes = (88 ÷ 100) × 25,000 = 22,000

⇒ Sum of votes for A and B = 40% + 30% = 70%

⇒ Votes for Candidate C = 100% − 70% = 30%

⇒ Number of votes for C = 30% of 22,000

⇒ Number of votes for C = (30 ÷ 100) × 22,000

⇒ Number of votes for C = 3 × 2,200 = 6,600

∴ The correct answer is 6600.

Additional Information

Percentage Base Change

Always calculate the final percentage on the Valid Votes base unless the question specifies the Total Votes base.

Successive Percentage

Votes for C = Total × (Valid Ratio) × (Share of C). Here, 25,000 × 0.88 × 0.30 = 6,600.

Shortcut Trick

Surplus runs scored in 10^th inning = New Score − Old Average

Surplus = 191 − 54 = 137 runs

Increase in average = Surplus ÷ Total Innings

Increase = 137 ÷ 10 = 13.7 ≈ 14

∴ The correct answer is 14.

Alternate Method

Given:

Average of 9 innings = 54 runs

Score in 10^th inning = 191 runs

Calculations:

⇒ Total runs in 9 innings = 9 × 54 = 486 runs

⇒ Total runs in 10 innings = 486 + 191 = 677 runs

⇒ New average for 10 innings = 677 ÷ 10 = 67.7 runs

⇒ Increase in average = New average − Old average

⇒ Increase = 67.7 − 54 = 13.7 runs

⇒ Rounding to the nearest whole number ≈ 14

∴ The correct answer is 14.

Additional Information

Average (Arithmetic Mean)

Average is defined as the Sum of Observations ÷ Number of Observations.

Effect of New Entry

If the new value added is greater than the old average, the average increases. If it is less, the average decreases.

Weighted Average Formula

Combined Average = (n₁x₁ + n₂x₂) ÷ (n₁ + n₂), where n is the number of items and x is the average.

Shortcut Trick

Calculate the semi-perimeter (s) of the triangle: s = (80 + 44 + 52) ÷ 2 = 88 cm.

Apply Heron's Formula for Area: √(88 × (88−80) × (88−44) × (88−52)) = √(88 × 8 × 44 × 36).

Area = √(2 × 44 × 8 × 44 × 36) = 44 × 4 × 6 = 1056 cm².

Altitude (h) = (2 × Area) ÷ base = (2 × 1056) ÷ 44 = 48 cm.

∴ The correct answer is 48 cm.

Alternate Method

Given:

Sides of the triangle (a, b, c) are 80 cm, 52 cm, and 44 cm.

Formula Used:

Semi-perimeter (s) = (a + b + c) ÷ 2

Area of triangle (A) = √(s × (s−a) × (s−b) × (s−c))

Area of triangle (A) = 1/2 × base × height

Calculations:

⇒ s = (80 + 52 + 44) ÷ 2 = 176 ÷ 2 = 88 cm

⇒ Area = √(88 × (88 − 80) × (88 − 52) × (88 − 44))

⇒ Area = √(88 × 8 × 36 × 44)

⇒ Area = √((2 × 44) × (2 × 4) × (6 × 6) × 44)

⇒ Area = √(44² × 4 × 6² × 4)

⇒ Area = 44 × 2 × 6 × 2 = 1056 cm²

⇒ Now, for the base of 44 cm, 1056 = 1/2 × 44 × height

⇒ 1056 = 22 × height

⇒ height = 1056 ÷ 22 = 48 cm

∴ The correct answer is 48 cm.

Additional Information

Heron's Formula

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.

Altitudes and Area

A triangle has three altitudes. Area = 1/2 × b₁h₁ = 1/2 × b₂h₂ = 1/2 × b₃h₃. The longest side has the shortest altitude.

Obtuse Triangle Properties

In an obtuse-angled triangle, two of the altitudes lie outside the triangle on the extended base lines.