Quantitative Aptitude PYQs

Shortcut Trick

Total List Price = 625 × 94 = 58,750.

Net Multiplier = (80 ÷ 100) × (93 ÷ 100) × (80 ÷ 100) = 0.8 × 0.93 × 0.8 = 0.5952.

Cash Payment = 58,750 × 0.5952 = 34,968.

∴ The correct answer is ₹34,968.

Alternate Method

Given: Quantity = 625, Rate = ₹94, Discounts = 20%, 7%, 20%.
Formulas: Total Value = Quantity × Rate | SP = MP × (100 − D₁)% × (100 − D₂)% × (100 − D₃)%

⇒ Total List Price = 625 × 94 = ₹58,750.
⇒ After 20% discount = 58,750 × 0.80 = 47,000.
⇒ After 7% discount = 47,000 × 0.93 = 43,710.
⇒ Final Cash Payment (after 20% further discount) = 43,710 × 0.80 = ₹34,968.

∴ The correct answer is ₹34,968.

Additional Information

Successive Discounts

When multiple discounts are applied one after another, the net selling price is calculated on the reducing balance at each step.

Cash Discount

A cash discount is an incentive offered by a seller to a buyer in return for paying a bill before the scheduled due date.

Shortcut Trick

Age of new person (Teacher) = New Average + (Original Strength × Increase in Average)

Teacher age = 31 + (31 × (31 − 30))

⇒ 31 + (31 × 1) = 31 + 31 = 62 years.

Alternatively, Teacher age = Old Average + (New Strength × Increase in Average)

⇒ 30 + (32 × 1) = 30 + 32 = 62 years.

∴ The correct answer is 62 years.

Alternate Method

Given:

Number of students = 31

Initial average age = 30 years

New average after including teacher = 31 years

Total members (Students + Teacher) = 31 + 1 = 32

Formula Used:

Average = Total Sum of Ages ÷ Number of Persons

Total Sum = Average × Number of Persons

Calculations:

⇒ Total age of 31 students = 31 × 30 = 930 years

⇒ Total age of group including teacher = 32 × 31

⇒ Total age (32 members) = 992 years

⇒ Age of the teacher = Total age with teacher − Total age of students

⇒ Age of the teacher = 992 − 930

⇒ Age of the teacher = 62 years

∴ The correct answer is 62 years.

Additional Information

Basic Average Formula

Average = (Sum of all observations) ÷ (Total number of observations).

Exclusion Case Shortcut

Age of excluded person = Old Average + (Old Strength − 1) × (Decrease in Average).

Replacement Case Shortcut

Value of new person = Value of replaced person + (Total number × Increase in Average).

Weighted Average

Combined Average = (n₁ × A₁ + n₂ × A₂) ÷ (n₁ + n₂), where n is quantity and A is average.

Shortcut Trick

Net multiplier = (1 − 0.45) × (1 − 0.35) × (1 − 0.30) × (1 + 0.45)

Net multiplier = 0.55 × 0.65 × 0.70 × 1.45

Net multiplier = 0.3575 × 1.015 = 0.3628625

Net Change % = (0.3628625 − 1) × 100 = −63.71375%

∴ The correct answer is Decreases by 63.71%.

Alternate Method

Given: Successive changes of −45%, −35%, −30% and +45%.

Formula Used: Final Price = Initial Price × (100 ± r₁)/100 × (100 ± r₂)/100 ...

⇒ Let the Original Price of fuel be 100.

⇒ Price after 1^st month = 100 × (100 − 45)/100 = 55

⇒ Price after 2^nd month = 55 × (100 − 35)/100 = 55 × 0.65 = 35.75

⇒ Price after 3^rd month = 35.75 × (100 − 30)/100 = 35.75 × 0.7 = 25.025

⇒ Price after 4^th month = 25.025 × (100 + 45)/100 = 25.025 × 1.45 = 36.28625

⇒ Total change = 36.28625 − 100 = −63.71375

⇒ Percentage Decrease ≈ 63.71%

∴ The correct answer is Decreases by 63.71%.

Additional Information

Successive Percentage Change

For two successive changes a% and b%, the net change is given by [a + b + (ab/100)]%.

Multiplier Method

Final value = Initial value × (1 + r/100)ⁿ. For different rates, multiply each factor (1 ± r/100) consecutively.

Percentage Decrease

It is calculated as [(Initial Value − Final Value) / Initial Value] × 100.

Shortcut Trick

Apply the algebraic identity: (a + b)² + (a − b)² = 2(a² + b²).

Here, a = 4 and b = √3.

Sum of areas = 2 × [4² + (√3)²]

⇒ 2 × (16 + 3) = 2 × 19 = 38 cm².

∴ The correct answer is 38 cm².

Alternate Method

Given:

Side of the first square (s₁) = (4 + √3) cm

Side of the second square (s₂) = (4 − √3) cm

Formula Used:

Area of a square = (Side)²

Calculations:

⇒ Area of first square = (4 + √3)²

⇒ Area₁ = 4² + (√3)² + 2 × 4 × √3

⇒ Area₁ = 16 + 3 + 8√3 = (19 + 8√3) cm²

⇒ Area of second square = (4 − √3)²

⇒ Area₂ = 4² + (√3)² − 2 × 4 × √3

⇒ Area₂ = 16 + 3 − 8√3 = (19 − 8√3) cm²

⇒ Sum of areas = Area₁ + Area₂

⇒ Sum = (19 + 8√3) + (19 − 8√3)

⇒ Sum = 19 + 19 + 8√3 − 8√3 = 38 cm²

∴ The correct answer is 38 cm².

Additional Information

Algebraic Identities

(a + b)² = a² + b² + 2ab and (a − b)² = a² + b² − 2ab. These are fundamental for solving area-based problems involving irrational numbers.

Perimeter of a Square

The perimeter of a square is 4 × side. For the first square, perimeter = 4(4 + √3) = 16 + 4√3 cm.

Diagonal of a Square

The diagonal of a square with side 's' is given by s√2. Area can also be calculated as (Diagonal)² ÷ 2.

Shortcut Trick

In any quadrilateral PQRS, the angle formed by the bisectors of two adjacent angles (say, ∠R and ∠S) meeting at point T is always equal to half the sum of the other two angles (∠P and ∠Q):

∠STR = (∠P + ∠Q) ÷ 2

⇒ 22.5° = (∠P + ∠Q) ÷ 2

⇒ ∠P + ∠Q = 45°

Since the ratio of ∠P to ∠Q is 4 : 5, the difference between them is:

Difference = (5 − 4) ÷ (4 + 5) × 45° = 1/9 × 45° = 5°

∴ The correct answer is 5°.

Alternate Method

Given:

• In quadrilateral PQRS, the bisectors of ∠R and ∠S meet at T.
• ∠STR = 22.5°
• Ratio of ∠P to ∠Q = 4 : 5

Formula Used:

• Sum of angles in a triangle = 180°
• Sum of angles in a quadrilateral = 360°

Calculations:

In triangle STR:

⇒ ∠TSR + ∠TRS + ∠STR = 180°

⇒ ∠TSR + ∠TRS + 22.5° = 180°

⇒ ∠TSR + ∠TRS = 157.5°

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

⇒ ∠TSR = ∠S ÷ 2

⇒ ∠TRS = ∠R ÷ 2

⇒ (∠S ÷ 2) + (∠R ÷ 2) = 157.5°

⇒ ∠S + ∠R = 2 × 157.5° = 315°

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

⇒ ∠P + ∠Q + ∠R + ∠S = 360°

⇒ ∠P + ∠Q + 315° = 360°

⇒ ∠P + ∠Q = 45°

Given the ratio of ∠P to ∠Q is 4 : 5:

⇒ Let ∠P = 4x and ∠Q = 5x

⇒ 4x + 5x = 45°

⇒ 9x = 45°

⇒ x = 5°

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

⇒ ∠Q − ∠P = 5x − 4x = x

⇒ Difference = 5°

Explanation of Options:

• Option (1) is incorrect because a difference of 1° would mean the sum of angles is not 45° under the 4 : 5 ratio.
• Option (2) is incorrect because a difference of 2° does not satisfy the ratio and sum conditions.
• Option (3) is correct as the calculated difference is exactly 5°.
• Option (4) is incorrect because a difference of 6° is mathematically inconsistent with the given angle bisector value.

∴ The correct answer is 5°.

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 remaining two angles.

Sum of Interior Angles of a Polygon

The sum of the interior angles of an n-sided polygon is (n − 2) × 180°. For any quadrilateral, this sum is always 360°.

Opposite Angle Bisectors

If the bisectors of two opposite angles of a quadrilateral intersect, the angle between them is equal to half the difference of the other two angles.

Shortcut Trick

For a rate of 20% and time of 2 years, the effective compound interest rate is calculated as: 20 + 20 + (20 × 20)/100 = 44%.

Total Amount = 100% (Principal) + 44% (Interest) = 144% of the Principal.

Amount = 4,200 × 144/100 = 42 × 144 = ₹6,048.

∴ The correct answer is ₹6,048.

Alternate Method

Given:

Principal (P) = ₹4,200

Rate (R) = 20% per annum

Time (n) = 2 years

Formula Used:

Amount (A) = P(1 + R/100)ⁿ

Calculations:

⇒ A = 4200 × (1 + 20/100)²

⇒ A = 4200 × (1 + 1/5)²

⇒ A = 4200 × (6/5)²

⇒ A = 4200 × (36/25)

⇒ A = 168 × 36

⇒ A = 6048

∴ The correct answer is ₹6,048.

Additional Information

Effective Rate for Compound Interest

For 2 years, the net effective interest rate is given by the formula: x + y + (xy/100), where x and y are the rates for the first and second years respectively.

Difference between CI and SI

The difference (D) between compound interest and simple interest for 2 years is D = P(R/100)².

Half-Yearly Compounding

When interest is compounded half-yearly, the rate is halved (R/2) and the time is doubled (2n).

Shortcut Trick

Total distance covered = Speed × Time

Total distance = 63 × 28 = 1764 m

Train length (X) = Total distance − Platform length

X = 1764 − 213 = 1551 m

∴ The correct answer is 1551 m.

Alternate Method

Given:

Speed of train = 63 m/s

Time taken = 28 sec

Platform length = 213 m

Train length = X m

Formula Used:

Speed = (Length of Train + Length of Platform) ÷ Time

Calculations:

⇒ 63 = (X + 213) ÷ 28

⇒ X + 213 = 63 × 28

⇒ X + 213 = 1764

⇒ X = 1764 − 213

⇒ X = 1551

∴ The correct answer is 1551 m.

Additional Information

Crossing Objects with Length

When a train crosses a platform, bridge, or another train, the total distance covered is the sum of the lengths of the train and the object.

Relative Speed

If two objects move in the same direction, relative speed is S₁ − S₂. In opposite directions, it is S₁ + S₂.

Speed Conversion

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

Shortcut Trick
Given that Cost Price (CP) is 80% of Selling Price (SP).
Ratio of CP : SP = 80 : 100 = 4 : 5.
Profit = SP − CP = 5 − 4 = 1 unit.
Profit% = (Profit ÷ CP) × 100 = (1 ÷ 4) × 100 = 25%.
∴ The correct answer is 25%.

Alternate Method
Given: Selling Price (SP) = ₹100, Cost Price (CP) = 80% of SP.
Formulas: Profit = SP − CP | Profit% = (Profit ÷ CP) × 100

⇒ Cost Price (CP) = 80% of 100 = (80 ÷ 100) × 100 = ₹80.
⇒ Selling Price (SP) = ₹100.
⇒ Profit = SP − CP = 100 − 80 = ₹20.
⇒ Profit Percentage = (20 ÷ 80) × 100 = 1/4 × 100 = 25%.
∴ The correct answer is 25%.

Additional Information
Profit and Loss on CP
Unless otherwise stated, Profit% and Loss% are always calculated on the Cost Price.
Relation between CP and SP
If SP > CP, it is a Profit. If CP > SP, it is a Loss.

Shortcut Trick

Number of oranges = (Total − Difference) ÷ 2

⇒ Oranges = (42 − 6) ÷ 2

⇒ Oranges = 36 ÷ 2 = 18

∴ The correct answer is 18.

Alternate Method

Given: Total fruits = 42, Apple (A) − Orange (O) = 6

Let the number of oranges be x.

Then, the number of apples = x + 6.

⇒ (x + 6) + x = 42

⇒ 2x + 6 = 42

⇒ 2x = 42 − 6

⇒ 2x = 36

⇒ x = 18

∴ The correct answer is 18.

Additional Information

Sum and Difference Rule

If the sum (S) and difference (D) of two numbers are known, the larger number is (S + D) ÷ 2 and the smaller number is (S − D) ÷ 2.

Linear Equations in One Variable

Equations where the highest power of the variable is 1, used to find unknown quantities by establishing relationships based on given conditions.

Shortcut Trick

Combined efficiency of A, B, and C = 4 + 1 + 1 = 6 units/day (taking Total Work as 12 units).

Twice the work = 12 × 2 = 24 units.

Time taken = 24 ÷ 6 = 4 days.

∴ The correct answer is 4 days.

Alternate Method

Given:

Time taken by A = 3 days

Time taken by B = 12 days

Time taken by C = 12 days

Formula Used:

Total Work = LCM of individual days

Efficiency = Total Work ÷ Time Taken

Time Taken = Total Work ÷ Combined Efficiency

Calculations:

⇒ Total Work = LCM of 3, 12, 12 = 12 units

⇒ Combined Efficiency of A, B, and C = 4 + 1 + 1 = 6 units/day

⇒ Twice the Work = 2 × 12 = 24 units

⇒ Time taken to complete twice the work = 24 ÷ 6 = 4 days

∴ The correct answer is 4 days.

Additional Information

Total Work Concept

Total work is assumed as the LCM of individual times to make efficiency calculations integer-based and simpler.

Work and Efficiency Relation

Efficiency is inversely proportional to time taken when the total work is constant.

Shortcut Trick

Efficiency ratio A : B = 100 : 125 = 4 : 5.

Total Work = Efficiency of A × Time of A = 4 × 10 = 40 units.

Work to be completed = 60% of 40 = 24 units.

Combined efficiency = 4 + 5 = 9 units/day.

Time required = 24 ÷ 9 = 8/3 = 2(2/3) days.

∴ The correct answer is 2(2/3) days.

Alternate Method

Given: Time taken by A = 10 days; B is 25% more efficient than A.

Formula Used: Efficiency × Time = Work; Efficiency ∝ 1/Time

⇒ Efficiency of B = 4 + 25% of 4 = 4 + 1 = 5 units/day.

⇒ Total work = 10 × 4 = 40 units.

⇒ 60% of work = 40 × 0.6 = 24 units.

⇒ Combined Efficiency of (A + B) = 4 + 5 = 9 units/day.

⇒ Time taken = 24 ÷ 9 = 8/3 days = 2(2/3) days.

∴ The correct answer is 2(2/3) days.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a task. If efficiency ratio is x:y, time ratio is y:x.

Work and Wages

Wages are distributed in the ratio of the actual work done by each person, which simplifies to the ratio of their efficiencies if they work for the same duration.

Combined Work Formula

If A takes 'a' days and B takes 'b' days, time taken together is (a × b) ÷ (a + b) days.

Shortcut Trick

Total number of observations (n) = 11 (Odd).

Median is the 6^th observation in ascending order.

Replacing a value lower than the median (41 → 43, both < 58) and a value higher than the median (92 → 99, both > 58) does not change the middle element.

Since the relative position of the 6^th term remains the same, the median is unchanged.

∴ The correct answer is 58.

Alternate Method

Given observations: 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33

Step 1: Arrange data in ascending order.

33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92

Number of observations (n) = 11

Median = ((n + 1) ÷ 2)^th term = 6^th term = 58

Step 2: Apply the changes (Replace 92 with 99 and 41 with 43).

Step 3: New ascending order:

33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99

⇒ New n = 11

⇒ New Median = 6^th term

⇒ New Median = 58

∴ The correct answer is 58.

Additional Information

Median for Even Observations

If n is even, Median = Average of (n/2)^th and (n/2 + 1)^th observations.

Empirical Relationship

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

Outlier Resistance

Median is a measure of central tendency that is not affected by extreme outliers, unlike the arithmetic mean.

Shortcut Trick

Average Speed = Total Distance ÷ Total Time

Total Distance = 40 + 50 + 60 = 150 km

Total Time = (40/60) + (50/50) + (60/40) = 2/3 + 1 + 3/2 = 19/6 hours

Average Speed = 150 ÷ (19/6) = 900/19 ≈ 47.4 km/hr

∴ The correct answer is 47.4 km/hr.

Alternate Method

Given:

Distance 1 (d₁) = 40 km, Speed 1 (s₁) = 60 km/hr

Distance 2 (d₂) = 50 km, Speed 2 (s₂) = 50 km/hr

Distance 3 (d₃) = 60 km, Speed 3 (s₃) = 40 km/hr

Formula Used:

Time = Distance ÷ Speed

Average Speed = (Total Distance) ÷ (Total Time)

Calculations:

⇒ Time t₁ = 40 ÷ 60 = 2/3 hours

⇒ Time t₂ = 50 ÷ 50 = 1 hour

⇒ Time t₃ = 60 ÷ 40 = 3/2 hours

⇒ Total Time = 2/3 + 1 + 3/2 = (4 + 6 + 9) ÷ 6 = 19/6 hours

⇒ Total Distance = 40 + 50 + 60 = 150 km

⇒ Average Speed = 150 ÷ (19/6) = (150 × 6) ÷ 19

⇒ Average Speed = 900 ÷ 19 ≈ 47.368...

∴ The correct answer is 47.4 km/hr.

Additional Information

Equal Distance Case

If two equal distances are covered at speeds u and v, then Average Speed = (2 × u × v) ÷ (u + v).

Average Speed Concept

Average speed is NOT the simple arithmetic mean of speeds unless the time spent at each speed is identical.

Relative Speed

Same direction: S₁ − S₂ | Opposite direction: S₁ + S₂.

Shortcut Trick
Sum of odd-place digits: 7 + x + 9 + 5 = 21 + x
Sum of even-place digits: 4 + 8 + 3 + y = 15 + y
Difference (D) = (21 + x) − (15 + y) = 6 + x − y
For divisibility by 11, D must be 0, 11, or −11.
If D = 0, y − x = 6 ⇒ (x, y) = (1, 7), (2, 8), (3, 9) [3 pairs]
If D = 11, x − y = 5 ⇒ (x, y) = (5, 0) is not allowed as y is positive, so (6, 1), (7, 2), (8, 3), (9, 4) [4 pairs]
∴ The correct answer is 7.
Alternate Method
Given: Eight digit number 74x8935y is divisible by 11. x, y > 0.
Formula Used: Divisibility rule of 11: Sum of digits at odd places − Sum of digits at even places = 0 or multiple of 11.

⇒ Odd place sum S1 = 7 + x + 9 + 5 = 21 + x
⇒ Even place sum S2 = 4 + 8 + 3 + y = 15 + y
⇒ Difference D = S1 − S2 = (21 + x) − (15 + y) = 6 + x − y
⇒ Case 1: 6 + x − y = 0 ⇒ y − x = 6. Since x, y ≥ 1: (1, 7), (2, 8), (3, 9)
⇒ Case 2: 6 + x − y = 11 ⇒ x − y = 5. Since x, y ≥ 1: (6, 1), (7, 2), (8, 3), (9, 4)
⇒ Case 3: 6 + x − y = −11 ⇒ y − x = 17 (Not possible as max difference is 8 for non-zero integers)
⇒ Total pairs = 3 + 4 = 7
∴ The correct answer is 7.
Additional Information
Divisibility by 11
A number is divisible by 11 if the alternating sum of its digits is divisible by 11.
Positive Integers constraint
Since x and y are positive integers, they must be from the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. Zero is not included.
Ordered Pairs
An ordered pair (x, y) counts unique combinations where the order of x and y matters relative to their positions in the number.

Shortcut Trick

Discount% = [Free Articles ÷ (Paid Articles + Free Articles)] × 100

Discount% = [2 ÷ (6 + 2)] × 100

Discount% = (2 ÷ 8) × 100 = (1 ÷ 4) × 100 = 25%

∴ The correct answer is 25%.

Alternate Method

Given: Number of articles purchased = 6, Number of free articles = 2.
Formulas: Discount% = (Discount ÷ MP) × 100 | Total MP = MP of all articles (Bought + Free).

⇒ Let the Marked Price (MP) of 1 article be 1 unit.
⇒ Total MP of 8 articles (6 + 2) = 8 units.
⇒ The Selling Price (SP) is equal to the price of 6 articles = 6 units.
⇒ Discount = MP − SP = 8 − 6 = 2 units.
⇒ Discount% = (2 ÷ 8) × 100 = 25%.

∴ The correct answer is 25%.

Additional Information

Marked Price & Discount

The price printed on the label is the Marked Price. Discount is always calculated on this value.

Free Item Concept

In 'Buy X Get Y Free', the total quantity (X+Y) represents the Marked Price equivalent, and Y represents the Discount.

Shortcut Trick

Numbers divisible by 5 = 1000 ÷ 5 = 200

Numbers divisible by 7 = 1000 ÷ 7 = 142 (integer part)

Numbers divisible by both (LCM of 5 and 7 = 35) = 1000 ÷ 35 = 28 (integer part)

Total numbers = 200 + 142 − 28 = 314

∴ The correct answer is 314.

Alternate Method

Given: Total limit = 1000, Divisors = 5, 7

Formula Used: n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

⇒ Let A be the set of numbers divisible by 5: n(A) = floor(1000/5) = 200

⇒ Let B be the set of numbers divisible by 7: n(B) = floor(1000/7) = 142

⇒ Let A ∩ B be numbers divisible by LCM(5, 7) = 35: n(A ∩ B) = floor(1000/35) = 28

⇒ Total numbers divisible by 5 or 7 = 200 + 142 − 28

⇒ Total = 342 − 28 = 314

∴ The correct answer is 314.

Additional Information

Inclusion-Exclusion Principle

For two sets, the number of elements in their union is found by adding the individual sizes and subtracting the size of their intersection.

Divisibility in a Range

The number of integers divisible by 'k' in the range [1, N] is given by the floor function ⌊N/k⌋.

Least Common Multiple (LCM)

To find numbers divisible by both 'a' and 'b', we must find numbers divisible by their LCM.

Shortcut Trick

Since Surface Area is a 2-dimensional property (depends on side²), use the Successive Percentage formula:

Net Increase = x + x + (x × x)/100

Here, increase (x) = 10%

Total Increase = 10 + 10 + (10 × 10)/100 = 20 + 1 = 21%

∴ The correct answer is 21%.

Alternate Method

Given: Each edge of a cube is increased by 10%.

Formula Used: Total Surface Area of a cube = 6 × (edge)²

⇒ Let the original edge (a) = 10 units.

⇒ Original Surface Area (S₁) = 6 × (10)² = 6 × 100 = 600 sq. units.

⇒ New edge (a’) = 10 + (10% of 10) = 11 units.

⇒ New Surface Area (S₂) = 6 × (11)² = 6 × 121 = 726 sq. units.

⇒ Increase in Surface Area = 726 − 600 = 126 sq. units.

⇒ Percentage Increase = (126 ÷ 600) × 100 = 21%.

∴ The correct answer is 21%.

Additional Information

Volume Percentage Change

Since Volume = side³, an increase of x% leads to a volume increase of [ (1 + x/100)³ − 1 ] × 100%. For 10%, it is 33.1%.

Ratio Method for Area

If side ratio is 10 : 11, then Area ratio is 10² : 11² = 100 : 121. The increase is directly 21 over 100, which is 21%.

Applicability to Other Shapes

This rule applies to all 3D regular figures. If the linear dimension of a sphere or cone increases by x%, its surface area always increases by 2x + x²/100 %.

Shortcut Trick

Sum of present ages = 56 years.

Sum after 8 years = 56 + (8 × 2) = 72 years.

Ratio after 8 years (Mother : Daughter) = 3 :

1. Daughter's age after 8 years = 72 × (1/4) = 18 years.

Daughter's present age = 18 − 8 = 10 years.

∴ The correct answer is 10 years.

Alternate Method

Given: Sum of present ages of Mother (M) and Daughter (D) is 56.

Formula Used: Linear equation in one or two variables.

⇒ Let Daughter's present age be D and Mother's present age be M.

⇒ M + D = 56 → (i)

⇒ After 8 years: (M + 8) = 3 × (D + 8)

⇒ M + 8 = 3D + 24

⇒ M − 3D = 16 → (ii)

⇒ Subtracting (ii) from (i): (M + D) − (M − 3D) = 56 − 16

⇒ 4D = 40

⇒ D = 10 years.

∴ The correct answer is 10 years.

Additional Information

Sum of Ages in Future/Past

If the current sum of ages of 'n' people is S, then after 't' years, the sum will be S + (n × t).

Age Ratio Comparison

The difference between the ages of two people remains constant throughout their lives, regardless of the years passed.

Linear Equations in Ages

Always assume variables for present ages to avoid confusion while moving back or forward in time.

Shortcut Trick

In a 30°-60°-90° triangle, the ratio of sides is 1 : √3 : 2.

The side opposite to 30° (Base) = 30 m.

The side opposite to 60° (Height) = Base × √3.

Height = 30 × 1.73 = 51.9 m.

∴ The correct answer is 51.9 m.

Alternate Method

Given:

Distance from the base (Base) = 30 m

Angle of elevation (θ) = 60°

Value of √3 = 1.73

Formula Used:

tan θ = Perpendicular / Base

Calculations:

⇒ tan 60° = Height / 30

⇒ √3 = Height / 30

⇒ Height = 30 × √3

⇒ Height = 30 × 1.73

⇒ Height = 51.9 m

∴ The correct answer is 51.9 m.

Additional Information

Trigonometric Ratios

Standard values for calculations: tan 30° = 1/√3, tan 45° = 1, and tan 60° = √3.

Angle of Elevation vs Depression

The angle of elevation is measured upwards from the horizontal, while the angle of depression is measured downwards; both are numerically equal for the same line of sight due to alternate interior angles.

Pythagoras Theorem

In any right-angled triangle, Hypotenuse² = Base² + Perpendicular².

Shortcut Trick

Required percentage = (Value ÷ Base Value) × 100

Required percentage = (1/30 ÷ 1/90) × 100

Required percentage = (90 ÷ 30) × 100 = 3 × 100 = 300%

∴ The correct answer is 300%.

Alternate Method

Given:

Target value = 1/30

Base value = 1/90

Formula Used:

Percentage = (A / B) × 100

Calculations:

⇒ Percentage = [(1/30) / (1/90)] × 100

⇒ Percentage = (1/30) × (90/1) × 100

⇒ Percentage = 3 × 100

⇒ Percentage = 300%

∴ The correct answer is 300%.

Additional Information

Fraction to Percentage

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

Reciprocal Comparison

When comparing unit fractions (1/n), the ratio of 1/a to 1/b is equal to b/a.

Percentage Change

Percentage change = [(New value − Old value) / Old value] × 100.