Quantitative Aptitude PYQs

Shortcut Trick
A 45° sector represents exactly 1/8 of the circle's total circumference.
Circumference = 2 × (22/7) × 21 = 132 m.
Arc Length = 132 ÷ 8 = 16.5 m.
Total Perimeter = Arc Length + (2 × Radius) = 16.5 + 42 = 58.5 m.
∴ The correct answer is 58.5 m.
Alternate Method
Given:
Radius (r) = 21 m
Central Angle (θ) = 45°
π = 22/7
Formula Used:
Perimeter of Sector = Arc Length (l) + 2 × Radius (r)
Arc Length (l) = (θ ÷ 360) × 2 × π × r

Calculations:
⇒ Arc length (l) = (45 ÷ 360) × 2 × (22 ÷ 7) × 21
⇒ l = (1 ÷ 8) × 2 × 22 × 3
⇒ l = (1 ÷ 8) × 132 = 16.5 m
⇒ Perimeter = 16.5 + (2 × 21)
⇒ Perimeter = 16.5 + 42
⇒ Perimeter = 58.5 m
∴ The correct answer is 58.5 m.
Additional Information
Area of a Sector
The region bounded by two radii and an arc, calculated as Area = (θ/360) × πr2 or (1/2) × l × r.
Length of an Arc
The distance along the curved edge of a sector, given by the formula l = (θ/360) × 2πr.
Major and Minor Sectors
A minor sector has an angle θ < 180°, while a major sector has an angle of (360 − θ).

Shortcut Trick

Total amount to be deducted = 33 + 70 + 19 = ₹122

Remaining amount = 12,986 − 122 = ₹12,864

Ratio of remaining shares A : B : C = 14 : 8 : 2 (Total parts = 24)

Difference between B and C parts = 8 − 2 = 6 parts

Difference in shares = (12,864 ÷ 24) × 6 + (70 − 19) = 3216 + 51 = ₹3,267

∴ The correct answer is ₹3,267.

Alternate Method

Given: Total sum = ₹12,986. Deductions from A, B, and C are ₹33, ₹70, and ₹19 respectively. Ratio after deduction = 14 : 8 : 2.

⇒ Sum of remaining shares = 12,986 − (33 + 70 + 19)

⇒ Sum of remaining shares = 12,986 − 122 = ₹12,864

⇒ Total ratio parts = 14 + 8 + 2 = 24 units

⇒ 24 units = 12,864

⇒ 1 unit = 12,864 ÷ 24 = ₹536

⇒ B's original share = (8 × 536) + 70 = 4288 + 70 = ₹4,358

⇒ C's original share = (2 × 536) + 19 = 1072 + 19 = ₹1,091

⇒ Difference = 4,358 − 1,091 = ₹3,267

∴ The correct answer is ₹3,267.

Additional Information

Ratio and Proportion

Ratio is the comparison of two quantities of the same kind. If a : b = c : d, then ad = bc (Product of extremes = Product of means).

Deduction Rule

When values are deducted from a total sum to maintain a ratio, subtract the total deduction from the principal amount before applying the ratio division.

Shortcut Trick

Convert all to same base

4 = 2², 16 = 2⁴, 3⁴ and 9 = 3²

Compare powers of 2 and 3 separately

⇒ a = 1, b = 2

∴ Answer = (1, 2)

Alternate Method

Given:

(4²)^a × (16²)^2a × (3⁴)^b × (9²)^2b = 2²⁰ × (9²)⁶

Concept Used:

Laws of exponents

Calculation:

⇒ (4²)^a = (2⁴)^a = 2^4a

⇒ (16²)^2a = (2⁸)^2a = 2^16a

⇒ Total power of 2 = 2^20a

⇒ RHS = 2²⁰ ⇒ 20a = 20 ⇒ a = 1

⇒ (3⁴)^b = 3^4b

⇒ (9²)^2b = (3⁴)^2b = 3^8b

⇒ Total power of 3 = 3^12b

⇒ RHS = (9²)⁶ = (3⁴)⁶ = 3²⁴

⇒ 12b = 24 ⇒ b = 2

∴ The correct answer is (1, 2).

Additional Information

Exponent Rule

(a^m)ⁿ = a^mn

Same Base Rule

a^x = a^y ⇒ x = y

Conversion Trick

Convert numbers into prime base form

Separation Method

Compare powers of different bases separately

Shortcut Trick

Product of two numbers = HCF × LCM

216 × 120 = HCF × 1080

HCF = (216 × 120) ÷ 1080 = 216 ÷ 9 = 24

∴ The correct answer is 24.

Alternate Method

Given:

First Number (A) = 216

Second Number (B) = 120

LCM = 1080

Formula Used:

A × B = HCF(A, B) × LCM(A, B)

Calculations:

⇒ 216 × 120 = HCF × 1080

⇒ 25920 = HCF × 1080

⇒ HCF = 25920 ÷ 1080

⇒ HCF = 24

∴ The correct answer is 24.

Additional Information

HCF and LCM Relation

For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b. This property only holds true for two numbers.

Co-prime Numbers

If two numbers are co-prime, their HCF is always 1 and their LCM is the product of the two numbers.

Fractional LCM/HCF

LCM of fractions = (LCM of Numerators) ÷ (HCF of Denominators). HCF of fractions = (HCF of Numerators) ÷ (LCM of Denominators).

Shortcut Trick

Total Expenditure % = 25% + 15% + 10% = 50%

Savings % = 100% − 50% = 50%

Monthly Savings = 50% of ₹40,000 = ₹20,000

∴ The correct answer is ₹20,000.

Alternate Method

Given:

Total Income = ₹40,000

Rent = 25% | Food = 15% | Transportation = 10%

Calculations:

⇒ Total spending = 25% + 15% + 10% = 50%

⇒ Remaining savings percentage = 100% − 50% = 50%

⇒ Savings amount = (50 ÷ 100) × 40,000

⇒ Savings amount = 0.5 × 40,000 = 20,000

∴ The correct answer is ₹20,000.

Additional Information

Percentage Change

To find the percentage of a value, use (Percentage/100) × Base Value.

Successive Percentage

If expenses are calculated on the remaining balance, use the formula: Final = Initial × (1 − r₁/100) × (1 − r₂/100).

Income-Expenditure Relation

Income = Expenditure + Savings. This base formula is used to solve most budget-related problems.

Shortcut Trick

Total Surface Area of a cylinder = 2πr(r + h), where π = 22/7.

The final result must be a multiple of 11 as it contains the factor 22.

Check divisibility by 11 (Difference of sum of digits at odd and even places):

Option (1) 2244: (2+4) − (2+4) = 0 (Divisible by 11).

Other options are not divisible by 11, so 2244 is the answer.

∴ The correct answer is 2244 cm².

Alternate Method

Given: Volume (V) = 6776 cm³, Height (h) = 44 cm

Formula Used: Volume = πr²h, Total Surface Area (TSA) = 2πr(r + h)

⇒ Volume = πr²h

⇒ 6776 = (22/7) × r² × 44

⇒ r² = (6776 × 7) / (22 × 44)

⇒ r² = 154 × 7 / 22 = 7 × 7

⇒ r = 7 cm

⇒ TSA = 2πr(r + h)

⇒ TSA = 2 × (22/7) × 7 × (7 + 44)

⇒ TSA = 44 × 51

⇒ TSA = 2244 cm²

∴ The correct answer is 2244 cm².

Additional Information

Curved Surface Area of a Cylinder

It is the area of the curved surface only, excluding the top and bottom circles. Formula: CSA = 2πrh.

Volume of a Cylinder

It is the total space occupied by the cylinder. Formula: Volume = πr²h.

Total Surface Area of a Cylinder

It is the sum of the curved surface area and the areas of the two circular bases. Formula: TSA = 2πr(r + h).

Radius and Diameter Relation

The diameter (d) of the circular base is always twice the radius (r). Formula: d = 2r.

Shortcut Trick

Distance is constant ∴ Speed ratio is the inverse of Time ratio.

Time ratio (Forward : Return) = 6 : 9 = 2 : 3

Speed ratio (Forward : Return) = 3 : 2

Speed difference = 3 − 2 = 1 unit → 2 km/h

Return speed = 2 units = 2 × 2 = 4 km/h

∴ The correct answer is 4 km/h.

Alternate Method

Given:

Time taken while walking forward (T₁) = 6 h

Time taken while returning (T₂) = 9 h

Decrease in speed = 2 km/h

Formula Used:

Distance = Speed × Time

⇒ Let the initial speed while walking forward be S km/h.

⇒ Distance covered = Speed × Time = S × 6 = 6S

⇒ Returning speed = (S − 2) km/h

⇒ Distance covered while returning = (S − 2) × 9 = 9(S − 2)

⇒ Since the distance is the same for both journeys:

⇒ 6S = 9(S − 2)

⇒ 6S = 9S − 18

⇒ 18 = 9S − 6S

⇒ 18 = 3S

⇒ S = 18 ÷ 3 = 6 km/h

⇒ Returning speed = S − 2 = 6 − 2 = 4 km/h

∴ The correct answer is 4 km/h.

Additional Information

Speed-Time Relationship

When distance is constant, speed is inversely proportional to time (S₁ × T₁ = S₂ × T₂).

Average Speed for Equal Distances

If an object covers equal distances at speeds x and y, the average speed is 2xy ÷ (x + y).

Unit Conversion

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

Shortcut Trick

In an equilateral triangle ABC of side 'a', if a point D divides side BC such that BD = a/n,

The relation is given by the formula: n²AD² = (n² − n + 1)AB².

Here, 3BD = BC ⇒ BD = BC/3, so n = 3.

Substituting n = 3: 3²AD² = (3² − 3 + 1)AB²

9AD² = (9 − 3 + 1)AB² ⇒ 9AD² = 7AB².

∴ The correct answer is 9AD²=7AB².

Alternate Method

Given: ΔABC is equilateral, so AB = BC = CA. Point D is on BC such that BD = BC/3.

Formula Used: Pythagoras Theorem: Hypotenuse² = Base² + Perpendicular².

Let the side of the equilateral triangle AB = BC = CA = a.

Draw altitude AM ⊥ BC. In an equilateral triangle, altitude bisects the base.

⇒ BM = MC = BC/2 = a/2.

⇒ AM = √(AB² − BM²) = √(a² − (a/2)²) = √(3a²/4) = (√3/2)a.

Given BD = BC/3 = a/3.

⇒ DM = BM − BD = a/2 − a/3 = (3a − 2a)/6 = a/6.

In right-angled ΔADM, apply Pythagoras theorem:

⇒ AD² = AM² + DM²

⇒ AD² = [(√3/2)a]² + (a/6)²

⇒ AD² = 3a²/4 + a²/36

⇒ AD² = (27a² + a²) / 36

⇒ AD² = 28a² / 36 = 7a² / 9

⇒ 9AD² = 7a²

Since a = AB:

⇒ 9AD² = 7AB².

∴ The correct answer is 9AD²=7AB².

Additional Information

Height of Equilateral Triangle

The altitude or height (h) of an equilateral triangle with side 'a' is always (√3/2)×a.

Area of Equilateral Triangle

The area is given by (√3/4)×a², where 'a' is the length of the side.

Inradius and Circumradius

For side 'a', the Inradius (r) is a/(2√3) and the Circumradius (R) is a/√3. Note that R = 2r.

Apollonius Theorem

For any triangle with side AB, AC and median AD on BC: AB² + AC² = 2(AD² + BD²).

Shortcut Trick

LCM of 9 and 12 = 36 units (Total Capacity of the tank).

Efficiency of Filling pipe = 36 ÷ 9 = +4 units/min.

Efficiency of Emptying pipe = 36 ÷ 12 = −3 units/min.

Combined Efficiency = 4 − 3 = 1 unit/min.

Time to fill one-half (36 ÷ 2 = 18 units) = 18 ÷ 1 = 18 minutes.

∴ The correct answer is 18.

Alternate Method

Given:

Time taken by filling pipe = 9 minutes

Time taken by emptying pipe = 12 minutes

Formula used: Efficiency = Total Work ÷ Time

Calculations:

⇒ Total Capacity of tank = LCM(9, 12) = 36 units

⇒ Net efficiency of both pipes = 4 − 3 = 1 unit/min

⇒ Required capacity to fill = 1/2 × Total Capacity

⇒ Required capacity = 36 ÷ 2 = 18 units

⇒ Time required = Required capacity ÷ Net efficiency

⇒ Time = 18 ÷ 1 = 18 minutes

∴ The correct answer is 18.

Additional Information

Inflow and Outflow Concept

Efficiency of an inlet pipe is always taken as positive, while the efficiency of an outlet or emptying pipe is taken as negative.

Net Rate of Filling

If pipe A fills in x hours and pipe B empties in y hours (where y > x), the net part filled in 1 hour is (1/x − 1/y).

Time and Work Relation

Total Capacity = Time × Efficiency. For fractional work, Time = (Fraction × Total Capacity) ÷ Efficiency.

Shortcut Trick

Let assumed mean be 70.

Deviations from 70: −3, +6, +27, −36, −32, −7, +25, +25, +4

Sum of deviations = −3 + 6 + 27 − 36 − 32 − 7 + 25 + 25 + 4 = +9

Adjustment = Sum ÷ Count = 9 ÷ 9 = +1

Correct Mean = 70 + 1 = 71.

∴ The correct answer is 71.

Alternate Method

Given: Observations = 67, 76, 97, 34, 38, 63, 95, 95, 74

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

⇒ Total number of observations (n) = 9

⇒ Sum of observations = 67 + 76 + 97 + 34 + 38 + 63 + 95 + 95 + 74

⇒ Sum = 639

⇒ Arithmetic Mean = 639 ÷ 9

⇒ Arithmetic Mean = 71

∴ The correct answer is 71.

Additional Information

Properties of Arithmetic Mean

The sum of the deviations of a set of items from their arithmetic mean is always zero: Σ(x_i − mean) = 0.

Median and Mode

Median is the middle value of sorted data. Mode is the value that appears most frequently (here, 95).

Weighted Arithmetic Mean

Used when observations have different levels of importance: Mean = Σ(w_i × x_i) ÷ Σw_i.

Shortcut Trick

Net percentage increase for two successive years at 10% each = 10 + 10 + (10 × 10)/100 = 21%.

Total multiplying factor = 121% or 1.21.

Income in 2021 = 22,200 × 1.21 = 26,862.

∴ The correct answer is 26,862.

Alternate Method

Given:

Income in 2019 (P) = ₹22,200

Rate of increment (R) = 10% per annum

Time (n) = 2021 − 2019 = 2 years

Formula used: Final Income = P × (1 + R/100)ⁿ

⇒ Income in 2021 = 22200 × (1 + 10/100)²

⇒ Income in 2021 = 22200 × (110/100) × (110/100)

⇒ Income in 2021 = 22200 × 1.1 × 1.1

⇒ Income in 2021 = 22200 × 1.21

⇒ Income in 2021 = 26862

∴ The correct answer is 26,862.

Additional Information

Successive Percentage Change

When a value changes by a% and then by b%, the effective change is [a + b + (ab/100)]%.

Compound Growth Formula

The formula for population or income growth is A = P(1 + r/100)ⁿ, where P is the initial value and n is the number of years.

Multiplying Factor Method

For a 10% increase, the multiplying factor is (1 + 0.1) = 1.1. For n years, it becomes (1.1)ⁿ.

Shortcut Trick

The average of the first n odd natural numbers is always equal to n.

Here, the number of terms n = 10.

∴ Average = 10.

∴ The correct answer is 10.

Alternate Method

Given:

First 10 odd natural numbers are: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.

Formula Used:

Average = (Sum of all terms) ÷ (Total number of terms)

Sum of an Arithmetic Progression (AP) = (n/2) × [First term + Last term]

Calculations:

⇒ Total terms (n) = 10

⇒ Sum = (10/2) × [1 + 19]

⇒ Sum = 5 × 20 = 100

⇒ Average = 100 ÷ 10

⇒ Average = 10

∴ The correct answer is 10.

Additional Information

Sum of first n odd numbers

The sum of the first n odd natural numbers is given by the formula n².

Average of first n even numbers

The average of the first n even natural numbers is always (n + 1).

Average of first n natural numbers

The average of the first n natural numbers is calculated as (n + 1) ÷ 2.

Shortcut Trick

When distance is constant, Speed is inversely proportional to Time (S ∝ 1/T).

Ratio of Time (T₁ : T₂) = 8 : 5

Ratio of Speed (S₁ : S₂) = 5 : 8

Given: 5 units = 17 km/h ⇒ 1 unit = 3.4 km/h

Required Speed (8 units) = 8 × 3.4 = 27.2 km/h

∴ The correct answer is 27.2 km/h.

Alternate Method

Given:

Initial Speed (S₁) = 17 km/h

Initial Time (T₁) = 8 min = 8/60 hours

New Time (T₂) = 5 min = 5/60 hours

Formula Used:

Distance = Speed × Time

⇒ Distance = S₁ × T₁

⇒ Distance = 17 × (8/60) km

⇒ New Speed (S₂) = Distance ÷ New Time (T₂)

⇒ S₂ = [17 × (8/60)] ÷ (5/60)

⇒ S₂ = 17 × 8 ÷ 5

⇒ S₂ = 136 ÷ 5

⇒ S₂ = 27.2 km/h

∴ The correct answer is 27.2 km/h.

Additional Information

Constant Distance Relation

When the same distance is covered at different speeds, S₁ × T₁ = S₂ × T₂. This allows for direct substitution without calculating distance.

Average Speed

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

Unit Conversion

To convert speed from km/h to m/s, multiply by 5/18. To convert m/s to km/h, multiply by 18/5.

Shortcut Trick

Use the formula: Emptying time = (a × b) ÷ (b − a)

where a = filling time without leak and b = filling time with leak.

⇒ Emptying time = (6 × 8) ÷ (8 − 6)

⇒ Emptying time = 48 ÷ 2 = 24 hours.

∴ The correct answer is 24 hours.

Alternate Method

Given: Filling pipe (A) time = 6 hours | Filling pipe with leak (A + L) time = 8 hours.

Total capacity of cistern = LCM of (6, 8) = 24 units.

⇒ Efficiency of Leak (L) = Efficiency of (A + L) − Efficiency of A

⇒ Efficiency of L = 3 − 4 = −1 units/hour (Negative sign indicates emptying)

⇒ Time taken by leak to empty cistern = Total Capacity ÷ |Efficiency of L|

⇒ Time taken = 24 ÷ 1 = 24 hours.

∴ The correct answer is 24 hours.

Additional Information

Leakage Principle

A leak acts as an emptying pipe. Net efficiency = (Efficiency of filling pipe) − (Efficiency of leak).

Combined Rate Formula

If two pipes fill at rates x and y, combined rate = (x × y) ÷ (x + y). If one is a leak, the rate becomes (x × y) ÷ (y − x).

Pipes and Cisterns Basics

This topic follows the same logic as Time and Work, where work done is equivalent to cistern capacity.

Shortcut Trick

SP (excluding GST) = 1680 ÷ 1.2 = 1400

CP = 1400 ÷ 1.12 = 1250

∴ The correct answer is 1,250.

Alternate Method

Given:

Selling Price (including GST) = ₹1,680

GST Rate = 20%

Profit = 12%

Formula Used:

SP (excluding tax) = Total SP ÷ (1 + Tax%/100)

CP = SP (excluding tax) ÷ (1 + Profit%/100)

Calculations:

⇒ SP (without GST) = 1680 ÷ (1 + 20/100)

⇒ SP (without GST) = 1680 ÷ 1.2 = 1400

⇒ CP × (1 + 12/100) = 1400

⇒ CP × 1.12 = 1400

⇒ CP = 1400 ÷ 1.12

⇒ CP = 1250

∴ The correct answer is 1,250.

Additional Information

Profit and Loss Relationship

Profit % is always calculated on the Cost Price (CP) unless stated otherwise. SP = CP × (100 + Profit%)/100.

GST and Sales Tax

Taxes are applied to the net selling price. Total SP = Base SP × (100 + Tax%)/100.

Successive Percentage Change

If a value changes by a% then b%, the net multiplier is (1 + a/100) × (1 + b/100).

Shortcut Trick

Scheme A: 19 + 19 − 3.61 = 34.39%

Scheme B: Buy 2 get 10 → Discount = (8/10) × 100 = 80%

Scheme C: 11%

Scheme D: 25 + 17 − 4.25 = 37.75%

Comparing discounts: 80% > 37.75% > 34.39% > 11%

∴ The correct answer is B.

Alternate Method

Given:

Scheme A: Successive discounts 19% and 19%

Scheme B: Buy 2, get 10 items total

Scheme C: Single discount 11%

Scheme D: Successive discounts 25% and 17%

Formula Used:

Net Discount for successive a% and b% = [a + b − (ab/100)]%

Discount % for Buy X Get Y = [(Y − X) / Y] × 100

Calculations:

⇒ Net Discount A = 19 + 19 − (19 × 19)/100

⇒ Net Discount A = 38 − 3.61 = 34.39%

⇒ Net Discount B = [(10 − 2) / 10] × 100

⇒ Net Discount B = (8 / 10) × 100 = 80%

⇒ Net Discount C = 11%

⇒ Net Discount D = 25 + 17 − (25 × 17)/100

⇒ Net Discount D = 42 − 4.25 = 37.75%

⇒ Highest Discount = 80%

∴ The correct answer is B.

Additional Information

Successive Discounts

When two discounts a% and b% are given, the total discount is always less than the sum (a+b) because the second discount is applied to the already reduced price.

Buy X Get Y Free

The effective discount percentage is calculated as [Free items ÷ Total items (Paid + Free)] × 100.

Marked Price and Selling Price

Selling Price = Marked Price × (100 − D%)/100. For multiple discounts, SP = MP × (100 − d₁)/100 × (100 − d₂)/100.

Shortcut Trick

To find the quantity of the larger value when the sum and difference are known:

Quantity of Mangoes = (Sum of fruits + Difference) ÷ 2

Quantity of Mangoes = (96 + 42) ÷ 2

Quantity of Mangoes = 138 ÷ 2 = 69

∴ The correct answer is 69.

Alternate Method

Given:

Total fruits (Mangoes + Oranges) = 96

Difference (Mangoes − Oranges) = 42

Method: Linear Equation in One Variable

Let the number of Oranges be x.

Then, the number of Mangoes = x + 42

According to the question:

⇒ x + (x + 42) = 96

⇒ 2x + 42 = 96

⇒ 2x = 96 − 42

⇒ 2x = 54

⇒ x = 54 ÷ 2

⇒ x = 27 (Number of Oranges)

⇒ Number of Mangoes = 27 + 42 = 69

∴ The correct answer is 69.

Additional Information

Sum and Difference Formula

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

Equations in the form ax + b = c are solved by isolating the variable x through inverse operations on both sides.

Ratio from Sum and Difference

The ratio of the two numbers can be calculated directly as (S+D) : (S−D). Here, Ratio = (96+42) : (96−42) = 138 : 54 = 23 : 9.

Shortcut Trick

Apply the identity (a − b)(a + b) = a² − b² successively to the terms.

(x − 1/x)(x + 1/x) = x² − 1/x²

(x² − 1/x²)(x² + 1/x²) = x⁴ − 1/x⁴

(x⁴ − 1/x⁴)(x⁴ + 1/x⁴) = x⁸ − 1/x⁸

∴ The correct answer is (x⁸ − 1/x⁸).

Alternate Method

Given: (x − 1/x)(x + 1/x)(x² + 1/x²)(x⁴ + 1/x⁴)

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

⇒ Simplifying the first two brackets:

⇒ (x − 1/x) × (x + 1/x) = x² − (1/x)² = x² − 1/x²

⇒ Multiplying the result with the third bracket:

⇒ (x² − 1/x²) × (x² + 1/x²) = (x²)² − (1/x²)² = x⁴ − 1/x⁴

⇒ Multiplying the result with the fourth bracket:

⇒ (x⁴ − 1/x⁴) × (x⁴ + 1/x⁴) = (x⁴)² − (1/x⁴)² = x⁸ − 1/x⁸

∴ The correct answer is (x⁸ − 1/x⁸).

Additional Information

Difference of Two Squares

The identity (a − b)(a + b) = a² − b² is used to simplify products of binomials containing the same terms with opposite signs.

Power of a Power Rule

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

Algebraic Identities

Other common identities include (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b².

Shortcut Trick

The trigonometric ratio sin θ = 3/5 corresponds to Perpendicular / Hypotenuse.

Recognize the fundamental Pythagorean triplet: (3, 4, 5).

In this triplet, if Perpendicular = 3 and Hypotenuse = 5, then Base must be 4.

Using the ratio cos θ = Base / Hypotenuse, we get cos θ = 4/5.

∴ The correct answer is 4/5.

Alternate Method

Given: sin θ = 3/5

Formula: sin²θ + cos²θ = 1

⇒ cos²θ = 1 − sin²θ

⇒ cos θ = √(1 − (3/5)²)

⇒ cos θ = √(1 − 9/25)

⇒ cos θ = √((25 − 9) / 25)

⇒ cos θ = √(16/25)

⇒ cos θ = 4/5

∴ The correct answer is 4/5.

Additional Information

Trigonometric Identities

The primary identity is sin²θ + cos²θ = 1, from which we derive cos θ = √(1 − sin²θ).

Pythagorean Theorem

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

Pythagorean Triplets

Common triplets used in exams include (3, 4, 5), (5, 12, 13), (8, 15, 17), and (7, 24, 25).

Ratio Definitions

sin θ = P/H, cos θ = B/H, tan θ = P/B, cot θ = B/P, sec θ = H/B, cosec θ = H/P.

Shortcut Trick

Ratio of Anil (A) to Bhavya (B) present ages = 5 : 4

Let present ages be 5x and 4x.

According to question: (4x + 6) = 2 × (5x) − 6

⇒ 4x + 6 = 10x − 6

⇒ 6x = 12 ⇒ x = 2

Difference between their ages = 5x − 4x = x = 2 years.

∴ The correct answer is 2.

Alternate Method

Given:

4 × Anil = 5 × Bhavya

After 6 years, Bhavya = 2 × Anil (Present) − 6

Let Anil's age = A and Bhavya's age = B

⇒ 4A = 5B ⇒ B = 4A/5

⇒ (B + 6) = 2A − 6

⇒ (4A/5) + 6 = 2A − 6

⇒ 4A/5 + 12 = 2A

⇒ 12 = 2A − 0.8A

⇒ 1.2A = 12

⇒ A = 10 years

⇒ B = (4 × 10)/5 = 8 years

⇒ Difference = 10 − 8 = 2 years

∴ The correct answer is 2.

Additional Information

Constant Age Difference

The difference between the ages of two people remains the same regardless of how many years pass.

Ratio to Value Conversion

If two quantities are in the ratio x : y, they can be represented as kx and ky, where k is the common multiplier.

Future and Past Ages

If current age is X, age after n years is (X + n) and age n years ago was (X − n).