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Shortcut Trick

Initial Ratio (Acid : Water) = 5 : 3

Final Ratio (Acid : Water) = 5 : 4

Since only water is added, the ratio of acid (5 parts) correctly remains unchanged.

Increase in water ratio = 4 − 3 = 1 part.

1 part = 4 litres ⇒ Original Acid = 5 parts = 5 × 4 = 20 litres.

∴ The correct answer is 20 litres.

Alternate Method

Initial Acid = 5u Water = 3u

Final Acid = 5u Water = 4u

• 4 L Water

Water Change: 4u − 3u = 1u = 4 Litres Original Acid: 5u = 5 × 4 = 20 Litres

Given:

Initial ratio of acid to water = 5 : 3

Water added = 4 litres

New ratio of acid to water = 5 : 4

Calculations:

⇒ Let initial acid = 5x and initial water = 3x.

⇒ After adding 4 litres of water, new quantity of water = 3x + 4.

⇒ New ratio = 5x / (3x + 4) = 5 / 4

⇒ 20x = 15x + 20

⇒ 20x − 15x = 20

⇒ 5x = 20 ⇒ x = 4

⇒ Original quantity of acid = 5x = 5 × 4 = 20 litres.

∴ The correct answer is 20 litres.

Additional Information

Constant Component Concept

When one substance is added to a mixture, the quantity of the other substance remains strictly constant. Equating the ratio parts of the undisturbed component simplifies calculations directly without full equations.

Cross-Multiplication Method

If the initial ratio is a:b and final ratio becomes c:d after adding x and y respectively, the relation is: (a × d) − (b × c) = (y × c) − (x × d). Here, y = 4 and x = 0.

Replacement in Mixtures

When a portion of a mixture is withdrawn without adding anything, the ratio of its components remains identical to the original mixture. The ratio alters only upon adding a new quantity.

Shortcut Trick

Total work = LCM(12, 18, 5) = 180 units.

Efficiency of A = 180 ÷ 12 = 15, B = 180 ÷ 18 = 10, A+B+C = 180 ÷ 5 = 36.

Efficiency of C = 36 − (15 + 10) = 11.

Wage ratio A : B : C = 15 : 10 : 11.

C's share = (11 ÷ 36) × 7200 = ₹2,200.

∴ The correct answer is ₹2,200.

Alternate Method

Given:

Time taken by A alone = 12 days

Time taken by B alone = 18 days

Time taken by A, B, and C together = 5 days

Total wages = ₹ 7,200

Formula Used:

Total Work = LCM of time taken

Efficiency = Total Work ÷ Time

Wages are distributed in the ratio of efficiencies when workers work for the same total time.

Calculations:

⇒ Total efficiency of A, B, and C = 36 units/day

⇒ Efficiency of C = Total efficiency − (Efficiency of A + Efficiency of B)

⇒ Efficiency of C = 36 − (15 + 10) = 11 units/day

⇒ Since they worked together for all 5 days, wages are shared directly in the ratio of their individual efficiencies.

⇒ Ratio of wages A : B : C = 15 : 10 : 11

⇒ Total ratio parts = 15 + 10 + 11 = 36 parts

⇒ 36 parts = ₹ 7,200

⇒ 1 part = 7200 ÷ 36 = ₹ 200

⇒ C's share = 11 parts = 11 × 200 = ₹ 2,200

∴ The correct answer is ₹2,200.

Additional Information

Work and Wages

Wages are strictly proportional to the actual amount of work done by each individual. If workers work together for the same duration, their wages are distributed directly in the ratio of their daily efficiencies.

Different Working Days

If workers work for different numbers of days, their respective wage shares are calculated by multiplying their efficiency with the number of days they worked: Share = Efficiency × Days Worked.

Pipes and Cisterns

This concept mirrors Time and Work exactly. Filling pipes have a positive efficiency (doing work), while emptying pipes or leaks have a negative efficiency (undoing work).

Shortcut Trick

Let remaining mixture components A, B, and C be 7k, 8k, and 5k (Total = 20k).

Final quantity of A = 7k + 15 L and Final quantity of B = 8k.

Given, Final B − Final A = 25 ⇒ 8k − (7k + 15) = 25 ⇒ k = 40.

Total remaining mixture = 20 × 40 = 800 L.

Initial mixture = 800 L + 40 L (drained) = 840 L.

∴ The correct answer is 840 litres.

Alternate Method

Given:

⇒ Initial ratio of A : B : C = 7 : 8 : 5

⇒ Quantity of mixture drained = 40 litres

⇒ Added quantities: 15 litres of A and 5 litres of C

Remaining Mixture Components

7k

• 15 L Solution A

8k Solution B

5k

• 5 L Solution C

Final B − Final A = 8k − (7k + 15) = 25 L

Calculations:

⇒ Let the quantities of A, B, and C in the remaining mixture be 7k, 8k, and 5k respectively.

⇒ Total remaining mixture = 7k + 8k + 5k = 20k

⇒ After addition, Final quantity of A = 7k + 15

⇒ After addition, Final quantity of B = 8k

⇒ After addition, Final quantity of C = 5k + 5

⇒ We are given that Final A is 25 litres less than Final B:

⇒ 8k − (7k + 15) = 25

⇒ 8k − 7k − 15 = 25

⇒ k − 15 = 25

⇒ k = 40

⇒ Total remaining mixture = 20k = 20 × 40 = 800 litres

⇒ Initial quantity of mixture = Remaining mixture + Drained mixture

⇒ Initial quantity of mixture = 800 + 40 = 840 litres

∴ The correct answer is 840 litres.

Additional Information

Mixture Replacement Concept

When a part of a mixture is drained, the remaining mixture still retains the exact same ratio of ingredients as the original mixture before any subsequent pure addition is made.

Successive Replacement

If V is the initial volume of pure liquid, and x volume is replaced by water n times sequentially, the final volume of pure liquid is given by V × (1 − x/V)ⁿ.

Rule of Alligation

It allows calculating the required ratio in which two components of given prices or concentrations must be mixed to obtain a mixture of a desired average price or concentration.

Shortcut Trick

Time ratio of A : B = 3 : 2, so efficiency ratio of A : B = 2 : 3.

Total work = Efficiency of B × Time taken by B = 3 × 18 = 54 units.

Combined efficiency of A and B = 2 + 3 = 5 units/hour.

Time taken together = Total work ÷ Combined efficiency = 54 ÷ 5 = 10.8 hours.

∴ The correct answer is 10.8 hours.

Alternate Method

Given:

Ratio of time taken by A and B = 3 : 2

Time taken by B = 18 hours

Calculations:

⇒ Let the time taken by A be 3x and by B be 2x.

⇒ 2x = 18 hours

⇒ x = 9 hours

⇒ Time taken by A = 3 × 9 = 27 hours.

⇒ Total efficiency of A and B together = 2 + 3 = 5 units/hour.

⇒ Time taken by A and B working together = Total Work ÷ Total Efficiency

⇒ Time taken = 54 ÷ 5 = 10.8 hours.

∴ The correct answer is 10.8 hours.

Additional Information

Time & Efficiency Relationship

Efficiency is inversely proportional to the time taken. If the ratio of time taken by A and B is x : y, then the ratio of their efficiencies is y : x.

Work and Wages

Wages are generally distributed in the ratio of the total work done by each individual. If workers work for the same number of days, the wages are distributed directly in the ratio of their efficiencies.

Pipes & Cisterns

The concept of time and work directly applies to pipes. Inlet pipes have positive efficiency (filling work) while outlet pipes or leaks have negative efficiency (emptying work).

Shortcut Trick

Time taken for one round = Distance ÷ Speed.

T_x = 2/10 = 1/5 h, T_y = 2/12 = 1/6 h, T_z = 2/15 h.

Time to meet at start = LCM(1/5, 1/6, 2/15) = 2/1 = 2 hours.

Total time from 9:00 a.m. to 1:00 p.m. = 4 hours.

Number of meetings = Total time ÷ Meeting interval = 4 ÷ 2 = 2 times.

∴ The correct answer is 2 times.

Alternate Method

Given:

Circumference of the track (D) = 2 km

Speeds: X = 10 km/h, Y = 12 km/h, Z = 15 km/h

Cycling duration = 9:00 a.m. to 1:00 p.m. = 4 hours

Formula Used:

Time = Distance ÷ Speed

Meeting time at start point = LCM of individual times taken for one round

LCM of fractions = LCM of Numerators ÷ HCF of Denominators

Start PointD = 2 kmSpeeds:X = 10 km/hY = 12 km/hZ = 15 km/hTime to meet = LCM of individual times

Calculations:

⇒ T_x = 2 ÷ 10 = 1/5 h

⇒ T_y = 2 ÷ 12 = 1/6 h

⇒ T_z = 2 ÷ 15 = 2/15 h

⇒ Meeting time = LCM(1/5, 1/6, 2/15)

⇒ LCM of numerators (1, 1, 2) = 2

⇒ HCF of denominators (5, 6, 15) = 1

⇒ LCM(1/5, 1/6, 2/15) = 2 ÷ 1 = 2 hours

⇒ They all meet at the starting point every 2 hours.

⇒ Total duration from 9:00 a.m. to 1:00 p.m. = 4 hours.

⇒ Number of meetings = 4 ÷ 2 = 2 times

∴ The correct answer is 2 times.

Additional Information

Meeting at the Starting Point

When multiple runners or objects move around a circular path starting at the same time, the time required for all of them to meet at the starting point simultaneously is always the LCM of the times taken by each to complete exactly one round.

First Meeting on a Circular Track

If two bodies move in the same direction, they will meet for the very first time anywhere on the track after a time = Distance ÷ Relative Speed (Difference of their speeds). If moving in opposite directions, relative speed is the sum of their speeds.

LCM & HCF of Fractions

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

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

Shortcut Trick

Area of a circular path = π × w × (2r + w)

Where w = 3 m (width of path) and r = 8 m (inner radius)

Area = π × 3 × (2 × 8 + 3) = π × 3 × 19

Area = 57π ≈ 57 × 3.14 ≈ 179 m²

∴ The correct answer is 179 m².

Alternate Method

Given:

Inner radius of the park (r) = 8 m

Width of the path (w) = 3 m

Outer radius of the park with path (R) = 8 m + 3 m = 11 m

8 m

3 m

11 m

Circular Park with Path

Formula Used:

Area of circular path = Area of outer circle − Area of inner circle

Area = πR² − πr² = π(R² − r²)

Calculations:

⇒ Area = π × (11² − 8²)

⇒ Area = π × (121 − 64)

⇒ Area = 57π

⇒ Area ≈ 57 × 3.14159

⇒ Area ≈ 179.07 m²

By comparing the options, the nearest value is 179 m².

∴ The correct answer is 179 m².

Additional Information

Area of a Circle

The total region enclosed by a circle is given by Area = πr², where r is the radius.

Area of a Circular Path

For a circular path of width w around a circle of radius r, the area is calculated as π(R² − r²) or πw(2r + w), where the outer radius R = r + w.

Circumference

The perimeter or length of the outer boundary of a circle is calculated as C = 2πr.

Shortcut Trick

The slope of the given line in the form y = mx + c is m = 1/4.

The slope of a perpendicular line is always the negative reciprocal of the original line's slope.

Perpendicular slope = −1 ÷ (1/4) = −4.

∴ The correct answer is -4.

Alternate Method

Given:

Equation of the line: y = 1/4x + 7

Formula Used:

For two perpendicular lines with slopes m₁ and m₂: m₁ × m₂ = −1

y = 1/4x + 7 m₁ = 1/4

Perpendicular Line m₂ = -4

Calculations:

⇒ The given line is structured in the standard form y = mx + c, where m represents the slope of the line.

⇒ By comparing the equations directly, the slope of the given line (m₁) = 1/4

⇒ Let the slope of the required perpendicular line be defined as m₂.

⇒ Since the properties of perpendicular lines state that their product is −1, we have m₁ × m₂ = −1

⇒ (1/4) × m₂ = −1

⇒ m₂ = −1 × 4

⇒ m₂ = −4

∴ The correct answer is -4.

Additional Information

Slope-Intercept Form

The equation of a straight line is typically expressed as y = mx + c, where m is the slope defining the steepness and c is the y-intercept.

Parallel Lines

If two straight lines are distinctly parallel to each other, their slopes will always be exactly equal (m₁ = m₂).

Perpendicular Lines

When two non-vertical straight lines intersect perfectly at a right angle, the resulting product of their slopes evaluates to −1 (m₁ × m₂ = −1).

Shortcut Trick

We know the standard conversion where 1 radian is approximately equal to 57.3°.

To convert 2.5 radians to degrees, we simply multiply the value by 57.3°.

⇒ 2.5 × 57.3° = 143.25°

Looking at the given options, the closest exact value is 143.24°.

∴ The correct answer is 143.24°.

Alternate Method

Given:

Angle = 2.5 radians

Formula Used:

Angle in Degrees = Angle in Radians × (180° ÷ π)

2.5 rad 143.24°

Calculation:

⇒ Angle in Degrees = 2.5 × (180° ÷ π)

⇒ Angle in Degrees = 450° ÷ π

Using the precise value of π ≈ 3.14159:

⇒ Angle in Degrees = 450° ÷ 3.14159

⇒ Angle in Degrees ≈ 143.2395°

⇒ Rounding off to two decimal places, we get 143.24°.

∴ The correct answer is 143.24°.

Additional Information

Radian to Degree Conversion

To convert any given angle from radians to degrees, multiply the radian value by 180° ÷ π.

Degree to Radian Conversion

To convert an angle from degrees back to radians, multiply the degree value by π ÷ 180°.

Standard Conversions

Some common unit circle angles include: π radians = 180°, π ÷ 2 radians = 90°, and π ÷ 4 radians = 45°.

Shortcut Trick

Direct formula for the area of a sector in radians: Area = ½ × r² × θ

Substitute the given values: r = 10 cm and θ = π/6.

Area = ½ × (10)² × (π/6) = ½ × 100 × π/6 = 25π / 3

Using π ≈ 3.14, Area = (25 × 3.14) ÷ 3 = 78.5 ÷ 3 ≈ 26.17 cm².

∴ The correct answer is 26.17 cm².

Alternate Method

Given:

Radius of the sector (r) = 10 cm

Central angle (θ) = π/6 radians

O π/6 10 cm 10 cm Radius (r) = 10 cm Angle (θ) = π/6 rad Area = ?

Formula Used:

Area of a sector = ½ × r² × θ (where θ is in radians)

Calculations:

⇒ Area = ½ × (10)² × (π / 6)

⇒ Area = ½ × 100 × (π / 6)

⇒ Area = 50 × π / 6

⇒ Area = 25π / 3

Using π ≈ 3.14:

⇒ Area = (25 × 3.14) / 3

⇒ Area = 78.5 / 3

⇒ Area = 26.166... ≈ 26.17 cm²

∴ The correct answer is 26.17 cm².

Additional Information

Area of a Sector (in degrees)

If the central angle θ is given in degrees instead of radians, the formula to calculate the area becomes Area = (θ ÷ 360°) × πr².

Length of an Arc

The length of the arc (s) is calculated directly as s = rθ (if θ is in radians) or s = (θ ÷ 360°) × 2πr (if θ is in degrees).

Perimeter of a Sector

The total perimeter bounds the sector, including the curved arc length and the two straight radii. It is found using Perimeter = s + 2r.

Shortcut Trick

If tan x = cot y, then their angles are complementary: x + y = 90°

Given: tan A = cot(2A − 30°)

⇒ A + (2A − 30°) = 90°

⇒ 3A = 120°

⇒ A = 40°

∴ The correct answer is 40°.

Alternate Method

Given:

tan A = cot(2A − 30°)

Formula Used:

cot θ = tan(90° − θ)

A

2A − 30° Adjacent to A Opposite to A If tan x = cot y, then x + y = 90°

Calculations:

⇒ tan A = tan(90° − (2A − 30°))

Comparing the angles on both sides:

⇒ A = 90° − (2A − 30°)

⇒ A = 90° − 2A + 30°

⇒ A + 2A = 120°

⇒ 3A = 120°

⇒ A = 40°

∴ The correct answer is 40°.

Additional Information

Complementary Angle Identities

sin(90° − θ) = cos θ | tan(90° − θ) = cot θ | sec(90° − θ) = cosec θ

Supplementary Angle Identities

sin(180° − θ) = sin θ | cos(180° − θ) = −cos θ | tan(180° − θ) = −tan θ

Negative Angle Identities

sin(−θ) = −sin θ | cos(−θ) = cos θ | tan(−θ) = −tan θ