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Shortcut Trick

Price after 20% discount = 1250 − 250 = ₹1,000

Final Selling Price = ₹900

Amount of second discount = 1000 − 900 = ₹100

Second Discount percentage = (100 ÷ 1000) × 100 = 10%

∴ The correct answer is 10%.

Alternate Method

Given: Marked Price (MP) = ₹1,250, First Discount (D₁) = 20%, Final Selling Price (SP) = ₹900

Formula: Selling Price = MP × (100 − D₁)% × (100 − D₂)%

⇒ Price after first discount = 1250 × (100 − 20) ÷ 100

⇒ Price after first discount = 1250 × 0.8 = ₹1,000

⇒ Now, the second discount is applied on ₹1,000 to get ₹900

⇒ Second Discount amount = 1000 − 900 = ₹100

⇒ Second Discount % = (Discount Amount ÷ Previous Price) × 100

⇒ Second Discount % = (100 ÷ 1000) × 100 = 10%

∴ The correct answer is 10%.

Additional Information

Successive Discounts Formula

When two discounts 'a%' and 'b%' are given successively, the equivalent single discount is given by: [a + b − (a × b ÷ 100)]%.

Marked Price (MP) and Discount

Discount is always calculated on the Marked Price. The price obtained after subtracting the discount from the MP is the Selling Price (SP).

Relationship between SP and MP

SP = MP × [(100 − D₁) ÷ 100] × [(100 − D₂) ÷ 100] × ... × [(100 − D_n) ÷ 100].

Shortcut Trick

Simplify the second term: 10 − √75 = 10 − 5√3 = 5(2 − √3).

The numbers are (2 + √3) and 5(2 − √3).

Product = 5 × (2 + √3)(2 − √3) = 5 × (4 − 3) = 5.

Mean proportional = √Product = √5.

∴ The correct answer is √5.

Alternate Method

Given:

First number (a) = 2 + √3

Second number (b) = 10 − √75

Formula Used:

The mean proportional between two numbers a and b is given by x = √(a × b).

⇒ First, simplify b = 10 − √(25 × 3) = 10 − 5√3

⇒ Factor out 5: b = 5(2 − √3)

⇒ Now, x = √[(2 + √3) × 5(2 − √3)]

⇒ x = √[5 × (2 + √3)(2 − √3)]

⇒ Using algebraic identity (a + b)(a − b) = a² − b²:

⇒ x = √[5 × (2² − (√3)²)]

⇒ x = √[5 × (4 − 3)]

⇒ x = √[5 × 1] = √5

∴ The correct answer is √5.

Additional Information

Mean Proportional

The mean proportional of two numbers a and b is x if a:x = x:b, which results in the formula x = √(ab).

Third Proportional

For two numbers a and b, the third proportional c is such that a:b = b:c, given by c = b²/a.

Conjugate Surds

Surds of the form (p + √q) and (p − √q) are conjugates; their product is always a rational number (p² − q).

Shortcut Trick

For any closed cylinder, the ratio of Curved Surface Area (CSA) to Total Surface Area (TSA) is always h : (r + h).

Given Height (h) = 10 m and Radius (r) = 12 ÷ 2 = 6 m.

Required Ratio = 10 : (6 + 10) = 10 : 16 = 5 : 8.

∴ The correct answer is 5:8.

Alternate Method

Given values for the cylindrical tank:

⇒ Diameter = 12 m

⇒ Height (h) = 10 m

⇒ Radius (r) = Diameter ÷ 2 = 12 ÷ 2 = 6 m

⇒ Curved Surface Area (CSA) = 2 × π × r × h

⇒ Total Surface Area (TSA) = 2 × π × r × (r + h)

⇒ Ratio = [2 × π × r × h] ÷ [2 × π × r × (r + h)]

⇒ Ratio = h ÷ (r + h)

⇒ Ratio = 10 ÷ (6 + 10) = 10 ÷ 16

⇒ Ratio = 5 : 8

∴ The correct answer is 5:8.

Additional Information

Volume of a Cylinder

The total space occupied by a cylinder is given by V = πr²h.

Base Area of a Cylinder

The area of each circular end of the cylinder is πr².

Surface Area for Open Tanks

If a cylindrical tank is open at the top, its total surface area is 2πrh + πr².

Shortcut Trick

Net efficiency of all taps = 1/4 + 1/22 − 1/22 = 1/4 cistern/hr.

The question asks for the time to fill 1/4^th of the empty cistern.

Time required = (Work to be done) ÷ (Net efficiency)

Time = (1/4) ÷ (1/4) = 1 hr.

∴ The correct answer is 1 hr.

Alternate Method

Given:

Time taken by Tap 1 (filling) = 4 hr

Time taken by Tap 2 (filling) = 22 hr

Time taken by Tap 3 (emptying) = 22 hr

Formula Used:

Total Work = LCM of individual times

Efficiency = Total Work ÷ Individual Time

Calculations:

⇒ Total capacity of the cistern (LCM of 4 and 22) = 44 units.

⇒ Net efficiency when all taps are open = 11 + 2 − 2 = 11 units/hr.

⇒ Work to be completed (1/4^th of the cistern) = 44 × (1/4) = 11 units.

⇒ Time taken to fill 11 units = 11 ÷ 11 = 1 hr.

∴ The time required to fill one-fourth of the cistern is 1 hr.

∴ The correct answer is 1 hr.

Additional Information

Pipes and Cisterns Basics

If a pipe can fill a tank in x hours, then the part filled in 1 hour is 1/x. This is known as the efficiency of the pipe.

Inlet and Outlet Taps

An inlet tap performs positive work (filling), while an outlet or emptying tap performs negative work. The net rate is the algebraic sum of their individual efficiencies.

LCM Method

Using the Least Common Multiple (LCM) of the given time periods as the total capacity helps avoid working with complex fractions, making calculations faster.

Shortcut Trick

Assume Total Capacity of the tank = LCM(10, 15, 20) = 60 units.

Net efficiency of (A + B + C) = 6 + 4 − 3 = 7 units/hr.

Work done in 4 hours = 7 × 4 = 28 units.

Remaining work = 60 − 28 = 32 units.

Time for A and B to fill remaining = 32 ÷ (6 + 4) = 3.2 hr.

Total Time = 4 + 3.2 = 7.2 hr.

∴ The correct answer is 7.2 hr.

Alternate Method

Given: Pipe A fills in 10 hr, Pipe B fills in 15 hr, and Pipe C empties in 20 hr.

Formula Used: Total Work = Net Efficiency × Time

⇒ Net efficiency of all three pipes (A + B − C) = 6 + 4 − 3 = 7 units/hr.

⇒ Work completed in the first 4 hours = 7 × 4 = 28 units.

⇒ Remaining work to be completed after closing Pipe C = 60 − 28 = 32 units.

⇒ Efficiency of remaining pipes (A + B) = 6 + 4 = 10 units/hr.

⇒ Time taken to fill the remaining part = 32 ÷ 10 = 3.2 hr.

⇒ Total time required = Initial time + Remaining time = 4 hr + 3.2 hr = 7.2 hr.

∴ The correct answer is 7.2 hr.

Additional Information

Inlet and Outlet Pipes

An Inlet pipe fills the tank (positive work), while an Outlet pipe empties the tank (negative work).

Net Efficiency

If multiple pipes operate simultaneously, the Net Efficiency is the algebraic sum of their individual efficiencies: Net Eff = Eff_Inlet1 + Eff_Inlet2 − Eff_Outlet.

Time and Work Relation

Efficiency is inversely proportional to the Time taken to complete the work (Efficiency = 1 ÷ Time), assuming the work is constant.

Shortcut Trick

Length of train = (Difference in speeds × 5/18) × (t₁ × t₂) ÷ (t₂ − t₁)

Difference in speeds = 7 − 5 = 2 km/h

Length = (2 × 5/18) × (18 × 20) ÷ (20 − 18)

Length = (5/9) × 360 ÷ 2 = 5 × 20 = 100 m

∴ The correct answer is 100 m.

Alternate Method

Given: Speed of runner₁ = 5 km/h, Speed of runner₂ = 7 km/h, Time taken t₁ = 18 s, t₂ = 20 s.

Formula: Relative Speed (same direction) = Speed of train − Speed of object.

Let Speed of the train be V km/h.

⇒ Relative Speed with first runner = (V − 5) × 5/18 m/s

⇒ Relative Speed with second runner = (V − 7) × 5/18 m/s

⇒ Length L = (V − 5) × (5/18) × 18 = (V − 7) × (5/18) × 20

⇒ (V − 5) × 1 = (V − 7) × (20/18)

⇒ 18(V − 5) = 20(V − 7)

⇒ 18V − 90 = 20V − 140

⇒ 2V = 50 ⇒ V = 25 km/h

⇒ L = (25 − 5) × (5/18) × 18 = 20 × 5 = 100 m

∴ The correct answer is 100 m.

Additional Information

Relative Speed in Same Direction

When two objects move in the same direction, the relative speed is the difference between their speeds: (u − v).

Relative Speed in Opposite Direction

When two objects move towards each other, the relative speed is the sum of their speeds: (u + v).

Unit Conversion

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

Length of Train

When a train passes a point object (man, pole, runner), the distance covered equals the length of the train.

Shortcut Trick

Assume all 120 pens are of Type B (₹8 each): Total sale = 120 × 8 = ₹960.

Actual total sale is ₹1120, so the difference is 1120 − 960 = ₹160.

Difference in price per pen between Type A and Type B = 12 − 8 = ₹4.

Number of Type A pens = 160 ÷ 4 = 40.

∴ The correct answer is 40 pens.

Alternate Method

Given:

Total pens = 120 | Price of Type A = ₹12 | Price of Type B = ₹8 | Total Amount = ₹1120

Formula Used:

Weighted Average Total = (n₁ × Price₁) + (n₂ × Price₂)

Let the number of Type A pens be x and Type B pens be (120 − x).

⇒ 12 × x + 8 × (120 − x) = 1120

⇒ 12x + 960 − 8x = 1120

⇒ 4x + 960 = 1120

⇒ 4x = 1120 − 960

⇒ 4x = 160

⇒ x = 40

∴ The correct answer is 40 pens.

Additional Information

Rule of Alligation

It is a rule that enables us to find the ratio in which two or more ingredients at their respective prices must be mixed to produce a mixture of a specified price.

Linear Equation Method

Standard method where unknown quantities are represented as variables (x, y) and solved using algebraic manipulation.

Weighted Average

The average of a data set that contains values with different levels of importance or frequency: Average = (Sum of all values) ÷ (Total number of items).

Given:

(340 / 17) × (156 / 13) + (1484 / 28) − (14 / 10)

Formula Used:

The BODMAS rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction) is applied to simplify the expression.

Calculations:

(340 ÷ 17) × (156 ÷ 13) + (1484 ÷ 28) − (14 ÷ 10)

⇒ 20 × 12 + 53 − 1.4

⇒ 240 + 53 − 1.4

⇒ 293 − 1.4

⇒ 291.6

∴ The correct answer is option (1).

Shortcut Trick

Check the sum of the base numbers: 22 + (−18) + (−4) = 0.

When the sum of three numbers is zero, the sum of their cubes is 3 × Product of the numbers.

Required value = 3 × 22 × (−18) × (−4) = 66 × 72 = 4752.

∴ The correct answer is 4752.

Alternate Method

Given:

The expression to evaluate is 22³ + (−18)³ + (−4)³.

Let a = 22, b = −18, and c = −4.

Formula Used:

We use the algebraic identity: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

Calculations:

⇒ First, calculate the sum of the variables: a + b + c = 22 + (−18) + (−4)

⇒ a + b + c = 22 − 22 = 0

⇒ Since a + b + c = 0, the identity simplifies to: a³ + b³ + c³ = 3abc

⇒ Substituting the values: 22³ + (−18)³ + (−4)³ = 3 × 22 × (−18) × (−4)

⇒ Value = 66 × 72

⇒ Value = 4752

Therefore, the correct option is 2.

∴ The correct answer is 4752.

Additional Information

Cubic Identity Properties

The expression a³ + b³ + c³ is equal to 3abc only if a + b + c = 0 or if a = b = c. This is a fundamental conditional identity in algebra.

General Expansion

For any three numbers, a³ + b³ + c³ − 3abc = ½ (a + b + c)[(a − b)² + (b − c)² + (c − a)²].

Applications in Exams

This identity is frequently used to simplify large cubic calculations by checking if the sum of the bases equals zero before performing traditional multiplication.

Shortcut Trick

Mean proportional of 9 and 16 = √(9 × 16) = 12.

Given mean proportional of 12 and the number is twice this value.

Target MP = 2 × 12 = 24.

Using the property (MP)² = Product of numbers ⇒ 24² = 12 × Number.

Number = 576 ÷ 12 = 48.

∴ The correct answer is 48.

Alternate Method

Given:

First set of numbers = 9 and 16

Second set of numbers = 12 and x (let the number be x)

Formula Used:

Mean proportional (MP) between two numbers a and b is √(a × b).

Calculations:

⇒ Mean proportional between 9 and 16 = √(9 × 16)

⇒ Mean proportional_(9,16) = √144 = 12

⇒ According to the question, MP_(12,x) = 2 × 12 = 24

⇒ √(12 × x) = 24

⇒ Squaring both sides: 12 × x = 24²

⇒ 12 × x = 576

⇒ x = 576 ÷ 12

⇒ x = 48

∴ The correct answer is 48.

Additional Information

Mean Proportional

The mean proportional between two numbers a and b is √(a × b). If x is the mean proportional, then a : x :: x : b.

Third Proportional

For two numbers a and b, the third proportional c is such that a : b :: b : c, which gives c = b² ÷ a.

Fourth Proportional

For three numbers a, b, and c, the fourth proportional d is such that a : b :: c : d, giving d = (b × c) ÷ a.