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Shortcut Trick

The time after which all bells ring together is the LCM of their individual ringing intervals.

LCM of 6, 7, 9, and 12 = 252 seconds.

Since the bells ring together every 252 seconds, we check for 252 or its multiples in the options.

Next multiple = 252 × 2 = 504 seconds, which matches an option.

∴ The correct answer is 504 seconds.

Alternate Method

Given:

Intervals at which the four bells ring = 6, 7, 9, and 12 seconds.

Concept Used:

The bells will ring together again after a time that is exactly divisible by all the individual intervals. This specific time represents the Least Common Multiple (LCM) of the given numbers. If the exact LCM is not among the options, any integer multiple of the LCM is a valid answer since the bells continue to ring together periodically.

Calculations:

⇒ Prime factorization of 6 = 2 × 3

⇒ Prime factorization of 7 = 7

⇒ Prime factorization of 9 = 3²

⇒ Prime factorization of 12 = 2² × 3

⇒ LCM = Product of the highest powers of all present prime factors

⇒ LCM = 2² × 3² × 7

⇒ LCM = 4 × 9 × 7 = 252 seconds

⇒ This means the bells will ring together every 252 seconds (i.e., at 252s, 504s, 756s, etc.).

⇒ Analyzing the given options, we find that 504 is exactly divisible by 252 (252 × 2 = 504).

∴ The correct answer is 504 seconds.

Additional Information

LCM for Periodic Events

For events that repeat at different regular intervals (like bells tolling, traffic lights changing, or runners on a circular track), they will all occur together again at exactly the LCM of their individual intervals.

Multiples of LCM

If the events coincide for the first time at the LCM, they will continue to simultaneously coincide at every successive integer multiple of the LCM (e.g., 2 × LCM, 3 × LCM).

HCF for Maximum Grouping

While LCM is for simultaneous forward occurrences, the Highest Common Factor (HCF) is used when dividing items into the largest possible equal groups without leaving any remainder (e.g., cutting ropes into maximum equal lengths).

Shortcut Trick

Let CP = x and SP = y. Original Profit = y − x.

New SP = 2y. New Profit = 2y − x.

According to the question: 2y − x = 4 × (y − x).

⇒ 2y − x = 4y − 4x ⇒ 3x = 2y ⇒ y/x = 3/2.

Profit % = (3 − 2) ÷ 2 × 100 = 1/2 × 100 = 50%.

∴ The correct answer is 50%.

Alternate Method

Given: New SP is twice the original SP; New Profit is four times the original profit.

Formulas: Profit = SP − CP | Profit% = (Profit ÷ CP) × 100

⇒ Let CP = 100 and Profit = P. Then SP = 100 + P.

⇒ New SP = 2 × (100 + P) = 200 + 2P.

⇒ New Profit = New SP − CP = (200 + 2P) − 100 = 100 + 2P.

⇒ According to the condition: 100 + 2P = 4P.

⇒ 2P = 100 ⇒ P = 50.

⇒ Profit% = (50 ÷ 100) × 100 = 50%.

∴ The correct answer is 50%.

Additional Information

Profit and Loss Basic Concepts

Profit occurs when the selling price (SP) is greater than the cost price (CP). Profit = SP − CP.

Percentage Calculation

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

Shortcut Trick

Lateral Surface Area is proportional to the square of the edge (LSA ∝ edge²).

Using the successive percentage formula: x + y + (x × y / 100).

Here x = 50% and y = 50%.

Required percentage increase = 50 + 50 + (50 × 50 / 100) = 125%.

∴ The correct answer is 125%.

Alternate Method

Given:

Percentage increase in each edge of the cube = 50%

Formula Used:

Lateral Surface Area (LSA) of a cube = 4 × (edge)²

Edge = a Initial LSA = 4a²

+50%

Edge = 1.5a New LSA = 9a²

Calculations:

⇒ Let the initial edge of the cube be a.

⇒ Initial lateral surface area = 4 × a²

⇒ New edge = a + 50% of a = 1.5a

⇒ New lateral surface area = 4 × (1.5a)²

⇒ New lateral surface area = 4 × 2.25a² = 9a²

⇒ Increase in LSA = 9a² − 4a² = 5a²

⇒ Percentage increase = (Increase / Initial LSA) × 100

⇒ Percentage increase = (5a² / 4a²) × 100

⇒ Percentage increase = 1.25 × 100 = 125%

∴ The correct answer is 125%.

Additional Information

Total Surface Area (TSA)

The TSA of a cube is 6a². If the edge increases by x%, the percentage increase in TSA is the same as the increase in LSA.

Volume Increase

The volume of a cube is a³. If the edge increases by x%, the volume increase is [(1 + x/100)³ − 1] × 100%.

Successive Percentage Formula

For any 2D dimension change (Area), the net change is a + b + (a × b / 100). For a cube, a = b = edge change %.

Shortcut Trick

The efficiency (flow rate) of a pipe is directly proportional to the square of its diameter.

Since the new pipes have double the diameter, their efficiency is 2² = 4 times that of the original pipes.

Total capacity of the cistern = 15 pipes × 1 unit × 12 minutes = 180 units.

Total efficiency of 6 new pipes = 6 × 4 units = 24 units/minute.

Time required = Total capacity ÷ New efficiency = 180 ÷ 24 = 7.5 minutes.

∴ The correct answer is 7.5 minutes.

Alternate Method

Given:

⇒ Number of original pipes = 15

⇒ Time taken by original pipes = 12 minutes

⇒ Number of new pipes = 6

⇒ Diameter of new pipes = 2 × Original diameter

Formula Used:

⇒ Flow Rate (Efficiency) ∝ (Diameter)²

⇒ Total Work = Number of Pipes × Efficiency per pipe × Time

Calculations:

⇒ Let the efficiency of 1 original pipe be 1 unit/min.

⇒ The new pipe has twice the original diameter, so its cross-sectional area and efficiency become 2² = 4 units/min.

⇒ Total combined efficiency of 15 original pipes = 15 × 1 = 15 units/min.

⇒ Total capacity of the cistern = Total original efficiency × Time taken = 15 × 12 = 180 units.

⇒ Total combined efficiency of 6 new pipes = 6 × 4 = 24 units/min.

⇒ Time required by new pipes = Total Capacity ÷ Combined New Efficiency

⇒ Time required = 180 ÷ 24 = 7.5 minutes.

∴ The correct answer is 7.5 minutes.

Additional Information

Pipes and Cisterns Flow Rule

When the speed of the fluid is constant, the volume of liquid flowing out is strictly proportional to the cross-sectional area of the pipe. Hence, Flow Rate ∝ r² or d².

MDH Work Equivalence Formula

Total work remains constant and can be solved using the chain rule: M₁ × D₁ × E₁ = M₂ × D₂ × E₂, where M is the number of pipes, D is the time elapsed, and E is the specific flow efficiency.

Rate & Efficiency Trick

If a pipe's radius or diameter is scaled by a factor of 'k', its filling capacity per unit of time scales by a factor of 'k²'. Time needed will correspondingly reduce by a factor of 'k²' for a single pipe.

Given:

84 ÷ 32 × 8 - 15 ÷ 8 × (19 - 35)

Concept used:

Calculation:

84 ÷ 32 × 8 - 15 ÷ 8 × (19 - 35)

⇒ 84 ÷ 32 × 8 - 15 ÷ 8 × (- 16)

⇒ 21/8 × 8 - (15/8) × (- 16)

⇒ 21 + 30

⇒ 51

∴ The required answer is 51.

Shortcut Trick

The total capacity of the tank is the product of the number of taps and the time taken: 15 × 36 = 540 tap-minutes.

To find the number of taps for a new time of 60 minutes (1 hour), divide the total capacity by the new time.

Required taps = 540 ÷ 60 = 9.

∴ The correct answer is 9.

Alternate Method

Given:

Initial number of taps (M₁) = 15

Initial time (T₁) = 36 minutes

Required time (T₂) = 1 hour = 60 minutes

Formula Used:

For a constant work (filling the same tank), the formula is: M₁ × T₁ = M₂ × T₂

Calculations:

⇒ Total Work = 15 × 36 = 540 units

⇒ M₂ × 60 = 540

⇒ M₂ = 540 ÷ 60

⇒ M₂ = 9

∴ The correct answer is 9.

Additional Information

Inverse Proportion

For a constant task, the number of resources (taps/men) is inversely proportional to the time taken. If time increases, fewer taps are needed.

Pipes and Cisterns

This concept follows the same logic as Time and Work, where the rate of filling is the efficiency of the pipe or tap.

Chain Rule Formula

The general formula for such problems is M₁ × D₁ × H₁ ÷ W₁ = M₂ × D₂ × H₂ ÷ W₂, where W represents the work done.

Shortcut Trick

Deviation of the newly adopted girl's age from the current average = 11 − 15 = −4 years.

This deficit of 4 years will be shared equally among all 4 girls (3 current + 1 new).

Change in average = −4 ÷ 4 = −1 year.

New average = 15 − 1 = 14 years.

∴ The correct answer is 14 years.

Alternate Method

Given:

Number of current daughters = 3

Average age of the 3 daughters = 15 years

Age of the adopted girl = 11 years

Formula Used:

Average = Total Sum of Observations ÷ Total Number of Observations

Current Daughters & Adopted Girl:

15

15

15

11

Sum = 56

New Average (Equally Distributed):

14

14

14

14

Avg = 14

Calculations:

⇒ Total age of 3 daughters = 3 × 15 = 45 years

⇒ Total age of all girls after adoption = 45 + 11 = 56 years

⇒ Total number of girls after adoption = 3 + 1 = 4

⇒ New average age = 56 ÷ 4 = 14 years

∴ The correct answer is 14 years.

Additional Information

Weighted Average

If group 1 has n₁ items with an average of A₁, and group 2 has n₂ items with an average of A₂, then the Combined Average = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Average Replacement

If an existing person is replaced by a new person and the overall average increases by x, the new person's value = Old person's value + (Total persons × x).

Effect of Arithmetic Operations

If each given observation in a set is systematically increased, decreased, multiplied, or divided by a constant value 'k', then the average of the set will also undergo the same exact operation by 'k'.

Shortcut Trick

Perimeter of Square = Circumference of Circle

4 × Side = 2 × π × Radius

4 × 66 = 2 × (22/7) × r

264 = (44/7) × r

r = (264 × 7) ÷ 44 = 6 × 7 = 42 cm

∴ The correct answer is 42 cm.

Alternate Method

Given:

Side of the square (a) = 66 cm

Formula Used:

Perimeter of square = 4 × a

Circumference of circle = 2 × π × r

Calculations:

⇒ Length of wire = Perimeter of square

⇒ Length of wire = 4 × 66 = 264 cm

⇒ Since the same wire is bent into a circle:

⇒ 2 × π × r = 264

⇒ 2 × (22/7) × r = 264

⇒ (44/7) × r = 264

⇒ r = (264 × 7) ÷ 44

⇒ r = 6 × 7

⇒ r = 42 cm

∴ The correct answer is 42 cm.

Additional Information

Perimeter and Area of Square

Perimeter = 4 × side and Area = side². The diagonal of a square is side × √2.

Circumference and Area of Circle

Circumference = 2πr and Area = πr², where r is the radius and π ≈ 22/7 or 3.14.

Shape Transformation Rule

When a wire is reshaped from one 2D figure to another, its perimeter (length) remains constant, but its area usually changes.

Given:

The shopkeeper announced 20% discount after increasing the price by 30%.

Actual costs/ CP = Rs. 300

Calculation:

According to the question,

Increasing the price by 30% of the actual price.

So, Increasing price = 300 × \(\dfrac{130}{100}\) = Rs. 390

Now, the shopkeeper announced 20% discount.

Discount price/SP = 390 × \(\dfrac{80}{100}\) = Rs. 312

Profit = SP - CP

⇒ 312 - 300 = Rs. 12

∴ He will get Rs.12 profit for that item.

Shortcut Trick

Identify the largest number among 56, 84, and 112, which is 112.

Check the multiples of 112 to find the first one divisible by 56 and 84.

112 × 1 = 112 (Not divisible by 84)

112 × 2 = 224 (Not divisible by 84)

112 × 3 = 336 (336 ÷ 84 = 4 and 336 ÷ 56 = 6)

∴ The correct answer is 336.

Alternate Method

Given: Numbers are 56, 84, and 112.

Method: Prime Factorization Method.

⇒ Prime factors of 56 = 2 × 2 × 2 × 7 = 2³ × 7

⇒ Prime factors of 84 = 2 × 2 × 3 × 7 = 2² × 3 × 7

⇒ Prime factors of 112 = 2 × 2 × 2 × 2 × 7 = 2⁴ × 7

⇒ LCM = Product of the highest powers of all prime factors involved.

⇒ LCM = 2⁴ × 3¹ × 7¹

⇒ LCM = 16 × 3 × 7

⇒ LCM = 48 × 7 = 336

∴ The correct answer is 336.

Additional Information

Least Common Multiple (LCM)

The smallest positive integer that is perfectly divisible by each of the given numbers.

Relationship between HCF and LCM

For any two numbers 'a' and 'b', the product of the numbers is equal to the product of their HCF and LCM: a × b = HCF(a,b) × LCM(a,b).

Highest Common Factor (HCF)

The greatest number that divides each of the given numbers without leaving a remainder.