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Total population increase = 6500 − 6000 = 500

Base increase (if all 6000 were males) = 8% of 6000 = 480

Extra increase due to females = 500 − 480 = 20

Female percentage difference = 10% − 8% = 2%

∴ 2% of Females = 20 ⇒ Females = 1000

∴ The correct option is A.

Alternate Method

Given:

Present population = 6000

New population = 6500

Male increase = 8%, Female increase = 10%

Formula: Alligation Rule or Percentage Change

⇒ Total increase = 6500 − 6000 = 500

⇒ Overall percentage increase = (500 ÷ 6000) × 100 = 25/3 %

Male (8%) Female (10%) Mean (25/3 %) Diff: (10 − 25/3) = 5/3 Diff: (25/3 − 8) = 1/3 Ratio M : F = 5 : 1

⇒ Ratio of Male to Female = (10 − 25/3) : (25/3 − 8) = 5/3 : 1/3 = 5 : 1

⇒ Total units = 5 + 1 = 6

⇒ Present number of females = (1 ÷ 6) × 6000

⇒ Present number of females = 1000

∴ The correct option is A.

Additional Information

Percentage Change Formula

Percentage change = (Final Value − Initial Value) ÷ Initial Value × 100. It measures the degree of change over time.

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.

Weighted Average

Combined % increase = (n₁p₁ + n₂p₂) ÷ (n₁ + n₂), where n is the quantity and p is the percentage change.

Shortcut Trick

Initial Ratio M : W = 3 : 2 (Total = 5 units = 60 L ⇒ 1 unit = 12 L)

New Ratio M : W = 1 : 5 (Milk remains constant, so multiply by 3) → 3 : 15

Water increase = 15 − 2 = 13 units

Quantity of water added = 13 units × 12 L/unit = 156 L

∴ The correct answer is 156 L.

Alternate Method

Given: Total Volume = 60 L, Initial Ratio = 3 : 2, Final Ratio = 1 : 5.

Formula: Quantity = [Ratio Part ÷ Total Parts] × Total Volume

Initial (60 L) Milk: 36 L Water: 24 L

• Water (x)

Final Ratio 1 : 5 Milk: 36 L (1 unit) Water: 180 L (5 units) x = 180 − 24 = 156 L

⇒ Initial Milk = (3/5) × 60 = 36 L

⇒ Initial Water = (2/5) × 60 = 24 L

⇒ Let x be the amount of water added.

⇒ New Ratio = Milk / (Water + x) = 1 / 5

⇒ 36 / (24 + x) = 1 / 5

⇒ 36 × 5 = 24 + x

⇒ 180 = 24 + x

⇒ x = 180 − 24 = 156 L

∴ The correct answer is 156 L.

Additional Information

Constant Part Concept

When one component is added or removed from a mixture, the quantity of the other component remains constant. Equating the constant part in both ratios simplifies the calculation.

Alligation Method

For adding pure water to a mixture, the concentration of water in the added part is 100%. Alligation cross-difference helps find the ratio of initial mixture to added water.

Percentage Base Change

Initial % of milk × Initial Volume = Final % of milk × Final Volume. This formula works efficiently when only one component changes.

Given:

A, B and C together can finish a piece of work in 2 days

A alone can finish it in 6 days

B is thrice as good a workman as C

Concept:

Efficiency of a person is reverse of total time taken

Calculation:

Let per day efficiencies of A, B and C be a, b and c respectively

⇒ b = 3c

According to the question:

\(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2}\)

⇒ \(\frac{1}{6} + \frac{3}{c} + \frac{1}{c} = \frac{1}{2}\)

⇒ 4/c = \(\frac{1}{2} - \frac{1}{6}\)

⇒ 4/c = 1/3

⇒ c = 12

∴ C will complete the work alone in 12 days

Alternate MethodWe know time is inversely proportional to efficiency

B is thrice as good a workman as C 

So, If B takes x days, then C will take 3x days

So, 1/6 + 1/x + 1/3x = 1/2

x = 4

So, time take by C = 3 × 4 = 12 days

Single equivalent percent of two successive increases of 20% = 20 + 20 + (20 × 20)/100 = 44%

Single equivalent percent of two successive decreases of 20% = -20 – 20 + (20 × 20)/100 = 36%

∴ Required percentage = \(\left[ {\frac{{44 - 36}}{{36}}} \right]{\rm{}} \times {\rm{}}100{\rm{}} = {\rm{}}22\frac{2}{9}{\rm{% }}\)

Given:

Radius of hemisphere = 6 cm

Radius of cone = 3 cm

Height of cone = 4 cm

Concept used:

Volume of hemisphere = (2/3) × πr³

Volume of cone = (1/3) πR²h

Calculation:

Volume of hemisphere = N × Volume of cone

⇒ (2/3) × π (6)³ = N × (1/3) × π (3)² × 4

⇒ (2 × 216) = 36N

⇒ N = 2 × 6

⇒ N = 12

∴ The value of N is 12.

Given:

Increased marks = 3 marks 

New candidate score =  65 marks

Calculations:

The total score increased = (5 × 3) = 15 marks. 

Among the 5, one has scored 65 marks and it's replaced by the score of the new candidate. 

Then the score of the new candidate = (65 + 15) = 80 marks. 

∴ The marks of the new candidate is 80 marks. 

Alternate Method

Let the average marks of 5 candidates be x marks and marks of new candidate be y

Total marks of 5 candidates = x × 5 = 5x marks. 

⇒ New average = (x + 3)  

⇒ Total marks = 5x - 65 + y 

According to question,

⇒5x - 65 + y = 5 (x + 3)  

⇒ - 65 + y = 15

⇒ y = 65 + 15

⇒ y = 80

∴ The marks of the new candidate is 80 marks. 

​in place of one of them means interchange not addition of new candidate.

Formula Used:

Height = \(\frac{a\sqrt3 }{2}\)

Calculations:

Let the side of the triangle be a m

Height = \(\frac{a\sqrt3 }{2}\)

Side = \(\frac{H\times 2}{\sqrt3}\)

Side = \(\frac{6\sqrt3\times 2}{\sqrt3}\)

Side = 12 m

Perimeter

= 12 × 3

= 36 m 

Hence, The Required value is 36 m

Given:

Speed of A = 45 km/hr

Speed of B = 60 km/hr

A starts at 9 ∶ 00 AM and B starts at 10 ∶ 00 AM

They meet at half distance while travelling in the opposite direction

Concept used:

Distance = Speed × Time

Calculation:

Let the time taken by B to reach half the distance be x hours

Time taken by A to reach half the distance will be x hours + 1 hour = (x + 1) hours

As, they have traveled half distance, i.e they have traveled equal distance

Half distance travelled by A = 45 × (x + 1)

Half distance traveled by B = 60 × x

45x + 45 = 60x

⇒ 15x = 45

⇒ x = 3

Distance travelled by A = 45 × (3 + 1) = 45 × 4 = 180

∴ Distance traveled by A is 180 km

Shortcut Trick

Ratio of numbers = 3 : 4

LCM in units = 3 × 4 = 12 units

Given, 12 units = 72 ⇒ 1 unit = 6

Sum of numbers in units = 3 + 4 = 7 units

Sum = 7 × 6 = 42

∴ The correct answer is 42.

Alternate Method

Given:

Ratio of two numbers = 3 : 4

LCM of numbers = 72

Let the two numbers be 3x and 4x, where x is the HCF.

⇒ LCM of (3x, 4x) = 3 × 4 × x = 12x

⇒ 12x = 72

⇒ x = 72 ÷ 12

⇒ x = 6

⇒ First number = 3 × 6 = 18

⇒ Second number = 4 × 6 = 24

⇒ Sum = 18 + 24 = 42

∴ The correct answer is 42.

Additional Information

Product of Two Numbers

The product of two numbers is always equal to the product of their HCF and LCM. (Product = HCF × LCM)

HCF and Ratio Relationship

If two numbers are in the ratio a : b, their LCM is a × b × HCF, and their Sum is (a + b) × HCF.

Co-prime Numbers

If two numbers are co-prime (HCF = 1), then their LCM is simply the product of the two numbers.

Given:

Speed of 40% = 12 km/h

Speed for remaining distance = 8 km/h

Formula used:

Speed = Distance/Time

Average speed = Total Distance/Total Time

Calculation:

Let the distance be x km

Speed at which 0.4xkm distance covered = 12km/h

Time = 0.4x/12 = x/30 hours

Speed at which 0.6xkm distance covered = 8 km/h

Time = 0.6x/8 = 3x/40 hours

Total time = x/30 + 3x/40 = (4x + 9x)/120 = 13x/120 hours

Average speed = \(\frac{x}{\frac{13x}{120}}\) = 120/13 = 9.2 km/h

∴ The average speed is 9.2 km/h