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As per pie chart,

Percentage fund allotted for Drinking water = 20%

Percentage fund allotted for Toilet = 28%

The total fund allocation for Drinking Water and Toilet together = 20 + 28 = 48%

Fund allocated to a village under a special village development scheme = Rs. 1200 Crore

The total fund allocation for Drinking Water and Toilet together = 

= 1200 × (48/100)

= 576

∴ The total fund allocation for Drinking Water and Toilet together is Rs. 576.

Given:

Principal = Rs. 6000

Rate = 20%

Time = 18 months

Concept Used:

A = P ×  (1 + r/n)^(nt)

Where:

A = Amount

P = Principal

r = Annual interest rate 

n = Number of times the interest is compounded per year

t = Time period (in years)

Calculation:

The interest is compounded half-yearly, so n = 2 (twice a year).

The time period (t) is 18 months, which is equivalent to 1.5 years.

Rate, r = 20% = 0.20

Let's substitute these values into the formula:

A = 6000 ×  (1 + 0.20/2)^{(2 × 1.5)}

⇒ 6000 ×  (1 + 0.10)³

⇒  6000 ×  1.10³

⇒  6000 ×  1.331

⇒ A = Rs. 7986

Compound interest = A - P

⇒  7986 - 6000

⇒  Rs. 1986

Therefore, the compound interest for 18 months at a rate of 20% per annum, compounded half-yearly, would be Rs. 1986.

∴ Option 2 is the correct answer.

Given:

Selling Price (SP) = Rs. 2400

Loss Percentage = 20%

Formula Used:

Cost Price (CP) = SP / (1 - Loss Percentage)

Calculation:

Loss Percentage = 20% = 20/100 = 0.2

CP = 2400 / (1 - 0.2)

CP = 2400 / 0.8

⇒ CP = 2400 × (1/0.8)

⇒ CP = 2400 × (10/8)

⇒ CP = 2400 × (5/4)

⇒ CP = 2400 × 1.25

⇒ CP = 3000

The cost price of the watch is Rs. 3000.

Given:

A retailer gets a discount of 40% on the printed price of an article. 

Calculation:

Let assume marked price of an article = 100

After 40% discount on the marked price,

Cost price for the retailer = 40% of Marked price

= 60% of 100

= 60

So, cost price = Rs. 60

Now the retailer is selling at the marked price,

So, profit of the retailer = $\dfrac{100-60}{60} \times 100$

= $\dfrac{40}{60} \times 100$

= $66 \frac{2}{3} \%$

So, the percentage of profit of the retailer is $66 \frac{2}{3} \%$.

Calculation : 

⇒ 16 × 12 - 672 ÷ 21 = ? - 211

⇒ 192 - 32 = x - 211

⇒ 160 = x - 211

⇒ x = 211 + 160 

⇒ x = 371

Shortcut Trick

Let R's age be 1 unit.

Then Q's age = 2 units, and P's age = 2 units + 5 years.

Total = (1 + 2 + 2) units + 5 = 55 years

⇒ 5 units = 50 years ⇒ 1 unit = 10 years.

P's age = 2 units + 5 = 2 × 10 + 5 = 25 years.

∴ The correct answer is 25 years.

Alternate Method

Given:

Total age of P, Q, and R = 55 years

R

x

Q

2x

P

2x

+5

Total = 55

Let the age of R be x years.

Q is twice as old as R ⇒ Q = 2x

P is 5 years older than Q ⇒ P = 2x + 5

Calculations:

According to the sum of their ages:

⇒ P + Q + R = 55

⇒ (2x + 5) + 2x + x = 55

⇒ 5x + 5 = 55

⇒ 5x = 50

⇒ x = 10

Age of P = 2x + 5 = 2(10) + 5 = 25 years.

∴ The correct answer is 25 years.

Additional Information

Constant Age Difference

The age difference between any two individuals always remains constant over time. If A is 5 years older than B today, A will be 5 years older than B after 10 years as well.

Past and Future Ages

If the current age of a person is x, their age n years ago was (x − n) and their age n years from now will be (x + n).

Shortcut Trick

Net efficiency = Filling rate − Emptying rate.

Let Total capacity (LCM of 8 and 16) = 16 units.

Filling pipe efficiency = 16 ÷ 8 = 2; Emptying pipe efficiency = 16 ÷ 16 = 1.

Combined efficiency = 2 − 1 = 1 unit/hour.

Time for 3/4 filled = (3/4 × 16) ÷ 1 = 12 hours.

∴ The correct answer is 12 hours.

Alternate Method

Given: Filling time = 8 hours, Emptying time = 16 hours.

Formula Used: Work = Efficiency × Time; Net Efficiency = (1/Filling time) − (1/Emptying time).

⇒ Combined efficiency of both pipes = 2 − 1 = 1 unit/hour.

⇒ Target volume to be filled = 3/4 × Total Capacity

⇒ Target volume = 3/4 × 16 = 12 units.

⇒ Time taken = Target volume ÷ Combined efficiency

⇒ Time taken = 12 ÷ 1 = 12 hours.

∴ The correct answer is 12 hours.

Additional Information

Pipes and Cisterns Concept

Filling pipes (inlets) have positive efficiency, while emptying pipes (outlets) have negative efficiency.

Net Work Formula

If pipe A fills in 'x' hours and pipe B empties in 'y' hours, net work in 1 hour = (1/x − 1/y). If x < y, the tank will eventually fill.

Fractional Filling

To find time for a specific fraction of the tank, multiply the total fill time by that fractional value.

Calculation:

Let 50 paise, 25 paise, and 10 paise coins are

in ratio 3x ,4x & 5x respectively.

Total Value of 50 paise coins = $\frac{50}{100}$× 3x = 3x/2

Total Value of 25 paise coins = $\frac{25}{100}$× 4x = x

Total Value of 10 paise coins = $\frac{10}{100}$× 5x = x/2

Total Value of coins is Rs. 12

⇒ 3x/2 + x + x/2 = 12

⇒ 6x/2 = 12

⇒ x = 4

⇒ Number of 25 paise coins =  4x = 4 × 4 = 16

∴ The correct answer is 16

Given 

The capacity of water tank = 900 liters 

Height = 5 m

Formula Used 

Volume = Area × Height

1 m³ = 1000 liters

1000 cm³ = 1 litre

Calculation 

Capacity of the tank = 900 litre = 900 × 1000 cm³ = 900000 cm³

Capacity of tank = Surface Area × 5 

⇒ 900000 = Area × 500

⇒ Area = 1800 cm²

∴ The area of surface tank is 1800 cm²

Calculation:

Let efficiency of 1 man and 1 boy be x and y respectively.

According to the question

(6x + 8y) × 10 = (26x + 48y) × 2

⇒ 60x + 80y = 52x + 96y

⇒ 60x – 52x = 96y – 80y

⇒ 8x = 16y

∴ x : y = 2 : 1