And 1 girl's 1 day's work = y

Now,

(5 boys+2 girls)'s work = 5x+2y

Given,

5x+2y is equal to 4 times work done by a boy and a girl.

Thus,

5x+2y = 4(x+y)

5x+2y = 4x+4y

x = 2y

^x/_y = ²/₁

Hence, the required ratio is 2:1.

Part of the tank filled in 1 minute \begin{align} & = \left( {\frac{x}{{24}}} \right) + \left( {\frac{x}{{40}}} \right) - 30 \cr & {\text{Or}}, \left( {\frac{x}{{24}}} \right) + \left( {\frac{x}{{40}}} \right) - 30 = \frac{x}{{60}} \cr & \frac{x}{{24}} + \frac{x}{{40}} - \frac{x}{{60}} = 30 \cr & \left[ {\frac{{\left( {10x + 6x - 4x} \right)}}{{240}}} \right] = 30 \cr & {\text{Or}}, \, 12x = 30 \times 240 \cr & {\text{Or}}, \, x = 600\, {\text{gallons}} \cr\end{align}

Work done by all the tanks working together in 1 hour,

=> (¹/₁₀) + (¹/₁₂) - (¹/₂₀) = ²/₁₅.

Hence, tank will be filled in ¹⁵/₂ = 7.5 hour.

1200*45 = 1000*x;

Hence, x = 54 days.

Variation Method:

After 15 days 200 soldiers leaved.

Soldiers Food for days

1200 ↓ 45

1000 ↑ x (let)

Arrows show the opposite variation to each other.

¹²⁰⁰/₁₀₀₀ = ^x/₄₅;

Or, x = ^1200*45/₁₀₀₀ = 54 days.

Short-cut method:

If there is a provision of food for A men for a days and after b days if B men join them or leave

Work rate of (A + B + C) = ¹⁰⁰/₁₀ = 10% per minute.

A alone can fill the tank in 30 minutes.

Work rate of A = ¹⁰⁰/₃₀ = 3.33% per minute.

B alone can fill the tank in 40 minutes.

Work rate of B = ¹⁰⁰/₄₀ = 2.5%.

Work rate of (A + B) = 3.33 + 2.5 = 5.83% per minute.

Work rate of C,

= Work rate of (A +B +c) - (A +B).

= 10 - 5.83 = 4.17% per minute.

So, C takes = ¹⁰⁰/_4.17 = 24 minutes to fill the tank.