Solve the following problems:

Let the price of Uniform be U Taka

atq, 

12Month = 30000+U………….(i)

9Month = 22000+U ………….(ii)

From Equation (i)$\times$9 - (ii)$\times$12

We get , 270000+ 9U - 264000 - 12U = 0    so, 3U = 6000    so, U = 2000 Taka 

So, Required price of the Uniform = 2000 Taka 

Let's solve the problem step-by-step.

1. **Let the cost of one horse be \( H \) Tk and the cost of one cow be \( C \) Tk.**

2. **According to the problem:**
  - The price of 3 horses is the same as the price of 5 cows. So, we can write this relationship as:
    \[ 3H = 5C \]
  - The total cost of 4 horses and 6 cows is Tk 1900. This can be expressed as:
    \[ 4H + 6C = 1900 \]

3. **First, solve for \( C \) from the first equation:**
  \[ C = \frac{3H}{5} \]

4. **Substitute \( C = \frac{3H}{5} \) into the second equation:**
  \[ 4H + 6 \left(\frac{3H}{5}\right) = 1900 \]
  Simplify:
  \[ 4H + \frac{18H}{5} = 1900 \]

5. **Combine the terms:**
  To combine \( 4H \) and \( \frac{18H}{5} \), convert \( 4H \) to a fraction with the same denominator:
  \[ 4H = \frac{20H}{5} \]
  So:
  \[ \frac{20H}{5} + \frac{18H}{5} = 1900 \]
  \[ \frac{38H}{5} = 1900 \]

6. **Solve for \( H \):**
  Multiply both sides by 5 to clear the fraction:
  \[ 38H = 1900 \times 5 \]
  \[ 38H = 9500 \]
  Divide by 38:
  \[ H = \frac{9500}{38} \]
  \[ H = 250 \]

7. **The cost of one horse is \( \text{Tk } 250 \).**

Balance = 20,00,000

To wife = (2/5)*20,00,000 = 8,00,000

Remain =20,00,000-8,00,000=12,00,000

To son = (¼)*12,00,000 = 3,00,000

Remain 12,00,00-3,00,000 = 9,00,000

Donated = (2/9)* 9,00,000 = 2,00,000 (Ans.)

Ratio of (1 : P) = (432 : 1200) = (9 : 25)

According to the Question,

(9 × 100) = 25n²

⇒ n² = 36

⇒ n = 6% (Ans)

According to the Question,

a√2 = 12√2

⇒ a = 12 cm

So, Perimeter of Square = (4 × 12 ) = 48 cm

So, Each sides of Triangle = $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ of 48 = 16 cm

Area of Triangle = $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ × 16 × 8√3 = 64√3 (Ans)