Geometry, Mensuration (2D/3D Areas & Volumes), Trigonometry & Coordinate Geometry - Agri-Business Management | Agrigyan

Volume displacement computation:

• Total volume of water displaced by 40 men:
  $\text{Total Volume} = 40 \times 4\text{ m}^3 = 160\text{ m}^3$.
• Base area of the rectangular tank:
  $\text{Base Area} = \text{Length} \times \text{Breadth} = 30\text{ m} \times 20\text{ m} = 600\text{ m}^2$.
• Rise in water level ($h$):
  $h = \frac{\text{Total Volume}}{\text{Base Area}} = \frac{160}{600} = \frac{4}{15}\text{ m}$.
• Convert to centimeters:
  $h = \frac{4}{15} \times 100\text{ cm} = \frac{400}{15} \approx 26.67\text{ cm} \approx 27\text{ cm}$.

Area and dimension calculation:

• Area of one carpet = $6\text{ m} \times 4\text{ m} = 24\text{ m}^2$.
  Total floor area covered by 34 carpets = $34 \times 24 = 816\text{ m}^2$.
• Let breadth of the hall be $B\text{ m}$, then length $L = (B + 10)\text{ m}$.
  $B(B + 10) = 816$
  $B^2 + 10B - 816 = 0$
  $(B + 34)(B - 24) = 0$
  Since breadth must be positive, $B = 24\text{ m}$.
• Length $L = 24 + 10 = 34\text{ m}$.

Perimeter (circumference) of a circle is given by $C = 2\pi r$.
Given $C = 40\text{ m}$:
$2 \times \pi \times r = 40$
$r = \frac{40}{2\pi} = \frac{20}{\pi} = \frac{20}{3.14159} \approx 6.366\text{ m} \approx 6.36\text{ m}$.

Option analysis:

• 6.23 m: Inaccurate approximation.
• 6.20 m: Inaccurate approximation.
• 6.28 m: Value of $2\pi$, not the radius.
• 6.36 m: Correct radius.

Let the initial edge length of the cube be $a$.
Initial total surface area $A_1 = 6a^2$.
If each edge is increased by $50\%$, the new edge length becomes:
$a' = a + 0.5a = 1.5a$.
New surface area $A_2 = 6(a')^2 = 6(1.5a)^2 = 6(2.25a^2) = 2.25 \times (6a^2) = 2.25 A_1$.
Increase in surface area:
$\Delta A = 2.25 A_1 - A_1 = 1.25 A_1$.
Percentage increase:
$\frac{1.25 A_1}{A_1} \times 100\% = 125\%$.

Option analysis:

• 125: Correct percentage increase.
• 150: Linear scaling error.
• 175: Calculation error.
• 110: Calculation error.