Quantitative Aptitude & Data Interpretation

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.

Volume calculation of rectangular hall:

• Dimensions: Length $L = 15\text{ m}$, Breadth $B = 12\text{ m}$, Height $H$.
• Floor area = $15 \times 12 = 180\text{ m}^2$.
  Ceiling area = $15 \times 12 = 180\text{ m}^2$.
  Sum of floor and ceiling areas = $180 + 180 = 360\text{ m}^2$.
• Area of four walls = $2(L + B)H = 2(15 + 12)H = 2(27)H = 54H$.
• Setting areas equal:
  $54H = 360 \implies H = \frac{360}{54} = \frac{20}{3}\text{ m}$.
• Volume of hall:
  $V = L \times B \times H = 15 \times 12 \times \frac{20}{3} = 180 \times \frac{20}{3} = 60 \times 20 = 1200\text{ m}^3$.
  
  Option analysis:
• 720: Calculation error.
• 900: Calculation error.
• 1200: Correct volume in cubic meters.
• 1800: Calculated as $180 \times 10$.