Question

If the length and height of a brick increases by 10% each respectively, and the breadth reduces by 20%, what is the percentage change in the volume of the brick?

• A. 28
• B. 30
• C. 32
• D. 36

Hint:

Use percentage change formula for compound dimension problems: \[ \text{New Volume} = V \times (1 + \frac{a}{100})(1 + \frac{b}{100})(1 + \frac{c}{100}) \]

Answer 1

The Correct Option is A

Original Volume \( = L \times B \times H \) New dimensions:
Length increases 10% \( \Rightarrow L' = L \times 1.10 \)
Height increases 10% \( \Rightarrow H' = H \times 1.10 \)
Breadth decreases 20% \( \Rightarrow B' = B \times 0.80 \) New Volume: \[ V' = L' \times B' \times H' = (1.10L)(0.80B)(1.10H) = 1.10 \times 1.10 \times 0.80 \times L \times B \times H \] \[ V' = (1.21 \times 0.80) \times V = 0.968 \times V \] \[ \text{Percentage change} = (0.968 - 1) \times 100 = -3.2% \text{ (volume decreased by 3.2%)} \] No matching option. Let’s recheck. Use formula: Percentage change = \( a + b + c + \frac{ab}{100} + \frac{ac}{100} + \frac{bc}{100} + \frac{abc}{10000} \) \[ a = 10,\ b = -20,\ c = 10 \] \[ \text{Total} = 10 - 20 + 10 + \frac{10 \times (-20)}{100} + \frac{10 \times 10}{100} + \frac{-20 \times 10}{100} + \frac{10 \times (-20) \times 10}{10000} \] \[ = 0 - 2 + 1 -2 + (-0.2) = -3.2% \] \[ \text{Final Volume Change} = \boxed{-3.2%} \] No matching options — issue with choices.