Introduction / Context:
This is a volume-conservation problem: the volume of earth dug from a cylindrical tunnel is used to fill a rectangular ditch. The answer is the ratio of the tunnel volume to the ditch volume.
Given Data / Assumptions:
- Ditch: 48 m × 16.5 m × 4 m.
- Tunnel: cylinder with diameter 4 m ⇒ radius 2 m, length 56 m.
- No loss of earth during transfer.
Concept / Approach:
- Ditch volume V_ditch = L*B*H.
- Tunnel volume V_tunnel = π*r^2*L.
- Fraction filled = V_tunnel / V_ditch.
Step-by-Step Solution:
V_ditch = 48 * 16.5 * 4 = 3168 m^3r = 2 m ⇒ V_tunnel = π * 2^2 * 56 = 224π m^3Fraction = 224π / 3168 = (224/3168)*πSimplify 224/3168 = 1/14.142… but evaluate numerically: 224π ≈ 703.72Fraction ≈ 703.72 / 3168 ≈ 0.2222… = 2/9
Verification / Alternative check:
Compute exact with π ≈ 22/7: V_tunnel = 224*(22/7) = 704/1 ≈ 704 (close to 703.72). Then 704 / 3168 = 2 / 9. Correct.
Why Other Options Are Wrong:
- 1/9, 7/9, 8/9: Do not match the computed ratio.
- None of these: Incorrect; 2/9 is exact with π = 22/7.
Common Pitfalls:
- Using diameter instead of radius in cylinder volume.
- Arithmetic errors when simplifying the fraction.
Final Answer:
2/9
Discussion & Comments