In the year 2147, Rays Martian colony imports 3 types of hydroponic modules: A, B, and C. Module A supports 120 people, Module B supports 80, and Module C supports 50. If the colony plans to house 1,160 people using exactly 10 modules, with twice as many Module A units as Module C, how many Module B units are deployed?

Mon, 04 May 2026 00:54:03 +0000

In the year 2147, Rays Martian colony imports 3 types of hydroponic modules—A, B, and C—designed to sustain life beyond Earth. Module A supports 120 residents, Module B supports 80, and Module C supports 50. As Mars settlement scales to accommodate growing populations, innovative designs like these are increasingly relevant. Gaining interest across tech, space exploration, and sustainability communities, the model reflects real-world progress toward self-sustaining extraterrestrial habitats. With strict module limits—exactly 10 units—the colony balances space, efficiency, and nutrition requirements. This scenario challenges problem-solvers to decode a precise, data-driven puzzle central to future off-world infrastructure. Why In the year 2147, Rays Martian colony imports 3 types of hydroponic modules: A, B, and C. Module A supports 120 people, Module B supports 80, and Module C supports 50. If the colony plans to house 1,160 people using exactly 10 modules, with twice as many Module A units as Module C, how many Module B units are deployed? This question reflects modern curiosity about scalable space agriculture. As private companies and global space agencies plan for long-term habitation, detailed models help readers engage with complex engineering realities. The challenge lies in aligning modular output with realistic colony growth—making this puzzle not just educational, but timely in the evolving space economy. To unpack the puzzle: the colony uses exactly 10 hydroponic modules—A, B, and C—with 2:1 A-to-C ratio. Let Module C count be *x*, so Module A becomes *2x*. With B as the remainder, *x + 2x + B = 10*, so B = 10 – 3x. Module capacity totals 1,160 people:
120(2x) + 80(B) + 50(x) = 1,160
Simplify:
240x + 80(10 – 3x) + 50x = 1,160
240x + 800 – 240x + 50x = 1,160
50x + 800 = 1,160
50x = 360 → x = 7.2? Wait—this isn’t right. Instead, recheck: 240x (A) + 80(10–3x) (B) + 50x (C) = 1,160
Try solving correctly:
240x + 800 – 240x + 50x = 1,160 → (240x – 240x + 50x) + 800 = 1,160
50x + 800 = 1,160 → 50x = 360 → x = 7.2. Not integer.