- 4,058
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A slightly simpler way for problem 3: There is a (unknown) distance D between the trains. Let ## v_t ## be the speed of the trains and ## v_c ## the speed of the bicycle. ## \\ ## We have ## 30= \frac{D}{v_t-v_c} ## and ## 20=\frac{D}{v_t+v_c} ##. The problem is to solve for ## T=\frac{D}{v_t} ##. The frequency will be the inverse of that. ## \\ ## Rewriting: ## \frac{v_t}{D}-\frac{v_c}{D}=\frac{1}{30} ##, and ## \frac{v_t}{D}+\frac{v_c}{D}=\frac{1}{20} ##. ## \\ ## Let ## \frac{v_t}{D}=x ## and ## \frac{v_c}{D}=y ##. ## \\ ## This gives ## x-y=\frac{1}{30} ## and ## x+y=\frac{1}{20} ##. ## \\ ## Adding these two equations: ## 2x=\frac{1}{12} ##. ## \\ ## The result is ## x=\frac{1}{T}=\frac{1}{24} ## (in frequency per minute), or ## \frac{60 \, min/hr}{24 \, min}= 2.5 ## trains each hour.