Factory Production Problem: Solving 1/3 Ratio with 700-Unit Constraint

Given that factory A produces 13 \frac{1}{3} of the productivity of factory B. Both factories together produce less than 700 cartons of milk per day. What can be said about the production of factory A?

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1

Understand the problem

Given that factory A produces 13 \frac{1}{3} of the productivity of factory B. Both factories together produce less than 700 cartons of milk per day. What can be said about the production of factory A?

2

Step-by-step solution

To solve this problem, follow these steps:

  • Step 1: Identify that x=13y x = \frac{1}{3} y and substitute into the inequality x+y<700 x + y < 700 .
  • Step 2: Rewrite the inequality as 13y+y<700 \frac{1}{3} y + y < 700 .
  • Step 3: Combine and simplify terms: 43y<700 \frac{4}{3} y < 700 .
  • Step 4: Solve for y y by multiplying both sides by 34\frac{3}{4}: y<525 y < 525 .
  • Step 5: Substitute back to find x x : Since x=13y x = \frac{1}{3} y , x<13×525=175 x < \frac{1}{3} \times 525 = 175 .

Thus, we conclude that the production of factory A, x x , is less than 175 cartons per day.

3

Final Answer

Less than 175 cartons per day

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Solve the inequality:


\( 5-3x>-10 \)

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