Depth · one concept from first principles · step 2 of 4

Two thresholds on one mixture give a range

The table shows, for $1$ mL of each kind of food waste poured down the drain, the amount of water needed to purify it and its BOD (biochemical oxygen demand). A total of $1000$ mL of milk and ramen broth together was poured down the drain. The water needed to purify it was more than $800$ L, and its BOD was at least $0.95$ mg/mL. Find the range of the amount of milk, in mL, in the $1000$ mL that was poured.

Food waste	Water needed to purify (L)	BOD (mg/mL)
Milk	$2$	$0.05$
Ramen broth	$0.5$	$1.25$

ExpressionsRatios

Your answer

How to solve

Strategy Convert to Algebra — One unknown, the amount of milk, controls both quantities. Extend the table to totals, write one inequality for the water and one for the BOD, and keep the values that satisfy both.

1STEP 1

Choose the unknown and extend the table

Let the milk be x mL, so the broth is 1000 - x mL. Multiply each per-mL value by the amount to get totals.

water = 2x + 0.5(1000 - x) L, oxygen demand = 0.05x + 1.25(1000 - x) mg

2STEP 2

Water condition

The total water must be more than 800 L.

2x + 500 - 0.5x ＞ 800 → 1.5x ＞ 300 → x ＞ 200

3STEP 3

BOD condition

BOD is milligrams per mL, so the mixture's BOD is the total oxygen demand in mg divided by 1000 mL. It must be at least 0.95.

(0.05x + 1.25(1000 - x))/1000 ≥ 0.95 → 1250 - 1.2x ≥ 950 → 1.2x ≤ 300 → x ≤ 250

4STEP 4

Put the two limits together

Both conditions must hold, so take the overlap.

200 ＜ x ≤ 250

Answer

more than 200 mL and at most 250 mL (200 ＜ x ≤ 250)

At x = 250: water = 500 + 375 = 875 L, more than 800; BOD = (12.5 + 937.5)/1000 = 0.95, allowed. At x = 200: water = 400 + 400 = 800 L, not more than 800, so 200 itself is excluded.

Takeaway

One unknown can face two tests at once -- solve each test on its own, then keep only the values that pass both.

Choose the unknown and extend the table
Water condition
BOD condition
Put the two limits together

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