Daily Math Puzzle: 2026-06-28
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2026-06-28
In a biology experiment, a petri dish initially contains a suspension where 20% of the total volume is composed of microorganisms, and the remaining 80% is nutrient broth. First, a lab assistant adds more nutrient broth, increasing the *volume of the nutrient broth* by 25%. Then, observing significant growth, the lead scientist notes that the *volume of the microorganisms* has also increased by 25%. What is the final percentage of microorganisms in the total suspension?
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Solution
20% — Let the initial total volume of the suspension be 'V'.
Initial volume of microorganisms = 0.20V.
Initial volume of nutrient broth = 0.80V.
Step 1: The volume of the nutrient broth increases by 25%.
Increase in nutrient broth volume = 0.25 * 0.80V = 0.20V.
New volume of nutrient broth = 0.80V + 0.20V = 1.00V.
The volume of microorganisms remains 0.20V.
Total volume after Step 1 = 0.20V (microorganisms) + 1.00V (nutrient broth) = 1.20V.
Step 2: The volume of the microorganisms increases by 25%.
Increase in microorganism volume = 0.25 * 0.20V = 0.05V.
New volume of microorganisms = 0.20V + 0.05V = 0.25V.
The volume of nutrient broth remains 1.00V (from the end of Step 1).
Final total volume of the suspension = 0.25V (microorganisms) + 1.00V (nutrient broth) = 1.25V.
To find the final percentage of microorganisms:
(Final volume of microorganisms / Final total volume) * 100%
= (0.25V / 1.25V) * 100%
= (1/5) * 100%
= 20%.
Despite both components increasing by 25%, they started from different base volumes (0.20V and 0.80V), leading to absolute volume increases that, when combined and the new ratio is calculated, maintain the original percentage.
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