A table question with percentages of growth tests how you connect claims to counts. Read what the question asks, then check the table’s headings and units. Ask what each percentage measures, then multiply that amount by the percentage divided by 100. Add gains to starting counts, and track what is known: exact values give one count; limits show the highest or lowest possible count. A supported claim must hold for every value the text allows.
Hints
- Hint 1
In the sentences after the table, check what the percentages apply to.
- Hint 2
For nitrate, calculate the culture’s total increase and then R’s share of that increase. What else must you include to reach R’s final cell count?
- Hint 3
For ammonium, find R’s largest possible final count using the percentage limit. Compare that limit with R’s final count in nitrate.
Step-by-step
Turn growth shares into final counts
Step 1Identify what the percentages measure
Each culture began with 20 million from each strain, so R started at 20 million cells in both nutrient treatments. What does the 70% measure? It describes R’s share of the culture’s net increase, the number of cells gained during the experiment. We need the ending count, so R’s final count is 20 million plus its gain.
- Step 2
Calculate R’s nitrate count
The table’s columns give cell counts in millions: nitrate began at 120 and ended at 220. How much of that change belongs to R? The culture gained 220 − 120 = 100 million, and R supplied 70% of the nitrate culture's net increase, so R gained 70 million. Adding its initial 20 million gives 90 million R cells.
- Step 3
Find the limit on R’s ammonium count
What can we know about R in ammonium? The culture gained 200 − 120 = 80 million cells, but no individual strain accounted for more than 25% of that culture's net increase. That limit includes R, so it gained at most 0.25 × 80 = 20 million cells. Adding the starting 20 million gives at most 40 million R cells at the end. R could have gained less, so we don’t know its exact ending count.
- Step 4
Test what the counts must show
What is best supported by the text and table without an exact ammonium count? Twice ammonium’s largest possible R count is 2 × 40 = 80 million, which is below nitrate’s 90 million. So at least twice as many is true for every possible ammonium count. Choice C.