Data Interpretation: Totals, Growth and Denominator Traps

Read what the figures count

The figures below count applications processed in each calendar year, not distinct people. For this exercise, each application is counted under exactly one centre within its year, so the centre counts can be added. A value such as 120 means 120 applications, not 120 thousand applications or 120%. All figures are fictional training data.

Original practice dataset

• Centre A: 2024 = 120; 2025 = 150.
• Centre B: 2024 = 200; 2025 = 220.
• Centre C: 2024 = 80; 2025 = 130.

Worked questions

Total in 2024 = 120 + 200 + 80 = 400 applications. Total in 2025 = 150 + 220 + 130 = 500 applications. Increase = 500 − 400 = 100 applications. Overall growth = (100/400)×100 = 25%. The denominator is the starting-year total, not the final total.

A's growth = (30/120)×100 = 25%. B's growth = (20/200)×100 = 10%. C's growth = (50/80)×100 = 62.5%. C has the largest percentage growth, although B has the largest 2025 count, 220. The largest final count and the largest growth rate answer different questions.

A's share of the 2025 total = (150/500)×100 = 30%. Its growth rate uses A's own earlier count, whereas its share uses the same-year total across centres. B's share falls from (200/400)×100 = 50% to (220/500)×100 = 44%, a fall of 6 percentage points, even though B's count rises by 20 applications.

Combine growth rates using starting counts

The simple mean of 25%, 10% and 62.5% is (25 + 10 + 62.5)/3 = 32.5%, which differs from the overall 25%. Here the starting counts differ, so each centre contributes a different weight. Aggregate growth = total change / total starting count. As a percentage, this equals (120×25 + 200×10 + 80×62.5)/400 = 25%. Equal starting counts are sufficient for a simple mean to give the aggregate rate; they are not necessary, because equal rates or other exact balancing can also make the results coincide.

A centre's percentage growth requires a positive starting count. If its starting count is zero, its percentage growth is undefined, so do not insert an invented rate into the weighted formula. The combined growth can still be calculated directly from totals if the combined starting total is positive.

Missing data and inference

Counts of applications processed do not by themselves establish service quality, demand, approval rate or the number of distinct applicants. One person may submit several applications. An approval rate requires the number approved and a clearly specified eligible total; those data are absent here. A larger processed count alone therefore cannot prove better service or greater efficiency.

Practice with solutions

1. What is C's share of the 2024 total? Use the same-year total: (80/400)×100 = 20%.
2. What is the ratio A:B:C in 2025? It is 150:220:130. Dividing all three entries by 10 gives 15:22:13; the centre order stays A:B:C.
3. How much of the total increase came from C? C's increase is 130 − 80 = 50; total increase is 500 − 400 = 100. C's contribution is (50/100)×100 = 50%. This denominator is the total increase, not the 2025 total.
4. If 20% of A's 2025 applications required correction, how many is that? A processed 150 applications, so the number is (20/100)×150 = 30 applications.
5. Can these figures prove B is the most efficient centre? No. They provide output counts, but not staffing, resources, case complexity or processing time. B's larger count is insufficient to rank efficiency.

Percentage points and relative share change

B's share changes from 50% to 44%. Subtraction gives 44 − 50 = −6 percentage points. Relative to its original share, the change is ((44 − 50)/50)×100 = −12%, so its share falls by 12%. Yet B's application count grows by (20/200)×100 = 10%. These results are consistent: the overall count grows faster than B's count. A share can fall while the count rises.

Recover a missing count

Suppose B's 2025 entry is hidden, but the total 500 and A = 150, C = 130 are given. Because the centre categories do not overlap and exhaust the total, B = 500 − 150 − 130 = 220. Check: 150 + 220 + 130 = 500. If categories overlapped or an unlisted centre were included in the total, this subtraction would not identify B uniquely.

Compare output with a specified rate

Now add a separate fictional detail: the 2025 work used 75 staff-hours at A, 110 staff-hours at B and 65 staff-hours at C. A staff-hour means one person's work for one hour. Applications per staff-hour are A: 150/75 = 2, B: 220/110 = 2 and C: 130/65 = 2. B's larger count does not give it a higher rate by this measure. These added hours were not implied by the original counts. Even equal throughput does not establish equal quality or overall efficiency; case difficulty and outcomes may differ.