I1 · Table Analysis
Table Analysis looks like the easiest format on the exam. A table of numbers, three statements, Yes or No on each. No calculator needed for most of it, no hidden algebra, nothing you have not seen in a spreadsheet a thousand times. It is also the format where competent people most often score zero on an item they understood perfectly.
All or nothing. Three statements, one item, and you need all three right to get anything. Two out of three is worth exactly what zero out of three is worth. That changes the arithmetic of how you spend time here in a way that is not obvious until you have lost an item to it.
The data is not the difficulty. Everything you need is on the screen. The difficulty is entirely in how the statement is worded — which rows it is really asking about, which comparison it is really making, and whether "more than half" includes exactly half. The exam is testing your reading, using numbers as the vehicle.
That second point is what makes this chapter possible to teach quickly. There are, in practice, about five shapes of statement. Once you recognise the shape, the arithmetic is trivial. This chapter is a catalogue of the shapes.
You get a table, a short description of what the columns hold, and three statements. For each one you select Yes or No — where Yes means the statement is accurate based on the information provided, and No means it is not.
Note the wording carefully. The question is never whether the statement is true in the world. It is whether the table supports it. A statement can be perfectly plausible and still be No because the table does not establish it.
Every column header is sortable. Click it and the whole table reorders. Click again for the reverse. Most candidates never use it. They scan the table visually, comparing numbers by eye, and this is where errors come from — not from arithmetic, but from missing a row in a column of twelve.
Sorting turns almost every statement in this format into something you can read off directly:
| Statement asks about | Sort by | Then |
|---|---|---|
| The largest or smallest | that column | read the top or bottom row |
| The range | that column | subtract bottom from top |
| The median | that column | count to the middle |
| How many exceed a threshold | that column | count from the top until you cross it |
| A filtered subset | the filter column | the subset is now contiguous |
That last one is the most valuable and the least used. If a statement is about "reserves larger than 500 hectares", sorting by area puts all of them together in a block. You are no longer hunting; you are reading a block of rows.
Sorting changes what you see, and after two or three sorts it is easy to lose track of which order you are in. Before evaluating each statement, decide which column you need sorted, sort it, and answer. One sort per statement, deliberately chosen.
Every example in this chapter uses this table. It describes eight nature reserves in a regional network. This section is worth a bookmark — I1.2 through I1.6 all refer back to it, and each example restates the rows it needs so you can follow without scrolling.
| Reserve | Area (hectares) | Bird species | Annual visitors | Rangers |
|---|---|---|---|---|
| Alderwood | 340 | 82 | 12,400 | 4 |
| Brackenmoor | 1,250 | 118 | 8,900 | 6 |
| Cranefield | 210 | 64 | 31,200 | 5 |
| Dunmarsh | 890 | 141 | 6,300 | 3 |
| Elmshaw | 470 | 95 | 22,700 | 7 |
| Fernhollow | 1,610 | 127 | 4,100 | 5 |
| Grayling | 155 | 51 | 18,900 | 2 |
| Hazelbank | 720 | 103 | 15,600 | 6 |
This is the single most common shape, and the single most common error.
The wrong path takes four seconds: scan the visitors column, find the largest number — 31,200 at Cranefield — and answer Yes.
The right path notices that the statement said among the reserves larger than 500 hectares, and Cranefield is 210. It is not in the conversation at all.
Filter first: Brackenmoor (1,250), Dunmarsh (890), Fernhollow (1,610), Hazelbank (720). Within those four, visitors are 8,900 / 6,300 / 4,100 / 15,600. The most is Hazelbank at 15,600, which is more than 14,000. Yes — but for a completely different reason than the fast path gave.
And the exam writes it so that the wrong path often produces the wrong answer. Change the threshold to 20,000 and the fast path still says Yes while the correct answer becomes No.
Before you look at any number, find the filter. It is almost always the phrase beginning among, of the, for those, or a subordinate clause with a condition in it. Then sort by the filter column so the qualifying rows sit together, and only then look at what is being asked.
The absolute leader is a salient number — the biggest figure in a column jumps out before your reading finishes the sentence. So the error is not analytical and you cannot fix it by being more careful with the arithmetic. The fix is sequential: read the whole statement, to the end, before your eyes touch the table. That ordering is the entire defence.
Two superlatives, one claim: that the same row wins both columns.
Most species: Dunmarsh, at 141. Largest area: Fernhollow, at 1,610. Different rows. No.
Find each winner separately, then compare names. Two sorts, two glances, one comparison. The error people make is trying to hold both columns in view at once and reasoning about which reserve "seems biggest overall". There is no such thing — the statement is about two specific rankings and whether they name the same row.
- "…is also the one with the fewest…" — one maximum and one minimum. Same method, opposite end of the second sort.
- "Among X, the one with the most Y is also the one with the most Z" — a filter stacked on a cross-superlative. Filter first, per I1.2, then run the cross-superlative inside the subset.
That second variant is the hardest statement in this format, because both mistakes are available at once. When you see it, slow down deliberately: it is a statement worth thirty seconds rather than ten.
Here you cannot read an answer off a sorted column. You have to add.
Filter first: Alderwood (340), Cranefield (210), Elmshaw (470), Grayling (155). Their visitors: 12,400 + 31,200 + 22,700 + 18,900.
12,400 + 31,200 = 43,600
43,600 + 22,700 = 66,300
66,300 + 18,900 = 85,200
85,200 > 80,000. Yes.
Now notice the margin: 85,200 against a threshold of 80,000 is about 6 percent. That is deliberate. A candidate who rounds to 12 + 31 + 23 + 19 = 85 thousand happens to land right, but one who rounds more aggressively — 10 + 30 + 20 + 20 = 80 — lands exactly on the boundary and has to guess.
Filter, then add exactly, then compare. The threshold is usually set within a few percent of the true total precisely to punish rounding.
- Averages, not totals. "The average number of visitors across reserves smaller than 500 hectares exceeds 20,000." Same filter, same sum, then divide by the count. 85,200 ÷ 4 = 21,300, which exceeds 20,000 — Yes. The extra step is where errors enter: candidates divide by the total number of rows in the table rather than the number in the subset.
- Aggregate of a computed column. Sometimes the quantity you need is not in the table — visitors per ranger, species per hundred hectares. Compute it for the qualifying rows only. Computing it for all eight rows when you need four is the most common way to run out of time in this format.
A statement with every, each, all or no makes a claim about a whole group. And there is an asymmetry that makes these fast if you approach them correctly:
To prove it, you must check every qualifying row. To disprove it, you need exactly one. So do not verify — hunt.
Qualifying rows — more than 100 species: Brackenmoor (118), Dunmarsh (141), Fernhollow (127), Hazelbank (103). Their rangers: 6, 3, 5, 6.
Dunmarsh has 3. That is fewer than 5. No — and you were done the moment you saw it. There was no need to check Fernhollow or Hazelbank.
Where to hunt first. Sort by the column in the claim and start from the end that would violate it. Here the claim is "at least 5 rangers", so sort rangers ascending and start at the bottom — the smallest values are where a counterexample lives. You will find it in seconds or establish there is none.
And the mirror image. A statement beginning "No reserve with…" is also universal, just negated. Same method: one counterexample kills it. "No reserve with fewer than 300 hectares has more than 20,000 annual visitors" — Cranefield is 210 hectares with 31,200 visitors. Dead in one row.
The three shapes above are the ones most preparation material covers. This fourth family is the one candidates are least ready for, because it asks you to compute a property of the set rather than to look something up.
The simplest: largest minus smallest, in one column. "The range of bird species recorded is greater than 85." Sort species: 51 (Grayling) at the bottom, 141 (Dunmarsh) at the top. 141 − 51 = 90, which is greater than 85. Yes.
Statements comparing two ranges appear often — "the range of X is greater than the range of Y" — and they need two sorts and two subtractions. Nothing subtle, but four operations where candidates usually do two.
Median requires counting, and the count is where it goes wrong.
"The median number of annual visitors is greater than 15,000."
Sort visitors ascending: 4,100 · 6,300 · 8,900 · 12,400 · 15,600 · 18,900 · 22,700 · 31,200.
Eight values, so the median is the average of the fourth and fifth: (12,400 + 15,600) ÷ 2 = 14,000. That is not greater than 15,000. No.
The first error is taking the fourth value alone, or the fifth alone — with an even count there is no single middle row. The second is confusing median with mean: the mean here is 15,012.5, which is above 15,000, and would give the opposite answer.
"More than half the reserves have fewer than 100 bird species." Count them: Alderwood (82), Cranefield (64), Elmshaw (95), Grayling (51). That is four rows out of eight. Four out of eight is exactly half. "More than half" requires five. No.
This is the most common single error in the entire format, and it is not an arithmetic error. It is a reading error about one word. Build the reflex now:
| Phrase | Includes the boundary? |
|---|---|
| more than 100 | no — 100 does not qualify |
| at least 100 | yes |
| fewer than 100 | no |
| at most 100 / no more than 100 | yes |
| more than half of 8 | no — 4 is not more than half |
| at least half of 8 | yes |
Whenever a statement lands exactly on a boundary, that is not a coincidence. The item was built around it.
Every error this chapter has described, in one place. When a Table Analysis item costs you a point, it is almost always one of these eight. Open each to see the shape it takes and the defence against it.
The practice for this chapter is on its own page, and every statement is scored separately: you answer one, you find out whether you were right, and you read why before moving on.
That is not how the exam scores a Table Analysis — there it is all three statements or nothing, and the mixed sets later in the course work that way too. Here the point is to find out which of the three decisions you get wrong, and a single verdict on all three would hide exactly that.
Read the whole statement before you look at the table. Find the filter, sort the column that answers the question, and treat every boundary word as deliberate — because it is.
I2 covers Two-Part Analysis, where the all-or-nothing scoring of this chapter applies again — two columns instead of three statements, and the same rule about what partial effort is worth.