Unit 7
Data and Percentage
About this unit
Read and interpret tables, bar charts, pie charts, and surveys — then use the data to calculate percentages, make comparisons, and answer multi-step numerical questions. This unit covers everything from reading a simple graph to computing layered percentage breakdowns.
What types of questions will you face?
- 1Calculate the percentage of a group from a bar or pie chart
- 2Compute the overall percentage when different-sized groups have different rates
- 3Find the "most improved" student or item using a ranked comparison table
- 4Determine a minimum score needed to guarantee winning a competition given partial results
- 5Read multi-column survey data and perform arithmetic across groups to find totals or percentages
Skills you will build
- Reading and extracting data accurately from tables, bar charts, and sector graphs
- Calculating percentages from fractions and whole numbers
- Combining percentages from groups of different sizes (weighted averages)
- Applying arithmetic and comparison logic to ranked data
- Performing multi-step calculations without a calculator
By the end of this unit, you will be able to
- Accurately extract numerical data from any graph or table format
- Calculate percentages across different group sizes without confusion
- Solve "minimum score needed" type competition problems by working backwards
- Interpret data to draw valid conclusions and avoid misleading readings
Difficulty profile
Easy difficulty (avg 2.46). Most questions are straightforward percentage and reading problems. The harder questions involve combining data from multiple-sized groups.
Exam tip: Data and Percentage
For percentage of a combined group: always multiply the proportion by the GROUP SIZE before adding. Never average percentages directly — a group of 1,000 at 50% contributes far more than a group of 10 at 90%.
Sample Questions
11 students, 2 jumps each — the longer jump counts. Student 6 turned a modest 4.6m into a 5.7m second jump. Student 9 peaked at 5.5m on their first attempt. One slip in the ranking and you'll call the wrong 5th place.
Table-reading and ranking questions appear regularly at the easier end of NSW Selective TS — they reward students who read the scoring rule carefully and process every row systematically before ranking.
The examiner tests whether students can apply a simple rule (take the maximum of two values), process all rows without skipping, and correctly sort a list — especially when the best attempt is the first one (a common trap).
A table gives two values per student (two attempts, two scores, two measurements). The 'result' is the larger (or smaller) value. Students calculate each result, rank all students, and identify who is in a specific position. Options include students near the target rank to test careful reading.
Best approach: Step 1: Read the rule — here it's max(attempt 1, attempt 2). Step 2: Go through every row and write the result next to it (don't skip any student). Step 3: Sort all results from highest to lowest. Step 4: Count down to the target position. Watch out: the best jump may be the FIRST attempt (students 7, 9) — don't automatically use attempt 2.
Question
In a competition, 11 students were allowed two attempts each at jumping as far as possible. The longer distance in the two attempts was counted as their ‘result’. The attempts for the 11 students are shown in the table.
| student number | attempt 1 (in metres) | attempt 2 (in metres) |
|---|---|---|
| 1 | 4.9 | 5.0 |
| 2 | 5.7 | 5.8 |
| 3 | 4.8 | 4.9 |
| 4 | 4.6 | 4.7 |
| 5 | 5.0 | 5.1 |
| 6 | 4.6 | 5.7 |
| 7 | 5.6 | 5.4 |
| 8 | 5.1 | 5.3 |
| 9 | 5.5 | 4.8 |
| 10 | 4.4 | 4.5 |
| 11 | 5.9 | 5.2 |
Which student came 5th in the competition?
- Astudent 7
- Bstudent 8
- Cstudent 9
- Dstudent 11
Decided on your answer? Check how you went below.
Five made-up coins in Oceantown: azzes, bayos, cotts, duits and enits. Four conversion rates are given. How many of the smallest coin fit into the largest? Chain the multiplications one step at a time.
Chained unit conversion questions appear regularly on NSW Selective TS Mathematical Problem Solving. The values always chain neatly — each step is a single multiplication. The trap is skipping a step or multiplying in the wrong direction.
The examiner tests whether students can (a) read each conversion in the correct direction (bigger unit = more of the smaller), (b) chain four multiplications without error, and (c) avoid the trap of stopping too early (after only three steps, you get 16 — option A — which is wrong).
A fictional currency (or measurement) system gives a chain of conversions from smallest to largest unit. Students must find how many of the smallest unit fit into the largest by multiplying step by step.
Best approach: Work from the smallest unit up: 1 bayo = 4 azzes. 1 cott = 2 × 4 = 8 azzes. 1 duit = 2 × 8 = 16 azzes. 1 enit = 4 × 16 = 64 azzes. Multiply at each step and don't stop early.
Question
In Oceantown there are five different coins: azzes, bayos, cotts, duits and enits.
There are 4 azzes in 1 bayo, 2 bayos in 1 cott, 2 cotts in 1 duit, and 4 duits in 1 enit.
How many azzes are in 1 enit?
- A16
- B32
- C64
- D128
Decided on your answer? Check how you went below.
60 legs in a dog park, 8 owners — some with 1 dog, some with 2. No one tells you how many of each, but two simple equations crack it. The secret: every extra dog adds exactly 4 legs.
Leg-count word problems with two unknowns appear regularly in NSW Selective TS at medium difficulty — they are disguised simultaneous equations that reward students who can translate a story into algebra.
The examiner tests whether students can set up two equations from a word problem (one for totals, one for leg counts), solve them, and verify the answer — rather than guessing or trial-and-error through all four options.
A scene describes two groups of people/animals, each contributing a different number of legs. The total count and total legs are given. Students form a system of two equations (count + legs) to find how many are in each group.
Best approach: Step 1: name your two unknowns (x = owners with 1 dog, y = owners with 2 dogs). Step 2: write the count equation (x + y = 8). Step 3: write the legs equation — calculate owner legs separately (fixed), then add dog legs (4x + 8y). Step 4: subtract the two equations to eliminate x and find y. Step 5: check by plugging back in. Shortcut: assume all owners have 1 dog, compute legs, then see how many extra legs remain — divide by 4 to find how many had 2 dogs instead.
Question
Yesterday, in the dog park, I saw 8 dog owners exercising their dogs.
Some of the dog owners had 1 dog, and some dog owners had 2 dogs.
Counting the legs of the dogs and the legs of the dog owners, there were a total of 60 legs.
All dogs have 4 legs and all dog owners have 2 legs.
How many dog owners had only 1 dog?
- A3
- B4
- C5
- D6
Decided on your answer? Check how you went below.
A zoo has snakes (0 legs), horses (4 legs), and eagles (2 legs). Nine animals, 10 legs total — what is the highest possible number of snakes? The trick: maximising one type means minimising the others, and you can only try a handful of combinations.
Maximise/minimise questions with simultaneous constraints appear regularly on NSW Selective TS Mathematical Problem Solving. They are distinct from standard equation questions because students must not only satisfy the equations but also optimise — choosing values that push one variable as high (or as low) as possible.
The examiner checks whether students can (a) convert the word problem into two equations, (b) recognise that maximising snakes means minimising horses + eagles, (c) apply the constraint that each animal type must appear at least once, and (d) test valid combinations starting from the minimum rather than guessing from the answer options.
Three quantities are linked by two linear equations (total count and total legs/cost/weight). One quantity contributes zero to the second equation (snakes have no legs), making it the easiest to maximise. Students must find valid whole-number solutions and then identify which gives the largest (or smallest) value of the target quantity.
Best approach: Write two equations: S + H + E = 9 and 4H + 2E = 10. Simplify the second to 2H + E = 5. To maximise S, minimise H + E. Try H = 1 (E = 3, S = 5) then H = 2 (E = 1, S = 6). H = 3 gives E = -1, which is invalid. Answer: 6. Always verify by checking both equations and confirming all values are at least 1.
Question
At a zoo there are snakes, horses and eagles.
Snakes have no legs, horses have 4 legs, and eagles have 2 legs.
There are 9 of these animals altogether with 10 legs in total.
What is the highest number of snakes that can be at the zoo?
- A4
- B5
- C6
- D7
Decided on your answer? Check how you went below.
Three piles of tokens — 2-point, 5-point, 10-point — all with the same count. Two piles together make 96. Which two? The answer must give a whole number of tokens, so only one pairing works.
Equal-count pile problems requiring systematic case testing appear regularly in NSW Selective TS Mathematical Problem Solving. The key constraint (n must be a whole number) eliminates all but one case, giving a unique answer. Recognising this constraint is the main skill being tested.
The examiner tests whether students can (a) represent pile totals algebraically (2n, 5n, 10n), (b) systematically test all three possible pairs of piles against the sum constraint, (c) reject pairs where n is not a whole number, and (d) calculate the remaining pile's total using the valid n.
Three groups each have n items worth different amounts. The combined total of two groups is given. Students test all three pairs, reject those giving non-integer n, and find the remaining group's total.
Best approach: Let n = number of tokens per pile. Write pile totals as 2n, 5n, 10n. Test each pair: (i) 2n+5n=7n=96 → n=13.7 ✗; (ii) 2n+10n=12n=96 → n=8 ✓; (iii) 5n+10n=15n=96 → n=6.4 ✗. Only pair (ii) gives n=8. Remaining pile = 5×8 = 40.
Question
In a game called 'Splott' there are tokens that are worth 2 points, 5 points or 10 points.
I have one pile of 2-point tokens, one pile of 5-point tokens and one pile of 10-point tokens.
Each pile contains the same number of tokens.
The combined total of two of the piles of tokens is 96 points.
What is the total of the other pile of tokens?
- A32 points
- B40 points
- C48 points
- D56 points
Decided on your answer? Check how you went below.
21 people vote on 5 movies. Kate says the winner needs at least 5 votes. Samira says 11 votes is a guaranteed win. To check Kate, you need to prove it's impossible for all movies to get 4 or fewer votes — a classic pigeonhole argument.
Voting minimum/majority problems appear in NSW Selective TS Mathematical Problem Solving. They test two complementary skills: proving the minimum winning vote count (Kate — pigeonhole) and proving a majority vote guarantees a win (Samira — majority). Questions ask whose reasoning is correct.
The examiner tests whether students can (a) prove Kate's claim by contradiction (5 movies × 4 max votes = 20 < 21, so winner needs ≥ 5), and (b) verify Samira's claim by checking 11 votes leaves only 10 for all others combined, making 11 unbeatable.
A vote-counting scenario gives total voters and number of options. Two people make claims — one about the minimum required to win, one about a vote count that guarantees winning. Both claims are correct and must be verified by arithmetic reasoning.
Best approach: For Kate: assume winner has 4 votes → all 5 movies max 4 votes → total ≤ 20 < 21 → contradiction → winner needs ≥ 5. For Samira: 11 votes → only 10 remain for 4 others → no other movie can reach 11 → 11 guarantees a win.
Question
| At the start of the meeting, every member of the movie club gets one vote on the movie to be watched. Everyone must use their vote. The movie that gets the most votes is the one that is watched that evening. This evening, there are 21 people at the meeting, and there are 5 movies to choose from. |
|---|
Kate: "The movie that we watch will need to get at least 5 votes."
Samira: "If a movie gets 11 votes then we will definitely be watching it."
If the information in the box is true, whose reasoning is correct?
- AKate only
- BSamira only
- CBoth Kate and Samira
- DNeither Kate nor Samira
Decided on your answer? Check how you went below.
Data questions often start with a table: read every row carefully, then do one clear calculation — no guessing from the options.
Chart- and table-reading items appear regularly on Selective TS; many wrong answers come from misreading a single cell or skipping a row.
The examiner checks that you can extract the right numbers from a table and combine them accurately — separating “read the data” from “do the maths.”
A table lists values by category (e.g. books borrowed per month). You find a total, difference, or average using only the figures given.
Best approach: Pass 1: copy or underline each value you need. Pass 2: add (or compare) systematically. Double-check you included every row the question asks for.
Question
The table shows the number of library books borrowed each month (January–June):
| Month | Books borrowed |
|---|---|
| January | 18 |
| February | 12 |
| March | 15 |
| April | 9 |
| May | 18 |
| June | 6 |
What is the total number of books borrowed across all six months?
- A72
- B78
- C84
- D90
Decided on your answer? Check how you went below.
Harder data items chain several steps: group sizes, fractions or ratios, then different percentages on different totals.
Multi-step survey problems sit in the harder third of Selective mocks — distractors are often the result of stopping one step early.
You must infer a missing group size, apply unlike fractions to each group, find supporters vs non-supporters, then apply two percentages to the correct bases and add only the final categories asked for.
A population is split across districts with different support rates; one district’s size is “the remainder.” Two follow-up percentages apply to different subgroups.
Best approach: Work in columns: district → size → supporters → total supporters → non-supporters. Apply each % to its own base. Add only at the end; trap answers usually omit the second group.
Question
A survey of 5,400 people across four districts asked whether they supported building a new community centre.
- District A had 1,200 people, and 3 out of every 4 supported the proposal.
- District B had 1,800 people, and 5 out of every 6 supported the proposal.
- District C had 900 people, and 2 out of every 9 supported the proposal.
- District D had the remaining people, and 2 out of every 3 supported the proposal.
If 35% of those who supported the proposal were senior citizens and 60% of those who did NOT support the proposal were aged under 18, what is the total number of people in these two categories?
What is the total number of people in these two categories?
- A1 080
- B1 260
- C1 800
- D2 340
Decided on your answer? Check how you went below.
To guarantee a prize, your target student must be able to score high enough that even the best possible opponent score cannot beat them — and a tie doesn't count as winning.
Guarantee/minimum score questions appear in the harder half of Selective TS papers and combine table reading with reverse-calculation — a reliable source of marks once you know the method.
The examiner tests whether you can identify the one student who can mathematically guarantee the prize, find their worst-case rival, and calculate the exact minimum score needed — not forgetting the tie rule.
A table shows scores for multiple students across two completed tests. A third test is upcoming. A prize goes to the highest total; ties mean nobody wins. One student is flagged as able to guarantee the prize — find the minimum score they need in the final test.
Best approach: Step 1: sum all current totals. Step 2: the only candidate who can guarantee the prize is the student with the highest current total (everyone else faces a potential leader who could beat them even at 100). Step 3: find the second-highest current total, add 100 — that is the worst-case threat. Step 4: solve for the minimum score that beats it strictly (remember: ties are losses).
Question
Twelve students are studying Mathematics. They take three tests, each marked out of 100.
The results for the first two tests taken are shown below.
| Test 1 | Test 2 | |
|---|---|---|
| Tom | 48 | 53 |
| James | 83 | 86 |
| Alira | 61 | 67 |
| Connor | 73 | 54 |
| Edward | 66 | 59 |
| Heidi | 69 | 78 |
| Jo | 33 | 98 |
| Kai | 88 | 44 |
| Sarah | 100 | 53 |
| Piper | 75 | 78 |
| Layla | 81 | 79 |
| Isabelle | 57 | 65 |
A prize is awarded to the student with the highest combined total score from the three tests. If two students are tied for the highest score, nobody gets the prize.
As the scores currently stand, one student is able to be sure to win this prize as long as a high enough score is achieved in the final test.
What is the minimum score in the final test that will guarantee this student the prize?
- A92
- B94
- C96
- D98
Decided on your answer? Check how you went below.
A six-digit bike lock code. The middle two digits are the product of the first two. The last two are double the middle two. The first digit is 7 and all six digits are different. Only one value of the second digit makes everything work.
Digit constraint puzzles — where you must find unknown digits satisfying multiplication or doubling rules plus an all-different constraint — appear regularly in NSW Selective TS Mathematical Problem Solving.
The examiner checks whether students can (a) correctly interpret 'two-digit number formed by the middle two digits' as a two-digit positional number (not a sum), (b) set up the relationships as equations, (c) narrow the candidate range for the unknown digit, and (d) systematically test each candidate until all constraints are satisfied.
A multi-digit code is described using multiplication, doubling, or other arithmetic relationships between groups of digits. An all-different constraint eliminates most candidates. Students test each feasible value of the key unknown digit and identify the unique valid solution.
Best approach: Label digits D1–D6. D1=7 is given. D3D4 = 7 × D2 (must be a 2-digit number, so D2 ≥ 2). D5D6 = 14 × D2 (must also be 2-digit, so D2 ≤ 7, and D2 ≠ 7). Candidates: D2 ∈ {2,3,4,5,6}. Test each systematically and apply the all-different rule. Only D2=4 gives 7-4-2-8-5-6 with no repeats. Last digit = 6.
Question
I have a six-digit code for my bike lock.
The two-digit number formed by the middle two digits was the result of multiplying the first digit by the second digit.
The two-digit number formed by the last two digits of the code was double the number formed by the middle two digits.
The first digit was 7 and all six digits were different.
What is the last digit of the code for my bike lock?
- A2
- B4
- C6
- D8
Decided on your answer? Check how you went below.
Give Your Child the Best Chance at Selective Entry
Join NSW families preparing their children for the Selective Schools Placement Test with the most realistic online Selective practice tests available. First tests free—no credit card required.
Claim Your Free Selective Practice Tests