Read the units before doing the arithmetic. The test supplies a basic on-screen calculator.
Start the free questionsTables, charts, percentages, rates and multi-step calculations.
These are original Cutline questions. They are not taken from the official UCAT ANZ question bank or an exam. The clock below gives each set 43 seconds per question, based on the published 2026 timing. The 2027 format has not yet been published.
Read the Quantitative Reasoning tips or check how scaled scores compare with the 2026 deciles.
A fictional community library records loans in three months. Adult and child loans are separate counts.
march adult renewal rate: 20 %
march child renewal rate: 30 %
march open days: 20 days
The March renewal rates are the shares of loans in each group that were renewals.
What percentage of January loans were child loans?
January had 240 adult loans and 160 child loans, giving 400 loans in total. The child share is 160 / 400 x 100 = 40%. Option B calculates the adult share instead: 240 / 400 x 100 = 60%.
Common trap: of vs more than.
By what percentage did adult loans increase from January to February?
Adult loans rose from 240 in January to 270 in February, an increase of 30 loans. Divide that increase by the original January count: 30 / 240 x 100 = 12.5%. Option E uses the whole February count, 270 / 240 x 100 = 112.5%, rather than the increase.
Common trap: wrong base.
What percentage of all March loans were renewals, when the adult and child renewal rates are applied to their respective loan counts?
Apply the two renewal rates to their own groups: 330 x 0.20 = 66 adult renewals, and 270 x 0.30 = 81 child renewals. That is 147 renewals among 330 + 270 = 600 March loans, giving 147 / 600 x 100 = 24.5%. Option B's 25.0% is an unweighted average of the two rates, despite the different group sizes.
Common trap: chain step missed.
In March, how many loans were recorded per open day, to the nearest whole loan?
March's 330 adult and 270 child loans total 600 loans. Divide by 20 open days: 600 loans / 20 days = 30 loans per open day. Option C omits child loans and gives 330 / 20 = 16.5, which rounds to 17 loans per day.
Common trap: chain step missed.
A fictional packing workshop records four batches. Batch D's packing time is missing. Packaging costs AUD 1.50 per carton, plus one AUD 8 handling fee for each batch.
total minutes: 200 min
batch count: 4 batches
fee per carton: 1.50 AUD/carton
handling fee per batch: 8 AUD/batch
The four batches took 200 minutes in total. The packaging charge is AUD 1.50 per carton, plus one AUD 8 handling fee per batch.
How many units were packed per carton in batch B?
Use the unit count and carton count from the same batch: 252 / 21 = 12 units per carton. Option A uses batch A's carton count, giving 12.6 before rounding.
Common trap: chain step missed.
What was the mean number of units packed per batch?
A mean across batches uses all four unit counts and divides their sum by four. The total is 1,188 units, which gives 297 units per batch. Option B finds the median instead.
Common trap: wrong statistic.
What was the median packing time across the four batches?
First subtract the three recorded times from 200 minutes to find batch D's 68 minutes. Ordered times are 36, 44, 52 and 68 minutes, so average the two middle times. Option E gives the missing time rather than the median.
Common trap: wrong statistic.
What is the packaging charge for batch D?
Multiply 30 cartons by AUD 1.50, then add the handling fee once for the batch. The carton charge is AUD 45, so the total is AUD 53. Option A leaves out the handling fee.
Common trap: modifier skipped.
A fictional honey packer labels one batch of jars. The pie chart shows the number of jars with each label design.
honey grams per jar: 250 g/jar
grams per kilogram: 1000 g/kg
labels per jar: 2 labels/jar
rejected share: 0.2 1
labels per sheet: 25 labels/sheet
Each jar contains 250 g of honey.
There are 1000 g in 1 kg.
Each jar needs two labels. Each sheet prints 25 labels, and 20% of printed labels are rejected.
What percentage of the jars have a coastal label, to the nearest whole percent?
Add the five design counts to get 300 jars. Coastal accounts for 75 / 300 x 100 = 25% of them. Option A uses the 225 jars outside coastal as the denominator instead of the full batch.
Common trap: wrong base.
What is the combined mass of honey in the bush blossom and coastal jars, in kilograms?
Bush blossom and coastal account for 110 + 75 = 185 jars. At 250 g per jar, that is 46,250 g, or 46.25 kg after dividing by 1000. Option D uses meadow's 55 jars in place of coastal's 75.
Common trap: chain step missed.
How many sheets' worth of labels must be printed to leave two usable labels for each coastal and meadow jar, if 20% of printed labels are rejected?
The 130 coastal and meadow jars need 260 usable labels. Since 80% of printed labels remain usable, divide 260 by 0.8 to get 325 labels to print. At 25 labels per sheet, this is 13 sheets' worth. Option C adds 20% to 260, but the rejected 20% is a share of the printed total.
Common trap: modifier skipped.
What is the median of the five jar counts, one for each label design?
The five jar counts in ascending order are 20, 40, 55, 75 and 110. The middle count is 55 jars, option C. Option A finds the mean instead: the total of 300 divided by five is 60.
Common trap: wrong statistic.
Choose your approach before you answer.
A fictional community pool records booked lane-hours and swimmer visits on six days.
How many swimmer visits were there per booked lane-hour on Wednesday?
You chose to before answering.
Wednesday's point shows 60 swimmer visits and 20 booked lane-hours. Divide 60 by 20 to get 3.0 visits per lane-hour, option B. Option A uses Thursday's point instead.
A fictional community pool records booked lane-hours and swimmer visits on six days.
Across Thursday and Friday combined, how many swimmer visits were there per booked lane-hour, to 2 decimal places?
You chose to before answering.
Thursday and Friday have 84 + 104 = 188 visits and 24 + 26 = 50 booked lane-hours. Divide the combined totals: 188 / 50 = 3.76 visits per lane-hour, option C. Option B averages the two daily rates equally instead of pooling the visits and lane-hours.
A fictional community pool records booked lane-hours and swimmer visits on six days.
For the three days with the most booked lane-hours, what was the mean of their separate daily visits per booked lane-hour, to 2 decimal places?
You chose to before answering.
The three highest booked lane-hour values are 24, 26 and 30, on Thursday, Friday and Saturday. Their separate rates are 84 / 24 = 3.5, 104 / 26 = 4 and 90 / 30 = 3. The mean is (3.5 + 4 + 3) / 3 = 3.50 visits per lane-hour, option D. Option A pools the visits and lane-hours, which weights the days by their bookings.
A fictional community pool records booked lane-hours and swimmer visits on six days.
What was the median number of booked lane-hours across the six days, to 1 decimal place?
You chose to before answering.
The ordered lane-hour values are 12, 18, 20, 24, 26 and 30. With six values, average the middle two: (20 + 24) / 2 = 22.0 lane-hours, option E. Option C calculates the mean instead of the median.
Review the explanation for every item, including ones you got right. Keep track of the traps that caught you, then try the Quantitative Reasoning tips before another timed set. See all UCAT practice and guides.
