Welcome to our collection of UCAT Quantitative Reasoning practice questions. Designed to help you prepare for the actual UCAT exam format, these examples will help you improve your numerical reasoning skills, speed, and accuracy under time pressure. Each question has a clear, step-by-step solution, ensuring you understand the concepts and strategies needed to succeed. By practising regularly, you’ll develop the confidence and efficiency required to excel in this critical section of the UCAT and secure a place in your dream medical or dental school. Get started today and take a step closer to UCAT success!

Please note that you will have a choice of answers in the actual exam. These questions are designed to assist you in starting your revision. Blue Peanut does not have question banks – we will help you in our course with selecting question banks.

Scenario 1: Hospital Ward Occupancy

Data: A hospital ward records the following data over a 4-day period:

Note: Beds Occupied by Evening = Beds Occupied in Morning + New Admissions - Discharges

Question 1:

On which day did the ward have the highest daily turnover rate, defined as (New Admissions + Discharges) per occupied bed in the morning?

Worked Solution:

Monday: Turnover = (15 + 10) / 62 = 25/62 ≈ 0.403
Tuesday: Turnover = (12 + 9) / 67 = 21/67 ≈ 0.313
Wednesday: Turnover = (8 + 15) / 70 = 23/70 ≈ 0.329
Thursday: Turnover = (10 + 8) / 63 = 18/63 ≈ 0.286

The highest turnover is on Monday (approximately 0.403).

Answer: Monday.

Question 2:

What percentage of the total admissions over the 4-day period occurred on Wednesday?

Worked Solution:

Total admissions over 4 days = (15 + 12 + 8 + 10) = 45
Admissions on Wednesday = 8
Percentage = (8 / 45) × 100% ≈ 17.78%

Answer: About 17.8%.

Scenario 2: Travel Times Between Cities

Data: The table below shows travel times between three cities (A, B, and C) using two different modes of transport: Car and Train.

Assumptions:

Car speed and train speed remain constant.
Distances: A–B = 210 km, B–C = 300 km, A–C = 450 km.
Car speed: 70 km/h.
Train speed: 90 km/h.

Question 3:

You need to travel from A to C and have a connecting meeting in B for 30 minutes. If you start in A and must arrive in C as quickly as possible, what is the fastest total travel time if you choose optimally between car and train for each leg and include the 30-minute stop at B?

Worked Solution: Options to consider:

A–B by Car + 30 min + B–C by Car
= 3h (A–B car) + 0.5h (meeting) + 4.29h (B–C car)
= 7.79 hours total
A–B by Car + 30 min + B–C by Train
= 3h + 0.5h + 3.33h
= 6.83 hours total
A–B by Train + 30 min + B–C by Car
= 2.33h + 0.5h + 4.29h
= 7.12 hours total
A–B by Train + 30 min + B–C by Train
= 2.33h + 0.5h + 3.33h
= 6.16 hours total

Fastest = 6.16 hours (A–B by Train, then B–C by Train).

Answer: Approximately 6.16 hours.

Question 4:

If a traveller chooses to go directly from A to C by train without stopping at B, by what percentage is this journey shorter in time than making the trip A–B–C by train with the 30-minute stop?

Worked Solution:

Direct A–C by train = 5 hours
A–B–C by train with 30 min stop = 2.33h (A–B) + 0.5h (stop) + 3.33h (B–C) = 6.16h

Difference in time = 6.16h - 5h = 1.16h

Percentage reduction = (1.16 / 6.16) × 100% ≈ 18.83%

Answer: About 18.8%.

Scenario 3: Supermarket Discounts

Data: A supermarket runs a promotion on a range of items:

Item A: Original price = £6, Discount = 25%
Item B: Original price = £12, Discount = 10%
Item C: Original price = £20, Discount = Buy one at full price, get a second at 50% off
Item D: Original price = £3.50, Discount = 3 for the price of 2

Question 5:

A customer buys 2 units of Item A, 2 units of Item C, and 3 units of Item D during the promotion. What is the total cost?

Worked Solution:

Item A (2 units): Original price per unit = £6; 25% discount means price after discount = £6 × (1 - 0.25) = £4.50 per unit. For 2 units: 2 × £4.50 = £9.00.
Item C (2 units): First at full price (£20), second at 50% off (£20 × 0.5 = £10). Total for 2 units: £20 + £10 = £30.
Item D (3 units): Original price per unit = £3.50. The deal is “3 for the price of 2” means pay for only 2 units. Cost for 3 units = 2 × £3.50 = £7.00.

Total cost = £9.00 (A) + £30 (C) + £7.00 (D) = £46.00.

Answer: £46.00

Question 6:

What is the overall percentage saving on Item A compared to its original total cost (for the 2 units bought)?

Worked Solution:

Original total cost for 2 units of A without discount: 2 × £6 = £12.
Discounted total cost: £9 (as calculated above).
Savings: £12 - £9 = £3.

Percentage saving = (£3 / £12) × 100% = 25%.

Answer: 25%.

Scenario 4: Patients’ Diagnostic Test Results

Data: A clinic performed a new diagnostic test on 400 patients. Results are as follows

Definitions:

Sensitivity = True Positive / (True Positive + False Negative) = Test’s ability to detect illness.
Specificity = True Negative / (True Negative + False Positive) = Test’s ability to detect health.

Question 7:

What is the sensitivity of the test?

Worked Solution: True Positive (TP) = 60
False Negative (FN) = 15
Sensitivity = TP / (TP + FN) = 60 / (60 + 15) = 60/75 = 0.8 = 80%

Answer: 80%.

Question 8:

If the test is given to 1,200 new patients with the same prevalence of disease (75/400 = 18.75%), how many would you expect to test positive in total, assuming the same sensitivity and specificity?

Worked Solution:

Prevalence = 18.75% means out of 1,200 patients, expected sick = 18.75% of 1,200 = 0.1875 × 1200 = 225 patients sick, 975 healthy.
Sensitivity = 80%: True positives among 225 sick = 225 × 0.8 = 180.
Specificity: First find specificity. True Negative (TN) = 285 out of 325 healthy. Specificity = 285/325 ≈ 0.8769 or about 87.7%.
For 975 healthy patients, true negatives = 975 × 0.8769 ≈ 855, false positives = 975 - 855 = 120.
Total positives = True positives + False positives = 180 + 120 = 300.

Answer: About 300 patients.

MORE UCAT RESCOURCES