Excellent scores on all three sections of the BMAT (BioMedical Admissions Test) are required by students hoping to study medicine at Oxford, Cambridge, Imperial, Keele, UCL, Lancaster, Leeds, Brighton & Sussex, and Manchester

These are the simple solutions, with easy to follow steps, written using concise language, to the complex BMAT questions that IST students have found to be the most difficult.

If you have a specific question that you would like answered, you may send it to us, and we will try to post the solution as quickly as possible.

## BMAT 2018, Section 1

### Question 1 Solution – B – 42

3.2 km

Seat every 400 m

2 bins next to each seat

1 bin every 100 m

Seat b1 – 100m – b2 – 100m – b3 – 100m -b4 – 100m – b5 – seat

So for each segment there will be 5 bins x (3.2/0.4) = 40

Plus the first and last bins = 2

Total = 42

### Question 18 Solution – E – 50

Mathematics A-Level

Order does not matter, so we must use the Combination formula C(n,r) = (n!)/(r!*(n-r)!).

The maximum number of red skittles is 4, then 2 yellow skittles must be chosen.

If there are 3 red skittles, then 3 yellow skittles.

If 2 red, then 4 yellow.

We must calculate each combination separately and add their subtotals, as shown in the table below.

 Red Yellow C(n,r) = (n!)/(r!*(n-r)!) Sub total 4 2 (6!)/(4!*(6-4)!) 15 3 3 (6!)/(3!*(6-3)!) 20 2 4 (6!)/(2!*(6-2)!) 15 total 50

If you can not remember the Combination formula, it will take too much time to calculate this answer, so just guess randomly.

### Question 23 Solution – D – 4

Mathematics Logic Puzzle, Trial and Improvement

We should try one of the middle numbers because it may be possible to eliminate half the choices, if it is not correct.

 Trial 1st 2nd 3rd 4th 1 5 6 2 5 = 10 1 6 = 6 2 1 3 = 4 2 2 = 4 1 2 2 3

### Question 25 Solution – B – \$27

Mathematics Algebra Logic

200 – 100 = 100

100 / 4 = 25

 2nd highest lowest total 25 25 25 25 100 26 25 25 24 100 27 25 25 23 100 27 26 24 23 100

### Question 31 Solution – A – 10

Mathematics Algebra, Trial and Improvement

D + C + R + E = 48

D – C = 2

C = D – 2

D – E = 11

E = D – 11

D + C + R + E = 48

D + (D – 2) + R + (D – 11) = 48

3D + R – 13 = 48

3D + R = 61

Try 12

3D + 12 = 61

3D = 49

D = 49/3

NO

Try 10

3D + 10 = 61

3D = 51

D = 17

YES