BMAT Solutions |

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

Try 10

3D + 10 = 61

3D = 51

D = 17

YES