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BMAT 2003S1 
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This section is Section 1 of 3.
Speed as well as accuracy is important in this section. Work quickly, or you might not finish the paper. There are no penalties for incorrect responses, only marks for correct answers, so you should attempt questions. Each question is worth one mark.
You must complete the answers within the time limit. Calculators are NOT permitted.
Good Luck!
A farmer has an underground water tank which he decided to calibrate by adding known volumes of water and measuring the depth using a dipstick. His calibration graph is shown The horizontal cross section of the tank is circular at all points.
The correct answer is E.
Initially, the gradient of the graph is increasing, which means there is an increasingly greater change in depth per unit change in volume. This means the width of the tank must be decreasing from the bottom. The gradient of the graph then decreases as we fill to the the top of the tank, which means there is a smaller change in depth per unit change in volume, so the tank must be getting narrower. Only E shows the tank getting narrower and then wider (from bottom to top), so this must be the answer.
“Ready meals” should be labelled with health warnings in the same way as cigarettes There has been a big rise in the number of ready meals being sold by the supermarkets. These meals are quick and easy to use, but they tend to have high levels of salt, fat and preservatives in them, which are not good for our long term health.
The correct answer is C.
The passage concludes that ‘“Ready meals” should be labelled with health warnings,’ as they ‘tend to have high levels of salt, fat and preservatives in them, which are not good for our long term health.’
Statement C strengthens the conclusion of the passage as it states that people who buy ready made meals are unaware of their unhealthy properties, thus supporting
labelling such foods with health warnings.
Statements A, B, D & E are not explicitly mentioned and are therefore irrelevant to the argument.
A survey has been carried out of the methods of transport to school used by pupils. The results, broken down by year group, are shown below.

Year 7  Year 8  Year 9  Year 10  Year 11  Total 
Car  30  33  16  18  10 
102 
Bus 
14  16  13  15  18  76 
Bicycle  5  12  23  25  30 
95 
Walk 
101  89  100  92  108  490 
Total  150  145  152  150  166 
763 
One of the individual entries in the above table has been typed incorrectly, although the marginal totals are correct.
The correct answer is B.
Use the process of elimination of the answers AE given. For answer A, check that the two totals involving the number 14 are correct: 30+14+5+101=150, therefore all of these values (in the first column of the table) are correct. Now check the row involving the number 14: 14+16+13+15+18=76, so all values in this row must be correct. Use the same method to determine whether B (33) is correct. Horizontally: 33+30+16+18+10=107, but the total for this column given is 102, therefore one of the values in this column must be incorrect. To check whether the incorrect value is 33, find the total of the column containing 33: 33+16+12+89=150, but the total given is 145, therefore 33 is the incorrect value.
The UK government wishes to increase the number of young people from poorer families entering university However, it has recently changed the way in which it provides financial support for students. Whereas it used to provide grants that the student did not need to repay, it now provides loans that the student must repay when he or she enters employment. Research shows that students from poorer families are more likely to be deterred from going to university by the prospect of debt.
The correct answer is C.
The passage concludes by saying that ultimately ‘students from poorer families are more likely to be deterred from going to university by the prospect of debt.’ Therefore it can be inferred that students from richer families are less likely to be deterred from going to university compared to students from poorer families. This is summarised in statement C.
Statement A is incorrect because there is no mention of how poorer families would be ‘less interested’ in going to university based on the increase in debt that would have to be repaid.
Statement B is incorrect because there is no mention of how universities are less willing to help meet government targets.
Statement D is incorrect as the final sentence implies that students from poorer families are more likely to be affected rather than all students being deterred equally.
Statement E is also incorrect because there is no mention of maintenance grants specifically, just simply grants in general.
A gardener decides to lay out his cabbage patch as a square grid. He has a lot of cabbages so cannot be bothered to count them, but lays them out on the ground. On completing his square, he finds he has 9 left over. He then lays them out again with one extra cabbage on each side of the He then finds that he is 12 cabbages short.
The correct answer is C.
Use the answers AD provided. If you subtract 9 from the number of cabbages he has, you must have a square number; similarly if you add 12 to the same number, you must also get a square number. So, try this for each of the options given until you find a suitable answer. For A: 99=0, but we know that the farmer did not start with 0 cabbages, so eliminate A. For B: 1099=100, which is a square number; 109+12=121, which is also a square number. Therefore the answer is C.
A proposal to counteract low turnout by voters in elections is to introduce the option of voting via the Internet. There are a number of practical issues (such as loss of secrecy, fear of abuse of the system) which mean the proposal would have to be carefully controlled. It may be undemocratic by favouring some political parties more than others. Those parties whose voters tend to be young and better educated would be likely to gain more votes than other.
The correct answer is D.
The conclusion here is internet voting might be undemocratic and lead to certain political parties being favoured over others. Political parties and sections of societies are synonymous to each other. This statement is paraphrased in statement D.
The table below shows statistics for consultations by family doctors in the country of Santesia for the years 1995 and 2000. Santesia has a population of approximately 10 million and there are 5000 family doctors (these figures have not changed significantly over the five years between the two surveys).

1995  2000 
Average number of surgery sessions a week  8.5 
8.4 
Average length of a surgery session (min) 
140  165 
Average length of a consultation (min:sec)  9:59 
10:00 
Average time spent on home visits a week (min) 
408  412 
Average length of a home visit (min)  25.5 
25.0 
Average number of patients seen per week 
135 
155 
The correct answer is A.
On average, a surgery session is 140 minutes long, with consultation each lasting 9:59 minutes, which can be rounded to 10 minutes. So the number of consultations (=the number of patients seen) in a surgery session would be: 140/10=14 patients.
The table below shows statistics for consultations by family doctors in the country of Santesia for the years 1995 and 2000. Santesia has a population of approximately 10 million and there are 5000 family doctors (these figures have not changed significantly over the five years between the two surveys).

1995  2000 
Average number of surgery sessions a week  8.5 
8.4 
Average length of a surgery session (min) 
140  165 
Average length of a consultation (min:sec)  9:59 
10:00 
Average time spent on home visits a week (min) 
408  412 
Average length of a home visit (min)  25.5 
25.0 
Average number of patients seen per week 
135 
155 
The correct answer is C (27).
The question again refers to values in the 1995 column. The length of time spent on surgery sessions per week is given by the number of surgery sessions per week multiplied by the average length of a surgery session: 140×8.5 =1190 minutes. The average time spent on home visits per week is given as 408. So in total 408+1190=1598 minutes was spent per week on average on surgery and home visits. To convert this into hours: 1598/60=26.63, which can be rounded to 27 as the nearest hour.
The table below shows statistics for consultations by family doctors in the country of Santesia for the years 1995 and 2000. Santesia has a population of approximately 10 million and there are 5000 family doctors (these figures have not changed significantly over the five years between the two surveys).

1995  2000 
Average number of surgery sessions a week  8.5 
8.4 
Average length of a surgery session (min) 
140  165 
Average length of a consultation (min:sec)  9:59 
10:00 
Average time spent on home visits a week (min) 
408  412 
Average length of a home visit (min)  25.5 
25.0 
Average number of patients seen per week 
135 
155 
The correct answer is C.
The total number of patients seen in 2000 can be given by the average number of patients seen per week multiplied by the number of weeks per year that doctors work: 155×50=7750 patients. The number of patients per doctor on average would be the population divided by the number of doctors: 10000000/5000=2000. So the average patient saw their doctor: 7750/2000=3.875 times, which can be rounded to 4 times.
The table below shows statistics for consultations by family doctors in the country of Santesia for the years 1995 and 2000. Santesia has a population of approximately 10 million and there are 5000 family doctors (these figures have not changed significantly over the five years between the two surveys).

1995  2000 
Average number of surgery sessions a week  8.5 
8.4 
Average length of a surgery session (min) 
140  165 
Average length of a consultation (min:sec)  9:59 
10:00 
Average time spent on home visits a week (min) 
408  412 
Average length of a home visit (min)  25.5 
25.0 
Average number of patients seen per week 
135 
155 
The correct answer is D.
If the number of doctors is cut but the number of consultations stay the same, the patients that were previously seen by the 500 doctors that are no longer working would have to be seen by the 4500 remaining doctors (in addition to their original patients), which means the doctors would have to work for longer on average (in order to see more people). This means we can eliminate AC because they give lengths of time less than the average length of a surgery session in 2000, when we know this actually needs to be more.
There are 500 fewer doctors whose consultations need to be shared between the remaining 4500 doctors, so that means on average each of the doctors remaining will get 500/4500=1/9th of an extra consultation. So the average length of a doctors surgery would increase by 1/9th: (10/9)x165=183.3. So the answer is D
The table below shows statistics for consultations by family doctors in the country of Santesia for the years 1995 and 2000. Santesia has a population of approximately 10 million and there are 5000 family doctors (these figures have not changed significantly over the five years between the two surveys).

1995  2000 
Average number of surgery sessions a week  8.5 
8.4 
Average length of a surgery session (min) 
140  165 
Average length of a consultation (min:sec)  9:59 
10:00 
Average time spent on home visits a week (min) 
408  412 
Average length of a home visit (min)  25.5 
25.0 
Average number of patients seen per week 
135 
155 
The correct answer is A.
Find the fraction of the values given of the value in 2000 over the value in 1995 for each of the options AC, and then do the reciprocal of this to check D. The largest value is the answer. A: 165/140=1.18. B: 10/9.59=1.04. C: 412/408=1.01. D: 25.5/25=1.02. Therefore A is the answer.
The table below shows statistics for consultations by family doctors in the country of Santesia for the years 1995 and 2000. Santesia has a population of approximately 10 million and there are 5000 family doctors (these figures have not changed significantly over the five years between the two surveys).

1995  2000 
Average number of surgery sessions a week  8.5 
8.4 
Average length of a surgery session (min) 
140  165 
Average length of a consultation (min:sec)  9:59 
10:00 
Average time spent on home visits a week (min) 
408  412 
Average length of a home visit (min)  25.5 
25.0 
Average number of patients seen per week 
135 
155 
The correct answer is A.
The average number of patients seen per week in home visits in 1995 is: 408/25.5=16, so eliminate graphs D and C because they show the value to be more than this. The height of the bar charts for home visits in 2000 is the same in graphs A and B, so it would be a waste of time calculating this figure. The number of patients seen in surgery per week in 2000 is given by: 165/10×8.4=138.6, which is shown on graph A but not B.
A commuter bus taking people home from work loads up in town and then stops at various places on its way to the edge of town. Nobody gets on the bus after the place where it loads up, but 1/3 of the people on the bus get off at the first stop and at each subsequent stop up to the fourth, where the last 8 people get off.
The correct answer is B.
Use trial by error for this question, by inputting the options AD given until you find one that works. For A: at the first stop 18(18/3)=12, at the second stop 12(12/3)=8, but the question states that a third of the remaining people also get off at the third stop, but 8 people cannot be divided by 3, so A is not the answer. For B: 27(27/3)=18, 18(18/3)=12, 12(12/3)=8, and at the final stop the remaining 8 people get off, so there were 27 people on the bus initially and B is correct.
While the Internet brings undoubted advantages to young people, the effects of excessive use are serious. Some teenagers are spending as much as 8 hours per day using the Internet. Isolation and obesity are increasing amongst children. Parents must enforce stricter controls over their children, to make sure that the Internet is not causing their children to have long term physical and emotional ill health.
The correct answer is E.
Twelve teams will take part in the Pitchball World Cup next month. They will compete in two pools of six. Every team will play two matches against each of the other teams in the same pool and one match against each team in the other pool. The winners of each pool will then contest the final.
The correct answer is C.
Each team will play two matches against the 5 other teams in its pool, and since there are 6 teams in each pool the number of intrapool matches played would be: 6x5x2=60 matches. Each team also plays 1 match against each team in the other pool: 6×6=36. Not to forget, the winners of each pool will play each other in the final, which gives 1 additional match. Altogether the number of matches played would be: 60+36+1=97 matches.
In the graphs below, the x and y scales are identical, and the same for each graph. On which one does the shaded area correctly represent the conditions:
The correct answer is C.
For 0<y<5, the shaded area would be between the two horizontal lines traversing the values 0 and 5 on the yaxis, so E can be eliminated because the shaded area shows no upper bound on the yaxis. For y>2x, we know that the line y=2x must have a gradient 2 times greater than the line y=x, therefore B and D can also be eliminated because the gradients of their diagonal lines are less than this. This leaves A and C, but because the value of y must be greater than 2x, the shaded area must be above the line y=2x, therefore the answer is C.
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TI108
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