Only for patient - 3 questions

• Mar 10th 2008, 10:25 PM
Snowboarder
Only for patient - 3 questions
1.(a) (i) Choose any four consecutive even numbers. (For example: 6, 8, 10, 12). Multiply the two middle numbers together. (e.g. 8 x 10 = 80) Multiply the first and last numbers. (e.g. 6 x 12 = 72) Now subtract your second answer from the first. (e.g. 80 — 72 = 8). What is the result each time? Give two more examples to convince someone else of
this result

(ii) Show algebraically that this is true for all sets of four consecutive even numbers.

(b) A Maths lecturer's commute into work is described below.
He left home and walked up a long hill. Halfway up he realized he had left something behind and returned home. When he began the walk for the second time he walked at a faster speed hoping he would not miss his usual bus. Once he reached the top he had a short walk along the flat to the bus stop. At the bus stop he had a short wait until his bus arrived. The bus journey was uninterrupted bringing him straight to the university.

(i) Give at least one (or more) assumption(s) which will be useful in drawing a distance time graph for the lecturer's journey to work.

(ii) Sketch a graph of Distance from home against Time to model the lecturer's journey into work.
You should label the axes and use an appropriate scale.

2. A guide in a European road code uses a mathematical model to calculate stopping
distances in a motor vehicle. It assumes there are two components when it comes to stopping:

the distance travelled in metres before the driver reacts = 0.1875V and
the braking distance in metres is given by O.006V²

where V is the speed the vehicle is travelling at in km/hr.
Use the model(s) used by the guide to answer the following:
(a) What is the distance travelled before a driver reacts if she is travelling at 100 km/hr ?

(b)Find the difference in braking distances between travelling at 50km/hr and at 100km/hr

(c)By combining the two components to stopping we get a 'total stopping distance',D in metres. D = 0.1875V + 0.006 V²
How much further is the stopping distance for a car travelling at 11 Okm/hr compared with
one travelling at I 00kin/hr?

(d)A Subaru Impreza car is 4.5 m long. How fast are you going if the guide suggests your total stopping distance is 10 car lengths?

3. A tyre manufacturing company has 1000 units of raw rubber to use in producing tyres for passenger cars and for truck tyres. Each car tyre needs 5 units and each truck tyre needs 20 units. Labour costs are \$8 for a car tyre and \$12 for a truck tyre. The manufacturer does not want to pay more than \$1500 in labour costs. To meet the requirements of a research grant he must produce at least 35 truck tyres and at least 48 car tyres.
Let c be the number of car tyres, and t the number of truck tyres that are produced.
(a)Write down the linear constraint inequations that model the above conditions.

(b)Sketch a graph of r against c (i.e r is vertical and c is horizontal) Draw each of your constraints from (i) on this graph, and clearly indicate the area of feasible solutions for the number of tyres the manufacturer might produce.

(c)Find the coordinates of each of the vertices of the area of feasible solutions.

(d)The manufacturer will make a profit of \$10 per car tyre and \$25 per truck tyre.

How many of each kind of tyre should the manufacturer produce if he wishes to maximize
the profit?
• Mar 11th 2008, 09:40 AM
Mathnasium
I.a.ii.

Represent your four consecutive even numbers as $x, x+2, x+4, x+6$. Then, multiply the middle two and the first and last, simplify (using FOIL, the distributive property), subtract, and see what you get.
• Mar 11th 2008, 10:06 AM
frankdent1
Quote:

Originally Posted by Mathnasium
I.a.ii.

Represent your four consecutive even numbers as $x, x+2, x+4, x+6$. Then, multiply the middle two and the first and last, simplify (using FOIL, the distributive property), subtract, and see what you get.

Here saying the numbers are any real number, but they should be even. So the sequence should go $2x, 2x+2, 2x+4, 2x+6$ or $2x-2, 2x, 2x+2, 2x+4$ as it would be easier to calculate. It turns out that a sequence of consecutive odd numbers gives the same answer of 8.