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ITSU2011 Computational Mathematics Assignment Sample

Question

Q1) 16 teams enter a competition. They are divided up into four Pools (A, B, C and D) of four teams each.

Every team plays one match against the other teams in its Pool. After the Pool matches are completed:

- the winner of Pool A plays the second placed team of Pool B
- the winner of Pool B plays the second placed team of Pool A
- the winner of Pool C plays the second placed team of Pool D
- the winner of Pool D plays the second placed team of Pool C

The winners of these four matches then play semi-finals, and the winners of the semi-finals play in the final.

How many matches are played altogether?
[5 marks]

Q2) A restaurant offers 5 choices of appetizer, 10 choices of main meal and 4 choices of dessert. A customer can choose to eat just one course, or two different courses, or all three courses. Assuming all choices are available, how many different possible meals does the restaurant offer?
[5 marks]

Q3) Find the number of spanning trees in the following graph.
[5 marks]

Q4) Suppose a delivery person needs to deliver packages to three locations and return to the home office A. Using the graph shown below, find the shortest route if the weights on the graph represent distance in kilometers.
[5 marks]

Solution

Question 1

As per given data, there are four Pools named as A, B, C and D of four teams each for assignment help
Let us assume that Pool A has team 1, 2, 3, and 4 respectively.
Number of games in Pool A = (4 – 1)! = 3! = 3 * 2 = 6
So, total number of games of all four Pools = 4 * 6 = 24
Now, as per given instructions,
Number of quarter finals = 4
Number of semi-finals = 2
And, number of finals = 1
Therefore, total number of matches played together = 24 + 4 + 2 + 1 = 31

Question 2

As per given data, the summary of choices are:
Number of choices of appetizer = 5
Number of choices of appetizer = 10
Number of choices of appetizer = 4
If customer eats only 1 meal, then
Number of ways = 5 + 10 + 4 = 19
If customer has 2 course meal, then
Number of ways = 5 * 10 + 5 * 4 + 10 * 4 = 50 + 20 + 40 = 110
If customer has 3 course meal, then
Number of ways = 5 * 10 * 4 = 200
Therefore, total number of possible meals = 19 + 110 + 200 = 329

Question 3

The given graph is as follows:

For the above graph, the Laplacian matrix can be written as:


Since from graph theory, we know that number of spanning trees is equal to the co-factor of Laplacian matrix.

So,

Co-factor of the above matrix

Therefore, number of spanning trees is 3.

Question 4

The provided graph is

Let the delivery person start from the home office A.

From the graph, it is clear that three edges AB, AD and AC are belongs to vertex A having weightage 4, 6 and 8 respectively.

Since, minimum weight of the edge from A is 4 (i.e. AB). So, the person goes to B from A.

In the similar way, the edge BD and BC have the minimum weight as 3. So, the person goes to C from B.

Similarly from C, edge CD have the minimum weight as 5. So, the person goes to D from C and then goes to home office A.

Therefore, the shortest route is

A – B – C – D – A with the weights 4 + 3 + 5 + 6 = 18
Hence, the optimal solution is 18 and path is ABCDA.

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 MATH11247 Foundation of Mathematics Assignment Sample

1. Without using a calculator, evaluate 16 log 8

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MATH11247 Foundation Mathematics Assignment 2 Sample

Objectives

This assessment item relates to the following unit learning outcomes, as detailed in the MATH11247 Foundation Mathematics Unit Profile:

- Develop solutions to problems through application of area and volume equations • Select appropriate mathematical methods, use them to investigate and solve problems, and interpret the results

- Use mathematics as a language to communicate results, concepts and ideas in context • Document the solution to problems in a way that demonstrates a clear, logical and precise approach.

1. A metal bracket is of the shape shown below. Dimensions are in centimeters. Determine the area of the bracket. (2 marks)

To find the area of the bracket, we can divide it into two shapes: a rectangle and a trapezoid.

The rectangle has sides of 6 and 25, so its area is:

A_rect = 6 x 25 = 150 square units

The trapezoid has bases of 9 and 6, a height of 6, and an average of the two bases (9 and 6) of 7.5. The area of the trapezoid is:

A_trap = 0.5 x (9 + 6) x 6 x 7.5 = 67.5 square units

To find the total area of the bracket, we add the areas of the rectangle and trapezoid:

A_bracket = A_rect + A_trap = 150 + 67.5 = 217.5 square units

Therefore, the area of the bracket is 217.5 square units.

2. The area of a plot of land on a map is 600 mm2. If the scale of the map is 1 to 20000, determine the true area of the land in hectares (2 1 hectare 10000 = m).

Given: Area of the plot on the map = 600 mm²
Scale of the map = 1 : 20000

To find: True area of the land in hectares

Solution

We know that the scale of the map is 1 : 20000. This means that 1 unit on the map represents 20000 units in real life. Therefore, we need to convert the area of the plot on the map from mm² to m².

1 mm² = (1/1000)² m² [Converting mm² to m²]
1 mm² = 1/1000000 m²

Area of the plot in m² = (600/1000000) m² = 0.0006 m²

Now, we can calculate the true area of the land in hectares:

1 hectare = 10000 m²
True area of the land in m² = Area of the plot in m² / Scale of the map
True area of the land in m² = 0.0006 / 20000
True area of the land in m² = 0.00000003 m²

True area of the land in hectares = 0.00000003 m² / 10000
True area of the land in hectares = 0.000000003 hectares

Therefore, the true area of the land in hectares is 0.000000003 hectares.
3. The following shape is comprised of two concentric circles. If the shaded area is 2 311 mfind
the diameter of the internal circle, correct to the nearest meter.

Let the radius of the smaller circle be "r" and the radius of the larger circle be "R" (where R = r + 3 cm).

Area of the shaded region = Area of the larger circle - Area of the smaller circle

311 sq m = πR^2 - πr^2

Substituting R = r + 3 cm, we get:

311 sq m = π(r + 3)^2 - πr^2

Simplifying this equation, we get:

311 sq m = 6πr + 9π

r ≈ 48.28 m / 6π

The diameter of the smaller circle is 2r ≈ 16.21 meters or approximately 16 meters when rounded to the nearest meter.


4. Determine the diameter and area of a circle whose circumference is 472.724 cm.

The formula for the circumference of a circle is C = 2πr, where C is the circumference and r is the radius. We can use this formula to find the radius of the circle:

C = 2πr

472.724 = 2πr

r = 472.724 / (2π) = 75.000 cm (approx)

The diameter of the circle is twice the radius:

d = 2r = 2(75.000) = 150.000 cm (approx)

The formula for the area of a circle is A = πr^2. We can use this formula to find the area of the circle:
A = πr^2
A = π(75.000)^2
A = 17,671.458 cm^2 (approx)

Therefore, the diameter of the circle is 150.000 cm (approx) and the area of the circle is 17,671.458 cm^2 (approx).

5. Convert (a) 115°35’ to radians (b) 0.875 radians to degrees and minutes

(a) To convert 115°35' to radians, we first convert the degree measure to radians by using the formula:
radians = (π/180) x degrees

So, for 115°, we have:
radians = (π/180) x 115 = 2.0071 radians (rounded to four decimal places)
Next, we convert the minute measure to radians by using the fact that there are 60 minutes in 1 degree:
radians = (π/180) x minutes/60

So, for 35 minutes, we have:
radians = (π/180) x 35/60 = 0.0102 radians (rounded to four decimal places)
Therefore, 115°35' is equivalent to 2.0071 + 0.0102 = 2.0173 radians (rounded to four decimal places).

(b) To convert 0.875 radians to degrees and minutes, we first convert the radians to degrees by using the formula:
degrees = radians x (180/π)
So, for 0.875 radians, we have:
degrees = 0.875 x (180/π) = 50.2655° (rounded to four decimal places)
Next, we convert the decimal portion of the degrees to minutes by multiplying by 60:
minutes = 0.2655 x 60 = 15.93 minutes (rounded to two decimal places)
Therefore, 0.875 radians is equivalent to 50°15.93'.

6. Three semicircles of the same radii are positioned next to each other and border on the diameter of a larger semicircle. Find the ratio for the assignment helpline of the white area to the black area in the shape below.

Hint: You will need to recall and apply ratios from Week 1 studies.

To solve this problem, we need to calculate the areas of the white and black regions separately and then find their ratio.

Let's label the radius of each smaller semicircle as "r". The diameter of the larger semicircle will be 2r + 2r + 2r = 6r. The radius of the larger semicircle is therefore 3r.

The area of each smaller semicircle is (1/2)πr², so the total area of the three smaller semicircles is (3/2)πr².
The area of the larger semicircle is (1/2)π(3r)² = (9/2)πr².

The white area is the area of the larger semicircle minus the area of the three smaller semicircles. Therefore:
White area = (9/2)πr² - (3/2)πr² = 3πr²

The black area is the remaining area inside the larger semicircle but outside the three smaller semicircles. It is easiest to calculate this area as the area of the larger semicircle minus the area of the circle with radius r (which is the same as the area of one of the smaller semicircles). Therefore:
Black area = (1/2)π(3r)² - (1/2)πr² = (8/2)πr² = 4πr²

The ratio of the white area to the black area is:
White area / Black area = (3πr²) / (4πr²) = 3/4

Therefore, the ratio of the white area to the black area is 3:4.

7. A car has wheels of radius 35 cm. Calculate the number of complete revolutions a wheel makes (correct to the nearest revolution) when the car travels for 18 minutes at a speed of 90 km/hour. (2 marks)

First, we need to convert the speed from km/h to cm/min, since we have the radius in cm and the time given in minutes:
90 km/h = 90000 cm/h = 1500 cm/min

The circumference of the wheel can be calculated as:
C = 2πr = 2π(35) = 70π cm

The distance traveled by the car during 18 minutes can be calculated as:
d = v × t = 1500 × 18 = 27000 cm

The number of complete revolutions the wheel makes is given by:
n = d / C = 27000 / (70π) ≈ 121.7
Rounding to the nearest revolution, we get:
n ≈ 122

Therefore, the wheel makes approximately 122 complete revolutions when the car travels for 18 minutes at a speed of 90 km/hour

8. The equation of a circle is:2 2 x y x y + − + + = 20 4 40 0. Determine the

(a) coordinates of the centre of the circle, and

(b) diameter of the circle.

The standard form equation of a circle is:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) is the center of the circle and r is the radius.

Rewriting the given equation in the standard form:
2(x^2 + 2xy + y^2) - 40x - 40y + 80 = 0
(x^2 + 2xy + y^2) - 20x - 20y + 40 = 0
(x + y)^2 - 20(x + y) + 40 = 0
Let's make a substitution: u = x + y

Then we get:
u^2 - 20u + 40 = 0

Solving this quadratic equation using the quadratic formula, we get:
u = 10 ± 2√5

Now, let's substitute back to find x and y:
x + y = u
x + y = 10 ± 2√5

Let's call the first solution (10 + 2√5) and the second solution (10 - 2√5). We can solve for x and y by setting one variable to a constant and solving for the other:
When x + y = 10 + 2√5:
x = (10 + 2√5) - y

Substituting into the original equation:
2x^2 + 2xy + 2y^2 - 40x - 40y + 80 = 0
2[(10 + 2√5 - y)^2 + 2(10 + 2√5 - y)y + y^2] - 40(10 + 2√5 - y) - 40y + 80 = 0
Simplifying and solving for y using the quadratic formula, we get:
y ≈ 5.59 or y ≈ 14.41

Substituting back to solve for x, we get:
When y = 5.59:
x ≈ 10.41
When y = 14.41:
x ≈ -0.41

So the two possible centers of the circle are approximately (10.41, 5.59) and (-0.41, 14.41).
To find the diameter, we need to find the radius. We can do this by plugging in the center point into the original equation and solving for the radius:
2x^2 + 2xy + 2y^2 - 40x - 40y + 80 = 0
Plugging in (10.41, 5.59):
2(10.41)^2 + 2(10.41)(5.59) + 2(5.59)^2 - 40(10.41) - 40(5.59) + 80 = 0
Simplifying, we get:
(10.41 - 5.59)^2 + (10.41 - 14.41)^2 = r^2
r ≈ 3.16
So the diameter is:
d = 2r ≈ 6.32

 

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