BEA1009 Introduction To Quantitative Methods : Essay Fountain

Question:

  1. Expand and simplify the following expressions
  2. a) ????(5???? − 2) − 5???????? (2 marks)
  3. b) 2????(3???? + 4) + 3????(1 − 3????) − 6???? + 1 (3 marks)
  4. Solve the following equations:

Answers should be given as integers or simplified fractions (do NOT use decimals).

  1. a) 7???? − 8 = −120 (2 marks)
  2. b) 5(3 − 2????) = 4???? − 1 (3 marks)Page 3 of 9
  3. c)

2(2???? + 4) = −

(3???? + 6)

4

  1. A rigger is a person who specialises in lifting and moving large or heavy objects. The equipment available to riggers on construction sites is often limited, so they rely on equations to help them figure out how to safely lift and move objects. The rigging equation for calculating the mass of a hollow steel pipe is given below:

???? = 24????????(???? − ????)

Where ???? is the mass of the pipe in tonnes (t), ???? is the length, ???? is the outside diameter, and ???? is the wall thickness of the pipe, all measured in metres (m).

  1. a) What is the mass of a 6 m pipe with an outside diameter of 0.3 m and a wall thickness of 5 cm? Give your answer to the nearest tonne.
  1. b) Transpose the equation to make ???? the subject. (2 marks)
  2. c) A steel pipe needs to be moved into place on a construction site using a crane. The rigger has a sling available that is approved for use on pipes that have an outside diameter less than 1 m. If the steel pipe to be moved is 8 m long, has a wall thickness of 0.04 m, and has a mass of 7.5 t, can the rigger safely move the pipe?

Use mathematics to explain how you determined your answer.

  1. a) A mobile phone plan costs $50 per month plus 7 cents per text message.

(i) Complete the equation below to represent the total monthly cost, C, of this mobile plan. Define any other variables that you use.

???? =

(ii) Use your equation to find how many text messages were sent in the last month if the bill came to $67.50.

  1. b) (i) The River Leven, L, is 26 km longer than the River Tamar, T.

Write an equation to show the length of L in terms of T (e.g. L=…)

(ii) The River Derwent is 249 km long, and if you double this length you

have triple the length of the River Tamar and River Leven combined.

Convert this sentence into an algebraic equation using the same

variables as above.

(iii) Combine your equations and solve to find the length of the River Tamar.

To stimulate his daughter in the pursuit of problem solving, a maths teacher offered to pay her $10 for every question she solved correctly, and to fine her $7 for every incorrect solution. After she had completed 32 problems the maths teacher was starting to regret his decision, he now owed his daughter $133! Use algebraic techniques to find out how many problems the daughter solved correctly.

 

 

Answer:

Q1

  1. a(5b-2)-5ab

Opening the brackets,

5ab-2a-5ab

Collecting like terms together,

5ab-5ab-2a

= -2a

  1. 2a(3b+4)+3b(1-3a)-6b+1

Opening the brackets,

6ab+8a+3b-9ab-6b+1

Collecting like terms together,

6ab-9ab+8a+3b-6b+1

= -3ab+8a-3b+1

Q2

  1. 7e-8=-120

7e=-120+8

7e=-112

=e = -112/7 = -16

  1. 5(3-2f)=4f-1

= 15-10f = 4f-1

= 15+1=4f+10f

= 14f =16

Therefore f=16/14 = 8/7 = 1 1/7

  1. 2(2c+4) = -{(3c+6)/4}

4(4c+8) = -(3c+6)

16c+32 = -3c-6

16c+3c = -6-32

19c =-38

C=-2

Q3

  1. M=24LT(D-T)

L=6m, D=0.3m, T =5/100 = 0.05m

M = 24 x 6 x 0.05x (0.3-0.05)

     = 24x 6 x 0.05 x 0.25

     M = 1.8t

  1. Transpose and make D subject

M =24LT (D-T)

M/24LT = D-T

D = M/24LT + T

  1. Yes, the rigger can safely move the load

The load outside diameter is less than 1m

L=8m, T=0.04m and M=7.5t

 

 

Finding the maximum outside diameters D for the load which can be lifted,

7.5= 24 x 8 x 0.04(D-0.04)

D-0.04 = 7.5/(24x8x0.04)

D=0.9765625+0.04

Maximum outside diameter which can be lifted is 1.164625m. this means that the available diameter of less than 1m will be safely lifted.

Q4

  1. mobile plan

C= 50 +7/100m, where m is the number of sms sent per  month

total monthly cost = 67.50

67.50 = 50+0.07m

67.50-50=0.07m

17.50/0,07 = m

M = 250 sms

L= T + 26

D=  249km

249×2 = 3(sum of L and T)

249×2 = 3(T + T +26)

498=6T+78

iii.

solving for T

498=6T +78

498-78 =6T

T =70km

Q5

Let C represent correct answer, W represent wrong answers and E total earning by the student

Generally E = 10C-7W

C+W= 32

10C-7W =133

C= 32-W

Replacing C in E equation,

10(32-W)-7W = 133

320-10W-7W =133

320-133=17W

187=17W

W=11

C=32-11

C=21

Therefore the correct answers were 21

Q6

  1. 2d+e=7 and 3d+2e=10

Multiplying equation 1 with 2 and equation 2 with 1

2d+e=7 /2

3d+2e=10/1

We get

4d+2e=14

 3d+2e=10

Subtracting the two equations downwards

4d+2e=14

 3d+2e=10

D= 4

Replacing value of d in one of the equations; 2(4)+e=7

E=7-8= -1

Therefore d = 4 and e=-1

  1. Brand A=$7/kg and B=$11/kg

Let y represent kg in brand A and z kgs in brand B

Therefore; y+z=15

The total cost equation will be 7y+11z=129

Y=15-z

Replacing y in the total cost equation;

7(15-z)+11z=129

105-7z+11z=129

-7z+11z=129-105

4z=24; z=6kg

Y=15-6 = 9kg

Brand A kgs = 6kg and Brand B = 9kg

Q7

y=1/10 x -5

Considering the standard line equation of y=mx+c; where m is the gradient;

Therefore the lines gradient will be 1/10

  1. Coordinates of y intercept means that x=0. Replacing x with 0 in the equation gives y values as y=1/10x 0 -5. Therefore, y value = -5

Coordinates of y intercept = (0,-5)

  1. y intercept of a perpendicular line

The gradient of a perpendicular is usually negative reciprocal

The negative reciprocal of 1/10 = -10. The two points which will be used to achieve this gradient are (-1,7) and (0,c)

m =

-10 =

-10(+7) = c-7

-70+7=c = -63

Therefore y intercept coordinates will be (0,-63)

  1. gradient

m =

m =

  1. equation of line

m =

m =

y=-4x+c

 

The y intercept will happen at point (0,c), picking one point of the line to find value of c and using the gradient, we have

-4 =

-4x-5 = 14-c

20=14-c

C=-6

The lines equation will be y = -4x-6

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