8 Mathematics for Construction

TASK 2

Scenario 1

Revenue	Number of customers (£1000)
	January	July
Less than 5	27	22
5 and less than 10	38	39
10 and less than 15	40	69
15 and less than 20	22	41
20 and less than 30	13	20
30 and less than 40	4	5

Solution

a) The given table can be organized in the following manner -

Revenue	Number of customers (£1000)
Â	January	July
0 to 5	27	22
5 to 10	38	39
10 to 15	40	69
15 to 20	22	41
20 to 30	13	20
30 to 40	4	5

 

The Histogram of each distribution can be calculated after converting the unequal class interval into an equal class interval –

Revenue	Number of customers (£1000)
Â	January
0 to 5	27
5 to 10	38
10 to 15	40
15 to 20	22
20 to 30	13
30 to 40	4

Equal class interval -

Revenue	Number of customers (£1000)
Â	January
0 to 10	27 + 38 = 65
10 to 20	40 + 22 = 62
20 to 30	13
30 to 40	4

From this histogram, it has been analyzed that groups 0 to 10 have the highest frequency, so, taking this class to find mode oa f data of grouped frequency in following the ing way –

Mode (z) = l + f₁   –    f₀       x    h

                         2 f₁ – f₀  -  f₂

Here,  f₁ is the highest frequency = 65

          f₀ is the above frequency = 0 and,

          f₂ is the below frequency = 62

          h is the class range  = 10 and,

          l is the lower interval of mode class = 0,

So, mode the  can be calculated as -

Mode (z) =  0 + 65 – 0    x 10

                =   2x65 – 0 – 62

                =   0 + 65 x 10        

                =   130 – 62

                =   0 + 650 / 68

                =   9.55

For July month -

Revenue	Number of customers (£1000)
Â	July
0 to 5	22
5 to 10	39
10 to 15	69
15 to 20	41
20 to 30	20
30 to 40	5

 

Revenue	Number of customers (£1000)
Â	July
0 to 10	22 + 39 = 61
10 to 20	69 + 41 = 110
20 to 30	20
30 to 40	5

From this histogram, 10 the o 20 group has high est frequency, so, taking this group as the modal  class of grouped frequency, the cathode can be calculated in following the wing way-

Mode (z) = l + f₁   –    f₀  x h

                   2 f₁ – f₀  -  f₂

here,  f₁ is the highest frequency = 110

          f₀ is the above frequency = 61 and,

          f₂ is the below frequency = 20

          h is the class range = 10 and,

          l is the lower interval of mode class = 10,

so, the ode can be calculated as -

Mode (z) = 10 + 110 – 61  x 10

                    2x110 – 61 – 20

                 = 10 + 49 x 10        

                     220 – 81

                  = 10 + 490 / 139

                  = 13.5

b) Cumulative frequency curve or O-give curve of January data

Revenue	Number of customers (£1000)	Less than O-give curve	Cumulative Frequency	More  than O-give curve	Cumulative Frequency
Â	January	Â	Â	Â	Â
0 to 10	65	Less than 10	65	More than 0	144
10 to 20	62	Less than 20	127	More than 10	79
20 to 30	13	Less than 30	140	More than 20	17
30 to 40	4	Less than 40	144	More than 30	4

 

Revenue	Number of customers (£1000)	Cumulative frequency
Â	January	Â
0 to 10	65	65
10 to 20	62	127
20 to 30	13	140
30 to 40	4	144

Now, the median of the data can be calculated by -

Median (M) = l + N/2 – cf   x  h

                                 f

Here, N/2 = sum of total freq / 2

                = 144 / 2  = 72

so, the main class will be from 10 to 20 

  l is the lowest interval = 10

  h is class difference = 10

  So, Median (M) = 10 + 72 – 65 x 10

                                       62

                         = 10 + 70/62

                        = 10 + 1.1 = 11.1

Cumulative frequency curve or O-give curve of July data

Revenue	Number of customers (£1000)	Less than O-give curve	Cumulative Frequency	More  than O-give curve	Cumulative Frequency
Â	July	Â	Â	Â	Â
0 to 10	61	Less than 10	61	More than 0	196
10 to 20	110	Less than 20	171	More than 10	135
20 to 30	20	Less than 30	191	More than 20	25
30 to 40	5	Less than 40	196	More than 30	5

 

Revenue	Number of customers (£1000)	Cumulative frequency
Â	July	Â
0 to 10	61	61
10 to 20	110	171
20 to 30	20	191
30 to 40	5	196

 

Now, the media ian of the data can be calculated by -

Median (M) = l + N/2 – cf   x  h

                                 f

Here, N/2 = sum of total freq / 2

                = 196 / 2  = 98S

So, the dian class will be from 10 to 20 

l is lowest interval = 10  and frequency = 110

h is class difference = 10

So, Median (M) = 10 + 98 – 61 x 10

                                       110

                         = 10 + 370/110

                        = 10 + 3.4 = 13.4

 c) Mean, Range an,d Standard deviation -

January data

Revenue	Number of customers (£1000)	X= middle term	fx
Â	January	Â	Â
0 to 10	65	5	325
10 to 20	62	15	930
20 to 30	13	25	325
30 to 40	4	30	120
Total	144	Â	1700

Mean = ∑fx / ∑f

Here,  ∑fx is the m of the product of frequency and middle-termed  ∑f is the total frequency

therefore, mean = 1700 / 144 = 11.8

July data

Revenue	Number of customers (£1000)	X= middle term	fx
Â	July	Â	Â
0 to 10	61	5	305
10 to 20	110	15	1650
20 to 30	20	25	500
30 to 40	5	35	175
Total	196	Â	2630

Mean = ∑fx / ∑f

Here,  ∑fx is the sum of the f produce ct of frequency and middle middle-term∑f is tota the l frequency

therefore, mean = 2630 / 196 = 13.4

Scenario 2

Solution

Given – Number of bulbs = 5000

Mean of lengths of life of bulb at normal distribution = 360 days

Standard deviation = 60 days

a) Assumption test if bulb life is normally distributed,

Let   H₀ : μ = 360

and, H₁ : μ Ç‚ 360

To calculate the test statistic, let's assume that normal distribution is valid when there is no change in μ at the null of the hypothesis. As, the lue of μ is 360 days, then to perform a hypothesis test with standard deviation, then z score -

                                                      z  = x¯ - μ

                                                     σ/√n

here, let the average life value of bulthe b is 360.5 then,

                                                      z =  x¯- 360

                                                      60 / √5000

                                                      z = 0.59

Using, the significant level at 5% and taking a right-sided one-tailed test, α = 0.05
Z _{1- α} = 1.645 is obtained as a test of the test statistic Therefore,

as 0.59 < 1.645 so, test the statistic is not in the critical cal region, thus, the null hypothesis cannot be rejected. Therefore, mean the under the given assumption is normally distributed.

b) Probability

(z₁ < Z < Z₂)

here, z₁ will be calculated at half of the standard deviation and z₂ at double of standard deviation i.e.

  z₁  = x¯ - μ   and,   z₁  = x¯ - μ

      2σ/√n                 σ/2√n

then,

         z_1   =  0.29    and,  z₂ = 1.16

B.SA simple random sample taken from a certain population is 10 people, with a mean of 27 years

the average mean is above or less than 30 years

variance is known to be 20, so standard deviation = √20 = 4.47

Hypothesis assumption -

Let   H₀ : μ = 30

and,  H₁ : μ Ç‚ 30

  z  = x¯ - μ

         σ/√n

variance, 

so, z = 30 – 27 / 4.47√9

so, z = 2.01

At 5% sa significant level and a right-sided one-tailed test, α = 0.05
Z _{1- α} = 1.645,

as 2.01 > 1.645 so, test the statistic is in the critical cal region, thus, the null hypothesis is rejected. Therefore, the age of the population is not 30 years.

So, two-tailed test, Probability can be defined as {1.645 ≤ Z  ≤ 2.01}

at, one-tailed test -

TASK 3

Scenario 1

Solution – Given equalization of wave

                 x₁ = 3.75 sin (100 Ï€ t + 2 Ï€/9)             ….(i)

and,          x₂ = 4.42 sin (100 Ï€ t - 2 Ï€/5)               ….(ii)

these equations can further be written as -

                  x₁ = 3.75 sin (100 Ï€ (t + 0.002))             ….(iii)

and,           x₂ = 4.42 sin (100 Ï€ (t – 0.004))               ….(iv)

a) Apmlitude in the first t equation is 3.75 and in Seco, the nd equation is 4.42

          Phase from the third equation is obtained as = 0.002 to the left

and,   Phase from fourth the equation is obtained as = -0.004 to the right

          Periodic times = 2 π / 100 π = 0.002.

While frequency will be 1/period

            i.e.     F = 1 / 0.002 = 500

b) When both machines are switched on, the time taken by each machine for displacement, can be calculated by -

first, differentiate equation (i) concerning -

           x₁ = 3.75 sin (100 Ï€t + 2 Ï€/9)    

           dx₁/dt = 3.75 cos (100 Ï€ t + 2 Ï€/9) x 100 π         ......... (v)

           at maximum,  dx₁/dt  = 0

           3.75 cos (100 πt + 2 π/9) x 100 π = 0

or,       cos (100 πt + 2 π/9) = 0

           100 πt + 2 π/9 = 90

           100 πt  = 90 – 2π/9

           t = 0.002 seconds

Now, differentiate equation (ii) concerning -

x₂ = 4.42 sin (100 Ï€ t - 2 Ï€/5)     

dx₂/dt = 4.42 cos (100 Ï€ t - 2 Ï€/5) x 100 π   ...........(vi)

at maximum, dx₁/dt  = 0

       4.42 cos (100 π t - 2 π/5) x 100 π = 0   

or,   cos (100 π t - 2 π/5) = 0

       100 π t - 2 π/5 = 90

       100  π t  =  90 + 2π/5

        t = 0.009 seconds 

c) Time taken each machine when displacement recreates-2mm

     from equation ..............(v)

    dx₁/dt = 3.75 cos (100 Ï€ t + 2 Ï€/9) x 100 π 

    at -2mm,

    3.75 cos (100 πt + 2 π/9) x 100 π  = -2

    375  π cos (100 πt + 2 π/9) = -2

    cos (100 π t + 2 π/9) = -0.00002963

    100 πt + 2 π/9 = 90.01

    100 πt  = 90.05 – 2 π/9

    t           =  0.0027sec

While, second machine will take -

dx₂/dt = 4.42 cos (100 Ï€ t - 2 Ï€/5) x 100 π                    

4.42 cos (100 π t - 2 π/5) x 100 π = -2

cos (100 π t - 2 π/5)  = 0.000025138

100 π t - 2 π/5  = 89.99

t  = 0.0089 seconds

(iv) Compound angle formula for expansion of above two equations -

       x₁ = 3.75 sin (100 Ï€ t + 2 Ï€/9)           

using, sin (A+B)  = sin A cos B + cos A sin B

then,

       x₁ = 3.75 (sin 100 Ï€ t . cos 2 Ï€/9 + cos 100 Ï€ t . sin 2 Ï€/9)

or,   x₁  = 3.75 ( sin 100 Ï€ t  x 0.76  + cos 100 Ï€ t x 0.64)

or,   x₁ = 2.85  sin 100 Ï€ t + 2.4  cos 100 Ï€ t (Ans)

similarly, sec the ond equation -

      x₂ = 4.42 sin (100 Ï€ t - 2 Ï€/5)           

      x₂ = 4.42 (sin 100 Ï€ t . cos 2 Ï€/5 - cos 100 Ï€ t . sin 2 Ï€/5)

or,  x₂  = 4.42 ( sin 100 Ï€ t  x 0.31 + cos 100 Ï€ t x 0.95)

or,  x₂ = 1.3702 sin 100 Ï€ t + 4.199  cos 100 Ï€ t (Ans) 

v) Expression of x₁ and x₂ in combined form -

       x₁ +  x₂ = (2.85  sin 100 Ï€ t + 2.4  cos 100 Ï€ t) + (1.3702 sin 100 Ï€ t + 4.199  cos 100 Ï€ t)

       x₁ +  x₂ = 4.2202  sin 100 Ï€ t + 6.599  cos 100 Ï€ t

        x₁ +  x₂  = 6 (sin 100 Ï€ t +  Ï€/4) (Ans). 

Scenario 2

(i) The distance between AB

Distance between two points on The artesian plane of 3D can be calculated by

AB = √ (x₁ – x₀)² + (y₁ – y₀)² + (z₁ – z₀)²

here, coordinates of A are (0, -40,0) and B are (40, 0, -20)

then, distance between AB =  √ ( 0 – 40)²  + (-40 – 0)² + (0 + 20)²

                                           = √ 1600 + 1600 + 400

                                           = √3600 = 60 unit.

Angle between 

Vector equation for the line passing through two points A and B with position vectors

→        →

a   and  b is represented by -

→  →      →  →

 r = a + λ (b – a)

As, B (40, 0, -20)  and C (a, b, 0)

The position of B (40, 0, -20) = ( 40ˆi + 0ˆj -20ˆk )

The position of C (a, b, 0) = ( aˆi + bˆj + 0ˆk )

So, →

 r  = ( 40ˆi + 0ˆj -20ˆk ) +  λ [  ( aˆi + bˆj + 0ˆk ) - ( 40ˆi + 0ˆj -20ˆk ) ]

    = ( 40ˆi + 0ˆj -20ˆk ) +  λ [  (a – 40) ˆi + bˆj + 20ˆk)  ]

   →

Given , r = 3i + 4j + k

then, on comparing both sides abthe ove equation,

 λ = 1.05, so, a = 75 (approx) and b = 4 (approx)

Now, the angle between two lines can be calculated by                               

cos Ó¨ =      u . v   

                ||u||   ||v||

AB = Position of B – position vector of A

      = (40, 40, -20)

BC = Position of C – position vector of B

      = (a - 40, b, 20)

Then,

 cos Ó¨ =       u . v   

                 ||u||   ||v||

                             

 cos Ó¨ = (40i + 40j -20k) . ((a-40)i + bj + 20k)

                √(40² + 40² + 20²).  √((a-40)² + b² + 20²)

                 cos Ó¨ = 40(a-40) + 40b -40

                               √3600 .    √((a-40)² + b² + 20²)

                             =  2(a+b-101)               

                                 3.√((a-40)² + b² + 20²)

TASK 4

The bending moment, M of a beam is given by

M = 3000 – 550x – 20x²

Bending moment M	variable x
2430	1
1810	2
1170	3
480	4
-250	5

on plotting the bending moment M, as shown in the graph above, the value of x at which it becomes zero will be 4.8 approx.

(ai For determining, the maximum or minimum value of Bendithe ng Moment Function, differentiate the given equation concerning ring/dx (M) =  d/dx (3000 – 550x – 20x²⁾

= – 550 – 40x

here, d/dx < 0,

so, the bending Function will attain its maximum value at -

d/dx (M)     = 0

– 550 – 40x = 0

x = -13.75 or -14 approx.

now, put this value of x in bend the ing moment function, to get maxi the mum value as -

M =  3000 – 550 (-14) – 20(-14)²

^{     Â}  = 3000 + 7700 – 3920  = 6780 (ans)

b) The temperature  Ó¨ (ºC), at time t (mins) of a body is given by

                                 Ó¨ = 300 + 100 e^{-0.1 t}

value of  Ó can be determined at the value of 0, 1, 2, and, 5 as -

at t = 0,

                 Ó¨ = 300 + 100 e^{-0.1 x 0}

                          = 300 + 100 x 1

                          = 400 ºC

at t = 1,

                 Ó¨ = 300 + 100 e^{-0.1 x 1}

                          = 300 + 100 x 0.90

                          = 390 ºC

at t = 2,

                 Ó¨ = 300 + 100 e^{-0.1 x 2}

                          = 300 + 100 x 0.81

                          = 381 ºC

at t = 5,

                 Ó¨ = 300 + 100 e^{-0.1 x 5}

                          = 300 + 100 x 0.60

                          = 360 ºC

Temperature (ºC)	Time (t)
400	0
390	1
381	2
360	5

c) In a thermodynamic system, the related tionship between pressure (P), Volume (V), and, constant C is given by,

           log (P) + n log (V) – log (C)

to show that PVⁿ  = C

Let,     log (P) + n log (V) – log (C)  = 0

           log (P) + n log (V) = log (C)

using product rule, i.e. log (a x b) = log a + log b

            log (PVⁿ) = log (C)

            PVⁿ = C   Hence Proved.

Now, to determine the rate of the f change of V when value the of P changes from 10N/m² from 60 to 100N/m² with variable n = 2, differentiate the above equation with concerning-

d/dV PVⁿ = d/dv C

dP/dV Vⁿ + nV^n-1 .P = 0 

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