Mathematics for Construction

This document provides solutions for various scenarios in Mathematics for Construction. It includes calculations for revenue, histograms, mode, cumulative frequency curves, mean, range, standard deviation, hypothesis tests, probability, and wave equations.

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 organised in following manner -
Revenue Number of customers
(£1000)
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
1

Histogram of each distribution can be calculated after converting the unequal class interval
into equal class interval –
Revenue
Number of
customers
(£1000)
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)
January
0 to 10 27 + 38 = 65
10 to 20 40 + 22 = 62
20 to 30 13
2

30 to 40 4
From this histogram, it has been analysed that group 0 to 10 has highest frequency, so, taking
this class to find mode of data of grouped frequency in following 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 can be calculated as -
Mode (z) = 0 + 65 – 0 x 10
2x65 – 0 – 62
= 0 + 65 x 10
130 – 62
= 0 + 650 / 68
3
0 to 10 10 to 20 20 to 30 30 to 40
0
10
20
30
40
50
60
70 65 62
13
4
Number of customers (£1000)
January

= 9.55
For July month -
Revenue
Number of
customers
(£1000)
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)
July
0 to 10 22 + 39 = 61
10 to 20 69 + 41 = 110
20 to 30 20
30 to 40 5
4

From this histogram, 10 to 20 group has highest frequency, so, taking this group as modal class
of grouped frequency, then mode can be calculated in following 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,
5
0 to 10 10 to 20 20 to 30 30 to 40
0
20
40
60
80
100
120
Number of customers (£1000)
July

so, mode 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
6

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
January
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
7
0 to 10 10 to 20 20 to 30 30 to 40
0
20
40
60
80
100
120
140
160
Less tha Cumulative
Frequency
More than Cumulative
Frequency

Revenue
Number of
customers
(£1000)
Cumulative
frequency
January
0 to 10 65 65
10 to 20 62 127
20 to 30 13 140
30 to 40 4 144
Now, 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, median class will be 10 to 20
l is 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
8

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
July
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
9
0 to 10 10 to 20 20 to 30 30 to 40
0
50
100
150
200
250
less tha Cumulative Frequency
More than Cumulative
Frequency

Revenue
Number of
customers
(£1000)
Cumulative
frequency
July
0 to 10 61 61
10 to 20 110 171
20 to 30 20 191
30 to 40 5 196
Now, 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
= 196 / 2 = 98
so, median class will be 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
10

c) Mean, Range and Standard deviation -
January data
Revenue
Number of
customers
(£1000) X= middle term fx
January
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 1700
Mean = ∑fx / ∑f
Here, ∑fx is sum of product of frequency and middle term
and ∑f is total frequency
therefore, mean = 1700 / 144 = 11.8
July data
Revenue
Number of
customers
(£1000) X= middle term fx
July
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 2630
11

Mean = ∑fx / ∑f
Here, ∑fx is sum of product of frequency and middle term
and ∑f is total frequency
therefore, mean = 2630 / 196 = 13.4
12

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 ǂ 360
To calculate the test statistic, let assume that normal distribution is valid when there is not
change in μ at null hypothesis. As, value of μ is 360 days, then to perform a hypothesis test with
standard deviation, then z score -
z = x¯ - μ
σ/√n
here, let average life value of bulb is 360.5 then,
z = x¯- 360
60 / √5000
z = 0.59
Using, significant level at 5% and taking a right-sided one-tailed test, α = 0.05
Z 1- α = 1.645 is obtained as test statistic. Therefore,
as 0.59 < 1.645 so, test statistic is not in critical region, thus, null hypothesis cannot reject.
Therefore, mean under given assumption is normally distributed.
b) Probability
(z1 < Z < Z2)
here, z1 will be calculated at half of standard deviation and z2 at double of standard deviation i.e.
z1 = x¯ - μ and, z1 = x¯ - μ
2σ/√n σ/2√n
then,
z1 = 0.29 and, z2 = 1.16
B. Simple random sample taken from a certain population is 10 people, with a mean of 27 years
average mean is above or less than 30 years
variance is known to be 20, so standard deviation = √20 = 4.47
Hypothesis assumption -
13

Let H0 : μ = 30
and, H1 : μ ǂ 30
z = x¯ - μ
σ/√n
variance,
so, z = 30 – 27 / 4.47√9
so, z = 2.01
At 5% significant level and a right-sided one-tailed test, α = 0.05
Z 1- α = 1.645,
as 2.01 > 1.645 so, test statistic is in critical region, thus, null hypothesis is rejected. Therefore,
mean age of population is not 30 years.
So, at two-tailed test, Probability can be defined as-
p{1.645 ≤ Z ≤ 2.01}
14

at, one-tailed test -
15

TASK 3
Scenario 1
Solution – Given equation of wave
x1 = 3.75 sin (100 π t + 2 π/9) ….(i)
and, x2 = 4.42 sin (100 π t - 2 π/5) ….(ii)
these equations can further be written as -
x1 = 3.75 sin (100 π (t + 0.002)) ….(iii)
and, x2 = 4.42 sin (100 π (t – 0.004)) ….(iv)
a) Apmlitude in first equation is 3.75 and in second equation is 4.42
Phase from third equation is obtained as = 0.002 to the left
and, Phase from fourth 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, then time taken by each machine for displacement, can
be calculated by -
first differentiate equation (i) with respect to t -
x1 = 3.75 sin (100 π t + 2 π/9)
dx1/dt = 3.75 cos (100 π t + 2 π/9) x 100 π (v)
at maximum, dx1/dt = 0
3.75 cos (100 π t + 2 π/9) x 100 π = 0
or, cos (100 π t + 2 π/9) = 0
100 π t + 2 π/9 = 90
100 π t = 90 – 2π/9
t = 0.002 seconds
Now, differentiate equation (ii) with respect to t -
x2 = 4.42 sin (100 π t - 2 π/5)
dx2/dt = 4.42 cos (100 π t - 2 π/5) x 100 π (vi)
at maximum, dx1/dt = 0
16

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 by each machine when displacement reach at -2mm
from equation (v),
dx1/dt = 3.75 cos (100 π t + 2 π/9) x 100 π
at -2mm,
3.75 cos (100 π t + 2 π/9) x 100 π = -2
375 π cos (100 π t + 2 π/9) = -2
cos (100 π t + 2 π/9) = -0.00002963
100 π t + 2 π/9 = 90.01
100 π t = 90.05 – 2 π/9
t = 0.0027sec
while, second machine will take -
dx2/dt = 4.42 cos (100 π t - 2 π/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 π t + 2 π/9)
using sin (A+B) = sin A cos B + cos A sin B
then,
x1 = 3.75 (sin 100 π t . cos 2 π/9 + cos 100 π t . sin 2 π/9)
or, x1 = 3.75 ( sin 100 π t x 0.76 + cos 100 π t x 0.64)
or, x1 = 2.85 sin 100 π t + 2.4 cos 100 π t (Ans)
17

similarly, second equation -
x2 = 4.42 sin (100 π t - 2 π/5)
x2 = 4.42 (sin 100 π t . cos 2 π/5 - cos 100 π t . sin 2 π/5)
or, x2 = 4.42 ( sin 100 π t x 0.31 + cos 100 π t x 0.95)
or, x2 = 1.3702 sin 100 π t + 4.199 cos 100 π t (Ans)
v) Expression of x1 and x2 in combined form -
x1 + x2 = (2.85 sin 100 π t + 2.4 cos 100 π t) + (1.3702 sin 100 π t + 4.199 cos 100 π t)
x1 + x2 = 4.2202 sin 100 π t + 6.599 cos 100 π t
x1 + x2 = 6 (sin 100 π t + π/4) (Ans).
18

Scenario 2
(i) The distance between AB
Distance between two points on Cartesian 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,
19

→ →
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 side of above equation,
λ = 1.05, so, a = 75 (approx) and b = 4 (approx)
Now, angle between two lines can be calculated by –
→ →
cos Ө = 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 Ө = u . v
||u|| ||v||
cos Ө = (40i + 40j -20k) . ((a-40)i + bj + 20k)
√(402 + 402 + 202). √((a-40)2 + b2 + 202)
cos Ө = 40(a-40) + 40b -400
√3600 . √((a-40)2 + b2 + 202)
= 2(a+b-101)
3.√((a-40)2 + b2 + 202)
20

TASK 4
ai) 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 above graph, the value of x at which it becomes
zero will be 4.8 approx.
21
1 2 3 4 5
-500
0
500
1000
1500
2000
2500
3000
Graph
bending moment M
variable x
variable x
M = 3000 – 550x – 20x²

(aii) For determining, the maximum or minimum value of Bending Moment Function,
differentiate the given equation with respect to x,
d/dx (M) = d/dx (3000 – 550x – 20x2)
= – 550 – 40x
here, d/dx < 0,
so, Bending Moment 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 bending moment function, to get maximum value as -
M = 3000 – 550 (-14) – 20(-14)2
= 3000 + 7700 – 3920 = 6780 (ans)
b) The temperature Ө (ºC), at time t (mins) of a body is given by
Ө = 300 + 100 e-0.1 t
value of Ө, can be determined at the value of 0, 1, 2 and 5 as -
at t = 0,
Ө = 300 + 100 e-0.1 x 0
= 300 + 100 x 1
= 400 ºC
at t = 1,
Ө = 300 + 100 e-0.1 x 1
= 300 + 100 x 0.90
= 390 ºC
at t = 2,
Ө = 300 + 100 e-0.1 x 2
= 300 + 100 x 0.81
22

= 381 ºC
at t = 5,
Ө = 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, relationship 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 rate of change of V when value of P changes from 10N/m2 from 60 to 100
N/m2 with variable n = 2, differentiate the above equation with respect to V as -
d/dV PVn = d/dv C
dP/dV Vn + nVn-1 .P = 0
Vn (dp/dV + n/V . P) = 0
dp/dV + n/V . P = 0
dp/dV = -n/V . P
23

Scenario 2
Given, C = 16 t-2 + 2t
variation of cost over a range of production times from 1 to 8 minutes as -
at t = 1,
C = 16 x 1 + 2 x 1
= 18 £
at t = 2,
C = 16 x 2-2 + 2 x 2
= 4 + 4
= 8 £
.
.
.
at t = 8,
C = 16 x 8-2 + 2 x 8
= 0.25 + 16
= 16.25 £
Cost (£) Time (in minutes)
18 1
8 2
7.78 3
9 4
10.64 5
12.44 6
14.33 7
16.25 8
24

To find optimal solution to this problem of optimisation, differentiate the cost function with
respect to t as
d/dt (C) = d/dt (16 t-2 + 2t)
= -32 t -3 + 2
therefore, by again differentiating the above equation with respect, the function attains its
maximum value,
now, put dC/dt = 0
-32 t -3 + 2 = 0
t -3 = 1/16
t 3 = 16
25
0 1 2 3 4 5 6 7 8 9
0
2
4
6
8
10
12
14
16
18
20
Graph
Cost (£)
Time (in minutes)
Cost Function (£) 16 t ² + 2t

or, t = (16)1/3
26

27