Mathematics for Construction

This document provides solutions for various scenarios in Mathematics for Construction. It includes calculations for histograms, mode, median, mean, range, standard deviation, hypothesis testing, 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
Firstly, organised the above table into grouped frequency 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

Since, this table is not in regular interval of group, so convert it into equal grouped intervals by
combining first two rows then, second two rows as 0 – 10 and 10 – 20, so that all group ranges
from 10 in following way -
Equal class interval -
Revenue
Number of customers
(£1000)
January July
0 to 10 65 61
10 to 20 62 110
20 to 30 13 20
30 to 40 4 5
Now, to construct histogram of each and find other central tendencies, separate above table into
two categories, firstly for January and secondly to July -
Revenue
Number of customers
(£1000)
January
0 to 10 65
10 to 20 62
20 to 30 13
30 to 40 4
Revenue Number of customers
2

(£1000)
July
0 to 10 61
10 to 20 110
20 to 30 20
30 to 40 5
a) Histogram of January month -
Now, mode can be calculated by using following formula -
Mode (z) = l + f1 – f0 x h
2 f1 – f0 - f2
here, f1 refers highest frequency, which is 65 as per above table
f0 refers to previous frequency from maximum, which is 0 and,
f2 is the next frequency, i.e. 62
3
0 to 10
10 to 20
20 to 30
30 to 40
0 10 20 30 40 50 60 70
Number of customers (£1000)
January

h is the class difference = 10 and,
l is the lower-bound interval of mode class = 0,
so, mode can be calculated as -
Mode (z) = 0 + 65 – 0 x 10
2 x 65 – 0 – 62
= 0 + 65 x 10
130 – 62
= 0 + 650 / 68
= 9.55
a) Histogram of July month -
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 = 110
4
0 to 10
10 to 20
20 to 30
30 to 40
0 20 40 60 80 100 120
Number of customers (£1000)
July

f0 = 61 and,
f2 = 20
h = 10 and,
l = 10,
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
5

b)
Cumulative frequency curve of January data
Revenue
Number of
customers
(£1000)
Less than O-
give curve
Cumulati
ve
Frequenc
y
More than
O-give curve
Cumulative
Frequency
0 to 10 65 10 65 0 144
10 to 20 62 20 127 10 79
20 to 30 13 30 140 20 17
30 to 40 4 40 144 30 4
6
0 to 10 10 to 20 20 to 30 30 to 40
0
20
40
60
80
100
120
140
160
65
127
140 144144
79
17
4
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
7

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
0 to 10 61 10 61 0 196
10 to 20 110 20 171 10 135
20 to 30 20 30 191 20 25
30 to 40 5 40 196 30 5
8
0 to 10 10 to 20 20 to 30 30 to 40
0
50
100
150
200
250
61
171
191 196196
135
25
5
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
To calculate median, use given formula
Median (M) = l + N/2 – cf x h
f
Here, N/2 = 196 / 2 = 98
Therefore, 10 to 20, will be considered as median class,
so, l = 10 and f = 110
h = 10
so, Median (M) = 10 + 98 – 61 x 10
110
= 10 + 370/110
= 10 + 3.4 = 13.4
9

c) To calculate central tendencies as Mean, Range and Standard deviation, the following
formulation has done -
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
10

Total 196 2630
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
11

Scenario 2
Solution
Given – Number of bulbs as per given scenario is 5000
the Mean of lengths of bulb life with normal distribution = 360 days
Standard deviation = 60 days
a) Set hypothesis of bulb life at normally distribution -
Null hypothesis, H0 : μ = 360
Alternative hypothesis, H1 : μ ≠ 360
For the validation of hypothesis, take test statistic in following way -
z = x¯ - μ
σ/√n
Now, μ at 360.5 then,
z = x¯- 360
60 / √5000
z = 0.59
At 5% significant level of variance, at right-sided one-tailed test, α = 0.05
then Z 1- α = 1.645 is obtained as test statistic. Therefore, from above calculation, 0.59 is less than
1.645, so, null hypothesis is accepted in the present case, i.e. mean 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 -
Let H0 : μ = 30
and, H1 : μ ǂ 30
12

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}
at, one-tailed test -
13

14

TASK 3
Scenario 1
Solution – Given equation of wave
x1 = 3.75 sin (100 π t + 2 π/9)
and, x2 = 4.42 sin (100 π t - 2 π/5)
Rewrite the above equations, in the form of x = A sin (Bt) + C
x1 = 3.75 sin (100 π (t + 0.002))
Amplitude = 3.75
Phase = 0.002 to the left
and, x2 = 4.42 sin (100 π (t – 0.004))
Amplitude = 4.42
Phase = - 0.004 to the right
Now, Periodic times = 2 π / B
= 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)
15

dx2/dt = 4.42 cos (100 π t - 2 π/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 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)
16

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).
17

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.
Now, angle between two lines can be calculated by –
→ →
cos Ө = u . v
||u|| ||v||
→
AB = Position of B – position vector of A
18

= (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)
Vector equation for the line passing of two position vectors can be determined by
→ → → →
r = a + λ (b – a)
As, B (40, 0, -20) and C (a, b, 0)
The position of B (40, 0, -20) = ( 40 i + 0 j -20 k )
The position of C (a, b, 0) = ( a i + b j + 0 k )
So, →
r = ( 40 i + 0 j – 20k ) + λ [ ( a i + b j + 0 k ) - ( 40 i + 0 j -20 k ) ]
= ( 40 i + 0 j -20 k ) + λ [ (a – 40) i + b j + 20 k) ]
→
as , r = 3i + 4j + k
then, on comparing both side of above equation,
λ = 1.05, so, a = 75 (approx) and b = 4 (approx)
19

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
As shown in above graph, by plotting different values of x to calculate bending moment M, it has
been estimated that at 4.8 approx, it will become zero.
20
1 2 3 4 5
-500
0
500
1000
1500
2000
2500
3000
2430
1810
1170
480
-250
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
21

= 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
22

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
23

The optimal solution to above 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
now, put dC/dt = 0
-32 t -3 + 2 = 0
t -3 = 1/16
t 3 = 16
24
0 1 2 3 4 5 6 7 8 9
0
2
4
6
8
10
12
14
16
18
20 18
8 7.78 9
10.64
12.44
14.33
16.25
Graph
Cost (£)
Time (in minutes)
Cost Function (£) 16 t ² + 2t

25