Building Mathematics

Building Mathematics 1
BUILDING MATHEMATICS
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Building Mathematics 2
Building Mathematics
Task 1
Scenario 1
1. Length and width
Area of rectangle = length x width
Let length be L m; width = (L – 3.2) m
Area = 26.5m2
L x (L – 3.2) = 26.5m2; L2 – 3.2L = 26.5m2
L2 – 3.2L – 26.5 = 0
The above equation is quadratic and can be solved using the quadratic formula
x=−b ± √ b2−4 ac
2 a as follows
L=− (−3.2 ) ± √(−3.2)²−(4 x 1 x−26.5)
2 x 1 = 3.2± √10.24 +106
2 = 3.2± √116.24
2 = 3.2± 10.7815
2
L = 6.9907 or -3.79075
Since length cannot be positive, it means that L = 6.99m
Width = L – 3.2m = 6.99m – 3.2m = 3.79m
2. Daily forfeit and original contract
Let the daily forfeit be x and the original contract be y
The amount the contract gets paid in case of late completion is expressed as p = y – xn; where p
= amount paid, y = original contract, x = daily forfeit and n = number of late days.
Forming the equations based on information given:
4250 = y – 5x ...................... (1)
2120 = y – 12x .................... (2)

Building Mathematics 3
These are simultaneous equations and can be solved simultaneously by subtracting the second
equation from the first one to give
2130 = 7x → x = £304.3
Substituting the value of x in the first equation gives
£4250 = y – (5*304.3)
£4250 = y – 1521.5; y = £5771.5
Therefore daily forfeit = £304.3 and original contract = £5,771.5
Scenario 2
a) Dimensional parameters
 Speed in m/s
First is to convert the 65 miles into meters.
1 mile = 1760 yards
65 miles = 65 miles x 1760 yards
1 miles =114,400 yards
Then convert the yards into meters by multiplying by 0.91 to give
114,400 x 0.91 = 104,104m
So the speed is 104,104m/hour
The next step is to convert one hour into seconds by multiplying it by 3,600 to give
1hr x 3,600 sec/hr. = 3,600 seconds
Speed is 104,104m/3600 seconds meaning
3600 s = 104,104m
1 s = 1 x 104,104 m
3600 s =28.92m/s
 Time taken

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Time= Distance
Speed = 100 miles
65 miles/hour =1.5385 hours ≈1 hour 32 minutes
 Fuel consumption
First is to convert 30 miles into kilometers by multiplying it by 1.61 to give
30 x 1.61 = 48.3 km
This means that the consumption is 48.3km/gallon. The consumption per kilometer is calculated
as follows:
48.3 km = 1 gallon
1 km = 1 x 1
48.3 =0.0207 gallons
Hence the consumption is 0.0207 gallons/kilometer.
The next step is to convert 0.0207 gallons into liters by multiplying it by 3.78 to give
0.0207 x 3.78 = 0.078246 liters
Thus the consumption is 0.078 liters/kilometer.
 Fuel quantity
The fuel consumption rate is 0.078 liters/kilometer and the distance to be travelled is 100 miles.
The miles can be converted into kilometers by multiplying by 1.61 to give 100 x 1.61 = 161 km.
Therefore
1 km = 0.078 liters
161 km = 161km x 0.078 liters
1 km =12.6 liters
b) Units of lift
Lift = k x ρ x V2 x A
Where k = dimensionless, ρ = kg/m3, v = m/s and A = m2.

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In terms of unit, lift = [kg/m3] x [m2/s2] x m2 = kgm/s2.
Scenario 3
1. Arithmetic Progression (AP) sequence
 Sixth term
Since it is an AP sequence, the Tn term = a + (n – 1)d; where a = first term, n = number of terms
and d = common difference
In this case, a = b, n = 6 and d = −b
3
6th term = b + (6 – 1) −b
3 = b + −5 b
3 = b – 5 b
3 = −2 b
3
 kth term
As aforementioned, kth term = a + (n – 1)d; a = b, n = k and d = and d =−b
3
kth term = b + (k – 1) −b
3 = b− bk
3 + b
3 = 4 b
3 −bk
3 =b
3 ( 4−k )
 Value of b and sum of AP
The kth term is b
3 ( 4−k ). This means that the 20th term is b
3 ( 4−20 )=−16 b
3
Therefore −16 b
3 =15 → -16b = 45; b = -2.8125
Sum of first 20 terms = n
2 ( 2 a+ ( n−1 ) d ); where n = number of terms, a = first term and d =
common difference.
In this case, n = 20, a = b = -2.8125 and d=−b
3 =2.8125
3 =0.9375
Hence sum of first 20 terms = 20
2 (2 (−2.8125 ) + ( 20−1 ) (0.9375) )

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= 10(-5.625 + 19(0.9375))
= 10(-5.625 + 17.8125)
= 10(12.1875)
= 121.875
2. Geometric Progression (GP) sequence
 20th term
For GP sequence, Tn = arn-1 where a = first term, r = common ratio and n = number of terms
In this case, a = 1, r = ½ and n = 20
20th = 1 x (½)20-1 = (½)19 = 0.000001907
 Value of terms to infinity
Since the absolute value of r is less 1 and greater than 0, the sum of infinite terms of a GP series
is calculated using the formula S= a1
1−r ; where a1 = first term and r = common ratio (Hit
Bullseye, 2019).
In this case, a1 = 1 and r = ½
HenceS= 1
1− 1
2
= 1
1
2
=2
The reason why the sequence of geometric series of infinite terms tends to the value Sn= ∑
n=0
n → ∞
a rn
is because the common ratio r is 0<r<1 (not equal to -1, 0 or 1) hence the sequence show
exponential decay (Lumen, (n.d.)).
3. Equations for x
a) 2Log (3x) + Log (18x) = 27

Building Mathematics 7
Using quotient rule,
Log(3x)2 + Log (18x) = 27
Log(9x2) + Log (18x) = 27
Using product rule,
Log [(9x2) (18x)] = 27
Log 162x3 = 27
162x3 = 1027
x3= 1027
162 ; x=( 1027
162 )1
3
x= 109
5.4514 =183,440,402.7=183.44 x 106
b) 2LOGe(3x) + LOGe(18x) = 9
Removing the parentheses gives
2log e * 3x + log e * 18x = 9
Multiplying 2 by 3 in the first term gives
6log e * x + 18log e * x = 9
Adding the like terms log e * x on the left hand side gives
24log e * x = 9
Dividing both sides by 24 log e gives
x= 9
24 log e = 9
24 x 0.4342945 = 9
10.4231 =0.8635
c) Hyperbolic equations
i) Cosh(X) + Sinh(X) = 5
This is solved using the identity sinh ( x )=¿ −e−x +e x
2 ¿ and cosh ( x )=¿ e−x +ex
2 ¿

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Substituting these identities in the equation gives
e−x +ex
2 + −e−x+ ex
2 =5
e−x
2 + e x
2 +−e−x
2 + ex
2 =5
ex = 5
log ex = log 5
xlog e = log 5
x= log 5
log e = 0.69897
0.4342945 =1.60944
Alternatively, ex = 5 means x = ln 5 = 1.60944
ii) Cosh(2Y) – Sinh(2Y) = 3
This is solved using the identity sinh ( y )=¿ −e− y +e y
2 ¿ and cosh ( y )=¿ e− y +e y
2 ¿
Substituting these identities in the equation gives
e−2 y+ e2 y
2 −−e−2 y +e2 y
2 =3
e−2 y
2 + e2 y
2 − (−e−2 y
2 + e2 y
2 )=3
e−2 y
2 + e2 y
2 + e−2 y
2 − e2 y
2 =3
e−2 y
2 + e−2 y
2 =3
e-2y = 3
-2y = ln 3
y=−1
2 ln 3 = -0.54931