HNCB 8 Mathematic for Construction - Desklib

Problem Solution
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HNCB 8 MATHEMATIC FOR
CONSTRUCTION

Problem Solution
Contents
TASK 1 SCENARIO 1....................................................................................................................4
Solution 1.....................................................................................................................................4
Solution 2.....................................................................................................................................4
TASK 1 SCENARIO 2....................................................................................................................5
Solution a)....................................................................................................................................5
Solution b)...................................................................................................................................5
TASK 1 SCENARIO 3....................................................................................................................6
Solution 1.....................................................................................................................................6
Solution 2.....................................................................................................................................7
Solution 3a...................................................................................................................................7
Solution 3b...................................................................................................................................8
Solution 3c...................................................................................................................................8
TASK 2 SCENARIO 1..................................................................................................................11
TASK 2 SCENARIO 2..................................................................................................................17
Solution A..................................................................................................................................17
Solution B..................................................................................................................................19
TASK 3 SCENARIO 1..................................................................................................................19
TASK 3 SCENARIO 2..................................................................................................................24
TASK 4 SCENARIO 1..................................................................................................................26
Solution a.i)................................................................................................................................26
Solution a.ii)..............................................................................................................................26
Solution b).................................................................................................................................26
Solution c)..................................................................................................................................27
TASK 4 SCENARIO 2..................................................................................................................27
Solution a)..................................................................................................................................27
Solution b).................................................................................................................................28
Solution c)..................................................................................................................................28
Solution d).................................................................................................................................28
TASK 4 SCENARIO 3..................................................................................................................29
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Problem Solution
TASK 1 SCENARIO 1
Solution 1
As given in problem,
Since there is no description of height of the tank, in this condition, Area given must be area of
the bottom of the tank.
Suppose the width is b, The length l will = b + 3.2 m
Area of the bottom od the surface = b(b+3.2) = 26.5 m2
b2+3.2 b−26.5=0
The root of the equation will be given rectangular tank,
The roots of the equation will be
x1,2=−b ± √ b2−4 ac
2 a =3.2 ± √ 3.22−4 x 26.5
2 = 3.2± √ 10.24+106
2
x1,2=−3.2 ± √ 10.24+106
2 =−3.2 ± 10.78
2
x1,2=−1.6−5.39 ,∧−1.6 +5.39
x1,2=−6.990733 ,∧3.7907
Since negative length is not accepted, the width is = b = 3.7907
Then length will be l = 3.7907+3.2 = 6.990733 m Ans
Solution 2
Suppose daily forfeited amount = £ x,
And Contract amount =£ y
For 5 days late, the contractor receives, y – 5x = £ 4250............. (i)
For 12 days late the contractor receives, y-12x =£ 2120.......... (ii)
Subtracting the equation (ii) from equation (i)
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Problem Solution
y- 5x -y+12x = 4250-2120, 7x = 2130, or x = £ 304.29,
The late penalty amount = £ 304.29
The original contract amount = y = 2120 + 7x304.29 =£ 2130.3 +£ 2120 = £ 4250.03
TASK 1 SCENARIO 2
Solution a)
The speed of the car is 65 miles /hr
 Speed in meter / second
65 miles/hr = 65*1760/3600 = 31.77 yards /sec,
The speed in meter /sec = 31.77 * 0.91 = 28.917 m/sec
 The time taken to drive 100 mile = 100/65 = 1.54 hrs Ans
Fuel consumption, 30 mile /gallon
= 30*1760 yards / gallon = 30*1760*0.91 / 3.78 litres
= 48048 / 3.78 = 12711 / litres = 12.711 km /litres
The amount in litres per kilometre = 0.0786 litres / km Ans
 The completed journey of 100 mile = 100*1760*0.91/1000 = 160.16 km
Since, the consumption of fuel = 0.0786 litres / km
Then, fuel required for complete journey = 0.0786 x 160.16 = 12.59 litres
Solution b)
As given in question, lift is directly proportional to air density, air speed, and surface area,
All in MKS unit
Units of area = m2
Unit of density = kg/m3
Unit of velocity = m/s
Then, Lift =k x ρx V 2 x A
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Problem Solution
Lift =k x kg
m3 x m
s x m2=kg/ sec
The unit for lift = kg/sec Ans
TASK 1 SCENARIO 3
Solution 1
1 the Athematic sequence as given b, 2b/3, b/3, 0
The difference in the term = 2b
3 −b=2 b−3 b
3 =−b
3
The common difference = −b
3
The first term = b,
 The 6th term will be = a1 + (6-1) d Putting the values,
= b− 6 b
3 − b
3 = 3 b−6 b−b
3 =−4 b
3 Ans
 The kth term will be = a1 + (k-1) d Putting the values,
= b− kb
3 − b
3 = 3 b−kb−b
3 = 2 b−kb
3 Ans
 The 20th term will be = 2b−kb
3 = 2b−20∗b
3 =−18 b
3
As given in question, −18 b
3 =15 ,
-18b = 45, b = -45/18 = -5/2
The value of b = -5/2 Ans
 The sum of first 20th term,
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Problem Solution
S20th=2 a1+ ( n−1 ) d
2 . n
S20th=2 b− ( 20−1 ) b /3
2 .20=6 b−19 b
6 x 20=−13 b
3 x 10
Putting the value of b,
= −13
3 x− 5
2 x 10= 20∗13
3 =260/3 Ans
Solution 2
First, we must fine common ration of the given =
1
2
1 =1
2
The nth term is given as = arn-1
The 20th term = 1. ( 1
2 )
20−1
= 1
219 =1.90735 x 10-06 Ans
The sum of the geometric progression = Sn= a (1−r n)
1−r
If the series goes to infinity,
S∝= a(1−r∝)
1−r ......(i)
As we can see that, the common ratio of the given GP is ½, in this condition, value rn will
become smaller and smaller and if n = infinite, then rn will be equal to zero.
In this condition, the equation (i) will become
S∝= a(1−r∝)
1−r = a
1−r Ans
Solution 3a
2log(3x) + log(18x) = 27
2log(3x) + log (32 Xx 2) = 27
2log(3x) +2log(3x) +log (2) = 27
4log(3x) = 27-0.301 = 26.699, or log(3x) = 26.699/4 = 6.67475
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Problem Solution
Log(3x) = 6.67475
Removing log from above
3x = 106.67475 = 550321470
X = 183440490 Ans
Solution 3b
2Log.e(3x) + Loge(18x) = 9
This is the function of natural logarithm, first we will resolve the equation as above, then
wo will solve it.
2loge (3x) + loge (18x) = 9
2loge (3x) + loge (32X x 2)
 2loge (3x) + 2loge (3x) +loge (2)
 4loge (3x) + loge (2) = 9
 4loge (3x) = 9-0.693
 4loge (3x) = 8.306853
 loge (3x) = 2.077
 3x = e2.077 = 11.056
 X = 3.685316 Ans
Solution 3c
i) Cosh(X) + Sinh(X) = 5
Since the value of cosh (x )= 1
2 (ex+ e−x)
And sinh( x)= 1
2 (ex−e− x)
Putting the value in question
1
2 ( ex +e− x )+ 1
2 ( ex−e− x )=5
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Problem Solution
⟹ ( ex +e− x ) + ( ex−e− x )=10
⟹ ( ex ) + ( ex )=10
⟹ 2 ex=10
⟹ ex=5 Taking natural log on both side
X = loge (5) = 1.609 Ans
(ii) Cosh(2Y) – Sin.h(2Y) = 3
Since the value of cosh (x )= 1
2 (ex+ e−x)
And sinh( x)= 1
2 (ex−e− x)
The value of cosh ( 2Y ) = 1
2 (e2 Y + e−2Y )
Similarly, sinh ( 2 Y ) = 1
2 (e2Y −e−2 Y )
cosh ( 2Y ) – sinh ( 2Y )=3
⇒ 1
2 ( e2 Y +e−2 Y ) −1
2 ( e2Y −e−2Y ) =3
⇒ ( e2Y +e−2 Y )− ( e2 Y −e−2 Y ) =6
⇒ e−2Y +e−2 Y =6
⇒ e−2Y =3 Putting natural log on both side
⇒ Lo ge e−2 Y =loge 3 Same base is cancelled
-2Y = 1.098612
Y = -1.098612/2 = -0.55 Ans
(iii) Cosh(K) * Sinh(K) = 2
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