Engineering Mathematics: Solved Problems and Solutions

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Unit 2: Engineering Math

Contents
LO1..................................................................................................................................................3
A) Solution:............................................................................................................................3
A) Solution:............................................................................................................................3
B) Solution.............................................................................................................................4
C) Answer..............................................................................................................................5
D) Solution.............................................................................................................................6
E) Solution.............................................................................................................................7
LO2..................................................................................................................................................9
A) Solution:............................................................................................................................9
B) Answer:.............................................................................................................................9
C) Solution:..........................................................................................................................10
D) Solution...........................................................................................................................11
E) Solution:..........................................................................................................................12
LO3................................................................................................................................................13
A) Solution:...............................................................................................................................13
B) Solution:................................................................................................................................15
C) Solution:................................................................................................................................15
D) Solution:...............................................................................................................................16
E)Solution..................................................................................................................................17
F) Solution:..........................................................................................................................17
G) Solution:..........................................................................................................................18
LO4................................................................................................................................................19
A) Solution:..........................................................................................................................19
B) Solution:..........................................................................................................................19
C) Solution:..........................................................................................................................20
D) Solution...........................................................................................................................21
E) Solution...........................................................................................................................22
References......................................................................................................................................24

LO1
A) Solution:
As per the question, Speed (v) = C px py Vz (Speed for the sound)
And, C is constant value as per the question (i.e. Dimensional)
On solving different dimensional equations, the resultant formula is:
After comparison of values on exponent;
x + 3y= 3z - 1
0-y = x
-1+ 2x = 0
Now, after solving all the equation, resultant solutions are;
y is -0.5, x is 0.5 and z is 0
At last, all the solved value is used for final result,
i.e. (Speed) v=C √ p/q is final solution.
A) Solution:
Here assumptions has been done for the solution;
i.e. F ∝ vx is first assumption
F ∝ ry is second assumption
F ∝ ηz is third assumption
Combining all the three assumptions togather,

The new equation is F (Force) = Kvxry ηz which will be used and have a constant
value for K
Further,
F (Force) = [M1L1T-2] is for force.
V (Velocity) = [L1T-1] is for velocity or speed
r (Length) = [L1] is for length
η = [M1L-1T-1] is for constant
Combining all togather;
{[L1T-1]a [L1]b} = [M1L1T-2] /{ [M1L-1T-1]c} which will gave result for the solution;
i.e. = [Mc La+b+c T-a-c]
Hence, solving the value on the basis of homogeneity principle;
0 = 1- c
A - c = 1 – b
-a –c = -2
After solving both a and c will have “1” value
Using a& c, b -1 = 0 (i.e. b = 1)
Lastly, all the values of a & c along with b has been taken for final solution;
F = kηrv is final solution.
B) Solution
As per the question, S8 = 2S5
a = 3
The formula for Sn is;
Sn ¿ { n
2∗(2 a+ ( n−1 ) d)} in standard form

After inserting S8 & A value in the formula for Sn;
8
2∗6=¿5(6 + 4d) – 8
2 (8 -1)d
(24) - 30 = 20d - (4*7d)
24 + 28d = 20d + 30
Once solved completely, d = 0.
As per the question, 8 − 4 + 2 − 1 + . . .
Here, n is 5 & a is 8
Inserting n & a in the formula for Sn;
Sn = [8 – 8(3/4)5)]
Further, S5 = ¿ ¿ ¿
( 4−12
4 )= 24.40
Lastly; S5 = 24.40 is final solution.
C) Answer
As per the question,
The solved value of d is;
d = 10 (i.e. Miles)

Further, angle of elevation tangentially has been considered;
i.e., tan (200)* (d + x) = h
Similarly, tan (600)*x = h
Both angles has been compared to exclude x from solution;
h (height) = 4.6 (i.e. Miles)
D) Solution
(1) solution: Graph for the solved value of radioactivity has been developed below:
0 1 2 3 4 5 6 7 8 9 10 11
0
5
10
15
20
25
Amount of radioactivity
Amount of radioactivity
Time (Half life a week)
2) Solution:
Total value for radioactivity solved on the time value has been shown below:
i.e. B (sub t) / B (sub 0) = (0.5) ̂ ( time
half −life )
In which, B(t) = Radioactivity count at t
Also, B (sub 0) = Radioactivity count at zero time
3) Solution:
As per the question,

i.e., B (sub 0) = 20
And, Time (t) = 27 days (i.e. 3*7 days)
Considering calculated value;
B (sub t) / B (sub 0) = (0.5) ̂ ( time
half −life ) = 20 ( 1
2) ̂ ( 21
7 )
= 2.5 (i.e. Radioactive count for each second)
E) Solution
a) Solution:
The equation for exponential for is;
Y/(C) = (Ax)
And, Y = 2000 (i.e. Initializing year)
x = no. of person with phone (i.e. in thousands)
As per the question;
A is (375/225) ̂ (1/2) = 1.29
Solution found after inserting A in the equation;
Now, Calculating with the help of value for A, we have;
Y/ C = (1.29x)
Further, C = 225
Considering all the solved values;
Y = (225)(1.29x) is the final solution.
b) Solution:
Finding years for solution (x) = 2025-2000 = 25 (i.e. Years)
Considering x = 25;
Y = (225)*(1.2925)

Y = 130895.669 is final solution.
c) Solution:
As per the question,
Taking value of year for solution = 2006 & 2010
Solution after considering x = 6 & 10;
Y1 / (225) = 1.296
i.e. Y1 = 1036.86
Similarly, Y2 / (225) = (1.2910)
Y2 = 2971. 307
Further, solving average change rate, final result will be;
(Y1 - Y2) / (x2-x1)
i.e. (Y1 - Y2) / (x2-x1) =1934.447 / (4) = 483.61 is final solution.
d) Solution:
As per the question year for consideration is 2016,
i.e. x = 6
Considering the closest value of 6 for solution;
i.e. 5.999 & 6.001
Doing solution on the basis of considered x;
Y1 / (225) = 1.295.999
Y1 = 1036.597
Similarly, Y2 / (225) = 1.296.001
Y2= 1037.125
Inserting all the values, average rate of change = (Y1 - Y2) / (x2-x1)

= (0.528)
(6.001−5.999) on solving
Average rate of change = 264.02 is final solution.
LO2
A) Solution:
As per the question;
Samples = 10
Solution for finding mean value has been done using;
x=(1/n)∑
i=1
n
x
i.e. x = Solved Mean
x = mid sample
n = total samples
So, as per the question;
x=(0.1)∑
i=1
n
x
i.e. x = 49.2 (solved value)
Taking, formula for standard deviation;
i.e. S= √ [ 1
1−n ]∗∑
i=1
n
( x−x)2
Inserting value of n which is 10;
Final value of S= 17 (approximate value for solution)
B) Answer:
Different values and data with variables given or taken randomly is called discrete which results
in complex calculation as well as analysis of the data for the recorded value due to which

different simple methods has been developed in order to generate simple and better results on the
randomly considered data which has been done using approximation of binomial distribution for
the provided random data (Taylor, 2018)
As per the question; n= is 20 & p is 0.25
Considering formula for the solution;
μ/n= p
Inserting provided value, μ/ 20 = (0.25)
After solving, μ = 5
In the same way, σ2 = np- np *p
Once solved, σ2 = 3.75
i.e., σ = 1.94
(Feller, 2015)
C) Solution:
As per the question;
μ = 10 (i.e. Average life which is provided in years)

σ = 2 (i.e. value for standard deviation)
Further;
Z* σ = X −μ
Considering, P (Z≤z) = 0.03
Once solved, Z will be 1.88
Now, inserting solved Z value, μ value & σ value;
i.e. -1.88*2 + 10 = X
Further, (X – 10)/2 = 1.88
Finally, X = 6.24 is final solution.
D) Solution
1) Solution:
As per the question, n is 10 which is total no. of pistons
q is 0.88 (i.e. success probability)
p is 0.12 (i.e. average rejection probability)
Considering, x as no of rejected pistons;
Taking, x = 0 (i.e. zero rejection)
Similarly, x = 1 (i.e. 1 rejected)
Lastly, x = 2 (i.e. 2 rejected)
Taking formula for Probability;
i.e. P = P(X ≤ x)
Further, inserting value for x and solving;
P (at x = 0) is 10C0 (0.12)0 (0.88)10-0
P (at x = 0) is 0.2785

Similarly, P (at x = 1) is 10C1 (0.12)1 (0.88)10-1
P (at x = 1) is 0.3797
Similarly, P (at x = 2) is 10C2 (0.12)2 (0.88)10-2
P (at x = 2) is 0.2330
Considering, x which is not more than 2;
P (at x ≤ 2) is P (at x=0) + P (at x=1) + P (at x=2)
i.e. P (at x ≤ 2) = 0.8912
2) Solution:
Considering x which is at least 2;
After inserting x= 2, P = 1 – P (i.e. at x ≤ 1)
P = 1 – (Probability at x=0 + Probability at x=1)
P + (0.2785 + 0.3797) = 1
Finally, P = 0.341 which is final solution.
3) Solution:
After solving all the results, the value for the probability of rejections for the pistons which is
0.8912 and refers to the beneficial value for the company because the average value for the
rejection for pistons after solution is 0.341 which represents towards the benefit for the company.
E) Solution:
Considering standard formula,
i.e. Z *σ = X −μ
As per the question;
At 1250 hours = 250
125
i.e. Z = 2