Mathematical Foundations for Computing: Semester 1

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Maths for Computing
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Contents
LO1..................................................................................................................................................4
Part 1............................................................................................................................................4
1.1)...........................................................................................................................................4
Part 2............................................................................................................................................5
2.1)...........................................................................................................................................5
2.2)...........................................................................................................................................7
2.3)...........................................................................................................................................8
LO2................................................................................................................................................10
Part 1..........................................................................................................................................10
1.1).........................................................................................................................................10
1.2).........................................................................................................................................11
Part 2..........................................................................................................................................12
2.1).........................................................................................................................................12
LO3................................................................................................................................................14
Part 1..........................................................................................................................................14
3. 1 ).......................................................................................................................................14
3.2).........................................................................................................................................15
Part 2......................................................................................................................................16
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3.1..........................................................................................................................................16
3.5).........................................................................................................................................17
LO4................................................................................................................................................19
Part 1..........................................................................................................................................19
1.1).........................................................................................................................................19
1.2).........................................................................................................................................19
Part 2..........................................................................................................................................21
2.1).........................................................................................................................................21
2.2).........................................................................................................................................21
2.3).........................................................................................................................................22
References......................................................................................................................................26
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LO1
Part 1
1.1)
Provided- Set of values: 2, 3, 5, 7 and 11.
To evaluate- GCD and LCM among the given set of numbers:
Answer: It can be clearly seen, each of the values present in the given set is a prime number;
therefore the GCD for the provided set of numbers = 1
For LCM of Provided set of values: (2)(3)(5)(7)(11) = 2310
G.C.D. = 1
L.C.M.= 2310
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Part 2
2.1)
Provided- Initial value of the A.P. = 9
The 10th value of the given A.P. = 40. 5
The last value of the A.P. = 425. 5
To evaluate - Total no. of terms, addition of every term and the value of the 70th term.
Answer:
It is already known that
The ‘n’ th term of an A. P. = a+(n–1)d
a=9 and 10 th value is=40. 5
Hence we can state that
a+(10–1)d=40. 5
9+(10–1)d=40. 5
9+9*d=40. 5
9*d=40.5–9
9*d=31. 5
d=31.5/9
d=3. 5
The common difference (d) of the given A.P. is =3. 5
As per the earlier findings
5

The ending value of the given A.P. series is=425. 5
We also know that the last term of an A. P. is =a+(n–1)d
Therefore we can say that
a+(n–1)d=425. 5
9+(n–1)*3.5=425. 5
(n–1)*3. 5=425. 5–9
(n–1)*3. 5=416. 5
n–1=416. 5/3. 5
n–1=119
n=119+1
n=120__________ (Answer ( i ) )
Hence the total number of values in this arithmetic progression is 120.
It is already know that,
Addition of all the values in A. P is =Sn=(n/2)*[2*a+(n–1)*d]
Sn =(120/2)*[2*9+(120–1)*3.5]
Sn =60*[18+(120–1)*3.5]
Sn = 60*[18+119*3.5]
Sn = 60*[18+416.6]
Sn =60*[434.5]
Sn =26070_______________ (Answer(ii))
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The addition of all values of the given A. P.=26070
The 70th value of the A. P.=
a70=a+(70–1)*d
a70=9+(69)*3. 5
a70=9+241. 5
a70=250. 5_____________________(Answer (ii) )
The 70th value of A. P.=250. 5
2.2)
Provided- The basic salary=7200
Annual Incrementing value =350
To evaluate –
i) Salary in 9th year
ii) The money he earned in 12th year.
Answer:
1st year income=7200
As per the given question
2nd year income=72350
3rd year income=72700
It makes the scenario, an AP of d = 350 and a = 7200
To evaluate income in the ninth year, means to evaluate the ninth value of this A.P.
7

To find ninth value of A.P.
a9 =a+(9–1)*d
a9 =7200+(8)*350
a9=7200+2800
a9 =10,000______ (Answer (i) )
The income of the worker in the ninth year=10,000
Now, total amount received after 12th year = Addition of the A.P. up to 12values.
S12 =12/2(2*7200+(12–1)*350)
S12 =12/2(2*7200+(11)*350)
S12 =12/2(2*7200+3850)
S12 =12/2(14400+3850)
S12=18250*6
S12=109500___________________ (Answer (ii) )
To total amount after 12 years=109,500
2.3)
Provided –Velocity variation of the machine = 50 rpm – 750 rpm
Number of velocities the device can be varied in=6
The given velocities in the range are in a form of a G. P.
To evaluate- All the velocities
Answer:
8

Let the values of the G. P=a,ar,ar2…arn−1
Here n=6
The 1st term a1 = 50rev / min
The 6th value =a * r 6-1
i. e.
ar5=750
Which r5=750 / a=750 / 50=15
Hence,
The ratio (r) of the Geometric Progression =5 √15=1. 7188
The 1st of all the values
a=50rev/min
Therefore is can be seen,
ar=(50)*(1.7188)=85. 94,
a r2=(50)*(1. 7188) *2=147. 71,
a r3=(50)*(1. 7188)*3=253. 89,
a r4=(50)*(1. 7188)4=436. 39,
a r5=(50)*(1. 7188)*5=750. 06
Hence, velocities of the machine are 50, 86, 148, 254, 436 and 750 rev / min
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LO2
Part 1
1.1)
Provided: Number of persons = 500
Mean height of persons = 170 cm
Standard Deviation=9cm
To evaluate: Number of persons probably with height ranging among the values=150cm–195cm
Answer:
Finding the z value,
I.e. z= (x–Mean)/Standard Deviation
z1=(150–170)/9
z1=-2.22
These values correspond to region 0. 4868 that is present at the left side of z=0 ordinates
for z values of 195
z2 = (195–170)/9=2. 78
These values correspond to region 0. 4973 that is present at the left side of z=0 ordinates
Total area for the range =z1+z2=0. 4868+0. 4973=0. 9841
Since the obtained region is directly proportional to the probability(P),
So, the probability of an individual having height 150–195=0.9841
Number of individuals with height in this range = 0. 9841*500=492 _________ (Answer 4)
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1.2)
Provided- Number of male candidates in the group =20
Number of female candidates in the group=33
To evaluate: Probability of a male candidate getting chose
Probability of a female candidate getting chose
Answer:
Total number of persons in the collective =20+33=53
Probability ( Pm ) of a male candidate is chosen = Number of male candidates/Total Candidates
P m=20/53=0. 378______Answer ( i )
Similarly Probability (P w ) of a female candidate is chosen = Number of Female Candidates /
Total Candidates
Pw =33/53=0. 622__________ Answer(ii)
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Part 2
2.1)
The expectation of any event is defined as the average occurring of any event and the evaluated
by having the product of the total probability and the number of attempts.
For this case,
Provided: Probability of obtaining a 4 in rolling 1 dice=
No. of 4 on a dice/ Total no. of outcomes= 1/6
If 3 attempts are made, then in that case Expectation= P*n
Here P= 1/6 and n= 3
Expectation = 1/6*3 = 0.5________________(Answer)
2.2)
Provided – No. of children=4
To evaluate – Probability of at least one female
Probability of a male and a female
Answer:
Probability of a male=½
So, probability of 4 male = ½*½*½*½ =1/16
So the probability of having at least a female=1–1 / 16
=15 / 16=0. 9375____________ (Answer( i ))
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