Mathematics Foundations: University Assignment 02 - Solutions

This document presents detailed solutions to a math assignment. The assignment covers a wide range of mathematical concepts, including calculations involving ratios, percentages, and ordering of numbers. It also includes problems on population ratios, and percentage increases. Further solutions are provided for algebraic problems, such as solving linear equations using determinants, and matrix operations like addition, multiplication, finding determinants, and inverses. Additionally, the assignment addresses arithmetic progressions, determining whether a number is a term in a sequence, and solving algebraic expressions with consecutive terms. The solutions are comprehensive, providing step-by-step explanations for each problem.

Assignment 02
1.1)
= 33
50=0.66
68.75%=0.6875
0.673
2:3=0.6
Therefore the numbers in descending order is,
0.6875>0.673>0.6>0.66
That is,
68.75%>0.673>2:3> 33
50
Hence the answer
1.2)given,
men:women:children = 6:7:3
total women = 42000
let the ratio be x
the ratio becomes 6x:7x:3x
7x=42000
x=6000
1.2.1)
total people living together in Outapi is,
6x+7x+3x
=16x
Where x =6000
Total people=16*6000

=96000
Hence the answer.
1.2.2)
Number of children living in outapi=3x
=3*6000=18000
Hence the answer.
1.2.3)
42000=20% increase in 10 years.
That is 120% increase.
42000=120x%
X=42000/1.2
=35000
Number of women lived 10 years ago=35000
Hence the answer.
1.2.4)
Given 12000 attend school and 48% of them are boys.
12000*48%=5760
Therefore number of boys attending the school are=5760
So number of girls attending school are = 12000-5760
=6240
Hence the answer.
2)
Given, -5≤x≤-3 and -1≤y≤2
To find the largest possible value of
2.1) x+y

Largest value of x = -3
Largest value of y= 2
So x+y = -3+2
=-1
Hence the answer.
2.2) xy
Smallest value of x = -5
Smallest value of y = -1
So largest value of xy to be positive= -5*-1
=5
Hence the answer.
2.3) x2 y
Smallest value of x=-5
Square of smallest value= −52
=25
Largest value of y =2
Therefore x2 y=50
Hence the answer.
3.1)
Given, 2 x−5 y
3 =3
2x-5y=9…………………………..1)
And, 7 x
3 =5+ y
3
7x-y=15……………………………2)

The solution is x= Dx
D and y= Dy
D
Where D is the determinant,
D= |2 −5
7 −1|=-2+35=33
And Dx=| 9 −5
15 −1|=-9+75=66
And Dy=|2 9
7 15|=30-63=-33
Therefore x= Dx
D = 66
33 =2
And y= Dy
D =−33
33 =-1
Hence the answer.
3.2)
Given, A= [2 −4
1 3 ]
B= [ 3 1
−1 2 ]
C= [ 4
3 ]
D= [ 3 1 ]
E= [−3 2 0
1 −1 −2 ]
3.2.1)
A+B=[2 −4
1 3 ]+ [ 3 1
−1 2 ]
=[ 5 −3
0 5 ]

Hence the answer.
3.2.2)
AC=[2 −4
1 3 ]∗[4
3 ]
=[ −4
13 ]
Hence the answer.
3.2.3)
B= [ 3 1
−1 2 ]
Determinant |B|=3∗2− (−1∗1 )
=7
Hence the answer.
3.2.4)
= B2= [ 3 1
−1 2 ]* [ 3 1
−1 2 ]=[ 8 5
−5 3 ]
Hence the answer
3.2.5)
= A−1
= adjA
|A|
adjA= [3 −1
4 2 ]T
=[ 3 4
−1 2 ]
determinant A, | A|=10
Therefore A−1 = 1
10 [ 3 4
−1 2 ]
Hence the answer.
4)

4.1)
T=number of people who like treadmill=935
E= number of people who like elliptical machine. =846
E-T= number of people who only like elliptical machine. =321

4.2)
Number of people who liked both machines, = 525.
%= 525
2050*100=26.609%
4.3)number of people who only liked treadmill = 935-525=410
4.4)number of people who liked neither of the machine = 2050-(410+525+321)
=1256
%age= 1256
2050 *100=61.268%
Hence the answer.
5.1)
Interest= Principal*rate*time(in years)
Here interest=17$
Rate=6%
Time= 4 months=4/12 yrs=1/3 yrs
Substituting in the equation,
I=Prt
17=P*0.06* 1
3
17=P*0.02
P=850$
Hence the answer
5.2)
∑
i=3
6
(i+1 )2
Terms= 42 +52 +62 +72
=16+25+36+49=126

Hence the answer.
6.1)
given
11, 8, 5, 2……
The given sequence is an arithmetical progression.
Here a=11
Common difference = -3
General term in an A.P is given by, an
=an=a + (n-1)d
Where n is the number of terms.
To find whether -150 is a term of this progression,
Substitute an=−150
That is -150=a + (n-1)d
=-150=11 + (n-1)*-3
=-161 = (n-1)*-3
Which leads to n is a fraction which cannot be true.
So -150 is not a term in the sequence.
6.2)
Given,
A,B,C,D are consecutive terms in an arithmetic progression.
Therefore let A be the first term,
B be the 2nd term,
C be the 3rd term, and
D be the 4th term.
Let the common difference be x.

The relation between A ,B,C,D=
B=A + x
C=A + 2x
D=A + 3x
To find,
D2− A2
C2−B2
D2=( A+3 x)2
A2= A2
C2= ( A +2 x ) 2
B2= ( A+ x ) 2
Substituting in the above equation…
D2− A2
C2−B2 = ( A+ 3 x )2− A2
( A +2 x ) 2− ( A + x ) 2
= A2 +6 Ax+9 x2− A2
A2 +4 Ax+ 4 x2− A2 −2 Ax−x2
= 6 Ax +9 x2
2 Ax+ 3 x2
=3 ( 2 Ax +3 x2
2 Ax +3 x2 )
=3
Hence the answer.