DEPARTMENT OF GENERAL STUDIES
YANBU INDUSTRIAL COLLEGE
MATH – 001 (Algebra)
Mid–Term Examination – Semester I , 2014 - 2015
ANSWER KEY
Question 1 ( 2 + 4 + 4 = 10 Marks)
(a)
Write the expression without absolute value symbol:   4
(2 Marks)
Solution:
(b)
 4   4
 2 Marks
Given A = { 1 , 2 , 3 } , B = { 1 , 3 , 5 , 7 }. Find
(i) A U B , (ii) A ∩ B
(4 Marks)
Solution:
(ii) A ∩ B = { 1 , 3 }
(c)
 2 Marks
(i) A U B = { 1 , 2 , 3 , 5 , 7 }
Graph and write in interval notation:
x|
 2 Marks
x  7
(4 Marks)
Solution:
Graph:
-

7
Interval:
 
, 7
 2 Marks
(Page 1 of 5)
 2 Marks
Question 2
(a)
Simplify:
( 3 + 3 + 4 = 10 Marks)
3x y 6 x y 
3
2
2
5
(3 Marks)
Solution:
(b)
3x y 6 x y  
3
2
2
5
18x 5 y 7
 3 Marks
Write the polynomial 5x2  3x  1  9x3 in standard form.
Identify the degree of the polynomial.
(3 Marks)
Solution:
Standard form:
Degree: 3
(c)
9x 3  5x 2  3x  1
 2 Marks
 1 Mark
Evaluate the polynomial x 2  5x  7
for x   2
(4 Marks)
Solution:
x 2  5x  7

 22  5  2
 7
 4  10  7
 21
 1 Mark
 2 Marks
 1 Mark
(Page 2 of 5)
Question 3
(a)
( 6 + 4 = 10 Marks)
Factor: ( i ) 4a2  9
,
x 2  8x  15
( i i)
(6 Marks)
Solution:
4a 2  9
( i)
 2a  3 2
2
 1 Mark
 2a  3 2a  3 
x 2  8x  15
( ii )
 x  3x  5
(b)
 2 Marks
 3 Marks
x 3
Solve the absolute value equation:
 8
(4 Marks)
Solution:
x 3
 8
 x3  8
or
x  3  8
 x  11,  5
 2 Marks
(Page 3 of 5)
 2 Marks
Question 4
(a)
( 2 + 5 + 3 = 10 Marks)
Write the complex number in standard form:
25 
(2 Marks)
Solution:
(b)
Simplify:
25 
 49  5  7i
 2 Marks
3  4i 5 2 i 
(5 Marks)
Solution:
3  4i 5  2 i 
 15  6i  20i  8i 2
 15  14i  8 1
 15  14i  8
 23  14i
(c)
Evaluate the power of i :
 2 Marks
 1 Mark
 1 Mark
 1 Mark
i
40
(3 Marks)
Solution:
i 40

i 

 120
2 20
 1
 1 Mark
 1 Mark
 1 Mark
(Page 4 of 5)
 49
Question 5
(a)
Solve:
( 3 + 4 + 3 = 10 Marks)
2
x 1  7
3
(3 Marks)
Solution:
2
x 1  7
3
2
x  6
3
 2x  18
 1 Mark

 x  9
(b)
Convert the decimal number
 1 Mark
 1 Mark
841 0
to its binary equivalent.
(4 Marks)
Solution:
2
2
2
2
2
2
8410
(c)
84
42
21
10
5
2
1
0
0
1
0
1
0
 10101002
3 Marks

1 Mark
A triangle has a perimeter of 91 centimeters. Each of the two longer sides of the triangle
is three times as long as the shortest side. Find the length of each side of the triangle.
(3 Marks)
Solution:
Let Shortest side = x , Then Each of Two Longer sides = 3x
Perimeter = 91
 3x  3x  x  91
 0.5 Mark
 1 Mark
 7x  91
91
 13  1 Mark
7
 Shortest side  13 cms , Each of Two Longer sides  313  39 cms  0.5 Mark
 x 
(Page 5 of 5)