Math IV Solution Set:
1. Function defined as a relation
with one-to-one correspondence.
It means that for every value of
X in the given equation, there
corresponds a unique value of Y.
y  x , is not a function
because given a value of x say
x=4, there corresponds two
values for Y, 2 and -2 C
2.
C
3. Solving for the domain, take note
that division by zero is
undefined. From the given
equation, x-3 must not be equal
to zero and must not be included
into the domain. Therefore the
domain of f(x) are all real
number except x = 3. D
4. Solving for the range is also the
same as solving for the domain.
2x 2  1
y=
5
5y = 2x2-1
5y + 1 = 2x2
5y 1
 x2
2
5y 1
x
2
Range is defined as all possible
values of y correspond to value
of x. From the given equation
5y 1
must be greater than or
2
equal to zero.
5y 1
0
2
5y +1  0
5y
1
1

y
5
5
5
Therefore, range for given
equation: all real numbers greater
than or equal to zero. D
5. Correction to the question:
How much will he save on the
last week of the third month?
a1= 10.00
d = 4.00
an= 4 * 3 = 12wks
a12 = a1+(12-1)d
= 10.00 + (11) 4.00
= 10.00 + 44.00
= 54.00
B
6. 1st Quiz = 3
2nd Quiz = 6
3rd Quiz = 12
a1 = 3
r=2
a1 (1  r n ) 3(1  2 5 )
Sn 

1 r
1 2
Sn 
3(1  32) 3(31)

1 r
1
Sn 
 93
 93
1
93/5 = 18.16
A
7. Statement I is false because it
shows
geometric
sequence.
Statement II is false because the
next term is 42. C
8. Given: a1= 6
d = -3
an= -51
n=?
an = a1+ (n-1)d
-51 = 6+(n-1)(-3)
-51 = 6 +-3n+3
-51 = 9 + -3n
-51-9 = -3n
-60 = -3n
n = 20
D
9. Given :
a7 = 20
a20 = 59
a7 = a1 +(7-1)d
a20 = a1 +(20-1)d
20 = a1+6d
59 = a1+19d
Solving by elimination, we can
solve the value of d.
a1+19d = 59
- a1+6d = 20
a1+19d = 59
+- a1-6d = -20
13d 39

13 13
d=3
Solving for a1: Solving for a25:
20=a1+6(3)
a25 = a1+24d
20 = a1+18
= 2 + 24(3)
a1 = 20-18
= 2+72
a1 = 2
a25 = 74 D
10. Getting the distance between the
two poles, we must get the
number of between the two
poles: 51-1 = 50
2,250 / 50 = 45 meters B
11. a1= 5,600 * 12
= 67,200 (starting annual salary)
n=1
d = 250
a11 = a1+(11-1)d
= 67,200 +10 (250)
= 67,200 +2,500
= 69,700 B
b. 26
c. 25
d. 24
Sum of first 6 terms:
a1= -1
a6 = -1 + 5(2)
=9
n
sn  (a1  a6 )
2
6
s6  (1  9)
2
= (3)8
D
= 24
14. Given: a1 =1
sn = 127
a2 =2
n=?
a
2
r 2  2
a1 1
sum of nth terms in geometric
progression:
sn 
a1 (1  r n )
1 r
1(1  2 n )
1 2
-127 = 1-2n
-127-1 = -2n
-27= -2n
B
n=7
127 
15. After 6 hrs = 4 amoebas
After 12 hrs = 4*4= 16 amoebas
After 18 hrs =16*4= 64 amoebas
After 24 hrs =64*4=256 amoebas
12. Solving for common ration:
a
a
a
r 2  3  4
a1 a 2 a3
12 36 108
r


4 12 36
16.  x1  x1  x 2  x3  x 4  x5
r=3 B
13. Correction in the choices:
a. 25
17. Array is the arrangement of data
according to size or magnitude
A
5
i 1
= 5+6+9+13+14
= 47
B
B
  1224
18. n = 30
1
average: 1224/30 = 40.8
B
19. To solve for the median, arrange
the set of data in increasing
order.
21 24 25 28 29 31 31 33 35 35 36
36 37 38 40 42 43 45 45 46 47
49 50 50 50 51 52 55 58 62
15th = 40
16th = 42
N  1 30  1 31
median =
=
=
2
2
2
= 15.5th score
x15 = 40
x16 = 42
25. Solving for permutation for
different objects
nPn =n(n-1)(n-2)...(3)(2)(1) or n!
5! = 5*4*3*2*1
= 120 ways B
26. Solving by circular permutation
P = (n-1)!
= (5-1)!
C
= 4! = 24
27. Solving by combination
n!
C=
(n  r )!r!
n=8
r=4
8!
8 * 7 * 6 * 5 * 4!

(8  4)!4!
4!4!
40  42 82
=
= 41
2
2
B

20. Frequent data is 50
B
C = 70
21. Correction in choices:
a. 55 b.50 c. 41
d. 40
Range = Highest value – Lowest Value
= 62 – 21 = 41 B
22. B
23. By fundamental principle of
counting
1st Dice
2nd Dice
1
1
2
2
3
3
4
4
5
5
6
6
n1 * n2
6*6
= 36 ways B
24. 3! = 3*2*1 = 6
C
4* 2*7 *6*5
 2*7 *5
4 * 3 * 2 *1
D
28. Let n = Total number of houses
n-1 = Odd number of house
n 1
= Houses that are painted blue
2
C
29. Total number of delegates = 8
Total number of boys = 5
5
D
Probability =
8
53
12
8
=
12
2
=
B
3
30. Probability not blue =