Math IV Solution Set

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

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 a (1  r n ) 3(1  2 5 ) Sn  1  1 r 1 2 3(1  32) 3(31) Sn   1 r 1 Sn 

4. Solving for the range is also the same as solving for the domain. y=

2x 2  1 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  5 5

y

1 5

 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

n (a1  a 6 ) 2 6 s 6  (1  9) 2

sn 

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

= (3)8 = 24

D

14. Given: a1 =1 sn = 127 a2 =2 n = ? a 2 r 2  2 a1 1 sum of nth terms in geometric progression: sn 

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

a1 (1  r n ) 1 r

127 

1(1  2 n ) 1 2

-127 = 1-2n -127-1 = -2n -27= -2n B n=7

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

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

16.

A 5

x i 1

 x1  x 2  x3  x 4  x5 = 5+6+9+13+14 = 47 B

12 36 108 r   4 12 36

r=3 B 13. Correction in the choices: a. 25 b. 26 c. 25 d. 24 Sum of first 6 terms: a1= -1 a6 = -1 + 5(2) =9

1

17. Array is the arrangement of data according to size or magnitude B

18. n = 30



1

 1224

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 40  42 82 = = 41 2 2

20. Frequent data is 50

B B

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 st 1 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

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 C=

n! (n  r )! r!

n=8

r=4

8! 8 * 7 * 6 * 5 * 4!  (8  4)!4! 4!4!



4* 2*7 *6*5  2*7 *5 4 * 3 * 2 *1

C = 70

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 Probability =

5 8

D

53 12 8 = 12 2 = B 3

30. Probability not blue =