need help in 2006 maths paper sc (1 Viewer)

[palozrule4lyf]

i need help with these questions and can u plz put the answers in detail (how to do them)
19
36
48
51
55
73
79
80

2006 maths sc ( thnkssssss)

http://www.boardofstudies.nsw.edu.au/schoolcertificate/sc2006/mathematics_sctest_06.pdf

 

[maths94]

i need help with these questions and can u plz put the answers in detail (how to do them)
19

my answer was 80 seconds im not quite sure if correct.

Greg:

20sec, 40 sec, 60 sec, 80 sec, 100 sec<--everytime i say sec[seconds] is one lap

Megan:

16 sec, 32sec, 48sec, 64sec, 80 sec<--everytime i say sec[seconds] is one lap

hope thats helps

 

[supre]

36. 15 chocolates cost $9.75 and she has $10
the change she will receive is 25 cents (B)

51. craig has 1/3 chance of winning but as stated dick has 50% chance of winning therefore chance of craig winning is 1/6 (D)

55. divide the rhombus in 8 sections and see there are 3 sections shaded therefore its 3/8 (D)

 

[pqd123]

48)
let x = karishmas age now
in 2 yrs
x + 2 = 3(t + 2)
x = 3t+4

73)
3 + 3 for the sides
1/2 x 2 x 2 x (pi) for the bottom semicircle
1/2 x 2 x 1 x (pi) for the top middle semicircle
2(1/4 x 2 x 1 x (pi)) for the two quarter circles
adding those together you get
6 + 2(pi) + (pi) + (pi)
=6 +4(pi)

80)
let x be the number of pages he's already read
there's 2x pages left in the book
so the total number of pages is x + 2x = 3x
this means that the number of pages in the book must be divisible by 3
therefore the only answer is 303

 

[supre]

48)


80)
let x be the number of pages he's already read
there's 2x pages left in the book
so the total number of pages is x + 2x = 3x
this means that the number of pages in the book must be divisible by 3
therefore the only answer is 303

this answer is wrong because it says there are twice as many pages after the book than before the book.
so u need to minus 1 from every value and divide it by 3 to see if it is a whole number or not.
so for example
A)202= 202-1=201 (201/3=67)
B)303=303-1=302 (302/3=100.6666....)
C)404=404-1=403 (403/3=134.333...)
D)505=505-1=504 (504/3=168)
therefore the two possible answers are A and D.

Does it make sense?

 

[palozrule4lyf]

thnks man and yea i understooooood it

 

[Amogh]

With my capacity as that nooby member, Welcome to BOS
I believe 79 is teh only unanswered question:

From that cumulative graph, we can understand the frequncy table as:
no. of matches--------------------frequency
47----------------------------------------2
48----------------------------------------6
49----------------------------------------4
50----------------------------------------2
51----------------------------------------2
52----------------------------------------4
(for eg. frequency of 49 was foudn by subtracting the cumulative frequncy of 50 and that of 49. By doing so we know the change, ie, the frequncy of 49)
i may sound a bit ambiguous and do tell me if this needs more clarification

from this table we can quite easily know that the
range is 52-47=5
median is 21/2 th number = 49
mode is 48 (most common number)
number of matches counted are 20 (sum of all frequencies)

 

[palozrule4lyf]

thnks for every ones help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!@@@@@@@@@@@@@##########