2004 school certificate..help (1 Viewer)

[Riot09]

can you please help me with these 2004 school certificate questions because i have no idea how to solve them,please include working out and a explanation with the answer,Thank you.

www.boardofstudies.nsw.edu.au/schoolcertificate/sc_mathematics.html

(2004)


the only ones left unswered are:

49,58,69,,72,73,75,80,82,83 (i got 73% for b is this right because im not sure?) 83 d and e and question 84 (is there a way to fiqure out the pattern without spending too much time on it?) <?xml:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" /><o></o>
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[Amogh]

For the benefit of the wider community, do you mind compiling all the SC questions and answers? It would be really beneficial for others with similar problems. Im sure that the thread would be stickied and be conciliatory for the years to come.
im starting from the bottom, as someone else(ninetypercent maybe? =P) is probably doing the first bit right now...

84. (a) try to work out the pattern by experimenting..it should take less than a minute...the numbers are being multiplied by 3. so its 810/3 = 270
(b) 2430 (read of the table)
(c) Let the missing value be x. So its 7290= x/3. Solve for x.
(d) Not sure, but it probably is the 3rd one (straight line down, as n is decreasing at a constant rate)
(e) keep dividing n by 3 till you get a value less than 3. The number of times you needed to divided is the number of minutes. 9 minutes.


soz i dont have time now for the others.

 

[supre]

49. divide 140/60 =2.33
75*2.33=175 therefore its B
58. wat i do is divide the price by the amount for example
A) $2.49/0.750=$3.32
B) $3.28/1= $3.28
C) $5.65/1.7= $3.32
D) $6.75/2= $3.38
therefore the best buy is option B

 

[Riot09]

For the benefit of the wider community, do you mind compiling all the SC questions and answers? It would be really beneficial for others with similar problems. Im sure that the thread would be stickied and be conciliatory for the years to come.

thats actually a good idea:

for now i know their all over the place.but ill organise them after the school certificates over:into one thread ill separate all the questions into their corresponding topics and which past paper they belong to and include the,explanations,solutions and post them online in the sc section of bos.i cant believe no one else ever thought or have done it earlier.nice thinking amogh.

i could also organise my assignments,notes and my short answers online aswell for history,geography,science i dont know about english will help that much but the short answer section would proberly help aswell.


i'll

 

[Riot09]

also thanx for the answers and working out supre and amogh(what a strange name)is it even audible?.for what its worth i think that Supre sounds really nice.

 

[supre]

oh ya u were asking if there is a short way to do question 84. yes, there is, each time the number has been divided by 3. so 21870/3=7290 and 7290/3= 2430 and then 810 and the final answer for n is 270.
u just have to consider the pattern...:balloon:

 

[Riot09]

love the colourful balloons,i guess you cant say the same about my tag its reallynot that flattering i guess.

 

[Amogh]

also thanx for the answers and working out supre and amogh(what a strange name)is it even audible?.for what its worth i think that Supre sounds really nice.

lol what do you mean by "is it audible?" any sound perceptible by the ear is audible, so of course any word spoken, is...
its pronounced "uh-mogh"

69. Measure the area of the page (length X breadth)
Measure the area of the coin. (Use a ruler and find the radius, then use the formula pi*r^2)
Divide the area of the page by the area of the coin
so those many $2 coins can be placed on the page. Now find their value, by multiplying by 2.

72. The answer is just double the area of triangle DEC
The following proof is most probably NOT the most efficient one:
AE = 10-BE
AD=6^2-(10-BE)^2
BC=8^2-BE^2
AD=BC
so 36-(10-BE)^2=64-BE^2
Solve for BE and you should get, 6.4
Now find AD or BC and then multiply by 10.
This explanation is ambiguous and will be expanded if required

73. 14+8=22 boys
but there are only 19 boys.
Thus, 3 must be palying both Tennis and Rugby.
So now we know, 11 plays tennis, 5 plays rugby, 3 paly both
P(plays tennis)=11/19

75. I dont know how to explain this to you.

 

[ninetypercent]

oh god. this seems quite hard

80. Guess and check.
11 => 11 x 1 is the only that gives an area of 11. calculate the area = 2(11+1) = 24
therefore, A is the answer

82. a) x = 45 + 90 = 135 (exterior angle of a triangle is equal to the sum of the two opposite interior angles)

b)3AE = EB
triangle ADE is similar to triangle ACB (equiangular)
AE/AB = 1/4
BC is 4 times the length of ED

c) triangle ABC = 50 x 4 = 200
trapezium = 200 - 50 = 150

83. b) area of ellipse: pi.a.b
= pi.3.5
= 15pi

area of rectangle = 60

percentage = 15pi/60 = 78.54%

c) P (0,5)

d) Area of large circle - area of ellipse/2 = (pi x 5^2 - pi x 3 x 5)/2
= (25pi - 15pi)/2
= 15.7

e) Perimeter of large circle = 2x pi x r
= 2 x pi x 5
= 31.42

80 grams for every meter 80 x 31.42 = 2513.27 grams

 

[Riot09]

thanx guys,(i like the thing where it says atar aim 98 with the gun aimed at it.clever pun ninety.

 

[ninetypercent]

nx guys,(i like the thing where it says atar aim 98 with the gun aimed at it.clever pun ninety.

lol