Mathematics solutions (1 Viewer)

[umz93]

Thanks, my answers were largely the same as yours. (Which is a good sign)



I don't really remember them exactly, I took the test paper home and after comparing the answers with the questions, it sort of jogs the memory a bit and you have an idea whether you got it or not!

What is 6(b)iii asking?

 

[kev-is-red]

I can't remember half of these questions, never mind what my answers were.

On a brighter note, Harold and Kumar 3 opens in cinemas on November 4 (3 days after I finish).

 

[asianese]

My proof for 9aii)

If AB = AC, and AD = AE (given)
Then BC is parallel to DE (equal intercept theory)

RTP triangle BCF is congruent to triangle DFE
Angle BFC = DFE (vertically opposite)
Angle FBC = FED (alternate angles in parallel lines)
Therefore triangle BCF is congruent to triangle DFE (AAA)

We know BC : DE = 1 : 2,

Hence in similar triangles BCF and DFC

BC : DE = BF : FE (corresponding sides)
Therefore BF : FE = 1 : 2

Is this reasoning correct?

I did this. I think its correct!

 

[mdunn94]

are you sure Q 5 a iii is rounded off correctly? i got 552.8

agreed

 

[nephh22]

u are fucking shit

Stop dissing him, at least he spent his TIME to post all the solutions up for US to see, if u dont like it, CLOSE the page! -_-

 

[OmmU]

Ok with those I worked out about 107/120 :\ I can't believe I screwed up Q7 stat pt >.< OMG so easy. I copied the equation into my book wrong I think EHHHHHh!!! Ok I have to get off here and study ext1 >.<

 

[umz93]

Ok with those I worked out about 107/120 :\ I can't believe I screwed up Q7 stat pt >.< OMG so easy. I copied the equation into my book wrong I think EHHHHHh!!! Ok I have to get off here and study ext1 >.<

Nice, im around 100-106 because i dont remember all my answers :/

 

[nephh22]

that was the easiest 2marks in the whole paper (though it's hard to judge, pretty much 118 out of the 120 marks were REALLY easy, the other 2 were slightly harded)

haha yeh true... it was so easy, took like 2 seconds, so i wasnt sure if i was right lol (cause it took like 2 seconds). :L

 

[someth1ng]

What is 6(b)iii asking?

Does not exist.

 

[nephh22]

I can't remember half of these questions, never mind what my answers were.

On a brighter note, Harold and Kumar 3 opens in cinemas on November 4 (3 days after I finish).

loool cant WAIT!!! I bet ur watching Harold and Kumar: Go to White Castle, as I am...

 

[o_0]

My proof for 9aii)

If AB = AC, and AD = AE (given)
Then BC is parallel to DE (equal intercept theory)

RTP triangle BCF is congruent to triangle DFE
Angle BFC = DFE (vertically opposite)
Angle FBC = FED (alternate angles in parallel lines)
Therefore triangle BCF is congruent to triangle DFE (AAA)

We know BC : DE = 1 : 2,

Hence in similar triangles BCF and DFC

BC : DE = BF : FE (corresponding sides)
Therefore BF : FE = 1 : 2

Is this reasoning correct?

Heya this is all good, except for the first part... i don't think assuming that AB=AC and going from there is a valid proof,


the proper proof is AB:AD = 1:2, and AC:AE = 1:2, and angle[DAE] is shared. therefore, triangles ABC and ADE are similar (two pairs of sides in proportion, with included equal angle)

 

[Gillywilly]

can i ask how you came to that answer in question 5b(iii)

 

[Proteus]

What is 6(b)iii asking?

There is no 6 (B) iii)!

 

[cheschirecat]

can i ask how you came to that answer in question 5b(iii)

The probability of her not wearing the same clothes is equal to:
P(RYR)+P(YRY)= 3/5*2/4*2/3+2/5*3/4*1/3=0.3

 

[weirdguy99]

FFFUUUUUUUUUUU...I got ln{(2I/10^-12)-(I/10^-12)} and then I proceeded to bloody cross multiply. If only I had solved it properly, I would have gotten 10ln2 instead of ln2I^2...

 

[Gillywilly]

hmm i thought they meant conscutive days, including 2 consecutive days as well
so i did 1 minus (probability of wearing the same shirt in 2 or 3 consecutive days)
i.e. 1 - [P(RR) + P(BB) + P(RRR)] siiiighh was a bit unclear for me :/

 

[damo_g]

fuked up on 4(e) but i realised that when i first looked back over the paper. Will still get 1 mark for 1 correct inequality. Other than that everything looks good. Confident that i am 100+/120, after seeing this i think my mark should be up near 110.

 

[weirdguy99]

Can someone shed some light on 8(b)(ii) pleaseee

 

[damo_g]

Can someone shed some light on 8(b)(ii) pleaseee

A1 = 29227(1.005) + M(1.005)

A2 = 29227(1.005)^2 + M(1.005)^2 + M(1.005)
= 29227(1.005)^2 + M(1.005 + 1.005^2)

A3 = you get the point


An = 29227(1.005)^n + M(1.005+1.005^2+...+1.005^n)
= 29227(1.005)^n + M(a(r^n-1)/(r-1))

After 240 months An = 800000. So solve the equation when n = 240 and An = 800000. Ends up with the answer in the OP.

 

[vaish94]

Is 6 b) v) right ??? I thought it was 6 ?

lol that's what i thought at first, but remember the area underneath the x axis has a negative value, so, -4 + 2 = - 2.