***2015 LC Maths Paper 2 - Higher Level - June 8th***

Peregrine

Can someone describe this question? The paper won't be up online til 6pm.

Ansum wrote: »

What was wrong with that question with the golf ball through the air. That question was like something for applied maths students. I don't like this at all. Maths should be a subject (mostly) taught on its own, just like any other subject.

Doctorhopeful

The_N4sir wrote: »

That was a hard question even some of the best guys in my class did not get it. I don't even remember what i did but i should get some decent attempt marks

Yep you had to equate the two perpendicular distances, i got 3/4 and -48.

PotmBottom

Ansum wrote: »

What was wrong with that question with the golf ball through the air. That question was like something for applied maths students. I don't like this at all. Maths should be a subject (mostly) taught on its own, just like any other subject.

It kind of is, but we'd have to account for gravity and boring things like vertical and horizontal components of gravity and it's a bunch of fun!. What did you get stuck on?

The_N4sir

Ansum wrote: »

What was wrong with that question with the golf ball through the air. That question was like something for applied maths students. I don't like this at all. Maths should be a subject (mostly) taught on its own, just like any other subject.

Yeah I do applied maths and i was thinking the same. I was trying to do it like an applied maths question but i had to restart the question again since my answers were definitely wrong

fin709

The basketball question was very confusing, the first two bits we didnt know were the throws in seperate groups or straight after each other so didnt know whether to use bernoulli trials or not. Used bernoulli trials in the end, im pretty sure i did it wrong

Benm123

Doctorhopeful wrote: »

Radius was definitely 20cm guys. Thought it was a really good paper. Only thing i couldn't get was showing how pn+1=0.6 +0.2pn, but using that equation i could prove the next parts. Questions 7, the final area was about 28,000cm^2. Very happy, but it was definitely harder than p2.

How did you arrive at that radius? I did the correct method, I think, squaring 3r and 8r and letting it equal to square (value for hyp) and i got 2root10?

Doctorhopeful

Nim wrote: »

Can someone describe this question? The paper won't be up online til 6pm.

Given the equation of the flight path for a golf ball. Then given the horizontal speed, had to work out the angle of elevation between where it landed and where it was hit from. Was very like a projectile question from app maths

PotmBottom

Nim wrote: »

Can someone describe this question? The paper won't be up online til 6pm.

we were given a slope, and the height of the ball was given by a quadratic equation. Think it was 6t^2+22t+8 or something, we were asked to find the height of the ball at t=0. Which was 8. Then it told us that the horizontal speed of the ball is 38m/s. We were then asked for the angle of inclination. so I let the equation equal zero and got t =4 so I multipled 38(4) to get 152. then the tan inverse of 8/152 was equal to the angle of inclination = 3 degrees.

Doctorhopeful

Benm123 wrote: »

How did you arrive at that radius? I did the correct method, I think, squaring 3r and 8r and letting it equal to square (value for hyp) and i got 2root10?

9r^2 plus 64r^2 = 29200
R^2 = 400
r= 20

EDIT: oh and 20root73 squared is 29200

Ansum

Nim wrote:

Can someone describe this question? The paper won't be up online til 6pm.

.
Sorry Nim but the question was is confusing to me that it's a blur, but hopefully someone will post it.

Doctorhopeful

Benm123 wrote: »

How did you arrive at that radius? I did the correct method, I think, squaring 3r and 8r and letting it equal to square (value for hyp) and i got 2root10?

I think you forgot a zero as root40 is 2root10 but root400 is 20

Kremin

simplyno1 wrote: »

can anyone tell me how you found the value of k in the line question? it said P(10, K) was a bisector of the angle between line 1 and line 2... does that mean the perpendicular distance from P->line 1 was equal to perp. distance from P->line 2? and you let the two equal and get 2 values for K?

Find point intersection 2 lines. Find angle between then. Divide angle by two. Use tan formula again to find angle between bisector (1/2 angle) and any of the lines. This will give you the slope of the bisector. Use then y-y1 formula on the point if intersection. Sub 10, k in.

Fiona G

Nim wrote:

Can someone describe this question? The paper won't be up online til 6pm.

It gave a formula for the height of a golf ball above the flat, something like h=-6tsquared + 30t + 20 (can't remember values) where t is the seconds since the ball was hit. It wasn't really like an applied maths question, just had to let h=0 to find t etc.

Doctorhopeful

fin709 wrote: »

The basketball question was very confusing, the first two bits we didnt know were the throws in seperate groups or straight after each other so didnt know whether to use bernoulli trials or not. Used bernoulli trials in the end, im pretty sure i did it wrong

Pretty sure you couldn't use benoulli trials because the trials weren't independent ie the previous throws affected the one you're doing

Kremin wrote: »

Find point intersection 2 lines. Find angle between then. Divide angle by two. Use tan formula again to find angle between bisector (1/2 angle) and any of the lines. This will give you the slope of the bisector. Use then y-y1 formula on the point if intersection. Sub 10, k in.

I did this and it got a really horrible answer, but then equating two perpendicular equations together got k= 3/4 and -48

fin709

Doctorhopeful wrote: »

Pretty sure you couldn't use benoulli trials because the trials weren't independent ie the previous throws affected the one you're doing

But it said first throws so i presumed they were seperate groups of first throws so every one of them was 0.7

Peregrine

Doctorhopeful wrote: »

Given the equation of the flight path for a golf ball. Then given the horizontal speed, had to work out the angle of elevation between where it landed and where it was hit from. Was very like a projectile question from app maths

PotmBottom wrote: »

we were given a slope, and the height of the ball was given by a quadratic equation. Think it was 6t^2+22t+8 or something, we were asked to find the height of the ball at t=0. Which was 8. Then it told us that the horizontal speed of the ball is 38m/s. We were then asked for the angle of inclination. so I let the equation equal zero and got t =4 so I multipled 38(4) to get 152. then the tan inverse of 8/152 was equal to the angle of inclination = 3 degrees.

Fiona G wrote: »

It gave a formula for the height of a golf ball above the flat, something like h=-6tsquared + 30t + 20 (can't remember values) where t is the seconds since the ball was hit. It wasn't really like an applied maths question, just had to let h=0 to find t etc.

Now, I haven't actually seen it but it sounds similar to a question in one of the Text and Tests books.

expiiplus1

simplyno1's solution also works and is a lot nicer than the other way of solving it. I was kicking myself for doing it the horrid way.
By the way Kremin, things work out a lot easier if you convert tan to angles.

Benm123

Doctorhopeful wrote: »

9r^2 plus 64r^2 = 29200
R^2 = 400
r= 20

EDIT: oh and 20root73 squared is 29200

Ah balls I did that same way must have made a numerical error

Fiona G

Doctorhopeful wrote:

I did this and it got a really horrible answer, but then equating two perpendicular equations together got k= 3/4 and -48

I got this too!

Doctorhopeful

fin709 wrote: »

But it said first throws so i presumed they were seperate groups of first throws so every one of them was 0.7

Why would they give the other information then? Seems like an oversimplification but obviously i can't be certain

Ansum

Can anyone tell me if I used the right method. So for the circle question instead of forming equations the way the question told me to. I formed my own equations. Just here me out.
[What went through my head]
(7,12) is on the circle c.
Therefore (-7,12),(-7,-12),(7,-12) and because all these points are on the circle I used the, let c(eqn. Of circle) be
X^2+Y^2+2gx+2fy+c, and with this formed 4 eqns and did simultaneous eqns to find the centre of the circle.

Dkey

I didnt understand any of the long questions should have revised it, I am so scared that I will get an F not an E ;x

Rimiko

Did anyone use the a:b ratio formula to find the co-ordinates for the centre (h,k) of the smaller circle? If so, did you get something like (5,6) ? (can't remember exactly!)

Like the radius of the smaller circle was 2root10, then the diameter would have been 4root10, and the radius of the bigger circle was 6root10, so 6root10 minus 4root10 = 2root10, so therefore the centre (h,k) split the distance been the 2 points (centre of big circle and point of intersection) in the ratio 2:1?

(sorry if that's a bit confusing..)

Oh! And was the tangent something like x^2+3y^2-43?

Overall, I think it was an okay paper. I definitely think I have come a long way since the mocks! Hoping for a B

Rimiko

Did anyone use the a:b ratio formula to find the co-ordinates for the centre (h,k) of the smaller circle? If so, did you get something like (5,6) ? (can't remember exactly!)

Like the radius of the smaller circle was 2root10, then the diameter would have been 4root10, and the radius of the bigger circle was 6root10, so 6root10 minus 4root10 = 2root10, so therefore the centre (h,k) split the distance been the 2 points (centre of big circle and point of intersection) in the ratio 2:1?

(sorry if that's a bit confusing..)

Oh! And was the tangent something like x^2+3y-43?

Overall, I think it was an okay paper. I definitely think I have come a long way since the mocks! Hoping for a B

tahtguy

Guys what was the right value for k? I got 0.75 and -48, but I dont think thats right.

skippy1977

Rimiko wrote: »

Did anyone use the a:b ratio formula to find the co-ordinates for the centre (h,k) of the smaller circle? If so, did you get something like (5,6) ? (can't remember exactly!)

Like the radius of the smaller circle was 2root10, then the diameter would have been 4root10, and the radius of the bigger circle was 6root10, so 6root10 minus 4root10 = 2root10, so therefore the centre (h,k) split the distance been the 2 points (centre of big circle and point of intersection) in the ratio 2:1?

(sorry if that's a bit confusing..)

Oh! And was the tangent something like x^2+3y-43?

Overall, I think it was an okay paper. I definitely think I have come a long way since the mocks! Hoping for a B

Yep looks right. Method is good. The tangent doesn't have a x^2 though...(typo?)
x+3y-43=0 I think

simplyno1

Rimiko wrote: »

Did anyone use the a:b ratio formula to find the co-ordinates for the centre (h,k) of the smaller circle? If so, did you get something like (5,6) ? (can't remember exactly!)

Like the radius of the smaller circle was 2root10, then the diameter would have been 4root10, and the radius of the bigger circle was 6root10, so 6root10 minus 4root10 = 2root10, so therefore the centre (h,k) split the distance been the 2 points (centre of big circle and point of intersection) in the ratio 2:1?

(sorry if that's a bit confusing..)

Oh! And was the tangent something like x^2+3y-43?

Overall, I think it was an okay paper. I definitely think I have come a long way since the mocks! Hoping for a B

Yeah I got 5,6 and x+3y-43=0 or +43 cant remember

dazzadazza

PotmBottom wrote: »

we were given a slope, and the height of the ball was given by a quadratic equation. Think it was 6t^2+22t+8 or something, we were asked to find the height of the ball at t=0. Which was 8. Then it told us that the horizontal speed of the ball is 38m/s. We were then asked for the angle of inclination. so I let the equation equal zero and got t =4 so I multipled 38(4) to get 152. then the tan inverse of 8/152 was equal to the angle of inclination = 3 degrees.

I did the same thing and got 3 too. I thought I was wrong because it seemed too small. Hope we are right.

Rimiko wrote: »

Did anyone use the a:b ratio formula to find the co-ordinates for the centre (h,k) of the smaller circle? If so, did you get something like (5,6) ? (can't remember exactly!)

Like the radius of the smaller circle was 2root10, then the diameter would have been 4root10, and the radius of the bigger circle was 6root10, so 6root10 minus 4root10 = 2root10, so therefore the centre (h,k) split the distance been the 2 points (centre of big circle and point of intersection) in the ratio 2:1?

(sorry if that's a bit confusing..)

Oh! And was the tangent something like x^2+3y^2-43?

Overall, I think it was an okay paper. I definitely think I have come a long way since the mocks! Hoping for a B

Yeah I did that but instead of a 2:1 ratio I said it was 2/3 of the way up. So then I translated.

simplyno1

For the probability question where you had to prove P(n+1) = 0.6 + 0.2P(n),

i let n=1, and got P(2) = 0.6 + 0.2P(1). But P1= probabilty on first trial which was 0.7, therefore
P(2) = 0.6+(0.2)(0.7) = 0.74. So the probability of scoring on the 2nd throw is equal to 0.74.

I tested this: Probability of scoring on the 2nd throw is either 0.6, or 0.8, depends if he missed or scored. So I said the probability of scoring on the 2nd throw is (0.3x0.6) + (0.7 x 0.8) which equals to 0.74, therefore proving the equation...

is that right?

MmmPancakes

What did you get for the total area of the machine? IIRC I got around 32000cm^2 or something around that

dazzadazza wrote: »

Yeah I did that but instead of a 2:1 ratio I said it was 2/3 of the way up. So then I translated.

What I done for that question was let 2root10 equal the distance from the centre (x,y) to the tangent point, then found the equation of the perpendicular line to the tangent (don't quote me on that, can't really remember a lot of the paper :pac:) and wrote y in terms of x, subbed it in to the distance thing, solved for y. Got (5,6).