Prelim. marathon? (1 Viewer)

[nifkeh]

Start by drawing a triangle and naming the sides a, b and
Then write down the givens in the question and try and manipulate an expression in order to get which is the area of the triangle.

Also I will add that this question is considered 'very tough' by some GMAT tutoring organisation (an American thing)



You cannot use discriminant for this, it has a sine function which cannot be considered a co-efficient of x (since its a function), since its multiple choice its best to do this:



So we have two sides now. Graph both sides of the equation the number of intersections is the number of solutions.

lol you are smart. I didn't get both at all.. maybe it's time to study more maths -_-' Thanks for the solutions btw, I'm really clueless lol

 

[likeaBOS]

You cannot use discriminant for this, it has a sine function which cannot be considered a co-efficient of x (since its a function), since its multiple choice its best to do this:



So we have two sides now. Graph both sides of the equation the number of intersections is the number of solutions.

Yeah I see what you mean, thanks.

 

[likeaBOS]

Here is a good one that can be attempted with only year 10 knowledge:

Did anyone else get A = (12+6root3) units squared?

 

[Capt Rifle]

1 + 1 ? (Must Show Working Out)

 

[HSC2014]

Did anyone else get A = (12+6root3) units squared?

Hm.. I got 6root3 units^2.

Here's my working out (Please point out any mistakes! I'm unsure about the final part where I say 2ab = "):

a + b + root(a^2 + b^2) = 6 + 6root3 (equation one)
a^2 + b^2 + (a^2 + b^2) = 96
which simplifies to a^2 + b^2 = 48 (equation two)

Subbing two into one,
a + b + root48 = 6 + 6root3
a + b + 4root3 = 6 + 6root3 (simplified)
a + b = 6 + 2root3

Taking the squares of both sides,
a^2 + 2ab + b^2 = 36 + 24root3 + 12

Therefore 2ab = 24root3
ab = 12root3
ab/2 = 6root3 units^2

Edit: Sorry, don't know how to use the equation thing.. LaTeX?

 

[nifkeh]

I got -3 + 9/2 root 3 units squared.. probs wrong lol

 

[nifkeh]

Hm.. I got 6root3 units^2.

Here's my working out (Please point out any mistakes! I'm unsure about the final part where I say 2ab = "):

a + b + root(a^2 + b^2) = 6 + 6root3 (equation one)
a^2 + b^2 + (a^2 + b^2) = 96
which simplifies to a^2 + b^2 = 48 (equation two)

Subbing two into one,
a + b + root48 = 6 + 6root3
a + b + 4root3 = 6 + 6root3 (simplified)
a + b = 6 + 2root3

Taking the squares of both sides,
a^2 + 2ab + b^2 = 36 + 24root3 + 12

Therefore 2ab = 24root3
ab = 12root3
ab/2 = 6root3 units^2

Edit: Sorry, don't know how to use the equation thing.. LaTeX?

yeah your answer makes sense, I was following through the working

 

[likeaBOS]

Hm.. I got 6root3 units^2.

Here's my working out (Please point out any mistakes! I'm unsure about the final part where I say 2ab = "):

a + b + root(a^2 + b^2) = 6 + 6root3 (equation one)
a^2 + b^2 + (a^2 + b^2) = 96
which simplifies to a^2 + b^2 = 48 (equation two)

Subbing two into one,
a + b + root48 = 6 + 6root3
a + b + 4root3 = 6 + 6root3 (simplified)
a + b = 6 + 2root3

Taking the squares of both sides,
a^2 + 2ab + b^2 = 36 + 24root3 + 12

Therefore 2ab = 24root3
ab = 12root3
ab/2 = 6root3 units^2

Yeah that seems right. I had a similar line of working to you although I made a careless error along the way which caused the difference in answer. Nice job

 

[HSC2014]

Haha cool. My first tackle at it I tried to do something with surd identities (if a + brootc = x + yrootz, then a = x, b = y, c = z). But I forgot how that works so I just went with long way :L Probably wouldn't have worked out anyway.

 

[Sy123]

Hm.. I got 6root3 units^2.

Here's my working out (Please point out any mistakes! I'm unsure about the final part where I say 2ab = "):

a + b + root(a^2 + b^2) = 6 + 6root3 (equation one)
a^2 + b^2 + (a^2 + b^2) = 96
which simplifies to a^2 + b^2 = 48 (equation two)

Subbing two into one,
a + b + root48 = 6 + 6root3
a + b + 4root3 = 6 + 6root3 (simplified)
a + b = 6 + 2root3

Taking the squares of both sides,
a^2 + 2ab + b^2 = 36 + 24root3 + 12

Therefore 2ab = 24root3
ab = 12root3
ab/2 = 6root3 units^2

Edit: Sorry, don't know how to use the equation thing.. LaTeX?

Correct, nice job :]

Another question for everyone. This is a little harder.



Basically a triangle is formed with vertices (corners) (0,4) (10,0) (0,0)
Find probability that if you pick a point in this triangle that the x-coordinate is bigger than y-coordinate (just in case the question wasnt clear)

The solution I did involved co-ordinate geometry. But if you can think of the first step correctly you have the rest in the bag

This relaly should be merged with the 2U marathon, sure this is prelim but that one is dead and its more efficient to have 1 whole thread for one topic (and impose a rule on labelling questions whether a prelim student can do it)

 

[Shazer2]

I've done no probability and I'm still trying to figure out how to answer nightweaver's question.

 

[Timske]

Correct, nice job :]

Another question for everyone. This is a little harder.



Basically a triangle is formed with vertices (corners) (0,4) (10,0) (0,0)
Find probability that if you pick a point in this triangle that the x-coordinate is bigger than y-coordinate (just in case the question wasnt clear)

The solution I did involved co-ordinate geometry. But if you can think of the first step correctly you have the rest in the bag

This relaly should be merged with the 2U marathon, sure this is prelim but that one is dead and its more efficient to have 1 whole thread for one topic (and impose a rule on labelling questions whether a prelim student can do it)

5/7

 

[HSC2014]

^ Imposterrrrrr!
Edit: Yep I got 5/7 too.
Find the point of intersection between the two lines y = x and y = -2x/5 + 4. And you can work it out from there by comparing areas ^^
These are pretty fun I have to admit.

 

[Timske]

^ Imposterrrrrr!
Edit: Yep I got 5/7 too.
Find the point of intersection between the two lines y = x and y = -2x/5 + 4. And you can work it out from there by comparing areas ^^
These are pretty fun I have to admit.

I did diff way

 

[HSC2014]

I did diff way

How'd you do it? O:

 

[Timske]

How'd you do it? O:

fked up way lool dont bother

 

[Sy123]

I've done no probability and I'm still trying to figure out how to answer nightweaver's question.

No need for any real logic of probability, just general logic of dividing the valid area for the whole area. For nightweaver's question do something to the numerator and denominator of your combined fraction

5/7

^ Imposterrrrrr!
Edit: Yep I got 5/7 too.
Find the point of intersection between the two lines y = x and y = -2x/5 + 4. And you can work it out from there by comparing areas ^^
These are pretty fun I have to admit.

Correct.

Another question to keep marathon rolling (other people need to post questions too lol)

 

[likeaBOS]

Correct.

Another question to keep marathon rolling (other people need to post questions too lol)

Is the answer 3?

 

[Shazer2]

I've seen that question before, but I don't know how to answer it. :/

 

[Sy123]

Is the answer 3?

Yes it is, I think someone else should post a question while I think of a new one