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MAT Prep Thread - 2nd November 2016

This is n't even a particularly difficult question but I keep getting 90% which is obviously wrong...I made the triangles side equal to 2 then got area of circle as pi/3 and area of triangle as 2/sqrt3

Reply 801

joodaa

5

Original post by Someboady

Hey
Use part (i) in reverse
take a=3 and b = 2. take your n as 2014.
Then you can make that equal to the brackets given in part (i)

i.e.

(3-2)(3^2014 + .... + 2 ^ 2014)

The first bracket is equal to 1. This implies that the number is prime as a prime number is only divisible by one and itself.
Hope this helps :smile:

Hey thanks for your help, your way makes sense to me but the mark scheme writes it as (3^5-2^5)(3^2010.........2^2010) could you explain how they reached this?? thanks again

Reply 802

theaverage

2

Original post by Mystery.

Yeah I am just doing it, what did you get?

a,b,c,a,c,b,d,d,b,b

Posted from TSR Mobile

Reply 803

Someboady

17

Original post by joodaa

Hey thanks for your help, your way makes sense to me but the mark scheme writes it as (3^5-2^5)(3^2010.........2^2010) could you explain how they reached this?? thanks again

Oh balls, sorry my answer was incorrect (I deleted the post)... I'm having a look at it now xD.. I just did the question and I thought I'd gotten it right without checking the solutions. Sorry! Working on it right now!

Reply 804

username2447825

7

Original post by Mystery.

This is n't even a particularly difficult question but I keep getting 90% which is obviously wrong...I made the triangles side equal to 2 then got area of circle as pi/3 and area of triangle as 2/sqrt3

Rather than giving the triangle a side length, express the area of the triangle as a function of the radius of the circle

Reply 805

Mystery.

21

Oh and how is 9^log3x = 4^log2x????

Reply 806

Mystery.

21

Original post by AlphaQuark

Rather than giving the triangle a side length, express the area of the triangle as a function of the radius of the circle

ok i'll try

Reply 807

Nonamebzja

13

Original post by Mystery.

This is n't even a particularly difficult question but I keep getting 90% which is obviously wrong...I made the triangles side equal to 2 then got area of circle as pi/3 and area of triangle as 2/sqrt3

I got area of triangle as x^2/4 where x is side of traingle

Reply 808

joodaa

5

Original post by Someboady

Oh balls, sorry my answer was incorrect (I deleted the post)... I'm having a look at it now xD.. I just did the question and I thought I'd gotten it right without checking the solutions. Sorry! Working on it right now!

ooo okay no worriess

Reply 809

some-student

7

Original post by Mystery.

This is n't even a particularly difficult question but I keep getting 90% which is obviously wrong...I made the triangles side equal to 2 then got area of circle as pi/3 and area of triangle as 2/sqrt3

This is a guess but...

The radius of the circle seems to be about 1/3 of the height of the triangle.
Assume the triangle has length

l

. Then the height of the triangle is

\sqrt{l^2 - (\frac{1}{2}l)^2} = \frac{\sqrt{3}}{2}l

The radius of the circle:

r \approx \frac{\sqrt{3}}{{6}}l

and hence the area

A_c \approx \frac{3\pi}{36}l^2= \frac{\pi}{12}l^2

The area of the triangle:

A_t \approx \frac{\sqrt{3}}{4}l^2

Dividing through we get

\frac{A_c}{A_t} \approx \frac{4\pi}{12\sqrt{3}} = \frac{\pi}{3\sqrt{3}} \approx 60\%

Reply 810

Mystery.

21

Original post by Nonamebzja

I got area of triangle as x^2/4 where x is side of traingle

how?

Reply 811

Mystery.

21

Original post by some-student

This is a guess but...

The radius of the circle seems to be about 1/3 of the height of the triangle.
Assume the triangle has length

l

. Then the height of the triangle is

\sqrt{l^2 - (\frac{1}{2}l)^2} = \frac{\sqrt{3}}{2}l

The radius of the circle:

r \approx \frac{\sqrt{3}}{{6}}l

and hence the area

A_c \approx \frac{3\pi}{36}l^2= \frac{\pi}{12}l^2

The area of the triangle:

A_t \approx \frac{\sqrt{3}}{4}l^2

Dividing through we get

\frac{A_c}{A_t} \approx \frac{4\pi}{12\sqrt{3}} = \frac{\pi}{3\sqrt{3}} \approx 60\%

How do you do the actual percentage without a calculator?

Reply 812

Nonamebzja

13

Original post by Mystery.

how?

Sorry (root3 × x^2)/4

Reply 813

Mystery.

21

Original post by Mystery.

ok i'll try

Ok so I did that but I still have pi from the area of circle formula

Reply 814

Nonamebzja

13

Original post by Mystery.

How do you do the actual percentage without a calculator?

Pie/3root3 i just did 1/root3 which is root3/3
(1.7×100)÷3 approx 60%😂

Reply 815

some-student

7

Original post by Mystery.

Oh and how is 9^log3x = 4^log2x????

9^{\log_3x} = 3^{2\log_3x} = 3^{\frac{2\log x}{\log 3}} = (3^{\frac{1}{\log 3}})^{2\log x} = 10^{2\log x}

(10 assuming that log is to the base of 10... the reason for this is because

3^{\frac{1}{\log 3}}

is like taking the

\log 3

root of 3, which will go in 10 times, assuming log is to the base of 10)

4^{\log_2x} = 2^{2\log_2x} = 2^{\frac{2\log x}{\log 2}} = (2^{\frac{1}{\log 2}})^{2\log x} = 10^{2\log x}

They are equal

Reply 816

11234

8

Original post by Mystery.

How do you do the actual percentage without a calculator?

The circle is the incircle of the triangle. There is a formula saying that area = (a+b+c/2) x radius of incircle (I'm pretty sure u could get a solution without these formulas but I'm lazy)

thus the ratio is pi*r^2/area = pi*r^2/sr where s is the semiperimeter (a+b+c/2)

r's cancel then we get pi*area/s^2. s=3a/2 area = sqrt3 / 4 * a^2

put back into formula then after lots of cancelling sqrt3*pi/9 which is roughly 3sqrt3/9 = sqrt3/3 which is about 2/3 so 60%