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Mathematics: Post your doubts here!

M

Maayee

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can anyone help m with part v and vi?
the answer for v is : Elizabeth wins by 0.05s and 0.5 m
the answer for vi is: Andrew wins
 

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Rizwan Javed

Rizwan Javed

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Can anyone please solve this?

Eight people sit in a minibus: four on sunny side and four on the shady side. If two people want to sit on opposite sides to each other and another two people want to sit on the shady side, in how many ways can this be done?
 
Anum96

Anum96

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Rizwan Javed said:
Can anyone please solve this?

Eight people sit in a minibus: four on sunny side and four on the shady side. If two people want to sit on opposite sides to each other and another two people want to sit on the shady side, in how many ways can this be done?
Click to expand...
Answer?
 
Anum96

Anum96

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Rizwan Javed said:
How did you get it? :p I didn't understand :p
Click to expand...

two people want to sit on opposite side. They can do this in 2P2 ways or 2!

There are 4 seats to choose from or in other word there are 4C1 seats the first person can choose to sit on one seat of the opposite side & 4C1 choices for the second person who wishes to sit on the opposite side.

another two want to sit on shady side. Remeber one is already sitting on shady side so we have 3 seats left on shady side. So the two people can arrange themselves in 3P2 ways.

Remaining 4 have 4 choices to sit in any order. They can sit in 4P4 ways.


Therefore,
(4C1)(4C1)*2P2*3P2*4P4 = 4608
 
Rizwan Javed

Rizwan Javed

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Anum96 said:
two people want to sit on opposite side. They can do this in 2P2 ways or 2!

There are 4 seats to choose from or in other word there are 4C1 seats the first person can choose to sit on one seat of the opposite side & 4C1 choices for the second person who wishes to sit on the opposite side.

another two want to sit on shady side. Remeber one is already sitting on shady side so we have 3 seats left on shady side. So the two people can arrange themselves in 3P2 ways.

Remaining 4 have 4 choices to sit in any order. They can sit in 4P4 ways.


Therefore,
(4C1)(4C1)*2P2*3P2*4P4 = 4608
Click to expand...
Oh thanks! :)
 
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Anum96

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Sarosh Jameel said:
PLZ help with part (I) !!
Click to expand...
Make a triangle ABO. Make a line in between. This line will bisect angle AOB

Label the end of this point D

Find AD using cos.

cos(x) = AD/r

AD= rcos(x)

So AB will twice of AD

Which is 2rcosx

AB =AC =2rcosx


Now make another triangle ABN; N being the second corner of shaded region on the line OA.

You have AB which is equal to AN(AN is the radius of the circle with center A)

So Find the length of arc BN

Which will be r(theta)

r =2rcosx

theta =2x

r(theta) = 2rcosx*2x =4xrcosx


Now the question says...

perimeter of shaded region = 1/2(circumference of circle

AB +AC +BC =1/2(2*pi*r)

2rcosx+ 2rcosx+ 4xrcosx = pi*r

All r's cancel out

Now you have 4cosx +4xcosx =pi

Take cos common

cosx(4+ 4x) = pi

x= cos^-1 [(pi/(4 +4x)] Shown
 
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