Approximation

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callmemo: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Sun Sep 28, 2008 11:22 am

by callmemo » Sun Jan 04, 2009 8:23 pm

If the area of square S and the area of circle C are equal, then the ratio of the perimeter of S to the circumference of C is closest to

a. 7/9
b. 8/9
c. 9/8
d. 4/3
e. 2

OA: C

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Source: Beat The GMAT — Problem Solving | Sun Jan 04, 2009 8:23 pm

rajataga: Senior | Next Rank: 100 Posts; Posts: 87; Joined: Tue Dec 23, 2008 11:06 pm; Thanked: 2 times

by rajataga » Sun Jan 04, 2009 8:34 pm

Let the side of square b s

Let the radius of circle be r

s^2 = pi.r^2

now, ratio of perimeter of both,

4s/2pi.r

squaring

= 2s/pi.r

4s^2/(pi.r)^2

4/pi..

taking square root,

2/root of pi..............answer

(now you can approximate this considering value of pi = 3.14

and root of 3 = 1.7)

now you know numerator is higher than denominator, but very minimally...

so, only options left are C and D.

Check both of by multiplying, and you will find that 9/8 servers better

However, i do not think a sum which requires this level of approximation would be likely in the gmat

Last edited by rajataga on Sun Jan 04, 2009 8:39 pm, edited 1 time in total.

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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Sun Jan 04, 2009 8:37 pm

Areas equal

s^2 = pi * r^2

Approx pi=3

s/r = sqrt(3)

Ratio of perimeters = 4 s / 2 * pi * r
= 2 s/ pi*r
= 2 sqrt(3) / 3
= 2 /sqrt(3)
= 4 /3

I am getting D). May be missing something here.

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rajataga: Senior | Next Rank: 100 Posts; Posts: 87; Joined: Tue Dec 23, 2008 11:06 pm; Thanked: 2 times

by rajataga » Sun Jan 04, 2009 8:41 pm

Cramya,

your solution is pretty much correct, until the last step....

You squared the ratio....hence, the answer you will get is also the sqaure of the ratio you want...

Note, the sqaure of a ratio is not the same as the ratio itself.

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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Sun Jan 04, 2009 8:46 pm

Note, the sqaure of a ratio is not the same as the ratio itself

U r correct.

Overlooked multiplying the ratios of numerator and denominator of a ratio by a constant versus squaring a ratio...Thanks for cathing that.

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Sachindh: Senior | Next Rank: 100 Posts; Posts: 37; Joined: Wed Sep 17, 2008 8:17 pm; Location: India; Thanked: 4 times

by Sachindh » Mon Jan 05, 2009 8:40 am

Cramya,

I dont think you did anything wrong in first time. I am also getting answer "d"

Can you write the corrected steps as suggested by rajat. I am not able to understand those steps. What shall be the final answer?

Thanks

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rajataga: Senior | Next Rank: 100 Posts; Posts: 87; Joined: Tue Dec 23, 2008 11:06 pm; Thanked: 2 times

by rajataga » Mon Jan 05, 2009 9:44 am

Hey Sachin,

You can refer to my earlier solution. Also, below is Cramya's solution corrected :-

Areas equal

s^2 = pi * r^2

Approx pi=3

s/r = sqrt(3)

Ratio of perimeters = 4 s / 2 * pi * r
= 2 s/ pi*r
= 2 sqrt(3) / 3
= 2 /sqrt(3)

= 2/1.75

Multiple by 4.5/4.5,

and you will get 9/7.8, which is approximately 9/8

Hence, D.

If you can, please post your solution here also, so we can determine where you went wrong.

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Sachindh: Senior | Next Rank: 100 Posts; Posts: 37; Joined: Wed Sep 17, 2008 8:17 pm; Location: India; Thanked: 4 times

by Sachindh » Mon Jan 05, 2009 10:18 am

Thanks Rajat,

I got it now.

actually my approach was same till last step. I was getting 3.54/3(2*sqrt3/3). Upon approximation i was getting 4/3 which is less accurate than yours approximation.

Thanks

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vivek.kapoor83: Legendary Member; Posts: 833; Joined: Mon Aug 04, 2008 1:56 am; Thanked: 13 times

by vivek.kapoor83 » Mon Jan 05, 2009 10:21 am

i did as cramya did and got it..
but[ i also applied 1 more a[/color]pproach[/color], but im not getting ans
can any1 explain...am i making correct approch as alternate sol.
i just wonder..it s bit silly but u co-op with [color=#444444]to see alternate sol is possible or not[/color]

s^2 = pi r^2
s^2/pi r^2 =1
4s*s /4*pi*r*r =1
4 s*s / 2r*2pi*r=1
P *s /C*2r=1
P/C = 2r/s
..So,stuck here..i know this looks silly but this came into my mind.. wht mistake i am making. i know i am doing blunder bt dont know wht.

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rajataga: Senior | Next Rank: 100 Posts; Posts: 87; Joined: Tue Dec 23, 2008 11:06 pm; Thanked: 2 times

by rajataga » Mon Jan 05, 2009 10:33 am

Hey Vivek,

You aren't making any mistake

, and its always good to try out other methods....

I will just finish your solution,

s^2 = pi r^2
s^2/pi r^2 =1
4s*s /4*pi*r*r =1
4 s*s / 2r*2pi*r=1
P *s /C*2r=1
P/C = 2r/s

so we need to find 2r/s

s^2 = pi r^2
(r^2)/(s^2) = 1/pi

Dividing by 2,

r/s = 1/root of pi

multiplying by2,

2r/s = 2/root of pi = P/C

There you go...you just had to go a couple of steps further...

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vivek.kapoor83: Legendary Member; Posts: 833; Joined: Mon Aug 04, 2008 1:56 am; Thanked: 13 times

by vivek.kapoor83 » Mon Jan 05, 2009 10:01 pm

nice.............

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