BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Approximation

Expert replies
Source: — Problem Solving |

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.
Join the discussion

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.
Join the discussion

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.
Join the discussion

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.
Join the discussion

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
Join the discussion

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.
Join the discussion

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
Join the discussion

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

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.
Join the discussion

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...:)
Join the discussion

by vivek.kapoor83 » Mon Jan 05, 2009 10:01 pm
nice............. :P
Join the discussion