BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Featured Course
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
Pond Problem
Source: Beat The GMAT — Problem Solving |
-
kevincanspain
- GMAT Instructor
- Posts: 613
- Location: madrid
- Thanked: 171 times
- GMAT Score:790
Do you see that we are dealing with a right-isosceles triangle?
Kevin Armstrong
GMAT Instructor
Gmatclasses
Madrid
GMAT Instructor
Gmatclasses
Madrid
-
Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Location: Toronto
- Thanked: 1710 times
- GMAT Score:800
Since it's a Kaplan problem, let's apply the Kaplan method!
Step 1 of the Kaplan Method for PS: analyze the problem
We see that AB is the diameter of the circle. We see that each bridge is equal, creating an isosceles triangle with the diameter. We know that any triangle that has the diameter at one side and the opposite vertex on the side of the circle will be right (i.e. the angle opposite the diameter is 90 degrees).
So, we have a 45/45/90 triangle, with sides of x, x and x(root2).
Step 2 of the Kaplan Method for PS: state the task
Here, we need to find the ratio of the sum of the bridges to the distance from A to B along the outside of the pond, i.e. 1/2 the circumference of the circle (or 1/2(pi)d).
So, the answer will be:
(AC + BC)/(1/2)(pi)d
Step 3 of the Kaplan Method for PS: approach strategically
We could work with "x"s, but we prefer numbers; so, to make life simple, let's set the diameter of the triangle equal to root2, giving us bridge lengths of 1 and 1.
Accordingly, the distance via bridge is 1+1=2.
d=root2, so circumference = (pi)d = (pi)root2
Plugging into (AC + BC)/(1/2)(pi)d:
(1+1)/(1/2)(pi)(root2)
2 /(1/2)(pi)(root2)
2 * 2/(pi)(root2)
4 / (pi)(root2)
On the GMAT, it's very unlikely you'll see a root in the denominator of a fraction; accordingly, to match up to a choice, we need to "rationalize the denominator", which we accomplish by multiplying the top and bottom by the root in question:
4 / (pi)(root2) * root2/root2
4(root2)/2(pi)
2(root2)/pi
choose (A).
Step 1 of the Kaplan Method for PS: analyze the problem
We see that AB is the diameter of the circle. We see that each bridge is equal, creating an isosceles triangle with the diameter. We know that any triangle that has the diameter at one side and the opposite vertex on the side of the circle will be right (i.e. the angle opposite the diameter is 90 degrees).
So, we have a 45/45/90 triangle, with sides of x, x and x(root2).
Step 2 of the Kaplan Method for PS: state the task
Here, we need to find the ratio of the sum of the bridges to the distance from A to B along the outside of the pond, i.e. 1/2 the circumference of the circle (or 1/2(pi)d).
So, the answer will be:
(AC + BC)/(1/2)(pi)d
Step 3 of the Kaplan Method for PS: approach strategically
We could work with "x"s, but we prefer numbers; so, to make life simple, let's set the diameter of the triangle equal to root2, giving us bridge lengths of 1 and 1.
Accordingly, the distance via bridge is 1+1=2.
d=root2, so circumference = (pi)d = (pi)root2
Plugging into (AC + BC)/(1/2)(pi)d:
(1+1)/(1/2)(pi)(root2)
2 /(1/2)(pi)(root2)
2 * 2/(pi)(root2)
4 / (pi)(root2)
On the GMAT, it's very unlikely you'll see a root in the denominator of a fraction; accordingly, to match up to a choice, we need to "rationalize the denominator", which we accomplish by multiplying the top and bottom by the root in question:
4 / (pi)(root2) * root2/root2
4(root2)/2(pi)
2(root2)/pi
choose (A).
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Thanks, Stuart! Little tidbits of knowledge such as "On the GMAT, it's very unlikely you'll see a root in the denominator of a fraction" are very helpful.
<ACB will always be 90°
i.e double of the < AdB cos <AdB is a st line = 180°
AC²+BC² = d² given that AC = BC
2AC² = d² AC = d/√2
Semi circle ACB = Pi d/2
Ratio of 2AC / (Pi D/2)
(2*d/√2)/(Pi d/2)
2*2/(Pi √2)
(2√2)/Pi
i.e double of the < AdB cos <AdB is a st line = 180°
AC²+BC² = d² given that AC = BC
2AC² = d² AC = d/√2
Semi circle ACB = Pi d/2
Ratio of 2AC / (Pi D/2)
(2*d/√2)/(Pi d/2)
2*2/(Pi √2)
(2√2)/Pi
