GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
In the diagram above, points A, O, and B all lie on a diameter of the
[#permalink]
21 Apr 2021, 16:00
21 Apr 2021, 16:00
Expert Reply
3
Bookmarks
00:00
Question Stats:
60% (03:38) correct
40% (03:13) wrong based on 5 sessions
Hide Show timer Statistics
In the diagram above, points A, O, and B all lie on a diameter of the large circle, and D and d are the diameters of the two smaller circles. If the sum of the areas of the two smaller circles is 5/8 of the large circle’s area, what is the ratio of D to d ?
A. 3:2
B. 5:3
C. 2:1
D. 3:1
E. 4:1
Show: ::
Attachment:
3092.png [ 3.99 KiB | Viewed 2181 times ]
Re: In the diagram above, points A, O, and B all lie on a diameter of the
[#permalink]
22 Apr 2021, 02:07
22 Apr 2021, 02:07
option d
lets assume are of outer circle as 100pi
equate 100pi=pi*r^2 after that we get radius as 10 units
from fig. we know that R+r=10...........(1)
from problem we have given relationship of area of outer and inner circle is 5/8
(area of two inner circle )/area of outer circle=5/8
x/100pi=5/8
x=62.5pi
for inner circles,
Area of larger circle +area of smaller circle=62.5pi
R^2+r^2=62.5
after simplifying and substituting value of R=10-r from equation (1)
we will get two values of r=7.5,2.5
after getting these two value we have to take smaller value
and substitute in equation 1 and get value of R=7.5
R/r=7.5/2.5=3:1
lets assume are of outer circle as 100pi
equate 100pi=pi*r^2 after that we get radius as 10 units
from fig. we know that R+r=10...........(1)
from problem we have given relationship of area of outer and inner circle is 5/8
(area of two inner circle )/area of outer circle=5/8
x/100pi=5/8
x=62.5pi
for inner circles,
Area of larger circle +area of smaller circle=62.5pi
R^2+r^2=62.5
after simplifying and substituting value of R=10-r from equation (1)
we will get two values of r=7.5,2.5
after getting these two value we have to take smaller value
and substitute in equation 1 and get value of R=7.5
R/r=7.5/2.5=3:1
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: In the diagram above, points A, O, and B all lie on a diameter of the
[#permalink]
22 Apr 2021, 03:06
22 Apr 2021, 03:06
1
GeminiHeat wrote:
In the diagram above, points A, O, and B all lie on a diameter of the large circle, and D and d are the diameters of the two smaller circles. If the sum of the areas of the two smaller circles is 5/8 of the large circle’s area, what is the ratio of D to d ?
A. 3:2
B. 5:3
C. 2:1
D. 3:1
E. 4:1
Area of smaller circle with diameter \(D\) \(= \frac{π}{4}D^2\)
Area of smaller circle with diameter \(d\) \(= \frac{π}{4}d^2\)
Diameter of the bigger circle = \(D + d\)
So, Area of bigger circle \(= \frac{π}{4}(D + d)^2\)
Given;
\(\frac{π}{4}D^2 + \frac{π}{4}d^2 = (\frac{5}{8})\frac{π}{4}(D + d)^2\)
\(8(D^2 + d^2) = 5(D^2 + d^2 + 2dD\)
\(3D^2 + 3d^2 = 10dD\)
Plug in the option choices;
A. \(D = 3, d = 2\)
\(3(3)^2 + 3(2)^2 ≠ 10(3)(2)\)
B. \(D = 5, d = 3\)
\(3(5)^2 + 3(3)^2 ≠ 10(5)(3)\)
C. \(D = 2, d = 1\)
\(3(2)^2 + 3(1)^2 ≠ 10(2)(1)\)
D. \(D = 3, d = 1\)
\(3(3)^2 + 3(1)^2 = 10(3)(1)\)
E. \(D = 4, d = 1\)
\(3(4)^2 + 3(1)^2 ≠ 10(4)(1)\)
Hence, option D
