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.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
PQRS is a rectangle with PQ < 65 and PQ + QR + RS = 100
[#permalink]
27 Sep 2021, 09:06
27 Sep 2021, 09:06
1
3
Bookmarks
00:00
Question Stats:
7% (02:27) correct
92% (02:32) wrong based on 28 sessions
Hide Show timer Statistics
PQRS is a rectangle with PQ < 65 and PQ + QR + RS = 100
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Maximum possible area of the rectangle PQRS |
1250 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
PQRS is a rectangle with PQ < 65 and PQ + QR + RS = 100
[#permalink]
05 Oct 2021, 03:35
05 Oct 2021, 03:35
4
2
Bookmarks
Explanantion:
PQ = RS = L
QR = SP = B
2L + B = 100
L = 50 - \(\frac{B}{2}\)
Area of rectangle PQRS = (L)(B) = (50 - \(\frac{B}{2}\))(B)
i.e. A = -\(\frac{B^2}{2}\) + 50B
NOTE: we have a quadratic equation and we can find its maximum by finding its vertex
x-coordinate of the vertex in a quadratic equation: \(ax^2 + bx + c\) is given by \(\frac{-b}{2a}\)
\(x_v = \frac{-(50)}{2(-0.5)} = 50\)
Maximum value of A = -\(\frac{50^2}{2}\) + 50(50) = -1250 + 2500 = 1250
Col. A: 1250
Col. B: 1250
Hence, option C
PQ = RS = L
QR = SP = B
2L + B = 100
L = 50 - \(\frac{B}{2}\)
Area of rectangle PQRS = (L)(B) = (50 - \(\frac{B}{2}\))(B)
i.e. A = -\(\frac{B^2}{2}\) + 50B
NOTE: we have a quadratic equation and we can find its maximum by finding its vertex
x-coordinate of the vertex in a quadratic equation: \(ax^2 + bx + c\) is given by \(\frac{-b}{2a}\)
\(x_v = \frac{-(50)}{2(-0.5)} = 50\)
Maximum value of A = -\(\frac{50^2}{2}\) + 50(50) = -1250 + 2500 = 1250
Col. A: 1250
Col. B: 1250
Hence, option C
General Discussion
Re: PQRS is a rectangle with PQ < 65 and PQ + QR + RS = 100
[#permalink]
04 Oct 2021, 12:01
04 Oct 2021, 12:01
solution pleasee.
