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)
A rectangular tank with outer dimensions 8cm x 10cm x 12cm has
[#permalink]
22 Sep 2021, 02:23
22 Sep 2021, 02:23
1
00:00
Question Stats:
66% (02:52) correct
33% (01:56) wrong based on 15 sessions
Hide Show timer Statistics
A rectangular tank with outer dimensions 8cm x 10cm x 12cm has a uniform thickness of 0.5cm. A right circular cylinder is to be placed inside the box so that it stands upright when the box rests on one of its sides and occupies the maximum volume?
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 |
Outer radius of the cylinder (in cm) |
4 |
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)
A rectangular tank with outer dimensions 8cm x 10cm x 12cm has
[#permalink]
28 Sep 2021, 11:25
28 Sep 2021, 11:25
1
2
Bookmarks
Explanation:
Since the thickness is 0.5cm, we need to subtract it twice (for both sides) from each dimension of the box
i.e. L = 18 - 1 = 17
B = 10 - 1 = 9
H = 12 - 1 = 11
Now the cylinder must stand upright and must have maximum volume.
NOTE: Only the smaller dimension of the face can be taken as the diameter
Case I: Base of cylinder at face LB
Diameter = B = 9
Height = H = 11
Volume = \(\frac{891π}{4}\)
Case II: Base of cylinder at face BH
Diameter = B = 9
Height = L = 17
Volume = \(\frac{1377π}{4}\)
Case I: Base of cylinder at face LH
Diameter = H = 11
Height = B = 9
Volume = \(\frac{1089π}{4}\)
Maximum volume is given for Case II with diameter as 9 and radius = \(\frac{9}{2} = 4.5\)
Col. A: 4.5
Col. B: 4
Hence, option A
Since the thickness is 0.5cm, we need to subtract it twice (for both sides) from each dimension of the box
i.e. L = 18 - 1 = 17
B = 10 - 1 = 9
H = 12 - 1 = 11
Now the cylinder must stand upright and must have maximum volume.
NOTE: Only the smaller dimension of the face can be taken as the diameter
Case I: Base of cylinder at face LB
Diameter = B = 9
Height = H = 11
Volume = \(\frac{891π}{4}\)
Case II: Base of cylinder at face BH
Diameter = B = 9
Height = L = 17
Volume = \(\frac{1377π}{4}\)
Case I: Base of cylinder at face LH
Diameter = H = 11
Height = B = 9
Volume = \(\frac{1089π}{4}\)
Maximum volume is given for Case II with diameter as 9 and radius = \(\frac{9}{2} = 4.5\)
Col. A: 4.5
Col. B: 4
Hence, option A
gmatclubot
A rectangular tank with outer dimensions 8cm x 10cm x 12cm has [#permalink]
28 Sep 2021, 11:25
Moderators:
