Events & Promotions
|
|
GMAT 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.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
06 Oct 2021, 06:15
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
61% (02:37) correct
39%
(02:20)
wrong
based on 36
sessions
History
Date
Time
Result
Not Attempted Yet
To cover the floor of an entry hall, a 1′ * 12′ strip of carpet is cut into two pieces, shown as the shaded strips in the figure above, and each piece is connected to a third carpet piece as shown. If the 1′ strips run parallel to each other, what is the total area of the carpeted floor, in square feet?
(A) 46
(B) 48
(C) 52.5
(D) 56
(E) 60
Attachment:
1.jpg [ 14.21 KiB | Viewed 1743 times ]
16 Oct 2021, 03:51
Kudos
Bookmarks
Bunuel
Understanding the question is the most hard part of this question, however once that is clear , we can understand the solution.
Please refer to the attached fig.
Attachment:
carpet.png [ 6.47 KiB | Viewed 1431 times ]
We can see an isosceles trapezium in the center, area of which is given by
Avg of parallel sides *height = \(\frac{ x+y}{2} * 8 = \frac{12}{2}*8 = 48\)
Also the \(1' * 12'\) strip has an area of \(12\) sq. feet
So total area ->\(48 +12 = 60 \) sq. feet
Ans- E
Hope it's clear.
17 Oct 2021, 11:30
Kudos
Bookmarks
Kudos for the correct answer.
I’m not sure whether you can infer that the middle figure is an Isosceles Trapezoid. It is a trapezoid.
However, it is irrelevant.
The Area of the trapezoid formula is as you described it: (sum of the 2 parallel sides) * (1/2) * (perpendicular height between the 2 parallel sides equal to 8 ft.)
Then, add in the area of the 2 rectangular strips that each have a width of 1 ft. and a combined length of 12 inches.
Posted from my mobile device
I’m not sure whether you can infer that the middle figure is an Isosceles Trapezoid. It is a trapezoid.
However, it is irrelevant.
The Area of the trapezoid formula is as you described it: (sum of the 2 parallel sides) * (1/2) * (perpendicular height between the 2 parallel sides equal to 8 ft.)
Then, add in the area of the 2 rectangular strips that each have a width of 1 ft. and a combined length of 12 inches.
stne
Posted from my mobile device
