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.
10 Jun 2018, 02:33
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
57% (02:31) correct
43%
(02:31)
wrong
based on 197
sessions
History
Date
Time
Result
Not Attempted Yet
A rectangular field of dimension 100 m x 80 m is to be covered by artificial grass. The cost of implementing artificial grass is at least $1 per sq. m. Will the total cost incurred to put grass only in the field be more than $7100?
Mastering Important Concepts tested by GMAT in Triangles - I
Mastering Important Concepts tested by GMAT in Triangles - II
Mastering Important Concepts tested by GMAT in Triangles - III
- Statement 1. In the field there are two restricted paths– of uniform width 5 m – which spread along the centre of the field, and parallel to the length and breadth of the field respectively.
Statement 2. The cost of putting grass in 25% of the total field is not more than $1800.
Articles on Geometry
Common Mistakes in Geometry QuestionsMastering Important Concepts tested by GMAT in Triangles - I
Mastering Important Concepts tested by GMAT in Triangles - II
Mastering Important Concepts tested by GMAT in Triangles - III
Consolidated list of all the articles
Must Read Articles and Practice Questions to score Q51 !!!!
10 Jun 2018, 06:30
Kudos
Bookmarks
d
Question stem: Will the total cost incurred to put grass only in the field be more than $7100? (Y/N)
Statement 1
Area of the portion of rectangular field where grass to be covered= Area of rectangular field-(Area of horizontal restricted rectangular path+Area of vertical restricted rectangular path)
=100*80-[(100*5)+(5*75)]
=8000-875
=7125 sq. m
The cost of implementing artificial grass is at least $1 per sq. m.
So, (cost of implementing artificial grass in 7125 sq. m) \(\geq{$7125}\)
Therefore, st1 is sufficient.( Answer to question stem is Yes)
Statement 2:-
25% of grass area=\(\frac{1}{4}\)*total area=\(\frac{1}{4}\)*80*100=2000 sq. m
(cost of covering grass in 2000 sq. m) \(\leq{1800}\)=X (say)-----(a)
(Cost of covering grass in the rest portion of field i.e, 6000 sq. m) \(\geq{$6000}\)=Y(say)...........(b)
from (a) and (b), we can have
when X=1000 & Y=6000, X+Y=7000, answer to question is No
when X=1800 & Y=6100, X+Y=7900,answer to question is Yes
Therefore, St2 is not sufficient.
Ans. A
EgmatQuantExpert
Question stem: Will the total cost incurred to put grass only in the field be more than $7100? (Y/N)
Statement 1
Area of the portion of rectangular field where grass to be covered= Area of rectangular field-(Area of horizontal restricted rectangular path+Area of vertical restricted rectangular path)
=100*80-[(100*5)+(5*75)]
=8000-875
=7125 sq. m
The cost of implementing artificial grass is at least $1 per sq. m.
So, (cost of implementing artificial grass in 7125 sq. m) \(\geq{$7125}\)
Therefore, st1 is sufficient.( Answer to question stem is Yes)
Statement 2:-
25% of grass area=\(\frac{1}{4}\)*total area=\(\frac{1}{4}\)*80*100=2000 sq. m
(cost of covering grass in 2000 sq. m) \(\leq{1800}\)=X (say)-----(a)
(Cost of covering grass in the rest portion of field i.e, 6000 sq. m) \(\geq{$6000}\)=Y(say)...........(b)
from (a) and (b), we can have
when X=1000 & Y=6000, X+Y=7000, answer to question is No
when X=1800 & Y=6100, X+Y=7900,answer to question is Yes
Therefore, St2 is not sufficient.
Ans. A
Attachments
File comment: Figure not drawn to scale.
Statement.JPG [ 22.96 KiB | Viewed 5107 times ]
10 Jun 2018, 08:11
Kudos
Bookmarks
A rectangular field of dimension 100 m x 80 m is to be covered by artificial grass. The cost of implementing artificial grass is at least $1 per sq. m. Will the total cost incurred to put grass only in the field be more than $7100?
Statement 1. In the field there are two restricted paths– of uniform width 5 m – which spread along the centre of the field, and parallel to the length and breadth of the field respectively.
Statement 2. The cost of putting grass in 25% of the total field is not more than $1800.
Firstly, let's figure out the area of the field. I always like to figure out values that may be used throughout the question. So 100 x 80 = 8000m^2. It is also valuable to think before hand how this would relate to potential answers. Since the minimum cost of the artificial grass is $1 per m^2, we know then that to cover the entire field, it would be at least $8000. Whether this will be the case after considering statements (1) and (2) is a different matter, but it is good to think of it beforehand.
Statement (1): We now have restricted paths, where there will be no grass. The wording is a bit confusing, but it could be understood that the paths intersect and form a cross. There's two ways to figure this out. Either calculate the parts of each path that does not intersect, and then add on the intersected area; or calculate the areas for both path, then subtracting the area of the intersect once (because of the intersection, calculating the areas for both path would have doubled up on the intersected area, hence we need to subtract that area once). First method: 5*(100-5) + 5*(80-5) + 5*5 = 5*95 + 5*75 + 5*5 = 5*(95+75+5) = 5*(175) = 875m^2. Second method: 5*100 + 5*80 - 5*5 = 5*(100+80-5) = 5*175 = 875m^2. Both give us the same answer. Note that I kept the '5' as is for as long as possible. Always avoid making calculations where not necessary. Anywayssss let's figure out the area to be grassed. 8000 - 875 = 7125m^2. Because the minimum cost is $1 per m^2, then 7125m^2 is AT LEAST $7,125. Which is >$7,100. So, sufficient.
Statement (2): This is really, really, really confusing. The cost of covering 25% of the field (i.e. 2,000m^2) is <=$1,800, although somehow this is less than the 'at least $1 per sq. m' thing in the question... but okay. I assume then, that specifically for that 25%, the cost can be any positive number less than or equal to $1,800. If that's the case, we now just need to test to see if various costs of the 25% combined with the cost of the remainder can a.) be <= $7,100 and b.) be > $7,100. So let's get started testing! With extremes. What if 25% of the field costed $1 to cover with grass, and the remainder $6,000? Then the total cost would be $6,001. Now what if 25% of the field costed $1,800? Then even if the remaining 75% cost $6,000, that'd be $7,800. To be honest, even if 25% of the field costed $1, the remaining 75% can still cost $7,101 on its own, and in and of itself make the cost of covering the field with grass be > $7,100. So (2) is not sufficient.
Thus, answer is A.
Statement 1. In the field there are two restricted paths– of uniform width 5 m – which spread along the centre of the field, and parallel to the length and breadth of the field respectively.
Statement 2. The cost of putting grass in 25% of the total field is not more than $1800.
Firstly, let's figure out the area of the field. I always like to figure out values that may be used throughout the question. So 100 x 80 = 8000m^2. It is also valuable to think before hand how this would relate to potential answers. Since the minimum cost of the artificial grass is $1 per m^2, we know then that to cover the entire field, it would be at least $8000. Whether this will be the case after considering statements (1) and (2) is a different matter, but it is good to think of it beforehand.
Statement (1): We now have restricted paths, where there will be no grass. The wording is a bit confusing, but it could be understood that the paths intersect and form a cross. There's two ways to figure this out. Either calculate the parts of each path that does not intersect, and then add on the intersected area; or calculate the areas for both path, then subtracting the area of the intersect once (because of the intersection, calculating the areas for both path would have doubled up on the intersected area, hence we need to subtract that area once). First method: 5*(100-5) + 5*(80-5) + 5*5 = 5*95 + 5*75 + 5*5 = 5*(95+75+5) = 5*(175) = 875m^2. Second method: 5*100 + 5*80 - 5*5 = 5*(100+80-5) = 5*175 = 875m^2. Both give us the same answer. Note that I kept the '5' as is for as long as possible. Always avoid making calculations where not necessary. Anywayssss let's figure out the area to be grassed. 8000 - 875 = 7125m^2. Because the minimum cost is $1 per m^2, then 7125m^2 is AT LEAST $7,125. Which is >$7,100. So, sufficient.
Statement (2): This is really, really, really confusing. The cost of covering 25% of the field (i.e. 2,000m^2) is <=$1,800, although somehow this is less than the 'at least $1 per sq. m' thing in the question... but okay. I assume then, that specifically for that 25%, the cost can be any positive number less than or equal to $1,800. If that's the case, we now just need to test to see if various costs of the 25% combined with the cost of the remainder can a.) be <= $7,100 and b.) be > $7,100. So let's get started testing! With extremes. What if 25% of the field costed $1 to cover with grass, and the remainder $6,000? Then the total cost would be $6,001. Now what if 25% of the field costed $1,800? Then even if the remaining 75% cost $6,000, that'd be $7,800. To be honest, even if 25% of the field costed $1, the remaining 75% can still cost $7,101 on its own, and in and of itself make the cost of covering the field with grass be > $7,100. So (2) is not sufficient.
Thus, answer is A.
