Forum Home > GMAT > Quantitative > Problem Solving (PS)
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.
Dec 1511:00 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.
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!- Dec 17
09:00 AM PST
-10:00 AM PST
Join Manhattan Prep instructor Whitney Garner for a fun—and thorough—review of logic-based (non-math) problems, with a particular emphasis on Data Sufficiency and Two-Parts. - Dec 18
10:00 AM PST
-11:00 AM PST
Here is the essential guide to securing scholarships as an MBA student! In this video, we explore the various types of scholarships available, including need-based and merit-based options.
Updated on: 15 May 2017, 04:55
Originally posted by GMATinsight on 15 May 2017, 03:23.
Last edited by GMATinsight on 15 May 2017, 04:55, edited 2 times in total.
Last edited by GMATinsight on 15 May 2017, 04:55, edited 2 times in total.
Kudos
Bookmarks
D
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:
65% (02:25) correct
35%
(02:45)
wrong
based on 120
sessions
History
Date
Time
Result
Not Attempted Yet
In a workshop, 10 machines working simultaneously can finish a work of manufacturing 3000 components in 10 days while working 9 hours a day. What is the minimum number of additional machines required at the same workshop to finish the work of manufacturing 5000 similar components if work needs to be delivered in 6 days and maximum number of hours that machines can operate in a day is 10?
A) 4
B) 5
C) 10
D) 15
E) 25
SOURCE: https://www.GMATinsight.com
A) 4
B) 5
C) 10
D) 15
E) 25
SOURCE: https://www.GMATinsight.com
15 May 2017, 05:03
Kudos
Bookmarks
GMATinsight
In a workshop, 10 machines working simultaneously can finish a work of manufacturing 3000 components in 10 days while working 9 hours a day. What is the minimum number of additional machines required at the same workshop to finish the work of manufacturing 5000 similar components if work needs to be delivered in 6 days and maximum number of hours that machines can operate in a day is 10?
A) 4
B) 5
C) 10
D) 15
E) 25
SOURCE: https://www.GMATinsight.com
A) 4
B) 5
C) 10
D) 15
E) 25
SOURCE: https://www.GMATinsight.com
OE
CONCEPT: \(\frac{(Machine_Power * Time)}{Work} = Constant\)
i.e. \(\frac{(M_1 * T_1)}{W_1} = \frac{(M_2 * T_2)}{W_2}\)
i.e. \(\frac{(10 * (10*9)}{3000} = \frac{(M_2 * (6*10))}{5000}\)
i.e. \(T_2 = 25\)
i.e. Additional Machines required = 25-10 = 15
Answer: option D
General Discussion
15 May 2017, 04:38
Kudos
Bookmarks
GMATinsight
If 10 machines working simultaneously can finish a work of manufacturing 3000 components in 10 days while working 9 hours a day. Then what is the minimum number of additional machines required to finish the work of manufacturing 5000 similar components if work needs to be delivered in 6 days and maximum number of hours that machines can operate in a day is 6.
A) 4
B) 5
C) 10
D) 15
E) 25
SOURCE: https://www.GMATinsight.com
A) 4
B) 5
C) 10
D) 15
E) 25
SOURCE: https://www.GMATinsight.com
Hi
When you say addl machines, does it mean these 10 are working for 9 hrs as earlier or these also wk for 6hrs.
Irrespective of either 6 or 9 ans cannot be D..
Let's take all for 9 hrs, so these 10 machine will cater for days 10 to 6 by 10*10/6..
Now to cater for increase in components from 3000 to 5000.... 10*10/6*5000/3000=500/18~28..
So ATLEAST 18 even if we take time as 9 hrs for all...
1) If we take all to be in 6 hrs, 500/18 * 9/6=1500/36~42 so addl 42-10=32..
2) if these 10 continue for 9 hrs, we can see how many these 10 make and rest can be seen with 6h.
After the necessary amendments...
To cater for change from 3000 to 5000 components
10 machines will become \(\frac{10*5000}{3000}\)
This will further cater for change of 10days to 6days..... Machines required \(\frac{10*5000*10}{3000*6}\)
Finally to cater for extra 1hr from 9 to 10hr...... Machines becomes \(\frac{10*5000*10*9}{3000*6*10}\)=25..
So addl required=25-10=15
D
