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.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
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
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
21 Aug 2024, 01:30
Kudos
Bookmarks
C
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:
68% (02:49) correct
32%
(03:08)
wrong
based on 151
sessions
History
Date
Time
Result
Not Attempted Yet
A shop produces sarongs. The daily average production is given by 5n + 20, where n is the number of workers aside from the owner. In the first k days, 500 units are produced, and then 5 workers are added to the team. After another k days, the cumulative total is 1250. How many workers were part of the latter production run?
A) 6
B) 10
C) 11
D) 15
E) 23.5
A) 6
B) 10
C) 11
D) 15
E) 23.5
21 Aug 2024, 02:01
Kudos
Bookmarks
First k days:
\(k(5n+20)=500\)
\(5nk + 20k = 500\)
\(nk + 4k = 100\) [equation 1]
Second k days:
As one is adding 5 more workers, the initial 5n+20 daily average becomes 5(n+5)+20
\(k[5(n+5)+20]= 750\)
\(k(5n + 45)=750\)
\(5nk + 45k = 750\)
\(nk + 9k = 150\) [equation 2]
Solving for k:
Subtract [equation 1] from [equation 2]:
\(5k = 50\)
\(k = 10\)
Solving for n:
Plugging k = 10 into [equation 1]:
\(10n + 40 = 100\)
\(10n = 60\)
\(n = 6\)
Plus the 5 workers which were added in the second set of k days:\(6+5=11\)
ANSWER C
\(k(5n+20)=500\)
\(5nk + 20k = 500\)
\(nk + 4k = 100\) [equation 1]
Second k days:
As one is adding 5 more workers, the initial 5n+20 daily average becomes 5(n+5)+20
\(k[5(n+5)+20]= 750\)
\(k(5n + 45)=750\)
\(5nk + 45k = 750\)
\(nk + 9k = 150\) [equation 2]
Solving for k:
Subtract [equation 1] from [equation 2]:
\(5k = 50\)
\(k = 10\)
Solving for n:
Plugging k = 10 into [equation 1]:
\(10n + 40 = 100\)
\(10n = 60\)
\(n = 6\)
Plus the 5 workers which were added in the second set of k days:\(6+5=11\)
ANSWER C
21 Aug 2024, 06:20
Kudos
Bookmarks
5n + 20 is Average production / day, where n = no. of workers
In k days 500 units were produced,
500/k = 5n + 20 .................................(1)
5 workers added, in another k days the cumulative total was 1250 units, therefore 750 more in the another k days were produced
750/k = 5*(n+5) + 20 .........................(5 workers added)
750/k = 5n + 25 +20
750/k = 500/k + 25 ...........................(from 1)
k = 10
Subs. k in (1), we get n = 6
5 workers were added later,
n + 5 = 11
Answer. C
In k days 500 units were produced,
500/k = 5n + 20 .................................(1)
5 workers added, in another k days the cumulative total was 1250 units, therefore 750 more in the another k days were produced
750/k = 5*(n+5) + 20 .........................(5 workers added)
750/k = 5n + 25 +20
750/k = 500/k + 25 ...........................(from 1)
k = 10
Subs. k in (1), we get n = 6
5 workers were added later,
n + 5 = 11
Answer. C
