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.
Jun 0712:00 PM PDT
-01:00 PM PDT
In this GMAT experience talk show, we talk to Tavishi, a young MBA aspirant from India who recently scored 725 in her latest GMAT Focus exam. It’s a 99.9 percentile score on the new GMAT Focus edition and she achieved this feat in her first GMAT attempt.
Jun 0811:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.- Jun 09
11:00 AM IST
-01:00 PM IST
Attend this session to evaluate your current skill level, learn process skills, and solve through tough Arithmetic questions - Jun 11
05:40 AM PDT
-12:00 PM PDT
Register for the GMAT Club Virtual MBA Spotlight fair, the biggest MBA fair of the year. You will have a chance to hear Admissions directors from almost every Top 20 program speak, network with peers, and more. 
Jun 1111:00 AM PDT
-12:00 PM PDT
Magoosh students get into top business schools like MIT Sloan, HBS, Stanford GSB, Wharton and more! See why over 6 million students trust us for exam prep.
A group of ten candy bars has an average cost of $0.89 per candy bar.
[#permalink]
19 May 2019, 21:33
19 May 2019, 21:33
5
Bookmarks
Expert Reply
Show timer
00:00
A
B
C
D
E
Difficulty:
55%
(hard)
Question Stats:
58% (02:08) correct
42%
(01:59)
wrong
based on 67
sessions
A group of ten candy bars has an average cost of $0.89 per candy bar. How many additional bars must be bought at the cost of $0.72 a piece to bring the average below $0.80?
A. 8
B. 9
C. 10
D. 11
E. 12
A. 8
B. 9
C. 10
D. 11
E. 12
GMAT Club Legend
Joined: 18 Aug 2017
Status:You learn more from failure than from success.
Posts: 7900
Given Kudos: 242
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1:
545 Q79 V79 DI73
545 Q79 V79 DI73
GPA: 4
WE:Marketing (Energy and Utilities)
Re: A group of ten candy bars has an average cost of $0.89 per candy bar.
[#permalink]
20 May 2019, 02:11
20 May 2019, 02:11
Bunuel wrote:
A group of ten candy bars has an average cost of $0.89 per candy bar. How many additional bars must be bought at the cost of $0.72 a piece to bring the average below $0.80?
A. 8
B. 9
C. 10
D. 11
E. 12
A. 8
B. 9
C. 10
D. 11
E. 12
solve
0.79=8.9+.72*x / 10+x
x = ~11 so value has to be >11
IMO E 12
CrackVerbal Representative
Joined: 13 May 2019
Posts: 66
Given Kudos: 4
Re: A group of ten candy bars has an average cost of $0.89 per candy bar.
[#permalink]
21 May 2019, 00:53
21 May 2019, 00:53
Expert Reply
The average cost for 10 candies = $0.89
Let the additional number of bars that must be bought to bring the average below $0.80 to be "\(x\)".
The average cost of the candies after x number of candies have been bought = \(\frac{(0.89*10 + 0.72*x)}{10 + x}\)
= \(\frac{(8.9 + 0.72x)}{10+x}\)
As per the question, the new average cost per candy should be less than $0.80
Hence, \((8.9 + 0.72x)10+x < 0.8\)
\(8.9 + 0.72x < 8 + 0.8x\)
\(0.9 < 0.08x\)
\(0.08x > 0.90\)
\(8x > 90\)
\(x > 90/8\)
\(x > 11.25\)
Since the number of candies will be a natural number, the least possible value of \(x\) is 12.
The correct answer is E.
Let the additional number of bars that must be bought to bring the average below $0.80 to be "\(x\)".
The average cost of the candies after x number of candies have been bought = \(\frac{(0.89*10 + 0.72*x)}{10 + x}\)
= \(\frac{(8.9 + 0.72x)}{10+x}\)
As per the question, the new average cost per candy should be less than $0.80
Hence, \((8.9 + 0.72x)10+x < 0.8\)
\(8.9 + 0.72x < 8 + 0.8x\)
\(0.9 < 0.08x\)
\(0.08x > 0.90\)
\(8x > 90\)
\(x > 90/8\)
\(x > 11.25\)
Since the number of candies will be a natural number, the least possible value of \(x\) is 12.
The correct answer is E.
