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.
Jul 0901:00 PM EDT
-02:00 PM EDT
More Target Test Prep students are earning 100th percentile GMAT scores. Don’t settle for less when the Target Test Prep course gives you everything you need to score high on the GMAT. Start your 5-day free trial today.
Jul 1410:00 AM PDT
-11:59 PM PDT
GMAT Preparation Online - A limited time sale use code: EXPERTSGLOBAL10- Jul 15
08:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations! 
Jul 1609:30 AM EDT
-11:30 AM EDT
Is INSEAD your dream b-school and you're looking for solid prep and application strategies to get there? Check out Barkavi’s end-to-end journey to know how she did! Get insights into the INSEAD admissions process, know what Adcoms look at, and more!
Jul 1611:00 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.
Jul 1911:00 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
05 Mar 2016, 12:08
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
56% (02:41) correct
44%
(02:41)
wrong
based on 407
sessions
History
Date
Time
Result
Not Attempted Yet
The average (arithmetic mean) cost of three computer models is $900. If no two computers cost the same amount, does the most expensive model cost more than $1,000?
(1) The most expensive model costs 25% more than the model with the median cost.
(2) The most expensive model costs $210 more than the model with the median cost.
(1) The most expensive model costs 25% more than the model with the median cost.
(2) The most expensive model costs $210 more than the model with the median cost.
The OA uses the concept of assigning variables a, b, c, and plugging in "LT" or "less than" to set up multiple equations. Is there an easier method to accomplish this?
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,328
Given Kudos: 816
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
05 Mar 2016, 12:39
Kudos
Bookmarks
GinGMAT
The average (arithmetic mean) cost of three computer models is $900. If no two computers cost the same amount, does the most expensive model cost more than $1,000?
(1) The most expensive model costs 25% more than the model with the median cost.
(2) The most expensive model costs $210 more than the model with the median cost.
The OA uses the concept of assigning variables a, b, c, and plugging in "LT" or "less than" to set up multiple equations. Is there an easier method to accomplish this?
(1) The most expensive model costs 25% more than the model with the median cost.
(2) The most expensive model costs $210 more than the model with the median cost.
The OA uses the concept of assigning variables a, b, c, and plugging in "LT" or "less than" to set up multiple equations. Is there an easier method to accomplish this?
A good question to understand the limits of the information given.
Try to learn the "best" method rather than the "easiest" method.
Average of 3 unequally priced computers = 900 --->a+b+c =2700 such that a<b<c. The limiting value asked in the question is 1000$
Per statement 1, c=1.25b ---> now let us look at the case when c=1000, b = 1000/1.25 = 800 ----> a= 2700-800-1000=900 and this makes a>b which is NOT allowed as b is the median value. So you can clearly see that the value of c MUST be >1000 for the values to make sense with this statement. Hence the answer to the question asked is a definite yes. Sufficient.
Per statement 2, c=210+b . Again use the same logic as above to use the limiting value of 1000$. This gives us c=1000, b=790 and a=910 , again a>b and as such this isnt allowed as the median value of 3 data points must be the centermost value. Hence c > 1000 in order for this statement to be true. Sufficient.
Hence D is the correct answer.
Sometimes in a DS question, you have to manipulate the given question to your advantage as is done above.
Hope this helps.
General Discussion
25 Jan 2019, 11:38
Kudos
Bookmarks
GinGMAT
The average (arithmetic mean) cost of three computer models is $900. If no two computers cost the same amount, does the most expensive model cost more than $1,000?
(1) The most expensive model costs 25% more than the model with the median cost.
(2) The most expensive model costs $210 more than the model with the median cost.
\({\rm{costs}}\,\,:\,\,a < b < c\,\,\,\,\,\,\left[ \$ \right]\)(1) The most expensive model costs 25% more than the model with the median cost.
(2) The most expensive model costs $210 more than the model with the median cost.
\(a + b + c = 3 \cdot 900 = 2700\,\,\,\left( * \right)\)
\(c\,\,\mathop > \limits^? \,\,1000\)
\(\left( 1 \right)\,\,c = {5 \over 4}b\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,a + {9 \over 5}c = 2700\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\)
\(\left( {**} \right)\,\,\,\,c \le 1000\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{\\
\,{9 \over 5}c\,\, \le \,\,1800\,\,\,\,\, \Rightarrow \,\,\,\,\,a = 2700 - {9 \over 5}c\,\, \ge \,\,2700 - 1800 = 900 \hfill \cr \\
\,b = {4 \over 5}c\,\, \le \,\,800 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,b \le 800 < 900 \le a\,\,\,{\rm{impossible}}\)
\(\left( 2 \right)\,\,c - b = 210\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,a + \left( {c - 210} \right) + c = 2700\,\,\,\,\, \Rightarrow \,\,\,\,\,a + 2c = 2910\,\,\,\,\mathop \Rightarrow \limits^{\left( {***} \right)} \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\)
\(\left( {***} \right)\,\,\,\,c \le 1000\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{\\
\,a = 2910 - 2c\,\, \ge \,\,2910 - 2000 = 910 \hfill \cr \\
\,b = c - 210\,\, \le \,\,790 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,b \le 790 < 910 \le a\,\,\,{\rm{impossible}}\)
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
