It is currently 14 Dec 2018, 02:20
My Tests

Close

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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Carla has 1/4 more sweaters than cardigans, and

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Moderator
Moderator
User avatar
Joined: 18 Apr 2015
Posts: 5166
Followers: 77

Kudos [?]: 1033 [0], given: 4655

CAT Tests
Carla has 1/4 more sweaters than cardigans, and [#permalink] New post 22 Aug 2017, 02:10
Expert's post
00:00

Question Stats:

43% (03:37) correct 56% (02:32) wrong based on 16 sessions


Carla has \(\frac{1}{4}\) more sweaters than cardigans, and \(\frac{2}{5}\) fewer cardigans than turtle­ necks. If she has at least one of each item, what is the minimum total number of turtlenecks plus sweaters that Carla could have?

enter your value

[Reveal] Spoiler: OA
35

_________________

Get the 2 FREE GREPrepclub Tests

1 KUDOS received
Director
Director
Joined: 03 Sep 2017
Posts: 521
Followers: 1

Kudos [?]: 330 [1] , given: 66

Re: Carla has 1/4 more sweaters than cardigans, and [#permalink] New post 25 Sep 2017, 07:50
1
This post received
KUDOS
I am a bit stuck on this one. I tried to rewrite the statements as formulas and I get that
\(S=(1+\frac{1}{4})C\)
\(C=(1-\frac{2}{5}T\)
where S are sweaters, C are cardigans and T are turtlenecks.

Then, I adjusted the expression in order to build a combined one as \(S=\frac{5}{4}C=\frac{3}{4}T\).
And I don't know how to go on.

Any hint?
1 KUDOS received
Director
Director
Joined: 20 Apr 2016
Posts: 758
Followers: 6

Kudos [?]: 514 [1] , given: 94

CAT Tests
Re: Carla has 1/4 more sweaters than cardigans, and [#permalink] New post 25 Sep 2017, 08:52
1
This post received
KUDOS
Carcass wrote:
Carla has \(\frac{1}{4}\) more sweaters than cardigans, and \(\frac{2}{5}\) fewer cardigans than turtle­ necks. If she has at least one of each item, what is the minimum total number of turtlenecks plus sweaters that Carla could have?


Given

S = \((\frac{1}{4} +1)*C\)

= \((\frac{5}{4})*C\)

and C = \((1-\frac{2}{5} )*T\)

= \((\frac{3}{5})*T\)

or T = \((\frac{5}{3})*C\)

Now

Let C = 12 (since it is a factor of 4 and 3. It should be the minimum value)

therefore S= \((\frac{5}{4})*12\)

or S = 15

and T =\((\frac{5}{3})*12\)

or T = 20

Therefore the sweater and turtle neck = 15+20 =35
_________________

If you found this post useful, please let me know by pressing the Kudos Button

2 KUDOS received
Moderator
Moderator
User avatar
Joined: 18 Apr 2015
Posts: 5166
Followers: 77

Kudos [?]: 1033 [2] , given: 4655

CAT Tests
Re: Carla has 1/4 more sweaters than cardigans, and [#permalink] New post 26 Sep 2017, 02:00
2
This post received
KUDOS
Expert's post
\(S = ( C + \frac{1}{4} C)\) this is the exact way to rephrase the sentence.

\(C = ( T - \frac{2}{5} T )\)

So you do have \(\frac{5}{4} C\) and \(\frac{5}{3} T\)

Now the LCM between 4 and 3 is 12. Then substitute and you have the minimum quantity 35

Another approach is thinking real numbers, for instance: the number you are looking for must be integers.

If you think you do have 10 of C, 25% more is 2.5 so not possible. 11 you do have 2.75 BUT 12, 25% of it is 3. So it possible 12 + 3 = 15.

This is also an easy approach to follow.

Regards
_________________

Get the 2 FREE GREPrepclub Tests

1 KUDOS received
Intern
Intern
Joined: 20 Mar 2018
Posts: 34
GRE 1: Q164 V150
Followers: 1

Kudos [?]: 24 [1] , given: 2

Re: Carla has 1/4 more sweaters than cardigans, and [#permalink] New post 11 Apr 2018, 20:02
1
This post received
KUDOS
Since the number of Cardigans, sweaters and turtlenecks should be integers.
Given: S = C + C/4 and C = T - 2T/5 --> S = 5C/4 and T = 5C/3

Now C/4 and 5C/3 have to be integers, we would take C as an LCM of 4 and 3 i.e 12 and work out the values of C and T
Re: Carla has 1/4 more sweaters than cardigans, and   [#permalink] 11 Apr 2018, 20:02
Display posts from previous: Sort by

Carla has 1/4 more sweaters than cardigans, and

  Question banks Downloads My Bookmarks Reviews Important topics  


GRE Prep Club Forum Home| About| Terms and Conditions and Privacy Policy| GRE Prep Club Rules| Contact

Powered by phpBB © phpBB Group

Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.