Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 49320

Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent
[#permalink]
Show Tags
05 Nov 2017, 02:04
Question Stats:
85% (01:49) correct 15% (01:23) wrong based on 94 sessions
HideShow timer Statistics



Manager
Joined: 05 Nov 2014
Posts: 112
Location: India
Concentration: Strategy, Operations
GPA: 3.75

Re: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent
[#permalink]
Show Tags
05 Nov 2017, 02:15
Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent more than John’s meal, and Mary’s meal cost as much as Karen’s meal. If Mary paid $2 more than John, how much was the total that the three of them paid?
(A) $28.33 (B) $30.00 (C) $35.00 (D) $37.50 (E) $40.00
Hello @Bunnel, I have tried solving this question algebraically, Am getting $16 as the answer. I don't see flaw in my approach. Can you please check the Question. Correct me if I am wong.



Math Expert
Joined: 02 Sep 2009
Posts: 49320

Re: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent
[#permalink]
Show Tags
05 Nov 2017, 02:17



BSchool Forum Moderator
Joined: 26 Feb 2016
Posts: 3137
Location: India
GPA: 3.12

Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent
[#permalink]
Show Tags
05 Nov 2017, 02:21
Bunuel wrote: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent more than John’s meal, and Mary’s meal cost 5/6 as much as Karen’s meal. If Mary paid $2 more than John, how much was the total that the three of them paid?
(A) $28.33 (B) $30.00 (C) $35.00 (D) $37.50 (E) $40.00 Lets assume John's meal costs 120$. Now Karen's meal will cost 180$ Since Mary's meal costs \(\frac{5}{6}\)th as much, it would be 150$. The prices of their meals are in the ratio 120:180:150 or 4x:6x:5x(John:Karen:Mary) Since Mary pays 2$ more than John, 5x  4x = 2 => x=2 Therefore, the three of them would have paid 15x or 30$(Option B)
_________________
You've got what it takes, but it will take everything you've got



Manager
Joined: 05 Nov 2014
Posts: 112
Location: India
Concentration: Strategy, Operations
GPA: 3.75

Re: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent
[#permalink]
Show Tags
05 Nov 2017, 02:23
Solution:
k=1.5 J1 m= (5/6J)2 mJ=23
Substituting 1 and 2 in 3, We get k=12, m=10 and J=8. So the total cost of the lunch is $30. Option B.



Manager
Joined: 11 Jul 2016
Posts: 53
Location: India
Concentration: Technology, General Management
GPA: 3.65
WE: Information Technology (Consulting)

Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent
[#permalink]
Show Tags
08 Nov 2017, 03:40
Let John's share=J Karen's share= 50% more of John = (3/2)J Mary's share = 5/6th of Karen = (5/6)(3/2J) = (5/4)J
Also given, (5/4)J=J+2 or,5J=4J +8 J=8
Therefore,Karen = (3/2)J=12 Mary= (5/4)J= 10
Total expenditure=8+12+10=30



Intern
Joined: 22 Aug 2017
Posts: 6

Re: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent
[#permalink]
Show Tags
08 Nov 2017, 09:05
let John's meal cost be J. karen (K) = 0.5J + J= 1.5J Mary (M) = 5/6*1.5J
As given, 5/6*1.5J = 2+J => J=8
We get, K = 1.5*8 = 12 and M = 5/6*1.5*8 = 10
Hence, total cost = 8+12+10= 30



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2835

Re: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent
[#permalink]
Show Tags
12 Nov 2017, 08:19
Bunuel wrote: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent more than John’s meal, and Mary’s meal cost 5/6 as much as Karen’s meal. If Mary paid $2 more than John, how much was the total that the three of them paid?
(A) $28.33 (B) $30.00 (C) $35.00 (D) $37.50 (E) $40.00 We can let M = the cost of Mary’s meal, K = the cost of Karen’s meal, and J = the cost of John’s meal. Thus: K = 1.5J and M = 5K/6 and M = J + 2 So, we can first use the first two equations and we have: M = 5(1.5J)/6 M = 7.5J/6 Finally, we have: 7.5J/6 = J + 2 Multiplying the entire equation by 6, we get: 7.5J = 6J + 12 1.5J = 12 J = 8 So, M = 8 + 2 = 10, and K = 1.5 x 8 = 12. So, the total for all 3 meals is 8 + 10 + 12 = $30. Answer: B
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4033
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)

Re: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent
[#permalink]
Show Tags
12 Nov 2017, 09:16
Bunuel wrote: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent more than John’s meal, and Mary’s meal cost 5/6 as much as Karen’s meal. If Mary paid $2 more than John, how much was the total that the three of them paid?
(A) $28.33 (B) $30.00 (C) $35.00 (D) $37.50 (E) $40.00 Let cost of John’s meal be 60 So, cost of Karen’s meal will be 90 And,cost of Mary’s meal will be 75 Total cost is 225 Quote: If Mary paid $2 more than John, how much was the total that the three of them paid \(\frac{2}{7560}*225\) =30 , Answer must be (B) 30
_________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
How to use Search Function in GMAT Club  Rules for Posting in QA forum  Writing Mathematical Formulas Rules for Posting in VA forum  Request Expert's Reply ( VA Forum Only )




Re: Mary, John, and Karen ate lunch together. Karen’s meal cost 50 percent &nbs
[#permalink]
12 Nov 2017, 09:16






