Find all School-related info fast with the new School-Specific MBA Forum

It is currently 19 Jun 2013, 05:46
Customize  |  Hide

m21 #8

  Question banks Downloads My Bookmarks Reviews  
Author Message
Manager
Manager
Joined: 11 Feb 2011
Posts: 91
Followers: 1

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

CAT Tests
Re: m21 #8 [#permalink] New post 14 Sep 2011, 18:15
easy one...C is the answer
Manager
Manager
User avatar
Affiliations: NABE
Joined: 25 Apr 2010
Posts: 67
Location: United States
Concentration: Strategy
GPA: 3.1
WE: Marketing (Energy and Utilities)
Followers: 1

Kudos [?]: 14 [0], given: 8

Re: m21 #8 [#permalink] New post 25 Oct 2011, 21:32
jallenmorris wrote:
I'm not sure what they mean by "system of equations" and I honestly don't think it's absolutely necessary to solve this problem.

We know that we originally had a ratio of 9:8. With ratios, the total number of items must be a multiple of the sum of the numbers in the ratio. (i.e., Apples:Oranges = 2:1 means we have a total of 3 items 2 apples + 1 orange). So 9:8 means we must have a multiple of 17.

Multiples of 17 are 17, 34, 51, 68, 85, etc

Now we know that once we have an addition of 3 sedans and 1 hatchback that we get a multiple of 11 (because 6:5 we need 6+5=11). Now, when we add 4 to a multiple of 17, which one gives us a multiple of 11?

17+4=21, not a multiple of 11
34+4 = 38, not either
51+4 = 55, YES! A multiple of 11. If the ratio is 6:5, and 55/11 = 5, means we have 5 groups of each, so # of sedans = 6*5, and # of hatchbacks = 5*5. Check it 30 + 25 = 55....still good.

How many more sedans than hatchbacks does the agency now have? 30-25 = 5 more sedans than hatchbacks.

It's much quicker to just work the problem than to explain it.

dakhan wrote:
In a fleet of a certain car-renting agency the ratio of the number of sedans to the number of hatchbacks was 9:8. After the addition of 3 sedans and 1 hatchback, this ratio rose to 6:5. How many more sedans than hatchbacks does the agency have now?

OA asks to setup a system of equations which i get. what is confusing is how the system of equations is leading to the answer choice provided. can someone help please? thanks.


I did it the same way, took me about 40 seconds.
Manager
Manager
User avatar
Joined: 16 Sep 2010
Posts: 230
Location: United States
Concentration: Finance, Real Estate
GMAT 1: 740 Q48 V42
Followers: 6

Kudos [?]: 42 [0], given: 2

GMAT Tests User
Re: m21 #8 [#permalink] New post 13 Nov 2011, 22:07
My approach is as follows:

1. Identify which multiple of 8 we can add 1 to, making it a multiple of 5. Ans - 24.

Then multiply each proportion by 3 and add the respective deltas as mentioned in the text.

Easy.
Manager
Manager
User avatar
Joined: 01 Jan 2011
Posts: 89
Schools: INSEAD,IIMA,IIMB
Followers: 0

Kudos [?]: 0 [0], given: 2

GMAT Tests User
Re: m21 #8 [#permalink] New post 20 Nov 2011, 23:21
simple eq solving..ans is C
_________________

_________________________
Try and you will succeed !

Director
Director
User avatar
Status: Final Countdown
Joined: 17 Mar 2010
Posts: 566
Location: India
GPA: 3.82
WE: Account Management (Retail Banking)
Followers: 10

Kudos [?]: 71 [0], given: 75

Re: m21 #8 [#permalink] New post 14 Sep 2012, 06:44
Initial Ratio:
s/h=9/8
=>s=(9/8)h........(i)

Then,we add more cars to already existing cars so the ratios becomes:
(s+3) / (h+1) =6/5
=> 5s+15=6h+6
=>5{(9/8) * h}=6h+6
=>h=24

so, from (i) s=27

later;
(s+3)/(h+1)=(27+3)/(24+1)=30/25

Hence the difference in no. of the new value of cars;
30-25=5

C wins
_________________

" Make more efforts "
Press Kudos if you liked my post

Manager
Manager
Joined: 07 Sep 2011
Posts: 71
GMAT 1: 660 Q41 V40
GMAT 2: 720 Q49 V39
WE: Analyst (Mutual Funds and Brokerage)
Followers: 0

Kudos [?]: 14 [0], given: 13

Re: m21 #8 [#permalink] New post 17 Sep 2012, 08:06
the original ratio of sedans to hatchbacks was 9:8
3 sedans and 1 hatcback is added, giving a new ratio of 6:5
remember that ratios can also be expressed as fractions

let x be the multiple of the original ratio so that 9x+8x=11x=number of original cars
(9x+3)/(8x+1)=6/5
cross multiply: 5(9x+3)=6(8x+1)
45x+15=48x+6
9=3x
x=3

plugging back in..
what is 9x+3-(8x+1)?
30-25=5
Answer is C
Re: m21 #8   [#permalink] 17 Sep 2012, 08:06
    Similar topics Author Replies Last post
Similar
Topics:
Popular new posts 13 M21 #28 arjtryarjtry 22 23 Aug 2008, 23:50
Popular new posts 7 Experts publish their posts in the topic M21#15 ventivish 16 23 Nov 2008, 12:43
Popular new posts 7 Experts publish their posts in the topic M21#29 ventivish 19 23 Nov 2008, 13:43
Popular new posts 28 Experts publish their posts in the topic m21 #20 dczuchta 24 27 Feb 2009, 11:40
This topic is locked, you cannot edit posts or make further replies. New M21 - #2 kt00381n 1 30 Sep 2009, 07:06
Display posts from previous: Sort by

m21 #8

  Question banks Downloads My Bookmarks Reviews  

Go to page   Previous    1   2   [ 26 posts ] 

Moderator: Bunuel



GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.