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# m21 #8

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01 Jun 2009, 19:41
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In a fleet of a certain car-renting agency the ratio of the number of sedans to the number of hatchbacks was 9:8. If 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?

(A) 2
(B) 3
(C) 5
(D) 7
(E) 9

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

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.
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Re: m21 #8 [#permalink]

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01 Jun 2009, 20:33
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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.

x/y = 9/8
8x - 9y = 0 ...................i

(x+3)/ (y+1) = 6/5
5x - 6y +9 = 0 ...................ii

Solve for x and y.

looks like y = 24
x = 27

so the diff = x+3 - y+1 = 27 + 3 - 24+1 = 5
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Re: m21 #8 [#permalink]

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01 Jun 2009, 20:53
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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.

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Intern Joined: 17 Mar 2008 Posts: 12 Followers: 0 Kudos [?]: 6 [0], given: 5 Re: m21 #8 [#permalink] ### Show Tags 01 Jun 2009, 21:23 thanks guys...both the system of equations and the backsolve obviously work. i see your point jalen. its always a better option to see if you dont have to solve the problem. SVP Joined: 30 Apr 2008 Posts: 1888 Location: Oklahoma City Schools: Hard Knocks Followers: 39 Kudos [?]: 524 [2] , given: 32 Re: m21 #8 [#permalink] ### Show Tags 01 Jun 2009, 21:38 2 This post received KUDOS Some questions can be answered quickly by doing simple math. Others may take 6 min with the number of steps required in simple math. One thing I've learned on the GMAT is that there is often a very simple way of working the problem to solve it within the < 2 min required. Sometimes we HAVE to use alegebra or something like that. Other questions reward creative thinking. For example: Whenever you get a question that asks you to sum a sequence of numbers, you can often add the ends and come up with a series of the same number. Adding all odd numbers in a sequnce from 13 to 39. 13 15 17 19 21 23 25 27 29 31 33 35 37 39 Realize that 13+37=50; 15 + 35 = 50; 17 + 33, 19+31, 21+29, 23+27 each = 50. There are 6 "groups" of these, so 6*50=300. Now you have 300 + 25 + 39. This is easier to add than 13 numbers. 25+39 = 64; 64+300 = 364. Or you can rememebr that when adding the first N odd numbers, you take $$N^2$$. Then to find out that 39 is the Nth odd number, (39+1)/2 = 20th odd integer. 20^2 = 400. But you started with 13, so the previous odd would be 11, which is the (11+2)/2=6th odd integer. 6^2 = 36. So 400-36 = 364. I think my way is easier to remember and less likely to get messed up on G-Day. It's to easy to think odd sequence. N^2 = total, so 39^2...but that's not right. dakhan wrote: thanks guys...both the system of equations and the backsolve obviously work. i see your point jalen. its always a better option to see if you dont have to solve the problem. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: m21 #8 [#permalink]

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10 Sep 2010, 05:49
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S:H = 9:8 or S/H= 9/8
8S=9H, S=9/8H--------(1)

S+3/H+1 =6/5, 5S+15 = 6H+6, 5S-6H=-9....(2) SUBSTITUTING S INTO THE equation gives us 5(9/8) -6h=-9.... FROM WHERE H=24 and from equation one we can find S=27.. so the answer is 3 .. Sedan is 3 more cars than hathbacks so the answer is B
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Re: m21 #8 [#permalink]

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10 Sep 2010, 06:10
This is a straight forward question; the answer is reflected in option (C), which
states the current [(S+3) - (h+1)] difference.

"S - h" only gives the previous difference before the additions; so, (B) is out.
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Re: m21 #8 [#permalink]

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10 Sep 2010, 07:08
solved it the long way

possible
X : 9 , 18, 27, 36, 45
Y : 8 , 16, 24, 32, 40

X+3: 12, 21, 30, 39, 48
Y+1: 9, 17, 25

at 30 - 25, the ratio is 6:5 therefore the diff is 5

not a bad method IMO, but will not work with big numbers
can someone provide a link with more questions like these? I would like to practise more with some difficult questions and test a few different ways to figure out the best method to solve these babies : )
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Re: m21 #8 [#permalink]

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10 Sep 2010, 08:06
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. If 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?

(A) 2
(B) 3
(C) 5
(D) 7
(E) 9

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

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.

The original ratio 9:8 gets 1 added to the hatchbacks (and 3 to the sedans), and the ratio becomes 6:5. What multiple of 8 is a multiple of 5 when 1 is added? 24.

If the original number of hatchbacks is 24, then the ratio is scaled by 3: 27:24. Adding 3 sedans and 1 hatchback = 30:25 = 6:5. So now there are 5 more sedans than hatchbacks.

This can be solved with a system of equations, but I believe this method is faster.
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Re: m21 #8 [#permalink]

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10 Sep 2010, 16:31
I went around the long way.

9 sedans : 8 hatchbacks + 3 sedans, 1 hatchback = 12:9 ratio. not 6:5
18 sedans : 16 hatchbacks + 3 sedans, 1 hatchback = 21:17 ratio, not 6:5
27 sedans : 24 hatchbacks + 3 sedans, 1 hatchback = 30:25 ratio, which = 6:5.

Thanks for the explanation J Allen! I'll be sure to use it next time. Need to brush up on quant after my SC cram.
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Re: m21 #8 [#permalink]

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10 Sep 2010, 21:18
Ayo's method is very simple and quick.

Bottom line - get the 2 ratios in the form of equations and substitute!
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Re: m21 #8 [#permalink]

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11 Sep 2010, 00:56
i like alternative solutions. faster. ghee.... i wish i can see that.
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Re: m21 #8 [#permalink]

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11 Sep 2010, 19:04
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If the ratio is 9:8,then each category would be having 9x or 8x cars..
If 3 sedans are added and 1 hatchback is added,the new ratio would become 9x+3 and 8x+1..now this new ratio becomes 6:5..

So,9x+3/8x+1=6/5..
Solve for value of x..Thus x=3..

Now,we can easily find the difference amongst cars by putting the value of x which comes out be 5..

So answers is C..
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Re: m21 #8 [#permalink]

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12 Sep 2010, 23:40
The initial ratios of sedans to hatchbacks is 9:8. Let's 17x be the total number of vehicles in the fleet, the percentage of sedans is given by 9x/17x.

Now we now that when we add 3sedans and 1 hatchback the initial ratios changes to 6:5. This can be written as: (9x+3)/(17x+4)=6/5
-->99x+33=102x+24 -->x=3

Hence the initial numbers of sedans is 27 (9*3) and hatchbacks is 24(8*3). After adding it becomes 30 and 25. Hence answer c.
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Re: m21 #8 [#permalink]

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13 Sep 2010, 02:08
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ratio is 9:8
after adding 3 sedan and 1 hatchbacks we have
(9k+3)/(8k+1)=6/5
k=3
so, 9k+3=30
8k+1=25
difference= (30-25)=5
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Re: m21 #8 [#permalink]

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13 Sep 2010, 07:35
And how do you know k=3? I realize you're just solving for "k", but it might help others if you explain that process.

piku9290dgp wrote:
ratio is 9:8
after adding 3 sedan and 1 hatchbacks we have
(9k+3)/(8k+1)=6/5
k=3
so, 9k+3=30
8k+1=25
difference= (30-25)=5

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Re: m21 #8 [#permalink]

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15 Sep 2010, 03:05
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Since the ratio of sedans to hatchbacks is 9:8, we can take actual number of sedans and hatchbacks as 9x and 8x respectively. Now according to problem statement,

(9x+3)/(8x+1) = 6/5

Since we have one equation and there is only one variable to find, we can solve for x that is equal to 3. But the question asks for the value of (9x+3)-(8x+1), which gives the value of 5 upon substituting the value of x.
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Re: m21 #8 [#permalink]

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14 Sep 2011, 05:24
Sedans with addition 9x+3
Hatchbacks with addition 8x+1
(9x+3)/(8X+1) = 6/5
cross-multiplying => 45x+15=48x+6
9 = 3x
X =3
difference between Sedans and Hatchbacks 30 – 25 = 5
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Re: m21 #8 [#permalink]

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14 Sep 2011, 05:30
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S/H = 9/8

&

S+3 / H+1 =6/5

ON SOLVING ,
H=24, H+1 = 25
S=27, S+3 = 30

DIFFERENCE = 5

easy one
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Re: m21 #8 [#permalink]

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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.
Re: m21 #8   [#permalink] 13 Nov 2011, 22:07

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# m21 #8

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