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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?

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.

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'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.

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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.

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$$.

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

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 : )

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?

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.

Answer: C) 5

This can be solved with a system of equations, but I believe this method is faster.

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..

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.

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. _________________