On a certain farm the ratio of horses to cows is 7:3. If the

Algebraic way:

Old ratio: horses/cows = 7x/3x.
New ratio: horses/cows = (7x-15)/(3x+15)=13/7 --> x=30.

New difference: (7x-15)-(3x+15)=4x-30=90.

Answer: D.

Hope it helps.
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We can also solve this using TWO VARIABLES

Let H = present number of horses
Let C = present number of cows

ratio of horses to cows is 7:3
So, H/C = 7/3
Cross multiply to get 3H = 7C
Rearrange to get: 3H - 7C = 0

If the farm were to sell 15 horses and buy 15 cows, the ratio of horses to cows would then be 13:7
So, (H-15)/(C+15) = 13/7
Cross multiply to get 7(H-15) = 13(C+15)
Expand: 7H - 105 = 13C + 195
Rearrange to get: 7H - 13C = 300

We now have a system of two equations and two variables:
3H - 7C = 0
7H - 13C = 300

Solve to get: H = 210 and C = 90
NOTE: These are the values BEFORE livestock was sold and bought.

After selling 15 horses, there are 195 horses
After buying 15 cows, there are 105 cows

After the transaction, how many more horses than cows would the farm own?
195 - 105 = 90

Answer:

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That is great, thanks - make it look so easy. It probably is... I was using the Veritas way of 7x-15 = 13y and 3x+15 = 7y but for some reason that did not work (after substituting), I guess it just me...

Hi
even if you are using the Veritas way
7x-15= 13 y
and 3x+15=7y

solving the equations as
7(3x+15) - 3( 7x-15) = 7*7y - 3*13y
=> 150 = 10y
y=15

now the difference between no of horses and cows after transaction is 13 y- 7y = 6y= 6*15 = 90

Answer: D

Kudos if you find this post helpful

what about a non-algebraic way? )
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There are almost always lots of non-algebraic ways to solve a GMAT question. Though in this question, algebraic way in quite straight forward so you may want to stick to it. If you can clearly see an equation in a single variable, it may be wise to use it. Usually, questions involving addition/subtraction of constants (add 15 cows, subtract 15 horses) are best done using algebra.

Anyway, other methods are easy too.

On a certain farm the ratio of horses to cows is 7:3. If the farm were to sell 15 horses and buy 15 cows, the ratio of horses to cows would then be 13:7. After the transaction, how many more horses than cows would the farm own?

A. 30
B. 60
C. 75
D. 90
E. 105

We want the difference between number of horses and cows after the transaction. We know that the ratio of horses:cows = 13:7. The difference between the numbers will be a multiple of 6 (= 13 - 7).
Since options (C) and (E) are not divisible by 6, ignore them.

Try option (B). If diff is 60, number of horses and cows is 130 and 70. Initial number must have been 145 and 55 which is not in the ratio 7:3.
Similarly, try other options.

Another method using integral solutions to equations involving two variables:

Consider horses:
You know that 7a - 15 = 13b
7a - 13b = 15

First solution a = 4, b = 1 (by hit and trial). This doesn't work since if initial no. of horses is 7*4 = 28 and cows is 3*4 = 12, when you take away 15 horses, you get 13 horses and when you add 15 cows, you get 27 cows. The new ratio is not 13:7.

Next solution will be a = 4+13 = 17, (and b = 1+7 = 8)

If initial number of horses = 7*17 = 119 and initial number of cows is 3*17 = 51, new no of horses = 104 and new number of cows = 66. The ratio is again not 13:7

Next solution will be a = 4+13+13 = 30
If initial number of horses = 7*30 = 210 and initial number of cows is 3*30 = 90, new no of horses = 195 and new number of cows = 105. The ratio is 13:7. So these must be the right numbers. The diff between number of horses and cows = 195 - 105 = 90

Alternatively, notice that you can make the other equation (using cows) as 3a + 15 = 7b.
You will solve the two simultaneously to get the answer but this is just your algebra approach!
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H/C = (7x-15)/(3x+15) = 13/7

Cross multiply.

7(7x-15)=13(3x+15)
49x-105=39x+195
10x=300
x=30

H/C = 195/105
195-105 = 90

D

original horse/total ratio=7/10➡14/20
changed horse/total ratio=13/20
15 horses=14/20-13/20=1/20 of total
20*15=300 total
13/20*300=195 horses
7/20*300=105 cows
195-105=90
D

After we get x=30

Multiplied 13*30=390 number of horses
Multiplied 7*30=210 number of cows
390-210=180 difference

What is wrong in what I did here?

Stuck at the same place.

If x is the multiplier, can't we just multiply both the 13 and the 7 in 13/7 (new ratio) and then subtract the difference to find how many more horses there are?

calappa1234 and Zoser - the multiplier x = 30 pertains to the original ratio.

So plug x = 30 and you get

original number of horses is (7)(30) = 210

original number of cows is (3)(30) = 90

15 horses were sold, 15 cows were purchased, so in the end,

H = (210 - 15) = 195
C = (90 + 15) = 105

195 - 105 = 90

In general, when given a ratio to which things are added or from which things are subtracted, and the ratio plus arithmetic operations equal a new ratio, the multiplier is for the original ratio. (See gracie , above, for a slightly shorter way).

The multiplier for the second ratio, 13/7, is x = 15.

Hope that helps.
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We are given the ratio of horses to cows is 7 to 3, or 7x to 3x. We can create the following equation:

(7x - 15)/(3x + 15) = 13/7

7(7x - 15) = 13(3x + 15)

49x - 105 = 39x + 195

10x = 300

x = 30

Thus, there are 7(30) = 210 horses and 3(30) = 90 cows in the farm before the transaction. After the transaction, there are 210 - 15 = 195 horses and 90 + 15 = 105 cows. So, there are 195 - 105 = 90 more horses than cows after the transaction.

Answer: D
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This is how i solved .

Let the total no of animals be x originally

Old ratio: horses/cows = 7x/3x.
since farm sold 15 horses and brought 15 cows after which we get the new ratio therfore
New ratio: horses/cows = (7x-15)/(3x+15)=13/7 cross multiply to get --> x=30.

the original no of horses = 7*30 = 210 and cows = 3*30 = 90

now reduce 15 horses and add 15 cows to the old no = 210 - 15 = 195 an 90+15 = 105

Now reduce 105 from 195 to get the ans = 90 ans option D

On a certain farm the ratio of horses to cows is 7:3. If the farm were to sell 15 horses and buy 15 cows, the ratio of horses to cows would then be 13:7. After the transaction, how many more horses than cows would the farm own?

A. 30
B. 60
C. 75
D. 90
E. 105

My Approach:
Final Ratio of Horse and Cow, H:C = 13:7 = 13:7
Initial Ratio of Horse and Cow, H:C = 7:3 = 14:6
(As the sum of final ratio is (13+7)= 20, to equal the initial sum to the final sum, initial sum (7+3)= 10 is multiplied by 2)

Now, 7-6 = 15
So, 20 = 20*15 = 300 (This 300 is the total number of horses and cows)
After the transaction, Horses= (13*300)/20 = 13*15 = 195
and Cows = (7*300)/20 = 7*15 = 105

After the transaction, Horses more than Cows = 195-105 = 90

Answer: D. 90

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