Problem Solving

BTGmoderatorDC wrote: ↑

Fri Feb 24, 2023 8:50 pm

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


OA D

Source: Princeton Review

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

BTGmoderatorDC wrote: ↑

Fri Feb 24, 2023 8:50 pm

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


OA D

Source: Princeton Review

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: D