The slaughterhouse brought a number of sheep at sh. 1200 each and a number of oxen at sh.15, 000 each.

Question

They paid a total of sh.135, 000. If they had brought twice as many sheep and three oxen less. They
would have saved sh. 15000.
a) Write two simultaneous equations presently the information.
b) Find the number of each type of animal they brought
The slaughterhouse sold the entire animal at a profit of 30% per sheep and 35% per oxen.
c) Determine the total profit they made.

Answer 1

Let's represent the number of sheep as S and the number of oxen as O.

a) The two simultaneous equations are:
1) 1200S + 15000O = 135000 (equation 1)
2) 1200(2S) + 15000(O-3) = 135000 + 15000 (equation 2)

b) Solving equation 2, we can simplify it to:
2400S + 15000O - 36000 = 150000

Now we can substitute this equation into equation 1 to solve for S:
1200S + 15000O = 135000
2400S + 15000O - 36000 = 150000

Subtracting equation 1 from equation 2, we get:
1200S - 36000 = 150000 - 135000
1200S - 36000 = 15000
1200S = 51000
S = 42

Substituting S = 42 into equation 1, we can solve for O:
1200(42) + 15000O = 135000
50400 + 15000O = 135000
15000O = 84600
O = 5.64 (approx.)

Therefore, the slaughterhouse brought 42 sheep and 6 oxen.

c) The total profit made can be calculated by finding the profit from selling sheep and oxen separately and then adding them together.

Profit from selling sheep:
Total cost of sheep = 42 * 1200 = 50400
Profit from selling sheep = 50400 * 0.30 = 15120

Profit from selling oxen:
Total cost of oxen = 6 * 15000 = 90000
Profit from selling oxen = 90000 * 0.35 = 31500

Total profit = Profit from selling sheep + Profit from selling oxen = 15120 + 31500 = 46620

Therefore, the slaughterhouse made a total profit of sh. 46,620.