Question 269374: A farmer spends $4,000 to obtain 100 head of livestock.Prices are: calves – $120 each; lambs – $50 each; piglets – $25 each. If he purchased at least one animal of each type and fewer than 10 calves, how many lambs did he buy?
(1) 44 (2) 46 (3) 48 (4) 50 (5) None of the above
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Let x=number of calves
Let y=number of lambs
And let z=number of piglets
Now we are told that:
x+y+z=100---------------------eq1
120x+50y+25z=4000
Divide each term by 5 just to deal in smaller numbers
24x+10y+5z=800----------eq2
We have only two equations but three unknowns. Hmmmmmmm.
BUT WE CAN STILL SOLVE IT!!
Multiply eq1 by 5 and then subtract it from eq2 and we get:
19x+5y=300----------eq1a
solve for y
y=(300-19x)/5---eq1b Now we have an equation that relates the number of lambs to the number of calves
We know several things about this problem that will help us solve it:
(1) We are told that the number of calves,(x),is less than 10
(2) We know that we can't have a half a calf, a leg of lamb or a piece of pork. We must deal in whole numbers.
(3) We can't have negative calves, lambs or piglets. We must deal in whole, positive numbers.
We will now solve eq1b by trial and error, starting with x=9.
x=9 y=(300-171)/5--------------no good not a whole number
x=8 y=(300-152)/5--------------no good not a whole number
x=7 y=(300-133)/5--------------no good not a whole number
x=6 y=(300-114)/5--------------ditto
x=5--------y=(300--95)/5=41-----------BINGO ---POSSIBLE SOLUTION
BY INSPECTION WE CAN SEE THAT WHEN x IS 4, 3, 2 OR 1. WE DO NOT GET ANY MORE WHOLE NUMBERS FOR Y
Now we have the following:
x=Calves=5
y=Lambs=41 ----so your answer is "none of the above"
Substitute these values into eq1 and we get
5+41+z=100
z=Piglets=54
Now
5*120+41*50+54*25=4000
600+2050+1350
4000=4000
Hope this helps---ptaylor
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