Question 109328: Hi sir, i am really confused with the sum and need immediate help .
Q.A rabbit costs $3.50 , ducklings are sold at 3 for $1 and chicks cost 50 cents each . A farmer paid $100 exactly for 100 of these animals . How many of each did he buy ? (There are 5 possible answers.)
I am getting (20,60,20)
Sir pls help me . THANK YOU
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Let x=number of rabbits
y=number of ducklings
z=number of chicks
Now we are told that:
x+y+z=100------------------------------eq1
We are also told that:
3.50x+(1/3)y+(1/2)z=100------------------eq2
First, multiply eq1 by 1/2 and then subtract it from eq2. This will give us a relationship between the rabbits and ducklings:
(1/2)x+(1/2)y+(1/2)z=50 subtracting this from eq2, we get
3x-(1/6)y=50 Now, multiplying each term by 6, we get:
18x-y=300 -----eq2
Since one equation with two unknowns is the best we can do, it's time now for some educated trial and error:
First, we know that we are dealing in whole, positive numbers. We cannot have fractions of animals or negative animals.
Second, eq2a tells us that 18 times the number of rabbits minus the number of ducks equals three hundred. So clearly, 18 times the number of rabbits has to be over 300 and that means that x>or=to 17. Also, we cannot have more than $100 worth of rabbits and that means that x < or=28. Now we know that:
17 < or=x < or=28.
Now, between these limits, we need to start looking at what works:
If x=17, then y=6 substituting into eq1, we see that z=77--------------BINGO!
If x=18, then y=24 substituting into eq1, we see that z=58-------------BINGO!
If x=19, then y=42 substituting into eq1, we see that z=39-------------BINGO!
If x=20, then y=60 substituting into eq1, we see that z=20-------------BINGO!
(YOUR ANSWER!!!!!!!!!!!!!!!!)
If x=21, then y=78 substituting into eq1, we see that z=1---------------BINGO!
If x=22, then y=96-------NOW WE HAVE BROKEN THE BANK IN TERMS OF NUMBERS OF ANIMALS.
So the answers are:
x=17, y=6, z=77
x=18, y=24, z=58
x=19, y=42, z=39
x=20, y=60, z=20---------------------your answer!!
x=21, y=78, z=1
Hope this helps-----ptaylor
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