SOLUTION: Of the animal on a farm 60% are cows and the rest are sheep. When 260 more cows and sheep are added to the farm, the percent of cows increases by 20% and the number of sheep double

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Of the animal on a farm 60% are cows and the rest are sheep. When 260 more cows and sheep are added to the farm, the percent of cows increases by 20% and the number of sheep double      Log On


   

Question 824918: Of the animal on a farm 60% are cows and the rest are sheep. When 260 more cows and sheep are added to the farm, the percent of cows increases by 20% and the number of sheep doubles. Find the number of sheep originally on the farm.
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Of the animal on a farm 60% are cows and the rest are sheep. When 260 more cows and sheep are added to the farm, the percent of cows increases by 20% and the number of sheep doubles. Find the number of sheep originally on the farm.
Let x = the number of animals originally on farm

Of the animal on a farm 60% are cows...
Then, since 60% = 0.6
0.6x = the number of cows originally on the farm

and the rest are sheep.
so the other 100%-60%=40% are sheep, and since 40% = 0.4

0.4x = the number of sheep originally on the farm

When...more cows and sheep are added to the farm...the number of sheep doubles
So the same number of sheep were added as were originally on the farm,
therefore,

0.4x = the number of sheep added

260 more cows and sheep are added to the farm
cows added = 260-sheep added = 260-0.4x

x+260 = the new number of animals on the farm

the percent of cows increases by 20%

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The word "percent" here should be "number".  For if you take what
is written literally, it would mean that after the 260 animals are 
added, then 60%+20%=80% of the animals are cows.  That interpretation 
will not work!  For if that were the case, then

   (original number of cows) + (cows added) = 80% of (new number of animals)    
                          0.6x + (260-0.4x) = 0.8(x+20)
                              0.6x+260-0.4x = 0.8x+16
                                   0.2x+260 = 0.8x+16
                                        244 = 0.6x
                                     406.67 = x
But the number of animals originally on the farm cannot be a fraction.
So I will change the wording to the only thing that could be meant:
----------------------------------------------------

the NUMBER of cows increases by 20% (not "the percent of cows increases by 20%"!)

             %22%22%2B%22%22%28matrix%285%2C1%2C%0D%0A+++++++++++++the%2Cnumber%2Cof%2Ccows%2Cadded%29%29 %22%22=%22%22 %22%22%2B%22%22 

            0.6x + (260-0.4x) = 0.6x + 0.2(0.6x)
            0.6x+260-0.4x = 0.6x+0.12x
                 0.2x+260 = 0.6x+0.12x
                 0.2x+260 = 0.72x
                      260 = 0.52x
                     260%2F0.52 = x
                      500 = x

So the farm originally contained 500 animals
0.6x = the number of cows originally on the farm = 0.6(500) = 300 cows
0.4x = the number of sheep originally on the farm = 0.4(500) = 200 sheep
0.4x = the number of sheep added = 200 sheep added
260-0.4x = the number of cows added = 260-200 = 60 cows added
x+260 = the new number of animals on the farm = 500+260 = 760
The new number of cows = 300+60 = 260

See how nicely it comes out when the word "percent" is changed to "number".
The new percent of cows is 34%262%2F19%, not 80%!

Answer: 200 sheep originally.

[You might point out this error in wording to your teacher.]

Edwin