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Question 1069914: there were 200 dogs in an compound, 99% of which were brown, how many brown dogs must be removed to make the percentage of brown dogs 98%??
Found 3 solutions by josgarithmetic, Theo, ikleyn:
Answer by josgarithmetic(39630)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! there are 200 dogs in the compound.
99% of them are brown.
that means there are 198 brown dogs and 2 dogs that are not brown.
let x equal the number of brown dogs that are removed.
if you remove x brown dogs, then you are also removing x dogs from the total.
i believe your formula should be:
198 - x = .98 * (200 - x).
simplify this to get 198 - x = 196 - .98 * x
subtract 196 from both sides of this equation and add x to both sides of this equation to get:
2 = .02 * x
divide both sides of this equation by .02 to get:
2/.02 = x
solve for x to get x 2/.02 = 200/2 = 100.
you would have to remove 100 dogs to make the percent equal to 98.
198 - 100 = 98.
200 - 100 = 100
98/100 = .98 * 100 = 98%.
as strange as it sounds, you would have to remove 100 brown dogs to make the percentage of brown dogs equal to 98%.
this is because, every time you remove a brown dog, the total number of dogs is reduced by 1.
to see if this makes sense, try some numbers less than 100.
if you remove 2 brown dogs, then you have 196 out of a total of 198.
the percentage is 98.9.....
if you remove 50 brown dogs, then you have 148 out of a total of 150.
the percentage is 98.67.
if you remove 100 brown dogs, then you have 98 / 100 = 98%.
note that the number of dogs that are not brown never changes.
you still have 2 dogs that are not brown.
2 / 100 = .02 = 2%.
this means that the percent of brown dogs is 98%.
Answer by ikleyn(52914)  (Show Source):
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