SOLUTION: If 1/4 of a mixture rice and corn was replaced by corn alone, the resulting mixture becomes 55% corn. What was the percentage of corn in the original mixture?

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Question 1097515: If 1/4 of a mixture rice and corn was replaced by corn alone, the resulting mixture becomes 55% corn. What was the percentage of corn in the original mixture?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
100 units of corn at p percent corn

Remove 25 units of original mixture and replace with 25 units of 100% corn.
%2875p%2B25%2A100%29%2F100=55
-
75p%2B25%2A100=55%2A100
75p=55%2A100-25%2A100
75p=30%2A100
p=%2830%2A100%29%2F%283%2A25%29
highlight%28p=40%29--------------40% corn, the original mixture

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let "M" be the the total amount/(mass) of the original mixture,

and let "x" be the fraction  of corn in the original mixture (the unknown under the question).


Then the amount of corn after replacing is  (0.75*xM + 0.25*M),
and the equation for the given corn percentage is


%280.75%2AxM+%2B+0.25%2AM%29%2FM = 0.55.


Cancel the factor M in the numerator and denominator. You will get


0.75x + 0.25 = 0.55  ====>  0.75x = 0.55 - 0.25 = 0.3  ====>  x = 0.3%2F0.75 = 2%2F5 = 0.4.


Answer.  The percentage of corn in the original mixture was 40%.