Question 1121497
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Let M be the mass of the coal before drying.


Since it contains 2.4% of water, after drying the mass will be  M - 0.024*M.


Let C be the mass of carbon in this coal.


Then the condition says  


    {{{C/(M - 0.024*M)}}} = 0.71.


Therefore,


    C = 0.71*(M-0.024*M) = 0.71*M*(1-0.024),   or


    {{{C/M}}} = 0.71*(1-0.024) = 0.693 (approximately),   or  69.3%.


<U>Answer</U>.  Before drying, the percentage of carbon in the coal is  69.3%, approximately (rounded to one decimal place).
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