Math, asked by sparksha6485, 1 year ago

A container has a mixture of kerosene and water in ratio of 7:5. when 9 litres of ixture are taken off and the container is filled with 9 litres of water, the ratio between kerosene and water becomes 7:9. how many litres of kerosene were iniially in the container

Answers

Answered by Anonymous
2
Ratio of Kerosene : water = 7:5Let the quantity of Kerosene = 7x && Water = 5xTotal = 12x
Now, 9 L of the mixture was removed:Hence, Kerosene = 7/12 * (12x - 9)Water = 5/12 * (12x -9)
Now, 9 L of Water was addedWater = 5/12 *(12x -9 ) + 9The new ratio (kero:Water) = 7:9[7/12*(12x-9)]/[5/12 *(12x -9 ) + 9] = 7/9[(84x -  63)/12] / [(60x - 45 + 9*12)/12] = 7/9[84x - 63]/[60x + 63] = 7/99*[84x - 63] = 7*[60x + 63]756x - 567 = 420x + 441756x - 420x = 441 + 567336x = 1008x = 3
Initial quantity of kerosene = 7x = 7*3 = 21 Litres
Hope it helps.
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