You are making 10 tons of concrete that is 47.6% cement by mixing a 20% cement mixture with an 89% cement mixture.a) How much of the 89% cement mixture must you use?b) How much of the 20% cement mixture must you use?Answer in units of tons
m+c=10 .20m+.89c=4.76 6 tons of 89% cement mixture 4 tons of 20% cement mixture
Let m represent the number of tons of the 20% cement mix Let n represent the number of tons of the 89% cement mix Total batch: m + n = 10 Cement amount: 0.20m + 0.89n = 10×0.476 Solve for m and n: m = 6 and n = 4 Check this result in the original equation(s), I did! The final batch should contain 6 tons of 20% cement mix with 4 tons of 89% cement mix to have 10 tons of the 47.6% cement mix. ------------------------------------- What other units are there?
a) Let the amount of 89% cement mixture used be X tons Then the amount of 20% cement mixture used will be ( 10 - X) tons 0.89X + 0.20 ( 10 - X) = 10* 0.476 0.89X + 2 - 0.20X = 4.76 0.69X = 2.76 X = 2.76 / 0.69 = 4 ANSWER 4 tons of 89% mixture and 6 tons of 20% mixture
