\(\frac{3}{4}\) t + \(\frac{1}{3}\) (21 - t) = 11
Multiply through by the LCM of 4 and 3 which is 12
12 x(\(\frac{3}{4}\) t) + 12 x (\(\frac{1}{3}\) (21 - t)) = (11 x 12)
9t + 4(21 - t) = 132
9t + 84 - 4t = 132
5t + 84 = 132
5t = 132 - 84 = 48
t = \(\frac{48}{5}\)
t = 9 \(\frac{3}{5}\)
Answer is D
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