\(\frac{1}{3}\)
1\(\frac{1}{2}\)
1\(\frac{1}{6}\)
\(\frac{1}{2}\)
Correct answer is B
The sum of 2 \(\frac{1}{6}\) and 2\(\frac{7}{12}\)
= \(\frac{13}{6}\) + \(\frac{31}{12}\)
= \(\frac{13 \times 2 + 31}{12}\)
= \(\frac{26 + 31}{12}\)
= \(\frac{57}{12}\)
What should be subtracted from \(\frac{57}{12}\) to give 3\(\frac{1}{4}\)
\(\frac{57}{12}\) - y = 3\(\frac{1}{4}\)
: y = \(\frac{57}{12}\) - 3\(\frac{1}{4}\) = \(\frac{57}{12}\) - \(\frac{13}{4}\)
y = \(\frac{57 - 3 \times 13}{12}\) = \(\frac{57 - 39}{12}\)
y = \(\frac{18}{12}\)
y = \(\frac{3}{2}\) or 1\(\frac{1}{2}\)
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