Question : Three fractions $x, y$ and $z$ are such that $x > y > z$. When the smallest of them is divided by the greatest, the result is $\frac{9}{16}$, which exceeds $y$ by 0.0625. If $x+y+z=2 \frac{3}{12}$, then what is the value of $x + z$?
Option 1: $\frac{5}{4}$
Option 2: $\frac{3}{4}$
Option 3: $\frac{7}{4}$
Option 4: $\frac{1}{4}$
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Correct Answer: $\frac{7}{4}$
Solution :
Given: $x > y > z$
According to the question,
$\frac{z}{x} = \frac{9}{16}$
⇒ $y = \frac{9}{16}-0.0625=\frac{9}{16}-\frac{1}{16} = \frac{1}{2}$
⇒ $x+y+z=2 \frac{3}{12}=\frac{27}{12}$
⇒ $x + \frac{1}{2} + z = \frac{27}{12}$
$\therefore x + z = \frac{27}{12} - \frac{1}{2}= \frac{27-6}{12}= \frac{21}{12}= \frac{7}{4}$
Hence, the correct answer is $\frac{7}{4}$.
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