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Question 2
Three fractions $$x, y$$ and $$z$$ are such that $$x > y > z$$.When small of them divided by the greatest, the result is $$\frac{9}{16}$$, which exceeds $$y$$ by 0.0625.If $$x + y + z = 1 \frac{13}{24}$$,then the value of $$x + z$$ is
Solution
$$\frac{z}{x} = \frac{9}{16}$$
$$\frac{9}{16} = y + 0.00625$$
y = $$\frac{1}{z}$$
x +Â y +Â z = $$1 \frac{13}{24}$$
x + $$\frac{1}{z} +Â z =Â \frac{37}{24}$$
x + z =Â $$\frac{37}{24} - \frac{12}{24} = \frac{25}{14}$$
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