$$3\frac{6}{17}\div 2\frac{7}{34}-1\frac{9}{25}=(?)^2$$

Solution

Let the missing number be y

$$3\frac{6}{17}\div 2\frac{7}{34}-1\frac{9}{25}=(y)^2$$

$$3\frac{6}{17}$$ = $$\frac{57}{17}$$

$$2\frac{7}{34}$$ = $$\frac{75}{34}$$

$$1\frac{9}{25}$$ = $$\frac{34}{25}$$

using BODMAS rule

$$\frac{57}{17} \div \frac{75}{34}$$ - $$\frac{34}{25}$$

$$\frac{38}{25}$$ - $$\frac{34}{25}$$

= $$\frac{4}{25} = (y)^2$$

y = $$\frac{2}{5}$$

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