Question

Refer to the following diagram determine the length by which the wingspan of Albatross is greater than that of Sea Gull.

Golden eagle Albatross Sea Gull \( \longleftarrow 1 \frac{7}{10} m ^{\longrightarrow} \longrightarrow \)

Answer 1

The length by which the wingspan of Albatross is greater than that of Sea Gull.

The wingspan of Albatross is given as $3\frac{3}{5}$ m.

Convert the given length $3\frac{3}{5}$ in mixed fraction to normal fraction, that is, $3\frac{3}{5}=\frac{18}{5}$.

The wingspan of Sea Gull is given as $1\frac{7}{10}$m.

Again convert the mixed fraction to normal fraction, that is, $1\frac{7}{10}=\frac{17}{10}$.

The length that the wingspan of Albatross is greater than wingspan of Sea Gull given by finding the difference of the fraction $\frac{17}{10}$ from $\frac{18}{5}$.

Thus the required length is $\frac{18}{5}-\frac{17}{10}$.

Step 2: Calculate the subtraction of the fraction, $\frac{18}{5}-\frac{17}{10}$.

To subtract the fraction make the denominator of both the fraction same, thus, divide and multiply the first fraction $\frac{18}{5}$ by $2$, this gives, $\frac{36}{10}$.

Now, subtract $\frac{17}{10}$from $\frac{36}{10}$.as follows:

$$\begin{gathered} \frac{36}{10}-\frac{17}{10}=\frac{36-17}{10} \\ =\frac{19}{10} \end{gathered}$$

Thus, the subtraction is $\frac{19}{10}$or in terms of mixed fraction is $1\frac{9}{10}$.

Hence, the length by which the wingspan of Albatross is greater than that of Sea Gull is $1\frac{9}{10}$m.