Question

do with method I Mark you branist

Answer 1

There are 5 questions here. Let me solve them one by one.

7) Let the other number be 'x'.
The product of the two numbers is equal to $\frac{16}{9}$.
Hence, $\frac{5}{2}$ x = $\frac{16}{9}$
Cross multiplying, we get 45x = 32
x = $\frac{32}{45}$

Therefore, the other number is $\frac{32}{45}$

8) Let each piece be of length 'x'.
Then, 6x = 3 $\frac{3}{5}$
Converting into an improper fraction,
6x = $\frac{18}{5}$

x = $\frac{18}{5 * 6}$

x = $\frac{3}{5}$

Therefore, each piece will be of length $\frac{3}{5}$ m.

9) Let 'n' be the number of notebooks that can be purchased.
Then, n * 6 $\frac{1}{3}$ = 126 $\frac{2}{3}$
Converting to improper fractions,
$\frac{19}{3}$ * n = $\frac{380}{3}$
Canceling out 3 in both the denominators,
19n = 380
n = $\frac{380}{19}$
n = 20

Therefore, 20 notebooks can be purchased.

10) Let the number be 'x'.
So, 10 $\frac{2}{3}$ ÷ x = 20

Converting to improper fraction, 
$\frac{32}{3}$  ÷ x = 20
$\frac{32}{3x}$  = 20
32 = 60x
x = $\frac{32}{60}$
x = $\frac{8}{15}$

Therefore, the number is $\frac{8}{15}$

11) Let the number be 'x'.
We have $\frac{2}{3}$ * x = $\frac{5}{4}$
x = $\frac{5}{4}$ *  $\frac{3}{2}$ 
x = $\frac{15}{8}$ or 1 $\frac{7}{8}$

Therefore, 1  $\frac{7}{8}$ is the required number.