Problem 5: If some soy sauce is used, the remainder is $\frac{3}{8}$ L. This amount is $\frac{1}{6}$ of the initial amount of soy sauce. What was the initial amount of soy sauce in liters? Problem 6: A rectangle has an area of $6\frac{3}{7} cm^2$ and a width of $3\frac{8}{9} cm$. What is the length of the rectangle in cm?

1. Problem Description

Problem 5: If some soy sauce is used, the remainder is $\frac{3}{8}$ L. This amount is $\frac{1}{6}$ of the initial amount of soy sauce. What was the initial amount of soy sauce in liters?

Problem 6: A rectangle has an area of $6\frac{3}{7} cm^2$ and a width of $3\frac{8}{9} cm$. What is the length of the rectangle in cm?

2. Solution Steps

Problem 5:

Let $x$ be the initial amount of soy sauce in liters.

The remainder is $\frac{3}{8}$ L, which is $\frac{1}{6}$ of the initial amount.

So, we have the equation:

$\frac{1}{6}x = \frac{3}{8}$

Multiply both sides by 6:

$x = \frac{3}{8} \times 6$

$x = \frac{3 \times 6}{8}$

$x = \frac{18}{8}$

$x = \frac{9}{4}$

$x = 2\frac{1}{4}$

Problem 6:

Let $A$ be the area of the rectangle, $l$ be the length, and $w$ be the width.

The area of a rectangle is given by the formula:

$A = l \times w$

Given that the area is $6\frac{3}{7} cm^2$ and the width is $3\frac{8}{9} cm$.

$A = 6\frac{3}{7} = \frac{6 \times 7 + 3}{7} = \frac{42 + 3}{7} = \frac{45}{7}$

$w = 3\frac{8}{9} = \frac{3 \times 9 + 8}{9} = \frac{27 + 8}{9} = \frac{35}{9}$

We have the equation:

$\frac{45}{7} = l \times \frac{35}{9}$

$l = \frac{45}{7} \div \frac{35}{9}$

$l = \frac{45}{7} \times \frac{9}{35}$

$l = \frac{45 \times 9}{7 \times 35}$

$l = \frac{405}{245}$

$l = \frac{81}{49}$

$l = 1\frac{32}{49}$

3. Final Answer

Problem 5: $2\frac{1}{4}$ L

Problem 6: $1\frac{32}{49}$ cm