The problem consists of two parts. Part 1: Evaluate $\frac{1}{4} \div \frac{1}{2}$ Part 2: Evaluate $\frac{5\frac{2}{7} + \frac{1}{14} \times \frac{2}{3} - 1\frac{1}{4}}{\frac{3}{8} \div \frac{1}{16}}$

1. Problem Description

The problem consists of two parts.

Part 1: Evaluate $\frac{1}{4} \div \frac{1}{2}$

Part 2: Evaluate $\frac{5\frac{2}{7} + \frac{1}{14} \times \frac{2}{3} - 1\frac{1}{4}}{\frac{3}{8} \div \frac{1}{16}}$

2. Solution Steps

Part 1:

To divide fractions, we multiply by the reciprocal of the second fraction.

$\frac{1}{4} \div \frac{1}{2} = \frac{1}{4} \times \frac{2}{1} = \frac{1 \times 2}{4 \times 1} = \frac{2}{4} = \frac{1}{2}$

Part 2:

First, we convert mixed numbers to improper fractions.

$5\frac{2}{7} = \frac{5 \times 7 + 2}{7} = \frac{35+2}{7} = \frac{37}{7}$

$1\frac{1}{4} = \frac{1 \times 4 + 1}{4} = \frac{4+1}{4} = \frac{5}{4}$

Now, we evaluate the expression $\frac{5\frac{2}{7} + \frac{1}{14} \times \frac{2}{3} - 1\frac{1}{4}}{\frac{3}{8} \div \frac{1}{16}}$.

Numerator:

$\frac{37}{7} + \frac{1}{14} \times \frac{2}{3} - \frac{5}{4}$

First, multiply $\frac{1}{14} \times \frac{2}{3}$:

$\frac{1}{14} \times \frac{2}{3} = \frac{1 \times 2}{14 \times 3} = \frac{2}{42} = \frac{1}{21}$

Now, the numerator is:

$\frac{37}{7} + \frac{1}{21} - \frac{5}{4}$

The least common multiple of 7, 21, and 4 is

8

4. $\frac{37}{7} = \frac{37 \times 12}{7 \times 12} = \frac{444}{84}$

$\frac{1}{21} = \frac{1 \times 4}{21 \times 4} = \frac{4}{84}$

$\frac{5}{4} = \frac{5 \times 21}{4 \times 21} = \frac{105}{84}$

So, the numerator becomes:

$\frac{444}{84} + \frac{4}{84} - \frac{105}{84} = \frac{444+4-105}{84} = \frac{448-105}{84} = \frac{343}{84}$

Denominator:

$\frac{3}{8} \div \frac{1}{16} = \frac{3}{8} \times \frac{16}{1} = \frac{3 \times 16}{8 \times 1} = \frac{48}{8} = 6$

The expression now becomes:

$\frac{\frac{343}{84}}{6} = \frac{343}{84} \div 6 = \frac{343}{84} \times \frac{1}{6} = \frac{343}{84 \times 6} = \frac{343}{504}$

Since $343 = 7^3$ and $504 = 2^3 \times 3^2 \times 7$, we can simplify the fraction.

$\frac{343}{504} = \frac{7^3}{2^3 \times 3^2 \times 7} = \frac{7^2}{2^3 \times 3^2} = \frac{49}{8 \times 9} = \frac{49}{72}$

3. Final Answer

Part 1: $\frac{1}{2}$

Part 2: $\frac{49}{72}$