In order to solve this problem, identify that the common portion of this Venn diagram represents Kayla's friends who play sports and video games. Since \(\displaystyle \frac{3}{10}\) of her friends play sports and video games, convert this fraction to a percent.

The solution is:

\(\displaystyle \frac{3}{10}=\frac{3\times 10}{10\times 10}=\frac{30}{100}=30\%\)


Note: the most efficient way to convert this fraction to a percent is to find an equivalent fraction to \(\displaystyle \frac{3}{10}\) with a denominator of \(\displaystyle 100\). 

To find the missing quantity for category \(\displaystyle y\), first calculate the sum from the common portion of the Venn diagram and category \(\displaystyle x\). Then, subtract that sum from \(\displaystyle 100\%\) , because the total percentage of respondents must equal \(\displaystyle 100\%\). 

The algebraic solution is:

\(\displaystyle 39 +12=51\)

\(\displaystyle 100-51=49\)

A ratio can be rewritten as a quotient; do this, and simplify it.

\(\displaystyle 10 \frac{1}{2} : 1 \frac{1}{2}\)

Rewrite as

 \(\displaystyle 10 \frac{1}{2} \div 1 \frac{1}{2} = \frac{21}{2} \div \frac{3}{2} = \frac{21}{2} \cdot \frac{2}{3} = \frac{7}{1} \cdot \frac{1}{1} = \frac{7}{1}\)

or \(\displaystyle 7:1\)

The ratio of wins to losses requires knowing the number of wins and losses.  The question says that there are 5 wins.  That means there must have been

\(\displaystyle 20-5=15\) losses. 

The ratio of wins to losses is thus 5 to 15 or 1 to 3.

Rewrite in fraction form for the sake of simplicity, then divide each number by \(\displaystyle GCF (90,63) = 9\):

\(\displaystyle \frac{90}{63} = \frac{90\div 9}{63\div 9} = \frac{10}{7}\)

The ratio, in simplest form, is \(\displaystyle 10:7\)

A ratio involving fractions can be simplified by rewriting it as a complex fraction, and simplifying it by division:

\(\displaystyle \frac{\frac{2}{3}}{\frac{5}{12}} = \frac{2}{3} \div \frac{5}{12}\)

Write as a product by taking the reciprocal of the divisor, cross-cancel, then multiply it out:

\(\displaystyle \frac{2}{3} \div \frac{5}{12} = \frac{2}{3} \times \frac{12}{5} = \frac{2}{1} \times \frac{4}{5} =\frac{8}{5}\)

The ratio simplifies to \(\displaystyle 8:5\)

A ratio involving fractions can be simplified by rewriting it as a complex fraction, and simplifying it by division:

\(\displaystyle \frac{\frac{4}{5}}{\frac{3}{10}} = \frac{4}{5} \div \frac{3}{10}\)

Write as a product by taking the reciprocal of the divisor, cross-cancel, then multiply it out:

\(\displaystyle \frac{4}{5} \div \frac{3}{10} = \frac{4}{5} \times \frac{10}{3} = \frac{4}{1} \times \frac{2}{3} =\frac{8}{3}\)

The ratio simplifies to \(\displaystyle 8:3\)

Rewrite in fraction form for the sake of simplicity, then divide each number by \(\displaystyle GCF (72,42) = 6\):

\(\displaystyle \frac{72}{42}= \frac{72\div 6}{42\div 6}= \frac{12}{7}\)

In simplest form, the ratio is \(\displaystyle 12:7\)

Since the answer to this question does not depend on the actual lengths of the sides, we will assume for simplicity that the larger square has sidelength 5; if this is the case, the smaller square has sidelength 3. The areas of the large and small squares are, respectively, \(\displaystyle 5^{2} = 25\) and \(\displaystyle 3^{2} = 9\). 

The white region is the small square and has area 9. The grey region is the small square cut out of the large square and has area \(\displaystyle 25-9=16\). Therefore, the ratio of the area of the gray region to that of the white region is 16 to 9.

Since the answer to this question does not depend on the actual lengths of the sides, we will assume for simplicity that the larger square has sidelength \(\displaystyle 4\); if this is the case, the smaller square has sidelength \(\displaystyle 3\). The areas of the large and small squares are, respectively, \(\displaystyle 4^{2} = 16\) and \(\displaystyle 3^{2} = 9\). 

The white region is the small square and has area \(\displaystyle 9\). The grey region is the small square cut out of the large square and has area \(\displaystyle 16-9=7\). Therefore, the ratio of the area of the gray region to that of the white region is \(\displaystyle 7\) to \(\displaystyle 9\).