\(\displaystyle B = A \div \frac{1}{2 } = A \cdot 2 = 2A\).

The reciprocal of \(\displaystyle B\) is therefore \(\displaystyle \frac{1}{B} = \frac{1}{2A} = \frac{1}{2} \cdot \frac{1}{A}\) - that is, \(\displaystyle \frac{1}{2}\) times the reciprocal of \(\displaystyle A\). The reciprocal of a positive number must be positive, and \(\displaystyle 1 > \frac{1}{2}\), so 

\(\displaystyle 1 \cdot \frac{1}{A} > \frac{1}{2} \cdot \frac{1}{A}\)

\(\displaystyle \frac{1}{A} > \frac{1}{B}\)

The reciprocal of \(\displaystyle A\) is the greater number.

We examine two scenarios to demonstrate that the given information is insufficient.

Case 1: \(\displaystyle Y = 2\frac{1}{2}\)

\(\displaystyle Y - \frac{1}{2} = 2 \frac{1}{2 }- \frac{1}{2} = 2\)

The reciprocal of this is \(\displaystyle \frac{1}{2}\).

\(\displaystyle Y + \frac{1}{2} = 2 \frac{1}{2 }+ \frac{1}{2} = 3\)

The reciprocal of this is \(\displaystyle \frac{1}{3}\). 

\(\displaystyle \frac{1}{2} > \frac{1}{3}\), so \(\displaystyle Y - \frac{1}{2}\) has the greater reciprocal.

Case 2: \(\displaystyle Y = \frac{1}{4}\).

\(\displaystyle Y - \frac{1}{2} = \frac{1}{4 }- \frac{1}{2} = \frac{1}{4 }- \frac{2}{4} = - \frac{1}{4 }\)

The reciprocal of this is \(\displaystyle -4\).

\(\displaystyle Y + \frac{1}{2} = \frac{1}{4 }+ \frac{1}{2} = \frac{1}{4 }+ \frac{2}{4} = \frac{3}{4 }\)

The reciprocal of this is \(\displaystyle \frac{4 }{3}\).

\(\displaystyle \frac{4 }{3} > -4\), so \(\displaystyle Y + \frac{1}{2}\) has the greater reciprocal.

\(\displaystyle Y\) is negative, so we will let \(\displaystyle X = -Y\), making \(\displaystyle X\) positive. It follows that \(\displaystyle Y = -X\).

The reciprocal of \(\displaystyle Y\) is \(\displaystyle \frac{1}{Y} = \frac{1}{ -X} = - \frac{1}{X}\);

the reciprocal of \(\displaystyle Y - 1\) is \(\displaystyle \frac{1}{Y-1} = \frac{1}{ -X-1 } = - \frac{1}{X+1}\).

\(\displaystyle X\) and \(\displaystyle X+ 1\) are both positive; \(\displaystyle X+1 > X\), so \(\displaystyle \frac{1}{X+1}< \frac{1}{X }\), and \(\displaystyle - \frac{1}{X+1}> -\frac{1}{X }\). Therefore, the reciprocal of \(\displaystyle Y-1\) is the greater quantity.

Rewrite the fractions in terms of their least common denominator, 12.

\(\displaystyle (LCD(2,3,4)=12)\) 

Add, then rewrite as a mixed fraction:

\(\displaystyle \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6}{12} + \frac{4}{12} + \frac{3}{12} =\frac{13}{12} = 1 \frac{1}{12}\)

Add the integers:

\(\displaystyle 3 + 5 + 4 = 12\)

Now add the sums:

\(\displaystyle 12 + 1 \frac{1}{12} = 13 \frac{1}{12}\)

In order to add fractions we must find a common denominator.  Since \(\displaystyle \small 12\) is a multiple of both \(\displaystyle \small 3\) and \(\displaystyle \small 4\), we must multiply the numerator and denominator of each fraction by a number to get a denomintor of \(\displaystyle \small 12\). 

Since \(\displaystyle \small 4\) times \(\displaystyle \small 3\) is \(\displaystyle \small 12\), we can multiply the numerator and denominator of the first fraction by \(\displaystyle \small 3\).

\(\displaystyle \small \frac{(3\cdot 3)}{(4\cdot 3)}= \frac{9}{12}\)

Since \(\displaystyle \small 3\) times \(\displaystyle \small 4\) is \(\displaystyle \small 12\), we can multiply the numerator and demonimator of the second fraction by \(\displaystyle \small 4\).

\(\displaystyle \small \frac{(1\cdot 4)}{(3\cdot 4)}=\frac{4}{12}\)

Now we add together the numerators.

\(\displaystyle \small \frac{9}{12}+\frac{4}{12}=\frac{13}{12}\)

The answer is \(\displaystyle \small \frac{13}{12}\).

In order to find the perimeter of a rectangle, you add together all the sides.  In this particular case, however, you must first find a common denominator for all of the fractions.  Luckily, \(\displaystyle \small 8\) is a multiple of \(\displaystyle \small 2\), so we can multiply the numerator and denominator of  \(\displaystyle \small \frac{1}2{}\) by \(\displaystyle \small 4\) to get a denominator of \(\displaystyle \small 8\).

\(\displaystyle \small \frac{(1\cdot 4)}{(2\cdot 4)}=\frac{4}{8}\)

Now we simply add all four sides.

\(\displaystyle \small \frac{4}{8}+\frac{4}{8}+\frac{1}{8}+\frac{1}{8}=\frac{10}{8}\)

Since \(\displaystyle \small \frac{10}{8}\) can be reduced by dividing the numerator and denominator by \(\displaystyle \small 2\), we must simplify.

\(\displaystyle \small \frac{(10/2)}{(8/2)}=\frac{5}{4}\)

The perimeter of the rectangle is \(\displaystyle \small \frac{5}{4}\).

Add both sides of the two equations:

\(\displaystyle x = \frac{1}{2} + \frac{1}{3}\)

\(\displaystyle y = \frac{1}{2} + \frac{2}{3}\)

\(\displaystyle x + y = \frac{1}{2} + \frac{1}{3}+ \frac{1}{2} + \frac{2}{3}\)

\(\displaystyle x + y = \frac{1}{2} + \frac{1}{2}+ \frac{1}{3} + \frac{2}{3}\)

\(\displaystyle x + y =1+1\)

\(\displaystyle x + y =2\)

(a) \(\displaystyle t - 4.5 = 8.9\)

\(\displaystyle t - 4.5 + 4.5 = 8.9 + 4.5\)

\(\displaystyle t = 13.4\)

(b) \(\displaystyle y -6 \frac{1}{10} = 7 \frac{2}{5}\)

\(\displaystyle y -6 \frac{1}{10} + 6 \frac{1}{10} = 7 \frac{2}{5} + 6 \frac{1}{10}\)

\(\displaystyle y = 7 \frac{4}{10} + 6 \frac{1}{10} = 13 \frac {5}{10} = 13 \frac{1}{2}\)

\(\displaystyle 13 \frac{1}{2} = 13.5 > 13.4\)

First, you must add the fractions in each column. When adding fractions, find the common denominator. The common denominator for Column A is 10. Then, change the numerators to reflect changing the denominators to give you \(\displaystyle \frac{4}{10}+\frac{5}{10}\). Combie the numerators to give you \(\displaystyle \frac{9}{10}.\)Then, add the fractions in Column B. The common denominator for those fractions is 72. Therefore, you get \(\displaystyle \frac{8}{72}+\frac{63}{72}\). Combine the numerators to get \(\displaystyle \frac{71}{72}\) . Compare those two fractions. Think of them as slices of pizza. There would be way more of Column B. Therefore, it is greater. Also, a little to trick to comparing fractions is cross-multiply. The side that has the biggest product is the greatest.

\(\displaystyle \frac{3}{5} = 3 \div 5 = 0.6\) and \(\displaystyle \frac{1}{4 } = 1 \div 4 = 0.25\), so 

\(\displaystyle \frac{3}{5} + \frac{1}{4} = 0.6 + 0.25 = 0.85\), the decimal equivalent of (A).

\(\displaystyle 0.4 + 0.44 = 0.84\), the value of (B).

(A) is the greater.