ANSWER
B is closer in value to [tex]\frac{1}{2}[/tex].
EXPLANATION
The first thing we need to do is to find each of the sums.
[tex]A=\frac{2}{15}+ \frac{1}{3}[/tex]
The least common multiple of [tex]3[/tex] and [tex]15[/tex] is [tex]15[/tex].
So our sum becomes;
[tex]A=\frac{2\times 1 + 1\times 5}{15}[/tex]
[tex]A=\frac{2 + 5}{15}[/tex]
[tex]A=\frac{7}{15}[/tex]
We now find the second sum;
[tex]B=\frac{6}{7}- \frac{1}{3}[/tex]
Again we find the least common multiple of [tex]7[/tex] and [tex]3[/tex], which is [tex]21[/tex].
Our sum now becomes;
[tex]B=\frac{6\times 3-1\times 7}{21}[/tex]
[tex]B=\frac{18-7}{21}[/tex]
[tex]B=\frac{11}{21}[/tex].
The next thing we have to do now is to figure out which of the two sums is nearer to [tex]\frac{1}{2}[/tex].
A more efficient way to do this is to express the two sums together with [tex]\frac{1}{2}[/tex] over the same denominator.
That way, we can see which one is closer.
The least common multiple of [tex]2,15,and\:21[/tex] is [tex]210[/tex].
[tex]A=\frac{7}{15}= \frac{98}{210}[/tex]
[tex]\frac{1}{2} =\frac{105}{210}[/tex]
[tex]B=\frac{11}{21}=\frac{110}{210}[/tex]
We can see that [tex]\frac{110}{210}[/tex] is closer to [tex]\frac{105}{210}[/tex] than [tex]\frac{98}{210}[/tex]
Hence we conclude that;
[tex]B=\frac{6}{7}- \frac{1}{3}[/tex] is closer in value to [tex]\frac{1}{2}[/tex] than
[tex]A=\frac{2}{15}+ \frac{1}{3}[/tex]
