Find the sum of 5

$\begin{aligned} X & = \frac 1{21} - \frac 1{70} - \frac 7{110} + \frac {19}{143} - \frac {46}{195} - \frac 1{255} + \frac {67}{323} + \frac {96}{437} \\ & = \small \frac 14\left(\frac 13 - \frac 17\right) - \frac 13\left(\frac 17 - \frac 1{10} \right) - 7\left(\frac 1{10} - \frac 1{11} \right) + \frac {19}2 \left(\frac 1{11} - \frac 1{13}\right) - 23\left(\frac 1{13}-\frac 1{15}\right) - \frac 12\left(\frac 1{15} - \frac 1{17} \right) + \frac {67}2 \left(\frac 1{17} - \frac 1{19}\right) + 24 \left(\frac 1{19} - \frac 1{23}\right) \\ & = \small \frac 1{12} - \left(\frac 13 + \frac 14\right)\frac 17 - \left(7-\frac 13\right)\frac 1{10}+\left(7+\frac {19}2\right)\frac 1{11} - \left(\frac {19}2+23\right)\frac 1{13} + \left(23-\frac 12\right)\frac 1{15} + \left(\frac 12+\frac {67}2 \right)\frac 1{17} - \left(\frac {67}2 -24 \right)\frac 1{19} - \frac {24}{23} \\ & = \small \frac 1{12} - \left(\frac 7{12}\right)\frac 17 - \left(\frac {20}3\right)\frac 1{10}+\left(\frac {33}2\right)\frac 1{11} - \left(\frac {65}2\right)\frac 1{13} + \left(\frac {45}2\right)\frac 1{15} + \left(34 \right)\frac 1{17} - \left(\frac {19}2 \right)\frac 1{19} - \frac {24}{23} \\ & = \small \cancel{\frac 1{12} - \frac 1{12}} - \frac 23 +\blue{\cancel{\frac 32}} - \blue{\cancel{\frac 52}} + \blue{\cancel{\frac 32}} + 2 - \blue{\cancel{\frac 12}} - \frac {24}{23} \\ & = \frac 43 - \frac {24}{23} = \frac {20}{69} \approx \boxed{0.2899} \end{aligned}$

(((((( $\frac{1}{21}$ − $\frac{1}{70}$ ) − $\frac{7}{110}$ ) + $\frac{19}{143}$ ) − $\frac{46}{195}$ ) − $\frac{1}{255}$ ) + $\frac{67}{323}$ ) + $\frac{96}{437}$

(((((( $0.047619047619047616$ − $\frac{1}{70}$ ) − $\frac{7}{110}$ ) + $\frac{19}{143}$ ) − $\frac{46}{195}$ ) − $\frac{1}{155}$ ) + $\frac{67}{323}$ ) + $\frac{96}{437}$

(((((( $0.047619047619047616 - 0.014285714285714285$ )− $\frac{7}{110}$ + $\frac{19}{143}$ ) − $\frac{46}{195}$ ) - $\frac{1}{255}$ ) + $\frac{67}{323}$ ) + $\frac{96}{437}$

((((( $0.03333333333333333$ − $\frac{7}{110}$ ) + $\frac{19}{143}$ ) − $\frac{46}{195}$ ) − $\frac{1}{255}$ ) + $\frac{67}{323}$ ) + $\frac{96}{437}$

((((( $0.03333333333333333-0.06363636363636363$ ) + $\frac{19}{143}$ ) − $\frac{46}{195}$ ) − $\frac{1}{255}$ ) + $\frac{67}{323}$ ) + $\frac{96}{437}$

(((( $-0.030303030303030297$ + $\frac{19}{143}$ ) − $\frac{46}{195}$ ) − $\frac{1}{255}$ ) + $\frac{67}{323}$ ) + $\frac{96}{437}$

(((( $-0.030303030303030297 + 0.13286713286713286$ ) − $\frac{46}{195}$ ) − $\frac{1}{255}$ ) + $\frac{67}{323}$ ) + $\frac{96}{437}$

((( $0.10256410256410256$ − $\frac{46}{195}$ ) − $\frac{1}{255}$ ) + $\frac{67}{323}$ ) + $\frac{96}{437}$

((( $0.10256410256410256 - 0.2358974358974359$ ) − $\frac{1}{255}$ ) + $\frac{67}{323}$ ) + $\frac{96}{437}$

(( $-0.13333333333333333$ − $\frac{1}{255}$ ) + $\frac{67}{323}$ ) + $\frac{96}{437}$

(( $-0.13333333333333333 - 0.00392156862745098$ ) + $\frac{67}{323}$ ) + $\frac{96}{437}$

( $-0.13725490196078433$ + $\frac{67}{323}$ ) + $\frac{96}{437}$

( $-0.13725490196078433 + 0.20743034055727555$ ) + $\frac{96}{437}$

$0.07017543859649122$ + $\frac{96}{437}$

$0.07017543859649122 + 0.21967963386727687$

$0.28985507246376807$

Therefore the answer to this problem is $0.2898$ .