150 three step recursive equation calculation problem urgent! There must be three steps

150 three step recursive equation calculation problem urgent! There must be three steps

1.3/7 × 49/9 - 4/3 2.8/9 × 15/36 + 1/27 3.12× 5/6 – 2/9 ×3 4.8× 5/4 + 1/4 5.6÷ 3/8 – 3/8 ÷6 6.4/7 × 5/9 + 3/7 × 5/9 7.5/2 -（ 3/2 + 4/5 ） 8.7/8 + （ 1/8 + 1/9 ） 9.9 × 5/6 + 5/6 10.3/4 × 8/9 - 1/3 0.12χ＋1.8×0.9=7.2 (9－5χ)×0.3=1.02 6.4χ-χ=28+4.4 11.7 × 5/49 + 3/14 12.6 ×（ 1/2 + 2/3 ） 13.8 × 4/5 + 8 × 11/5 14.31 × 5/6 – 5/6 15.9/7 - （ 2/7 – 10/21 ） 16.5/9 × 18 – 14 × 2/7 17.4/5 × 25/16 + 2/3 × 3/4 18.14 × 8/7 – 5/6 × 12/15 19.17/32 – 3/4 × 9/24 20.3 × 2/9 + 1/3 21.5/7 × 3/25 + 3/7 22.3/14 ×× 2/3 + 1/6 23.1/5 × 2/3 + 5/6 24.9/22 + 1/11 ÷ 1/2 25.5/3 × 11/5 + 4/3 26.45 × 2/3 + 1/3 × 15 27.7/19 + 12/19 × 5/6 28.1/4 + 3/4 ÷ 2/3 29.8/7 × 21/16 + 1/2 30.101 × 1/5 – 1/5 × 21 31.50＋160÷40 （58+370）÷（64-45） 32.120-144÷18+35 33.347+45×2-4160÷52 34（58+37）÷（64-9×5） 35.95÷（64-45） 36.178-145÷5×6+42 420+580-64×21÷28 37.812-700÷（9+31×11） （136+64）×（65-345÷23） 38.85+14×（14+208÷26） 39.（284+16）×（512-8208÷18） 40.120-36×4÷18+35

Number a and number B are integers. Their sum is 671. The zero at the end of number a is equal to number B. what's number B?
There's an answer in 20 minutes,

The number a is 10 times of the number B, and the number B is 671 (1 + 10) = 61,

1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 +.. + 1 / 25) + (2 / 3 + 2 / 4 + 2 / 5 +.. + 2 / 25) + (3 / 4 + 3 / 5 + 3 / 6 +.. + 3 / 25) +
1/2+1/3+1/4+1/5+...+1/25)+(2/3+2/4+2/5+.+2/25)+(3/4+3/5+3/6+.+3/25)+)+（23/24+23/25）+24/25

（1/2+1/3+1/4+… +1/25）+（2/3+2/4+2/5+… 2/25）+… +(23/24+23/25)+24/25
=1/2+(1/3+2/3)+(1/4+2/4+3/4)+(1/5+2/5+3/5+4/5)+.+（1/25+2/25+.+24/25)
=1/2+2*3/(2*3)+3*4/(2*4)+4*5/(2*5)+.+24*25/(2*25)
=1/2+2/2+3/2+4/2+.+24/2
=1/2(1+2+3+4+.+24)
=1/2*24*25/2
=150