Solucionario completo de Aritmetica de Baldor (Por Leonardo F. Apala T.)

SOLUCIONARIO DE ARITMETICA DE BALDOR
-20. 34 136/7 432
-25. 184 286/17 189
-14. 2 134/289
1 563
315 = 4 303
315
34 136
7 432 = 4 4 408
7 432
-21. 54 137/189
18 4286 12 396
= 10
17 189 17 189
EJERCICIO 104
2 134
289 = 7 111
289
Reducir:
-15. 3 115/417
-1. 2 = 2
2 = 2 × 2
2
= 4 2
-16. 4 200/954
3 115
417 = 7 196
417
-22. 60 185/419
54 137 83
= 286
189 189
-2. 3 = 2
3 = 3 × 2 = 6 2 2
-3. 4 = 3
4 = 4 × 3
3
= 12
3
-17. 8 632/1 115
4 200
954 = 4 384
954
60 185 268
= 143
419 419
-4. 5 = 1
5 = 5 × 1 = 5 1 1
-5. 5 = 8
5 = 5 × 8
-23. 89 356/517
8
= 40
8
-18. 9 732/2 164
8 632
1 115 = 7 827
1 115
-6. 6 = 4
6 = 6 × 4
4
= 24
4 = 6
-19. 12 485/3 284
9 732
2 164 = 4 1 076
2 164
-24. 102 102/1 111
89 356 432
= 172
517 517
-7. 7 = 2
7 = 7 × 2
2
-8. 8 = 5
8 = 8 × 5
5
= 14
2
= 40
5
12 485
3 284 = 3 2 633
3 284
102 102 1 001
= 91
1 111 1 111
-9. 9 = 6
9 = 9 × 6
-10. 7 = 11
6
= 54
6
LEONARDO F. APALA TITO 136