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

SOLUCIONARIO DE ARITMETICA DE BALDOR
-20.
16 1 4 = 16 × 4 + 1
4
= 65
4
-12.
65 7 80
65 × 80 + 7
= = 5 207
80 80
-5. 108/12
81
9 = 9
EJERCICIO 101
18 2 3 = 18 × 3 + 2
3
= 56
3
-13.
5 3
106 = 5 × 106 + 3
106
= 533
106
-6. 125/25
108
12 = 9
Convertir en quebrados:
-1.
15 3 8 = 15 × 8 + 3 = 123
8 8
-14.
-15.
8 1
102 = 8 × 102 + 1
102
= 817
102
-7. 7/2
125
25 = 5
-2.
-3.
12 3 11
12 × 11 + 3
= = 135
11 11
16 7 8 = 16 × 8 + 7 = 135
8 8
-16.
-17.
25 7 73
90 19
31
25 × 73 + 7
= = 1 832
73 73
90 × 31 + 19
= = 2 809
31 31
-8. 5/2
7
2 = 3 1 2
-4.
-5.
19 3 11
20 3 19
19 × 11 + 3
= = 212
11 11
20 × 19 + 3
= = 383
19 19
-18.
-19.
90 19
37
101 13
18
90 × 37 + 19
= = 3 349
37 37
101 × 18 + 13
= = 1 831
18
18
-9. 8/5
5
2 = 2 1 2
-6.
-7.
17 5 18
23 4 23
17 × 18 + 5
= = 311
18 18
23 × 23 + 4
= = 533
23 23
-20.
102 15
17
500 8 67
EJERCICIO 102
102 × 17 + 15
= = 1 749
17
17
500 × 67 + 8 33 508
= =
67
67
-10. 19/7
8
5 = 1 3 5
-8.
-9.
31 5 31
42 7 25
31 × 31 + 5
= = 966
31 31
42 × 25 + 7
= = 1 057
25 25
Hallar por simple inspección, los enteros
contenidos en:
-1. 12/3
-2. 21/7
12
3 = 4
-11. 25/8
19
7 = 2 5 7
-10.
-11.
53 9 17
60 3 17
59 × 17 + 9
= = 910
17 17
60 × 17 + 3
= = 1 023
17 17
-3. 32/8
-4. 81/9
21
7 = 3
32
8 = 4
-12. 31/4
25
8 = 3 1 8
LEONARDO F. APALA TITO 134