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

SOLUCIONARIO DE ARITMETICA DE BALDOR
301 =
301 × 27
= 8 127
27 27
-8. 405 a quebrado equivalente de
denominador 28.
405 =
405 × 28 11 340
=
28 28
-9. 999 a quebrado equivalente de
denominador 14.
999 =
999 × 14 13 986
=
14 14
-10. 1 000 a quebrado equivalente de
denominador 56.
1 000 =
1 000 × 56 56 000
=
56 56
-11. 2 356 a quebrado equivalente de
denominador 19.
2 356 =
2 356 × 19 44 764
=
19 19
-12. 3 789 a quebrado equivalente de
denominador 17.
3 789 =
3 789 × 17 64 413
=
17 17
-13. 4 444 a quebrados equivalente de
denominador 15
4 444 =
4 444 × 15 66 660
=
15 15
-14. 8 888 a quebrados equivalente de
denominador 11
8 888 =
EJERCICIO 107
8 888 × 11 97 768
=
11 11
Reducir, por simple inspección:
-1. 1 2 = 4
-2. 1 3 = 6
-3. 1 4 = 12
-4. 1 5 = 20
1
2 = 1 × 2
2 × 2 = 2 4
1
3 = 1 × 2
3 × 2 = 2 6
1
4 = 1 × 3
4 × 3 = 3 12
-5. 2 3 = 12
-6. 3 4 = 20
-7. 3 5 = 25
-8. 1 6 = 18
-9. 2 7 = 21
-10. 1 8 = 24
-11. 2 9 = 36
-12. 1
10 = 40
-13. 2
11 = 33
-14. 5
12 = 24
-15. 1
13 = 39
-16. 1
14 = 56
1
5 = 1 × 4
5 × 4 = 4 20
2
3 = 2 × 4
3 × 4 = 8 12
3
4 = 3 × 5
4 × 5 = 15/20
3
5 = 3 × 5
5 × 5 = 15
25
1
6 = 1 × 3
6 × 3 = 3 18
2
7 = 2 × 3
7 × 3 = 6 21
1
8 = 1 × 3
8 × 3 = 3 24
2
9 = 2 × 4
9 × 4 = 8 36
1
10 = 1 × 4
10 × 4 = 4 40
2
11 = 2 × 3
11 × 3 = 6 33
5
12 = 5 × 2
12 × 2 = 10
24
1
13 = 1 × 3
13 × 3 = 3 39
1
14 = 1 × 4
14 × 4 = 4 56
-17. 2
15 = 45
-18. 7
16 = 80
-19. 11
20 = 100
-20. 13
30 = 180
EJERCICIO 108
Reducir:
-1. 3/5 a 35avos
-2. 1/6 a 42avos
-3. 6/7 a 63avos
-4. 7/8 a 96avos
-5. 5/11 a 121avos
-6. 4/13 a 130avos
-7. 8/17 a 102avos
2
15 = 2 × 3
15 × 3 = 6 45
7
16 = 7 × 5
16 × 5 = 35
80
11
20 = 11 × 5
20 × 5 = 55
100
13
30 = 13 × 6
30 × 6 = 78
180
3
5 = 3 × 7
5 × 7 = 21
35
1
6 = 1 × 7
6 × 7 = 7 42
6
7 = 6 × 9
7 × 9 = 54
63
7
8 = 7 × 12
8 × 12 = 84
96
5
11 = 5 × 11
11 × 11 = 55
121
4
13 = 4 × 10
13 × 10 = 40
130
-8. 12/19 a 133avos
-9. 8/21 a 105avos
8
17 = 8 × 6
17 × 6 = 48
102
12
19 = 12 × 7
19 × 7 = 84
133
LEONARDO F. APALA TITO 138