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

SOLUCIONARIO DE ARITMETICA DE BALDOR
23
8 ÷ 1 8 9 = 23
8 ÷ 17
9 = 23
8 × 9 17
59 59
× 12 =
24 2 = 29 1 2
239
× 12 = 239
12
207
136 = 1 71
136
-20.
-27.
-13.
(6 − 3 5 + 1 10 ) ÷ 5 1 2
( 1 2 − 1 3 ) × (2 − 1 5 ) ÷ (1 − 1 3 )
(7 + 3 1 8 ) ÷ (14 + 6 1 4 ) = 10 1 8 ÷ 20 1 4
( 60 − 6 + 1 ) ÷ 11
10 2 = 55
10 ÷ 11
2
( 3 − 2
6 ) × (1 5 5 − 1 5 ) ÷ (3 3 − 1 3 )
81
8 ÷ 81
4 = 81
8 × 4 81 = 1 2
55
10 × 2 11 = 5 5 = 1
1
6 × 1 4 5 ÷ 2 3 = 1 6 × 9 5 × 3 2
-14.
(60 − 1 8 ) ÷ (30 − 1 16 )
-21.
(150 1 8 ÷ 1 8 ) ÷ (4 × 2 7 8 )
-28.
9
2 × 5 × 2 = 9 20
-15.
-16.
-17.
(59 8 8 − 1 16
) ÷ (29
8 16 − 1 16 )
59 7 15
÷ 29
8 16 = 479
8 ÷ 479
16
479
8 × 16
479 = 2
( 5 8 × 10
50 ) ÷ 10 1 12 = ( 5
4 × 10 ) ÷ 121
12
1
8 × 12
121 = 3
2 × 121 = 3
242
(10 ÷ 5 6 ) ÷ 10 9 32 = (10 × 6 5 ) ÷ 329
32
12 × 32
329 = 384
329 = 1 55
329
( 3 5 × 10
9 × 3 4 ) ÷ 3 1 2 = 2 4 ÷ 7 2 = 1 2 × 2 7 = 1 7
-18.
( 1 2 + 3 4 − 1 8 ) ÷ 1 3 5 = (4 + 6 − 1 ) ÷ 8 8 5
-19.
9
8 × 5 8 = 45
64
(2 1 3 + 3 1 4 − 3 1 8 ) ÷ 1 12
( 7 3 + 13
4 − 25
8 ) × 12
56 + 78 − 75
( ) × 12
24
( 1 201
8
-22.
× 8) ÷ (4 × 23
23
) = 1 201 ÷
8 2
1 201 × 2 23 = 2 402 10
= 104
23 23
( 7 30 + 7 90 + 1 3 ) ÷ 1 9
-23.
+ 7 + 30
= (21 ) × 9
90
58
90 × 9 = 58
10 = 29
5 = 5 4 5
( 1 6 + 1 3 − 1 45 ) ÷ 1 1 90
15 + 30 − 2
( ) ÷ 91
90 90 = 43
90 × 90
91 = 43
91
-24.
-25.
(2 × 6 5 ) ÷ (2 + 3 8 ) = 12
5 ÷ 2 3 8
12
5 ÷ 19
8 = 12
5 × 8 19 = 96
95 = 1 1 95
(5 ÷ 1 5 ) ÷ (2 ÷ 1 ) = (5 × 5) ÷ (2 × 3)
3
-26.
25 ÷ 6 = 4 1 6
(19 2 3 + 1 4 ) ÷ (4 1 5 × 5 42 × 1 6 )
( 59
3 + 1 4 ) ÷ (21 5 × 5 42 × 1 6 )
( 236 + 3 ) ÷ ( 1
12 2 × 6 ) = 239
12 ÷ 1 12
-29.
(4 − 1 4 ) × (5 − 1 5 ) ÷ 1 18
(3 4 4 − 1 4 ) × (4 5 5 − 1 5 ) × 18
3 3 4 × 4 4 15
× 18 =
5 4 × 24
5 × 18
3 × 6 × 18 = 324
( 1 2 × 4 3 ) ÷ (1 2 ÷ 6) ÷ (1 2 + 1 4 )
2
3 ÷ 1
2 × 6 ÷ (4 + 2
8 ) = 2 3 ÷ 1 12 ÷ 6 8
2
3 × 12 × 8 6 = 2 × 2 × 8 = 32
3 3 = 10 2 3
-30.
(2 1 3 − 1 1 6 ) ÷ (3 1 4 + 2 1 8 ) ÷ 28
129
( 7 3 − 7 6 ) ÷ (13 4 + 17
8 ) × 129
28
( 14 − 7 ) ÷ (
6
26 + 17
8
) × 129
28
7
6 ÷ 43
8 × 129
28 = 7 6 × 8 43 × 129
28
-31. 3 5 de (8 9 ÷ 1 6 )
4 × 3
3 × 4 = 1
3
5 × (8 9 × 6) = 8 × 6
5 × 3 = 8 × 2
5
-32. 5 de los 6 (2 ÷ 3 ) de 72
3 2
= 16
5 = 3 1 5
5
6 × (2 3 ÷ 3 2 ) × 72 = 5 6 × (2 3 × 2 3 ) × 72
LEONARDO F. APALA TITO 171