Problemas matmaticos

P= 20000 I = (20000)(0.12)(1)
Tea = 12% I = 2400
n = 1

Solución caso 2
P= 4000 I = (4000)(0.12)(150/360)
i = 12% I = 200
n = 150

Solución caso 3
P= 2000 I = (2000)(0.08)(3)i m = 8% I = 4800
n = 3

Solución caso 4
Julio 28
Agosto 31
Setiembre 18
……………………………..
77 días

Solución caso 5
P= 25000 I = (25000)(0.04)( 287/30)
i = 4% I = 9566.67
n =287/30

Solución caso 6
P= ? 2000 = (P)(0.2)(72/360)
tea = i = 20% P = 50000
I = 2000
n = 72/360

Solución caso 7
P= 2000 2300 = (2000)(i)(45/30)
I = 2300 i = 0.76
n = 45/30 tem =7.6%

Solución caso 8
2a = a (0.1)(n)
20 = n

Solución caso 9
P= 4000 I = (4000) [(1+0.15)7 - 1]
i = 15% I = 6640.08
n = 7

Solución caso 10
P= 6000 I = (6000) [(1+0.14)1/4- 1]tea = 14% I = 119.80
n = 1 trimestre

Solución caso 11
P= 6000 8100 = (6000) [(1+i)6- 1]
I = 8100 i = 0.15
n = 6 tea = 15%

Solución caso 12
2a = (a) [(1+0.08)n- 1]
3 =(1+0.08)n
log 3= (n) log 1.08
n = 14.27

Solución caso 13
P= 5000 I = (5000) [(1+0.025)7.5- 1]
n = 1 trim = 90/12 I = 1017.26
tna = 156
m = 6
i = 0.15/6 = 0.025

Solución caso 14
I =800 800 = (P) [(1+0.20)0.06- 1]
n = 20/360=0.06 P = 72731.60
tea = 20%
P = ?

Solución caso 15
Tem = X 4550 = (4000) [(1+i)1.5- 1]
P = 4000 2.1375 = (1+i)1.5
I = 4550 i = 0.6593m = 45/30 = 1.5 tem = 6.6%
Solución caso 16
P= 45000 46865 = (45000) [(1+0.08)n- 1]
I = 1865 + 45000 = 46865 1.041444444 = [(1+0.08)n- 1]
Tem = 8% 2.041444444 = (1.08)n
i = 0.08Iog(2.041444444) = n log (1.08)
n = ? 0.3099375655 = 0.03342375549(n)
n = 9.2729725

26/02
21/01

20/02
31/01
Solución caso 17
P = 200000
Tem = i = 0.05
20
10
6 días de mora
im =0.15(0.05) = 0.0075

Ic = (200000) [(1+0.05)6/30- 1] = 1961.16

Im = (200000) [(1+0.0075)6/30- 1] = 299.10

Liquidacion

Capital final | 200000.00 |
Interés compensatorio(Ic) | 1961.16 |...