Given A = 10,000 r = \(\cfrac { 8 }{ 100 } \times \cfrac { 1 }{ 4 } \) = 2% = 0.02, n = 5 X 4 =20
A = \(\cfrac { a }{ i } \left( 1+i \right) \left[ \left( 1+i \right) ^{ n }-1 \right] \)
10,000 = \(\cfrac { a }{ 0.02 } \) (1.02)[(1.02)²⁰-1]
\(\cfrac { 10,000\times 0.02 }{ 1.02 } \) = a[1.4859-1]
196.08 = a[0.4859]
a =\(\cfrac { 196.08 }{ 0.4859 } \) = Rs.403.53
a = Rs.404
(1.02)²⁰ =20 log(1.02)
= 20(0.0086)
= 0.172
Antilog of 0.172 is 1.4859

For machine A:
a = 8000, i =0.1, n = 5
P = \(\cfrac { a }{ i } \left[ 1-\left( 1+i \right) ^{ -n } \right] \)
= \(\cfrac { 8000 }{ 0.1 } \) [1-(1.1)-5]
= 80,000[1-0.6209]
= 80,000(0.3791)
= 30,328
\(\Rightarrow\) Net inflow 30,328-2500 = Rs.5328
(1.1)^-5 = -5 log (1.1)
= -5(0.0413)
= -0.2065
=-0.02065-1+1
=\(\bar { 1 } .7930\)
Antilog of \(\bar { 1 } .7930\) is 0.6209
For machine B:
a = Rs.10,000, i = 0.1,n = 7
P = \(\cfrac { 10,000 }{ 0.1 } \)[1-(1.1)-7]
= 1,00,000[1-(1.1)-7]
= 1,00,000[0.4868]
= Rs.48,680
(1.1)^-7= -7 log (1.1)
= -7(0.0413)
= -0.28973 + 1-1
=\(\bar { 1 } .71027\)
Antilog of \(\bar { 1 } .7103\) is 0.5132
\(\therefore\) Net inflow = 48,680 - 40,000 = 8680
Net inflow of machine B is more than machine A
Hence, machine B can be purchased.

Given a = Rs.75, i=\(\cfrac { 8 }{ 12 } \)% = 4% = 0.004,n = 10 X 2 =20
A = \(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
=\(\cfrac { 75 }{ 0.04 } \) [(1.04)²⁰-1]
=1875 (2.1878-1)
= 1875(1.878)
= Rs.2227
(1.04)⁴⁰ = 20 log (1.04)
= 20(0.0170)
= 0.34
Antilog of 0.34 is 2.1878

Given A = Rs.4,00,000,r =\(\cfrac { 15 }{ 100 } \times \cfrac { 1 }{ 12 } \) = 0.0125,n = 3 x 12 = 36
A = \(\cfrac { a }{ i } \left( 1+i \right) \left[ \left( 1+i \right) ^{ n }-1 \right] \)
4,00,000 = \(\cfrac { a }{ 0.0125 } \left( 0.125 \right) \left[ \left( 1.0125 \right) ^{ 36 }-1 \right] \)
\(\cfrac { 4,00,000\times 0.0125 }{ 1.0125 } \) = a[(1.0125)³⁶-1]
4938.27 = a[1.578-1]
4938.7 = a[0.578]
a = \(\cfrac { 4938.27 }{ 0.578 } \)
a = Rs.8543.71
a = Rs.8544(app)
(1.0125)³⁶ = 36 log (1.0125)
= 36(0.0055)
= 0.1980
Antilog of 0.1980 is 1.578

Given A = Rs.3000,r =\(\cfrac { 8 }{ 100 } \times \cfrac { 1 }{ 4 } \) = 0.02,n = 3 x 4 =12
A=\(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
3000=\(\cfrac { a }{ 0.02 } \left[ \left( 1.02 \right) ^{ 12 }-1 \right] \)
3000 x 0.02 = a[1.2690-1]
60 = a[0.2690]
a=\(\cfrac { 60 }{ 0.2690 } =223.04\)
a = Rs.223 (app)
(1.02)¹² = 12 log (1.02)
= 12 (0.0086)
= 0.1032
Antilog of 0.1032 is 1.2690