Q. 55.0( 1 Vote )
Using antil
Answer :
(i) Characteristic is 3
Mantissa is 0.072
From the antilog table antilog 0.072 = 1.180
Now as the characteristic is 3, therefore we will place the decimal after 3+1=4 numbers in 1180
∴ antilog 3.072 = 1180
(ii) Characteristic is 1
Mantissa is 0.759
From the antilog table antilog 0.759 = 5.741
Now as the characteristic is 1, therefore we will place the decimal after 1+1=2 numbers in 5741
∴ antilog 1.759 = 57.41
(iii) Characteristic is ̅1 = –1
Mantissa is 0.3826
From the antilog table antilog 0.382 = 2.410
Mean Value of 6 is 0.003
Thus, antilog 0.3826 = 2.410+0.003 = 2.413
Now as the characteristic is –1, therefore we will move decimal
–1+1=0 places left in 2.413
∴ antilog ̅1.3826 = 0.2413
(iv) Characteristic is ̅3 = –3
Mantissa is 0.6037
From the antilog table antilog 0.603 = 4.009
Mean Value of 7 is 0.006
Thus, antilog 0.6037 = 4.009+0.006 = 4.015
Now as the characteristic is –3,
therefore we will move decimal
–3+1=2 places left in 4.015
∴ antilog ̅3.6037 = 0.004015
(v) Characteristic is 0
Mantissa is 0.2732
From the antilog table antilog 0.273 = 1.875
Mean value 2 is 0.001
Thus, antilog 0.2732 = 1.875+0.001 = 1.876
Now as the characteristic is 0, therefore we will place the decimal after 0+1=1 numbers in 1876
∴ antilog 0.2732 = 1.876
(vi) Characteristic is ̅2 = –2
Mantissa is 0.1798
From the antilog table antilog 0.179 = 1.510
Mean Value of 8 is 0.003
Thus, antilog 0.1798 = 1.510+0.003 = 1.513
Now as the characteristic is –2, therefore we will move decimal
–2+1=1 places left in 1.513
∴ antilog ̅2.1798 = 0.01513
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