F.2 MATH about error ~ 10marks

A boy uses a ruler with finest markings 0.1 cm to measure the length and width of his mathematics textbook. He finds that the length and width are 18.8 and 26.1 cm respectively.
a1
the greatest absolute error=0.1/2=0.05cm

Upper limit of the length = 18.8+0.05=18.85cm
a2 Lower limit of the length = 18.8-0.05=18.75cm
a3 Upper limit of the width = 26.1+0.05=26.15cm


a4 Lower limit of the width = 26.1-0.05=26.05
b1 What is the area of the textbook?
18.8x26.1=490.68cm^2
b2 What is the maximum value of the area of the textbook?
(18.85x26.15)cm^2
=492.9275cm^2
b3 What is the maximum absolute error of the area of the textbook?
the minimum value of the area of the textbook
=18.75x26.05=488.4375
=492.9275-490.68=2.2475
b4 What is the percentage error of the area of the textbook? (Give the answer correct to 3 significant figures if necessary.)
2.2475/490.68x100%=0.458%

2008-06-25 18:18:23 補充：
re:003
if the maximum value of length is 18.9cm, it is a problem that you are over the limit
Because the question talks that a boy measures that the length is 18.8cm, so I am afraid that your answer is incorrect......

2008-06-26 17:01:02 補充：
之前有人答咗(第3位)，不過佢知道衰咗之後佢刪咗咋~~~~~~

(a1-a4)to Use formula:(1/2*scale interval of the measureing tool)

then use ''Lower limit of the actual value''

=measured value-maximum absolute error

or use''Upper limit of the actual value''

==measured value+maximum absolute error

a1.Upper limit of the length =18.8+(1/2*0.1) =18.85

a2.Lower limit of the length = 18.8-(1/2*0.1)=18.75

a3.Upper limit of the width = 26.1+(1/2*0.1)=26.15

a4.Lower limit of the width = 26.1-(1/2*0.1)=26.05
-----------------------------------------------------------------------------------------
B1 of the formula is length*width

b1.The area of the text book:

26.1*18.8

=490.68cm^2

B2 of the formula is use Upper limit of the actual value at length*upper limit of the actual value at width''

b2.the maximum value of the area of the textbook

18.75*26.15

=492.9275

B3.Absolute error =actual value -approximation

b3.the maximum absolute error of the area of the textbook

492.9275-490.68

=2.2475cm

B4.Percentage error = relative error *100%

b4. the percentage error of the area of the textbook:

(2.2475/490.68)x100%

=0.458%

2008-06-25 18:48:18 補充：
re 001

Are you write wrong number for your supplement?!