easy f.2 maths question (only 1 )

2007-12-02 7:26 am
a)Mary measures that the length of a cube is 10cm.If the percentage error of the measured length is 1%,what are the lower limit and the upper limit of the actual length?
b)Mary uses this length to calculate the area of one face of the cube.What is the percentage error of the area of the face?
c)Mary also uses the measurement found in (a) to calculate the volume of the cube.What is the percentage error of the volume?

回答 (1)

2007-12-03 4:14 pm
✔ 最佳答案
a) the lower limit of the actual length
=10x10x(1-1%)
=99mm
the upper limit of the actual length
=10x10x(1+1%)
=101mm
b)the area of the face
=100x100
=10000mm^2
the lower limit of the actual area
=99x99
=9801mm^2
the percentage error of the area of the face
=100%-(9801/10000)x100%
=100%-98.99%
=1.99%
the upper limit of the actual area
=101x101
=10201mm^2
the percentage error of the area of the face
=(10201/10000)x100%-100%
=102.01%-100%
=2.01%
c)the volume of the cube
=100x100x100
=1000000mm^3
the lower limit of the actual volume of the cube
=99x99x99
=970299mm^3
the percentage error of the volume of the cube
=100%-(970299/1000000)x100%
=100%-97.0299%
=2.9701%
the upper limit of the actual volume of the cube
=101x101x101
=1030301mm^3
the percentage error of the volume of the cube
=(1030301/1000000)x100%-100%
=103.0301%-100%
=3.0301%


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