rectangle a has a width of 8 cm and a length of 16 cm. rectangle b has the same area as the first but its width is 62.5% of the width of the first rectangle. express the length of rectangle b as a percent of the length of rectangle a. what percent more or less is the length of rectangle b than the length of rectangle a? !

Area A = 8 * 16 = 128 cm^2 ;
Area B = 128 cm^2 ;
But Area B = L *  [ (62.5 / 100) * 8 ] ; where L = the length of rectangle B ;
128 = L * ( 625 / 1000 ) * 8 ;
128 / 8 = L * ( 625 / 1000 );
16 = L * ( 5 / 8 ) ;
L = ( 16 * 5 ) / 8 ;
L = 80 / 8 ;
L = 10 cm ;
p / 100 = 10 / 16 ;
p / 100 = 5 / 8 ;
p = 500 / 8 ;
p = 62.5 ;
p% = 62.5% (  the length of rectangle B is 62.5% of the length of rectangle A ) ;
q / 100 = ( 16 - 10 ) / 16 ;
q / 100 = 6 / 16 ;
q / 100 = 3 / 8 ;
q = 300 / 8 ;
q = 37.5 ;
q% = 37.5 % ( 37.5% more the length of Rectangle B than the length of Rectangle A );

127 millimeters is the answer

Ican't do any of the other stuff, but i can get you started. the answer, i believe, is $2.50.