document.write( "Question 280523: The front wheel of a wagon makes 110 more revolutions than the rear wheel in going a mile . if the circumference of the front wheel is increased by 25% and that of the rear wheel decreased 25%, the rear wheel then revolves 88 more times than the front wheel in going a mile . the original circumference of the front wheel in feet is\r
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document.write( "A 8 b 10 c 12 d 14 e 16 \n" );
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Algebra.Com's Answer #203917 by richwmiller(17219)![]() ![]() You can put this solution on YOUR website! (x+110)*a=5280 \n" ); document.write( "x*b=5280\r \n" ); document.write( "\n" ); document.write( "(y)*1.25*a=5280 \n" ); document.write( "(y+88)*.75b=5280 \n" ); document.write( "(x+110)*a=5280 and x*b=5280 and (y)*1.25*a=5280 and (y+88)*.75b=5280 \n" ); document.write( "a = 12, b = 16, x = 330, y = 352 \n" ); document.write( "the original front wheel circumference is 12 \n" ); document.write( "check \n" ); document.write( "original\r \n" ); document.write( "\n" ); document.write( "330+110=440 \n" ); document.write( "440 front wheel revolutions \n" ); document.write( "12 is the circumference \n" ); document.write( "440*12=5280 \n" ); document.write( "rear wheel revolutions 330 \n" ); document.write( "440-330=110 more in front wheel\r \n" ); document.write( "\n" ); document.write( "330*16=5280 \n" ); document.write( "rear wheel circumference 16 \n" ); document.write( "rear wheel revolutions 330 \n" ); document.write( "second arrangement \n" ); document.write( "352+88=440 \n" ); document.write( "front wheel circumference 15 \n" ); document.write( "front wheel revolutions 352 \n" ); document.write( "15*352=5280 \n" ); document.write( "rear wheel circumference 12 \n" ); document.write( "rear wheel revolutions 440 \n" ); document.write( "rear wheel revolutions 88 more than front \n" ); document.write( "12*440=5280 \n" ); document.write( "ok \n" ); document.write( " |