document.write( "Question 1202919: Kira had a roll of blue paper and a roll of red paper. The length of the roll of blue paper is 1/2 the length of the roll of red paper. She cut the roll of blue paper into equal parts of length 9 cm and on each part she drew 3 star shapes. After that, she cut the roll of red paper into equal parts of length 7 cm and on each part she drew 5 heart shapes. What fraction of the shapes Kira drew were star shapes? \n" ); document.write( "
Algebra.Com's Answer #838134 by Theo(13342)![]() ![]() You can put this solution on YOUR website! let the length of the blue roll equal x. \n" ); document.write( "then the length of the red roll equals 2x. \n" ); document.write( "the number of stars in the blue roll = s = x/9 * 3 = x/3 \n" ); document.write( "the number of hearts in the red roll = h = 2x/7 * 5 = 10x/7 \n" ); document.write( "you get s = x/3 and h = 10x/7 \n" ); document.write( "solve for x in both equations to get: \n" ); document.write( "x = 3s \n" ); document.write( "x = 7h/10 \n" ); document.write( "since they both equal to x, you can set them equal to each other to get: \n" ); document.write( "3s = 7h/10 \n" ); document.write( "solve for s to get s = 7h/30 \n" ); document.write( "the ratio of star to total is equal to s/(s + h) \n" ); document.write( "the ratio of hearts to total is equal to h/(s + h) \n" ); document.write( "since s = 7h/30, then s + h = 7h/30 + h which is equal to 7h/30 + 30h/30 which is equal to 37h/30. \n" ); document.write( "s / (s+h) is equal to s / (37h/30) which is equal to (7h/30) / (37h/30) which is equal to 7h/30 * 20/37h which is equal to 7h/37h which is equal to 7/37. \n" ); document.write( "h / (s+h) is equal to h / (37h/30) which is equal to h * 30/37h which is equal to 30h / 37h which is equal to 30/37. \n" ); document.write( "the fraction of the star shapes to the total shapes that kira drew was 7/37. \n" ); document.write( "that's your solution.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "note that the x for the blue roll and 2x for the red roll had to be multiples of 63 because x had to be evenly divisible by 9 and 2x had to be evenly divisible by 7. \n" ); document.write( "the least common multiple of 7 and 9 is 63.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |