priyasom Posted June 19, 2005 Share Posted June 19, 2005 I found this problem in a Powerprep test that i gave recently : The jewels in a crown consist of diamonds,rubies and emeralds. If the ratio of diamonds to rubies is 5 : 6 and ratio of rubies to emeralds is 8 : 3, what is the least number of jewels that could be in the tiara ? 16 22 40 53 67 Can someone provide with an answer and an explanation ? Priya Quote Link to comment Share on other sites More sharing options...
samy_dg Posted June 19, 2005 Share Posted June 19, 2005 The jewels in a crown consist of diamonds,rubies and emeralds. If the ratio of diamonds to rubies is 5 : 6 and ratio of rubies to emeralds is 8 : 3, what is the least number of jewels that could be in the tiara ? 16 22 40 53 67 Let the no. of diamonds, rubies, emeralds be d,r,e. d/r=5/6 ====> d=(5/6)*r r/e=8/3 ====> e=(3/8)*r So the ratio of diamonds, rubies and emeralds is: 5/6 : 1 : 3/8 ==> 20 : 24 : 9 -----> this is the same ratio expressed in terms of the smallest integers possible. For the least no. of jewels, d=20,r=24,e=9 So, least possible total no. of jewels=20+24+9=53. Hey Priya, could you follow my solution? Quote Link to comment Share on other sites More sharing options...
priyasom Posted June 19, 2005 Author Share Posted June 19, 2005 thanks sam for the nice explanation. That was the answer given in PP too :-) The jewels in a crown consist of diamonds,rubies and emeralds. If the ratio of diamonds to rubies is 5 : 6 and ratio of rubies to emeralds is 8 : 3, what is the least number of jewels that could be in the tiara ? 16 22 40 53 67 Let the no. of diamonds, rubies, emeralds be d,r,e. d/r=5/6 ====> d=(5/6)*r r/e=8/3 ====> e=(3/8)*r So the ratio of diamonds, rubies and emeralds is: 5/6 : 1 : 3/8 ==> 20 : 24 : 9 -----> this is the same ratio expressed in terms of the smallest integers possible. For the least no. of jewels, d=20,r=24,e=9 So, least possible total no. of jewels=20+24+9=53. Hey Priya, could you follow my solution? Quote Link to comment Share on other sites More sharing options...
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