Question
A man distributed some candies to his three sons A, B
and C. A, being the eldest got two times the number of candies that C got while A and B together have 62 candies. B and C have 20/33 times the number of candies that A and C together have. Find the number of candies that B has.Solution
A = 2C B + C = 20/33 (A + C) ∴B + C = 20/33 ( 2C + C ) ⇒B + C = 20/11C ⇒B = 9/11C Now, it’s given that A +B = 62 ∴2C + 9/11C = 62 31/11C = 62 ⇒C = 22 ⇒B = 9/11 X 22 = 18 candies
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