akagboy Posted May 18, 2005 Share Posted May 18, 2005 Ann, Bob, Chris, Dana, and Eric are different ages such that their average age is 16. Bob is twice as old as Ann was three years ago. Eric is nine years younger than Bob. The sum of Chris and Ann's ages is equal to twice Dana's age. Dana is one year older than Ann. How old is Eric? A) 13 B) 14 C) 15 D) 22 E) 23 Can some one help Solve this problem B Quote Link to comment Share on other sites More sharing options...
txaggie Posted May 19, 2005 Share Posted May 19, 2005 I get A as the answer. Maybe..I maybe making a mistake in adding the numbers....so I will check and post my response. Quote Link to comment Share on other sites More sharing options...
stick boy Posted May 20, 2005 Share Posted May 20, 2005 I get A) 13 sorry for the long winded solution: A = Ann B = Bob C = Chris D = Dana E = Eric (A+B+C+D+E)/5 = 16 , Therefore total age = 80 B=2(A-3) E=B-9 C+A = 2D D= A+1 What is age of E? A + B + C + D + E = 80 relate each age in terms of A: B = 2A - 6 C+A = 2(A+1) C+A=2A+2 C=A+2 E = (2A-6) -9 E = 2A-15 A+B+C+D+E = A + (2A - 6) + (A+2) + (A+1) + (2A-15) = 80 7A = 98 A = 14 If A = 14 Then B = 22 And E = 13 Eric is 13 years old. Quote Link to comment Share on other sites More sharing options...
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