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The sum of Joe's and Sheila's ages is 115. Fourteen years ago, Joe was twice as old as Sheila. How old is Sheila now?

User Ritratt
by
4.3k points

1 Answer

5 votes

Answer: Sheila today =
46(1)/(3) yrs old

Explanation:

J + S = 115 v⇒ J = 115 - S

Current Ages Ages 14 years ago

Joe (J) = 115 - S J - 14 = 2(S - 14)

Sheila (S) = S

Substitute J = 115 - S into the "14 years ago" equation

J - 14 = 2(S - 14)

(115 - S) - 14 = 2(S - 14)

111 -S = 2S - 28

111 = 3S - 28

139 = 3S

46
(1)/(3) = S

It is odd that the result was not an integer. I wonder if you meant to type "Joe was twice as old as Sheila is today. That would change the equation to:

J - 14 = 2S

111 - S = 2S

111 = 3S

37 = S

User Chansuk
by
4.4k points