SOLUTION: 5 years ago, bob was 2/3 mary's age. In 10 years, he will be 5/6 her age. How old is bob now?
a. 10 b. 15 c. 20 d. 21
How is this set up and how is it solved?
Algebra.Com
Question 295232: 5 years ago, bob was 2/3 mary's age. In 10 years, he will be 5/6 her age. How old is bob now?
a. 10 b. 15 c. 20 d. 21
How is this set up and how is it solved?
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
I'll give you the basics and let's see how you do.
Bob is age b now
Mary is age m now
5 years ago Bob was b-5 and Mary was m-5
In ten years Bob will be b+10 and Mary will be m+10
The biggest mistake that students make is forgetting to keep the ages balanced by adding or subtracting the same number form both.
I'll set up the first equation
5 years ago, bob was 2/3 mary's age
(b-5)=2/3*(m-5)
Now you set up the second equation.
In 10 years, he will be 5/6 her age
(b-5)=2/3*(m-5)
When you write the other equation, remember, they are asking for bob's age not mary's so solve for b
bob will be the younger because he is 2/3 of her age at one point and later 5/6 of her age.
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