SOLUTION: Loraine is eighteen years older than her cousin. She was three times as old one year ago. How old are they now?

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Question 1190874: Loraine is eighteen years older than her cousin. She was three times as old one year ago.
How old are they now?
Found 3 solutions by math_tutor2020, greenestamps, MathTherapy:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

c = cousin's current age
c+18 = Loraine's current age, since she is 18 years older.

Rewinding the clock 1 year ago, we have these past ages
c-1 = cousin's past age
(c+18)-1 = c+17 = Loraine's past age

Last year she was 3 times as old as her cousin, so,
her past age = 3*(cousin's past age)
c+17 = 3*(c-1)
c+17 = 3c-3
17+3 = 3c-c
20 = 2c
2c = 20
c = 20/2
c = 10

Her cousin is currently 10 years old.
Loraine is currently c+18 = 10+18 = 28 years old.

One year ago, her cousin was 10-1 = 9 years old and Loraine was 28-1 = 27 years old.
The jump from 9 to 27 is "times 3" to help fully confirm we have the correct ages.

Answers:
Loraine is currently 28 years old
Her cousin is currently 10 years old
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Let x and 3x be their ages 1 year ago.

The difference in their ages (now and always) is 18:





Their ages 1 year ago were x=9 and 3x=27.

ANSWER: Their ages now are 10 and 28
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Loraine is eighteen years older than her cousin. She was three times as old one year ago.
How old are they now?
Difference in ages will ALWAYS be 18, with Loraine being older

Let the cousin's age be C
Then Loraine's is C + 18
We then get: C + 18 - 1 = 3(C - 1)
                 C + 17 = 3C - 3
                 C - 3C = - 3 - 17
                   - 2C = - 20
The cousin's age, or 

You should now be able to find Loraine's age.
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