SOLUTION: Hello.
I have been struggling with a question for a while and I am getting frustrated. The question is: "The sum of the ages of John and Mary is 32. Four years ago, John was twice
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I have been struggling with a question for a while and I am getting frustrated. The question is: "The sum of the ages of John and Mary is 32. Four years ago, John was twice
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Question 1052923: Hello.
I have been struggling with a question for a while and I am getting frustrated. The question is: "The sum of the ages of John and Mary is 32. Four years ago, John was twice as old as Mary. Find the present age of each."
I have tried sum of 32, then divide by 2 = 16. Then 16 - 4 = 12 and then I get stuck. Please can you help? Thank you :) Found 3 solutions by ewatrrr, MathTherapy, addingup:Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! x John's age..the (32-x) is Mary's age NOW
4 yrs ago...
(x-4) = 2[(32-x) - 4}
x - 4 = 64 - 2x - 8
3x = 60
x = 20, John's age. Mary's age is 12
And...4 yrs ago...
16 = 2*8 CHECKS!
You can put this solution on YOUR website!
Hello.
I have been struggling with a question for a while and I am getting frustrated. The question is: "The sum of the ages of John and Mary is 32. Four years ago, John was twice as old as Mary. Find the present age of each."
I have tried sum of 32, then divide by 2 = 16. Then 16 - 4 = 12 and then I get stuck. Please can you help? Thank you :)
Let John’s age be J
Then Mary’s is: 32 – J
We then get: J – 4 = 2(32 – J – 4)
J – 4 = 2(28 – J)
J – 4 = 56 – 2J
J + 2J = 56 + 4
3J = 60
J, or John is:
Mary is:
You can put this solution on YOUR website! The sum of the ages:
J+M = 32 Subtract M on both sides
J = 32-M (1)
Four years ago John was 2*Mary:
J-4 = 2(M-4) (2)
In equation (2) substitute the value of J per (1):
32-M-4 = 2M-8
28-M = 2M-8
36-M = 2M
36 = 3M flip the equation so we have the unknown of the left
3M = 36 divide both sides by 3
M = 12
and J = 32-M = 32-12 = 20
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Check
20-4 = 2(12-4)
16 = 16 correct