SOLUTION: 5 year ago a man was 7 times of his son 5year hence he was 3 times of his son find their present age

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Question 821759: 5 year ago a man was 7 times of his son 5year hence he was 3 times of his son find their present age
Found 2 solutions by TimothyLamb, jsmallt9:
Answer by TimothyLamb(4379) About Me  (Show Source):
You can put this solution on YOUR website!
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x - 5 = 7(y - 5)
x + 5 = 3(y + 5)
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put the system of linear equations into standard form
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x - 5 = 7y - 35
x + 5 = 3y + 15
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x - 7y = -30
x - 3y = 10
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copy and paste the above standard form linear equations in to this solver:
https://sooeet.com/math/system-of-linear-equations-solver.php
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answer:
x= man = 40
y= son = 10
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Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
When I solve word problems I often like to set my variable to the smallest unknown (if I know what that is). This way, the other expressions I use will have additions and/or multiplications (which are often easier to work with than subtractions and/or divisions).

The smallest unknown in this problem would be:
Let x = the son's age 5 years ago

Also, this problem can be solved using either 1 or 2 variables. If we use 1 variable we only need one equation. If we use two variables then we need two equations. I will solve this using 1 variable.

From "5 year ago a man was 7 times of his son" we can say
7x = the man's age five years ago.
And, if the son was x years old 5 years ago, then:
x + 5 = son's age now
And, if the father was 7x years old 5 years ago, then:
7x + 5 = man's age right now
And we can rewrite "5year hence he was 3 times of his son" ("hence" means later) as: "Now, the man's age is three times his son's age". And from this we can write the equation:
7x + 5 = 3(x + 5)

Now we solve. Simplifying the right side:
7x + 5 = 3x + 15
Subtracting 3x from each side:
4x + 5 = 15
Subtracting 5 from each side:
4x = 10
Dividing by 4:
x+=+10%2F4+=+5%2F2+=+2.5

Looking back we can see that x is the son's age 5 years ago. So his present age would be x + 5 or 7.5. The father's present age, 7x + 5, would be 7(2.5) + 5 = 17.5 + 5 = 22.5. So their present ages are:
son: 7.5
father: 17.5

I strongly suspect that there was something wrong with what you posted:
  • Usually these problems don't end up with fractions/decimals for answers. (I have checked and the answers above are correct if the problem was posted accurately.)
  • The father is quite young! These answers show that the father must have been 15 years old when the son was born!?
I hope that what I've done here will help you solve the correct problem.

P.S. The other tutor's answer is not correct. After all, 40 is not 3 times 10!!

P.P.S. If you want to use two variables:
Let x = the son's age 5 years ago.
Let y = the father's age 5 years ago.
This makes their current ages:
x + 5 = son's current age
y + 5 = father's current age

From "5 year ago a man was 7 times of his son" we can write:
y = 7x
From "5year hence he was 3 times of his son" we can write:
y + 5 = 3(x + 5)

So we end up with the system:
y = 7(x + 5)
y + 5 = 3(x + 5)
which has the same solution as we found earlier.