SOLUTION: Four years ago,a mother was 4 times as old as her son.In 4 years time, the sum of their ages will be 56years .Find the age of the mother when the son was born

Algebra.Com
Question 1199026: Four years ago,a mother was 4 times as old as her son.In 4 years time, the sum of their ages will be 56years .Find the age of the mother when the son was born
Found 5 solutions by Theo, math_tutor2020, ikleyn, MathTherapy, josgarithmetic:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
x = the mother's age today.
y = the son's age today.

4 years ago, the mother was 4 times as old as her son.
the equation for that is:
(x - 4) = 4 * (y -4)
simplify to get:
x - 4 = 4 * y - 16
solve for x to get:
x = 4 * y - 12

in 4 years time, the sum of their ages will be 56.
the equation for that is:
x + 4 + y + 4 = 56
combine like terms to get:
x + y + 8 = 56
solve for x + y to get:
x + y = 48

replace x in the second equation with 4 * y - 12 from the first equation to get:
4 * y - 12 + y = 48
combine like terms and add 12 to both sides of the equation to get:
5 * y = 60
solve for y to get:
y = 12

since x + y = 48, then x = 36.
the mother's age is currently 36
the son's age is currently 12

4 years ago, the mother was 32 and the son was 8.
32 is 4 * 8.
in 4 years time, the mother will be 40 and the son will be 16.
40 + 16 = 56.
the requirements of the problem are satisfied when the mother's current age is 36 and the son's current age is 12.

the son was born 12 years ago.
12 years ago, the mother was 24.
the mother was 24 years old when the son was born.
that's your solution.



Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

x = the son's age 4 years ago
4x = the mother's age 4 years ago

Fast-forward 4 years to the present day
x+4 = the son's present age
4x+4 = the mother's present age

Then fast-forward another 4 years
(x+4)+4 = x+8 = the son's age 4 years into the future
(4x+4)+4 = 4x+8 = the mother's age 4 years into the future

These future ages add to 56
(x+8)+(4x+8) = 56
5x+16 = 56
5x = 56-16
5x = 40
x = 40/5
x = 8

The son was 8 years old four years ago.
His mother was 4x = 4*8 = 32 years old four years ago.

Fast-forward 8 years and the son and mother will be 8+8 = 16 and 32+8 = 40 respectively.
We can see that 16+40 = 56 to help confirm we have the correct ages.

Since the son was 8 years old when she was 32, we can rewind 8 years to find that the mother was 32-8 = 24 years old when her son was born.

Answer: 24 years old
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
Four years ago, a mother was 4 times as old as her son.
In 4 years time, the sum of their ages will be 56 years.
Find the age of the mother when the son was born
~~~~~~~~~~~~~~~~~~


            Below is the mental solution. 


Since in 4 years the sum of their ages will be 56 years,
      it means that the sum of their ages now is 56 - 4 - 4 = 48.


Since now the sum of their ages is 48 years,
      it means that the sum of their ages four years ago was 48 - 4 - 4 = 40.


Since four years ago the sum of their ages was 40 and the mother was 4 times as old as her son,
      it implies that at that time the son was 8 years old and the mother was 32 years old
      (it is the only place, where minimal mental step is required,
       but for most people it is a quite affordable effort).


Since the difference of their ages was 32-8 = 24 at that time, it means that the mother is 24 years older than the son.


In other terms, it means that the mother was 24 years old when the son was born.    ANSWER

Solved.

The total time to solve this problem in this way is about and around of 10 seconds.


/////////////////


Regarding the post by @josgarithmetic, I recommend you to tape your eyes and plug your ears

in order for do not see and do not hear that gibberish, which this person writes in his posts.


In more soft terms, simply ignore his writing, for safety of your mind.



Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
Four years ago,a mother was 4 times as old as her son.In 4 years time, the sum of their ages will be 56years .Find the age of the mother when the son was born 
Since the sum of their ages in 4 years' time will be 56, sum of their ages now is 56 - 2(4) = 56 - 8 = 48.

Four years ago, the sum of their ages was 48 - 2(4) = 48 - 8 = 40. Since at that time, the mother was 4 times
as old as the son, and with the son's age being S at that time, we get: S + 4S = 40
5S = 40
Son's age then (4 years ago) was , and mother's was, 40 - 8 = 32.
Difference in their ages, 4 years ago, and as long as they're alive: 32 - 8 = 24

With the difference in their ages being 24, the mother was 24 when the son was born.
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
m Mother
c Son

Description becomes system:


The question means, find m when c is 0.



Mother was 48 y.o. when son was born.
RELATED QUESTIONS

1) A mother is five times as old as her daughter. If the sum of their ages 6 years ago... (answered by addingup)
Five years ago a mother was twice as old as her son in six years time the sum of their... (answered by stanbon)
Five years ago, a mother was twice as old as her son. In 6 years time the sum of their... (answered by Alan3354)
Five years ago, a mother was twice as old as her son. In 6 years time, the sum of the... (answered by algebrahouse.com)
Five years ago, a mother was twice as old as her son. In 6years time,the sum of their... (answered by mark_henrhiz)
five years ago, a mother was thrice as old as her son. in 6 yrs time the sum of their... (answered by ikleyn)
Five years ago a mother was twice as old as her own son in 6 years time the sum of their... (answered by ikleyn,josgarithmetic,greenestamps)
1) A father is five times as old as his son. If the difference in their ages is 32 years, (answered by ikleyn)
five years ago a mother was twice as old ss her son. in 6 yrs time, the sum of their age (answered by jorel555,advanced_Learner)