Posted by: so s. on 26.11.2020
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Let the man's age be x ans son's age be y
Man's age 18 years ago = (x-18)
Son's age 18 years ago = (y-18)
x= 2y -----------------(1)
(x-18) = 3(y-18) ----------(2)
Substituting value of x from (1) in (2)
2y-18 = 3y - 54
y = 54-18 = 36
y = 36
x= 72
Man' age = 72
Son's age = 36
Sum of age = 72+36 = 108
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Let us start from the first statement given, Age of man is 3 times that of his son
i.e, m=3s ----------(1)
(note that, m- age of man 18 years ago and s-age of son 18 years ago)
Also given in the question that, Now age of man is twice of son
present age of man= m+18, similiarly present age of son=s+18
Now, m+18=2(s+18)
i.e, m+18=2s+36 --------------(2)
Now substituting(1) in(2)
3s+18=2s+36
then we get s=18, and m=3s=54
then the present age of man m+18=72
present age of son s+18=36
then sum of ages=72+36=108
Let, their present ages be x yrs & 2x yrs respectively.
So, the sum of their ages = 3x yrs.
18 yrs ago, their ages were (x-18) yrs & (2x - 18) yrs respectively.
ATP, 2x - 18 = 3.(x - 18)
=> 3x - 2x = 54 - 18 => x = 36
Thus, 3x = 3x18 = 108.
Hence, the sum of the ages of the man and his son is 108 years. (Ans.)
Let x be the present age of the father and y be the present age of the son.
According to the first statement,
x-18 = 3* (y-18) ------ (1)
According to the second statement,
x = 2y -------- (2)
Substitute (2) in (1), it becomes
2y-18=3y - 3*18
3y-2y = 3*18 - 18
y = 2*18
y = 36.
Substitute y = 36 in Eq. (2)
x= 2*36 = 72
So the father present age is 72 years and the Son's age is 36 years.
Sum of their present ages = 72+36 = 108 years.
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Let the age of the father be be f and that of his son be s
Then f-18 = 3(s-18) and f= 2s
Solving we get f = 72 and m= 36 years
So currently f + s = 108 years
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Let, the present age of man = x years
and the present age of the son = y years
According to 1st condition, we can write
(x-18) = 3(y-18)
or x-3y = -36........(1)
Now, according to second condition, we can write
x = 2y.........(2)
using it in Eq. (1), we get
2y-3y = -36
or y = 36
Using it in Eq. (2), we get
x = 2* 36 = 72
Hence, the sum of the present ages of father and son = x+y = 72+36 = 108 years.
Let Present age of man = m
The present age of son = s
18 Years ago man was thrice as of son, So,
m-18 = 3(s-18) ---------Equation1
Present age of man is twice of son, So , m = 2s
Substituting m = 2s in equation1
We get s = 36, m = 72
Their sum, m+ s = 108
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Suppose the present age of the son is=x, So the present age of the father is=2x.
18 years ago the age of the son was=x-18. So the age of the father was =2x-18.
According to the condition, 2x-18=3(x-18).
Solving the eqation we get x=36.
So the present age of the son is 36 years and the present age of the father is 2x that is 72 years.
The sum of present ages of son and father is 36+72=108 years.
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