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Question: 18 years ago a man was 3 times as old as his son now the man is twice as old as his son the sum of the present ages of the man and his son is

Posted by: so s. on 26.11.2020

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Answer:
If present age of son is X Then present age of father is 2x 18 years ago Son was X - 18 Father was 2x - 18 Given that 2x - 18 = 3(X - 18) Hence 2x - 18 = 3x - 54 Hence 3x - 2x = 54 - 18 Hence X = 36 Sum of their present ages = X + 2x = 36 + 72 = 108
Answer:
Solve
Answer:
Let, son's and man's present ages be x and 2x respectively, 18 years ago, son's age was (x-18)yrs and man's age was (2x-18) yrs, ATP, (2x-18)=3(x-18)=>3x-2x=54-18=>x=36, 2x=72, therefore, son's present age=36yrs and man's present age=72 yrs
Answer:

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

Answer:
माना कि पुत्र की वर्तमान आयु =x वर्ष है, अतः पिता की वर्तमान आयु 2x है 18 वर्ष पहले पुत्र की आयु x-18 वर्ष थी और पिता की आयु 2x-18 वर्ष थी प्रश्न अनुसार 2x-18 = 3(x-18) 2x-18 = 3x-54 54-18 = 3x-2x 36 = x अतः पुत्र की वर्तमान आयु 36 वर्ष होगी और पिता की वर्तमान आयु 2x जो कि 72 वर्ष होगी अतः उन दोनों की वर्तमान आयु का योग 36+72 = 108 है
Answer:

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

Answer:
Let assume present son age is = × So father present age is =2x Now 18 years ago Father age was= 2x -18 Son age was = x-18 Now, according to Q .. 18 years ago father age is three times of son age So , 2x-18 = 3(×-18) 2x-18=3x-54 ×= 36 , So son present age is 36 . And father present age is 2x= 2×36 =72 So sum of ages is = 36+72 =108 .
Answer:

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.)

Answer:
Let the present age of son be x years Then according to given condintion, Father's present age will be 2x years 18 years ago, son's age was( x-18) years And father's age was (2x-18) years According to given condition, 2x-18=3(x-18) Solving this equation, we get x=54-18 x=36 years So, son's present age is 36 years and father's present age is 72 years So the required sum is 36+72=108 years.
Answer:

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.

Answer:

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

Answer:

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.

 

Answer:

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

Answer:

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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