Question

Ten years ago, the age of a man was 20 years less
than 6 times his son's age. Ten years hence, his
age will be 30 years less than thrice his son's age.
After how many years from now will their combined
age be 90 years?
(A) 5 (B). 10 (C) 15
(D) 20​

Answer 1

Step-by-step explanation:

Let the man's age be m

let the son's age be s

according to first condition

m-10+20 = 6(s-10)

m+10+30 = 3(s+10)

on solving these equations we get,

m = 40

s = 20

let after x years their age will be 90 years

(m+x) + (s+x) = 90

40+20+2x = 90

x= 15 years

so after 15 years their age will have a sum 90

Answer 2

Answer:

er will be twice his age goes up by one year, but since the son was a year older at the start, the age at which the father is twice as old always happen 35 years later, or 25 years from now.

There are actually two sets of equations: One for ten years ago and one for the future date when the father is twice the age of the son.

Let S be the son’s age ten years ago and F be the father’s age ten years ago.

The first equation is:

[1] F = 2S + 35

The second set of equations are for the future, when the father’s age is twice that of the son.

Let n be the number of years from now. n=0 means now, n=1 is a year from now, etc.

For the second set of equations, there are three equations:

The son’s age S’:

[2] S’ = S + 10 + n

The father’s age F’:

[3] F’ = F + 10 + n

The relationship between them:

[4] F’ = 2S’

Write [4] in terms of [2] and [3]:

[5] F + 10 + n = 2 (S + 10 + n)

Simplify:

F + 10 + n = 2S + 20 + 2n

[6] F = 2S + 10 + n

Substitute F from [1]:

[7] 2S + 35 = 2S + 10 + n

2S on both sides cancel out:

[8] 35 = 10 + n

n = 35 - 10 = 25

Step-by-step explanation:

hope it helps you dear ❤️❤️