Question

solve question no 5 it is easy or not

Answer 1

let fathers age=x

let sons age=y

fathers age before 10 yrs= x-10

sons age before 10 yrs=y-10

according to question,

x-10=12(y-10)

x=12y-120+10

x=12y-110-----(1)

age of father after 10yrs=x+10

age of son after 10 yrs=y+10

according to question,

x+10=2(y+10)

x+10=2y +20

subsitituting (1)

12y-110+10=2y+20

12y-100=2y+20

12y-2y=20+100

10y=120

y=120/10

y=12

therefore,x= 12y-110

x=12(12)-110

x=144-110

x=  34

So answer is 34.

Hope it helpzz let me know

Answer 2

Ten years ago, A father was 12 times as old as his son and 10 years hence he would be twice as old as his son will be then.
Find their present ages.

$\rule{200}{2}$

Let the age of father be x years and age of son be y years.

$\rule{200}{2}$

CASE 1 :-

10 years ago :-

Age of father = x - 10

Age of son = y - 10

So,

As per question we have :-

$x - 10 = 12(y - 10) \\ \\ \\ x - 10 = 12y - 120 \\ \\ \\ x - 12y = - 120 + 10 = - 110 \\ \\ \\ x - 12y = ( - 110) \cdots(i)$

$\rule{200}{3}$

CASE 2 :-

10 years hence :-

Age of father = x +10

Age of son = y +10

As per question :-

$x + 10 = 2(y + 10) \\ \\ \\ x + 10 = 2y + 20 \\ \\ \\ x - 2y = 20 - 10 = 10 \\ \\ \\ x - 2y = 10 \cdots(ii)$

$\rule{200}{2}$

Subtracting the two equations we have :-

$x - 12y = - 110 \\ \\ x - 2y \: = \: \: \: \: \: \: \: 10 \\ - \: \: \: + \: \: \: \: \: \: \: \: \: \: - \\ - - - - - - - - - \\ - 10y = - 120 \\ \\ \\ \\y = 12 \\ \\ x - 2y = 10 \\ x = 10 + 2y \\ x = 10 + 2(12) \\ x = 10 + 24 \\ x = 34$

$\rule{200}{2}$

$\boxed{x=34}$

$\boxed{y = 12}$

$\rule{200}{2}$