Question

2 years ago ,the age of a father was 3 and a half times the age of his daughter then. 6 years hence ,the age of father will be 10 years more than twice the age of his daughter then. Find the answer
1 let the present age of father be x and that of daughter be y years 2 form 2 equations from the given condition 3 solve the equation and find the answer

Answer 1

Answer:

father's age = 44 years

daughters age = 14 years

Step-by-step explanation:

let father's present age be x

two years ago father's age was x - 2

let daughters age be y

two years ago daughters age was y - 2

father's age was 3 1/2 of daughters age two years ago

x - 2 = 7/2(y - 2)

x - 2 = 7y/2 - 7

x - 2 = (7y - 14)/2

2(x - 2) = 7y - 14

2x - 4 = 7y - 14

2x - 7y = - 14 + 4

2x - 7y = - 10 equ (1)

six years after father's age will be 10 more than twice daughters age

x + 6 = 10 + 2(y + 6)

x + 6 = 10 + 2y + 12

x - 2y = 10 + 12 - 6

x - 2y = 16 equ (2)

multiply equ (2) with 2

2x - 4y = 32. equ (3)

subtract equ (3) - (1)

2x - 4y - (2x - 7y) = 32 - (-10)

2x - 4y - 2x + 7y = 32 + 10

3y = 42

y = 42/3

y = 14

substitute y = 14 in equ (2)

x - 2y = 16

x - 2(14) = 16

x - 28 = 16

x = 16 + 28

x = 44

therefore father's age = x = 44 years

daughters age = y = 14 years

hope you get your answer