One year ago, a father was four times as old as his son. After six years his age exceeds twice his sons age by 9 years. The ratio of their present age is

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Q: 80 (CAPF/2017)
One year ago, a father was four times as old as his son. After six years his age exceeds twice his son’s age by 9 years. The ratio of their present age is

question_subject: 

Maths

question_exam: 

CAPF

stats: 

0,10,6,1,10,4,1

keywords: 

{'ratio': [1, 0, 1, 12], 'age': [2, 1, 1, 2], 'present age': [0, 0, 3, 1], 'father': [15, 3, 10, 14], 'years': [1, 0, 0, 2], 'times': [5, 2, 7, 3], 'year': [27, 4, 33, 49]}

To solve this problem, let`s represent the present ages of the father and son as f and s, respectively.

According to the first condition, one year ago the father`s age was four times the son`s age. This can be expressed as (f-1) = 4(s-1).

According to the second condition, after six years, the father`s age exceeds twice the son`s age by 9 years. Mathematically, this can be written as f+6 = 2(s+6) + 9.

Now, let`s solve these two equations to find the values of f and s.

From the first equation, we get f - 1 = 4s - 4, which simplifies to f = 4s - 3.

Substituting this value in the second equation, we have 4s - 3 + 6 = 2(s + 6) + 9.

Simplifying further, we get 4s + 3 = 2s + 12 + 9.

Combining like terms, we have 4s - 2s = 12 + 9 - 3.

This simplifies to 2s = 18, which means s =

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