I know the answer through trial and error but how do you set it out in a

L

Learning Student

New member
proper working out!!!(algebra)? here;s the question i got on a practise test. A father is 36 yrs old and his son is a quarter of his. IN how many years will the fathers age be twice and 7 yrs more than the sons age. the options were 11, 13, 10, 14, and none of these. I dodnt know how to set it up to work it out so i just did trial and error but i would like to know the proper algebraic way to solve it for my actual test!! PLEASE HELP
 
????
A father is 36 yrs old and his son is a quarter of his;

so, the son's age is 1/4 of 36 = 9.

let y = number of years

36 + y = 2 • ( 9 + y ) + 7
36 + y = 18 + 2y + 7
36 + y = 25 + 2y
............................subtract y & 25 from both sides of the equation;
36 + y - y - 25 = 25 + 2y - y - 25
11 = y
.....or.....
y = 11
?????

?????let's check:
in 11 years,
the father's age would be 36 + 11 = 47.
the son's age would be 9 + 11 = 20

twice the son's age is 40
47 is 7 more than 40.
so, his father's age is 7 years more than twice the age of his son.
?????
 
Let the no. of yrs. in which the father's age will be twice and 7 yrs more than the sons age be x yrs.

Father's age: now is 36 yrs.
Father's age: then will be (36+x) yrs.

Son's age: now is (36/4 =) 9 yrs.
Son's age: then will be (9+x) yrs.

Then,
Father's age = 2*(Son's age then) + 7
36+x = 2(9+x) + 7
x = 11
 
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