Question

In 1985 Maria was 3 times as old as Sue. Five years earlier the sum of their ages was 18. When was Maria born?

Answer 1

Answer: 1964

Explanation:

to solve this question we should assign an equation for each year :

let x be Maria's age and y Sue's age.

• 1985 x+5=3(y+5) wich means: x-3y-10=0
• 1980 x+y= 18

Now let's solve each equation by trying to minimise the number of variables we are dealing with

how ?

let's multiply x+y= 18 by -1

we get -x-y=-18

Now let's add this last equation to x-3y-10 = 0 to get rid of x

• x-3y-10-x-y = -18
• -4y-10= -18 wich means : 4y+10 = 18
• so y= 2

so Sue's age is 2 .

let's calculate x the age of Maria by replacing y by 2 in x+y= 18

• x+2 = 18
• so x = 16

Let's check : in 1985 Maria's age is 21 and sue's age is 7

• 7*3 = 21

so it's true.

Now let's calculate Maria's year of birth :

In 1985 Maria was 21 so :

• 1985-21 = 1964

Maria was born in 1964

• The trick is to imaginate the situation and represent it :

Answer 2

Answer: Maria was born in 1964

Explanation:

Let Sue's age be x

Maria's age be 3x

Five years ago, their ages were

x-5, 3x-5

So

(x-5) + (3x-5) = 18

x-5 + 3x - 5 = 18

4x -10 = 18

4x = 18+10

x = 28/4 = 7

Sue's age x-5 = 7-5 = 2

Maria's age = 3x-5 = 3(7)-5 = 21-5=16

Maria was born in 1980 - 16 = 1964