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Alex is 12 years older than George, Carl is three times older than Alex, The sum of their ages is 68. Find the ratio of George's age to Carl's age to Alex's age.

Sagot :

Firstly, let x represent Alex's age, y represent George's age and z represent Carl's age.

from the question;

Alex is 12 years older than George, So;

[tex]x=y+12\ldots\ldots\ldots\ldots.1[/tex]

Carl is three times older than Alex, So;

[tex]z=3x\ldots\ldots\ldots..2[/tex]

The sum of their ages is 68, So;

[tex]x+y+z=68\ldots\ldots\ldots\ldots\ldots3[/tex]

Now we have three equations and three unknowns, so it is solvable.

Let us substitute equation 2 into equation 3; that is replace z with 3x in equation 3.

[tex]\begin{gathered} x+y+3x=68 \\ 4x+y=68\ldots\ldots\ldots\ldots\ldots\ldots4 \end{gathered}[/tex]

Next, let us substitute equation 1 into equation 4. that is replace x with y+12 in equation 4.

[tex]\begin{gathered} 4(y+12)+y=68 \\ 4y+48+y=68 \\ 5y+48=68\ldots\ldots\ldots.5 \end{gathered}[/tex]

we can now solve for the value of y from equation 5.

[tex]\begin{gathered} 5y+48=68\ldots\ldots\ldots.5 \\ \text{subtract 48 from both sides.} \\ 5y+48-48=68-48 \\ 5y=20 \\ y=\frac{20}{5} \\ y=4 \end{gathered}[/tex]

let us now replace y with 4 in equation 1 to get the value of x. since y = 4;

[tex]\begin{gathered} x=y+12\ldots\ldots\ldots\ldots.1 \\ x=4+12 \\ x=16 \end{gathered}[/tex]

then, since x =16 let us replace x with 16 in equation 2 to get z.

[tex]\begin{gathered} z=3x\ldots\ldots\ldots..2 \\ z=3(16) \\ z=\text{ 48} \end{gathered}[/tex]

so we have;

[tex]\begin{gathered} \text{Alex's age = x = 4 years} \\ George^{\prime}sage_{}=y=16\text{ years} \\ Carl^{\prime}sage=z=48\text{ years} \\ \end{gathered}[/tex]

We now need to find the ratio of George, Carl and Alex's age.

[tex]\begin{gathered} 16\colon48\colon4 \\ \text{dividing through by 4 we have;} \\ 4\colon12\colon1 \end{gathered}[/tex]

So the ratio of their ages are;

[tex]4\colon12\colon1[/tex]

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