Question

Jenny is $9$ years younger than Sammie.

Five years ago Sammie was $4$ times older than Jennie.

How old is each now?

Answer 1

Step-1: Form an algebraic equation:

Given that Jenny is $9$ years younger than Sammie.

Let present age of Sammie be $\mathrm{x}$ then age of Jennie will be $\mathrm{x}-9$.

Also given that $5$ years ago Sammie was $4$years older than Jennie.

Age of Sammie $5$ years ago will be $\mathrm{x}-5$ .

Age of Jenny will be $\mathrm{x}-9-5=\mathrm{x}-14$.

$5$ years ago Sammie was $4$ times older than Jennie.

Therefore , $x-5=4\times \left(x-14\right)$ is the required equation.

Step 2.Solve the equation.

$$\begin{gathered} \mathrm{x}-5=4\times \left(\mathrm{x}-14\right) \\ \mathrm{x}-5=4\mathrm{x}-56 \\ \mathrm{x}-4\mathrm{x}=-56+5 \\ -3\mathrm{x}=-51 \\ \mathrm{x}=\frac{-51}{-3} \\ \mathrm{x}=17 \end{gathered}$$

The present age of Sammie is $17$ years.

Substitute the value of $\mathrm{x}$ in Jennie's age that is $\mathrm{x}-9$.

$17-9=8$

Hence, Present age of Jammie is $8$ years.