six years ago Albert was 1/5 as old as his mother. his mother is now 3 times as old as alert. how old is his mother?

Guest Nov 13, 2015

#1**+5 **

Let his mother's age 6 years ago = x and his age then was (1/5)x

And now we know that :

3 times his age now = his mother's age now....so........

3 [ (1/5)x + 6 ] = x + 6 simplify

(3/5)x + 18 = x + 6 subtract (3/5)x , 6 from both sides

12 = (2/5)x multiply both sides by 5/2

12 [ 5/2] = x = 60/2 = 30

This was his mother's age 6 years ago......so.....she must be 36, now

CPhill
Nov 13, 2015

#1**+5 **

Best Answer

Let his mother's age 6 years ago = x and his age then was (1/5)x

And now we know that :

3 times his age now = his mother's age now....so........

3 [ (1/5)x + 6 ] = x + 6 simplify

(3/5)x + 18 = x + 6 subtract (3/5)x , 6 from both sides

12 = (2/5)x multiply both sides by 5/2

12 [ 5/2] = x = 60/2 = 30

This was his mother's age 6 years ago......so.....she must be 36, now

CPhill
Nov 13, 2015

#2**+5 **

**six years ago Albert was 1/5 as old as his mother. his mother is now 3 times as old as alert. how old is his mother?**

**\(\begin{array}{|l|r|r|} \hline & Albert(a) & Mother(x) \\ \hline \text{Now} & a & \quad (1) \quad x = 3a \\ -6\text{ years} & \quad (2) \quad a-6 = \frac15(x-6) & x -6 \\ \hline \end{array}\\ \begin{array}{lrcl} (1)& x &=& 3a \\ & a &=& \frac{x}{3} \\ \\ \hline \\ (2)& a-6 &=& \frac15 (x-6) \quad & | \quad a = \frac{x}{3}\\\\ & \frac{x}{3}-6 &=& \frac15 (x-6)\\\\ & \frac{x}{3}-6 &=& \frac{x}{5} - \frac{6}{5} \quad & | \quad -\frac{x}{5}\\\\ & \frac{x}{3}- \frac{x}{5}-6 &=& -\frac{6}{5}\\\\ & \frac{x}{3}- \frac{x}{5}-6 &=& -\frac{6}{5} \quad & | \quad +6 \\\\ & \frac{x}{3}- \frac{x}{5} &=& 6-\frac{6}{5} \\\\ & \frac{5x-3x}{15} &=& \frac{30-6}{5} \\\\ & \frac{2x}{15} &=& \frac{24}{5} \\\\ & \frac{2x}{15} &=& \frac{24}{5} \quad & | \quad :2 \\\\ & \frac{x}{15} &=& \frac{12}{5} \quad & | \quad \cdot 15 \\\\ & x &=& \frac{12\cdot 15}{5} \\\\ & x &=& 12\cdot 3 \\\\ & x &=& 36 \\\\ \end{array}\\ \text{His mother is } \mathbf{36} \text{ years old.}\\\\ \begin{array}{lrcl} \\ \hline \\ (1)& a &=& \frac{x}{3} \quad & | \quad x = 36\\\\ & a &=& \frac{36}{3} \\\\ & a &=& 12 \\\\ \end{array}\\ \text{Albert is } \mathbf{12} \text{ years old.} \)**

heureka
Nov 13, 2015