Three numbers whose sum is 54 are such that one is double another and triple the other. What are the three numbers?

Dec 8, 2016

9, 18, 27

Explanation:

Let the unknown numbers be a, b, c where $a < b < c$

The question is not explicit enough to have no doubt about all the relationships.

No doubt about this one: $\text{ } a + b + c = 54$

Assumption

$b = 2 a$
$c = 3 a$

$a \text{ "+" "b" "+" "c" } = 54$
$\downarrow \text{ "darr" } \downarrow$
$a \text{ "2a" "3a" } = 54$

$\implies 6 a = 54$

Divide both sides by 6

$\frac{6}{6} a = \frac{54}{6} = 9$

$\textcolor{b l u e}{a \text{ "=" } \textcolor{w h i t e}{2} 9}$
$\textcolor{b l u e}{b = 2 a = \text{ } 18}$
ul(color(blue)(c=3a=" "27) larr" add"
$\text{ } 54$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Suppose you meant

$b = 2 a$
$c = 3 b = 3 \left(2 a\right) = 6 a$

Then proceed using the same approach as the solution above.

Dec 8, 2016

$\frac{324}{11}$, $\frac{162}{11}$, $\frac{108}{11}$

Explanation:

I suspect that the question is incorrectly posed. For example, if two of the numbers were respectively double and triple the other number, then the numbers would be $9 , 18 , 27$.

With the question as posed, the three numbers take the form:

$x$, $\frac{x}{2}$ and $\frac{x}{3}$ for some constant $x$ to be determined.

So their sum is:

$x + \frac{x}{2} + \frac{x}{3} = \frac{6 x + 3 x + 2 x}{6} = \frac{11 x}{6} = 54$

Hence:

$x = \frac{6 \cdot 54}{11} = \frac{324}{11}$

and the three numbers are:

$\frac{324}{11}$, $\frac{162}{11}$, $\frac{108}{11}$