December 3, 2024 · 6:54 am

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Puzzle Page 1

If $\frac{a+b+2c}{a+b-c}=\frac{31}{15}$ , what does $\frac{a+b}{c}$ equal?

$\begin{equation*}15(a+n+2c)=31(a+b-c)\end{equation}$

$\begin{equation*}15a+15b+30c=31a+31b-31c)\end{equation}$

$\begin{equation*}61c=16a+16b)\end{equation}$

$\begin{equation*}\frac{61}{16}=\frac{a+b}{c}\end{equation}$

Two positive numbers are such that their difference, their sum, and their product are in the ratio $2:5:21$ . What is the smaller of the two numbers?

Let $x$ and $y$ be the two numbers. Then

(1) $\begin{equation*}x-y=2k\end{equation*}$

(2) $\begin{equation*}x+y=5k\end{equation*}$

(3) $\begin{equation*}xy=21k\end{equation*}$

Add equation $1$ and $2$ together to eliminate the $y$

$\begin{equation*}2x=7k\end{equation}$

(4) $\begin{equation*}x=\frac{7k}{2}\end{equation*}$

From $2$ $=5k-x$ , substitute for $y$ into equation $3$ .

(5) $\begin{equation*}x(5k-x)=21k\end{equation*}$

Substitute $x=\frac{7k}{2}$ into equation $5$ .

$\begin{equation*}\frac{7k}{2}(5k-\frac{7k}{2})=21k\end{equation}$

$\begin{equation*}\frac{35k^2}{2}-\frac{49k^2}{4}=21k\end{equation}$

$\begin{equation*}\frac{70k^2}{4}-\frac{49k^2}{4}=\frac{84k}{4}\end{equation}$

$\begin{equation*}21k^2-84k=0\end{equation}$

$\begin{equation*}21k(k-4)=0\end{equation}$

Hence, $k=0$ or $k=4$ .

When $k=4$ , $x=\frac{7\times 4}{2}=14$ and $y=5\times 4-14=6$

Therefore the smaller number is $6$ .

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Filed under Algebra, Puzzles, Ratio, Solving Equations

Tagged as problem solving, ratio, Solving Equations

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