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Always Two

Find all the triples of numbers a, b, c such that each one of them plus the product of the other two is always 2.

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Not Continued Fractions

Which rational numbers cannot be written in the form x + 1/(y + 1/z) where x, y and z are integers?

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Coffee

To make 11 kilograms of this blend of coffee costs £15 per kilogram. The blend uses more Brazilian, Kenyan and Mocha coffee... How many kilograms of each type of coffee are used?

Alien Currency

Age 14 to 16 Short Challenge Level:

Answer: $19$


Elimination multiplying equations by $3$ and $8$
               $\begin{align}3g+8b=46\quad\Rightarrow\quad&24g+64b=368\\
8g+3b=31\quad\Rightarrow\quad&\ \ \underline{24g+9b=\ \ 93}\\
\text{subtracting gives:}\quad&\ \ \ \qquad 55b=275\\
&\ \ \ \Rightarrow\ \ 11b=55\\
& \ \ \ \Rightarrow\quad\ \ b=5\end{align}$
And so $3g+40=46\Rightarrow g=2$
$\therefore 2g+3b= 4+15=19$


Elimination by combining equations more than once
$$\begin{align} 3g+8b=46&\\
\underline{+\qquad 8g+3b=31}&\\
11g+11b=77&\\
\Rightarrow \quad g+b=\ \ 7&\\
\Rightarrow 3g+3b=21\end{align}$$
                     $\begin{align} 3g+8b=46&\\
\underline{-\qquad 3g+3b=21}&\\
5b=25&\\
\Rightarrow \quad b=\ 5\ &\end{align}$              $\begin{align} 8g+83=31&\\
\underline{-\qquad 3g+3b=21}&\\
5g=10&\\
\Rightarrow \quad g=\ 2\ &\end{align}$

Therefore $2g + 3b = 19$.


Getting $2g+3b$ without finding $g$ and $b$
$\begin{align} 3g+8b=46&\\
\underline{+\qquad 8g+3b=31}&\\
11g+11b=77&\\
\Rightarrow \quad g+b=\ \ 7& \end{align}$


        $3g+8b$
$+ \quad (\ g\ +\ b\ )\times$ some    to get blues and greens in ratio $2:3$
                                    difference will be $5$ since $8$ and $3$ have difference of $5$
                                    $10:15 = 3+7:8+7$

$\begin{align} 3g+8b=46&\\
\underline{+\qquad 7g+7b=49}&\\
10g+15b=95&\\
\Rightarrow \ \ 2g+3b=19& \end{align}$


This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.