Question 39

Challenge question #39 (05/12/2016)

Junior Question

A motorist, after having his car overhauled, finds that he can travel an additional $5$ km per litre of petrol. He is now using $2$ litres less for a journey of $200$ km than previously.

How many km per litre was he getting previously?

Answer

The motorist was previously getting $20$ km per litre of petrol.

Worked Solution

Suppose the car was originally travelling $x$ km per L before the overhaul.

So, it travels $(x+5)$ km per L after the overhaul.

The journey of $200$ km required $\footnotesize\dfrac{200}{x}$ L of petrol before the overhaul, and $\footnotesize\dfrac{200}{(x+5)}$ L afterwards.

Hence ${\footnotesize\dfrac{200}{x}-\dfrac{200}{(x+5)}}=$

$\ 2$

$\therefore\enspace\ 200(x+5)-200x=$

$\ 2x(x+5)$

$\therefore\enspace\ 1000=$

$\ 2x^{2}+10x$

$\therefore\enspace\ x^{2}+5x-500=$

$\ 0$

$\therefore\enspace\ (x-20)(x+25)=$

$\ 0$

$\therefore\enspace\ x=$

$\ 20$ $\{$as $x\gt0\}$

Thus the motorist was previously getting $20$ km per litre of petrol.

Senior Question

Suppose $a+b=1$ and $a^{2}+b^{2}=3$.

Find the value of $a^{5}+b^{5}$.

Answer

$a^{5}+b^{5}=11$

Worked Solution

$(a+b)^{3}=$

$\ a^{3}+3a^{2}b+3ab^{2}+b^{3}$

$\therefore\enspace\ (a+b)^{3}=$

$\ a^{3}+b^{3}+3ab(a+b)\quad....$ (1)

Now $(a+b)^{2}=$

$\ a^{2}+b^{2}+2ab$

$\therefore\enspace\ 1^{2}=$

$\ 3+2ab$

$\therefore\enspace\ 2ab=$

$\ -2$

$\therefore\enspace\ ab=$

$\ -1\quad....$ (2)

Thus in (1), $13=$

$\ a^{3}+b^{3}+3(-1)(1)$

$\therefore\enspace\ 1=$

$\ a^{3}+b^{3}-3$

$\therefore\enspace\ a^{3}+b^{3}=$

$\ 4\quad....$ (3)

Now $(a+b)^{5}=$

$\ a^{5}+5a^{4}b+10a^{3}b^{2}+10a^{2}b^{3}+5ab^{4}+b^{5}$

$\therefore\enspace\ 1^{5}=$

$\ (a^{5}+b^{5})+5ab(a^{3}+b^{3})+10a^{2}b^{2}(a+b)$

$\therefore\enspace\ 1=$

$\ a^{5}+b^{5}+5(-1)(4)+10(-1)^{2}(1)\quad\{$using (2) and (3)$\}$

$\therefore\enspace\ a^{5}+b^{5}-20+10=$

$\ 1$

$\therefore\enspace\ a^{5}+b^{5}-10=$

$\ 1$

$\therefore\enspace\ a^{5}+b^{5}=$

$\ 11$

Question 39

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