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

\{

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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