Maths Olympiad Prep

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, 2024

Algebra Difficulty 2.0 Junior Prove it Canada

At Radford Motors, 40504050
trucks were sold. Of the trucks sold, 32%32\% were white, 24%24\% were grey, and 44%44\% were black.

How many white trucks were sold?
If \text{}
14$ of the grey trucks sold were electric, how many trucks sold
were both grey and electric?
In addition to the 40504050 trucks that were sold, there were
kk unsold trucks, all of which were
black. In total, 46%46\% of all
trucks, sold and unsold, were black. Determine the value of kk.

Solution

Of the 40504050 trucks sold,
32%32\% were white or 321004050=1296\dfrac{32}{100}\cdot4050=1296 were
white.
Solution 1:

Of the 40504050 trucks sold, 24%24\% were grey or 241004050=972\dfrac{24}{100}\cdot4050=972 were
grey.

Since 14\dfrac14 of the grey trucks
sold were electric, then 14972=243\dfrac14\cdot972=243 trucks sold were
both grey and electric.

Solution 2:

Since 24%24\% of the trucks sold
were grey, and 14\dfrac14 of those
were electric, then 2410014=6100\dfrac{24}{100}\cdot\dfrac14=\dfrac{6}{100}
(or 6%6\%) were both grey and
electric.

Thus, of the 40504050 trucks sold,
61004050=243\dfrac{6}{100}\cdot4050=243 were
both grey and electric.
Solution 1:

Of the 40504050 trucks sold, 44%44\% were black or 441004050=1782\dfrac{44}{100}\cdot4050=1782 were
black.

Thus, the total number of black trucks, sold and unsold, was 1782+k1782+k, and the total number of trucks,
sold and unsold, was 4050+k4050+k.

Since 46%46\% of all trucks, sold and
unsold, were black, then 1782+k4050+k=46100\dfrac{1782+k}{4050+k}=\dfrac{46}{100}.

Solving, we get 1782+k4050+k=461001782+k4050+k=235050(1782+k)=23(4050+k)89100+50k=93150+23k27k=4050k=150\begin{align*} \dfrac{1782+k}{4050+k}&=\dfrac{46}{100} \\ \dfrac{1782+k}{4050+k}&=\dfrac{23}{50} \\ 50(1782+k)&=23(4050+k) \\ 89\,100+50k&=93\,150+23k \\ 27k&=4050\\ k&=150\end{align*} and so there were 150150 unsold trucks, all of which were
black.

Solution 2:

Of the 40504050 trucks sold, 44%44\% were black and so 100%44%=56%100\%-44\%=56\% were not black.

Therefore, 561004050=2268\dfrac{56}{100}\cdot4050=2268 trucks sold
were not black.

Since all unsold trucks were black, then there were 22682268 trucks, sold and unsold, that were
not black.

Since 46%46\% of all trucks, sold and
unsold, were black, then 100%46%=54%100\%-46\%=54\% of all trucks, sold and
unsold, were not black.

The total number of trucks, sold and unsold, was 4050+k4050+k and 54%54\% of these trucks were not black, thus
22684050+k=54100\dfrac{2268}{4050+k}=\dfrac{54}{100}.

Solving, we get 22684050+k=5410022684050+k=275050(2268)=27(4050+k)113400=109350+27k4050=27kk=150\begin{align*} \dfrac{2268}{4050+k}&=\dfrac{54}{100} \\ \dfrac{2268}{4050+k}&=\dfrac{27}{50} \\ 50(2268)&=27(4050+k) \\ 113\,400&=109\,350+27k \\ 4050&=27k\\ k&=150\end{align*} and so there were 150150 unsold trucks, all of which were
black.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.