In a car park, there are 2 white car for every 3 blue cars and for every 2 blue cars there are 5 silver cars. What is the least number of cars in the park?

I am a bit confused about my approach to the question, according to my thinking...

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But I still think that I am making a wrong approach. Need help.

Thanks in advance!

  • $\begingroup$ well if the cars are distinguishable then for the blue ones you need to have 15 silver ones, because you can pick 2 blue cars out of 3 in 3 ways if the order doesn't count. So over all 20. Hope this this is right. $\endgroup$ – randomname Jan 21 '14 at 18:11
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    $\begingroup$ I thought this was a simple question, but so far you have four different answers: $20$, $25$, $30$ and $36$. $\endgroup$ – Henry Jan 21 '14 at 18:19

If there are $w$ white cars, $b$ blue cars and $s$ silver cars then you know $$\frac{w}{2} =\frac{b}{3},$$ $$\frac{b}{2} =\frac{s}{5}.$$

If you are not allowed fractions of a car or striped cars, then $b$ must be a multiple of $3$ and a multiple of $2$, which means $b$ is a multiple of $6$. If $b=6$ then $w=4$ amd $s=15$ making a total of $25$ cars.

Or perhaps there are no cars at all.

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    $\begingroup$ agree with Henry, the LCM restriction applies to blue car relation between white and silver (which is LCM(2,3)=6. If you have 6 Blue cars, the minimum total number of cars is 25 $\endgroup$ – CAGT Jan 21 '14 at 18:41

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