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 Post subject: Tricky Question
PostPosted: Fri Oct 21, 2011 9:01 pm 
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Joined: Tue May 10, 2011 1:12 pm
Posts: 117
Hi,
My DD got this question in her tuition. I am unable to solve it.
I hope someone can help.

Five garages A, B, C, D & E have different numbers of cars for sale.
No garage has less than 3 cars or more than 12 cars. The total number of cars is 36.
Only one garage has an even number of cars for sale.

The cars od D and E added together give 2 more than C's cars.
D has 3 times as many cars as B who has less than A

How many cars has each garage for sale?

Thanks in advace.


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 Post subject: Re: Tricky Question
PostPosted: Fri Oct 21, 2011 10:00 pm 
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Joined: Mon Jan 30, 2006 4:07 pm
Posts: 2660
Facts are...

Total 36
Least 3
Most 12
One even number
All different numbers
D has 3 x B
B is less than A
D + E = 2 more than C

1] D and B would be the 1st port of call as we know they must be in the 3 times table. D can only be 9.
3 and 6 would make B 1 or 2 [less than 3]
12 would make B 4 [but only one even number allowed]

Therefore D is 9 and B is 3

2] D + E minus 2 = C
E can only be 4 or 5 as 3 [the lowest] already used anything higher than 5 would make C higher than 12 [not allowed]

If you applied 4 that would give 11 for C and therefore 9 for A [not allowed as we have already used the number 9]
Therefore must be 5


A 7

B 3

C 12

D 9

E 5

Patricia


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 Post subject: Re: Tricky Question
PostPosted: Mon Oct 24, 2011 11:52 am 
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Joined: Tue May 10, 2011 1:12 pm
Posts: 117
Thanks a lot Patricia.


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