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Bright Houses (Posted on 2005-10-08) Difficulty: 3 of 5
Five children: Jane, Kevin, Leo, Mike and Nina live in the same street in consecutively numbered houses.
(Addresses do not alternate from one side of the road to another so they are all on the same side of the road).
The ages of the children are 5, 6, 9, 10 and 12 and the colours of their houses are red, yellow, green, blue and purple.

Using the following information determine each child’s information.

1.) The sum of their addresses is 55.

2.) The person who lives in the blue house has an age equal to half her address.

3.) The 9 year old lives in house No. 13.

4.) Leo’s age is the same as the number on the purple house.

5.) The colour of Jane’s house, combined with the colour of house No. 12, gives the colour of the 6 year old’s house.

6.) The Blue house is the same distance from the purple house as it is from the yellow house.

7.) 4 years ago Nina’s age was equal to twice as much as the difference between Kevin’s age (today) and the number on Mike’s house.

8.) One boy’s age is the same as his house number.

9.) The difference between Kevin and Mike’s house No.s is greater than the difference between Nina and Kevin’s age.

See The Solution Submitted by scott    
Rating: 3.0000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
solution | Comment 12 of 13 |
Age Name Colour House#
5  Jane   Blue     10
6  Mike   Purple    9
9  Leo     Red      13
10 Nina   Yellow  11
12 Kevin Green  12

That was a good one...  finally something productive to do at work...
:)

  Posted by Morgan on 2005-11-18 18:53:10
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