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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 Solution | Comment 1 of 13
Jane 5 years old, blue house address 10
Kevin 12 years old, red house address 12
Leo 9 years old, green house address 13
Mike 6 years old, purple house address 9
Nina 10 years old, yellow house address 11
  Posted by Justin on 2005-10-08 03:55:26
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