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.

By (1), since the addresses are in consecutive numbers, it is clear that the numbers are 9, 10, 11, 12 and 13. (By obtaining 11 as the mean)

Considering (2), the possibilites are 5years old/No. 10 and 6years old/No. 12. By (6), since the colours are no white, the info shows that the 6year old house is not the same one as No. 12. Hence Blue house is No. 10, and a 5 year old girl lives in it.

Consider (4), the possibilities are 9, 10 and 12. Blue house is No. 10 and hence 10 is eliminated. the possibilities are 9 / 12.

Consider (6), since Blue house is No. 10, yellow/purple has to be either 9/11. Consider (4) again, since purple house is either 9/11, combining the results, Leo is 9 years old and the purple house has the No. 9. Yellow is hence No. 11. By (3), Leo lives in house 13.

Consider (7),  ignore the numbers first, it is clear that Nina is at least 6 years old (assuming she is 2 yr old 4 yrs ago). Hence the 5 year old girl in (2) must be Jane.

Consider (5), Blue + colour of house 12 = colour of house of the 6yr old. colour of house 12 could not be yellow (which combines with blue to green)since yellow is No. 11. Combining blue with green or purple would not result in any colour of the five. hence the colour of House 12 will be red, and the house of the 6 yr old is purple( No. 9). The last house will be No. 13 -> Green, where Leo lives.

Consider (9), the house that Kevin and Mike could be living in are 12, 11 and 9. Max difference is 3. Hence difference bet'n Nina and Kevin's age is 1 or 2. the possibilites are : 5, 6 / 9, 10/ 10, 12. yet Jane is 5 and Leo is 9. Hence, The age of Nina and Kevin is 10 and 12, whoever's which.  

Consider (8), the age of the boy is 12 since house 9 and 10 are already occupied. And since by (9) only Nina and Kevin can be 12, Kevin is. He lives in the red house. Also, Nina is hence 10. By elimination, Mike is 6 and he lives in the purple house. Again by elimination, Nina lives in the yellow house.

Tidied results:

Jane - No. 10 - 5 years old - Blue house

Kevin - No. 12 - 12 years old - Red house

Leo - No. 13 - 9 years old - Green house

Mike - No. 9 - 6 years old - Purple house

Nina - No. 11 - 10 years old - Yellow house

The solution seems to satisfy the 9 hints upon checking. A probable solution.

 

 

Edited on October 8, 2005, 10:45 am

Edited on October 8, 2005, 10:46 am

Posted by Terence on 2005-10-08 10:43:44