Question

Cody, Brandon, Jasmine, Ryan, Abigail, and Zachary each have a different number of books (118, 140, 115, 149, 137, and 145).

• A digit in the number of books that Jasmine has is odd.
• Ryan has an odd number of books.
• Ryan is not the one with 145 books.
• A digit in the number of books that Abigail has is even.
• A digit plus another digit in the number of books Cody has is even.
• Zachary is not the one with 140 books.
• Cody has an odd number of books.
• Jasmine does not have an odd number of books.
• A digit in the number of books that Abigail has is odd.
• Cody is not the one with 137 books.
• A digit in the number of books that Zachary has is even.
• A digit plus another digit in the number of books Brandon has is odd.
• A digit in the number of books that Brandon has is odd.
• Brandon does not have an even number of books.
• A digit in the number of books that Zachary has is odd.
• Brandon is not the one with 149 books.
• A digit in the number of books that Cody has is odd.
• A digit plus another digit in the number of books Abigail has is odd.
• Abigail does not have an even number of books.
• Jasmine is not the one with 149 books.

How many books each of the people has?

Answer 1

Sol:

Ans:

Cody has 115 books

Brandon has 145 books

Jasmine has 140 books

Ryan has 137 books

Abigail has 149 books

Zachary has 118 books

Solution : there are two even number books.

From the above condition zachary and jasmine has even books.

Zachary does not have 140 number,so zachary occupied118 and remaining jasmine occupied 140.

Remaining four has remaining odd books

Abigail has one even and one odd digit so he have two option 149,145

145 occupied by Brandon according to the condition so remaining 149 is for aabigail

Cody does not have137 so he has 115 number and remaining last 137 number is for ryan