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08 May 2019, 17:32
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
38% (02:52) correct
62%
(03:17)
wrong
based on 125
sessions
History
Date
Time
Result
Not Attempted Yet
Verna decided to sell her collection of books. To Fred, she sold 2 books, and one fifth of what was left. Later to Joan she sold 6 books, and one fifth of what then remained. If she sold more books to Fred than to Joan, what was the least possible number of books in her original collection?
1. 22
2. 47
3. 72
4. 97
5. 122
1. 22
2. 47
3. 72
4. 97
5. 122
08 May 2019, 22:31
Kudos
Bookmarks
nick1816
Make use of choices
Let us start from C..
So initial = 72,
Fred gets = 2+(72-2)/5=2+14=16
J gets = 6+(72-16-6)/5=6+(50/5)=16
Since both get equal books, we will have to increase the initial books..
So initial = 97
Fred gets = 2+(97-2)/5=2+19=21
J gets = 6+(97-21-6)/5=6+(70/5)=6+14=20..
Here F gets more book as 21>20
D
08 May 2019, 22:51
Kudos
Bookmarks
nick1816
Algebraic way
Let initial be x..
So, Fred gets \(2+\frac{x-2}{5}=\frac{x+8}{5}\)..
Remaining = \(x-\frac{x+8}{5}=\frac{4x-8}{5}\)
Next Joan gets ---> \(6+(\frac{4x-8}{5}-6)/5=6+(\frac{4x-38}{5})/5=\frac{150+4x-38}{25}\)..
Now..\(\frac{x+8}{5}>\frac{112+4x}{25}.....5x+40>4x+112.......x>72....\)
So, next higher value is 97, and we can also substitute x as 97 and see if it works..
D
