Verna decided to sell her collection of books. To Fred, she sold 2

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
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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
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Given: 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.

Asked: If she sold more books to Fred than to Joan, what was the least possible number of books in her original collection?

Let the number of books in Verna's collection be x.

Books sold to Fred = 2 + (x-2)/5 = 2 + k
Since books sold is an integer: x -2 is a multiple of 5: x-2 = 5k; x = 5k + 2
Books sold to Joan = 6 + (x - 8-k)/5 = 6 + (5k+2-8-k)/5 = 6 + (4k-6)/5
2 + k > 6 + (4k-6)/5
10 + 5k > 30 + 4k - 6
k > 14
x = 5k + 2 > 72

Since 4k - 6 is a multiple of 5
Minimum k = 19
mininum x = 5k +2 = 5*19 + 2 = 97

IMO D
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