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# hard word problem

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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?

May 4, 2020

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Let N be the  number of books she started with

To Fred she sells  2 books  and then 1/5 of what is left  =    2 + (1/5)(N - 2)    books

To Joan....she sells  6  books and what remains

What remains  is  what Verna started with  - what she sells to Fred  - 6 books she sold to Joan

So she sells Joan  6 + (1/5)(N - [2 + (1/5)(N-2)]  - 6)

So  we have this inequality

2 + (1/5)(N - 2)  >  6 + (1/5)(N - (2 + (1/5)(N - 2)) - 6)    mutiply through by 5

10  + (N - 2)  >  30 +  (N - (2 + (1/5)(N - 2)) - 6)

N + 8  >   30 + ( N - (2 + (1/5)N - 2/5) - 6)

N + 8 > 30 + (N  - 2 - (1/5)N  + 2/5  - 6)

N + 8 >  30  + (4/5)N - 38/5      multiply through by 5

5N + 40  > 150 + 4N - 38

5N + 40 > 150 + 4N  - 38

N > 150 - 38 - 40

N > 72

Then...she had  to  have at  least 73 books in her collection

Check....if she started with just 72 books

She sells  to Fred    2 + (1/5)(72 - 2)  =  2 + 14  =  16 books

She has  56 books left

She sels to Joan   6  + (1/5) (56 -6)  =  6 + 10 = 16 books

But....she sells to Fred more books.....so....N must be greater than 72

May 4, 2020