Books Puzzle

Let the initial number of books in her collection be x . For Fred , she sold; ${2+\frac{1}{5}({x-2})}$ Note that after she sold 2 books, what remained is (x-2) *Thus, the number of books that remained after this transaction equals $x-(2+\frac{1}{5}({x-2}))$ which gives $\frac{4x}{5}-\frac{8}{5}$ So for Joan , she sold ; ${6+\frac{1}{5}({\frac{4x}{5}-\frac{38}{5}})}$ Note also that after she sold 6 books what remained is $(\frac{4x}{5}-\frac{8}{5})-6=\frac{4x}{5}-\frac{38}{5}$ We are told that she sold more books to Fred than Joan . That is $2+\frac{1}{5}({x-2})>6+\frac{1}{5}({\frac{4x}{5}-\frac{38}{5}})$ Solving the inequality we get ${x>72}$ *Thus any number from 73 on wards is a possible number of books in Verna's collection. But she cannot sell a fraction of a book. That in mind, the least possible number of books in her original collection that can enable her sell whole number of books is $\boxed{97}$

Let the number of books that Verna has be X. Then she gave (2 + (X-2)/5) books to Fred and got left with (X-2) (4/5) books. Then she gave (6 + ((X-2) (4/5) - 6)/5) books to Joan. The first number is greater than the second number. Solving the inequality, you get that X must be greater than 72.

Now if the number of books originally with Verna was 72, then she would have given 14 + 2 = 16 books to Fred, and 10 + 6 = 16 books to Joan, which would have been equal. Now, as the number of total books increases, the variable parts in the two calculations (14 and 10) increase by 1/5 or 20% and (4/5)*(1/5) or 16%. Hence, their difference increases by 4%. As the next difference also must be an integer, we must increase the number of books by 25. So the answer is 72 + 25 = 97.

Algebraically, it looks like this:

$2 + \frac{(x-2)}{5} > 6 + \frac{1}{5}(x - 8 - \frac{(x - 2)}{5})$

And cleaning it up a bit, we get

$\frac{x + 8}{5} > \frac{4x + 112}{25} \rightarrow x > 72$

However, this is not our final answer. Since she must have sold an integer number of books (you can't sell a fifth of a book!), then both sides of the equation must yield an integer. In particular,

$x \equiv 2 \pmod 5$

$4x \equiv -112 \pmod {25} \rightarrow x \equiv 22 \pmod {25}$

We can see that our solution set $S = \{x | x > 72, x = 25k + 22\}$ for some $k \in \mathbb{Z}$ .

The least solution is $97$ , and that is our answer.

Let The Number Of Books Be --> x

Therefore Fred buys 2 books now (x-2) books remain he buys one fifth of the remaining books therefore total number of ---------->books fred buys ------> 2+(x-2)/5----------->

remaining books -----> x - 2 - (x-2)/5 -----> 4(x-2)/5---------> similarly joan buys ----> 6 + ( 4(x-2)/5 - 6 )/5 -------->(24/5+4(x-2)/25)

now as fred buys more books then joan therefore -----------------> 2+(x-2)/5 > (24/5+4(x-2)/25) --------------> solving the inequality we will get x > 72 paragraph 4 now x must be such that books bought by fred and joan are in whole numbers --------------> therefore (x-2) must be completely divisible by 5 so that books bought by fred is an integer and 4(x-2)/25+24/5 must be an integer ---------------> i.e 4(x-2)+120 must be completely divisible by 25 therefore by solving this the least integer we get is x=97

Let the least possible number of books in her original collection be a.

Fred gets 2 + (a-2)/5 no. of books.---------------(1)

Joan gets 6 + (4a-38)/25 no. of books.----------(2)

So, 2 + (a-2)/5 > 6 + (4a-38)/25

Solving we get, a > 72

But if we put a as 73, we get a decimal value for the no. of books Joan gets.

Clearly, a shall assume values like 77,82,87,92,97,102..... so that Fred gets a whole no. as the no. of books.

Putting the values in (2), we get the no. of books as a whole no. when a = 97.

So, the least possible number of books in her original collection is 97.

If we find it awkward to work and reason with inequality, we can also do this instead. You can also write (x+8)/5= (4x+112)/25+1 Because we know that books sold to Fred is more than that sold to Joan, therefore, we can assume the minimum that it is ONE more.

Hence, the +1 to the amount on the right.

Solving this will give x=97.

initially total books=x

books given to fred = 2+ {(x-2)/5} = (x+8)/5

books left with verna after giving books to fred = initial no. of books - books given to fred

                                                                                     = x - (x+8)/5 = (4x-8)/5

now books given to joan = 6 + [{(4x-8)/5}-6)}]/5=(4x+112)/25

required condition
(x+8)/5 > (4x+112)/25 ....(1)

x>72 by solving this eq. (1)

(if we took initially 72 books both will get 16-16 books but we want that fred must get more books so 72 can't be the answer)

as the required answer must be practical and possible i.e. a whole book must be given to a person i.e. not in fraction it should be a integer

so how to find the correct answer we know in the second last activity after giving 6 books to joans the last activity is that, that we hav to give 1/5 of the left books, means the no. must be a multiple of 5 ahead run your heads and a bit hit and trial you'll get the answer 97 THEREFORE we get the answer 97

x>72 then ensure that x-2/5 is an integer then check if Joan is getting an integral value!