Let P be the price of one book and N be the number of books that the boy buys (the basic scenario).


Then at the first scenario he pays (P-10) per book and buys (N+50) books, so

    (P-10)*(N+50) = NP.    (1)


Similarly, under the second scenario, you get the equation

    (P+5)*(N-10) = NP.     (2)



You can simplify the first equation in this way

    PN - 10N + 50P - 500 = NP,   or

   -10N + 50P = 500.            or

    10N - 50P = -500        (1')



You can simplify the second equation in this way

    PN + 5N - 10P - 50 = NP,     or

    5N - 10P = 50.          (2')         



Thus you have this system of two equations

    10N - 50P = -500        (1')
     5N - 10P =   50.       (2')         


To solve it, use the elimination method. For it, multiply the eq(2') by 2 (both sides) and then subtract it from eq(1'). You will get

    -50P - (-20P) = -500 - 2*50,    or

    -30P          = -600

which implies   P =  = 20.


Now from eq(2') you have

     5N = 50 + 10P = 50+10*20 = 250,

hence,  N = 250/5 = 50.


Answer.  P = 20,  N = 50  and NP (which is under the question)  is NP = 50*20 = 1000.

         What is the amount of money that the boy was able to use ?   - It is 1000.