Question #2817d

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Answer 1 · 2017-10-14T11:31:04

16 and 5

Let shopkeeper buys x books for Rs. 80.
So, 1 book costs Rs. 80/x.

If he buys (x + 4) books and then each book costs ( 80/x - 1).
So, (x + 4) books cost = Rs. ( x + 4 ).(80/x - 1)

As per question, $(x+4).(80/x-1) = 80$

$rArr x. 80/x-x+4. 80/x-4 =80$

$rArr 80-x+320/x-4=80$

$rArr -x+320/x=80-80+4$

$rArr -x+320/x=4$

$rArr [-x^2+320]/x=4$

$rArr -x^2+320 = 4x$

$rArr -x^2+320-4x=0$

$rArr -(x^2+4x-320)=0$

$rArr x^2+4x-320 = 0$ [Divide by -1 in both sides]

$rArr x^2+20x-16x-320=0$

$rArr x(x+20)-16(x+20)=0$

$rArr (x+20)(x-16)=0$

$rArr x+20 =0, x-16=0$

$rArr x = -20, 16$ [ x cannot be - 20 ]

$rArr x = 16$

No. of books = 16 and Each book buys Rs.80/16 = Rs. 5/-