Question

A shopkeeper buys a certain no. of books for Rs. 960. If the cost per book was Rs. 8 less, the no. of books that could be bought for Rs. 960 would be 4 more. Taking the original cost of each book to be Rs. x, write an equation in x and solve it.​

Answer 1

$\sf \Large Solution:$

$\sf \because Original\;cost\;of\;each\;book=Rs.\:x$

$\sf \therefore Number\;of\;books\;bought\;for\;Rs.\:960=\frac{960}{x}$

$\sf \large In\;2nd\;case:$

$\sf The\;cost\;of\;each\;book=Rs.\:(x-8)$

$\sf \therefore Number\;of\;books\;bought\;for\;Rs.\:960=\frac{960}{x-8}$

$\sf \large Given:$

$\sf \frac{960}{x-8}-\frac{960}{x}=4\quad i.e.\;\frac{960x-960x+7680}{x(x-8)}=4$

$\sf \to 7680=4(x^{2}-8x)$

$\sf \to x^{2}-8x=1920$

$\sf \to x^{2}-8x-1920=0$

$\sf \large Now,$

$\sf \to x^{2}-8x-1920=0$

$\sf \to x^{2}-48x+40x-1920=0$

$\sf \to x(x-48)+40(x-48)=0$

$\sf \to (x-48)(x+40)=0$

$\sf \to x=48\;or\;x=-40$

$\sf Rejecting\;x=-40,\;we\;get,$

$\sf \boxed{\bigstar x=48}$

$\sf \Large Alternative\;method:$

$\sf \because Cost\;of\;each\;book=Rs.\:x$

$\sf \therefore Number\;of\;books\;bought\;for\;Rs.\:960=\frac{960}{x}$

$\sf \large In\;2nd\;case:$

$\sf The\;cost\;of\;each\;book=Rs.\:(x-8)$

$\sf Number\;of\;books\;bought=(\frac{960}{x}+4)$

$\sf \because Cost\;of\;each\;book\times Number\;of\;books=Cost\;of\;all\;the\;books$

$\sf \to (x-8)(\frac{960}{x}+4)=960$

$\sf \to (x-8)(\frac{960+4x}{x})=960$

$\sf \to 960x+4x^{2}-7680-32x=960x$

$\sf \to 4x^{2}-32x-7680=0$

$\sf \to x^{2}-8x-1920=0$

$\sf \to (x-48)(x+40)=0$

$\sf \to x=48\;or\;x=-40$

$\sf \boxed{\bigstar x=48}$