# linear equations

A shopkeeper buys a number of books for Rs.1200. If he had bought 5 less books for the same amount, each book would have cost him Rs.20 more. How many books did he buy?

I have this question found in my brother's tenth class book and unable to understand it.

Update:

I could not give a suitable title and since the problem is of Quadratic Equation, so the current title is obviously wrong. I shall appreciate if someone can give a suitable title and edit the same.

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Doing this step by step: Let $x$ be the amount of books and $p$ the price. We have $$xp=1200$$ If he bought 5 less books at the price $p+20$ then $$(x-5)(p+20)=1200$$ Working this out we have $$x=5+\frac{p}{4}$$ Substituting this in the first equation and working it out gives $$p^2+20p-4800=0$$ which only positive solution is $p=60$, and we can conclude that $x=20$.

Number of books = 20

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I think you should have $x=20=\frac{1200}{60}$. ($x=-15$ is the other solution.) –  copper.hat Feb 24 '13 at 16:11
It was a typo, sorry. –  Marra Feb 24 '13 at 16:12

If $n$ is the initial number of books, and $P$ is the initial price, then we have: $n P = 1200$, $(n-5)(P+20)=1200$.

Equating the two equations (since they both equal $1200$) gives $4n-P=20$, which gives $P=4n-20$. Substituting this into the equation $n P =1200$ gives $n^2-5n-300=0$ which has solutions $\{-15,20\}$. Since we are looking for positive solutions, this gives $n=20$.

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Let $x$ be the number of books.

Each Book costs : $\frac{1200}{x}$

Solving the equation: $\frac{1200}{x-5} = \frac{1200}{x}+20$ gives the solution the solution as $x=20$.

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