linear equations

Question

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.

score 3 · Answer 1 · edited Feb 24 '13 at 16:14

3

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

edited Feb 24 '13 at 16:14

Julian Kuelshammer

9,670

answered Feb 24 '13 at 16:01

Marra

4,839

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

score 0 · Answer 2 · edited Feb 24 '13 at 16:18

0

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 $x=20$.

edited Feb 24 '13 at 16:18

answered Feb 24 '13 at 16:03

Inceptio

7,881

score 0 · Answer 3 · answered Feb 24 '13 at 16:07

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 $n=20$.

linear equations

3 Answers3