# algebra

• Jun 26th 2009, 09:29 AM
abhishek arora
algebra
two consecutive no's outof 1,2,3,...,n-1,n are removed. the average of remaining no's is 105/4.find n and the removed numbers
• Jun 26th 2009, 09:52 AM
pankaj
Quote:

Originally Posted by abhishek arora
two consecutive no's outof 1,2,3,...,n-1,n are removed. the average of remaining no's is 105/4.find n and the removed numbers

Let the removed numbers be $\displaystyle p$ and $\displaystyle p+1$

$\displaystyle \frac{(1+2+...+n)-(p+p+1)}{n-2}=\frac{105}{4}$

$\displaystyle \frac{\frac{n(n+1)}{2}-(2p+1)}{n-2}=\frac{105}{4}$

$\displaystyle 8p=2n^2-103n+206$

Now $\displaystyle 1\leq p\leq n-1$

$\displaystyle 8\leq 8p\leq 8n-8$

$\displaystyle 8\leq 2n^2-103n+206\leq 8n-8$

On solving we have two inequalities

$\displaystyle 2n^2-103n+198\geq 0;(n-2)(2n-99)\geq 0;n\geq \frac{99}{2};n\geq 50$

Since $\displaystyle n\geq 2$ and $\displaystyle n$ must be an integer.

Similarly solving the inequality, $\displaystyle 2n^2-111n+214\leq 0; (n-2)(2n-107)\leq 0$ and thus $\displaystyle n\leq 53$

Therefore, $\displaystyle 50\leq n\leq 53$

$\displaystyle n=50,51,52,53$

It is only for $\displaystyle n=50$ that p is an integer.Thus $\displaystyle p=7$

Therefore the number of pages in the book are $\displaystyle 50$ and the removed pages are $\displaystyle 7$ and $\displaystyle 8$
• Jun 26th 2009, 10:01 AM
abhishek arora
thanks
thank you pankaj sir u are like an angel for me