Question

A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then k-20 is equal to:

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

The correct option is A 5

Clearly, $1+2+3+\cdots n-2\le 1224\le 3+4+\cdots n$
$\Rightarrow \frac{(n-2)(n-1)}{2}\le 1224\le \frac{n(n+10)}{2}-3$
$\Rightarrow \frac{(n-2)(n-1)}{2}\le 1224\le \frac{(n-2)(n+3)}{2}$
$\Rightarrow n^2-3n-2446\le 0and{n}^2+n-2454\ge 0$
$\Rightarrow 49<n<51$
$\Rightarrow n=50$
$\Rightarrow \frac{n(n+1)}{2}-k-(k+1)=1224$
$\Rightarrow k=25$
$\Rightarrow k-20=5$