JEE Advanced 2013 · Paper 1 · Mathematics Q57

← JEE Advanced 2013 · Paper 1

MathematicsInteger+4 / -1Question 57 of 60

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 $k-20=$

Your answer

Answer and worked solution

Official answer: 5

Sum. $\dfrac{n(n+1)}2-(2k+1)=1224$ with $1\le k<n$ .

Try n = 50: $1275-1224=51=2k+1$ , so $k=25$ . ( $n=49$ gives 1225, too small; $n=51$ gives $2k+1=102$ , impossible.)

So $k-20=5$ .

Answer: 5

Topics:Arithmetic progression Number systems

← Previous question Next question →

JEE Advanced 2013 · Paper 1, Mathematics question 57