Question

2. A pack Containon Cards wankered from 1 to n. Two Consecutive
numbered Cards are remoned from the pack and sum of the
numbers on the remaining Cards is 1224. If the smaller of the
numbers on the remoued Cards is k, then k-20=. 2012

Answer 1

Sum of 1st n natural nos. =

2

n(n+1)

Given:

2

n(n+1)

−(k+k+1)=1224

⇒

2

n(n+1)

−2k=1225....(1)

also

2

n(n+1)

−(2n−1) (when k=(n−1) $$< 1224< \cfrac{n(n+1)}{2}-3$$ (when k=1)

$$\cfrac{({n}^{2}-3n+2)}{2}< 1224< \cfrac{({n}^{2}+n-6)}{2}$$

$$(n-2)(n-1)< 1448< (n-2)(n+1)$$

So n=50 by wt and trial method

So in equation (1) (51)(

2

50

)−2k=1225

1275−2k=1225

k=25

k−20=5 (Ans)