Difficult Math Question #21 - Probability

[800guy]

OA coming after some people respond...

Each of the integers from 0 to 9, inclusive, is written on a separate slip of blank paper and the ten slips are dropped into a hat. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10?
3
4
5
6
7

[gamemaster]

My solution:

we have:

1 9
2 8
3 7
4 6

and also 0 and 5

so if we choose just one half of a pair (4 numbers for example 1,2,3,4) and 0 and 5, then only on the 7th number we'll succseed

the answer will be 7 also if we look for combination of more than 2 numbers to sum up to 10, because we can choose: 9,8,7,6,5,0 first

so: E

[800guy]

OA:

ok consider this

0 + 1+ 2 + 3 + 4 +5 +6..Stop --> Don’t go further. Why? Here’s why.
At the worst the order given above is how u could pick the out the slips. Until u add the slip with no. 6 on it, no two slips before that add up to 10 (which is what the Q wants)

the best u can approach is a sum of 9 (slip no. 5 + slip no. 4)

but as soon as u add slip 6. Voila u get your first sum of 10 from two slips and that is indeed the answer = 7 draws

[rahulg83]

I really don't understand what this question is asking for?
We can simply take out 2 chits numbered 1 and 9, and get the sum 10..there could be a lot of ways to get this 2 and 8, 3 and 7..so on..
can someone help me understand the language of this question?

[kaulnikhil]

what question is indeed asking is that whenever you pick a new chit the sum of other chits that you have taken out and this new chit must be 10. now if we go by ur case we take out 1. the next chict we can take out any other no except1 . So u cant guarantee that the sum will be 10.
Now consider this u take out 0 ,1, 2,3,4 ,5
now whenever u take out a new chict irrespective of the no u take out sum will be 10 ..
so in all u need 7 turns