VJesus12 wrote: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 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?

A. 3
B. 4
C. 5
D. 6
E. 7

Here are the PAIRS of numbers that yield a sum of 10:
(1 and 9)
(2 and 8)
(3 and 7)
(4 and 6)
Also, 0 and 5 have no other values to pair with to get a sum of 10

Now let's try to AVOID getting a sum of 10.
Notice that, if we choose the numbers 0, 1, 2, 3, 4, and 5, there are no pair of values that yield a sum of 10
Since these 6 values do NOT ensure that two numbers yield a sum of 10, we can conclude that the correct answer is GREATER THAN 6
So, the correct answer must be E

Cheers,
Brent $$$$

VJesus12 wrote: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 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?

A. 3
B. 4
C. 5
D. 6
E. 7

E is OA.

I would like to know how can I solve this PS question. Experts, may you give me an explanation? Thanks.

We can pull the following slips before getting a sum of 10:

0, 1, 2, 3, 4, 5

No matter what number (6, 7, 8 or 9) we pull on the next card , we are sure that we will obtain a sum of 10. Thus, the minimum number of cards drawn to ensure that the numbers on two of the slips drawn will have a sum of 10 is 7.

Answer: E