Three Math Problems

[Pravin]

How would you solve these problems?

Please explain solution of problem 1 and quick solution method for 2 and 3.

"10 tennis players are to play in a 'doubles' tournament, in which a team of 2 players on one plays another of 2 players on the other side. How many different 'doubles' games can be scheduled?" 1260

"A wild life preserve is being planned for 3,000 rhinoceroses. The preserve is to contain a total of 10,000 acres of watering area, and 100 acres of grazing area for each of the rhinoceros in the preserve. If the number of rhiniceros is expected to increase by 10%, how many thousand acres should the preserve have in order to provide for the increased population?"

"If X and Y are two different three digit integers, Is X - Y divisible by 9?"

(1) X and Y have the same digits but in different order.

(2) The units' digit of X and of Y are the same.

[Pravin]

Source - MLI Consulting... Hope I can post from different source.. instructions talk about OG... if not please delete the post, better to avoid any copyright issues :oops:

[ahistegt4]

1.10 tennis players are to play in a 'doubles' tournament, in which a team of 2 players on one plays another of 2 players on the other side. How many different 'doubles' games can be scheduled?

So, let's see this one:
You can choose first pair of a"double" in 10!/(2!8!) ways = 45
You can choose second pair of a "double" in 8!/(2!6!)=28
There are 45*28=1260 ways you can play doubles.

[JonathanSchneider]

Your math is perfect, ahistegt4, but you don't need to multiply it all out in this way.

Instead, line it up first:

10! 8!
---- x -----
8! 2! 6! 2!

Notice that the 8!'s cancel, leaving us with 10! / (6! 2! 2!), which can be rewritten as 10*9*8*7 / 2*2, or 10*9*2*7, or 630*2, 0r 1260. A bit simpler?

[JonathanSchneider]

As to the second problem, I'm not sure that the wording clearly expresses the issue. But from what I can tell, we need to say that there is a total of 10,000 watering acres for all the rhinocereses combined, and there will be 100 acres for each of the (now) 3,300 animals to graze. As such, we would need 10,000 acres (watering) and 330,000 acres (grazing), for a total of 340,000 acres. Either that or I'm misreading the question.

[JonathanSchneider]

As to the third problem, you can test numbers to see that (2) is not sufficient. Statement (1) is much easier if you assign variables to each of the three digits places. In cases like these, you might want to try x, y, and z. Just remember that these are in certain places, so the value of the first # is actually 100x + 10y + z. You can do the algebra from there to see that both statements together are sufficient.

[Guest]

For the third question why not one alone is suffiecient:

X=100A+10B+C
Y=100C+10B+A

X-Y=99A-99C

Thus x-y in all cases is divisible by 9.Pl. correct where I am wrong?

[JonathanSchneider]

Ah, good catch. I think I was moving a little too fast there. Yes, statement one on its own is sufficient.

However, you should still clarify something: you're assuming that they mean the exact opposite order. Were that the case, your logic above would be correct. However, the statement has only told us that they are in a "different order." We have three digits, so there are actually 3!, or 6, different ways that we could order those digits:

ABC
ACB
BAC
BCA
CAB
CBA

Fortunately, however, the difference between any two of these is still divisible by 9. The easiest way to see this is that for each letter, we can either have the letter occuring in the same digits place (in which case it will cancel), or in two different places. If the letter A appears in the 100's place and 1's place, for example, the difference between them will be either 99A or -99A. In either case, divisible by 9. Ditto for the 10's and 1's place, or the 100's and 10's place. And again, ditto for the other letters (B and C). Kind of a complicated explanation, I realize. A much trickier problem than I noticed first go-round.