1. If a + (1/(b+(1/c))) = 37/16
Then what does a + b + c = ?
2. What is the largest possible size of an angle of a triangle formed by joing the midpoints of three edges of a cube?
3. There are 10 couples at a party. Each pair of people shake hands, except that no woman shakes hands with either her husband or another woman. How many handshakes are there?
Originally posted by phgao1. 10 (2+3+5)
1. If a + (1/(b+(1/c))) = 37/16
Then what does a + b + c = ?
2. What is the largest possible size of an angle of a triangle formed by joing the midpoints of three edges of a cube?
3. There are 10 couples at a party. Each pair of people shake hands, except that no woman shakes hands with either her husband or another woman. How many handshakes are there?
Kody has sent me this:
Not seen this before and solved through a combination of workings and assumptions.
Workings would be:
37/16=2.3125
So, a+(1/(b+(1/c)))=2.3125
1/(b+(1/c))=2.3125-a
Now as the left is obviously going to be a fraction (as it is 1/something) I assumed a=2, so:
1/(b+(1/c))=0.3125
Again, 1/c is going to be a fraction, so 1/b must be b=3, as 0.3125 is just under 0.3333 (or 1/3), so:
1/(3+(1/c))=0.3125
1=0.3125(3+(1/c))
1=0.9375+0.3125(1/c)
0.0625=0.3125(1/c)
0.2=1/c
1=0.2c=5
Not very scientific, but it got there!
But the thing that I dont quite comprehend is how b=3 was got. You can get a=2 as 1/(b+(1/c)) is a fraction but how can you just say that b=3, i'm stuck to find the reasoning there as 5/16 is already a fraction, you cant just change it into 0.3125 and say if b=3 it is only slightly bigger than .3125.
If we did it that way, then when we get 1/(b+(1/c) = 5/16 we can just flip it to get: b+1/c = 16/5
b+1/c = 3.2
And you can just say that b=3 and c=5, and i'm sure that isnt the way to get the values. Can anyone else find a way that doesnt use guesses even though they are fairly obvious.
Originally posted by phgao1)
1. If a + (1/(b+(1/c))) = 37/16
Then what does a + b + c = ?
You should really define what type of numbers a,b,c are.
Are they reals, (positive) fractions, (positive) integers, etc?
2) I can't see one greater than 90 degrees.
3)
There are (20 x 19)/2 = 190 ways for everybody to shake hands with each other.
There are (10 x 9)/2 = 45 ways for the 10 women (not) to shake hands with each other.
There are 10 ways for the women (not) to shake hands with their respective husbands.
Hence required number of ways = 190 - 45 - 10 = 135
In general, with 2n couples number of ways = 3n(n-1)/2
Originally posted by phgao1. Yeah, THUDandBLUNDER is right about defining what kind of answer you're looking for. I assumed you wanted positive integers, because they're nice and round, and the stipulation usually gives you the info you need to answer the question. Here's how I did it:
1. If a + (1/(b+(1/c))) = 37/16
Then what does a + b + c = ?
2. What is the largest possible size of an angle of a triangle formed by joing the midpoints of three edges of a cube?
3. There are 10 couples at a party. Each pair of peo ...[text shortened]... ither her husband or another woman. How many handshakes are there?
a + 1/(b + 1/c)) = (abc + a + c)/(bc + 1) = 37/16
I assumed that the numerators and denominators were the same, and equated them.
bc + 1 = 16
bc = 15
abc + a + c = 37
a(15) + a + c = 37
16a + c = 37
Then I tried values for "a". Letting a = 1 gave c = 21, which seemed odd for this type of question. However, letting a = 2 gave c = 5, and subsequently b = 3. Nice round numbers, so I was pretty sure I had your solution in mind.
a + b + c = 2 + 3 + 5 = 10.
2. I'm not sure, but I think the maximum angle is 90 degrees. You can create a right angle triangle by starting at any mid-point and moving to the mid-point opposite the first on the same face, and then connecting point 1 to point 3. Picking a different point for point 2 results in a longer triangle with the same base. If you imagine a right angle triangle laying on it's hypotenuse like so --> /\, stretching the triangle upwards will shrink the angle at the top (the 90 degree angle). So any elongation will result in a smaller angle. That's why I think the largest possible angle is 90 degrees.
3. All 10 men will shake hands with each other, which results in (10 choose 2) handshakes, and each woman will shake hands with 9 men (not her husband). So the total is:
(10 choose 2) + 10*9 = 45 + 90 = 135 handshakes.
Originally posted by phgaoIt's impossible to tell, because there's no information on whether all couples consist of one man and one woman.
3. There are 10 couples at a party. Each pair of people shake hands, except that no woman shakes hands with either her husband or another woman. How many handshakes are there?
Originally posted by PBE6We missed it because the obvious implication is that we are speaking of husbands and wifes only.
That's true! I wonder how we missed that?
Or should we show our puzzles to a bunch of hair-splitting, nit-picking forensic lawyers before posting them?
Originally posted by THUDandBLUNDERTechnically, the terms "husband" and "wife" apply to married homosexuals, too. And I've watched enough "L Word" and "Queer As Folk" that I should have snarked out Nordlys' post myself.
We missed it because the obvious implication is that we are speaking of husbands and wifes only.
Or should we show our puzzles to a bunch of hair-splitting, nit-picking forensic lawyers before posting them?
As far as consulting with RHP legal staff before posting, I prefer boxers to briefs, legal or otherwise. Besides, I think the hair-splitting, nit-picking forensic lawyers only tell you if you have crabs or not.
Originally posted by PBE6I prefer to apply the principle of Occam's razor, especially if the alternatives leads to no answer at all.
Technically, the terms "husband" and "wife" apply to married homosexuals, too. And I've watched enough "L Word" and "Queer As Folk" that I should have snarked out Nordlys' post myself.
I didn't particularly mean RHP lawyers.
Any lawyer would be happy to sniff out double meanings etc.
At the right price, of course.
Originally posted by THUDandBLUNDERWell, if you can do the hair-splitting and nit-picking yourself, that would be fine, too. 😉 Seriously, you have to be prepared for trick questions in a forum like this, so if I encounter a puzzle which seems trivial, I'll start looking for less obvious answers. So in this context, I think nit-picking is indeed a good idea.
We missed it because the obvious implication is that we are speaking of husbands and wifes only.
Or should we show our puzzles to a bunch of hair-splitting, nit-picking forensic lawyers before posting them?
One more thing b4 I go,
The answer to Q2, is 120 degrees not 90 degrees, note: the largest possible size of an angle of a triangle in a Square is 90 degrees!
Also thanks to PEB6 for explaining the question.
The answer to C is correct.
Plus the 'couple' thing is explained in the Q- ...no woman shakes hands with either her husband...!! So why the confusion?
Here's another
There are n prisoners in a row of locked cells. With the return of the King from the Crusades, a partial amnesty is declared and it works like this. When the prisoners are still asleap, the jailer walks past the cells n times, each time he turns the lock in every cell (so that every cell is now open). On the second pass he turns the lock on every secod cell (meaning that these cells are now locked again). On the third pass, he turns the lock on every third cell, and so on. In general, on the kth pass, he turns the lock on every kth cell. The question is, which cells are unlocked at the ned of the process so that the prisoner is free to go?