Here is a math problem that will be fun to see the various answers given . Just give your answer , don't explain how you arrived at it , just your answer to see what most come up with .
10 :yesnod:
14 :scratchchin:
I'm going with 14 as well.
30 :shruggy:
Edit LOL... 28!
10
14
10 and 14, both are correct. 14 is more correct.
You did not provide sufficient or complete algerbraic expressions on the final line, so both can be correct.
Hope you don't work on dam inspections here in Northern California.
::)
there is only one answer, which is 10
the rule of math is that multiplication & division is done before addition & subtaction
There is an accepted order of operations (PEMDAS) that gives 10 (ditto the post above). 14 would follow some other rule. I do not see the 28 (please explain), but I suppose in base 9, it would be 11. I cannot figure out if there is some other answer that has escaped us all.
Sorry, Did not see the please don't post how you got the answer until after I did this one and the next one. (Guess you should also have reading test.)
14
I would like to see the explanation of how to get to 10 when 14 seems clearly (to me at least) the only possible logical answer...as a screwdriver is 1, pliers is three and a drill is two. Four screwdrivers (4x1=4) plus pliers (3) times drill (2) is clearly 14??????
I think the 10 comes from multiply first (3*2) then add 4 from the order of operations rule. 14 would be following another order of operations rule that I have not seen, but may be there. I still cannot figure out the 28.
Sorry, Did not see the please don't post how you got the answer until after I did this one and the last one. (Guess you should also have reading test.)
Quote from: alfaitalia on February 14, 2017, 05:17:54 PM
I would like to see the explanation of how to get to 10 when 14 seems clearly (to me at least) the only possible logical answer...as a screwdriver is 1, pliers is three and a drill is two. Four screwdrivers (4x1=4) plus pliers (3) times drill (2) is clearly 14??????
Please Excuse My Dear Aunt Sally ( PEMDAS) , order of operations .
Quote from: XH29N0G on February 14, 2017, 05:25:08 PM
I think the 10 comes from multiply first (3*2) then add 4 from the order of operations rule. 14 would be following another order of operations rule that I have not seen, but may be there. I still cannot figure out the 28.
Really? Must be US thing!...over here if you wanted the multiplication done first....it would have to be first in the line from the left.......pretty obvious really imo.
PEMDAS is an acronym for the words Parenthesis Exponents Multiplication Division Addition Subtraction. For any expression, all exponents should be simplified first, followed by multiplication and division from left to right and finally addition and subtraction from left to right
10
O.K , since you've solved that one ........
1.....unless you have some other shady maths rule over there!!! :lol:
hmmm :think: , I've come up with 13 or 12 for the second one ! :shruggy: :P not even sure 14 was right on the first one now :lol: , unless the first answer should of been a hammer !! thinking if a selection of four screwdrivers , pliers & a drill don't work , use a hammer :P i'm going to wait for my bus :cryin: :shortbus: ;D
Since only the last line is solvable.. 2
..and Alf, maybe check with your grade 3 teacher on the rules of math. :pity:
Quote from: XH29N0G on February 14, 2017, 04:43:13 PM
There is an accepted order of operations (PEMDAS) that gives 10 (ditto the post above). 14 would follow some other rule. I do not see the 28 (please explain), but I suppose in base 9, it would be 11. I cannot figure out if there is some other answer that has escaped us all.
Sorry, Did not see the please don't post how you got the answer until after I did this one and the next one. (Guess you should also have reading test.)
I see what I did. I had the value of the drill as 4 vs 2.
20? unless I cannot count and it is
5
5
2=2
I was wondering what the first one would be if the third line was a 0 e.g. the screw driver also equaled 2? would it be 2^4+6 or 2*4+6????? The drills work as 2+2 or 2^2.
10
and 12 for the second ,Assuming you keep adding as you drop from one line to the next........what happens when you assume?
the spacing looks funny for the 1x0 , are they looking for 120,130,140,150,160,170,180,190,200, or 210?
10 for the first one.
12 for the second if that 1x0 can be placed within brackets. The spacing is weird. If it's all that precedes 0 in brackets then the answer is 1.
I cannot count :slap: 19+10 = 29 and then 30 or 130 to 220 by 10's
How are you getting to those numbers? I don't see it.
Ok, XH saw something I didn't! The answer is 30. Thanks for the pm and the explanation. :cheers:
Quote from: flyinlow on February 14, 2017, 06:32:13 PM
10
and 12 for the second ,Assuming you keep adding as you drop from one line to the next........what happens when you assume?
Give the man a cigar! :2thumbs:
(Meaning I agree....) ;)
...and me!
Dan
https://youtu.be/6blD-cfko1A
What the Fu** ?????
Quote from: birdsandbees on February 14, 2017, 06:04:57 PM
Since only the last line is solvable.. 2
..and Alf, maybe check with your grade 3 teacher on the rules of math. :pity:
True only the last line is solvable....its still one. 1 plus 1 is two times 0 equals 0 plus one equals one. My teacher would have agreed!!! Nothing wrong with my maths!! You guys just play by different rules over there!!!
Here , I could would only be allowed to do the multiplication part first (or any other part first) if it was in brackets like so;
1+(1x0)+1 = .....in which case the answer would indeed be 2. No brackets? ....then the answer is 1 ! I believe this would be the way its done in the vast majority of the world.....but hey I know you guys like to be different!!
I failed trigonometry in college 3 times before I finally passed, so I am horrible at math.
My guesses:
First pic is 9
Second pic is 1
There are no addition signs between the rows so it's 1+1+1+1+11+1+1+1+11+(1x0)+1=30
First one is still 10.
Quote from: DeltaV on February 14, 2017, 07:34:33 PM
https://youtu.be/6blD-cfko1A
Good clean humor. Love Ma and Pa Kettle.
At first I missed the two drills in the first problem, hard to make out.
Quote from: Dino on February 15, 2017, 05:44:00 AM
There are no addition signs between the rows so it's 1+1+1+1+11+1+1+1+11+(1x0)+1=30
First one is still 10.
Lol....you cant have the two 1s in eleven on different lines!!....so its one! And the other is 14 :nana:. Its fast becoming apparent that the correct answer will depend on where you are from and which Math rules you apply....but you still should not do the multiply first unless its in brackets.....Oh hell I give up!! :lol:
Quote from: alfaitalia on February 15, 2017, 08:00:41 AM
Quote from: Dino on February 15, 2017, 05:44:00 AM
There are no addition signs between the rows so it's 1+1+1+1+11+1+1+1+11+(1x0)+1=30
First one is still 10.
Lol....you cant have the two 1s in eleven on different lines!!....so its one! And the other is 14 :nana:. Its fast becoming apparent that the correct answer will depend on where you are from and which Math rules you apply....but you still should not do the multiply first unless its in brackets.....Oh hell I give up!! :lol:
That's why the second problem is so tricky because the lines have been cut up, but it's still the correct answer.
And order of operation still dictates multiplication first, and that's pretty much universal. So it's 10. :lol:
The first answer he is asking how many pliers, drills and screwdrivers he needs for his tool box; and the second is posted on a door of a young mans college dorm room. He is saying that he had 5 girls in his room the first day, 5 on the second and on the third he was so drunk he couldn't remember and that is why he is asking.
Quote from: Dino on February 15, 2017, 08:02:28 AM
Quote from: alfaitalia on February 15, 2017, 08:00:41 AM
Quote from: Dino on February 15, 2017, 05:44:00 AM
There are no addition signs between the rows so it's 1+1+1+1+11+1+1+1+11+(1x0)+1=30
First one is still 10.
Lol....you cant have the two 1s in eleven on different lines!!....so its one! And the other is 14 :nana:. Its fast becoming apparent that the correct answer will depend on where you are from and which Math rules you apply....but you still should not do the multiply first unless its in brackets.....Oh hell I give up!! :lol:
That's why the second problem is so tricky because the lines have been cut up, but it's still the correct answer.
And order of operation still dictates multiplication first, and that's pretty much universal. So it's 10. :lol:
Interesting....but we don't use PEMDAS much in Europe we use BODMAS.....where the B stands for brackets. Multiplication and division still come first (well second...after Orders (or powers as they are sometimes known)) but there they will (nearly) always be in brackets.
Quote from: alfaitalia on February 15, 2017, 10:36:03 AM
Quote from: Dino on February 15, 2017, 08:02:28 AM
Quote from: alfaitalia on February 15, 2017, 08:00:41 AM
Quote from: Dino on February 15, 2017, 05:44:00 AM
There are no addition signs between the rows so it's 1+1+1+1+11+1+1+1+11+(1x0)+1=30
First one is still 10.
Lol....you cant have the two 1s in eleven on different lines!!....so its one! And the other is 14 :nana:. Its fast becoming apparent that the correct answer will depend on where you are from and which Math rules you apply....but you still should not do the multiply first unless its in brackets.....Oh hell I give up!! :lol:
That's why the second problem is so tricky because the lines have been cut up, but it's still the correct answer.
And order of operation still dictates multiplication first, and that's pretty much universal. So it's 10. :lol:
Interesting....but we don't use PEMDAS much in Europe we use BODMAS.....where the B stands for brackets. Multiplication and division still come first (well second...after Orders (or powers as they are sometimes known)) but there they will (nearly) always be in brackets.
I was born in Europe so I get the bracket thing. They use it here as well but not always.
Wrong from the start cuz math is never fun. :lol:
Quote from: Dino on February 15, 2017, 05:44:00 AM
There are no addition signs between the rows so it's 1+1+1+1+11+1+1+1+11+(1x0)+1=30
Yup, one of them there optical conclusions. ::)
Quote from: Dino on February 15, 2017, 05:44:00 AM
There are no addition signs between the rows so it's 1+1+1+1+11+1+1+1+11+(1x0)+1=30
First one is still 10.
Dino is correct on both !!!!!!
10 for the first problem
30 for the second
XH29N0G gets credit for the second one, not me. :yesnod:
In the interest of keeping /giving this thread a boost back into view. Someone posted this --I forgot where--cause I'm old.
I like the puzzel part of this one.
BTW --I found the problem/answer and so did -A friend of mine- but neither of us did so in a clean concise way to answer.
Let that be the task here. Tells us what the deal is clear and simple.
If you look at these two figures closely you can see that the big figure( the one that is 13 blocks wide and five tall) is not a right triangle at all. The two smaller triangles are not proportional to each other so what you are actually comparing are two different objects. So what looks like a straight line for the hypotenuse is not straight at all but is two sections. I'll try to make an exaggerated picture and post.
Mathematically, the larger triangular piece(red) has a tangent of 3/8 while the smallest triangle (green) has a tangent of 2/5. So the smallest angle of both triangles is not equal, therefore the triangle are not similar.
GreenRT,
Here is what you suggested in case it is a pain.
Now we need to find some more problems.
I just wonder, and don't know the answer, whether the size of the block has to be the extra space and the missing space between each line, or whether it can be done with just one? Maybe that should be the next question.
God, don't let the bean counters at work see that.
Quote from: XH29N0G on February 15, 2017, 07:44:49 PM
GreenRT,
Here is what you suggested in case it is a pain.
Now we need to find some more problems.
I just wonder, and don't know the answer, whether the size of the block has to be the extra space and the missing space between each line, or whether it can be done with just one? Maybe that should be the next question.
No you can't take those four pieces and make a proper right triangle, just won't happen unless you bend space :lol:
So I finally figured out my question about how the space adds up.
Looking at the line for the large triangle there is one side at 13 and another at 5 (13:5 aspect ratio). There is a gap that is barely visible in the top diagram and an overlap that is barely visible in the lower diagram. This is because the 5:2 and the 8:3 triangles are close to 13:5, but not exactly 13:5 aspect ratio.
Thinking about the areas:
The red triangle is 8 x 3 with an area of 8x3x1/2 = 12.
The green triangle is 5x2x1/2 =5.
An 8 side 13:5 aspect ratio would have a vertical side of 3.08, giving it an area of 8x3.08x1/2 = 12.308 boxes.
A 5 side 13:5 aspect ratio would have a vertical side of 1.92, giving it an area of 5x1.92x1/2=4.808 boxes.
The gap in the other is the difference in area 12.308-12 = 0.308 and 5-4.808=.192 which added together make up 1/2 of a box.
The other 1/2 box is made up by the overlap, which is calculated the same way. So both the overlap and the space need to be considered to explain the empty box in the lower diagram. I find this very strange, and don't mind being corrected if someone sees an error.
So I have got a problem (to pay back into the pot).
Say we have a 20 black dots and 20 white dots and we randomly split them into groups of two. We will get some groups with two black dots, some groups with two white dots, and some groups with one of each.
If we use all 40 dots to make all possibilities in a fully randomized proportions, How many will we get of
2 black dots:
2 white dots:
1 of each:
Why?
Quote from: XH29N0G on February 15, 2017, 09:00:25 PM
So I finally figured out my question about how the space adds up.
Looking at the line for the large triangle there is one side at 13 and another at 5 (13:5 aspect ratio). There is a gap that is barely visible in the top diagram and an overlap that is barely visible in the lower diagram. This is because the 5:2 and the 8:3 triangles are close to 13:5, but not exactly 13:5 aspect ratio.
Thinking about the areas:
The red triangle is 8 x 3 with an area of 8x3x1/2 = 12.
The green triangle is 5x2x1/2 =5.
An 8 side 13:5 aspect ratio would have a vertical side of 3.08, giving it an area of 8x3.08x1/2 = 12.308 boxes.
A 5 side 13:5 aspect ratio would have a vertical side of 1.92, giving it an area of 5x1.92x1/2=4.808 boxes.
The gap in the other is the difference in area 12.308-12 = 0.308 and 5-4.808=.192 which added together make up 1/2 of a box.
The other 1/2 box is made up by the overlap, which is calculated the same way. So both the overlap and the space need to be considered to explain the empty box in the lower diagram. I find this very strange, and don't mind being corrected if someone sees an error.
I like your effort to do the math. I did the same thing when I first saw this puzzel.
The next thing I did was look at as if it was presented to me to fabricate.
That is when I realized that important information detail was missing.
The missing stuff is assumed by the picture but would be spelled out on a fabrication drawing.
I guess that should be a part of---explaning the missing square in a concise way?
5 groups both black, 5 groups both white. 10 groups mixed
because I said so.....er ,wait, law of probability
Quote from: flyinlow on February 15, 2017, 09:37:16 PM
5 groups both black, 5 groups both white. 10 groups mixed
because I said so.....er ,wait, law of probability
Me too-- :slap:
While you guys think about your balls, here's a little design/logic problem that I had to solve in the Navy electrician "A" school.
You have three single pole switches, three light bulbs and a battery.
Design a circuit that will do the following.
1. All switches open and then close switch #1 and only light #1 comes on.
2. All switches open and then close switch #2 and only light #2 comes on.
3. All switches open and then close switch #3 and lights #1, #2 and #3 come on at the same time.
4. You are not allowed to use any other devices in building the circuit.
The picture shows how to represent the items.
I think I got it, but don't know much about wiring. I'll try to draw it up, but will wait a little before posting so as not to spoil the fun.
Regarding the probability question, what I always find difficult to explain is why the one each has twice as much as the double black or double white. I had thought it was just that you counted each side separately, so there were 5 black white combos and 5 white black combo in addition to the 5 white white and 5 black black.
I brought it up because I have been wrestling with a quantum mechanics problem someone else has been explaining to me where this does not appear to be the case.
Almost a day, This is what I thought might work. Don't know what rules I broke.
I think that works and it is the same as my answer but I've had a few years to make it look pretty.
That would work. With switch 3 closed , lights 1,2&3 would be in series and very dim. The original question said nothing about the light bulbs brightness....so congratulations :2thumbs:
I wanted to use diodes, but the question prohibits additional devices.
Yours is much cleaner than mine, :2thumbs:
but here's what I came up with.
(http://i1213.photobucket.com/albums/cc464/funkyjonx/circuit.jpeg)
Here is a problem/puzzle that is an old favorite. I think I posted this one a couple of years back.
This is a 3D object that can be fabricated.
This view of it is a top view. It also is a front view.
What it the third (side) view?
BTW this is a true drawn view as it would be on a a fabrication drawing.---The only thing left out here are dimentions. Also material is a non issue with this.
Quote from: funknut on February 19, 2017, 01:14:42 PM
Yours is much cleaner than mine, :2thumbs:
but here's what I came up with.
(http://i1213.photobucket.com/albums/cc464/funkyjonx/circuit.jpeg)
I could get picky on yours. It looks like all the switches would light all the lights unless the lights had some kind of way to block reverse flow of the current. As a hint, the lights do not have three connections.
WS, I think it has to be the same from the side but might be wrong. I'll keep thinking. I am assuming that the circle is some part of a cylinder (or a shape made by intersecting a cylinder - in this case a sphere) and that the square is some type of rectangular prism which ends up being a cube on intersection.
WS , the side would be a square.....you could add dashed ghost lines to show the hidden 90* hole (like a 90* plumbing elbow)
Quote from: green69rt on February 19, 2017, 08:18:22 PM
I could get picky on yours. It looks like all the switches would light all the lights unless the lights had some kind of way to block reverse flow of the current. As a hint, the lights do not have three connections.
Interesting! Thanks for the feedback. I see now that it is sloppily drawn, sorry. I intended for the switches to connect to the same terminal on all the lights, the battery on the other.
I will rethink this...
Quote from: flyinlow on February 19, 2017, 10:04:27 PM
WS , the side would be a square.....you could add dashed ghost lines to show the hidden 90* hole (like a 90* plumbing elbow)
Didn't think about it that way.
I was thinking shape in shape. I also realized I was too limited for the internal shape of a shape in shape option because the intersection of two cylinders could be anything from a the limit of a + to a square. A sphere would be but one option.
Maybe this is the side elevation. The dotted lines represent the hole drilled through the wedge.
Quote from: flyinlow on February 19, 2017, 10:04:27 PM
WS , the side would be a square.....you could add dashed ghost lines to show the hidden 90* hole (like a 90* plumbing elbow)
If a side view had dshed lines to show a hole those lines would also show in the other views.
Quote from: ws23rt on February 20, 2017, 10:35:04 AM
Quote from: flyinlow on February 19, 2017, 10:04:27 PM
WS , the side would be a square.....you could add dashed ghost lines to show the hidden 90* hole (like a 90* plumbing elbow)
If a side view had dshed lines to show a hole those lines would also show in the other views.
Not in the top view but yes they would be there in the side view, so back to the drawing board :slap:
Edit: oops, I see we have two conversations going but my dashed line comment still applies.
ok, the side view is a square
Quote from: flyinlow on February 20, 2017, 01:44:02 PM
ok, the side view is a square
If the side view is a square - -where are items shown in the top and front views---? This can be fabricated to include what is seen from the top and front.
This a 3D object.
A cube with circles drawn on two sides is not it.
Now I feel completely lost. I suppose it could be something like this, but I do not know how the third dimension is represented.
PS. This should be my #1970 post...
top view is the same as the front. so the object is a cube.
It is a cube with a hole turning 90* encased in it, or it is a cube with a tee shaped hole encased in it or it is a cube with two or three holes bored thru intersecting at the center.
So the view is a square or the same view you have shown as the top.
In the words of Mr. Spook, "insufficient data" to decide between the two choices.
Quote from: XH29N0G on February 20, 2017, 04:44:45 PM
Now I feel completely lost. I suppose it could be something like this, but I do not know how the third dimension is represented.
PS. This should be my #1970 post...
If you project a top view of your (side view drawing) -- the circle will be below center. ---And a front view will show it to be above center. :scratchchin:
Sorry I miss spoke a bit above and fixed it. :shruggy:
Also maybe my mentioning 3D was misleading. It was not intentional. I just ment to say that this can be fabricated (in 3D :shruggy:) with dimentions added for size.
The third view (the question) just shows the missing information needed.
Quote from: flyinlow on February 20, 2017, 05:11:08 PM
top view is the same as the front. so the object is a cube.
It is a cube with a hole turning 90* encased in it, or it is a cube with a tee shaped hole encased in it or it is a cube with two or three holes bored thru intersecting at the center.
So the view is a square or the same view you have shown as the top.
In the words of Mr. Spook, "insufficient data" to decide between the two choices.
There is enough data here. In your proposals hidden lines would identify just where holes start and end in the top and front views.
As far as I know their is only one answer.
??
Weird. I would not have gotten there if you hadn't pointed out the high low thing.
Now I have a 1970 charger and am on post 1972 :icon_smile_big:
possible solution
The sphere and the curve of the modified prism would be tangential and only share one molecule. Therefore you want to use a very strong material such as solid Neutronium (Star Trek the original series)
In the last view , if the the object was oriented with the arrow pointing toward a very strong source of gravity, say near a singularity in the range of lets say 10,000 G's The sphere and the prism could be separate ,just held in position by the high coefficient of friction from 10,000 G's
I think. :shruggy:
Try figuring out math problems the new common core way....
(http://media.salon.com/2015/11/ccmath3.jpg)
https://www.youtube.com/watch?v=2YMbKh9a0Sk
Quote from: Mytur Binsdirti on March 02, 2017, 01:09:05 PM
Try figuring out math problems the new common core way....
(http://media.salon.com/2015/11/ccmath3.jpg)
https://www.youtube.com/watch?v=2YMbKh9a0Sk
HAHAHA perfect Mark!
It is pretty bad that I need to check with a friend who was a math major in college to help me help my kids with their homework.. F that!!
If they'd continued the number line to the left, it would have been easy to solve.
And yes, it's a dumb way to have to do math homework.
I remember getting lectured about doing math the wrong way, and that was in the 1970's...........
Ma and Pa Kettle are turning in their graves.
https://www.youtube.com/watch?v=t8XMeocLflc