Hey guys,
I have a math problem I can't get...
Could anyone help me?
This is the math problem:
A commuter jet has 60 seats available for passengers. If 35% of the seats are taken, how many seats remain?
Full Version: The Math Thread
is this including the pilot, co-pilot, and stewards' seats?
Moved to help.
Come on! You all know this is a help thread really. Stop looking at me like that, you know I'm right!
Come on! You all know this is a help thread really. Stop looking at me like that, you know I'm right!
What is the circumference of morbs fig fat belly
QUOTE(tvshez @ Aug 4 2008, 08:09 PM) 
What is the circumference of morbs fig fat belly
the circumference of your fig fat head
QUOTE(Schizo @ Aug 4 2008, 08:49 PM)
the circumference of your fig fat head
lmao
QUOTE(Schizo @ Aug 4 2008, 08:49 PM)
the circumference of your fig fat head
which is?
QUOTE(jimboy @ Aug 4 2008, 02:24 PM)
Well the answer according to them is 39..... ?
QUOTE(gotmilklol @ Aug 5 2008, 02:22 PM)
Well the answer according to them is 39..... ?
21 is the number of seats taken, maybe jimboy misread the question
60 - 21 = 39
Is this a math or metaphysics test?
39 is the maths answer.
If it's a metaphysics question, all 60 of the seats are left, because they haven't been moved anywhere, just sat in. Even if they had been moved to another area (ambiguous using of 'taken' in the question) then they would still remain, just elsewhere.
Erm... am I overthinking this?
*returns to the documentaries->pedantry section*
39 is the maths answer.
If it's a metaphysics question, all 60 of the seats are left, because they haven't been moved anywhere, just sat in. Even if they had been moved to another area (ambiguous using of 'taken' in the question) then they would still remain, just elsewhere.
Erm... am I overthinking this?
*returns to the documentaries->pedantry section*
work this bad boy out
(9^62773 + 2)^83721
or
Three men walk inside a hotel and decide to stay for the night. They pay the price of $30 by splitting it so they each pay $10. They go up their room, but then the manager realizes that the price that night is $25 dollars so he sends $5 dollars up with the bell boy. The bellboy realizes that $5 can't be split by 3 so he pockets $2 and gives the men $1 each. But if you calculate it, the 3 men each payed $9, and 9 times 3 is 27, then add the $2 from the bellboy you got $29. Where did the last $1 go?
(9^62773 + 2)^83721
or
Three men walk inside a hotel and decide to stay for the night. They pay the price of $30 by splitting it so they each pay $10. They go up their room, but then the manager realizes that the price that night is $25 dollars so he sends $5 dollars up with the bell boy. The bellboy realizes that $5 can't be split by 3 so he pockets $2 and gives the men $1 each. But if you calculate it, the 3 men each payed $9, and 9 times 3 is 27, then add the $2 from the bellboy you got $29. Where did the last $1 go?
the answer is 39 to the first problem
QUOTE(FlyZzer @ Aug 6 2008, 12:53 PM)
work this bad boy out
(9^62773 + 2)^83721
or
Three men walk inside a hotel and decide to stay for the night. They pay the price of $30 by splitting it so they each pay $10. They go up their room, but then the manager realizes that the price that night is $25 dollars so he sends $5 dollars up with the bell boy. The bellboy realizes that $5 can't be split by 3 so he pockets $2 and gives the men $1 each. But if you calculate it, the 3 men each payed $9, and 9 times 3 is 27, then add the $2 from the bellboy you got $29. Where did the last $1 go?
(9^62773 + 2)^83721
or
Three men walk inside a hotel and decide to stay for the night. They pay the price of $30 by splitting it so they each pay $10. They go up their room, but then the manager realizes that the price that night is $25 dollars so he sends $5 dollars up with the bell boy. The bellboy realizes that $5 can't be split by 3 so he pockets $2 and gives the men $1 each. But if you calculate it, the 3 men each payed $9, and 9 times 3 is 27, then add the $2 from the bellboy you got $29. Where did the last $1 go?
...there is no missing dollar,its just the way this problem is worded,the men paid $25 for the room,which dived by 3 =$8.33' each...
...so,3x$8.33=$24.99'plus the $3 they got back and add $2 the thieving bum took gives you the grand total of $29.99',(0.01 was lost in the rounding up)...
QUOTE(mando123456 @ Aug 6 2008, 06:39 PM)
the answer is 39 to the first problem
no
(9^62773 + 2)^83721= 83721 x log (9^62773 + 2)
log (9^62773 + 2) = log (9^62773) +log ( 2)
log (9^62773 + 2) = 62773 log (9) + log ( 2)
log (9^62773 + 2) = 62773 (0.954242509) + ( 0.301029996)
log (9^62773 + 2) = 59900.966
So:
83721 x log (9^62773 + 2)= (83721)(59900.966)
83721 x log (9^62773 + 2) =5.01496877 × 10^9
Digital sum is
5 + 0+ 1+4+9+6+8+7+7= 47
4+7=11
1+1=2
QUOTE(FlyZzer @ Aug 7 2008, 11:46 AM)
no
(9^62773 + 2)^83721= 83721 x log (9^62773 + 2)
log (9^62773 + 2) = log (9^62773) +log ( 2)
log (9^62773 + 2) = 62773 log (9) + log ( 2)
log (9^62773 + 2) = 62773 (0.954242509) + ( 0.301029996)
log (9^62773 + 2) = 59900.966
So:
83721 x log (9^62773 + 2)= (83721)(59900.966)
83721 x log (9^62773 + 2) =5.01496877 × 10^9
Digital sum is
5 + 0+ 1+4+9+6+8+7+7= 47
4+7=11
1+1=2
(9^62773 + 2)^83721= 83721 x log (9^62773 + 2)
log (9^62773 + 2) = log (9^62773) +log ( 2)
log (9^62773 + 2) = 62773 log (9) + log ( 2)
log (9^62773 + 2) = 62773 (0.954242509) + ( 0.301029996)
log (9^62773 + 2) = 59900.966
So:
83721 x log (9^62773 + 2)= (83721)(59900.966)
83721 x log (9^62773 + 2) =5.01496877 × 10^9
Digital sum is
5 + 0+ 1+4+9+6+8+7+7= 47
4+7=11
1+1=2
Haha they weren't talking about your problem rtard
QUOTE(MorbidPig @ Aug 7 2008, 12:32 PM)
Haha they weren't talking about your problem rtard
Yeh i knew that
Yeh...
Just Shut up.
Q: A commuter jet has 60 seats available for passengers. If 35% of the seats are taken, how many seats remain?
60 x .65 =
A: 39
Three men walk inside a hotel and decide to stay for the night. They pay the price of $30 by splitting it so they each pay $10. They go up their room, but then the manager realizes that the price that night is $25 dollars so he sends $5 dollars up with the bell boy. The bellboy realizes that $5 can't be split by 3 so he pockets $2 and gives the men $1 each. But if you calculate it, the 3 men each payed $9, and 9 times 3 is 27, then add the $2 from the bellboy you got $29. Where did the last $1 go?
25 was paid in... 3 was returned... 2 was pocketed
25/3 = 8.3333333333333333333333333333333 not 9
60 x .65 =
A: 39
Three men walk inside a hotel and decide to stay for the night. They pay the price of $30 by splitting it so they each pay $10. They go up their room, but then the manager realizes that the price that night is $25 dollars so he sends $5 dollars up with the bell boy. The bellboy realizes that $5 can't be split by 3 so he pockets $2 and gives the men $1 each. But if you calculate it, the 3 men each payed $9, and 9 times 3 is 27, then add the $2 from the bellboy you got $29. Where did the last $1 go?
25 was paid in... 3 was returned... 2 was pocketed
25/3 = 8.3333333333333333333333333333333 not 9
A high school has a strange principal. On the first day, he has his students perform an odd opening day ceremony:
There are one thousand lockers and one thousand students in the school. The principal asks the first student to go to every locker and open it. Then he has the second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?
There are one thousand lockers and one thousand students in the school. The principal asks the first student to go to every locker and open it. Then he has the second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?
QUOTE(SADEYESBLEEDING @ Aug 7 2008, 01:15 PM)
Three men walk inside a hotel and decide to stay for the night. They pay the price of $30 by splitting it so they each pay $10. They go up their room, but then the manager realizes that the price that night is $25 dollars so he sends $5 dollars up with the bell boy. The bellboy realizes that $5 can't be split by 3 so he pockets $2 and gives the men $1 each. But if you calculate it, the 3 men each payed $9, and 9 times 3 is 27, then add the $2 from the bellboy you got $29. Where did the last $1 go?
25 was paid in... 3 was returned... 2 was pocketed
25/3 = 8.3333333333333333333333333333333 not 9
25 was paid in... 3 was returned... 2 was pocketed
25/3 = 8.3333333333333333333333333333333 not 9
trifke was right
QUOTE
..there is no missing dollar,its just the way this problem is worded,the men paid $25 for the room,which dived by 3 =$8.33' each...
...so,3x$8.33=$24.99'plus the $3 they got back and add $2 the thieving bum took gives you the grand total of $29.99',(0.01 was lost in the rounding up)...
...so,3x$8.33=$24.99'plus the $3 they got back and add $2 the thieving bum took gives you the grand total of $29.99',(0.01 was lost in the rounding up)...
...do i win anything?...
QUOTE(greeniz @ Aug 5 2008, 02:44 PM)
21 is the number of seats taken, maybe jimboy misread the question
60 - 21 = 39
60 - 21 = 39
O thats how
Thanks mate!
QUOTE(trifke @ Aug 7 2008, 02:17 PM)
...do i win anything?...
yes.
Math question..
If there are 10 flies buzzin on a picnic table.... I swap and kill two with my hand. How many remain?
If there are 10 flies buzzin on a picnic table.... I swap and kill two with my hand. How many remain?
QUOTE(mando123456 @ Aug 7 2008, 12:39 PM)
Math question..
If there are 10 flies buzzin on a picnic table.... I swap and kill two with my hand. How many remain?
If there are 10 flies buzzin on a picnic table.... I swap and kill two with my hand. How many remain?
None, cuz I bet the rest got the hell out of there.
Good Job Sad
QUOTE(SADEYESBLEEDING @ Aug 7 2008, 01:17 PM)
A high school has a strange principal. On the first day, he has his students perform an odd opening day ceremony:
There are one thousand lockers and one thousand students in the school. The principal asks the first student to go to every locker and open it. Then he has the second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?
There are one thousand lockers and one thousand students in the school. The principal asks the first student to go to every locker and open it. Then he has the second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?
im gonna take a mad guess at this..
there are no lockers open as the second student closed them all.
the key word in the second half of the question is "if" . so if all the lockers are closed by the second student then there is no need for anything to be changed...
EDIT: nevermind.. i read the question wrong...
Wow... this thread makes me feel we tall did
[quote name='SADEYESBLEEDING' date='Aug 7 2008, 02:17 PM' post='461722']
A high school has a strange principal. On the first day, he has his students perform an odd opening day ceremony:
There are one thousand lockers and one thousand students in the school. The principal asks the first student to go to every locker and open it. Then he has the second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?
[/quote
31 lockers
One thing we can do is to let the first 10 students go do their open/shut thing with the lockers. The students who come after them are not going to touch lockers 1-10, so we can see which ones in that first batch are still open and try to guess the pattern.
When we do that, we find that lockers 1, 4, and 9 are open and the others are closed. Now, that isn't much to go on, so maybe you could let the next 10 students go do their thing. Then the first 20 lockers are through being touched, and we find that lockers 1, 4, 9, and 16 are the only ones in the first 20 that are still open. So what is the pattern?
Let's take any old locker, like 48 for example. It gets its state altered once for every student whose number in line is an exact divisor of 48. Here is a chart of what I mean:
this Student leaves locker 48
1 open
2 shut
3 open
4 shut
6 open
8 shut
12 open
16 shut
24 open
48 shut
Notice that 48 has an even number (ten) of divisors, namely1,2,3,4,6,8,12,16,24,48. So the locker goes open-shut-open-shut ... and ends up shut. Any locker number that has an even number of divisors will end up shut.
Which numbers have an odd number of divisors? That's the answer to this problem. Just to help you along, here are the locker numbers up to 100 that are left open:
1,4,9,16,25,36,49,64,81,100.
See, if you can describe these numbers in a different way from "having an odd number of divisors." Think about multiplying numbers together. When you understand how to describe them, you will see that 31 of the 1000 lockers are still open (without having to work it all out!)
A high school has a strange principal. On the first day, he has his students perform an odd opening day ceremony:
There are one thousand lockers and one thousand students in the school. The principal asks the first student to go to every locker and open it. Then he has the second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?
[/quote
31 lockers
One thing we can do is to let the first 10 students go do their open/shut thing with the lockers. The students who come after them are not going to touch lockers 1-10, so we can see which ones in that first batch are still open and try to guess the pattern.
When we do that, we find that lockers 1, 4, and 9 are open and the others are closed. Now, that isn't much to go on, so maybe you could let the next 10 students go do their thing. Then the first 20 lockers are through being touched, and we find that lockers 1, 4, 9, and 16 are the only ones in the first 20 that are still open. So what is the pattern?
Let's take any old locker, like 48 for example. It gets its state altered once for every student whose number in line is an exact divisor of 48. Here is a chart of what I mean:
this Student leaves locker 48
1 open
2 shut
3 open
4 shut
6 open
8 shut
12 open
16 shut
24 open
48 shut
Notice that 48 has an even number (ten) of divisors, namely1,2,3,4,6,8,12,16,24,48. So the locker goes open-shut-open-shut ... and ends up shut. Any locker number that has an even number of divisors will end up shut.
Which numbers have an odd number of divisors? That's the answer to this problem. Just to help you along, here are the locker numbers up to 100 that are left open:
1,4,9,16,25,36,49,64,81,100.
See, if you can describe these numbers in a different way from "having an odd number of divisors." Think about multiplying numbers together. When you understand how to describe them, you will see that 31 of the 1000 lockers are still open (without having to work it all out!)
QUOTE(elcarnero @ Aug 7 2008, 09:01 PM)
31 lockers
One thing we can do is to let the first 10 students go do their open/shut thing with the lockers. The students who come after them are not going to touch lockers 1-10, so we can see which ones in that first batch are still open and try to guess the pattern.
When we do that, we find that lockers 1, 4, and 9 are open and the others are closed. Now, that isn't much to go on, so maybe you could let the next 10 students go do their thing. Then the first 20 lockers are through being touched, and we find that lockers 1, 4, 9, and 16 are the only ones in the first 20 that are still open. So what is the pattern?
Let's take any old locker, like 48 for example. It gets its state altered once for every student whose number in line is an exact divisor of 48. Here is a chart of what I mean:
this Student leaves locker 48
1 open
2 shut
3 open
4 shut
6 open
8 shut
12 open
16 shut
24 open
48 shut
Notice that 48 has an even number (ten) of divisors, namely1,2,3,4,6,8,12,16,24,48. So the locker goes open-shut-open-shut ... and ends up shut. Any locker number that has an even number of divisors will end up shut.
Which numbers have an odd number of divisors? That's the answer to this problem. Just to help you along, here are the locker numbers up to 100 that are left open:
1,4,9,16,25,36,49,64,81,100.
See, if you can describe these numbers in a different way from "having an odd number of divisors." Think about multiplying numbers together. When you understand how to describe them, you will see that 31 of the 1000 lockers are still open (without having to work it all out!)
One thing we can do is to let the first 10 students go do their open/shut thing with the lockers. The students who come after them are not going to touch lockers 1-10, so we can see which ones in that first batch are still open and try to guess the pattern.
When we do that, we find that lockers 1, 4, and 9 are open and the others are closed. Now, that isn't much to go on, so maybe you could let the next 10 students go do their thing. Then the first 20 lockers are through being touched, and we find that lockers 1, 4, 9, and 16 are the only ones in the first 20 that are still open. So what is the pattern?
Let's take any old locker, like 48 for example. It gets its state altered once for every student whose number in line is an exact divisor of 48. Here is a chart of what I mean:
this Student leaves locker 48
1 open
2 shut
3 open
4 shut
6 open
8 shut
12 open
16 shut
24 open
48 shut
Notice that 48 has an even number (ten) of divisors, namely1,2,3,4,6,8,12,16,24,48. So the locker goes open-shut-open-shut ... and ends up shut. Any locker number that has an even number of divisors will end up shut.
Which numbers have an odd number of divisors? That's the answer to this problem. Just to help you along, here are the locker numbers up to 100 that are left open:
1,4,9,16,25,36,49,64,81,100.
See, if you can describe these numbers in a different way from "having an odd number of divisors." Think about multiplying numbers together. When you understand how to describe them, you will see that 31 of the 1000 lockers are still open (without having to work it all out!)
I'm just gonna go ahead and say this, that solving that little problem was pretty bad ass. And nice explanation too, by the way!
I had to read over it a couple of times to get the whole gist of it, but yeah, once you learn how to describe the problem in a different way, it all comes together!
Kudos to you.
You are correct elcarnero:
The only lockers that remain open are perfect squares (1, 4, 9, 16, etc) because they are the only numbers divisible by an odd number of whole numbers; every factor other than the number's square root is paired up with another. Thus, these lockers will be "changed" an odd number of times, which means they will be left open. All the other numbers are divisible by an even number of factors and will consequently end up closed.
So the number of open lockers is the number of perfect squares less than or equal to one thousand. These numbers are one squared, two squared, three squared, four squared, and so on, up to thirty one squared. (Thirty two squared is greater than one thousand, and therefore out of range.) So the answer is thirty one.
The only lockers that remain open are perfect squares (1, 4, 9, 16, etc) because they are the only numbers divisible by an odd number of whole numbers; every factor other than the number's square root is paired up with another. Thus, these lockers will be "changed" an odd number of times, which means they will be left open. All the other numbers are divisible by an even number of factors and will consequently end up closed.
So the number of open lockers is the number of perfect squares less than or equal to one thousand. These numbers are one squared, two squared, three squared, four squared, and so on, up to thirty one squared. (Thirty two squared is greater than one thousand, and therefore out of range.) So the answer is thirty one.
This is a "lo-fi" version of our main content. To view the full version with more information, formatting and images, please click here.