I think the most efficient way to think about it is not in terms of 99% vs 98%. It is in terms 1% vs 2%. One person equals one percent of one hundred people. The same one person equals two percent of how many people? 50. So 50 people have to leave.
The answer is that 50 people must leave. We know that 49/50 is equal to 98%, thus, since there are 99 people who are left-handed currently and you need 49, 50 must leave.
Yep, it should be instantly obvious that 98/100 = 49/50 based on the obvious fact that if the numerator and denominator are both even you can reduce any fraction by dividing by two. Once you notice that, you just need the obvious fact that 99/100 left handed people means one right handed person, and therefore you just need to go from 99 left handed to 49 left handed in the room to get to the target 49/50.
Intuitively, it's like trying to increase the saltiness of saltwater by removing water rather than by adding salt. Since most of saltwater is water, you have to boil off a lot of water vs. adding a small amount of salt to change the concentration.
It should be easily seen that 49/50 = 98% So, in order to get 98%, 100-50=50 people must left, so that the remaining one is not left-handed still remains in the room. Quite a logical IQ puzzle.
Yes, that person not left-handed represents a 1%, in order for that person to represent a 2%, it means his "weight" has to double. Therefore all the rest must half. It is pretty obvious but at first I didn't realized 😮
In 30 seconds 49/50 might as well be a random number. The solution other people posted is much better imo. Which is to think out of how many people does 1 right-handed person make 2%?
You should have continued the graph till 99 left-handed people (Ls) left the room. While the percentage initially drops slowly from 99% to 98%, the decline sharpens till we reach 98 Ls leaving (at which point its 50% with 1 L and 1 Rleft in the room), and then drops very sharply to 0% as the last L leaves. It's kind of like the inverse of radioactive decay.
It never says that the non-left-handed person is right handed. You're making assumptions. They could be missing both arms or identify as ambidextrous or laterality fluid.
I think it helps, but I think the truly confusing part is that you could get to 2% by just swapping one left-handed person with a right-handed person. It the nature of *only removing* people that I think trips people up.
If you asked me if you missed one question on a 100 point test and got a 99% how many questions would there have to be to miss one question and get a 98% I would have immediately got this in less than 30 seconds. I wonder why my brain failed like this.
But this engineer would want the percentage to be less than 98.5%, otherwise it's still pretty much the same as 99%. 😊 One leaving would not be enough, but 34 leaving would be.
@@GodmanchesterGoblin A smarter engineer might say: well, if the 98% is not necessarily meant to be exact, then perhaps the 99% is also not exact. For example, if a person who is _mixed-handed_ is counted as "half left-handed, half right-handed", then the percentage of left-handed people in a room with 98 lefthanders, 1 righthander, and 1 mixed-hander is (98.5/100)*100% = 98.5% which is rounded up to 99% . In that case, only one (fully) lefthanded person needs to leave the room to bring the percentage to (97.5/99)*100% = 98.484848...%, which is rounded down to 98% . In other words, if both 99% and 98% are not necessarily meant to be exact, then it's possible to imagine a situation in which only one left-handed person leaving would be enough.
It's not paradoxical. Its recursive because the total number in the room shrinks at the same rate as the number being measured are, leaving only a tiny fraction of change.
So you need to get a 98% left handed, a 2% no left handed. Simply consider the transformation of the original 1% no left handed to the new 2%. You can't add left handed people, but you can extract left handed. 2% = 2/100 is the result, nevertheless the numerator needs to be 1 (for this is the no left handed people), so 2% = 1/(100/2)=1/50. And 50 = 100 - 50, as a consequence x = 50.
Man, they weren't kidding when they said the UK version was a lot tougher. The US version would've probably done something like "A farmer has 20 sheep and all but 9 die. How many are left?"
i feel this explanation he gave was more complicated than it needed to be. rather than focusing on the 99 left-handers just focus on the 1% or 1 right handed person. you ask yourself how do i get him to be 2% of the population. And that's doubling his percentage. so do that you need to half the room.
I thought of the question in terms of the frequency of the non left handed person. 1% is equivalent to a frequency of 1 in 100. 2% is equivalent to a frequency of 1 in 50. Therefore to accomplish a state where 98% of people are left handed, 49 left handed people must be in the room. 99 - 49 = 50
In general, most math contest folks know that only the factors of 100 lead to an integer percentage (between 1 and 100). So if x is a factor of 100, 1/x is an integer percentage. The two highest factors of 100 are 50 and 100, leading to 2% and 1%. Focusing on the left-handers gives the puzzle it's juice, but focusing on the single right-hander leads to the quick and elegant solution.
I worked it out by halfing 98 to get 49 then converting 49/50 back you get 98/100 which is 98% Therefore there are 50 people left of which 49 are left handed
I'd never seen the other questions, but the answer was obvious and only took a second or two, most of which was checking I'd not misread the question. 1% is 1 in 100, and 2% is 1 in 50, therefore 50 left-handers have to leave. It's a classic example of a question where you're given unhelpful figures, but the easy figures (1% and 2%) are easy to derive and trivial to work with.
I did watch the other two videos mentioned but didn’t recall them when thinking of today’s riddle. Was able to get the correct answer but only because my brain must have been calibrated to reach the correct solution. Watching your videos is working to help me find solutions to riddles and other math questions. Thanks for sharing.
The way I thought it through in my head was by expressing 98% in terms of 98/100. That simplifies to 49/50. You can quickly do other specific calculations that way too. 90% is 90/100 which reduces to 9/10. 75% is 75/100 which reduces to 3/4, etc... As long as the denominator is exactly one greater than the numerator, you can have a valid percentage to aim for.
2:18 at this moment I knew the answer 😂😂😂 I knew something was up, it wasn't 1, but I couldn't figure out how to easiely solve it until you said "only one is right handed".
1 out of 100 = 1% right handed 2 out of 100 = 2% right handed BUT, you can’t have 2 right handers, so instead, reduce the fraction: 2/100 = 1/50 Therefore 50 left handers need to leave the room.
If you had 2 right-handed in that 100, it would be 98% left-handed. Then split the 100 in 2 groups of 50 with each 1 right-handed to have the same percentage in both groups of 50
You have 1 right-hander equalling 1%. You need that 1 right-hander to equal 2%. Most people can quickly discern that 1 is 2% of 50, requiring the removal of 50 left-handers. This leaves a total of 50 people in the room, one of them being the right-hander. If you need to do the math: Let x equal the total number of people in the room. (2 / 100)x = 1 x = 1 * (100 / 2) x = 50
I didn't know the problem before. And the easiest approach to me is that the fraction of right handers needs to double. You could double the right handers or removing half of the left handers. (...and round properly)
I've seen this one before, probably either here or Numberphile. The trick is indeed to flip the question to focus on the right-handed person (the left-handers are sinister anyway) - how many people need to be in the room to make them 2%? 50.
Where does it say someone is right handed ? Did you just assume this person's laterality ? This person could identify as an ambidextrous or a non-armed person.
I did solve this in under 30 seconds with no problem, but I had access to pen and paper. I don't know if they have those on the show and I doubt I'd be able to calculate this in memory, and under time pressure to boot.
This is trivial to solve in your head (if you halve 100 to 50 then you've doubled the 1% to 2%). Many people in the comments have reported solving it in under 10 seconds.
@@tuomasronnberg5244Yeah, sure, believe a random commenter. It's a trivial question, but not under pressure. My chain of though was - when left handed person leaves, total number shrink, so we go there untill we reach 98%, the biggest ratio that suits us is 49/50. Not under 30 seconds, took around minute to find that number, and much easy on paper than in head.
50. 99 out of 100 is 99% 49 out of 50 is 98%, so half of the people in the room need to leave before the right handed individual could constitute 2% of the room's population
I don't understand what is paradoxical about this. The right hander needs to be double the percentage so the total number needs to be halved. What paradox?
@@taurasuzdila not 1 but 2 bcos we're interested in min amount at least 98%, it's an semantics problem of understanding. In real life we're more interested in degrees of error rather than using the exact value of pi.
Definitely, the folks looking at it as increasing from 1 to 2% are identifying the quickest accurate way to get it. I thought about it as an iterative search via an analogy of Newton's method with my intuition pointing towards something like 10 to 25 leaving the room. And, the mental math should be quick: 0% are left handers when 99 have left the room. So, the midpoint is about 50, let's look there, and that's the answer. But, that was just a happy coincidence. I was prepared for it to be too low again and needing to go to 25 leaving and finding that ratio might be too high or low and needing to either have more or less than 25 leave. .
In a room of 100 people, "x%" are some condition. How many people must leave, to bring that percentage down to "y%"? The answer is 100(x-y)/(100-y) In the question above, x = 99, and y = 98. Therefore, it is 100(99-98)/(100-98) = 100/2 = 50. This also means that the number of people that need to leave to bring it down to a certain percentage, can be swapped. I.e. 98 people need to leave to bring the percentage down to 50%. If 99 people left, then the percentage would be 0% obviously.
One way I like to think of it is "how many left-handers will need to stop being left-handed to get to 98%?" Obviously, that's 1, since that will give us 98 left-handers and 2 non-left-handers. Now just divide the room into 2 equally sized halves, and the solution is evident.
Similar to the algebraic setup in the video, I set it up as x/(x+1) = 0.98 -> x = 0.98x + 0.98 -> 0.02x = 0.98 -> x = 49. That means that 99 - 49 = 50 people have to leave the room.
It's easier, in my opinion, if you think that the 1% must become 2% instead of 99% becoming 98%. The percentage has to double so the total has to halve.
Think the other way around , 1/100 is righthanded, so 1%. In order to duble the percentage, you need to half the quantity, because 1/50 is 2%. So if you remove 50 lefthanders, then 1 righthanded among 50 people will be a 2%. That leaves a 98% of lefthanded people in the room.
Here's a problem I would like the solution to. So I have a group of 7 friends that does secret Santa every year. In previous years we simply drew from a hat, and if anyone got their own name we would redraw. However, this year 2 people were missing. With one person, I could just put all the names in the hat except the missing person and then draw their name, sealing it for them to see later. Then put their name in the hat and do the draw as usual. However, for two people, I'm not sure there is a way to draw for them while they are not present, with them still having the same chance to draw anyone's name, and for no one but them to know what name they drew. So the question is, Is there a way to draw for two people that are absent out of 7 in such a way that no one gets their own name, everyone has the chance to draw everyone else's name, and no one but the person who receives the name knows the name they have? If so, how many people can be absent from the drawing while still being able to meet the above requirements?
That game show's creators are extremely sneaky asking specific questions like that. You need a be a maths expert to answer that in 30 seconds. If that is in the spirit of the game then I am glad I haven't seen it before. They may as well ask the specific polymers contained within a random plastic. Unless you're a specialist you won't get it.
There are always one more person in one room than the number of left handers. That makes x/(x+1)=0.98 => x=49. So 50 of the 99 left handers must leave to get down to 49.
Haven't watched the video yet, but I came up with 50 in about 5 seconds. I think my math is right. Take away 1 person and you have 1/99 right-handed. To get that to 98% it needs to be 1/50, so take away 50 lefties.
This problem is encountered in the real world too. Oil filters have a beta rating indicating how efficiently they remove particles of a given size. To calculate the efficiency, you subtract one from the beta rating and divide it by the beta rating (beta-1/beta), so a beta 50 is 98% efficient and a beta 100 is 99% efficient.
I've seen this question several times this week, and I immediately knew the answer. The reason I know is that just like you said, I remember you covering the watermelon.
When you read the question and realise the requirement is 98% in the room, and is not interested in the number of people, its easy to see that finding 2% of non left handed people is the quickest way to find the answer. one person becomes 2% when the total number of people halve.
We have 1 right handed person, 99 left-handed. We need 98/100 people left handed, or 49/50 people left handed, so if 50 left-handed people leave, we have 49 left-handed, and 1 right-handed, achieving the required ratio.
The intuition is being confused by the left hander leaving the room not being replaced by a right hander, but merly reducing the number of left handers. Else: it's obvious when you think about it this way: 2% is dou le od 1%. You can reach that by doublng the number of right handers or halfing the number of left handers.
If you think of this the other way, “How many right-handers would you have to add to the room so that left-handers made up 98%. There isn’t an exact integer answer, but adding 1 right hander gets it very close to 98% left-handers.
If there is 1ppm fluoride in the water, you gotta remove a LOT of water (removing just water, not fluoride) to up the concentration, supposing you had a litre of solution. Even removing half the water still doesn't make it very fluoridated by ratio.
Great video - the question didn't say EXACTLY 98% though, so the 34th person would drop the percentage to 98 plus a rounding error. No need to have more left handers leave the room after that point.
Exactly, numbers are rounded. The video explaining the answer did when they showed 99.9, 99.8 and so on. The numbers wern't exactly 99.9, it was rounded. When 34 leave the number is 98.48. .48 rounds down.
It should be clear from these numbers that there is only 1 right handed person in the room and 99 left handed ones. But 1 is 1% of 100 and 2% of 50. So for the 1 right handed person to be 2% of the room - only left handed people leave, no right handed ones enter - there have to be 50 people in the room - the same 1 right handed person and 49 left handers. This means that 50 left handed people must leave.
My brain has always seen math differently. I never used a formula to go back and forth between a decimal and a percentage. I also always saw fractions as being the same as decimals. Converting between the three, I instantly see 1/50 as 2%. My brain sees that one can simplify 2/100 to 1/50 and that they are equal.
I think the most efficient way to think about it is not in terms of 99% vs 98%. It is in terms 1% vs 2%. One person equals one percent of one hundred people. The same one person equals two percent of how many people? 50. So 50 people have to leave.
Now, how many to make it 97% 🙄
1 / (1 + 1) = 50%
1 / (1 + 2) = 33%
1 / (1 + 9) = 10%
1 / (1 + 19) = 5%
1 / (1 + 49 ) = 2%
My approach was:
1? Nope. Brief thought. 50? Yup.
Going back to do it analytically, I did what you did:
1% = 1/100
2% = 1/50
Done.
Not a round number at least 67 would have to leave for 3%
Yeah, focus on remaining items, not on leaving ones 🙂
The way I thought of it is the 1 right hander has to be 2% of the room, so total needs to be 50, so 50 left handers need to leave.
This is how I got it too. Pretty straight forward.
Yes, by considering that 1 person is 2% I got the correct answer in less than ten seconds. Maybe I should start entering game shows...
Did you just assume that person's laterality?
Yeah, it was pretty easy tbh
I immediately thought of 1 divided by .02 which is 1 out of 50 people. So the answer is 50 people have to leave
I solved it, but I'm not confident that I would have solved it live on stage within 30 seconds.
Same here, easy to solve when on my own.. in front of people, no chance
But remember, this wasn’t the first question. I hope the contestant had acclimated to the pressures by the time this came up.
Same here, I solved it algebraically, but see now how it could be reasoned out quicker.
The answer is that 50 people must leave. We know that 49/50 is equal to 98%, thus, since there are 99 people who are left-handed currently and you need 49, 50 must leave.
Yep, it should be instantly obvious that 98/100 = 49/50 based on the obvious fact that if the numerator and denominator are both even you can reduce any fraction by dividing by two. Once you notice that, you just need the obvious fact that 99/100 left handed people means one right handed person, and therefore you just need to go from 99 left handed to 49 left handed in the room to get to the target 49/50.
would be clearer if he said exactly 98% then it's only 2 people to have >=98
Intuitively, it's like trying to increase the saltiness of saltwater by removing water rather than by adding salt. Since most of saltwater is water, you have to boil off a lot of water vs. adding a small amount of salt to change the concentration.
Such a great analogy
It should be easily seen that
49/50 = 98%
So, in order to get 98%, 100-50=50 people must left, so that the remaining one is not left-handed still remains in the room.
Quite a logical IQ puzzle.
There is one right handed person (1% of 100) which is 2% of 50.
Yes, that person not left-handed represents a 1%, in order for that person to represent a 2%, it means his "weight" has to double. Therefore all the rest must half.
It is pretty obvious but at first I didn't realized 😮
Now you know why getting 1 percentile higher becomes so tougher and tougher as you climb higher 🙄
In 30 seconds 49/50 might as well be a random number. The solution other people posted is much better imo. Which is to think out of how many people does 1 right-handed person make 2%?
You should have continued the graph till 99 left-handed people (Ls) left the room. While the percentage initially drops slowly from 99% to 98%, the decline sharpens till we reach 98 Ls leaving (at which point its 50% with 1 L and 1 Rleft in the room), and then drops very sharply to 0% as the last L leaves. It's kind of like the inverse of radioactive decay.
Flip it, in a room of 100 people, 1% are right handed so only 1 person. To raise that to 2% 50 people must leave.
Yep this is the best one, classic GMAT stuff kudos brother
It never says that the non-left-handed person is right handed. You're making assumptions.
They could be missing both arms or identify as ambidextrous or laterality fluid.
I think it helps, but I think the truly confusing part is that you could get to 2% by just swapping one left-handed person with a right-handed person. It the nature of *only removing* people that I think trips people up.
Quick mafs
Interestingly if 90 lefthanded people leave the room, the percentage of lefthanded people remaining is still 90%
The only time less than 50% are left-handed is when they've all left the room.
If you asked me if you missed one question on a 100 point test and got a 99% how many questions would there have to be to miss one question and get a 98% I would have immediately got this in less than 30 seconds. I wonder why my brain failed like this.
I got the riddle but putting it in this format makes it surprisingly so much easier
This question is way too sinister to be believable.
You're correct, it should say that most people are right-handed.
You win the comment of the day. This one was out of the left field.
But solving it required dexterity... 😌
(These are Latin jokes, in case anyone is wondering!)
Well done. That's a good comment.
boooo 😂
I am left handed ,I would like to know where in the world did they find 99 left handed people. Lol
Damn, I saw the question in the thumbnail, and solved it in less than the time limit before I even clicked the video. Where's my cash?
Time was up some years back.
Depends, if you are an engineer and want 98 point something, then 1 should be enough, if you want exactly 98, then 50....
But this engineer would want the percentage to be less than 98.5%, otherwise it's still pretty much the same as 99%. 😊 One leaving would not be enough, but 34 leaving would be.
@@GodmanchesterGoblin A smarter engineer might say: well, if the 98% is not necessarily meant to be exact, then perhaps the 99% is also not exact. For example, if a person who is _mixed-handed_ is counted as "half left-handed, half right-handed", then the percentage of left-handed people in a room with 98 lefthanders, 1 righthander, and 1 mixed-hander is (98.5/100)*100% = 98.5% which is rounded up to 99% . In that case, only one (fully) lefthanded person needs to leave the room to bring the percentage to (97.5/99)*100% = 98.484848...%, which is rounded down to 98% .
In other words, if both 99% and 98% are not necessarily meant to be exact, then it's possible to imagine a situation in which only one left-handed person leaving would be enough.
@yurenchu Of course.
The 'Watermelon paradox' immediately came to mind. You gave several great explanations of how to think about the solution.
It's not paradoxical. Its recursive because the total number in the room shrinks at the same rate as the number being measured are, leaving only a tiny fraction of change.
So you need to get a 98% left handed, a 2% no left handed. Simply consider the transformation of the original 1% no left handed to the new 2%. You can't add left handed people, but you can extract left handed. 2% = 2/100 is the result, nevertheless the numerator needs to be 1 (for this is the no left handed people), so 2% = 1/(100/2)=1/50. And 50 = 100 - 50, as a consequence x = 50.
Man, they weren't kidding when they said the UK version was a lot tougher. The US version would've probably done something like "A farmer has 20 sheep and all but 9 die. How many are left?"
A plane crashes exactly on the border of 2 countries, which side do they bury the survivors on ?
98% = 98/100
simplifying the fraction brings it to 49/50
admittedly, I didn't know the answer at first, but after it was revealed, it seemed obvious
i feel this explanation he gave was more complicated than it needed to be. rather than focusing on the 99 left-handers just focus on the 1% or 1 right handed person. you ask yourself how do i get him to be 2% of the population. And that's doubling his percentage. so do that you need to half the room.
I thought of the question in terms of the frequency of the non left handed person. 1% is equivalent to a frequency of 1 in 100. 2% is equivalent to a frequency of 1 in 50. Therefore to accomplish a state where 98% of people are left handed, 49 left handed people must be in the room.
99 - 49 = 50
The way I did it in my head
98% of the room needs to be left-handed. You have 1 person who is (presumably) right handed. 1 is 2 percent of what number?
It's just me, or these videos are becoming easier with the years?
Thanks, watermelon puzzle!
In general, most math contest folks know that only the factors of 100 lead to an integer percentage (between 1 and 100). So if x is a factor of 100, 1/x is an integer percentage. The two highest factors of 100 are 50 and 100, leading to 2% and 1%. Focusing on the left-handers gives the puzzle it's juice, but focusing on the single right-hander leads to the quick and elegant solution.
I worked it out by halfing 98 to get 49 then converting 49/50 back you get 98/100 which is 98%
Therefore there are 50 people left of which 49 are left handed
At a glance, I thought 2
After 10 seconds, I realized 50 people must leave
Possibly the simplest problem you have ever had. Thanks.
Look at the more recent video where he asks us to evaluate "3 × 3 - 3 ÷ 3 + 3 "...
Your *last* explanation is the most intuitive to me. Thanks.
It depends on whether the right-handed person sticks around to find out.
I'd never seen the other questions, but the answer was obvious and only took a second or two, most of which was checking I'd not misread the question. 1% is 1 in 100, and 2% is 1 in 50, therefore 50 left-handers have to leave. It's a classic example of a question where you're given unhelpful figures, but the easy figures (1% and 2%) are easy to derive and trivial to work with.
This is exactly like that watermelon question.
I saw this same riddle several days ago. It must be making the rounds.
The right hander is 1% of the room. To double his percentage value, you have to halve the room.
I did watch the other two videos mentioned but didn’t recall them when thinking of today’s riddle. Was able to get the correct answer but only because my brain must have been calibrated to reach the correct solution. Watching your videos is working to help me find solutions to riddles and other math questions. Thanks for sharing.
The way I thought it through in my head was by expressing 98% in terms of 98/100. That simplifies to 49/50. You can quickly do other specific calculations that way too. 90% is 90/100 which reduces to 9/10. 75% is 75/100 which reduces to 3/4, etc... As long as the denominator is exactly one greater than the numerator, you can have a valid percentage to aim for.
2:18 at this moment I knew the answer 😂😂😂
I knew something was up, it wasn't 1, but I couldn't figure out how to easiely solve it until you said "only one is right handed".
We want 98 percent .
So let us say x people left from 99 so we will get
(99-x/x)=98%
So we will get x=49.7 approximately 50. Can we go like this??
1 out of 100 = 1% right handed
2 out of 100 = 2% right handed
BUT, you can’t have 2 right handers, so instead, reduce the fraction:
2/100 = 1/50
Therefore 50 left handers need to leave the room.
The watermelon problem in one of its 1000 variants
If you had 2 right-handed in that 100, it would be 98% left-handed.
Then split the 100 in 2 groups of 50 with each 1 right-handed to have the same percentage in both groups of 50
this actually makes sense. 50 is one half of 100, meaning that 49/50 can be over-complicated to 98/100, which is 98%
This one was easier to follow than the watermelon one
There can be a 100 people in a room and 99 don’t believe in you…
Seemingly paradoxical, not truly paradoxical.
30 second time frame to answer and here is this guy explaining it from 1:24 to 8:30, way too long of a thought process :(
You have 1 right-hander equalling 1%.
You need that 1 right-hander to equal 2%.
Most people can quickly discern that 1 is 2% of 50, requiring the removal of 50 left-handers. This leaves a total of 50 people in the room, one of them being the right-hander.
If you need to do the math:
Let x equal the total number of people in the room.
(2 / 100)x = 1
x = 1 * (100 / 2)
x = 50
My solution went like: 98% equals 98/100. If you divide both numbers by two, you have 49/50.
So 50 persons have to leave.
98% is exactly 49/50.
Thus, 50 left handed people should leave to reach 98%.
While watching the show I got the answer to this question in less than 5 seconds, but I always manage to fail on one of the earlier rounds.
I didn't know the problem before. And the easiest approach to me is that the fraction of right handers needs to double. You could double the right handers or removing half of the left handers. (...and round properly)
I've seen this one before, probably either here or Numberphile. The trick is indeed to flip the question to focus on the right-handed person (the left-handers are sinister anyway) - how many people need to be in the room to make them 2%? 50.
Where does it say someone is right handed ? Did you just assume this person's laterality ?
This person could identify as an ambidextrous or a non-armed person.
I did solve this in under 30 seconds with no problem, but I had access to pen and paper. I don't know if they have those on the show and I doubt I'd be able to calculate this in memory, and under time pressure to boot.
This is trivial to solve in your head (if you halve 100 to 50 then you've doubled the 1% to 2%). Many people in the comments have reported solving it in under 10 seconds.
@@tuomasronnberg5244Yeah, sure, believe a random commenter. It's a trivial question, but not under pressure. My chain of though was - when left handed person leaves, total number shrink, so we go there untill we reach 98%, the biggest ratio that suits us is 49/50. Not under 30 seconds, took around minute to find that number, and much easy on paper than in head.
I immediately wondered if the question allowed us to round down. If that's the case, we need 34 people to leave to get it to 66/65 or 98.48%.
i love all these maths puzzles
50.
99 out of 100 is 99%
49 out of 50 is 98%, so half of the people in the room need to leave before the right handed individual could constitute 2% of the room's population
0 people need to leave, 1 left-handed and -1 right-handed.
If 1 person weighs 2% of total people, that total is equal to 50, since 1/50 = 2%. Simple ratio. No equation required.
50. I didn’t really have a ”system” for calculating it, I just imagined a fraction that would give 98% and came up with 49/50.
I don't understand what is paradoxical about this. The right hander needs to be double the percentage so the total number needs to be halved. What paradox?
It's paradoxical because for a lot of people, especially on a 30 second timer, their intuitive answer would be 1 but it isn't.
@@taurasuzdila not 1 but 2 bcos we're interested in min amount at least 98%, it's an semantics problem of understanding. In real life we're more interested in degrees of error rather than using the exact value of pi.
Definitely, the folks looking at it as increasing from 1 to 2% are identifying the quickest accurate way to get it. I thought about it as an iterative search via an analogy of Newton's method with my intuition pointing towards something like 10 to 25 leaving the room. And, the mental math should be quick: 0% are left handers when 99 have left the room. So, the midpoint is about 50, let's look there, and that's the answer. But, that was just a happy coincidence. I was prepared for it to be too low again and needing to go to 25 leaving and finding that ratio might be too high or low and needing to either have more or less than 25 leave. .
In a room of 100 people, "x%" are some condition.
How many people must leave, to bring that percentage down to "y%"?
The answer is 100(x-y)/(100-y)
In the question above, x = 99, and y = 98.
Therefore, it is 100(99-98)/(100-98) = 100/2 = 50.
This also means that the number of people that need to leave to bring it down to a certain percentage, can be swapped.
I.e. 98 people need to leave to bring the percentage down to 50%.
If 99 people left, then the percentage would be 0% obviously.
One way I like to think of it is "how many left-handers will need to stop being left-handed to get to 98%?" Obviously, that's 1, since that will give us 98 left-handers and 2 non-left-handers. Now just divide the room into 2 equally sized halves, and the solution is evident.
Flip it. There's 1% of right handed people, you need to DOUBLE that percentage, which means halving the number of people who aren't right-handed.
Similar to the algebraic setup in the video, I set it up as x/(x+1) = 0.98 -> x = 0.98x + 0.98 -> 0.02x = 0.98 -> x = 49. That means that 99 - 49 = 50 people have to leave the room.
It's easier, in my opinion, if you think that the 1% must become 2% instead of 99% becoming 98%. The percentage has to double so the total has to halve.
1% of 100 are 1 person. Twice higer percentage with only 1 person means 1/50 or 50 people have to be removed.
Think the other way around , 1/100 is righthanded, so 1%. In order to duble the percentage, you need to half the quantity, because 1/50 is 2%. So if you remove 50 lefthanders, then 1 righthanded among 50 people will be a 2%. That leaves a 98% of lefthanded people in the room.
Here's a problem I would like the solution to.
So I have a group of 7 friends that does secret Santa every year. In previous years we simply drew from a hat, and if anyone got their own name we would redraw. However, this year 2 people were missing.
With one person, I could just put all the names in the hat except the missing person and then draw their name, sealing it for them to see later. Then put their name in the hat and do the draw as usual. However, for two people, I'm not sure there is a way to draw for them while they are not present, with them still having the same chance to draw anyone's name, and for no one but them to know what name they drew.
So the question is, Is there a way to draw for two people that are absent out of 7 in such a way that no one gets their own name, everyone has the chance to draw everyone else's name, and no one but the person who receives the name knows the name they have?
If so, how many people can be absent from the drawing while still being able to meet the above requirements?
What happened to the outro music? I really miss it
That game show's creators are extremely sneaky asking specific questions like that. You need a be a maths expert to answer that in 30 seconds. If that is in the spirit of the game then I am glad I haven't seen it before.
They may as well ask the specific polymers contained within a random plastic. Unless you're a specialist you won't get it.
50 lefties have to leave. 1 right (1/100=1%)+ 99 left (99/100=99%) => 1 right (1/50=2%) and 49 left (49/50=98%)
There are always one more person in one room than the number of left handers. That makes x/(x+1)=0.98 => x=49. So 50 of the 99 left handers must leave to get down to 49.
* There *is* always
Haven't watched the video yet, but I came up with 50 in about 5 seconds. I think my math is right. Take away 1 person and you have 1/99 right-handed. To get that to 98% it needs to be 1/50, so take away 50 lefties.
solving a puzzle faster than someone else doesn't mean you have the higher IQ, fyi
This problem is encountered in the real world too. Oil filters have a beta rating indicating how efficiently they remove particles of a given size. To calculate the efficiency, you subtract one from the beta rating and divide it by the beta rating (beta-1/beta), so a beta 50 is 98% efficient and a beta 100 is 99% efficient.
if you want something to double its share without changing the number of that thing, cut the total by half.
I've seen this question several times this week, and I immediately knew the answer.
The reason I know is that just like you said, I remember you covering the watermelon.
50
let the number of left-handed people = n
n / n+ 1 = 98 /100
n/(n+1) = 98/100
100n = 98 (n+1) cross multiply
100n =98n + 98
2n = 98
n= 49
99 -49 = 50
Rh (right handed) = 1
Lh = x
Total = x+1
x/(x+1)×100=98
2x=98
x=49
Lh left =99-49 =50
Or just say 1 is 2 percent of 50
Sir, what do you do and where are you from?
When you read the question and realise the requirement is 98% in the room, and is not interested in the number of people, its easy to see that finding 2% of non left handed people is the quickest way to find the answer. one person becomes 2% when the total number of people halve.
50. 49:1 makes exactly 98% lefties. Anything over 49 lefties make it 98+some change.
We have 1 right handed person, 99 left-handed. We need 98/100 people left handed, or 49/50 people left handed, so if 50 left-handed people leave, we have 49 left-handed, and 1 right-handed, achieving the required ratio.
Ugh, it's just 1/100 right handers to 2/100 which is 1/50, so 50 out of 99 left handers has to leave, and that is the answer
I figured it out, but only as you explained the problem. It's 50. 49/50. 98/100 reduces to 49/50
My gut is telling me 50 but my brain is too sleepy to do the maths for me
The intuition is being confused by the left hander leaving the room not being replaced by a right hander, but merly reducing the number of left handers.
Else: it's obvious when you think about it this way: 2% is dou le od 1%. You can reach that by doublng the number of right handers or halfing the number of left handers.
I just thought "2% is 1 of 50, so.... 50 of the 99 need to leave."
Pre-video solving, another way to think of it is "what is 1 2% of?". You need to reduce the total to 50, so 50 left-handers should leave.
If 1=1% and you need to double it simply halve the total sample to 50, so 50 leave.
Answer: 50.
Or just 1, if we're allowed to bring one right-handed person in.
If you think of this the other way, “How many right-handers would you have to add to the room so that left-handers made up 98%. There isn’t an exact integer answer, but adding 1 right hander gets it very close to 98% left-handers.
God damn, 2 minutes to finally start solving a problem he already solved previously with a different "name"
If there is 1ppm fluoride in the water, you gotta remove a LOT of water (removing just water, not fluoride) to up the concentration, supposing you had a litre of solution. Even removing half the water still doesn't make it very fluoridated by ratio.
To be frank, I feel this question would fit in better on "Are You Smarter Than A Fifth Grader?"
Yup…same as the watermelon puzzle. Knew the answer straight away.
Great video - the question didn't say EXACTLY 98% though, so the 34th person would drop the percentage to 98 plus a rounding error. No need to have more left handers leave the room after that point.
Exactly, numbers are rounded. The video explaining the answer did when they showed 99.9, 99.8 and so on. The numbers wern't exactly 99.9, it was rounded. When 34 leave the number is 98.48. .48 rounds down.
98% means 98/100, ie. 49/50. So we have to make 100-50, =50 left-handed leave the room to make 98%.
It should be clear from these numbers that there is only 1 right handed person in the room and 99 left handed ones. But 1 is 1% of 100 and 2% of 50. So for the 1 right handed person to be 2% of the room - only left handed people leave, no right handed ones enter - there have to be 50 people in the room - the same 1 right handed person and 49 left handers. This means that 50 left handed people must leave.
My brain has always seen math differently. I never used a formula to go back and forth between a decimal and a percentage. I also always saw fractions as being the same as decimals. Converting between the three, I instantly see 1/50 as 2%. My brain sees that one can simplify 2/100 to 1/50 and that they are equal.